{"text": "Preface About OpenStax OpenStax is part of Rice University, which is a 501(c)(3) nonprofit charitable corporation. Our mission is to make an amazing education accessible for all. Through our partnerships with philanthropic organizations and our alliance with other educational resource companies, we’re breaking down the most common barriers to learning. Because we believe that everyone should and can have access to knowledge. About OpenStax Resources Customization Contemporary Mathematics is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY-NC-SA) license, which means that you can non-commercially distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors, and distribute all derivatives under the same license. Because our books are openly licensed, you are free to use the entire book or select only the sections that are most relevant to the needs of your course. 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If you reuse art from this text that does not have attribution provided, use the following attribution: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. Errata All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web-based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on OpenStax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past and pending errata changes on your book page on OpenStax.org. Format You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print. About Contemporary Mathematics Contemporary Mathematics is designed to meet the requirements for a liberal arts mathematics course. The textbook covers a range of topics that are typically found in a liberal arts course as well as some topics to connect mathematics to the world around us. The text provides stand-alone sections with a focus on showing relevance in the features as well as the examples, exercises, and exposition. Pedagogical Foundation Learning Objectives Every section begins with a set of clear and concise learning objectives, which have been thoroughly revised to be both measurable and more closely aligned with current teaching practice. These objectives are designed to help the instructor decide what content to include or assign and to guide student expectations of learning. After completing the section and end-of-section exercises, students should be able to demonstrate mastery of the learning objectives. Key Features Check Your Understanding: Concept checks to confirm students understand content at the end of every section immediately before the exercise sets are provided to help bolster confidence before embarking on homework. People in Mathematics: A mix of historic and contemporary profiles aimed to incorporate extensive diversity in gender and ethnicity. The profiles incorporate how the person’s contribution has benefitted students or is relevant to their lives in some way. Who Knew?: A high-interest feature designed to showcase something interesting related to the section contents. These features are crafted to offer something students might be surprised to find is so relevant to them. Work It Out: Offers some activity ideas in line with the sections to support the learning objectives. Tech Check: Highlights technologies that support content in the section. Projects: A feature designed to put students in the driver’s seat researching a topic using various online resources. It is intended to be primarily or wholly non-computational. Projects utilize online research and writing to summarize their findings. Section Summaries Section summaries distill the information in each section for both students and instructors down to key, concise points addressed in the section. Key Terms Key terms are bold and are followed by a definition in context. Answers and Solutions to Questions in the Book Answers for Your Turn and Check Your Understanding exercises are provided in the Answer Key at the end of the book. The Section Exercises, Chapter Reviews, and Chapter Tests are intended for homework assignments or assessment; thus, student-facing solutions are provided in the Student Solution Manual for only a subset of the exercises. Solutions for all exercises are provided in the Instructor Solution Manual for instructors to share with students at their discretion, as is standard for such resources. About the Authors Senior Contributing Author Donna Kirk, University of Wisconsin at Superior Donna Kirk received her B.S. in Mathematics from the State University of New York at Oneonta and her master’s degree from City University – Seattle in Educational Technology and Curriculum Design. After teaching math in higher education for more than twenty years, she joined University of Wisconsin’s Education Department in 2021, teaching math education for teacher preparation. She is also the director of a STEM institute focused on connecting underrepresented students with access to engaging and innovative experiences to empower themselves to pursue STEM related careers. Contributing Authors Barbara Boschmans, Northern Arizona University Brian Beaudrie, Northern Arizona University Matthew Cathey, Wofford College Valeree Falduto, Palm Beach State College Maureen Gerlofs, Texas State University Quin Hearn, Broward College Ian Walters, D’Youville College Reviewers Anna Pat Alpert, Navarro College Mario Barrientos, Angelo State University Keisha Brown, Perimeter College at Georgia State University Hugh Cornell, University of North Florida David Crombecque, University of Southern California Shari Davis, Old Dominion University Angela Everett, Chattanooga State Community College David French, Tidewater Community College Michele Gribben, McDaniel College Celeste Hernandez, Dallas College-Richland Trevor Jack, Illinois Wesleyan University Kristin Kang, Grand View University Karla Karstens, University of Vermont Sergio Loch, Grand View University Andrew Misseldine, Southern Utah University Carla Monticelli, Camden County College Cindy Moss, Skyline College Jill Rafael, Sierra College Gary Rosen, University of Southern California Faith Willman, Harrisburg Area Community College Additional Resources Student and Instructor Resources We’ve compiled additional resources for both students and instructors, including student solution manuals, instructor solution manuals, and PowerPoint lecture slides. Instructor resources require a verified instructor account, which you can apply for when you log in or create your account on OpenStax.org. Take advantage of these resources to supplement your OpenStax book. Academic Integrity Academic integrity builds trust, understanding, equity, and genuine learning. While students may encounter significant challenges in their courses and their lives, doing their own work and maintaining a high degree of authenticity will result in meaningful outcomes that will extend far beyond their college career. Faculty, administrators, resource providers, and students should work together to maintain a fair and positive experience. We realize that students benefit when academic integrity ground rules are established early in the course. To that end, OpenStax has created an interactive to aid with academic integrity discussions in your course. attribution: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license Visit our academic integrity slider . Click and drag icons along the continuum to align these practices with your institution and course policies. You may then include the graphic on your syllabus, present it in your first course meeting, or create a handout for students. At OpenStax we are also developing resources supporting authentic learning experiences and assessment. Please visit this book’s page for updates. For an in-depth review of academic integrity strategies, we highly recommend visiting the International Center of Academic Integrity (ICAI) website . Community Hubs OpenStax partners with the Institute for the Study of Knowledge Management in Education (ISKME) to offer Community Hubs on OER Commons—a platform for instructors to share community-created resources that support OpenStax books, free of charge. Through our Community Hubs, instructors can upload their own materials or download resources to use in their own courses, including additional ancillaries, teaching material, multimedia, and relevant course content. We encourage instructors to join the hubs for the subjects most relevant to your teaching and research as an opportunity both to enrich your courses and to engage with other faculty. To reach the Community Hubs, visit www.oercommons.org/hubs/openstax . Technology partners As allies in making high-quality learning materials accessible, our technology partners offer optional low-cost tools that are integrated with OpenStax books. To access the technology options for your text, visit your book page on OpenStax.org.", "section": "Preface", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction A flatware drawer is like a set in that it contains distinct objects. (credit: modification of work “silverware” by Jo Naylor/Flickr, CC BY 2.0) Think of a drawer in your kitchen used to store flatware. This drawer likely holds forks, spoons, and knives, and possibly other items such as a meat thermometer and a can opener. The drawer in this case represents a tool used to group a collection of objects. The members of the group are the individual items in the drawer, such as a fork or a spoon. The members of a set can be anything, such as people, numbers, or letters of the alphabet. In statistical studies, a set is a well-defined collection of objects used to identify an entire population of interest. For example, in a research study examining the effects of a new medication, there can be two sets of people: one set that is given the medication and a different set that is given a placebo (control group). In this chapter, we will discuss sets and Venn diagrams, which are graphical ways to show relationships between different groups.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Basic Set Concepts A spoon, fork, and knife are elements of the set of flatware. (credit: modification of work “Cupofjoy” Wikimedia CC0 1.0 Public Domain Dedication) Learning Objectives After completing this section, you should be able to: Represent sets in a variety of ways. Represent well-defined sets and the empty set with proper set notation. Compute the cardinal value of a set. Differentiate between finite and infinite sets. Differentiate between equal and equivalent sets. Sets and Ways to Represent Them Think back to your kitchen organization. If the drawer is the set , then the forks and knives are elements in the set. Sets can be described in a number of different ways: by roster, by set-builder notation, by interval notation, by graphing on a number line, and by Venn diagrams. Sets are typically designated with capital letters. The simplest way to represent a set with only a few members is the roster (or listing ) method , in which the elements in a set are listed, enclosed by curly braces and separated by commas. For example, if F represents our set of flatware, we can represent F by using the following set notation with the roster method: F = { fork, spoon, knife, meat thermometer, can opener } Writing a Set Using the Roster or Listing Method Write a set consisting of your three favorite sports and label it with a capital S . There are multiple possible answers depending on what your three favorite sports are, but any answer must list three different sports separated by commas, such as the following: S = { hockey, basketball, soccer } All the sets we have considered so far have been well-defined sets . A well-defined set clearly communicates whether an element is a member of the set or not. The members of a well-defined set are fixed and do not change over time. Consider the following question. What are your top 10 songs of 2021? You could create a list of your top 10 favorite songs from 2021, but the list your friend creates will not necessarily contain the same 10 songs. So, the set of your top 10 songs of 2021 is not a well-defined set. On the other hand, the set of the letters in your name is a well-defined set because it does not vary (unless of course you change your name). The NFL wide receiver, Chad Johnson, famously changed his name to Chad Ochocinco to match his jersey number of 85. Identifying Well-Defined Sets For each of the following collections, determine if it represents a well-defined set. The group of all past vice presidents of the United States. A group of old cats. The group of all past vice presidents of the United States is a well-defined set, because you can clearly identify if any individual was or was not a member of that group. For example, Britney Spears is not a member of this set, but Joe Biden is a member of this set. A group of old cats is not a well-defined set because the word old is ambiguous. Some people might consider a seven-year-old cat to be old, while others might think a cat is not old until it is 13 years old. Because people can disagree on what is and what is not a member of this group, the set is not well-defined. On January 20, 2021, Kamala Harris was sworn in as the first woman vice president of the United States of America. If we were to consider the set of all women vice presidents of the United States of America prior to January 20, 2021, this set would be known as an empty set ; the number of people in this set is 0, since there were no women vice presidents before Harris. The empty set, also called the null set, is written symbolically using a pair of braces, { } , or a zero with a slash through it, ∅ . The set containing the number 0 , { 0 } , is a set with one element in it. It is not the same as the empty set , { } , which does not have any elements in it. Symbolically: { 0 } ≠ { } . Representing the Empty Set Symbolically Represent each of the following sets symbolically. The set of prime numbers less than 2. The set of birds that are also mammals. A prime number is a natural number greater than 1 that is only divisible by one and itself. Since there are no prime numbers less than 2, this set is empty, and we can represent it symbolically as follows: { } or ∅ . These two different symbols for the empty set can be used interchangeably. The set of birds and the set of mammals do not intersect, so the set of birds that are also mammals is empty, and we can represent it symbolically as ∅ or { } . The Number Zero We use the number zero to represent the concept of nothing every day. The machine language of computers is binary, consisting only of zeros and ones, and even way before that, the number zero was a powerful invention that allowed our understanding of mathematics and science to develop. The historical record shows the Babylonians first used zeros around 300 B.C., while the Mayans developed and began using zero separately around 350 A.D. What is considered the first formal use of zero in arithmetic operations was developed by the Indian mathematician Brahmagupta around 650 A.D. Brahmagupta, Mathematician and Astronomer Another interesting feature of the number zero is that although it is an even number, it is the only number that is neither negative nor positive. For larger sets that have a natural ordering, sometimes an ellipsis is used to indicate that the pattern continues. It is common practice to list the first three elements of a set to establish a pattern, write the ellipsis, and then provide the last element. Consider the set of all lowercase letters of the English alphabet, A . This set can be written symbolically as A = { a, b, c, … , z } . The sets we have been discussing so far are finite sets . They all have a limited or fixed number of elements. We also use an ellipsis for infinite sets , which have an unlimited number of elements, to indicate that the pattern continues. For example, in set theory, the set of natural numbers , which is the set of all positive counting numbers, is represented as ℕ = { 1 , 2 , 3 , … } . Notice that for this set, there is no element following the ellipsis. This is because there is no largest natural number; you can always add one more to get to the next natural number. Because the set of natural numbers grows without bound, it is an infinite set. Writing a Finite Set Using the Roster Method and an Ellipsis Write the set of even natural numbers including and between 2 and 100, and label it with a capital E . Include an ellipsis. Write the label, E , followed by an equal sign and then a bracket. Write the first three even numbers separated by commas, beginning with the number two to establish a pattern. Next, write the ellipsis followed by a comma and the last number in the list, 100. Finally, write the closing bracket to complete the set. Write the label, E , followed by an equal sign and then a bracket. E = { Write the first three even numbers separated by commas, beginning with the number 2 to establish a pattern. E = { 2 , 4 , 6 , Next, write the ellipsis followed by a comma and the last number in the list, 100. E = { 2 , 4 , 6 , … , 100 Finally, write the closing bracket to complete the set. E = { 2 , 4 , 6 , … , 100 } Our number system is made up of several different infinite sets of numbers. The set of integers , ℤ , is another infinite set of numbers. It includes all the positive and negative counting numbers and the number zero. There is no largest or smallest integer. Writing an Infinite Set Using the Roster Method and Ellipses Write the set of integers using the roster method, and label it with a ℤ . Step 1: As always, we write the label and then the opening bracket. Because the negative counting numbers are infinite, to represent that the pattern continues without bound to the left, we must use an ellipsis as the first element in our list. Step 2: We place a comma and follow it with at least three consecutive integers separated by commas to establish a pattern. Step 3: Add an ellipsis to the end of the list to show that the set of integers continues without bound to the right. Complete the list with a closing bracket. The set of integers may be represented as follows: ℤ = { … , − 2 , − 1 , 0 , 1 , 2 , … } . A shorthand way to write sets is with the use of set builder notation , which is a verbal description or formula for the set. For example, the set of all lowercase letters of the English alphabet, A , written in set builder notation is: A = { x | x is a lowercase letter of the English alphabet. } This is read as, “Set A is the set of all elements x such that x is a lowercase letter of the English alphabet.” Writing a Set Using Set Builder Notation Using set builder notation, write the set B of all types of balls. Explain what the notation means. The verbal description of the set is, “Set B is the set of all elements b such that b is a ball.” This set can be written in set builder notation as follows: B = { b | b is a ball . } Writing Sets Using Various Methods Consider the set of letters in the word “happy.” Determine the best way to represent this set, and then write the set using either the roster method or set builder notation, whichever is more appropriate. Because the letters in the word “happy” consist of a small finite set, the best way to represent this set is with the roster method. Choose a label to represent the set, such as H . H = { h , a , p , y } . Notice that the letter “p” is only represented one time. This occurs because when representing members of a set, each unique element is only listed once no matter how many times it occurs. Duplicate elements are never repeated when representing members of a set. Computing the Cardinal Value of a Set Almost all the sets most people work with outside of pure mathematics are finite sets. For these sets, the cardinal value or cardinality of the set is the number of elements in the set. For finite set A , the cardinality is denoted symbolically as n ( A ) . For example, a set that contains four elements has a cardinality of 4. How do we measure the cardinality of infinite sets? The ‘smallest’ infinite set is the set of natural numbers, or counting numbers, ℕ = { 1 , 2 , 3 , … } . This set has a cardinality of ℵ 0 (pronounced \"aleph-null\"). All sets that have the same cardinality as the set of natural numbers are countably infinite . This concept, as well as notation using aleph, was introduced by mathematician Georg Cantor who once said, “A set is a Many that allows itself to be thought of as a One.” Computing the Cardinal Value of a Set Write the cardinal value of each of the following sets in symbolic form. F = { fork, spoon, knife, meat thermometer, can opener } The empty set. There are 5 distinct elements in set F : a fork, a spoon, a knife, a meat thermometer, and a can opener. Therefore, the cardinal value of set F is 5 and written symbolically as n ( F ) = 5. Because the empty set does not have any elements in it, the cardinality of the empty set is zero. Symbolically we write this as: n ( ∅ ) = 0. Now that we have learned to represent finite and infinite sets using both the roster method and set builder notation, we should also be able to determine if a set is finite or infinite based on its verbal or symbolic description. One way to determine if a set is finite or not is to determine the cardinality of the set. If the cardinality of a set is a natural number, then the set is finite. Differentiating Between Finite and Infinite Sets Classify each of the following sets as infinite or finite. E = { 2 , 4 , 6 , 8 , 10 } A is the set of lowercase letters of the English Alphabet, A = { a , b , c , … , z } . ℚ = { p q | p and q are integers and q ≠ 0 } n ( E ) = 5 . Since 5 is a natural number, the set is finite. n ( A ) = 26 . Since 26 is a natural number, the set is finite. Set ℚ is the set of rational numbers or fractions. Because the set of integers is a subset of the set of rational numbers, and the set of integers is infinite, the set of rational numbers is also infinite. There is no smallest or largest rational number. Equal versus Equivalent Sets When speaking or writing we tend to use equal and equivalent interchangeably, but there is an important distinction between their meanings. Consider a new Ford Escape Hybrid and a new Toyota Rav4 Hybrid. Both cars are hybrid electric sport utility vehicles; in that sense, they are equivalent . They will both get you from place to place in a relatively fuel-efficient way. In this example we are comparing the single member set {Toyota Rav4 Hybrid} to the single member set {Ford Escape Hybrid}. Since these two sets have the same number of elements, they are also equivalent mathematically, meaning they have the same cardinality. But they are not equal, because the two cars have different looks and features, and probably even handle differently. Each manufacturer will emphasize the features unique to their vehicle to persuade you to buy it; if the SUVs were truly equal, there would be no reason to choose one over the other. Now consider two Honda CR-Vs that are made with exactly the same parts, on the same assembly line within a few minutes of each other—these SUVs are equal . They are identical to each other, containing the same elements without regard to order, and the only differentiator when making a purchasing decision would be varied pricing at different dealerships. The set {Honda CR-V} is equal to the set {Honda CR-V}. Symbolically, we represent equal sets as A = B and equivalent sets as A ∼ B . Now, let us consider a Toyota dealership that has 10 RAV4s on the lot, 8 Prii, 7 Highlanders, and 12 Camrys. There is a one-to-one relationship between the set of vehicles on the lot and the set consisting of the number of each type of vehicle on the lot. Therefore, these two sets are equivalent, but not equal. The set {RAV4, Prius, Highlander, Camry} is equivalent to the set {10, 8, 7, 12} because they have the same number of elements. Equal and Equivalent Sets If two sets are equal, they are also equivalent, because equal sets also have the same cardinality. Differentiating Between Equivalent and Equal Sets Determine if the following pairs of sets are equal, equivalent, or neither. E = { 2 , 4 , 6 , 8 , 10 } and F = { fork, spoon, knife, meat thermometer, can opener } The empty set and the set of prime numbers less than 2. The set of vowels in the word happiness and the set of consonants in the word happiness. Sets E and F both have a cardinal value of 5, but the elements in these sets are different. So, the two sets are equivalent, but they are not equal: E ∼ F . The set of prime numbers consists of the set of counting numbers greater than one that can only be divided evenly by one and itself. The set of prime numbers less than 2 is an empty set, since there are no prime numbers less than 2. Therefore, these two sets are equal (and equivalent). The set of vowels in the word happiness is { a , i , e } and the set of consonants in the word happiness is { h , p , n , s } . The cardinal value of these sets two sets is n ( { a , i , e } ) = 3 and n ( { h , p , n , s } ) = 4 , respectively. Because the cardinality of the two sets differs, they are not equivalent. Further, their elements are not identical, so they are also not equal. Georg Cantor Georg Cantor (credit: Wikimedia, public domain) Georg Cantor, the father of modern set theory, was born during the year 1845 in Saint Petersburg, Russia and later moved to Germany as a youth. Besides being an accomplished mathematician, he also played the violin. Cantor received his doctoral degree in Mathematics at the age of 22. In 1870, at the age of 25 he established the uniqueness theorem for trigonometric series. His most significant work happened between 1874 and 1884, when he established the existence of transcendental numbers (also called irrational numbers) and proved that the set of real numbers are uncountably infinite—despite the objections of his former professor Leopold Kronecker. Cantor published his final treatise on set theory in 1897 at the age of 52, and was awarded the Sylvester Medial from the Royal Society of London in 1904 for his contributions to the field. At the heart of Cantor’s work was his goal to solve the continuum problem, which later influenced the works of David Hilbert and Ernst Zermelo. References: Wikipedia contributors. “Cantor.” Wikipedia, The Free Encyclopedia, 23 Mar. 2021. Web. 20 Jul. 2021. Akihiro Kanamori, “Set Theory from Cantor to Cohen,” Editor(s): Dov M. Gabbay, Akihiro Kanamori, John Woods, Handbook of the History of Logic , North-Holland, Volume 6, 2012. Check Your Understanding Key Terms set elements well-defined set empty set roster method finite set infinite set natural numbers integer set-builder notation cardinality of a set countably infinite equal sets equivalent sets Key Concepts Identify a set as being a well-defined collection of objects and differentiate between collections that are not well-defined and collections that are sets. Represent sets using both the roster or listing method and set builder notation which includes a description of the members of a set. In set theory, the following symbols are universally used: ℕ - The set of natural numbers, which is the set of all positive counting numbers. ℕ = { 1 , 2 , 3 , ... } ℤ - The set of integers, which is the set of all the positive and negative counting numbers and the number zero. ℤ = { ... , − 2 , − 1 , 0 , 1 , 2 , ... } ℚ - The set of rational numbers or fractions. ℚ = { p q | p and q are integers and q ≠ 0 } Distinguish between finite sets, infinite sets, and the empty set to determine the size or cardinality of a set. Distinguish between equal sets which have exactly the same members and equivalent sets that may have different members but must have the same cardinality or size. Video Equal and Equivalent Sets", "section": "Basic Set Concepts", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Subsets The players on a soccer team who are actively participating in a game are a subset of the greater set of team members. (Credit: “PAFC-Mezokovesd-108” by Puskás Akadémia/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Represent subsets and proper subsets symbolically. Compute the number of subsets of a set. Apply concepts of subsets and equivalent sets to finite and infinite sets. The rules of Major League Soccer (MLS) allow each team to have up to 30 players on their team. However, only 18 of these players can be listed on the game day roster, and of the 18 listed, 11 players must be selected to start the game. How the coaches and general managers form the team and choose the starters for each game will determine the success of the team in any given year. The entire group of 30 players is each team’s set. The group of game day players is a subset of the team set, and the group of 11 starters is a subset of both the team set and the set of players on the game day roster. Set A is a subset of set B if every member of set A is also a member of set B . Symbolically, this relationship is written as A ⊆ B . Sets can be related to each other in several different ways: they may not share any members in common, they may share some members in common, or they may share all members in common. In this section, we will explore the way we can select a group of members from the whole set. Every set is also a subset of itself, B ⊆ B Recall the set of flatware in our kitchen drawer from Section 1.1 , F = { fork, spoon, knife, meat thermometer, can opener } . Suppose you are preparing to eat dinner, so you pull a fork and a knife from the drawer to set the table. The set D = { knife, fork } is a subset of set F , because every member or element of set D is also a member of set F . More specifically, set D is a proper subset of set F , because there are other members of set F not in set D . This is written as D ⊂ F . The only subset of a set that is not a proper subset of the set would be the set itself. The empty set or null set, ∅ , is a proper subset of every set, except itself. Graphically, sets are often represented as circles. In the following graphic, set A is represented as a circle completely enclosed inside the circle representing set B , showing that set A is a proper subset of set B . The element x represents an element that is in both set A and set B . While we can list all the subsets of a finite set, it is not possible to list all the possible subsets of an infinite set, as it would take an infinitely long time. Listing All the Proper Subsets of a Finite Set Set L is a set of reading materials available in a shop at the airport, L = { newspaper, magazine, book } . List all the subsets of set L . Step 1: It is best to begin with the set itself, as every set is a subset of itself. In our example, the cardinality of set L is n ( L ) = 3 . There is only one subset of set L that has the same number of elements of set L : { newspaper , magazine , book } . Step 2: Next, list all the proper subsets of the set containing n ( L ) − 1 elements. In this case, 3 − 1 = 2 . There are three subsets that each contain two elements: { newspaper , magazine } , { newspaper , book } , and { magazine , book } . Step 3: Continue this process by listing all the proper subsets of the set containing n ( L ) − 2 elements. In this case, 3 − 2 = 1 . There are three subsets that contain one element: { newspaper } , { magazine } , and { book } . Step 4: Finally, list the subset containing 0 elements, or the empty set: { } . Determining Whether a Set Is a Proper Subset Consider the set of common political parties in the United States, P = { Democratic, Green, Libertarian, Republican } . Determine if the following sets are proper subsets of P . M = { Democratic, Republican } G = { Green } V = { Republican, Libertarian, Green, Democratic } M is a proper subset of P , written symbolically as M ⊂ P because every member of M is a member of set P , but P also contains at least one element that is not in M . G is a single member proper subset of P , written symbolically as G ⊂ P , because Green is a member of set P , but P also contains other members (such as Democratic) that are not in G . V is subset of P because every member of V is also a member of P , but it is not a proper subset of P because there are no members of V that are not also in set P . We can represent the relationship symbolically as V ⊆ P , or more precisely, set V is equal to set P , V = P . Expressing the Relationship between Sets Symbolically Consider the subsets of a standard deck of cards: S = { spades, hearts, diamonds, clubs } ; R = { hearts, diamonds } ; B = { spades, clubs } ; and C = { clubs } . Express the relationship between the following sets symbolically. Set S and set B . Set C and set B . Set R and R . B ⊂ S . B is a proper subset of set S . C ⊂ B . C is a proper subset of set B . R ⊆ R or R = R . R is subset of itself, but not a proper subset of itself because R is equal to itself. Exponential Notation So far, we have figured out how many subsets exist in a finite set by listing them. Recall that in , when we listed all the subsets of the three-element set L = { newspaper, magazine, book } we saw that there are eight subsets. In Your Turn 1.11 , we discovered that there are four subsets of the two-element subset, S = { heads, tails } . A one-element set has two subsets, the empty set and itself. The only subset of the empty set is the empty set itself. But how can we easily figure out the number of subsets in a very large finite set? It turns out that the number of subsets can be found by raising 2 to the number of elements in the set, using exponential notation to represent repeated multiplication. For example, the number of subsets of the set L = { newspaper, magazine, book } is equal to 2 3 = 2 ⋅ 2 ⋅ 2 = 8 . Exponential notation is used to represent repeated multiplication, b n = b ⋅ b ⋅ b ⋅ … ⋅ b , where b appears as a factor n times. The number of subsets of a finite set A is equal to 2 raised to the power of n ( A ) , where n ( A ) is the number of elements in set A : Number of Subsets of Set A = 2 n ( A ) . Note that 2 0 = 1 , so this formula works for the empty set, also. Computing the Number of Subsets of a Set Find the number of subsets of each of the following sets. The set of top five scorers of all time in the NBA: S = { LeBron James, Kareem Abdul-Jabbar, Karl Malone, Kobe Bryant, Michael Jordan } . The set of the top four bestselling albums of all time: A = { Thriller, Hotel California, The Beatles White Album, Led Zepplin IV } . R = { Snap, Crackle, Pop } . n ( S ) = 5 . So, the total number of subsets of S is 2 5 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 32 . n ( A ) = 4 . Therefore, the total number of subsets of A is 2 4 = 16 . n ( R ) = 3 . So, the total number of subsets of R is 2 3 = 8 . Equivalent Subsets In the early 17th century, the famous astronomer Galileo Galilei found that the set of natural numbers and the subset of the natural numbers consisting of the set of square numbers, n 2 , are equivalent. Upon making this discovery, he conjectured that the concepts of less than, greater than, and equal to did not apply to infinite sets. Sequences and series are defined as infinite subsets of the set of natural numbers by forming a relationship between the sequence or series in terms of a natural number, n . For example, the set of even numbers can be defined using set builder notation as { a | a = 2 n where n is a natural number } . The formula in this case replaces every natural number with two times the number, resulting in the set of even numbers, { 2 , 4 , 6 , … } . The set of even numbers is also equivalent to the set of natural numbers. Employment Opportunities You can make a career out of working with sets. Applications of equivalent sets include relational database design and analysis. Relational databases that store data are tables of related information. Each row of a table has the same number of columns as every other row in the table; in this way, relational databases are examples of set equivalences for finite sets. In a relational database, a primary key is set up to identify all related information. There is a one-to-one relationship between the primary key and any other information associated with it. Database design and analysis is a high demand career with a median entry-level salary of about $85,000 per year, according to salary.com. Writing Equivalent Subsets of an Infinite Set Using natural numbers, multiples of 3 are given by the sequence {3, 6, 9, …} . Write this set using set builder notation by expressing each multiple of 3 using a formula in terms of a natural number, n . { m | m = 3 n where n is a natural number} or { m | m = 3 n where n ∈ N } . In this example, m is a multiple of 3 and n is a natural number. The symbol ∈ is read as “is a member or element of.” Because there is a one-to-one correspondence between the set of multiples of 3 and the natural numbers, the set of multiples of 3 is an equivalent subset of the natural numbers. Creating Equivalent Subsets of a Finite Set That Are Not Equal A fast-food restaurant offers a deal where you can select two options from the following set of four menu items for $6: a chicken sandwich, a fish sandwich, a cheeseburger, or 10 chicken nuggets. Javier and his friend Michael are each purchasing lunch using this deal. Create two equivalent, but not equal, subsets that Javier and Michael could choose to have for lunch. The possible two-element subsets are: {chicken sandwich, fish sandwich}, {chicken sandwich, cheeseburger}, {chicken sandwich, chicken nuggets}, {fish sandwich, cheeseburger}, {fish sandwich, chicken nuggets}, and {cheeseburger, chicken nuggets}. One possible solution is that Javier picked the set {chicken sandwich, chicken nuggets}, while Michael chose the {cheeseburger, chicken nuggets}. Because Javier and Michael both picked two items, but not exactly the same two items, these sets are equivalent, but not equal. Creating Equivalent Subsets of a Finite Set A high school volleyball team at a small school consists of the following players: {Angie, Brenda, Colleen, Estella, Maya, Maria, Penny, Shantelle}. Create two possible equivalent starting line-ups of six players that the coach could select for the next game. There are actually 28 possible ways that the coach could choose his starting line-up. Two such equivalent subsets are {Angie, Brenda, Maya, Maria, Penny, Shantelle} and {Angie, Brenda, Colleen, Estella, Maria, Shantelle}. Each subset has six members, but they are not identical, so the two sets are equivalent but not equal. Check Your Understanding Key Terms subset proper subset equivalent subsets exponential notation Every member of a subset of a set is also a member of the set containing it. A ⊆ B A proper subset of a set does not contain all the members of the set containing it. There is a least one member of set B that is not a member of set A . A ⊂ B The number subsets of a finite set A with n ( A ) members is equal to 2 raised to the n ( A ) power. The empty set is a subset of every set and must be included when listing all the subsets of a set. Understand how to create and distinguish between equivalent subsets of finite and infinite sets that are not equal to the original set. Formulas The number of subsets of a finite set A is equal to 2 raised to the power of n ( A ) , where n ( A ) is the number of elements in set A : Number of Subsets of Set A = 2 n ( A ) .", "section": "Subsets", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Understanding Venn Diagrams When assembling furniture, instructions with images are easier to follow, just like how set relationships are easier to understand when depicted graphically. (credit: \"Time to assemble more Ikea furniture!\" by Rod Herrea/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Utilize a universal set with two sets to interpret a Venn diagram. Utilize a universal set with two sets to create a Venn diagram. Determine the complement of a set. Have you ever ordered a new dresser or bookcase that required assembly? When your package arrives you excitedly open it and spread out the pieces. Then you check the assembly guide and verify that you have all the parts required to assemble your new dresser. Now, the work begins. Luckily for you, the assembly guide includes step-by-step instructions with images that show you how to put together your product. If you are really lucky, the manufacturer may even provide a URL or QR code connecting you to an online video that demonstrates the complete assembly process. We can likely all agree that assembly instructions are much easier to follow when they include images or videos, rather than just written directions. The same goes for the relationships between sets. Interpreting Venn Diagrams Venn diagrams are the graphical tools or pictures that we use to visualize and understand relationships between sets. Venn diagrams are named after the mathematician John Venn, who first popularized their use in the 1880s. When we use a Venn diagram to visualize the relationships between sets, the entire set of data under consideration is drawn as a rectangle, and subsets of this set are drawn as circles completely contained within the rectangle. The entire set of data under consideration is known as the universal set . Consider the statement: All trees are plants. This statement expresses the relationship between the set of all plants and the set of all trees. Because every tree is a plant, the set of trees is a subset of the set of plants. To represent this relationship using a Venn diagram, the set of plants will be our universal set and the set of trees will be the subset. Recall that this relationship is expressed symbolically as: Trees ⊂ Plants . To create a Venn diagram, first we draw a rectangle and label the universal set “ U = Plants . ” Then we draw a circle within the universal set and label it with the word “Trees.” This section will introduce how to interpret and construct Venn diagrams. In future sections, as we expand our knowledge of relationships between sets, we will also develop our knowledge and use of Venn diagrams to explore how multiple sets can be combined to form new sets. Interpreting the Relationship between Sets in a Venn Diagram Write the relationship between the sets in the following Venn diagram, in words and symbolically. The set of terriers is a subset of the universal set of dogs. In other words, the Venn diagram depicts the relationship that all terriers are dogs. This is expressed symbolically as T ⊂ U . So far, the only relationship we have been considering between two sets is the subset relationship, but sets can be related in other ways. Lions and tigers are both different types of cats, but no lions are tigers, and no tigers are lions. Because the set of all lions and the set of all tigers do not have any members in common, we call these two sets disjoint sets , or non-overlapping sets. Two sets A and B are disjoint sets if they do not share any elements in common. That is, if a is a member of set A , then a is not a member of set B . If b is a member of set B , then b is not a member of set A . To represent the relationship between the set of all cats and the sets of lions and tigers using a Venn diagram, we draw the universal set of cats as a rectangle and then draw a circle for the set of lions and a separate circle for the set of tigers within the rectangle, ensuring that the two circles representing the set of lions and the set of tigers do not touch or overlap in any way. Describing the Relationship between Sets Describe the relationship between the sets in the following Venn diagram. The set of triangles and the set of squares are two disjoint subsets of the universal set of two-dimensional figures. The set of triangles does not share any elements in common with the set of squares. No triangles are squares and no squares are triangles, but both squares and triangles are 2D figures. Creating Venn Diagrams The main purpose of a Venn diagram is to help you visualize the relationship between sets. As such, it is necessary to be able to draw Venn diagrams from a written or symbolic description of the relationship between sets. Procedure To create a Venn diagram: Draw a rectangle to represent the universal set, and label it U = set name . Draw a circle within the rectangle to represent a subset of the universal set and label it with the set name. If there are multiple disjoint subsets of the universal set, their separate circles should not touch or overlap. Drawing a Venn Diagram to Represent the Relationship Between Two Sets Draw a Venn diagram to represent the relationship between each of the sets. All rectangles are parallelograms. All women are people. The set of rectangles is a subset of the set of parallelograms. First, draw a rectangle to represent the universal set and label it with U = Parallelograms , then draw a circle completely within the rectangle, and label it with the name of the set it represents, R = Rectangles . In this example, both letters and names are used to represent the sets involved, but this is not necessary. You may use either letters or names alone, as long as the relationship is clearly depicted in the diagram, as shown below. or The universal set is the set of people, and the set of all women is a subset of the set of people. Drawing a Venn Diagram to Represent the Relationship Between Three Sets All bicycles and all cars have wheels, but no bicycle is a car. Draw a Venn diagram to represent this relationship. Step 1: The set of bicycles and the set of cars are both subsets of the set of things with wheels. The universal set is the set of things with wheels, so we first draw a rectangle and label it with U = Things with Wheels . Step 2: Because the set of bicycles and the set of cars do not share any elements in common, these two sets are disjoint and must be drawn as two circles that do not touch or overlap with the universal set. The Complement of a Set Recall that if set A is a proper subset of set U , the universal set (written symbolically as A ⊂ U ), then there is at least one element in set U that is not in set A . The set of all the elements in the universal set U that are not in the subset A is called the complement of set A , A ' . In set builder notation this is written symbolically as: A ' = { x ∈ U | x ∉ A } . The symbol ∈ is used to represent the phrase, “is a member of,” and the symbol ∉ is used to represent the phrase, “is not a member of.” In the Venn diagram below, the complement of set A is the region that lies outside the circle and inside the rectangle. The universal set U includes all of the elements in set A and all of the elements in the complement of set A , and nothing else. Consider the set of digit numbers. Let this be our universal set, U = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . Now, let set A be the subset of U consisting of all the prime numbers in set U , A = { 2 , 3 , 5 , 7 } . The complement of set A is A ' = { 0 , 1 , 4 , 6 , 8 , 9 } . The following Venn diagram represents this relationship graphically. Finding the Complement of a Set For both of the questions below, A is a proper subset of U . Given the universal set U = { Billie Eilish, Donald Glover, Bruno Mars, Adele, Ed Sheeran} and set A = { Donald Glover, Bruno Mars, Ed Sheeran} , find A ' . Given the universal set U = { d|d is a dog } and B = { b ∈ U|b is a beagle } , find B ' . The complement of set A is the set of all elements in the universal set U that are not in set A . A ' = { Billie Eilish, Adele } . The complement of set B is the set of all dogs that are not beagles. All members of set B ′ are in the universal set because they are dogs, but they are not in set B , because they are not beagles. This relationship can be expressed in set build notation as follows: B ′ = { All dogs that are not beagles .} , B ′ = { d ∈ U | d is not a beagle .} , or B ′ = { d ∈ U | d ∉ B } . Check Your Understanding Key Terms Venn diagram universal set disjoint set complement of a set Key Concepts A Venn diagram is a graphical representation of the relationship between sets. In a Venn diagram, the universal set, U is the largest set under consideration and is drawn as a rectangle. All subsets of the universal set are drawn as circles within this rectangle. The complement of set A includes all the members of the universal set that are not in set A . A set and its complement are disjoint sets, they do not share any elements in common. To find the complement of set A remove all the elements of set A from the universal set U , the set that includes only the remaining elements is the complement of set A , A ′ . Determine the complement of a set using Venn diagrams, the roster method and set builder notation.", "section": "Understanding Venn Diagrams", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Set Operations with Two Sets A large, multigenerational family contains an intersection and a union of sets. (credit: “Family Photo Shoot Bani Syakur” by Mainur Risyada/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Determine the intersection of two sets. Determine the union of two sets. Determine the cardinality of the union of two sets. Apply the concepts of AND and OR to set operations. Draw conclusions from Venn diagrams with two sets. The movie Yours, Mine, and Ours was originally released in 1968 and starred Lucille Ball and Henry Fonda. This movie, which is loosely based on a true story, is about the marriage of Helen, a widow with eight children, and Frank, a widower with ten children, who then have an additional child together. The movie is a comedy that plays on the interpersonal and organizational struggles of feeding, bathing, and clothing twenty people in one household. If we consider the set of Helen's children and the set of Frank's children, then the child they had together is the intersection of these two sets, and the collection of all their children combined is the union of these two sets. In this section, we will explore the operations of union and intersection as it relates to two sets. The Intersection of Two Sets The members that the two sets share in common are included in the intersection of two sets . To be in the intersection of two sets, an element must be in both the first set and the second set. In this way, the intersection of two sets is a logical AND statement. Symbolically, A intersection B is written as: A ∩ B . A intersection B is written in set builder notation as: A ∩ B = { x | x ∈ A and x ∈ B } . Let us look at Helen's and Frank's children from the movie Yours, Mine, and Ours. Helen's children consist of the set H = { Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph } and Frank's children are included in the set F = { Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan, Joseph } . H intersection F is the set of children they had together. H ∩ F = { Joseph } , because Joseph is in both set H and set F . Finding the Intersection of Set A and Set B Set A = { 1 , 3 , 5 , 7 , 9 } and B = { 2 , 3 , 5 , 7 } . Find A intersection B . The intersection of sets A and B include the elements that set A and B have in common: 3, 5, and 7. A ∩ B = { 3 , 5 , 7 } . Notice that if sets A and B are disjoint sets, then they do not share any elements in common, and A intersection B is the empty set, as shown in the Venn diagram below. Determining the Intersection of Disjoint Sets Set A = { 0 , 2 , 4 , 6 , 8 } and set B = { 1 , 3 , 5 , 7 , 9 } . Find A ∩ B . Because sets A and B are disjoint, they do not share any elements in common. So, the intersection of set A and set B is the empty set. A ∩ B = ∅ . Notice that if set A is a subset of set B , then A intersection B is equal to set A , as shown in the Venn diagram below. Finding the Intersection of a Set and a Subset Set A = { 1 , 3 , 5 , … } and set B = ℕ = { 1 , 2 , 3 , … } Find A ∩ B . Because set A is a subset of set B , A intersection B is equal to set A . A ∩ B = A = { 1 , 3 , 5 , … } , the set of odd natural numbers. The Union of Two Sets Like the union of two families in marriage, the union of two sets includes all the members of the first set and all the members of the second set. To be in the union of two sets, an element must be in the first set, the second set, or both. In this way, the union of two sets is a logical inclusive OR statement. Symbolically, A union B is written as: A ∪ B . A union B is written in set builder notation as: A ∪ B = { x | x ∈ A or x ∈ B } . Let us consider the sets of Helen's and Frank's children from the movie Yours, Mine, and Ours again. Helen's children is set H = { Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph } and Frank's children is set F = { Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan, Joseph } . The union of these two sets is the collection of all nineteen of their children, H ∪ F = { Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph, Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan } . Notice, Joseph is in both set H and set F , but he is only one child, so, he is only listed once in the union. Finding the Union of Sets A and B When A and B Overlap Set A = { 1 , 3 , 5 , 7 , 9 } and set B = { 2 , 3 , 5 , 7 } . Find A union B . A union B is the set formed by including all the unique elements in set A , set B , or both sets A and B : A ∪ B = { 1 , 3 , 5 , 7 , 9 , 2 } . The first five elements of the union are the five unique elements in set A . Even though 3, 5, and 7 are also members of set B , these elements are only listed one time. Lastly, set B includes the unique element 2, so 2 is also included as part of the union of sets A and B . When observing the union of sets A and B , notice that both set A and set B are subsets of A union B . Graphically, A union B can be represented in several different ways depending on the members that they have in common. If A and B are disjoint sets, then A union B would be represented with two disjoint circles within the universal set, as shown in the Venn diagram below. A ∪ B If sets A and B share some, but not all, members in common, then the Venn diagram is drawn as two separate circles that overlap. If every member of set A is also a member of set B , then A is a subset of set B , and A union B would be equal to set B . To draw the Venn diagram, the circle representing set A should be completely enclosed in the circle containing set B . Finding the Union of Sets A and B When A and B Are Disjoint Set A = { 0 , 2 , 4 , 6 , 8 } and set B = { 1 , 3 , 5 , 7 , 9 } . Find A ∪ B . Because sets A and B are disjoint, the union is simply the set containing all the elements in both set A and set B . A ∪ B = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . Finding the Union of Sets A and B When One Set is a Subset of the Other Set A = { 1 , 3 , 5 , … } and set B = ℕ = { 1 , 2 , 3 , … } . Find A ∪ B . Because set A is a subset of set B , A union B is equal to set B . A ∪ B = ℕ = { 1 , 2 , 3 , … } = B . The Basics of Intersection of Sets, Union of Sets and Venn Diagrams Set Operation Practice Sets Challenge is an application available on both Android and iPhone smartphones that allows you to practice and gain familiarity with the operations of set union, intersection, complement, and difference. Google Play Store image of Sets Challenge game. (credit: screenshot from Google Play) The Sets Challenge application/game uses some notation that differs from the notation covered in the text. The complement of set A in this text is written symbolically as A ′ , but the Sets Challenge game uses A C to represent the complement operation. In the text we do not cover set difference between two sets A and B , represented in the game as A − B . In the game this operation removes from set A all the elements in A ∩ B . For example, if set A = { a , b , c , d } and set B = { b , d , f , h } are subsets of the universal set U = { a , b , c , … , z } , then A − B = { a , b , c , d } − { b , d } = { a , c } , and B − A = { b , d , f , h } − { b , d } = { f , h } . There is a project at the end of the chapter to research the set difference operation. Determining the Cardinality of Two Sets The cardinality of the union of two sets is the total number of elements in the set. Symbolically the cardinality of A union B is written, n ( A ∪ B ) . If two sets A and B are disjoint, the cardinality of A union B is the sum of the cardinality of set A and the cardinality of set B . If the two sets intersect, then A intersection B is a subset of both set A and set B . This means that if we add the cardinality of set A and set B , we will have added the number of elements in A intersection B twice, so we must then subtract it once as shown in the formula that follows. The cardinality of A union B is found by adding the number of elements in set A to the number of elements in set B , then subtracting the number of elements in the intersection of set A and set B . n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) or n ( A or B ) = n ( A ) + n ( B ) − n ( A and B ) . If sets A and B are disjoint, then n ( A ∩ B ) = n ( A and B ) = 0 and the formula is still valid, but simplifies to n ( A ∪ B ) = n ( A ) + n ( B ) . Determining the Cardinality of the Union of Two Sets The number of elements in set A is 10, the number of elements in set B is 20, and the number of elements in A intersection B is 4. Find the number of elements in A union B . Using the formula for determining the cardinality of the union of two sets, we can say n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) = 10 + 20 − 4 = 26. Determining the Cardinality of the Union of Two Disjoint Sets If A and B are disjoint sets and the cardinality of set A is 37 and the cardinality of set B is 43, find the cardinality of A union B . To find the cardinality of A union B , apply the formula, n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) . Because sets A and B are disjoint, A ∩ B is the empty set, therefore n ( A ∩ B ) = n ( ∅ ) = 0 and n ( A ∪ B ) = 37 + 43 − 0 = 80. Applying Concepts of “AND” and “OR” to Set Operations To become a licensed driver, you must pass some form of written test and a road test, along with several other requirements depending on your age. To keep this example simple, let us focus on the road test and the written test. If you pass the written test but fail the road test, you will not receive your license. If you fail the written test, you will not be allowed to take the road test and you will not receive a license to drive. To receive a driver's license, you must pass the written test AND the road test. For an “AND” statement to be true, both conditions that make up the statement must be true. Similarly, the intersection of two sets A and B is the set of elements that are in both set A and set B . To be a member of A intersection B , an element must be in set A and also must be in set B . The intersection of two sets corresponds to a logical \"AND\" statement. The union of two sets is a logical inclusive \"OR\" statement. Say you are at a birthday party and the host offers Leah, Lenny, Maya, and you some cake or ice cream for dessert. Leah asks for cake, Lenny accepts both cake and ice cream, Maya turns down both, and you choose only ice cream. Leah, Lenny, and you are all having dessert. The “OR” statement is true if at least one of the components is true. Maya is the only one who did not have cake or ice cream; therefore, she did not have dessert and the “OR” statement is false. To be in the union of two sets A and B , an element must be in set A or set B or both set A and set B . Applying the \"AND\" or \"OR\" Operation A = { 0 , 3 , 6 , 9 , 12 } , B = { 0 , 4 , 8 , 12 , 16 } , and C = { 1 , 2 , 3 , 5 , 8 , 13 } . Find the set consisting of elements in: A and B . A or B . A or C . ( B and C ) or A . A and B = A ∩ B = { 0 , 12 } , because only the elements 0 and 12 are members of both set A and set B . A or B = A ∪ B = { 0 , 3 , 4 , 6 , 8 , 9 , 12 , 16 } , because the set A or B is the collection of all elements in set A or set B , or both. A or C = A ∪ C = { 0 , 1 , 2 , 3 , 5 , 6 , 8 , 9 , 12 , 13 } , because the set A or C is the collection of all elements in set A or set C , or both. ( B and C ) or A = ( B ∩ C ) ∪ A . Parentheses are evaluated first: ( B and C ) = B ∩ C = { 8 } , because the only member that both set B and set C share in common is 8. So, now we need to find { 8 } or { 0 , 3 , 6 , 9 , 12 } , Because the word translates to the union operation, the problem becomes { 8 } ∪ { 0 , 3 , 6 , 9 , 12 } , which is equal to { 0 , 3 , 6 , 8 , 9 , 12 } . Determine and Apply the Appropriate Set Operations to Solve the Problem Don Woods is serving cake and ice cream at his Juneteenth celebration. The party has a total of 54 guests in attendance. Suppose 30 guests requested cake, 20 guests asked for ice cream, and 12 guests did not have either cake or ice cream. How many guests had cake or ice cream? How many guests had cake and ice cream? The total number of people at the party is 54, and 12 people did not have cake or ice cream. Recall that the total number of elements in the universal set is always equal to the number of elements in a subset plus the number of elements in the complement of the set, n ( U ) = n ( A ) + n ( A ′ ) . That means 54 = n ( cake or ice cream ) + n ( not ( cake or ice cream ) ) , or equivalently, n ( cake ∪ ice cream ) = 54 − n ( ( cake ∪ ice cream ) ′ ) = 54 − 12 = 42. A total of 42 people at the party had cake or ice cream. To determine the number of people who had both cake and ice cream, we need to find the intersection of the set of people who had cake and the set of people who had ice cream. From Question 1, the number of people who had cake or ice cream is 42. This is the union of the two sets. The formula for the union of two sets is n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) . Use the information given in the problem and substitute the known values into the formula to solve for the number of people in the intersection: 42 = 30 + 20 − n ( A ∩ B ) . Adding 30 and 20, the equation simplifies to 42 = 50 − n ( cake and ice cream) . Which means n ( cake and ice cream) = 50 − 42 = 8. The Real Inventor of the Venn Diagram John Venn, in his writings, references works by both John Boole and Augustus De Morgan, who referred to the circle diagrams commonly used to present logical relationships as Euler's circles. Leonhard Euler's works were published over 100 years prior to Venn's, and Euler may have been influenced by the works of Gottfried Leibniz. So, why does John Venn get all the credit for these graphical depictions? Venn was the first to formalize the use of these diagrams in his book Symbolic Logic, published in 1881. Further, he made significant improvements in their design, including shading to highlight the region of interest. The mathematician C.L. Dodgson, also known as Lewis Carroll, built upon Venn’s work by adding an enclosing universal set. Invention is not necessarily coming up with an initial idea. It is about seeing the potential of an idea and applying it to a new situation. References: Margaret E. Baron. \"A Note on the Historical Development of Logic Diagrams: Leibniz, Euler and Venn.\" The Mathematical Gazette , vol. 53, no. 384, 1969, pp. 113-125. JSTOR , www.jstor.org/stable/3614533. Accessed 15 July 2021. Deborah Bennett. \"Drawing Logical Conclusions.\" Math Horizons , vol. 22, no. 3, 2015, pp. 12-15. JSTOR , www.jstor.org/stable/10.4169/mathhorizons.22.3.12. Accessed 15 July 2021. Drawing Conclusions from a Venn Diagram with Two Sets All Venn diagrams will display the relationships between the sets, such as subset, intersecting, and/or disjoint. In addition to displaying the relationship between the two sets, there are two main additional details that Venn diagrams can include: the individual members of the sets or the cardinality of each disjoint subset of the universal set. A Venn diagram with two subsets will partition the universal set into 3 or 4 sections depending on whether they are disjoint or intersecting sets. Recall that the complement of set A , written A ′ , is the set of all elements in the universal set that are not in set A . Side-by-side Venn diagrams with disjoint and intersecting sets, respectively. Using a Venn Diagram to Draw Conclusions about Set Membership Find A ∪ B . Find A ∩ B . Find B ′ . Find n ( B ′ ) . A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 } , because A union B is the collection of all elements in set A or set B or both. Because A and B are disjoint sets, there are no elements that are in both A and B . Therefore, A intersection B is the empty set, A ∩ B = ∅ . The complement of set B is the set of all elements in the universal set that are not in set B : B ′ = { 0 , 1 , 3 , 5 , 7 , 9 } . The cardinality, or number of elements in set B ′ , is n ( B ′ ) = 6. Using a Venn Diagram to Draw Conclusions about Set Cardinality Venn diagram with two intersecting sets and number of elements in each section indicated. Find n ( A or B ) . Find n ( A and B ) . Find n ( A ) . The number of elements in A or B is the number of elements in A union B : n ( A ∪ B ) = n ( { 2 , 5 , 7 } ) = 14 . The number of elements in A and B is the number of elements in A intersection B : n ( A ∩ B ) = 5 . The number of elements in set A is the sum of all the numbers enclosed in the circle representing set A : n ( A ) = n ( { 7 , 5 } ) = 12 . Check Your Understanding Key Terms intersection of two sets union of two sets Key Concepts The intersection of two sets, A ∩ B is the set of all elements that they have in common. Any member of A intersection B must be is both set A and set B . The union of two sets, A ∪ B , is the collection of all members that are in either in set A , set B or both sets A and B combined. Two sets that share at least one element in common, so that they are not disjoint are represented in a Venn Diagram using two circles that overlap. The region of the overlap is the set A intersection B , A ∩ B . The regions that include everything in the circle representing set A or the circle representing set B or their overlap is the set A union B , A ∪ B . Apply knowledge of set union and intersection to determine cardinality and membership using Venn Diagrams, the roster method and set builder notation. Formulas The cardinality of A union B : n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) Video The Basics of Intersection of Sets, Union of Sets and Venn Diagrams", "section": "Set Operations with Two Sets", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Set Operations with Three Sets Companies like Google collect data on how you use their services, but the data requires analysis to really mean something. (credit: “Man holding smartphone and searches through google” by Nenad Stojkovic/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Interpret Venn diagrams with three sets. Create Venn diagrams with three sets. Apply set operations to three sets. Prove equality of sets using Venn diagrams. Have you ever searched for something on the Internet and then soon after started seeing multiple advertisements for that item while browsing other web pages? Large corporations have built their business on data collection and analysis. As we start working with larger data sets, the analysis becomes more complex. In this section, we will extend our knowledge of set relationships by including a third set. A Venn diagram with two intersecting sets breaks up the universal set into four regions; simply adding one additional set will increase the number of regions to eight, doubling the complexity of the problem. Venn Diagrams with Three Sets Below is a Venn diagram with two intersecting sets, which breaks the universal set up into four distinct regions. Next, we see a Venn diagram with three intersecting sets , which breaks up the universal set into eight distinct regions. Shading Venn Diagrams Venn Diagram is an Android application that allows you to visualize how the sets are related in a Venn diagram by entering expressions and displaying the resulting Venn diagram of the set shaded in gray. Google Play Store image of Venn Diagram app. (credit: screenshot from Google Play) The Venn Diagram application uses some notation that differs from the notation covered in this text. The complement of set A in this text is written symbolically as A ′ , but the Venn Diagram app uses A C to represent the complement operation. The set difference operation, − , is available in the Venn Diagram app, although this operation is not covered in the text. It is recommended that you explore this application to expand your knowledge of Venn diagrams prior to continuing with the next example. In the next example, we will explore the three main blood factors, A, B and Rh. The following background information about blood types will help explain the relationships between the sets of blood factors. If an individual has blood factor A or B, those will be included in their blood type. The Rh factor is indicated with a + or a − . For example, if a person has all three blood factors, then their blood type would be AB + . In the Venn diagram, they would be in the intersection of all three sets, A ∩ B ∩ R h + . If a person did not have any of these three blood factors, then their blood type would be O − , and they would be in the set ( A ∪ B ∪ R h + ) ′ which is the region outside all three circles. Interpreting a Venn Diagram with Three Sets Use the Venn diagram below, which shows the blood types of 100 people who donated blood at a local clinic, to answer the following questions. How many people with a type A blood factor donated blood? Julio has blood type B + . If he needs to have surgery that requires a blood transfusion, he can accept blood from anyone who does not have a type A blood factor. How many people donated blood that Julio can accept? How many people who donated blood do not have the Rh + blood factor? How many people had type A and type B blood? The number of people who donated blood with a type A blood factor will include the sum of all the values included in the A circle. It will be the union of sets A − , A + , A B − and A B + . n ( A ) = n ( A − ) + n ( A + ) + n ( A B − ) + n ( A B + ) = 6 + 36 + 1 + 3 = 46. In part 1, it was determined that the number of donors with a type A blood factor is 46. To determine the number of people who did not have a type A blood factor, use the following property, A ′ union is equal to U , which means n ( A ) + n ( A ′ ) = n ( U ) , and n ( A ′ ) = n ( U ) − n ( A ) = 100 − 46 = 54. Thus, 54 people donated blood that Julio can accept. This would be everyone outside the Rh + circle, or everyone with a negative Rh factor, n ( R h − ) = n ( O − ) + n ( A − ) + n ( A B − ) + n ( B − ) = 7 + 6 + 1 + 2 = 16. To have both blood type A and blood type B, a person would need to be in the intersection of sets A and B . The two circles overlap in the regions labeled A B − and A B + . Add up the number of people in these two regions to get the total: 1 + 3 = 4. This can be written symbolically as n ( A and B ) = n ( A ∩ B ) = n ( A B − ) + n ( A B + ) = 1 + 3 = 4. Blood Types Most people know their main blood type of A, B, AB, or O and whether they are R h + or R h − , but did you know that the International Society of Blood Transfusion recognizes twenty-eight additional blood types that have important implications for organ transplants and successful pregnancy? For more information, check out this article: Blood mystery solved: Two new blood types identified Creating Venn Diagrams with Three Sets In general, when creating Venn diagrams from data involving three subsets of a universal set, the strategy is to work from the inside out. Start with the intersection of the three sets, then address the regions that involve the intersection of two sets. Next, complete the regions that involve a single set, and finally address the region in the universal set that does not intersect with any of the three sets. This method can be extended to any number of sets. The key is to start with the region involving the most overlap, working your way from the center out. Creating a Venn Diagram with Three Sets A teacher surveyed her class of 43 students to find out how they prepared for their last test. She found that 24 students made flash cards, 14 studied their notes, and 27 completed the review assignment. Of the entire class of 43 students, 12 completed the review and made flash cards, nine completed the review and studied their notes, and seven made flash cards and studied their notes, while only five students completed all three of these tasks. The remaining students did not do any of these tasks. Create a Venn diagram with subsets labeled: “Notes,” “Flash Cards,” and “Review” to represent how the students prepared for the test. Step 1: First, draw a Venn diagram with three intersecting circles to represent the three intersecting sets: Notes, Flash Cards, and Review. Label the universal set with the cardinality of the class. Step 2: Next, in the region where all three sets intersect, enter the number of students who completed all three tasks. Step 3: Next, calculate the value and label the three sections where just two sets overlap. Review and flash card overlap . A total of 12 students completed the review and made flash cards, but five of these twelve students did all three tasks, so we need to subtract: 12 − 5 = 7 . This is the value for the region where the flash card set intersects with the review set. Review and notes overlap . A total of 9 students completed the review and studied their notes, but again, five of these nine students completed all three tasks. So, we subtract: 9 − 5 = 4 . This is the value for the region where the review set intersects with the notes set. Flash card and notes overlap . A total of 7 students made flash cards and studied their notes; subtracting the five students that did all three tasks from this number leaves 2 students who only studied their notes and made flash cards. Add these values to the Venn diagram. Step 4: Now, repeat this process to find the number of students who only completed one of these three tasks. A total of 24 students completed flash cards, but we have already accounted for 2 + 5 + 7 = 14 of these. Thus, 24 - 14 = 10 students who just made flash cards. A total of 14 students studied their notes, but we have already accounted for 4 + 5 + 2 = 11 of these. Thus, 14 - 11 = 3 students only studied their notes. A total of 27 students completed the review assignment, but we have already accounted for 4 + 5 + 7 = 16 of these, which means 27 - 16 = 11 students only completed the review assignment. Add these values to the Venn diagram. Step 5: Finally, compute how many students did not do any of these three tasks. To do this, we add together each value that we have already calculated for the separate and intersecting sections of our three sets: 3 + 2 + 4 + 5 + 10 + 7 + 11 = 42 . Because there 43 students in the class, and 43 − 42 = 1 , this means only one student did not complete any of these tasks to prepare for the test. Record this value somewhere in the rectangle, but outside of all the circles, to complete the Venn diagram. Applying Set Operations to Three Sets Set operations are applied between two sets at a time. Parentheses indicate which operation should be performed first. As with numbers, the inner most parentheses are applied first. Next, find the complement of any sets, then perform any union or intersections that remain. Applying Set Operations to Three Sets Perform the set operations as indicated on the following sets: U = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } , A = { 0 , 1 , 2 , 3 , 4 , 5 , 6 } , B = { 0 , 2 , 4 , 6 , 8 , 10 , 12 } , and C = { 0 , 3 , 6 , 9 , 12 } . Find ( A ∩ B ) ∩ C . Find A ∩ ( B ∪ C ) . Find ( A ∩ B ) ∪ C ′ . Parentheses first, A intersection B equals A ∩ B = { 0 , 2 , 4 , 6 } , the elements common to both A and B . ( A ∩ B ) ∩ C = { 0 , 2 , 4 , 6 } ∩ { 0 , 3 , 6 , 9 , 12 } = { 0 , 6 } , because the only elements that are in both sets are 0 and 6. Parentheses first, B union C equals B ∪ C = { 0 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 } , the collection of all elements in set B or set C or both. A ∩ ( B ∪ C ) = { 0 , 1 , 2 , 3 , 4 , 5 , 6 } ∩ { 0 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 } = { 0 , 2 , 3 , 4 , 6 } , because the intersection of these two sets is the set of elements that are common to both sets. Parentheses first, A intersection B equals A ∩ B = { 0 , 2 , 4 , 6 } . Next, find C ′ . The complement of set C is the set of elements in the universal set U that are not in set C . C ′ = { 1 , 2 , 4 , 5 , 7 , 8 , 10 , 11 } . Finally, find ( A ∩ B ) ∪ C ′ = { 0 , 2 , 4 , 6 } ∪ { 1 , 2 , 4 , 5 , 7 , 8 , 10 , 11 } = { 0 , 1 , 2 , 4 , 5 , 6 , 7 , 8 , 10 , 11 } . Notice that the answers to the Your Turn are the same as those in the Example. This is not a coincidence. The following equivalences hold true for sets: A ∩ ( B ∩ C ) = ( A ∩ B ) ∩ C and A ∪ ( B ∪ C ) = ( A ∪ B ) ∪ C . These are the associative property for set intersection and set union. A ∩ B = B ∩ A and A ∪ B = B ∪ A . These are the commutative property for set intersection and set union. A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) and A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) . These are the distributive property for sets over union and intersection, respectively. Proving Equality of Sets Using Venn Diagrams To prove set equality using Venn diagrams, the strategy is to draw a Venn diagram to represent each side of the equality, then look at the resulting diagrams to see if the regions under consideration are identical. Augustus De Morgan was an English mathematician known for his contributions to set theory and logic. De Morgan’s law for set complement over union states that ( A ∪ B ) ′ = A ′ ∩ B ′ . In the next example, we will use Venn diagrams to prove De Morgan’s law for set complement over union is true. But before we begin, let us confirm De Morgan’s law works for a specific example. While showing something is true for one specific example is not a proof, it will provide us with some reason to believe that it may be true for all cases. Let U = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } , A = { 2 , 3 , 4 } , and B = { 3 , 4 , 5 , 6 } . We will use these sets in the equation ( A ∪ B ) ′ = A ′ ∩ B ′ . To begin, find the value of the set defined by each side of the equation. Step 1: A ∪ B is the collection of all unique elements in set A or set B or both. A ∪ B = { 2 , 3 , 4 , 5 , 6 } . The complement of A union B , ( A ∪ B ) ′ , is the set of all elements in the universal set that are not in A ∪ B . So, the left side the equation ( A ∪ B ) ′ is equal to the set { 1 , 7 } . Step 2: The right side of the equation is A ′ ∩ B ′ . A ′ is the set of all members of the universal set U that are not in set A . A ′ = { 1 , 5 , 6 , 7 } . Similarly, B ′ = { 1 , 2 , 7 } . Step 3: Finally, A ′ ∩ B ′ is the set of all elements that are in both A ′ and B ′ . The numbers 1 and 7 are common to both sets, therefore, A ′ ∩ B ′ = { 1 , 7 } . Because, { 1 , 7 } = { 1 , 7 } we have demonstrated that De Morgan’s law for set complement over union works for this particular example. The Venn diagram below depicts this relationship. Proving De Morgan’s Law for Set Complement over Union Using a Venn Diagram De Morgan’s Law for the complement of the union of two sets A and B states that: ( A ∪ B ) ′ = A ′ ∩ B ′ . Use a Venn diagram to prove that De Morgan’s Law is true. Step 1: First, draw a Venn diagram representing the left side of the equality. The regions of interest are shaded to highlight the sets of interest. A ∪ B is shaded on the left, and ( A ∪ B ) ′ is shaded on the right. Step 2: Next, draw a Venn diagram to represent the right side of the equation. A ′ is shaded and B ′ is shaded. Because A ′ and B ′ mix to form A ′ ∩ B ′ is also shaded. Venn diagram of intersection of the complement of two sets. Step 3: Verify the conclusion. Because the shaded region in the Venn diagram for ( A ∪ B ) ′ matches the shaded region in the Venn diagram for A ′ ∩ B ' , the two sides of the equation are equal, and the statement is true. This completes the proof that De Morgan’s law is valid. Check Your Understanding Key Concepts A Venn diagram with two overlapping sets breaks the universal set up into four distinct regions. When a third overlapping set is added the Venn diagram is broken up into eight distinct regions. Analyze, interpret, and create Venn diagrams involving three overlapping sets. Including the blood factors: A, B and Rh To find unions and intersections. To find cardinality of both unions and intersections. When performing set operations with three or more sets, the order of operations is inner most parentheses first, then fine the complement of any sets, then perform any union or intersection operations that remain. To prove set equality using Venn diagrams the strategy is to draw a Venn diagram to represent each side of the equality or equation, then look at the resulting diagrams to see if the regions under consideration are identical. If they regions are identical the equation represents a true statement, otherwise it is not true. Projects Cardinality of Infinite Sets In set theory, it has been shown that the set of irrational numbers has a cardinality greater than the set of natural numbers. That is, the set of irrational numbers is so large that it is uncountably infinite. Perform a search with the phrase, “Who first proved that the real numbers are uncountable?” Who first proved that the real numbers are uncountable? What was the significance of this proof to the development of set theory and by extension other fields of mathematics? Recent discoveries in the field of set theory include the solution to a 70-year-old problem previously thought to be unprovable. To learn more read this article : What does it mean for two infinite sets to have the same size? The real numbers are sometimes referred to as what? Summarize your understanding of the problem known as the “Continuum Hypothesis.” Malliaris and Shelah’s proof of this 70-year-old problem is opening up investigation in what two fields of mathematics? Summarize your understanding of infinity. Define what it means to be infinite. Explain the difference between countable and uncountable sets. Research the difference between a discrete set and a continuous set, then summarize your findings. Set Notation In arithmetic, the operation of addition is represented by the plus sign, +, but multiplication is represented in multiple ways, including ⋅ , × , ∗ , and parentheses, such as 5(3). Several set operations also are written in different forms based on the preferences of the mathematician and often their publisher. Search for “Set Complement” on the internet and list at least three ways to represent the complement of a set. Both the Set Challenge and Venn Diagram smartphone apps highlighted in the Tech Check sections have an operation for set difference. List at least two ways to represent set difference and provide a verbal description of how to calculate the difference between two sets A and B . When researching possible Venn diagram applications, the Greek letter delta, Δ appeared as a symbol for a set operator. List at least one other symbol used for this same operation. Search for “List of possible set operations and their symbols.” Find and select two symbols that were not presented in this chapter. The Real Number System The set of real numbers and their properties are studied in elementary school today, but how did the number system evolve? The idea of natural numbers or counting numbers surfaced prior to written words, as evidenced by tally marks in cave writing. Create a timeline for significant contributions to the real number system. Use the following phrase to search online for information on the origins of the number zero: “History of the number zero.” Then, record significant dates for the invention and common use of the number zero on your timeline. Find out who is credited for discovering that the 2 is irrational and add this information to your timeline. Hint: Search for, “Who was the first to discover irrational numbers?” Research Georg Cantor’s contribution to the representation of real numbers as a continuum and add this to your timeline. Research Ernst Zermelo’s contribution to the real number system and add this to your timeline. Chapter Review Basic Set Concepts Subsets Understanding Venn Diagrams Set Operations with Two Sets Set Operations with Three Sets Chapter Test", "section": "Set Operations with Three Sets", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Logic is key to a well-reasoned argument, in both math and law. (credit: modification of work \"Lady Justicia holding sword and scale bronze figurine with judge hammer on wooden table\" by Jernej Furman/Flickr, CC BY 2.0) What is logic? Logic is the study of reasoning, and it has applications in many fields, including philosophy, law, psychology, digital electronics, and computer science. In law, constructing a well-reasoned, logical argument is extremely important. The main goal of arguments made by lawyers is to convince a judge and jury that their arguments are valid and well-supported by the facts of the case, so the case should be ruled in their favor. Think about Thurgood Marshall arguing for desegregation in front of the U.S. Supreme Court during the Brown v. Board of Education of Topeka lawsuit in 1954, or Ruth Bader Ginsburg arguing for equality in social security benefits for both men and women under the law during the mid-1970s. Both these great minds were known for the preparation and thoroughness of their logical legal arguments, which resulted in victories that advanced the causes they fought for. Thurgood Marshall and Ruth Bader Ginsburg would later become well respected justices on the U.S. Supreme Court themselves. In this chapter, we will explore how to construct well-reasoned logical arguments using varying structures. Your ability to form and comprehend logical arguments is a valuable tool in many areas of life, whether you're planning a dinner date, negotiating the purchase of a new car, or persuading your boss that you deserve a raise.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Statements and Quantifiers Construction of a logical argument, like that of a house, requires you to begin with the right parts. (credit: modification of work “Barn Raising” by Robert Stinnett/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify logical statements. Represent statements in symbolic form. Negate statements in words. Negate statements symbolically. Translate negations between words and symbols. Express statements with quantifiers of all, some, and none. Negate statements containing quantifiers of all, some, and none. Have you ever built a club house, tree house, or fort with your friends? If so, you and your friends likely started by gathering some tools and supplies to work with, such as hammers, saws, screwdrivers, wood, nails, and screws. Hopefully, at least one member of your group had some knowledge of how to use the tools correctly and helped to direct the construction project. After all, if your house isn't built on a strong foundation, it will be weak and could possibly fall apart during the next big storm. This same foundation is important in logic. In this section, we will begin with the parts that make up all logical arguments. The building block of any logical argument is a logical statement , or simply a statement. A logical statement has the form of a complete sentence, and it must make a claim that can be identified as being true or false. When making arguments, sometimes people make false claims. When evaluating the strength or validity of a logical argument, you must also consider the truth values , or the identification of true or false, of all the statements used to support the argument. While a false statement is still considered a logical statement, a strong logical argument starts with true statements. Identifying Logical Statements Not all roses are red. (credit: “assorted pink yellow white red roses macro” by ProFlowers/Flickr, CC BY 2.0) An example of logical statement with a false truth value is, “All roses are red.” It is a logical statement because it has the form of a complete sentence and makes a claim that can be determined to be either true or false. It is a false statement because not all roses are red: some roses are red, but there are also roses that are pink, yellow, and white. Requests, questions, or directives may be complete sentences, but they are not logical statements because they cannot be determined to be true or false. For example, suppose someone said to you, “Please, sit down over there.” This request does not make a claim and it cannot be identified as true or false; therefore, it is not a logical statement. Identifying Logical Statements Determine whether each of the following sentences represents a logical statement. If it is a logical statement, determine whether it is true or false. Tiger Woods won the Master’s golf championship at least five times. Please, sit down over there. All cats dislike dogs. This is a logical statement because it is a complete sentence that makes a claim that can be identified as being true or false. As of 2021, this statement is true: Tiger Woods won the Master’s in 1997, 2001, 2002, 2005 and 2019. This is not a logical statement because, although it is a complete sentence, this request does not make a claim that can be identified as being either true or false. This is a logical statement because it is a complete sentence that makes a claim that can be identified as being true or false. This statement is false because some cats do like some dogs. Representing Statements in Symbolic Form When analyzing logical arguments that are made of multiple logical statements, symbolic form is used to reduce the amount of writing involved. Symbolic form also helps visualize the relationship between the statements in a more concise way in order to determine the strength or validity of an argument. Each logical statement is represented symbolically as a single lowercase letter, usually starting with the letter p . To begin, you will practice how to write a single logical statement in symbolic form. This skill will become more useful as you work with compound statements in later sections. Representing Statements Using Symbolic Form Write each of the following logical statements in symbolic form. Barry Bonds holds the Major League Baseball record for total career home runs. Some mammals live in the ocean. Ruth Bader Ginsburg served on the U.S. Supreme Court from 1993 to 2020. p : Barry Bonds holds the Major League Baseball record for total career home runs. The statement is labeled with a p . Once the statement is labeled, use p as a replacement for the full written statement to write and analyze the argument symbolically. q : Some mammals live in the ocean. The letter q is used to distinguish this statement from the statement in question 1, but any lower-case letter may be used. r : Ruth Bader Ginsburg served on the U.S. Supreme Court from 1993 to 2020. When multiple statements are present in later sections, you will want to be sure to use a different letter for each separate logical statement. Mathematics is not the only language to use symbols to represent thoughts or ideas. The Chinese and Japanese languages use symbols known as Hanzi and Kanji, respectively, to represent words and phrases. At one point, the American musician Prince famously changed his name to a symbol representing love. BBC Prince Symbol Article Negating Statements Consider the false statement introduced earlier, “All roses are red.” If someone said to you, “All roses are red,” you might respond with, “Some roses are not red.” You could then strengthen your argument by providing additional statements, such as, “There are also white roses, yellow roses, and pink roses, to name a few.” The negation of a logical statement has the opposite truth value of the original statement. If the original statement is false, its negation is true, and if the original statement is true, its negation is false. Most logical statements can be negated by simply adding or removing the word not . For example, consider the statement, “Emma Stone has green eyes.” The negation of this statement would be, “Emma Stone does not have green eyes.” The table below gives some other examples. Logical Statement Negation Gordon Ramsey is a chef. Gordon Ramsey is not a chef. Tony the Tiger does not have spots. Tony the Tiger has spots. The way you phrase your argument can impact its success. If someone presents you with a false statement, the ability to rebut that statement with its negation will provide you with the tools necessary to emphasize the correctness of your position. Negating Logical Statements Write the negation of each logical statement in words. Michael Phelps was an Olympic swimmer. Tom is a cat. Jerry is not a mouse. Add the word not to negate the affirmative statement: “Michael Phelps was not an Olympic swimmer.” Add the word not to negate the affirmative statement: “Tom is not a cat.” Remove the word not to negate the negative statement: “Jerry is a mouse.” Negating Logical Statements Symbolically The symbol for negation, or not, in logic is the tilde, ~. So, not p is represented as ~ p . To negate a statement symbolically, remove or add a tilde. The negation of not (not p ) is p . Symbolically, this equation is ~ ( ~ p ) = p . Negating Logical Statements Symbolically Write the negation of each logical statement symbolically. p : Michael Phelps was an Olympic swimmer. r : Tom is not a cat. ~ q : Jerry is not a mouse. To negate an affirmative logical statement symbolically, add a tilde: ~ p . Because the symbol for this statement is r , its negation is ~ r . The symbol for this statement is ~ q , so to negate it we simply remove the ~, because ~ ( ~ q ) = q . The answer is q . Translating Negations Between Words and Symbols In order to analyze logical arguments, it is important to be able to translate between the symbolic and written forms of logical statements. Translating Negations Between Words and Symbols Given the statements: r : Elmo is a red Muppet. p : Ketchup is not a vegetable. Write the symbolic form of the statement, “Elmo is not a red Muppet.” Translate the statement ~ p into words. “Elmo is not a red muppet” is the negation of “Elmo is a red muppet,” which is represented as r . Thus, we would write “Elmo is not a red muppet” symbolically as ~ r . Because p is the symbol representing the statement, “Ketchup is not a vegetable,” ~ p is equivalent to the statement, “Ketchup is a vegetable.” Expressing Statements with Quantifiers of All, Some, or None A quantifier is a term that expresses a numerical relationship between two sets or categories. For example, all squares are also rectangles, but only some rectangles are squares, and no squares are circles. In this example, all, some, and none are quantifiers. In a logical argument, the logical statements made to support the argument are called premises , and the judgment made based on the premises is called the conclusion . Logical arguments that begin with specific premises and attempt to draw more general conclusions are called inductive arguments . Consider, for example, a parent walking with their three-year-old child. The child sees a cardinal fly by and points it out. As they continue on their walk, the child notices a robin land on top of a tree and a duck flying across to land on a pond. The child recognizes that cardinals, robins, and ducks are all birds, then excitedly declares, \"All birds fly!\" The child has just made an inductive argument. They noticed that three different specific types of birds all fly, then synthesized this information to draw the more general conclusion that all birds can fly. In this case, the child's conclusion is false. The specific premises of the child's argument can be paraphrased by the following statements: Premise: Cardinals are birds that fly. Premise: Robins are birds that fly. Premise: Ducks are birds that fly. The general conclusion is: “All birds fly!” All inductive arguments should include at least three specific premises to establish a pattern that supports the general conclusion. To counter the conclusion of an inductive argument, it is necessary to provide a counter example. The parent can tell the child about penguins or emus to explain why that conclusion is false. On the other hand, it is usually impossible to prove that an inductive argument is true. So, inductive arguments are considered either strong or weak. Deciding whether an inductive argument is strong or weak is highly subjective and often determined by the background knowledge of the person making the judgment. Most hypotheses put forth by scientists using what is called the “scientific method” to conduct experiments are based on inductive reasoning. In the following example, we will use quantifiers to express the conclusion of a few inductive arguments. Drawing General Conclusions to Inductive Arguments Using Quantifiers For each series of premises, draw a logical conclusion to the argument that includes one of the following quantifiers: all, some, or none. Squares and rectangles have four sides. A square is a parallelogram, and a rectangle is a parallelogram. What conclusion can be drawn from these premises? 1 + 2 = 3 , 6 + 7 = 13 , and 23 + 24 = 47. Of these, 1 and 2, 6 and 7, and 23 and 24 are consecutive integers; 3, 13, and 47 are odd numbers. What conclusion can be drawn from these premises? Sea urchins live in the ocean, and they do not breathe air. Sharks live in the ocean, and they do not breathe air. Eels live in the ocean, and they do not breath air. What conclusion can be drawn from these premises? The conclusion you would likely come to here is “Some four-sided figures are parallelograms.” However, it would be incorrect to say that all four-sided figures are parallelograms because there are some four-sided figures, such as trapezoids, that are not parallelograms. This is a false conclusion. From these premises, you may draw the conclusion “All sums of two consecutive counting numbers result in an odd number.” Most inductive arguments cannot be proven true, but several mathematical properties can be. If we let n represent our first counting number, then n + 1 would be the next counting number and n + ( n + 1 ) = 2 n + 1 . Because 2 n is divisible by two, it is an even number, and if you add one to any even number the result is always an odd number. Thus, the conclusion is true! Based on the premises provided, with no additional knowledge about whales or dolphins, you might conclude “No creatures that live in the ocean breathe air.” Even though this conclusion is false, it still follows from the known premises and thus is a logical conclusion based on the evidence presented. Alternatively, you could conclude “Some creatures that live in the ocean do not breathe air.” The quantifier you choose to write your conclusion with may vary from another person’s based on how persuasive the argument is. There may be multiple acceptable answers. It is not possible to prove definitively that an inductive argument is true or false in most cases. Negating Statements Containing Quantifiers Recall that the negation of a statement will have the opposite truth value of the original statement. There are four basic forms that logical statements with quantifiers take on. All A are B . Some A are B . No A are B . Some A are not B . The negation of logical statements that use the quantifiers all , some , or none is a little more complicated than just adding or removing the word not . For example, consider the logical statement, “All oranges are citrus fruits.” This statement expresses as a subset relationship. The set of oranges is a subset of the set of citrus fruit. This means that there are no oranges that are outside the set of citrus fruit. The negation of this statement would have to break the subset relationship. To do this, you could say, “At least one orange is not a citrus fruit.” Or, more concisely, “Some oranges are not citrus fruit.” It is tempting to say \"No oranges are citrus fruit,\" but that would be incorrect. Such a statement would go beyond breaking the subset relationship, to stating that the two sets have nothing in common. The negation of \" A is a subset of B \" would be to state that \" A is not a subset of B ,\" as depicted by the Venn diagram in . The statement, “All oranges are citrus fruit,” is true, so its negation, “Some oranges are not citrus fruit,” is false. Now, consider the statement, “No apples are oranges.” This statement indicates that the set of apples is disjointed from the set of oranges. The negation must state that the two are not disjoint sets, or that the two sets have a least one member in common. Their intersection is not empty. The negation of the statement, “ A intersection B is the empty set,” is the statement that \" A intersection B is not empty,\" as depicted in the Venn diagram in . The negation of the true statement “No apples are oranges,” is the false statement, “Some apples are oranges.” summarizes the four different forms of logical statements involving quantifiers and the forms of their associated negations, as well as the meanings of the relationships between the two categories or sets A and B . Logical Statements with Quantifiers Negation of Logical Statements w/Quantifiers Form: All A are B . Means: A is a subset of B , A ⊂ B . All zebras have stripes. (True) Form: Some A are not B . Means: A is not a subset of B , A ⊄ B . Some zebras do not have stripes. (False) Form: Some A are B . Means: A intersection B is not empty, A ∩ B ≠ ∅ . Some fish are sharks. (True) Form: No A are B . Means: A intersection B is empty, A ∩ B = ∅ . No fish are sharks. (False) Form: No A are B . Means: A intersection B is empty, A ∩ B = ∅ . No trees are evergreens. (False) Form: Some A are B . Means: A intersection B is not empty, A ∩ B ≠ ∅ . Some trees are evergreens. (True) Form: Some A are not B . Means: A is not a subset of B , A ⊄ B . Some horses are not mustangs. (True) Form: All A are B . Means: A is a subset of B , A ⊂ B . All horses are mustangs. (False) We covered sets in great detail in Chapter 1 . To review, \" A is a subset of B \" means that every member of set A is also a member of set B . The intersection of two sets A and B is the set of all elements that they share in common. If A intersection B is the empty set, then sets A and B do not have any elements in common. The two sets do not overlap. They are disjoint. Logic Part 1A: Logic Statements and Quantifiers Negating Statements Containing Quantifiers All , Some , or None Given the statements: p : All leopards have spots. r : Some apples are red. s : No lemons are sweet. Write each of the following symbolic statements in words. ~ p ~ r ~ s The statement “All leopards have spots” is p and has the form “All A are B .” According to , the negation will have the form “Some A are not B .” The negation of p is the statement, “Some leopards do not have spots.” The statement “Some apples are red” has the form “Some A are B .” This indicates that the categories A and B overlap or intersect. According to , the negation will have the form, “No A are B ,” indicating that A and B do not intersect. This results in the opposite truth value of the original statement, so the negation of “Some apples are red” is the statement: “No apples are red.” Because s is the statement: “No lemons are sweet,” s is asserting that the set of lemons does not intersect with the set of sweet things. The negation of s , ~ s , must make the opposite claim. It must indicate that the set of lemons intersects with the set of sweet things. This means at least one lemon must be sweet. The statement, “Some lemons are sweet” is ~ s . The negation of the statement, “No A are B ,” is the statement, “Some A are B ,” as indicated in . Check Your Understanding Key Terms logic logical statement truth values symbolic form negation of a logical statement quantifier premises conclusion inductive logical arguments Key Concepts Logical statements have the form of a complete sentence and make claims that can be identified as true or false. Logical statements are represented symbolically using a lowercase letter. The negation of a logical statement has the opposite truth value of the original statement. Be able to Determine whether a sentence represents a logical statement. Write and translate logical statements between words and symbols. Negate logical statements, including logical statements containing quantifiers of all, some, and none . video Logic Part 1A: Logic Statements and Quantifiers", "section": "Statements and Quantifiers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Compound Statements A person seeking their driver's license must pass two tests. A compound statement can be used to explain performance on both tests at once. (credit: modification of work “Drivers License -Teen driver” by State Farm/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Translate compound statements into symbolic form. Translate compound statements in symbolic form with parentheses into words. Apply the dominance of connectives. Suppose your friend is trying to get a license to drive. In most places, they will need to pass some form of written test proving their knowledge of the laws and rules for driving safely. After passing the written test, your friend must also pass a road test to demonstrate that they can perform the physical task of driving safely within the rules of the law. Consider the statement: \"My friend passed the written test, but they did not pass the road test.\" This is an example of a compound statement , a statement formed by using a connective to join two independent clauses or logical statements. The statement, “My friend passed the written test,” is an independent clause because it is a complete thought or idea that can stand on its own. The second independent clause in this compound statement is, “My friend did not pass the road test.” The word \"but\" is the connective used to join these two statements together, forming a compound statement. So, did your friend acquire their driving license?. This section introduces common logical connectives and their symbols, and allows you to practice translating compound statements between words and symbols. It also explores the order of operations, or dominance of connectives, when a single compound statement uses multiple connectives. Common Logical Connectives Understanding the following logical connectives, along with their properties, symbols, and names, will be key to applying the topics presented in this chapter. The chapter will discuss each connective introduced here in more detail. The joining of two logical statements with the word \"and\" or \"but\" forms a compound statement called a conjunction . In logic, for a conjunction to be true, all the independent logical statements that make it up must be true. The symbol for a conjunction is ∧ . Consider the compound statement, “Derek Jeter played professional baseball for the New York Yankees, and he was a shortstop.” If p represents the statement, “Derrick Jeter played professional baseball for the New York Yankees,” and if q represents the statement, “Derrick Jeter was a short stop,” then the conjunction will be written symbolically as p ∧ q . The joining of two logical statements with the word “or” forms a compound statement called a disjunction . Unless otherwise specified, a disjunction is an inclusive or statement, which means the compound statement formed by joining two independent clauses with the word or will be true if a least one of the clauses is true. Consider the compound statement, \"The office manager ordered cake for for an employee’s birthday or they ordered ice cream.” This is a disjunction because it combines the independent clause, “The office manager ordered cake for an employee’s birthday,” with the independent clause, “The office manager ordered ice cream,” using the connective, or . This disjunction is true if the office manager ordered only cake, only ice cream, or they ordered both cake and ice cream. Inclusive or means you can have one, or the other, or both! Joining two logical statements with the word implies , or using the phrase “if first statement , then second statement,” is called a conditional or implication . The clause associated with the \"if\" statement is also called the hypothesis or antecedent , while the clause following the \"then\" statement or the word implies is called the conclusion or consequent . The conditional statement is like a one-way contract or promise. The only time the conditional statement is false, is if the hypothesis is true and the conclusion is false. Consider the following conditional statement, “If Pedro does his homework, then he can play video games.” The hypothesis/antecedent is the statement following the word if , which is “Pedro does/did his homework.” The conclusion/consequent is the statement following the word then , which is “Pedro can play his video games.” Joining two logical statements with the connective phrase “if and only if” is called a biconditional . The connective phrase \"if and only if\" is represented by the symbol, ↔ . In the biconditional statement, p ↔ q , p is called the hypothesis or antecedent and q is called the conclusion or consequent. For a biconditional statement to be true, the truth values of p and q must match. They must both be true, or both be false. The table below lists the most common connectives used in logic, along with their symbolic forms, and the common names used to describe each connective. Connective Symbol Name and but ∧ conjunction or ∨ disjunction, inclusive or not ~ negation if … , then implies → conditional, implication if and only if ↔ biconditional These connectives, along with their names, symbols, and properties, will be used throughout this chapter and should be memorized. Associate Connectives with Symbols and Names For each of the following connectives, write its name and associated symbol. or implies but A compound statement formed with the connective word or is called a disjunction, and it is represented by the ∨ symbol. A compound statement formed with the connective word implies or phrase “ if …, then” is called a conditional statement or implication and is represented by the → symbol. A compound statement formed with the connective words but or and is called a conjunction, and it is represented by the ∧ symbol. Translating Compound Statements to Symbolic Form To translate a compound statement into symbolic form, we take the following steps. Identify and label all independent affirmative logical statements with a lower case letter, such as p , q , or r . Identify and label any negative logical statements with a lowercase letter preceded by the negation symbol, such as ~ p , ~ q , or ~ r . Replace the connective words with the symbols that represent them, such as ∧ , ∨ , → , or ↔ . Consider the previous example of your friend trying to get their driver’s license. Your friend passed the written test, but they did not pass the road test. Let p represent the statement, “My friend passed the written test.” And, let ~ q represent the statement, “My friend did not pass the road test.” Because the connective but is logically equivalent to the word and , the symbol for but is the same as the symbol for and ; replace but with the symbol ∧ . The compound statement is symbolically written as: p ∧ ~ q . My friend passed the written test, but they did not pass the road test. When translating English statements into symbolic form, do not assume that every connective “and” means you are dealing with a compound statement. The statement, “The stripes on the U.S. flag are red and white,” is a simple statement. The word “white” doesn’t express a complete thought, so it is not an independent clause and does not get its own symbol. This statement should be represented by a single letter, such as s : The stripes on the U.S. flag are red and white. Translating Compound Statements into Symbolic Form Let p represent the statement, “It is a warm sunny day,” and let q represent the statement, “the family will go to the beach.” Write the symbolic form of each of the following compound statements. If it is a warm sunny day, then the family will go to the beach. The family will go to the beach, and it is a warm sunny day. The family will not go to the beach if and only if it is not a warm sunny day. The family not go to the beach, or it is a warm sunny day. Replace “it is a warm sunny day” with p . Replace “the family will go to the beach.” with q . Next. Next, because the connective is if …, then place the conditional symbol, → , between p and q . The compound statement is written symbolically as: p → q . Replace “The family will go to the beach” with q . Replace “it is a warm sunny day.” with p . Next, because the connective is and , place the ∧ symbol between q and p . The compound statement is written symbolically as: q ∧ p . Replace “The family will not go to the beach. with ~ q . Replace “it is not a warm sunny day” with ~ p . Next, because the connective is or , if and only if , place the biconditional symbol, ↔ between ~ q and ~ p . The compound statement is written symbolically as: ~ q ↔ ~ p . Replace “The family will not go to the beach” with ~ q . Replace “it is a warm sunny day” with p . Next, because the connective is or , place the ∨ symbol between ~ q and p . The compound statement is written symbolically as: ~ q ∨ p . Translating Compound Statements in Symbolic Form with Parentheses into Words When parentheses are written in a logical argument, they group a compound statement together just like when calculating numerical or algebraic expressions. Any statement in parentheses should be treated as a single component of the expression. If multiple parentheses are present, work with the inner most parentheses first. Consider your friend’s struggles to get their license to drive. Let p represent the statement, “My friend passed the written test,” let q represent the statement, “My friend passed the road test,” and let r represent the statement, “My friend received a driver’s license.” The statement ( p ∧ q ) → r can be translated into words as follows: the statement p ∧ q is grouped together to form the hypothesis of the conditional statement and r is the conclusion. The conditional statement has the form “ if p ∧ q , then r . ” Therefore, the written form of this statement is: “If my friend passed the written test and they passed the road test, then my friend received a driver’s license.” Sometimes a compound statement within parentheses may need to be negated as a group. To accomplish this, add the phrase, “it is not the case that” before the translation of the phrase in parentheses. For example, using p , q , and r of your friend obtaining a license, let’s translate the statement ~ ( p ∧ q ) → ~ r into words. In this case, the hypothesis of the conditional statement is ~ ( p ∧ q ) and the conclusion is ~ r . To negate the hypothesis, add the phrase “it is not the case” before translating what is in parentheses. The translation of the hypothesis is the sentence, “It is not the case that my friend passed the written test and they passed the road test,” and the translation of the conclusion is, “My friend did not receive a driver’s license.” So, a translation of the complete conditional statement, ~ ( p ∧ q ) → ~ r is: “If it is not the case that my friend passed the written test and the road test, then my friend did not receive a driver’s license.” It is acceptable to interchange proper names with pronouns and remove repeated phrases to make the written statement more readable, as long the meaning of the logical statement is not changed. Logic Part 1B: Compound Statements, Connectives and Symbols Translating Compound Statements in Symbolic Form with Parentheses into Words Let p represent the statement, “My child finished their homework,” let q represent the statement, “My child cleaned her room,” let r represent the statement, “My child played video games,” and let s represent the statement, “My child streamed a movie.” Translate each of the following symbolic statements into words. ~ ( p ∧ q ) ( p ∧ q ) → ( r ∨ s ) ~ ( r ∨ s ) ↔ ~ ( p ∧ q ) Replace ~ with “It is not the case,” and ∧ with “and.” One possible translation is: “It is not the case that my child finished their homework and cleaned their room.” The hypothesis of the conditional statement is, “My child finished their homework and cleaned their room.” The conclusion of the conditional statement is, “My child played video games or streamed a movie.” One possible translation of the entire statement is: “If my child finished their homework and cleaned their room, then they played video games or streamed a movie.” The hypothesis of the biconditional statement is ~ ( r ∨ s ) and is written in words as: “It is not the case that my child played video games or streamed a movie.” The conclusion of the biconditional statement is ~ ( p ∧ q ) , which translates to: “It is not the case that my child finished their homework and cleaned their room.” Because the biconditional, ↔ translates to if and only if , one possible translation of the statement is: “It is not the case that my child played video games or streamed a movie if and only if it is not the case that my child finished their homework and cleaned their room.” The Dominance of Connectives The order of operations for working with algebraic and arithmetic expressions provides a set of rules that allow consistent results. For example, if you were presented with the problem 1 + 3 × 2 , and you were not familiar with the order of operation, you might assume that you calculate the problem from left to right. If you did so, you would add 1 and 3 to get 4, and then multiply this answer by 2 to get 8, resulting in an incorrect answer. Try inputting this expression into a scientific calculator. If you do, the calculator should return a value of 7, not 8. Scientific Calculator The order of operations for algebraic and arithmetic operations states that all multiplication must be applied prior to any addition. Parentheses are used to indicate which operation—addition or multiplication—should be done first. Adding parentheses can change and/or clarify the order. The parentheses in the expression 1 + ( 3 × 2 ) indicate that 3 should be multiplied by 2 to get 6, and then 1 should be added to 6 to get 7: 1 + ( 3 × 2 ) = 7. As with algebraic expressions, there is a set of rules that must be applied to compound logical statements in order to evaluate them with consistent results. This set of rules is called the dominance of connectives . When evaluating compound logical statements, connectives are evaluated from least dominant to most dominant as follows: Parentheses are the least dominant connective. So, any expression inside parentheses must be evaluated first. Add as many parentheses as needed to any statement to specify the order to evaluate each connective. Next, we evaluate negations. Then, we evaluate conjunctions and disjunctions from left to right, because they have equal dominance. After evaluating all conjunctions and disjunctions, we evaluate conditionals. Lastly, we evaluate the most dominant connective, the biconditional. If a statement includes multiple connectives of equal dominance, then we will evaluate them from left to right. See for a visual breakdown of the dominance of connectives. Let’s revisit your friend’s struggles to get their driver’s license. Let p represent the statement, “My friend passed the written test,” let q represent the statement, “My friend passed the road test,” and let r represent the statement, “My friend received a driver’s license.” Let's use the dominance of connectives to determine how the compound statement p ∧ ~ q → r should be evaluated. Step 1: There are no parentheses, which is least dominant of all connectives, so we can skip over that. Step 2: Because negation is the next least dominant, we should evaluate ~ q first. We could add parentheses to the statement to make it clearer which connecting needs to be evaluated first: p ∧ ~ q → r is equivalent to p ∧ ( ~ q ) → r . Step 3: Next, we evaluate the conjunction, ∧ . p ∧ ( ~ q ) → r is equivalent to ( p ∧ ( ~ q ) ) → r . Step 4: Finally, we evaluate the conditional, → , as this is the most dominant connective in this compound statement. When using spreadsheet applications, like Microsoft Excel or Google Sheets, have you noticed that core functions, such as sum, average, or rate, have the same name and syntax for use? This is not a coincidence. Various standards organizations have developed requirements that software developers must implement to meet the conditions set by governments and large customers worldwide. The OpenDocument Format OASIS Standard enables transferring data between different office productivity applications and was approved by the International Standards Organization (ISO) and IEC on May 6, 2006. Prior to adopting these standards, a government entity, business, or individual could lose access to their own data simply because it was saved in a format no longer supported by a proprietary software product—even data they were required by law to preserve, or data they simply wished to maintain access to. Just as rules for applying the dominance of connectives help maintain understanding and consistency when writing and interpreting compound logical statements and arguments, the standards adopted for database software maintain global integrity and accessibility across platforms and user bases. Applying the Dominance of Connectives For each of the following compound logical statements, add parentheses to indicate the order to evaluate the statement. Recall that parentheses are evaluated innermost first. p ∧ ~ q ∨ r q → ~ p ∧ r ~ ( p ∨ q ) ↔ ~ p ∧ ~ q Because negation is the least dominant connective, we evaluate it first: p ∧ ( ~ q ) ∨ r . Because conjunction and disjunction have the same dominance, we evaluate them left to right. So, we evaluate the conjunction next, as indicated by the additional set of parentheses: ( p ∧ ( ~ q ) ) ∨ r . The only remaining connective is the disjunction, so it is evaluated last, as indicated by the third set of parentheses. The complete solution is: ( ( p ∧ ( ~ q ) ) ∨ r ) . Negation has the lowest dominance, so it is evaluated first: q → ( ~ p ) ∧ r . The remaining connectives are the conditional and the conjunction. Because conjunction has a lower precedence than the conditional, it is evaluated next, as indicated by the second set of parentheses: q → ( ( ~ p ) ∧ r ) . The last step is to evaluate the conditional, as indicated by the third set of parentheses: ( q → ( ( ~ p ) ∧ r ) ) . This statement is known as De Morgan’s Law for the negation of a disjunction. It is always true. Section 2.6 of this chapter will explore De Morgan’s Laws in more detail. First, we evaluate the negations on the right side of the biconditional prior to the conjunction. Then, we evaluate the disjunction on the left side of the biconditional, followed by the negation of the disjunction on the left side. Lastly, after completely evaluating each side of the biconditional, we evaluate the biconditional. It does not matter which side you begin with. The final solution is: ( ~ ( p ∨ q ) ) ↔ ( ( ~ p ) ∧ ( ~ q ) ) . Logic Terms and Properties – Matching Game Materials: For every group of four students, include 30 cards (game, trading, or playing cards), 30 individual clear plastic gaming card sleeves, and 30 card-size pieces of paper. Prepare a list of 60 questions and answers ahead of time related to definitions and problems in Statements and Quantifiers and Compound Statements . Provide each group of four students with 20 questions and their associated answers. Instruct each group to select 15 of the 20 questions, and then, for each problem selected, create one card with the question and one card with the answer. Instruct the groups to then place each problem and answer in a separate card sleeve. Once they create 15 problem cards and 15 answer cards, have students shuffle the set of cards. To play the game, groups should exchange their card decks to ensure no team begins playing with the deck that they created. Split each four-person group into teams of two students. After shuffling the cards, one team lays cards face down on their desk in a five-by-six grid. The other team will go first. Each player will have a turn to try matching the question with the correct answer by flipping two cards to the face up position. If a team makes a match, they get to flip another two cards; if they do not get a match, they flip the cards face down and it is the other team’s turn to flip over two cards. The game continues in this manner until teams match all question cards with their corresponding answer cards. The team with the most set of matching cards wins. In the first module of this chapter, we evaluated the truth value of negations. In the following modules, we will discuss how to evaluate conjunctions, disjunctions, conditionals, and biconditionals, and then evaluate compound logical statements using truth tables and our knowledge of the dominance of connectives. Check Your Understanding Key Terms compound statement connective conjunction disjunction conditional hypothesis conclusion biconditional dominance of connectives Key Concepts Logical connectives are used to form compound logical statements by using words such as and, or, and if …, then . A conjunction is a compound logical statement formed by combining two statements with the words “and” or “but.” If the two independent clauses are represented by p and q , respectively, then the conjunction is written symbolically as p ∧ q . For the conjunction to be true, both p and q must be true. A disjunction joins two logical statements with the or connective. In, logic or is inclusive. For an or statement to be true at least one statement must be true, but both may also be true. A conditional statement has the form if p , then q , where p and q are logical statements. The only time the conditional statement is false is when p is true, and q is false. The biconditional statement is formed using the connective if and only if for the biconditional statement to be true, the true values of p and q , must match. If p is true then q must be true, if p is false, then q must be false. Translate compound statements between words and symbolic form. Connective Symbol Name and but ∧ conjunction or ∨ disjunction, inclusive or not ~ negation if … , then implies → conditional, implication if and only if ↔ biconditional The dominance of connectives explains the order in which compound logical statements containing multiple connectives should be interpreted. The dominance of connectives should be applied in the following order Parentheses Negations Disjunctions/Conjunctions, left to right Conditionals Biconditionals video Logic Part 1B: Compound Statements, Connectives and Symbols", "section": "Compound Statements", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Constructing Truth Tables Just like solving a puzzle, a computer programmer must consider all potential solutions in order to account for each possible outcome. (credit: modification of work “Deadline Xmas 2010” by Allan Henderson/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Interpret and apply negations, conjunctions, and disjunctions. Construct a truth table using negations, conjunctions, and disjunctions. Construct a truth table for a compound statement and interpret its validity. Are you familiar with the Choose Your Own Adventure book series written by Edward Packard? These gamebooks allow the reader to become one of the characters and make decisions that affect what happens next, resulting in different sequences of events in the story and endings based on the choices made. Writing a computer program is a little like what it must be like to write one of these books. The programmer must consider all the possible inputs that a user can put into the program and decide what will happen in each case, then write their program to account for each of these possible outcomes. A truth table is a graphical tool used to analyze all the possible truth values of the component logical statements to determine the validity of a statement or argument along with all its possible outcomes. The rows of the table correspond to each possible outcome for the given logical statement identified at the top of each column. A single logical statement p has two possible truth values, true or false. In truth tables, a capital T will represent true values, and a capital F will represent false values. In this section, you will use the knowledge built in Statements and Quantifiers and Compound Statements to analyze arguments and determine their truth value and validity. A logical argument is valid if its conclusion follows from its premises, regardless of whether those premises are true or false. You will then explore the truth tables for negation, conjunction, and disjunction, and use these truth tables to analyze compound logical statements containing these connectives. Interpret and Apply Negations, Conjunctions, and Disjunctions The negation of a statement will have the opposite truth value of the original statement. When p is true, ~ p is false, and when p is false, ~ p is true. Finding the Truth Value of a Negation For each logical statement, determine the truth value of its negation. p : 3 + 5 = 8. q : All horses are mustangs. ~ r : New Delhi is not the capital of India. p is true because 3 + 5 does equal 8; therefore, the negation of p , ~ p : 3 + 5 ≠ 8 , is false. q is false because there are other types of horses besides mustangs, such as Clydesdales or Arabians; therefore, the negation of q , ~ q , is true. ~ r is false because New Delhi is the capital of India; therefore, the negation of ~ r , r , is true. A conjunction is a logical and statement. For a conjunction to be true, both statements that make up the conjunction must be true. If at least one of the statements is false, the and statement is false. Finding the Truth Value of a Conjunction Given p : 4 + 7 = 11 , q : 11 - 3 = 7 , and r : 7 × 11 = 77 , determine the truth value of each conjunction. p ∧ q ~ q ∧ r ~ p ∧ q p is true, and q is false. Because one statement is true, and the other statement is false, this makes the complete conjunction false. q is false, so ~ q is true, and r is true. Therefore, both statements are true, making the complete conjunction true. ~ p is false, and q is false. Because both statements are false, the complete conjunction is false. The only time a conjunction is true is if both statements that make up the conjunction are true. A disjunction is a logical inclusive or statement, which means that a disjunction is true if one or both statements that form it are true. The only way a logical inclusive or statement is false is if both statements that form the disjunction are false. Finding the Truth Value of a Disjunction Given p : 4 + 7 = 11 , q : 11 - 3 = 7 , and r : 7 × 11 = 77 , determine the truth value of each disjunction. p ∨ q ~ q ∨ r ~ p ∨ q p is true, and q is false. One statement is true, and one statement is false, which makes the complete disjunction true. q is false, so ~ q is true, and r is true. Therefore, one statement is true, and the other statement is true, which makes the complete disjunction true. ~ p is false, and q is false. When all of the component statements are false, the disjunction is false. In the next example, you will apply the dominance of connectives to find the truth values of compound statements containing negations, conjunctions, and disjunctions and use a table to record the results. When constructing a truth table to analyze an argument where you can determine the truth value of each component statement, the strategy is to create a table with two rows. The first row contains the symbols representing the components that make up the compound statement. The second row contains the truth values of each component below its symbol. Include as many columns as necessary to represent the value of each compound statement. The last column includes the complete compound statement with its truth value below it. Logic Part 2: Truth Values of Conjunctions: Is an \"AND\" statement true or false? Logic Part 3: Truth Values of Disjunctions: Is an \"OR\" statement true or false? Finding the Truth Value of Compound Statements Given p : 4 + 7 = 11 , q : 11 - 3 = 7 , and r : 7 × 11 = 77 , construct a truth table to determine the truth value of each compound statement ~ p ∧ q ∨ r ~ p ∨ q ∧ r ~ ( p ∧ r ) ∨ q Step 1: The statement “ ~ p ∧ q ∨ r ” contains three basic logical statements, p , q , and r , and three connectives, ~ , ∧ , ∨ . When we place parentheses in the statement to indicate the dominance of connectives, the statement becomes ( ( ( ~ p ) ∧ q ) ∨ r ) . Step 2: After we have applied the dominance of connectives, we create a two row table that includes a column for each basic statement that makes up the compound statement, and an additional column for the contents of each parentheses. Because we have three sets of parentheses, we include a column for ~ p , the innermost parentheses, a column for ( ~ p ) ∧ q , the next set of parentheses, and ( ( ~ p ) ∧ q ) ∨ r in the last column for the third parentheses. Step 3: Once the table is created, we determine the truth value of each statement starting from left to right. The truth values of p , q , and r are true, false, and true, respectively, so we place T, F, and T in the second row of the table. Because p is true, ~ p is false. Step 4: Next, evaluate ~ p ∧ q from the table: ~ p is false, and q is also false, so ~ p ∧ q is false, because a conjunction is only true if both of the statements that make it are true. Place an F in the table below its heading. Step 5: Finally, using the table, we understand that ( ~ p ∧ q ) is false and r is true, so the complete statement ( ~ p ∧ q ) ∨ r is false or true, which is true (because a disjunction is true whenever at least one of the statements that make it is true). Place a T in the last column of the table. The complete statement ~ p ∧ q ∨ r is true. p q r ~ p ( ~ p ) ∧ q ( ( ~ p ) ∧ q ) ∨ r T F T F F T Step 1: Applying the dominance of connectives to the original compound statement ~ p ∨ q ∧ r , we get ( ( ( ~ p ) ∨ q ) ∧ r ) . Step 2: The table needs columns for p , q , r , ~ p , ( ~ p ) ∨ q , and ( ( ~ p ) ∨ q ) ∧ r . Step 3: The truth values of p , q , r , and ~ p are the same as in Question 1. Step 4: Next, ~ p ∨ q is false or false, which is false, so we place an F below this statement in the table. This is the only time that a disjunction is false. Step 5: Finally, ~ p ∨ q and r are the conjunction of the statements, ~ p ∨ q and r , and so the expressions evaluate to false and true, which is false. Recall that the only time an \"and\" statement is true is when both statements that form it are also true. The complete statement ~ p ∨ q ∧ r is false. p q r ~ p ( ~ p ) ∨ q ( ( ~ p ) ∨ q ) ∧ r T F T F F F Step 1: Applying the dominance of connectives to the original statement, we have: ( ( ~ ( p ∧ r ) ) ∨ q ) . Step 2: So, the table needs the following columns: p , q , r , p ∧ r , ~ ( p ∧ r ) , and ~ ( p ∧ r ) ∨ q . Step 3: The truth values of p , q , and r are the same as in Questions 1 and 2. Step 4: From the table it can be seen that p ∧ r is true and true, which is true. So the negation of p and r is false, because the negation of a statement always has the opposite truth value of the original statement. Step 5: Finally, ~ ( p ∧ r ) ∨ q is the disjunction of ~ ( p ∧ r ) with q , and so we have false or false, which makes the complete statement false. p q r p ∧ r ~ ( p ∧ r ) ~ ( p ∧ r ) ∨ q T F T T F F Construct Truth Tables to Analyze All Possible Outcomes Recall from Statements and Questions that the negation of a statement will always have the opposite truth value of the original statement; if a statement p is false, then its negation ~ p is true, and if a statement p is true, then its negation ~ p is false. To create a truth table for the negation of statement p , add a column with a heading of ~ p , and for each row, input the truth value with the opposite value of the value listed in the column for p , as depicted in the table below. Negation p ~ p T F F T Conjunctions and disjunctions are compound statements formed when two logical statements combine with the connectives “and” and “or” respectively. How does that change the number of possible outcomes and thus determine the number of rows in our truth table? The multiplication principle , also known as the fundamental counting principle , states that the number of ways you can select an item from a group of n items and another item from a group with m items is equal to the product of m and n . Because each proposition p and q has two possible outcomes, true or false, the multiplication principle states that there will be 2 × 2 = 4 possible outcomes: {TT, TF, FT, FF}. The tree diagram and table in demonstrate the four possible outcomes for two propositions p and q . A conjunction is a logical and statement. For a conjunction to be true, both the statements that make up the conjunction must be true. If at least one of the statements is false, the and statement is false. A disjunction is a logical inclusive or statement. Which means that a disjunction is true if one or both statements that make it are true. The only way a logical inclusive or statement is false is if both statements that make up the disjunction are false. Conjunction (AND) Disjunction (OR) p q p ∧ q p q p ∨ q T T T T T T T F F T F T F T F F T T F F F F F F Logic Part 4: Truth Values of Compound Statements with \"and\", \"or\", and \"not\" Logic Part 5: What are truth tables? How do you set them up? Constructing Truth Tables to Analyze Compound Statements Construct a truth table to analyze all possible outcomes for each of the following arguments. p ∨ ~ q ~ ( p ∧ q ) ( p ∨ ~ q ) ∧ r Step 1: Because there are two basic statements, p and q , and each of these has two possible outcomes, we will have 2 ( 2 ) = 4 rows in our table to represent all possible outcomes: TT, TF, FT, and FF. The columns will include p , q , ~ q , and p ∨ ~ q . Step 2: Every value in column ~ q will have the opposite truth value of the corresponding value in column q : F, T, F, and T. Step 3: To complete the last column, evaluate each element in column p with the corresponding element in column ~ q using the connective or . p q ~ q p ∨ ~ q T T F T T F T T F T F F F F T T Step 1: The columns will include p , q , p ∧ q , and ~ ( p ∧ q ) . Because there are two basic statements, p and q , the table will have four rows to account for all possible outcomes. Step 2: The p ∧ q column will be completed by evaluating the corresponding elements in columns p and q respectively with the and connective. Step 3: The final column, ~ ( p ∧ q ) , will be the negation of the p ∧ q column. p q p ∧ q ~ ( p ∧ q ) T T T F T F F T F T F T F F F T Step 1: This statement has three basic statements, p , q , and r . Because each basic statement has two possible truth values, true or false, the multiplication principal indicates there are 2 ( 2 ) ( 2 ) = 8 possible outcomes. So eight rows of outcomes are needed in the truth table to account for each possibility. Half of the eight possibilities must be true for the first statement, and half must be false. Step 2: So, the first column for statement p , will have four T’s followed by four F’s. In the second column for statement q , when p is true, half the outcomes for q must be true and the other half must be false, and the same pattern will repeat for when p is false. So, column q will have TT, FF, FF, FF. Step 3: The column for the third statement, r , must alternate between T and F. Once, the three basic propositions are listed, you will need a column for ~ q , p ∨ ~ q , and ( p ∨ ~ q ) ∧ r . Step 4: The column for the negation of q , ~ q , will have the opposite truth value of each value in column q . Step 5: Next, fill in the truth values for the column containing the statement p ∨ ~ q . The or statement is true if at least one of p or ~ q is true, otherwise it is false. Step 6: Finally, fill in the column containing the conjunction ( p ∨ ~ q ) ∧ r . To evaluate this statement, combine column p ∨ ~ q and column r with the and connective. Recall, that only time \"and\" is true is when both values are true, otherwise the statement is false. The complete truth table is: p q r ~ q p ∨ ~ q ( p ∨ ~ q ) ∧ r T T T F T T T T F F T F T F T T T T T F F T T F F T T F F F F T F F F F F F T T T T F F F T T F Logic Part 6: More on Truth Tables and Setting Up Rows and Column Headings Determine the Validity of a Truth Table for a Compound Statement A logical statement is valid if it is always true regardless of the truth values of its component parts. To test the validity of a compound statement, construct a truth table to analyze all possible outcomes. If the last column, representing the complete statement, contains only true values, the statement is valid. Determining the Validity of Compound Statements Construct a truth table to determine the validity of each of the following statements. ~ p ∧ q ~ ( p ∧ ~ p ) Step 1: Because there are two statements, p and q , and each of these has two possible outcomes, there will be 2 ( 2 ) = 4 rows in our table to represent all possible outcomes: TT, TF, FT, and FF. Step 2: The columns, will include p , q , ~ p and ~ p ∧ q . Every value in column ~ p will have the opposite truth value of the corresponding value in column p . Step 3: To complete the last column, evaluate each element in column ~ p with the corresponding element in column q using the connective and . The last column contains at least one false, therefore the statement ~ p ∧ q is not valid. p q ~ p ~ p ∧ q T T F F T F F F F T T T F F T F Step 1: Because the statement ~ ( p ∧ ~ p ) only contains one basic proposition, the truth table will only contain two rows. Statement p may be either true or false. Step 2: The columns will include p , ~ p , p ∧ ~ p , and ~ ( p ∧ ~ p ) . Evaluate column p ∧ ~ p with the and connective, because the symbol ∧ represents a conjunction or logical and statement. True and false is false, and false and true is also false. Step 3: The final column is the negation of each entry in the third column, both of which are false, so the negation of false is true. Because all the truth values in the final column are true, the statement ~ ( p ∧ ~ p ) is valid. p ~ p p ∧ ~ p ~ ( p ∧ ~ p ) T F F T F T F T Check Your Understanding Key Terms Truth table Multiplication principle Valid Key Concepts Determine the true values of logical statements involving negations, conjunctions, and disjunctions. The negation of a logical statement has the opposite true value of the original statement. A conjunction is true when both p and q are true, otherwise it is false. A disjunction is false when both p and q are false, otherwise it is true. Know how to construct a truth table involving negations, conjunctions, and disjunctions and apply the dominance of connectives to determine the truth value of a compound logical statement containing, negations, conjunctions, and disjunctions. Negation Conjunction (AND) Disjunction (OR) p ~ p p q p ∧ q p q p ∨ q T F T T T T T T F T T F F T F T F T F F T T F F F F F F A logical statement is valid if it is always true. Know how to construct a truth table for a compound statement and use it to determine the validity of compound statements involving negations, conjunctions, and disjunctions. Video Logic Part 2: Truth Values of Conjunctions: Is an \"AND\" statement true or false? Logic Part 3: Truth Values of Disjunctions: Is an \"OR\" statement true or false? Logic Part 4: Truth Values of Compound Statements with \"and\", \"or\", and \"not\" Logic Part 5: What are truth tables? How do you set them up? Logic Part 6: More on Truth Tables and Setting Up Rows and Column Headings", "section": "Constructing Truth Tables", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Truth Tables for the Conditional and Biconditional If-then statements use logic to execute directions. (credit: “Coding” by Carlos Varela/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Use and apply the conditional to construct a truth table. Use and apply the biconditional to construct a truth table. Use truth tables to determine the validity of conditional and biconditional statements. Computer languages use if-then or if-then-else statements as decision statements: If the hypothesis is true, then do something. Or, if the hypothesis is true, then do something; else do something else. For example, the following representation of computer code creates an if-then-else decision statement: Check value of variable i . If i < 1 , then print \"Hello, World!\" else print \"Goodbye\". In this imaginary program, the if-then statement evaluates and acts on the value of the variable i . For instance, if i = 0 , the program would consider the statement i < 1 as true and “Hello, World!” would appear on the computer screen. If instead, i = 3 , the program would consider the statement i < 1 as false (because 3 is greater than 1), and print “Goodbye” on the screen. In this section, we will apply similar reasoning without the use of computer programs. The Countess of Lovelace, Ada Lovelace, is credited with writing the first computer program. She wrote an algorithm to work with Charles Babbage’s Analytical Engine that could compute the Bernoulli numbers in 1843. In doing so, she became the first person to write a program for a machine that would produce more than just a simple calculation. The computer programming language ADA is named after her. Reference: Posamentier, Alfred and Spreitzer Christian, “Chapter 34 Ada Lovelace: English (1815-1852)” pp. 272-278, Math Makers: The Lives and Works of 50 Famous Mathematicians , Prometheus Books, 2019. Use and Apply the Conditional to Construct a Truth Table A conditional is a logical statement of the form if p , then q . The conditional statement in logic is a promise or contract. The only time the conditional, p → q , is false is when the contract or promise is broken. For example, consider the following scenario. A child’s parent says, “If you do your homework, then you can play your video games.” The child really wants to play their video games, so they get started right away, finish within an hour, and then show their parent the completed homework. The parent thanks the child for doing a great job on their homework and allows them to play video games. Both the parent and child are happy. The contract was satisfied; true implies true is true. Now, suppose the child does not start their homework right away, and then struggles to complete it. They eventually finish and show it to their parent. The parent again thanks the child for completing their homework, but then informs the child that it is too late in the evening to play video games, and that they must begin to get ready for bed. Now, the child is really upset. They held up their part of the contract, but they did not receive the promised reward. The contract was broken; true implies false is false. So, what happens if the child does not do their homework? In this case, the hypothesis is false. No contract has been entered, therefore, no contract can be broken. If the conclusion is false, the child does not get to play video games and might not be happy, but this outcome is expected because the child did not complete their end of the bargain. They did not complete their homework. False implies false is true. The last option is not as intuitive. If the parent lets the child play video games, even if they did not do their homework, neither parent nor child are going to be upset. False implies true is true. The truth table for the conditional statement below summarizes these results. p q p → q T T T T F F F T T F F T Notice that the only time the conditional statement, p → q , is false is when the hypothesis, p , is true and the conclusion, q , is false. Logic Part 8: The Conditional and Tautologies Constructing Truth Tables for Conditional Statements Assume both of the following statements are true: p : My sibling washed the dishes, and q : My parents paid them $5.00. Create a truth table to determine the truth value of each of the following conditional statements. p → q p → ~ q ~ p → q Because p is true and q is true, the statement p → q is, “If my sibling washed the dishes, then my parents paid them $5.00.” My sibling did wash the dishes, since p is true, and the parents did pay the sibling $5.00, so the contract was entered and completed. The conditional statement is true, as indicated by the truth table representing this case: T → T = T. p q p → q T T T p → ~ q translates to the statement, “If my sibling washed the dishes, then my parents did not pay them $5.00.” p is true, but ~ q is false. The sibling completed their end of the contract, but they did not get paid. The contract was broken by the parents. The conditional statement is false, as indicated by the truth table representing this case: T → F = F. p q ~ q p → ~ q T T F F ~ p → q translates to the statement, “If my sibling did not wash the dishes, then my parents paid them $5.00.” ~ p is false, but q is true. The sibling did not do the dishes. No contract was entered, so it could not be broken. The parents decided to pay them $5.00 anyway. The conditional statement is true, as indicated by the truth table representing this case: F → T = T. p q ~ p ~ p → q T T F T Determining Validity of Conditional Statements Construct a truth table to analyze all possible outcomes for each of the following statements then determine whether they are valid. p ∧ q → ~ q p → ~ p ∨ q Applying the dominance of connectives, the statement p ∧ q → ~ q is equivalent to ( p ∧ q ) → ( ~ q ) . So, the columns of the truth table will include p , q , p ∧ q , ~ q , and p ∧ q → ~ q . Because there are only two basic propositions, p and q , the table will have 2 ( 2 ) = 4 rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true. p q p ∧ q ~ q p ∧ q → ~ q T T T F F T F F T T F T F F T F F F T T Applying the dominance of connectives, the statement p → ~ p ∨ q is equivalent to ( p ) → ( ( ~ p ) ∨ q ) . So, the columns of the truth table will include p , q , ~ p , ~ p ∨ q , and p → ( ~ p ∨ q ) . Because there are only two basic propositions, p and q , the table will have 2 ( 2 ) = 4 rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true. p q ~ p ~ p ∨ q p → ( ~ p ∨ q ) T T F T T T F F F F F T T T T F F T T T Use and Apply the Biconditional to Construct a Truth Table The biconditional, p ↔ q , is a two way contract; it is equivalent to the statement ( p → q ) ∧ ( q → p ) . A biconditional statement, p ↔ q , is true whenever the truth value of the hypothesis matches the truth value of the conclusion, otherwise it is false. The truth table for the biconditional is summarized below. p q p ↔ q T T T T F F F T F F F T Logic Part 11B: Biconditional and Summary of Truth Value Rules in Logic Constructing Truth Tables for Biconditional Statements Assume both of the following statements are true: p : The plumber fixed the leak, and q : The homeowner paid the plumber $150.00. Create a truth table to determine the truth value of each of the following biconditional statements. p ↔ q p ↔ ~ q ~ p ↔ ~ q Because p is true and q is true, the statement p ↔ q is “The plumber fixed the leak if and only if the homeowner paid them $150.00.” Because both p and q are true, the leak was fixed and the plumber was paid, meaning both parties satisfied their end of the bargain. The biconditional statement is true, as indicated by the truth table representing this case: T ↔ T = T. p q p ↔ q T T T p → ~ q translates to the statement, “The plumber fixed the leak if and only if the homeowner did not pay them $150.” If the plumber fixed the leak and the homeowner did not pay them, the homeowner will have broken their end of the contract. The biconditional statement is false, as indicated by the truth table representing this case: T ↔ F = F. p q ~ q p ↔ ~ q T T F F ~ p ↔ ~ q translates to the statement, “The plumber did not fix the leak if and only if the homeowner did not pay them $150.” In this case, neither party—the plumber nor the homeowner—entered into the contract. The leak was not repaired, and the plumber was not paid. No agreement was broken. The biconditional statement is true, as indicated by the truth table representing this case: F ↔ F = T. p q ~ p ~ q ~ p ↔ ~ q T T F F T The biconditional, p ↔ q , is true whenever the truth values of p and q match, otherwise it is false. Logic Part 13: Truth Tables to Determine if Argument is Valid or Invalid Determining Validity of Biconditional Statements Construct a truth table to analyze all possible outcomes for each of the following statements, then determine whether they are valid. p ∧ q ↔ p ∧ ~ q p ∨ q ↔ ~ p ∨ q p → q ↔ ~ q → ~ p p ∧ q → ~ r ↔ p ∧ q ∧ r Applying the dominance of connectives, the statement p ∧ q ↔ p ∧ ~ q is equivalent to ( p ∧ q ) ↔ ( p ∧ ( ~ q ) ) . So, the columns of the truth table will include p , q , p ∧ q , ~ q , p ∧ ~ q and ( p ∧ q ) ↔ ( p ∧ ~ q ) . Because there are only two basic propositions, p and q , the table will have 2 ( 2 ) = 4 rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true. p q p ∧ q ~ q p ∧ ~ q ( p ∧ q ) ↔ ( p ∧ ~ q ) T T T F F F T F F T T F F T F F F T F F F T F T Applying the dominance of connectives, the statement p ∨ q ↔ ~ p ∨ q is equivalent to ( p ∨ q ) ↔ ( ( ~ p ) ∨ q ) . So, the columns of the truth table will include p , q , p ∨ q , ~ p , ~ p ∨ q , and ( p ∨ q ) ↔ ( ~ p ∨ q ) . Because there are only two basic propositions, p and q , the table will have 2 ( 2 ) = 4 rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true. p q p ∨ q ~ p ~ p ∨ q ( p ∨ q ) ↔ ( ~ p ∨ q ) T T T F T T T F T F F F F T T T T T F F F T T F Applying the dominance of connectives, the statement p → q ↔ ~ q → ~ p is equivalent to ( p → q ) ↔ ( ( ~ q ) → ( ~ p ) ) . So, the columns of the truth table will include p , q , p → q , ~ q , ~ p , ~ q → ~ p , and ( p → q ) ↔ ( ~ q → ~ p ) . Because there are only two basic propositions, p and q the table will have 2 ( 2 ) = 4 rows of truth values to account for all the possible outcomes. The statement is valid because the last column is all true. p q p → q ~ q ~ p ~ q → ~ p ( p → q ) ↔ ( ~ q → ~ p ) T T T F F T T T F F T F F T F T T F T T T F F T T T T T Applying the dominance of connectives, the statement p ∧ q → ~ r ↔ p ∧ q ∧ r is equivalent to ( ( p ∧ q ) → ( ~ r ) ) ↔ ( ( p ∧ q ) ∧ r ) . So, the columns of the truth table will include p , q , r , ~ r , p ∧ q , ( p ∧ q ) ∧ ~ r , ( p ∧ q ) ∧ r , and ( ( p ∧ q ) → ( ~ r ) ) ↔ ( ( p ∧ q ) ∧ r ) . Because there are three basic propositions, p , q , and r , the table will have 2 ( 2 ) ( 2 ) = 8 rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true. p q r ~ r p ∧ q ( p ∧ q ) → ~ r ( p ∧ q ) ∧ r ( p ∧ q → ~ r ) ↔ ( p ∧ q ∧ r ) T T T F T F T T T T F T T T F T T F T F F T F F T F F T F T F F F T T F F T F F F T F T F T F F F F T F F T F F F F F T F T F F Check Your Understanding Key Concepts The conditional statement, if p then q , is like a contract. The only time it is false is when the contract has been broken. That is, when p is true, and q is false. Conditional p q p → q T T T T F F F T T F F T The biconditional statement, p if and only if q , it true whenever p and q have matching true values, otherwise it is false. Biconditional p q p ↔ q T T T T F F F T F F F T Know how to construct truth tables involving conditional and biconditional statements. Use truth tables to analyze conditional and biconditional statements and determine their validity. Video Logic Part 8: The Conditional and Tautologies Logic Part 11B Biconditional and Summary of Truth Value Rules in Logic Logic Part 13: Truth Tables to Determine if Argument is Valid or Invalid", "section": "Truth Tables for the Conditional and Biconditional", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Equivalent Statements How your logical argument is stated affects the response, just like how you speak when holding a conversation can affect how your words are received. (credit: modification of work by Goelshivi/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Determine whether two statements are logically equivalent using a truth table. Compose the converse, inverse, and contrapositive of a conditional statement Have you ever had a conversation with or sent a note to someone, only to have them misunderstand what you intended to convey? The way you choose to express your ideas can be as, or even more, important than what you are saying. If your goal is to convince someone that what you are saying is correct, you will not want to alienate them by choosing your words poorly. Logical arguments can be stated in many different ways that still ultimately result in the same valid conclusion. Part of the art of constructing a persuasive argument is knowing how to arrange the facts and conclusion to elicit the desired response from the intended audience. In this section, you will learn how to determine whether two statements are logically equivalent using truth tables, and then you will apply this knowledge to compose logically equivalent forms of the conditional statement. Developing this skill will provide the additional skills and knowledge needed to construct well-reasoned, persuasive arguments that can be customized to address specific audiences. An alternate way to think about logical equivalence is that the truth values have to match. That is, whenever p is true, q is also true, and whenever p is false, q is also false. Determine Logical Equivalence Two statements, p and q , are logically equivalent when p ↔ q is a valid argument, or when the last column of the truth table consists of only true values. When a logical statement is always true, it is known as a tautology . To determine whether two statements p and q are logically equivalent, construct a truth table for p ↔ q and determine whether it valid. If the last column is all true, the argument is a tautology, it is valid, and p is logically equivalent to q ; otherwise, p is not logically equivalent to q . Determining Logical Equivalence with a Truth Table Create a truth table to determine whether the following compound statements are logically equivalent. p → q ; ~ p → ~ q p → q ; ~ p ∨ q Construct a truth table for the biconditional formed by using the first statement as the hypothesis and the second statement as the conclusion, ( p → q ) ↔ ( ~ p → ~ q ) . p q p → q ~ p ~ q ~ p → ~ q ( p → q ) ↔ ( ~ p → ~ q ) T T T F F T T T F F F T T F F T T T F F F F F T T T T T Because the last column it not all true, the biconditional is not valid and the statement p → q is not logically equivalent to the statement ~ p → ~ q . Construct a truth table for the biconditional formed by using the first statement as the hypothesis and the second statement as the conclusion, ( p → q ) ↔ ( ~ p ∨ q ) . p q p → q ~ p ~ p ∨ q ( p → q ) ↔ ( ~ p ∨ q ) T T T F T T T F F F F T F T T T T T F F T T T T Because the last column is true for every entry, the biconditional is valid and the statement p → q is logically equivalent to the statement ~ p ∨ q . Symbolically, p → q ≡ ~ p ∨ q . Compose the Converse, Inverse, and Contrapositive of a Conditional Statement The converse , inverse , and contrapositive are variations of the conditional statement, p → q . The converse is if q then p , and it is formed by interchanging the hypothesis and the conclusion. The converse is logically equivalent to the inverse. The inverse is if ~ p then ~ q , and it is formed by negating both the hypothesis and the conclusion. The inverse is logically equivalent to the converse. The contrapositive is if ~ q then ~ p , and it is formed by interchanging and negating both the hypothesis and the conclusion. The contrapositive is logically equivalent to the conditional. The table below shows how these variations are presented symbolically. Conditional Contrapositive Converse Inverse p q ~ p ~ q p → q ~ q → ~ p q → p ~ p → ~ q T T F F T T T T T F F T F F T T F T T F T T F F F F T T T T T T Writing the Converse, Inverse, and Contrapositive of a Conditional Statement Use the statements, p : Harry is a wizard and q : Hermione is a witch, to write the following statements: Write the conditional statement, p → q , in words. Write the converse statement, q → p , in words. Write the inverse statement, ~ p → ~ q , in words. Write the contrapositive statement, ~ q → ~ p , in words. The conditional statement takes the form, “if p , then q ,” so the conditional statement is: “If Harry is a wizard, then Hermione is a witch.” Remember the if … then … words are the connectives that form the conditional statement. The converse swaps or interchanges the hypothesis, p , with the conclusion, q . It has the form, “if q , then p .” So, the converse is: \"If Hermione is a witch, then Harry is a wizard.\" To construct the inverse of a statement, negate both the hypothesis and the conclusion. The inverse has the form, “if ~ p , then ~ q ,” so the inverse is: \"If Harry is not a wizard, then Hermione is not a witch.\" The contrapositive is formed by negating and interchanging both the hypothesis and conclusion. It has the form, “if ~ q , then ~ p ,\" so the contrapositive statement is: \"If Hermione is not a witch, then Harry is not a wizard.\" Identifying the Converse, Inverse, and Contrapositive Use the conditional statement, “If all dogs bark, then Lassie likes to bark,” to identify the following. Write the hypothesis of the conditional statement and label it with a p . Write the conclusion of the conditional statement and label it with a q . Identify the following statement as the converse, inverse, or contrapositive: “If Lassie likes to bark, then all dogs bark.” Identify the following statement as the converse, inverse, or contrapositive: “If Lassie does not like to bark, then some dogs do not bark.” Which statement is logically equivalent to the conditional statement? The hypothesis is the phrase following the if . The answer is p : All dogs bark. Notice, the word if is not included as part of the hypothesis. The conclusion of a conditional statement is the phrase following the then . The word then is not included when stating the conclusion. The answer is: q : Lassie likes to bark. “Lassie likes to bark” is q and “All dogs bark” is p . So, “If Lassie likes to bark, then all dogs bark,” has the form “if q , then p ,” which is the form of the converse. “Lassie does not like to bark” is ~ q and “Some dogs do not bark” is ~ p . The statement, “If Lassie does not like to bark, then some dogs do not bark,” has the form “if ~ q , then ~ p ,” which is the form of the contrapositive. The contrapositive ~ q → ~ p is logically equivalent to the conditional statement p → q . Determining the Truth Value of the Converse, Inverse, and Contrapositive Assume the conditional statement, p → q : “If Chadwick Boseman was an actor, then Chadwick Boseman did not star in the movie Black Panther ” is false, and use it to answer the following questions. Write the converse of the statement in words and determine its truth value. Write the inverse of the statement in words and determine its truth value. Write the contrapositive of the statement in words and determine its truth value. The only way the conditional statement can be false is if the hypothesis, p : Chadwick Boseman was an actor, is true and the conclusion, q : Chadwick Boseman did not star in the movie Black Panther , is false. The converse is q → p , and it is written in words as: “If Chadwick Boseman did not star in the movie Black Panther , then Chadwick Boseman was an actor.” This statement is true, because false → true is true. The inverse has the form “ ~ p → ~ q . ” The written form is: “If Chadwick Boseman was not an actor, then Chadwick Boseman starred in the movie Black Panther .” Because p is true, and q is false, ~ p is false, and ~ q is true. This means the inverse is false → true, which is true. Alternatively, from Question 1, the converse is true, and because the inverse is logically equivalent to the converse it must also be true. The contrapositive is logically equivalent to the conditional. Because the conditional is false, the contrapositive is also false. This can be confirmed by looking at the truth values of the contrapositive statement. The contrapositive has the form “ ~ q → ~ p .” Because q is false and p is true, ~ q is true and ~ p is false. Therefore, ~ q → ~ p is true → false, which is false. The written form of the contrapositive is “If Chadwick Boseman starred in the movie Black Panther , then Chadwick Boseman was not an actor.” Check Your Understanding Key Terms logically equivalent tautology inverse converse contrapositive Key Concepts Two statements p and q are logically equivalent if the biconditional statement, p ↔ q is a valid argument. That is, the last column of the truth table consists of only true values. In other words, p ↔ q is a tautology. Symbolically, p is logically equivalent to q is written as: p ≡ q . A logical statement is a tautology if it is always true. To be valid a local argument must be a tautology. It must always be true. Know the variations of the conditional statement, be able to determine their truth values and compose statements with them. The converse of a conditional statement, if p then q , is the statement formed by interchanging the hypothesis and conclusion. It is the statement if q then p . The inverse of a conditional statement if formed by negating the hypothesis and the conclusion of the conditional statement. The contrapositive negates and interchanges the hypothesis and the conclusion. Conditional Contrapositive Converse Inverse p q ~ p ~ q p → q ~ q → ~ p q → p ~ p → ~ q T T F F T T T T T F F T F F T T F T T F T T F F F F T T T T T T The conditional statement is logically equivalent to the contrapositive. The converse is logically equivalent to the inverse. Know how to construct and use truth tables to determine whether statements are logically equivalent.", "section": "Equivalent Statements", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "De Morgan’s Laws De Morgan’s Laws were key to the rise of logical mathematical expression and helped serve as a bridge for the invention of the computer. (credit: modification of work “Golden Gate Bridge (San Francisco Bay, California, USA)” by James St. John/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Use De Morgan’s Laws to negate conjunctions and disjunctions. Construct the negation of a conditional statement. Use truth tables to evaluate De Morgan’s Laws. The contributions to logic made by Augustus De Morgan and George Boole during the 19th century acted as a bridge to the development of computers, which may be the greatest invention of the 20th century. Boolean logic is the basis for computer science and digital electronics, and without it the technological revolution of the late 20th and early 21st centuries—including the creation of computer chips, microprocessors, and the Internet—would not have been possible. Every modern computer language uses Boolean logic statements, which are translated into commands understood by the underlying electronic circuits enabling computers to operate. But how did this logic get its name? George Boole Boole’s algebra of logic was foundational in the design of digital computer circuits. (credit: “Circuit Board” by Squeezyboy/Flickr, CC BY 2.0) George Boole was born in Lincolnshire, England in 1815. He was the son of a cobbler who provided him some initial education, but Boole was mostly self-taught. He began teaching at 16 years of age, and opened his own school at the age of 20. In 1849, at the age of 34, he was appointed Professor of Mathematics at Queens College in Cork, Ireland. In 1853, he published the paper, An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and Probabilities , which is the treatise that the field of Boolean algebra and digital circuitry was built on. Reference: Posamentier, Alfred and Spreitzer Christian, “Chapter 35 George Boole: English (1815-1864)” pp. 279-283, Math Makers: The Lives and Works of 50 Famous Mathematicians , Prometheus Books, 2019. Negation of Conjunctions and Disjunctions In Chapter 1, used a Venn diagram to prove De Morgan’s Law for set complement over union. Because the complement of a set is analogous to negation and union is analogous to an or statement, there are equivalent versions of De Morgan’s Laws for logic. De Morgan’s Law for negation of a conjunction: ~ ( p ∧ q ) ≡ ~ p ∨ ~ q De Morgan’s Law for the negation of a disjunction: ~ ( p ∨ q ) ≡ ~ p ∧ ~ q Negation of a conditional: ~ ( p → q ) ≡ p ∧ ~ q Writing conditional as a disjunction: p → q ≡ ~ p ∨ q Recall that the symbol for logical equivalence is: ≡ . De Morgan’s Laws allow us to write the negation of conjunctions and disjunctions without using the phrase, “It is not the case that …” to indicate the parentheses. Avoiding this phrase often results in a written or verbal statement that is clearer or easier to understand. Applying De Morgan’s Law for Negation of Conjunctions and Disjunctions Write the negation of each statement in words without using the phrase, “It is not the case that.” Kristin is a biomedical engineer and Thomas is a chemical engineer. A person had cake or they had ice cream. Kristin is a biomedical engineer and Thomas is a chemical engineer has the form “ p ∧ q ,” where p is the statement, “Kristin is a biomedical engineer,” and q is the statement, “Thomas is a chemical engineer.” By De Morgan’s Law, the negation of a conjunction, ~ ( p ∧ q ) , is logically equivalent to ~ p ∨ ~ q . ~ p is “Kristen is not a biomedical engineer,” and ~ q is “Thomas is not a chemical engineer.” By De Morgan’s Law, the solution has the form “ ~ p ∨ ~ q ,” so the answer is: “Kristin is not a biomedical engineer or Tom is not a chemical engineer.” A person had cake or they had ice cream has the form “ p ∨ q , ” where p is the statement, “A person had cake,” and q is the statement, “A person had ice cream.” By De Morgan’s Law for the negations of a disjunction, ~ ( p ∨ q ) ≡ ~ p ∧ ~ q . The solution is the statement: “A person did not have cake and they did not have ice cream.” Negation of a Conditional Statement The negation of any statement has the opposite truth values of the original statement. The negation of a conditional , ~ ( p → q ) , is the conjunction of p and not q , p ∧ ~ q . Consider the truth table below for the negation of the conditional. p q p → q ~ ( p → q ) T T T F T F F T F T T F F F T F The only time the negation of the conditional statement is true is when p is true, and q is false. This means that ~ ( p → q ) is logically equivalent to p ∧ ~ q , as the following truth table demonstrates. p q p → q ~ ( p → q ) ~ q p ∧ ~ q ~ ( p → q ) ↔ ( p ∧ ~ q ) T T T F F F T T F F T T T T F T T F F F T F F T F T F T Constructing the Negation of a Conditional Statement Write the negation of each conditional statement. If Adele won a Grammy, then she is a singer. If Henrik Lundqvist played professional hockey, then he did not win the Stanley Cup. The negation of the conditional statement, p → q , is the statement, p ∧ ~ q . The hypothesis of the conditional statement is p : “Adele won a Grammy,” and conclusion of the conditional statement is q : “Adele is a singer.” The negation of the conclusion, ~ q , is the statement: “She is not a singer.” Therefore, the answer is p ∧ ~ q : “Adele won a Grammy, and she is not a singer.” The hypothesis is p : “Henrik Lundqvist played professional hockey,” and the conclusion of the conditional statement is q : “He did not win the Stanley Cup.” The negation of q is the statement: “He won the Stanley Cup.” The negation of the conditional statement is equal to p ∧ ~ q : “Henrick Lundqvist played professional hockey, and he won the Stanley Cup.” Constructing the Negation of a Conditional Statement with Quantifiers Write the negation of each conditional statement. If all cats purr, then my partner’s cat purrs. If a penguin is a bird, then some birds do not fly. The negation of the conditional statement p → q is the statement p ∧ ~ q . The hypothesis of the conditional statement is p : “All cats purr,” and the conclusion of the conditional statement is q : “My partner’s cat purrs.” The negation of the conclusion, ~ q , is the statement: “My partner’s cat does not purr.” Therefore, the answer is p ∧ ~ q : “All cats purr, but my partner’s cat does not purr.” The hypothesis is p : “A penguin is a bird,” and the conclusion of the conditional statement is q : “Some birds do not fly.” The negation of q is the statement: “All birds fly.” Therefore, the negation of the conditional statement is equal to p ∧ ~ q : “A penguin is a bird, and all birds fly.” Many of the properties that hold true for number systems and sets also hold true for logical statements. The following table summarizes some of the most useful properties for analyzing and constructing logical arguments. These properties can be verified using a truth table. Property Conjunction (AND) Disjunction (OR) Commutative p ∧ q ≡ q ∧ p p ∨ q ≡ q ∨ p Associative ( p ∧ q ) ∧ r ≡ p ∧ ( q ∧ r ) ( p ∨ q ) ∨ r ≡ p ∨ ( q ∨ r ) Distributive p ∧ ( q ∧ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r ) p ∨ ( q ∧ r ) ≡ ( p ∨ q ) ∧ ( p ∨ r ) De Morgan’s ~ ( p ∧ q ) ≡ ~ p ∨ ~ q ~ ( p ∨ q ) ≡ ~ p ∧ ~ q Conditional ~ ( p → q ) ≡ p ∧ ~ q p → q ≡ ~ p ∨ q Negating a Conditional Statement with a Conjunction or Disjunction Write the negation of each conditional statement applying De Morgan’s Law. If mom needs to buy chips, then Mike had friends over and Bob was hungry. If Juan had pizza or Chris had wings, then dad watched the game. The conditional has the form “If p then q or r ,” where p is “Mom needs to buy chips,” q is “Mike had friends over,” and r is “Bob was hungry.” The negation of p → ( q ∧ r ) is p ∧ ~ ( q ∧ r ) . Applying De Morgan’s Law to the statement ~ ( q ∧ r ) the result is ~ q ∨ ~ r , so our conditional statement becomes p ∧ ( ~ q ∨ ~ r ) . By the distributive property for conjunction over disjunction, this statement is equivalent to ( p ∧ ~ q ) ∨ ( p ∧ ~ r ) . Translating the statement ( p ∧ ~ q ) ∨ ( p ∧ ~ r ) into words, the solution is: “Mom needs to buy chips and Mike did not have friends over, or Mom needs to buy chips and Bob was not hungry.” The conditional has the form “If p or q , then r ,” where p is “Juan had pizza,” q is “Chris had wings,” and r is “Dad watched the game.” The negation of ( p ∨ q ) → r is ( p ∨ q ) ∧ ~ r . By the distributive property for disjunction over conjuction, the statement is equivalent to ( p ∨ ~ r ) ∧ ( q ∨ ~ r ) . Translating the statement ( p ∨ ~ r ) ∧ ( q ∨ ~ r ) into words, the solution is: “Juan had pizza or dad did not watch the game, and Chris had wings or dad did not watch the game.” Evaluating De Morgan’s Laws with Truth Tables In Chapter 1, you learned that you could prove the validity of De Morgan’s Laws using Venn diagrams. Truth tables can also be used to prove that two statements are logically equivalent. If two statements are logically equivalent, you can use the form of the statement that is clearer or more persuasive when constructing a logical argument. The next example will prove the validity of one of De Morgan’s Laws using a truth table. The same procedure can be applied to any two logical statement that you believe are equivalent. If the last column of the truth table is a tautology, then the two statements are logically equivalent. Verifying De Morgan’s Law for Negation of a Conjunction Construct a truth table to verify De Morgan’s Law for the negation of a conjunction, ~ ( p ∧ q ) ≡ ~ p ∨ ~ q , is valid. Step 1: To verify any logical equivalence, you must first replace the logical equivalence symbol, ≡ , with the biconditional symbol, ↔ . The statement ~ ( p ∧ q ) ≡ ~ p ∨ ~ q becomes ~ ( p ∧ q ) ↔ ~ p ∨ ~ q . Step 2: Next, you create a truth table for the statement. Because we have two basic statements, p , and q , the truth table will have four rows to account for all the possible outcomes. The columns will be p , q , ~ p , ~ q , p ∧ q , ~ ( p ∧ q ) , ~ p ∨ ~ q , and the biconditional statement is ~ ( p ∧ q ) ↔ ~ p ∨ ~ q . p q p ∧ q ~ ( p ∧ q ) ~ p ~ q ~ p ∨ ~ q ~ ( p ∧ q ) ↔ ( ~ p ∨ ~ q ) T T T F F F F T T F F T F T T T F T F T T F T T F F F T T T T T Step 3: Finally, verify that the statement is valid by confirming it is a tautology. In this instance, the last column is all true. Therefore, the statement is valid and De Morgan’s Law for the negation of a conjunction is verified. Check Your Understanding Key Terms Boolean logic negation of a conditional Key Concepts De Morgan’s Law for the negation of a disjunction states that, ~ ( p ∨ q ) is logically equivalent to ~ p ∧ ~ q . De Morgan’s Law The negation of a conjunction states that, ~ ( p ∧ q ) ≡ ~ p ∨ ~ q . Use De Morgan’s Laws to negate conjunctions and disjunctions. The negation of a conditional statement, if p then q is logically equivalent to the statement p and not q . Use this property to write the negation of conditional statements. Use truth tables to evaluate De Morgan’s Laws.", "section": "De Morgan’s Laws", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Logical Arguments Not all logical arguments are valid, and the ongoing fight for equal rights proves that much progress has yet to be made. (credit: \"The Steam Roller\" by Library of Congress Prints and Photographs Division, public domain) Learning Objectives After completing this section, you should be able to: Apply the law of detachment to determine the conclusion of a pair of statements. Apply the law of denying the consequent to determine the conclusion for pairs of statements. Apply the chain rule to determine valid conclusions for pairs of true statements. The previous sections of this chapter provide the foundational skills for constructing and analyzing logical arguments. All logical arguments include a set of premises that support a claim or conclusion; but not all logical arguments are valid and sound. A logical argument is valid if its conclusion follows from the premises, and it is sound if it is valid and all of its premises are true. A false or deceptive argument is called a fallacy . Many types of fallacies are so common that they have been named. In 1936, Dale Carnegie published his first book, titled How to Win Friends and Influence People . It was marketed as training materials for the improvement of public speaking and negotiation skills, and the methods it presented are still used today. Carnegie famously said, “When dealing with people, remember you are not dealing with creatures of logic, but creatures of emotion.” People who put forth fallacious logical arguments often take advantage of our susceptibility to emotional appeals, to try to convince us that what they are saying is true. The study of logic helps us combat this weakness through recognition and learning to focus on the facts and structure of the argument. This section focuses on the two main forms that logical arguments can take. While inductive arguments attempt to draw a more general conclusion from a pattern of specific premises, deductive arguments attempt to draw specific conclusions from at least one or more general premises. Deductive arguments can be proven to be valid using Venn diagrams or truth tables. Inductive arguments generally cannot be proven to be true. They are judged as being strong or weak, but, like any opinion, whether you believe an argument is strong or weak often depends on your knowledge of the topic being discussed along with the evidence being provided in the premises. Hasty generalization is the name given to any fallacy that presents a weak inductive argument. Be careful! Premises may be true or false. If a premise is false, the claims made by the argument should be questioned. Law of Detachment The law of detachment is a valid form of a conditional argument that asserts that if both the conditional, p → q , and the hypothesis, p , are true, then the conclusion q must also be true. The law of detachment is also called affirming the hypothesis (or antecedent) and modus ponens. Symbolically, it has the form ( ( p → q ) ∧ p ) → q . Law of Detachment Premise: p → q Premise: p Conclusion: ∴ q The ∴ is read as the word, “therefore.” Looking at the truth table for the conditional statement, the only time the conditional is true is when the hypothesis p is also true. The only place this happens is in the first row, where q is also true, confirming that the law of detachment is a valid argument. p q p → q T T T T F F F T T F F T Another way to verify that the law of detachment is a valid argument is to construct a truth table for the argument ( ( p → q ) ∧ p ) → q and verify that it is a tautology. p q p → q ( p → q ) ∧ p ( ( p → q ) ∧ p ) → q T T T T T T F F F T F T T F T F F T F T Venn diagrams may also be used to verify deductive arguments, which include conditional premises. Consider the statement p → q : “If you play guitar, then you are a musician.” The set of guitarists is a subset of the set of musicians, p ⊂ q . To verify that an argument is valid using a Venn diagram, draw the Venn diagram representing all the premises in the argument only, as shown in . Then verify if the conclusion is also represented by the Venn diagram of the premises. If it is, the argument is valid. If it is not, the argument is not valid. The set of guitarists is drawn as a subset of the set of musicians to represent the premise, p → q . The × represents the premise: p is true. This completes the drawing of the premises. Now, examine the Venn diagram to verify if the conclusion is included in the picture. The conclusion is q . Because the × is in the set p , and p is a subset of q , × is also in q ; therefore, the law of detachment is a valid argument. Remember that an argument can be valid without being true. For the argument to be proven true, it must be both valid and sound. An argument is sound if all its premises are true. Logic Part 14: Common Argument Forms like Modus Ponens and Tollens Applying the Law of Detachment to Determine a Valid Conclusion Each pair of statements represents the premises in a logical argument. Based on these premises, apply the law of detachment to determine a valid conclusion. If Leonardo da Vinci was an artist, then he painted the Mona Lisa . Leonardo da Vinci was an artist. If Michael Jordan played for the Chicago Bulls, then Michael Jordan was not a soccer player. Michael Jordan played for the Chicago Bulls. If all fish have gills, then clown fish have gills. All fish have gills. The premises are p → q : If Leonardo da Vinci was an artist, then he painted the Mona Lisa , and p : Leonardo da Vinci was an artist. This argument has the form of the law of detachment, so, the conclusion is q : Leonardo da Vinci painted the Mona Lisa . The premises follow the form of the law of detachment, so a valid conclusion would be q . The premises are p → q : If Michael Jordan played for the Chicago Bulls, then Michael Jordan was not a soccer player, and p : Michael Jordan played for the Chicago Bulls. The conclusion that follows from the premises is q : Michael Jordan was not a soccer player. The premises are p → q : If all fish have gills, then clown fish have gills, and p : All fish have gills. This argument has the form of the law of detachment, so the conclusion is q : clown fish have gills. Law of Denying the Consequent Another form of a valid conditional argument is called the law of denying the consequent , or modus tollens. Recall, that the conditional statement, p → q , is logically equivalent to the contrapositive, ~ q → ~ p . So, if the conditional statement is true, then the contrapositive statement is also true. By the law of detachment, if ~ q is also true, then it follows that ~ p must also be true. Symbolically, it has the form ( ( p → q ) ∧ ~ q ) → ~ p . Law of Denying the Consequent Premise: p → q Premise: ~ q Conclusion: ∴ ~ p The conditional statement can also be described as, “If antecedent, then consequent.”This is where the law of denying the consequent gets its name. To verify if the law of denying the consequent is a valid argument, construct a truth table for the argument, ( ( p → q ) ∧ ~ q ) → ~ p , and verify that it is a tautology. p q ~ p ~ q p → q ( p → q ) ∧ ~ q ( ( p → q ) ∧ ~ q ) → ~ p T T F F T F T T F F T F F T F T T F T F T F F T T T T T To verify an argument of this form using a Venn diagram, again consider the premise: p → q : “If you play guitar, then you are a musician.” We will change the second premise to ~ q . In this case, the × represents the premise, ~ q . So, it will be placed inside the universal set of all people, but outside the set of musicians, as depicted in the Venn diagram in . Because the × is also outside the set of guitarists, the statement ~ p follows from the premises and the argument is valid. Applying the Law of Denying the Consequent to Determine a Valid Conclusion Each pair of statements represents the premises in a logical argument. Based on these premises, apply the law of denying the consequent to determine a valid conclusion. If Leonardo da Vinci was an artist, then he painted the Mona Lisa . Leonardo da Vinci did not paint the Mona Lisa . If Michael Jordan played for the Chicago Bulls, then Michael Jordan was not a soccer player. Michael Jordan was a soccer player. If all fish have gills, then clown fish have gills. Clown fish do not have gills. The premises are p → q : If Leonardo da Vinci was an artist, then he painted the Mona Lisa , and ~ q : Leonardo da Vinci did not paint the Mona Lisa . This argument has the form of the law of denying the consequent, so the conclusion is ~ p : Leonardo da Vinci was not an artist. The premises follow the form of the law of denying the consequent, so a valid conclusion would be ~ p . The premises are: p → q : If Michael Jordan played for the Chicago Bulls, then Michael Jordan was not a soccer player, and ~ q : Michael Jordan was a soccer player. The conclusion that follows from the premises is ~ p : Michael Jordan did not play for the Chicago Bulls. The premises are p → q : If all fish have gills, then clown fish have gills, and ~ q : Clown fish do not have gills. This argument has the form of the law denying the consequent, so the conclusion is ~ p : Some fish do not have gills. Chain Rule for Conditional Arguments The chain rule for conditional arguments is another form of a valid conditional argument. It is also called hypothetical syllogism or the transitivity of implication. Recall that the conditional statement p → q can also be read as p implies q . This is where the name transitivity of implication comes from. The transitive property for numbers states that, if 3 < 4 and 4 < 5 , then it follows that 3 < 5. The chain rule extends this property to conditional statements. If the premises of the argument consist of two conditional statements, with the form “ p → q ” and “ q → r , ” then it follows that p → r . Symbolically, it has the form ( ( p → q ) ∧ ( q → r ) ) → ( p → r ) . Chain Rule for Conditional Arguments Premise: p → q Premise: q → r Conclusion: ∴ p → r To verify the chain rule for conditional arguments, construct a truth table for the argument, ( ( p → q ) ∧ ( q → r ) ) → ( p → r ) , and verify that it is a tautology. p q r p → q q → r ( p → q ) ∧ ( p → r ) p → r ( ( p → q ) ∧ ( q → r ) ) → ( p → r ) T T T T T T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T T T T T T F T F T F F T T F F T T T T T T F F F T T T T T To verify an argument of this form using a Venn diagram, again consider the premise p → q : “If you play guitar, then you are a musician,” but change the second premise to q → r : “If you are a musician, then you are an artist.” In this case, the set p of guitarists is a subset of the set r of artists, and it follows that if you are a guitarist, then you are an artist. Therefore, the conclusion p → r follows from the premises and the chain rule for logical arguments is valid. See . Applying the Chain Rule for Conditional Arguments to Determine a Valid and Sound Conclusion Each pair of statements represents true premises in a logical argument. Based on these premises, apply the chain rule for conditional arguments to determine a valid and sound conclusion. If my roommate goes to work, then my roommate will get paid. If my roommate gets paid, then my roommate will pay their bills. If robins can fly, then some birds can fly. If some birds can fly, then we will watch birds fly. If Irma is a teacher, then Irma has a college degree. If Irma has a college degree, then Irma graduated from college. The premises are p → q : “If my roommate goes to work, then they will get paid,” and q → r : “If my roommate gets paid, then my roommate will pay their bills.” This argument has the form of the chain rule for conditional arguments, so the valid conclusion will have the form “ p → r . ” Because all the premises are true, the valid and sound conclusion of this argument is: “If my roommate goes to work, then my roommate will pay their bills.” The premises are p → q : “If robins can fly, then some birds can fly,” and q → r : “If some birds can fly, then we will watch them fly.” This argument has the form of the chain rule for conditional arguments, so, the valid conclusion will have the form “ p → r . ” Because all the premises are true, the valid and sound conclusion of this argument is: “If robins can fly, then we will watch birds fly.” The premises are p → q : (see line 1 of solution 1 and 2 above) “If Irma is a teacher, then Irma has a college degree,” and q → r : “If Irma has a college degree, then Irma graduated from college.” This argument has the form of the chain rule for conditional arguments, so the valid conclusion will have the form “ p → r . ” Because all the premises are true, the valid and sound conclusion of this argument is: “If Irma is a teacher, then Irma graduated from college.” Check Your Understanding Projects Logic Gates Logic gates are the basis for all digital circuits. Research and document the following terms: logic gate, OR gate, AND gate, and NOT gate. Construct a diagram of a NAND gate, NOR gate, and a XOR gate by using at least two of the following gates: AND, OR, and NOT. Digital electronics use a 1 for true or on, and a 0 for false or off. Create a truth table documenting all possible cases using 0s and 1s for the NAND gate, NOR gate and XOR gate. Use a truth table to explain how XOR is related to the biconditional statement. Logical Fallacies Fallacies are false or deceptive logical arguments. Research and document the structure of five of the following named fallacies: hasty generalization, limited choice, false cause, appeal to popularity, appeal to emotion, appeal to authority, personal attack, gamblers' ruin, slippery slope, and circular reasoning. Create a presentation highlighting one of the five fallacies researched in the previous question. The presentation must include an introductory slide with the title of the fallacy and the form or structure of the argument. The second slide must include an example of this fallacy as used in a commercial, a political cartoon or a current event or new article. The third slide must include an explanation of why the example on slide to is a representative example of the fallacy. The last slide must include citations for any materials used. No textbooks should be used as reference. Careers in Logic Lawyers, mathematicians, and computer programmers are a few of the careers that require knowledge of logic. What career are you interested in? Research how knowledge of logic applies to your chosen field of study. Then, write a cover letter for a position in your field you'd like to apply to. In the cover letter, include how your knowledge of logic qualifies you for the position you are applying for. If you do not think logic is important for your given career choice, find a position where logic is an essential element of the position and complete the project by pretending you are writing a cover letter for that job. Key Terms sound fallacy deductive arguments law of detachment law of denying the consequent chain rule for conditional arguments Key Concepts A logical argument uses a series of facts or premises to justify a conclusion or claim. It is valid if its conclusion follows from the premises, and it is sound if it is valid, and all of its premises are true. The law of detachment is a valid form of a conditional argument that asserts that if both the conditional, p → q is true and the hypothesis, p is true, then the conclusion q must also be true. Law of Detachment Premise: p → q Premise: p Conclusion: ∴ q Know how to apply the law of detachment to determine the conclusion of a pair of statements. The law of denying the consequent is a valid form of a conditional argument that asserts that if both the conditional, p → q is true and the negation of the conclusion, ~ q is true, then the negation of the hypothesis ~ p must also be true. Law of Denying the Consequent Premise: p → q Premise: ~ q Conclusion: ∴ ~ p Know how to apply the law of denying the consequent to determine the conclusion for pairs of statements. The chain rule for conditional arguments is a valid form of a conditional argument that asserts that if the premises of the argument have the form, p → q and q → r , then it follows that p → r . Chain Rule for Conditional Arguments Premise: p → q Premise: q → r Conclusion: ∴ p → r Know how to apply the chain rule to determine valid conclusions for pairs of true statements. Video Logic Part 14: Common Argument Forms like Modus Ponens and Tollens Chapter Review Statements and Quantifiers Compound Statements Constructing Truth Tables Truth Tables for the Conditional and Biconditional Equivalent Statements De Morgan’s Laws Logical Arguments Chapter Test", "section": "Logical Arguments", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Encryption of computers and messages use very large prime numbers. (credit: modification of work \"Jefferson cylinder cipher (replica)\" by Daderot/Wikimedia Commons, Public Domain) Encryption is used to secure online banking, for secure online shopping, and for browsing privately using VPNs (Virtual Private Networks). We need encryption (using prime numbers) for a secure exchange of information. For a prime number to be useful for encryption, though, it has to be large. Encryption uses a composite number that is the product of two very large primes. In order to break the encryption, one must determine the two primes that were used to form the composite number. If the two prime numbers used are sufficiently large, even the fastest computer cannot determine those two prime numbers in a reasonable amount of time. It would take a computer 300 trillion years to crack the current encryption standard.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Prime and Composite Numbers Computers are protected using encryption based on prime numbers. (credit: “Data Security” by Blogtrepreneur/Flickr, CC BY 2.0) After completing this section, you should be able to: Apply divisibility rules. Define and identify numbers that are prime or composite. Find the prime factorization of composite numbers. Find the greatest common divisor. Use the greatest common divisor to solve application problems. Find the least common multiple. Use the least common multiple to solve application problems. Encryption, which is needed for the secure exchange of information (i.e., online banking or shopping) is based on prime numbers. Encryption uses a composite number that is the product of two very large prime numbers. To break the encryption, the two primes that were used to form the composite number need to be determined. If the two prime numbers used are sufficiently large, even the fastest computer cannot determine those two prime numbers in a reasonable amount of time. It would take a computer 300 trillion years to crack the current encryption standard. Applying Divisibility Rules Before we begin our investigation of divisibility, we need to know some facts about important sets of numbers: The counting numbers are referred to as the natural numbers . This set of numbers, { 1 , 2 , 3 , 4 , … } , is denoted with the symbol ℕ . Another important set of numbers is the integers . The integers are the natural numbers, along with 0, and the negatives of the natural numbers. This set is often written as { … , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , … } . We denote the integers with the symbol ℤ. Notice that ℕ is a proper subset of ℤ, or, ℕ ⊂ ℤ. All the ideas of this section apply to the natural numbers, while only some apply to all the integers. Divisibility is when the integer n is divisible by m , if n can be written as m times another integer. Equivalently, there is no remainder when n is divided by m . There are many occasions when separating items into equal groups comes into play to ensure an equal distribution of whole items. For example, Francis, a preschool art teacher, has 15 students in one class. Francis has 225 sheets of construction paper and wants to provide each student with an equal number of pieces. To know if he will use all the construction paper, Francis is really asking if 225 can be evenly divided into 15 groups. Determining if a Number Divides Another Number Determine if 36 is divisible by 4. We could divide 36 by 4 and see if there is a remainder, or we could see if we can write 36 as 4 times another integer. If we divide 36 by 4, we see 36 ÷ 4 = 9 with no remainder. We see that 36 is divisible by 4. We can write 36 as 4 times another integer, 36 = 4 × 9 . By the definition of divisibility, 36 is divisible by 4. You can quickly check if a number is divisible by 2, 3, 4, 5, 6, 9, 10, and 12. Each has an easy-to-identify feature, or rule, that indicates the divisibility by those numbers, as shown in the following table. Divisor Rule 2 Last digit is even 3 Add the digits of the number together. If that sum is divisible by 3, then so is the original number 4 Look at only the last two digits. If this number is divisible by 4, so is the original number 5 Look at only the last digit. If it is a 5 or a 0, then the original number is divisible by 5 6 If the number passes the rule for divisibility by 2 and for 3, then the number is divisible by 6 9 Add the digits of the number together. If that number is divisible by 9, then so is the original number 10 Look at only the last digit. If it is a 0, then the original number is divisible by 10 12 If the number passes the rule for 3 and 4, the number is divisible by 12 Using Divisibility Rules Using divisibility rules, determine if 245 is divisible by 5. Since the last digit is a 5, the number 245 is divisible by 5 because the rule states if the last digit of the number is a 5 or a 0, then the original number is divisible by 5. Using Divisibility Rules Using divisibility rules, determine if 25,983 is divisible by 9. The divisibility rule for 9 is when the digits of the number are added, the sum is divisible by 9. So, we calculate the sum of the digits. 2 + 5 + 9 + 8 + 3 = 27 . Since 27 is divisible by 9, so is the original number 25,983. Using Divisibility Rules Can 298 coins be stacked into 6 stacks with an equal number of coins in each stack? In order for the coins to be in equal-sized stacks, 298 would need to be divisible by 6. The divisibility rule for 6 is that the number passes the divisibility rules for both 2 and 3. Since the last digit is even, 298 is divisible by 2. To determine if 298 is divisible by 3, we first add the digits of the number: 2 + 9 + 8 = 19 . Since 19 is not divisible by 3, neither is 298. Because 298 is not divisible by 3, it is also not divisible by 6, which means they cannot be put into 6 equal stacks of coins. Using Divisibility Rules Using divisibility rules, determine if 4,259 is divisible by 10. The divisibility rule for 10 is that the last digit of the number is 0. Since the last digit of 4,259 is not 0, then 4,259 is not divisible by 10. Using Divisibility Rules Using divisibility rules, determine if 936,276 is divisible by 4. The divisibility rule for 4 is to check the last two digits of the number. If the number formed by the last two digits of the original number is divisible by 4, then so is the original number. The last two digits make the number 76 and 76 is divisible by 4, since 76 = 4 × 19 . Since 76 is divisible by 4, so is 936,276. Divisibility Rules Prime and Composite Numbers Sometimes, a natural number has only two unique divisors, 1 and itself. For instance, 7 and 19 are prime . In other words, there is no way to divide a prime number into groups with an equal number of things, unless there is only one group, or those groups have one item per group. Other natural numbers have more than two unique divisors, such as 4, or 26. These numbers are called composite . The number 1 is special; it is neither prime nor composite. To determine if a number is prime or composite, you have to determine if the number has any divisors other than 1 and itself. The divisibility rules are useful here, and can quickly show you if a number has a divisor on that list. However, if none of those divide the number, you still have to check all other possible prime divisors. What are the prime numbers that are possibly divisors of the number you are checking? You need only check the prime numbers up to the square root of the number in question. For instance, if you want to know if 2,117 is prime, you need to determine if any primes up to the square root of 2,117 (which is 46.0 when rounded to one decimal place) divide 2,117. If any of those primes are divisors of the number in question, then the number is composite. If none of those primes work, then the number is itself prime. We can check divisibility with whatever tool we wish. Divisibility rules are quick for some prime divisors (2 and 5 come to mind) but aren't quick for other values (like 11). In place of divisibility rules, we could just use a calculator. If the prime number divides the number evenly (that is, there is no decimal or fractional part), then the number is divisible by that prime. is a quick list of the prime numbers up to 50. There are 15 prime numbers less than 50. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Prime Numbers Less than 50 Determining If a Number Is Prime or Composite Determine if 2,117 is prime or composite. The square root of 2,117 is 46.0 (rounded to one decimal place). So, we need to check if 2,117 is divisible by any prime up to 46. Step 1: First we’ll use the rules of divisibility we learned earlier: We can tell 2,117 is not divisible by 2, as the last digit isn't even. 2,117 is not divisible by 5 (the last digit isn't 0 or 5). Add the digits of 2,117 to get 11, which is not divisible by 3. So, 2,117 is also not divisible by 3. Step 2: Now we repeat the process for all the primes up to 46. Using a calculator, we find that 2,117 divided by the prime numbers 7, 11, 13, 17, 19, and 23 results in a remainder, a decimal part. So, we know that 2,117 is not divisible by these prime numbers. (You should check these results yourself.) Moving on, we check the next prime: 29. Using the calculator to divide 2,117 by 29 results in 73. Since there is no decimal part, 2,117 is divisible by 29. This means that 2,117 is not a prime number, but rather, a composite number. Writing 2,117 as the product of 29 and another natural number, 2,117 = 29 × 73 . Determining if a Number Is Prime or Composite Determine if 423 is prime or composite. The square root of 423 is 20.57 (rounded to two decimal places). So, we need to check if 423 is divisible by any prime up to 20. Step 1: Check 2. We can tell 423 is not divisible by 2, as the last digit isn't even. Step 2: Check 5. It is not divisible by 5 (the last digit isn't 0 or 5). Step 3: Check 3. To check if 423 is divisible by 3, we use the divisibility rule for 3. When we take the sum of the digits of 423, the result is 9. Since 9 is divisible by 3, so is 423. Since 423 is divisible by 3, then 423 is a composite number. Writing 423 as the product of 3 and another natural number, 423 = 3 × 141 . Determining if a Number Is Prime or Composite Determine if 1,034 is prime or composite A quick inspection of 1,034 shows it is divisible by 2 since the last digit is even, and so 1,034 is a composite number. Determining if a Number Is Prime or Composite Determine if 2,917 is prime or composite. The square root of 2,917 is 50.01 (rounded to two decimal places). So, we need to check if 2,917 is divisible by any prime up to 50. Step 1: Check 2. We can tell 2,917 is not divisible by 2, as the last digit isn't even. Step 2: Check 5. It is it divisible by 5 (the last digit isn't 0 or 5). Step 3: Check 3. Using the divisibility rule for 3, we take the sum of the digits of 2,917, which is 19. Since 19 is not divisible by 3, neither is 2,917. Step 4: Check the rest of the primes up to 50 using a calculator. When 2,917 is divided by every prime number up to 50, the result has a decimal part. Since no prime up to 50 divides 2,917, it is a prime number. ILLEGAL PRIMES Large primes are a hot commodity. Using two very large primes (some have more than 22 million digits!) is necessary for secure encryption. Anyone who has a new prime that is large enough can use that prime to create a new encryption. Of course, whoever discovers a large prime could sell it to a security company. These primes are so useful for encryption, it is necessary to protect that intellectual property. In fact, at least one prime number was declared illegal. Illegal Prime Number Sophie Germain Sophie Germain (credit: “Sophie Germain at 14 years,” Illustration from histoire du socialisme, approx. 1880, Wikimedia Commons, public domain) Born into a wealthy French family in 1776, Sophie Germain discovered and fell in love with mathematics by browsing her father’s books. Clandestine study, hard and tenacious work, and a mathematical mindset did not lead to college, however, as she was not allowed to attend. She did manage, through friends, to obtain problem sets and submit brilliant solutions under the name Monsieur LaBlanc. One of her great interests was number theory, which is the study of properties of integers. One of her theorems, titled “Sophie Germain’s Theorem,” partially solved one of the great mathematical mysteries, Fermat’s Last Theorem. From this she discovered what are now known as Sophie Germain Primes. A Sophie Germain Prime is a prime number that can be written in the form 2 p + 1 , where p is a prime number. For instance, 23 is prime: 2 ( 23 ) + 1 = 47 , which is prime, so 47 is a Sophie Germain Prime. It should be noted that 2 p + 1 , where p is a prime number, may or may not be prime (check for p = 7 !). Finding the Prime Factorization of Composite Numbers Before we can start with prime factorization, let’s remind ourselves what it means to factor a number. We factor a number by identifying two (or more) numbers that, when multiplied, result in the original number. For instance, 3 and 24, when multiplied, give 72. So, 72 can be factored into 3 × 24 . Notice that we could have factored the 72 differently, say as 72 = 6 × 12 , or 72 = 2 × 36 , or even as 72 = 3 × 4 × 6 . Finding the prime factorization of a composite number means writing the number as the product of all of its prime factors. For example, 80 = 2 × 2 × 2 × 2 × 5 . Notice that all the numbers being multiplied on the right-hand side are prime numbers. Sometimes prime numbers repeat themselves in the factorization. When prime factors do repeat, we may write the prime factorization using exponents , as in 80 = 2 4 × 5 . In that equation, the 2 is raised to the 4th power. The 4 is the exponent, and the 2 is the base . More generally, in the exponent notation a b , the number represented by a is the base, and the number represented by b is the exponent. One has to wonder if finding the prime factorization could result in different factorizations. The Fundamental Theorem of Arithmetic tells us that there is only one prime factorization for a given natural number. Fundamental Theorem of Arithmetic Every natural number, other than 1, can be expressed in exactly one way, apart from the arrangement, as a product of primes. The process of finding the prime factorization of a number is iterative, which means we do a step, then repeat it until we cannot do the step any longer. The step we use is to identify one prime factor of the number, then write the number as the prime factor times another factor. We repeat this step on the other, newly found, factor. This step is repeated until no more primes can be factored from the remaining factor. This is easier to see and explain with an example. Finding the Prime Factorization Find the prime factorization of 140. Step 1: Identify a prime number that divides 140. Since 140 is even (the last digit is even), 2 divides 140. We then factor the 2 out of the 140, giving us 140 = 2 × 70 . Step 2: With the other factor, 70, find a prime factor of 70. Since 70 is also even, 2 divides 70. We factor the 2 out of the 70 and the factorization is now 140 = 2 × 2 × 35 = 2 2 × 35 . Notice that we expressed the two factors of 2 as 2 2 . Step 3: Look to the remaining factor, 35. The last digit of 35 is 5, so 5 is a factor of 35. We factor the 5 out of the 35. The factorization is now 140 = 2 2 × 5 × 7 . Step 4: Look to the remaining factor, 7. Since 7 is prime, the process is complete. The prime factorization of 140 is 2 2 × 5 × 7 . Factor Trees A useful tool for helping with prime factorization is a factor tree . To create a factor tree for the natural number n (where n is not 1), perform the following steps: Step 1: If n is prime, you're done. If n is composite, continue to the next step. Step 2: Identify two divisors of n , call them a and b . Step 3: Write the number n down, and draw two branches extending down (or to the right) of the number n . Step 4: Label the end of one branch a , the other as b . See . Step 5: If a and b are prime, the tree is complete. When a number at this step is a prime number, we refer to it as a leaf of the tree diagram. Step 6: If either a or b are composite, repeat Steps 2 through 4 for a and b . Step 7: The process stops when the leaves are all prime. Step 8: The prime factorization is then the product of all the leaves. This is best seen in an example. Finding the Prime Factorization Find the prime factorization of 66. Since 66 is even, 2 is a factor. Step 1: Factor out the 2. The factorization is 66 = 2 × 33 . The factor tree is shown in . Since 2 is a factor, that branch is done, and 2 is a leaf. Step 2: The 33, though, is divisible by 3, and is the product of 3 and 11. We attach that to the factor tree ( ). Since the 2, 3 and 11 are all prime, the factor tree is done. The prime factorization of 66 is the product of the leaves, so 66 = 2 × 3 × 11 . The factorization is complete. Using a Factor Tree to Find the Prime Factorization Finding the Prime Factorization Find the prime factorization of 135. The number 135 is divisible by 3, and so 3 is a factor of 135. Step 1: Factor out the 3. The factorization is 135 = 3 × 45 . Using the factor tree ( ), 45 is also divisible by 3. Step 2: Factor out a 3 from 45. The other factor is 15. The factor tree is shown in . 15 is also divisible by 3. Step 3: The factors of 15 are 3 and 5. The factor tree is shown in . All the leaves are prime, so the process is complete. The prime factorization of 135 is 3 3 × 5 . Identifying Prime Factors How many different prime factors does 10,241 have? To know how many different prime factors 10,241 has, we need the prime factorization of 10,241. Step 1: Use divisibility rules, to see that the number 10,241 is not divisible by 2 or by 3 (the sum of the digits is 8), or by 5. However, it is divisible by 7. Step 2: After factoring the 7, the factorization is 10 , 241 = 7 × 1463 ; 1,463 is also divisible by 7. Step 3: After factoring the 7, the factorization is 10 , 241 = 7 × 7 × 209 = 7 2 × 209 . The number 209 is not divisible by 7. Step 4: Check the next prime number: 11; 11 does divide 1,463. Step 5: After factoring the 11, the factorization is 10 , 241 = 7 2 × 11 × 19 . Since 19 is prime, the prime factorization of 10,241 is complete. We see that 10,241 has three different prime factors: 7, 11, and 19. Finding the Prime Factorization of 168 Using Wolfram Alpha to Find Prime Factorizations The Wolfram Alpha website is a powerful resource available for free to use. It is designed using AI so that it understands natural language requests. For instance, typing the question “What is the prime factorization of 543,390?” gets a rather quick answer of 2 × 3 × 5 × 59 × 307 . So, if you want to find the prime factorization of a number, you can simply ask Wolfram Alpha to find the prime factorization of the number. Finding the Greatest Common Divisor Two numbers often have more than one divisor in common (all pairs of natural numbers have the common divisor 1). When listing the common divisors, it’s often the case that the largest is of interest. This divisor is called the greatest common divisor and is denoted GCD . It is also sometimes referred to as the greatest common factor (GCF). For instance, 6 is the greatest common divisor of 12 and 18. We can see this by listing all the divisors of each number and, by inspection, select the largest value that shows up in each list. The divisors of 12 1, 2, 3, 4, 6, 12 The divisors of 18 1, 2, 3, 6, 9 It is easy to see that 6 is the largest value that appears in both lists. Finding the Greatest Common Divisor Using Lists Find the greatest common divisor of 1,400 and 250 by listing all divisors of each number. We create a list of all the divisors of 1,400 and of 250, and choose the largest one. The divisors of 1,400 are 1, 2, 4, 5, 7, 8, 10, 14, 20, 25, 28, 35, 40, 50, 56, 70, 100, 140, 175, 200, 280, 350, 700, 1,400. The divisors of 250 are 1, 2, 5, 10, 25, 50, 125, 250. The largest value that appears on both lists is 50, so the greatest common divisor of 1,400 and 250 is 50. Listing all the divisors of the numbers in the set will always work, but for some relatively small numbers, the set of all divisors can become pretty big, and finding them all can be a chore. Another approach to finding the greatest common divisor is to use the prime factorization of the numbers. To do so, use the following steps: Step 1: Find the prime factorization of the numbers. Step 2: Identify the prime factors that appear in every number’s prime factorization. These are called the common prime factors. Step 3: Identify the smallest exponent of each prime factor identified in Step 2 in the prime factorizations. Step 4: Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. The result is the greatest common divisor. Finding the Greatest Common Divisor Using Prime Factorization Find the greatest common divisor of 1,400 and 250 by using their prime factorizations. Step 1: Find the prime factorizations of the numbers. The prime factorization of 1,400 is 2 3 × 5 2 × 7 . The prime factorization of 250 is 250 = 2 × 5 3 . Step 2: Identify the prime factors that appear in every number’s prime factorization. The common prime factors are 2 and 5. Step 3: Identify the smallest exponent of each prime identified in Step 2 in the prime factorizations. The exponent of common prime factor 2 in the prime factorization of 1,400 is 3, and in the prime factorization of 250 is 1. The smallest of those exponents is 1. The exponent of the common prime factor 5 in the prime factorization of 1,400 is 2 and is in the prime factorization of 250 is 3. The smallest of these exponents is 2. Step 4: Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. This gives 2 1 × 5 2 = 50 . The greatest common divisor of 1,400 and 250 is 2 × 5 2 = 50 . Notice that the answer matches the one we found in Example 3.15. Finding the Greatest Common Divisor Using Prime Factorization Find the greatest common divisor of 600 and 784 by using their prime factorizations. Step 1: Find the prime factorizations of the numbers. The prime factorization of 600 is 2 3 × 3 × 5 2 . The prime factorization of 784 is 2 4 × 7 2 . Step 2: Identify the prime factors that appear in every number’s prime factorization. There is only one common prime factor, 2. Step 3: Identify the smallest exponent of each prime from identified in Step 2 in the prime factorizations. The exponent of 2 in the prime factorization of 600 is 3. The exponent of 2 in the prime factorization of 784 is 4. So, the smallest exponent of 2 is 3. Step 4: Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. This gives 2 3 = 8 . The greatest common divisor of 600 and 784 is 8. Srinivasa Ramanujan Srinivasa Ramanujan (credit: Srinivasa Ramanujan, photo by Konrad Jacobs/Oberwolfach Photo Collection/public domain) Ramanujan was born in southern India in 1887, during British rule. He was a self-taught mathematician, who, while in high school, began working through a two-volume text of mathematical theorems and results. His work included examination of the distribution of primes. He eventually came to the attention of British mathematician, G.H. Hardy. During one visit, Hardy mentioned to Ramanujan that his taxicab number was 1,729, remarking that 1,729 appeared to be a rather dull number. To which Ramanujan responded, “It is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.” Using Desmos to Find the GCD To find the GCD of a set of numbers in Desmos, type gcd(first_number,second_number…) and Desmos will display the GCD of the numbers, as the numbers are typed! Using Desmos to Find the GCD Applications of the Greatest Common Divisor The greatest common divisor has uses that are related to other mathematics (reducing fractions) but also in everyday applications. We’ll look at two such applications, which have very similar underlying structures. In each case, something must divide the groups or measurements equally. Calculating Floor Tile Size Suppose you have a rectangular room that is 570 cm wide and 450 cm long. You wish to cover the floor of the room with square tiles. What’s the largest size square tile that can be used to cover this floor? Using squares means that the length and width of the tiles are equal. To ensure we are using full tiles, the side length of the square tiles must divide the length of the room and the width of the room. Since we want the largest square tiles, we need the GCD of the width and length of the room or the GCD of 570 and 450. Step 1: Find the prime factorizations of the numbers. The prime factorization of 570 is 2 × 3 × 5 × 19 . The prime factorization of 450 is 2 × 3 2 × 5 2 . Step 2: Identify the prime factors that appear in every number’s prime factorization. The common prime factors are 2, 3, and 5. Step 3: Identify the smallest exponent of each prime identified in Step 2 in the prime factorizations. The exponents of 2 in the prime factorizations of 570 and 450 are 1 and 1. So the smallest exponent for 2 is 1. The exponents of 3 in the prime factorizations of 570 and 450 are 1 and 2, so the smallest exponent for 3 is 1. The exponents of 5 in the prime factorizations of 570 and 450 are 1 and 2, so the smallest exponent for 5 is 1. Step 4: Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. This gives 2 1 × 3 1 × 5 1 = 30 . The GCD of 450 and 570 is 30, so the largest size square tile that can be used to cover the floor is 30 cm by 30 cm. Organizing Books Per Shelf Suppose you want to organize books onto shelves, and you want the shelves to hold the same number of books. Each shelf will only contain one genre of book. You have 24 sci-fi, 42 fantasy, and 30 horror books. How many books can go on each shelf? Since we want shelves that hold an equal number of books, and a shelf can only hold one genre of book, we need a number that will equally divide 24, 42, and 30. So, we need the GCD of the number of books of each genre or the GCD of 24, 42, and 30. Step 1: Find the prime factorizations of the numbers. The prime factorization of 24 is 2 3 × 3 . The prime factorization of 42 is 2 × 3 × 7 . The prime factorization of 30 is 2 × 3 × 5 . Step 2: Identify the prime factors that appear in every number’s prime factorization. The common prime factors are 2 and 3. Step 3: Identify the smallest exponent of each prime identified in Step 2 in the prime factorizations. The smallest exponent of 2 and 3 in the factorizations is 1. Step 4: Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. This gives 2 1 × 3 1 = 6 . The GCD of 24, 42, and 30 is 6, so the largest number of books that can go on a shelf is 6. Applying the GCD Finding the Least Common Multiple The flip side to a divisor of a number is a multiple of a number. For example, 5 is a divisor of 45 and so 45 is a multiple of 5. More generally, if the number a divides the number b , then b is a multiple of a . This drives the idea of the least common multiple of a set of numbers. A common multiple of a set of numbers is a multiple of each of those numbers. For instance, 45 is a common multiple of 9 and 5, because 45 is a multiple of 9 (9 divides 45) and 45 is also a multiple of 5 (5 divides 45). The least common multiple (LCM) of a set of number is the smallest positive common multiple of that set of numbers. There are (at least) three ways to find the LCM of a set of numbers, and we will explore two of them. One way is to create a list of multiples of each number in the set, and then identify the smallest multiple that appears in those lists. Finding the Least Common Multiple Using Lists Find the LCM of 24 and 90 by listing multiples and choosing the smallest common multiple. Create a list of the multiples of each number. Step 1: The first 15 multiples for 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360. Step 2: The first 15 multiples for 90: 90, 180, 270, 360, 450, 540, 630, 720, 810, 900, 990, 1,080, 1,170, 1,260, 1,350. There is one multiple common to these lists, which is 360. So, 360 is the LCM of 24 and 90. The second method we can use is to find the prime factorizations of the number in the set to build the LCM of the numbers based on the prime divisors of the numbers. The LCM can be built from the prime factorization of the numbers in the set in a similar way as when finding the greatest common divisor. Here are the steps for using the prime factorization process for finding the LCM. Step 1: Find the prime factorization of each number. Step 2: Identify each prime that is present in any of the prime factorizations. Step 3: Identify the largest exponent of each prime identified in Step 2 in the prime factorizations. Step 4: . Multiply the prime factors identified in Step 2 raised to the powers identified in Step 3. Finding the Least Common Multiple Using Prime Factorization Use the prime factorizations of 24 and 90 to identify their LCM. Step 1: Find the prime factorization of each number. 24 = 2 3 × 3 90 = 2 × 3 2 × 5 Step 2: Identify each prime that is present in any of the prime factorizations. The prime numbers present in the prime factorizations are 2, 3, and 5. Step 3: Identify the largest exponent of each prime identified in Step 2 in the prime factorizations. Prime 2 3 5 Largest exponent 3 2 1 Step 4: Compute the LCM by multiplying the prime factors identified in Step 2 raised to the powers identified in Step 3. The LCM for 24 and 90 is 2 × 2 × 2 × 3 × 3 × 5 = 2 3 × 3 2 × 5 1 = 360 . Finding the Least Common Multiple Using Prime Factorization Use the prime factorizations of 36, 66, and 250 to identify the LCM. Step 1: Find the prime factorization of each number. 36 = 2 2 × 3 2 66 = 2 × 3 × 11 250 = 2 × 5 3 Step 2: Identify each prime that is present in any of the prime factorizations. The prime numbers present in the prime factorizations are 2, 3, 5, and 11. Step 3: Identify the largest exponent of each prime identified in Step 2 in the prime factorizations. Prime 2 3 5 11 Largest exponent 2 2 3 1 Step 4: Compute the LCM by multiplying the prime factors identified in Step 2 raised to the powers identified in Step 3. The LCM for 36, 66, and 250 is 2 × 2 × 3 × 3 × 5 × 5 × 5 × 11 = 2 2 × 3 2 × 5 3 × 11 1 = 49,500 . Using lists for three or more numbers, particularly larger numbers, could take quite a bit of time. Frequently, as in this example, the prime factorization process is much quicker. In practice, you can use either listing or prime factorization to find the LCM. Finding the Least Common Multiple Using Both Methods Find the LCM of 20, 36, and 45 using lists and prime factorization. Step 1: Use listing. List the multiples: 20 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220 36 36, 72, 108, 144, 180, 216, 252, 288, 324, 360 45 45, 90, 135, 180, 225, 270, 315, 360, 405, 450, 495, 540 The smallest value that appears on all the lists is 180, so 180 is the LCM of 20, 36, and 45. Step 2: Find the prime factorization of each number. 20 = 2 2 × 5 36 = 2 2 × 3 2 45 = 3 2 × 5 Step 3: Identify each prime that is present in any of the prime factorizations. The prime numbers present in the prime factorizations are 2, 3, 5. Step 4: Identify the largest exponent of each prime identified in Step 3 in the prime factorizations. Prime 2 3 5 Largest exponent 2 2 1 Step 5: Compute the LCM by multiplying the prime factors identified in Step 3 raised to the powers identified in Step 4. The LCM for 20, 36, and 45 is 2 2 × 3 2 × 5 1 = 180 . Both listing and prime factorization produced the same result: the LCM is 180. Finding the LCM Using Desmos to Find the LCM To find the LCM of a set of numbers in Desmos, type lcm(first_number,second_number…) and Desmos will display the LCM of the numbers, as the numbers are typed! Using Desmos to find the LCM Applications of the Least Common Multiple Some applications of LCM involve events that repeat at fixed intervals, such as visits to a location. Other applications involve getting things to be of equal magnitude when using parts that are different sizes (see , for instance). In each case, we may be looking to determine when two processes “line up.” Determining Scheduling Overlap Using the Least Common Multiple Two students, João, and Amelia, meet one day at an assisted living facility where they volunteer. João volunteers every 6 days. Amelia volunteers every 10 days. How many days will it be until they are both volunteering on the same day again? If we list the days that each student will volunteer, it becomes clear that we could solve this problem using the LCM of 6 and 10. João will be at the assisted living facility 6, 12, 18, 24, 30, 36, 42, and 48 days later. Amelia will be at the assisted living facility 10, 20, 30, 40, 50, and 60 days later. The smallest number appearing on both lists is 30. João and Amelia will once more be volunteering together 30 days later. Determining the Minimum Height Using the LCM A team-building exercise has teams build a house of cards as high as possible. However, the cards for different teams are of different sizes. Team 1 uses 10 cm × 10 cm cards, while Team 2 uses 8 cm × 8 cm cards. What is the minimum height when the two teams will be tied? Ignore the width of the cards. This is an example where we want to put together objects with different sizes. We want to know the minimum height when they are tied, or when the houses of cards line up the first time. The heights of the towers built using the 10 cm × 10 cm cards will be 10, 20, 30, 40, 50, and 60 cm tall. When the 8 cm × 8 cm cards are used, the tower will be 8, 16, 24, 32, 40, 48, and 56 cm tall. The smallest number appearing on both lists is 40. The first time they are tied is when the two towers are 40 cm tall. Application of LCM Prime Number Life Cycles—Cicadas Cicadas are known to have life cycles of 13 or 17 years, which are prime numbers. Why would a prime number life cycle be an evolutionary advantage? To figure this out, we have to explore how common multiples work with prime numbers. Make a conjecture regarding the LCM of a prime number and another number. Test this conjecture with a few examples of your own making. Check Your Understanding Key Terms natural numbers factor of a number multiple of a number prime number composite number prime factorization greatest common divisor (GCD) least common multiple (LCM) Key Concepts The natural numbers can be categorized as 1, prime numbers, and composite numbers. Prime numbers have as their only factors 1 and themselves. Composite numbers have at least three distinct factors. Composite numbers can be written in their prime factorization form, which is found by repeatedly factoring prime factors from the number. The greatest common divisor (GCD) of a set of numbers is the largest integer that divides all of the numbers in the set. The prime factorizations of the numbers can be used to identify the greatest common divisor. The least common multiple (LCM) of a set of numbers is the smallest integer that is divisible by all of the numbers in the set. The prime factorizations of the numbers can be used to identify the least common multiple. There are various ways that the GCD and LCM are applied. Videos Divisibility Rules Illegal Prime Number Using a Factor Tree to Find the Prime Factorization Finding the Prime Factorization of 168 Using Desmos to find the GCD Applying the GCD Finding the LCM Using Desmos to find the LCM Application of LCM", "section": "Prime and Composite Numbers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Integers A ledger comparing assets to debts, resulting in net wealth. (credit: modification of work “Reviewing Financial Statements” by Mary Cullen/Flickr, CC BY 2.0) After completing this section, you should be able to: Define and identify numbers that are integers. Graph integers on a number line. Compare integers. Compute the absolute value of an integer. Add and subtract integers. Multiply and divide integers. Positive net wealth is when the total value of a person’s assets, such as their home, their 401(k), their car, and savings account balance, exceed that of their debts, such as car loans, mortgages, or credit card debt. However, when the total value of debt exceeds the total value of assets, then the person has negative net wealth. Expressing the negative net wealth as a negative number allows people to work with the positive net values and negative net values with the same mathematical processes, and in the same applications. This section introduces the integers and operations with integers. Defining and Identifying Integers Extending the counting numbers to include negative numbers and zero forms the integers . Any other number that cannot be written as { … − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , … } is not an integer. Identifying Integers Which of the following are integers and which are not? −3 Is an integer, as it is the negative of a counting number 24 This is not written as an integer. Entering the square root of 24 in a calculator, such as desmos, the result is 4.899 (rounded off). Since this is not an integer, then 24 is not an integer. 36/4 Since 36 divided by 4 is 9, and 9 is an integer, then 36/4 is an integer. 45 Is an integer, as it is a counting number 63.9 Is not an integer, because it is not a counting number and not the negative of a counting number. 2/7 Dividing 2 by 7 results in a number less than 1, but greater than 0, so is between two consecutive integers. So, 2/7 is not an integer. −16.0 Is an integer, since the decimal part is 0 Graphing Integers on a Number Line Integers are often imagined as steps along a path. You start at 0, and going to the left is going backward, or in the negative direction, while going to the right is going forward, or in the positive direction. A number line ( ) helps envision the integers. This also means that an integer gives magnitude (size) and direction (positive is to the right, negative is to the left). Graphing an integer on the number line means placing a solid dot at the integer on the number line. Graphing Integers on the Number Line Graphing Numbers on the Number Line Graph the following on the number line: 1 −4 3 Comparing Integers When determining if one quantity or size is larger than another, we know it means there is more of whatever is being discussed. In terms of positive integers, we can envision that larger integers are further to the right on the number line. This idea applies to negative integers also. This means that a is greater than b when a is to the right of b on the number line. We write a > b . When a is greater than b , we can also say that b is less than a . On the number line, b would be to the left of a . We write b < a . We need to recognize that a > b means the same thing as b < a . This can be seen on the number line in . On this number line, a is to the right of b , so a > b . But this means b is to the left of a , so b < a . Comparing Integers Using the Number Line Comparing Integers Using a Number Line Determine which of −6 and 4 is larger using a number line, and express that using both the greater than and the less than notations. To illustrate this, we use a number line ( ). Since −6 is to the left of 4, then −6 is less than 4. We can write this as −6 < 4. Another way of expressing this is that 4 is greater than −6. So we can also write 4 > − 6 . Comparing Negative Integers Determine which of −6 and −2 is larger, and express that using both the greater than and the less than notations. To illustrate this, we use a number line ( ). Since −6 is to the left of −2, then −6 is less than −2. We can write this as − 6 < − 2 . Another way of expressing this is that −2 is greater than −6. So we can also write − 2 > − 6 . Warning: People often ignore the negative signs, and think than since 6 is greater than 2, −6 is greater than −2. To avoid that error, remember that the greater number is to the right on the number line. Comparing Integers by Quantity Determine which of 27 and 410 is larger, and express that using both the greater than and the less than notations. When thinking about quantity, 410 is more than 27. So, 410 is greater than 27 and 27 is less than 410. We can write this as 410 > 27 or as 27 < 410 . The Absolute Value of an Integer When talking about graphing integers on the number line, one interpretation suggests it is like walking along a path. Negative is going to the left of 0, and positive going to the right. If you take 30 steps to the right, you are 30 steps away from 0. On the other hand, when you take 30 steps to the left, you are still 30 steps away from 0. So, in a way, even though one is negative and the other positive, these two numbers, 30 and −30, are equal since both are 30 steps away from 0. The absolute value of an integer n is the distance from n to 0, regardless of the direction. The notation for absolute value of the integer n is | n | . If we think of an integer as both direction and magnitude (size), absolute value is the magnitude part. Calculating the absolute value of an integer is very straightforward. If the integer is positive, then the absolute value of the integer is just the integer itself. If the integer is negative, then to compute the absolute value of the integer, simply remove the negative sign. Keeping in mind the number line as a path, when you’ve gone 10 steps to the left of 0, you have still taken 10 steps, and the direction does not matter. Evaluating the Absolute Value of an Integer Calculating the Absolute Value of a Positive Integer Calculate |19|. Since the number inside the absolute value symbol is positive, the absolute value is just the number itself. So |19| = 19. Calculating the Absolute Value of a Negative Integer Calculate |−435|. Since the number inside the absolute value is negative, the absolute value removes the negative sign. So |−435| = 435. Adding and Subtracting Integers You may recall having approached adding and subtracting integers using the number line from earlier in your academic life. Adding a positive integer results in moving to the right on the number line. Adding a negative integer results in moving to the left. Subtracting a positive integer results in a move to the left on the number line. But subtracting a negative integer results in a move to the right. This leads to a few adding and subtracting rules, such as: Rule 1: Subtracting a negative is the same as adding a positive. Rule 2: Adding two negative integers always results in a negative integer. Rule 3: Adding two positive integers always results in a positive integer. Rule 4: The sign when adding integers with opposite signs is the same as the integer with the larger absolute value. These rules are good to keep in the back of your mind, as they can serve as a quick error check when you use a calculator. Adding Integers Use your calculator to calculate 4 + (−7). Explain how the answer agrees with what was expected. Using a calculator, we find that 4 + (−7) = −3. Since we are adding integers with opposite signs, the sign of the answer matches the sign of the integer with the larger absolute value which |−7|=7. Subtracting Positive Integers Use your calculator to calculate 18 − 9. Explain how the answer agrees with what was expected. Using a calculator, we find that 18 − 9 = 9. Since 18 was larger than 9, we expected the difference to be positive. Subtracting with Negative Integers Use your calculator to calculate 27 − (−13). Explain how the answer agrees with what was expected. Using a calculator, we find that 27 – (−13) = 40. Since we’re subtracting a negative number, it is the same as adding a positive, so this is the same as 27 + 13 = 40. Adding Integers with Opposite Signs Use your calculator to calculate (−13) + 90. Explain how the answer agrees with what was expected. Using a calculator, we find that (−13) + 90 = 77. Since we are adding integers with opposite signs, the sign of the answer matches the sign of the integer with the larger absolute value, which is positive since 90 is positive. One use of negative numbers is determining net worth , which is all the weath someone owns less all that someone owes. Sometimes net worth is positive (which is good), and sometimes net worth is negative (which can be stressful). Calculating Net Worth Jennifer is owed $50 from her friend Janice, but owes her friend Pat $87. What is Jennifer’s net worth? Net worth is the amount that one is owed minus the amount one owes. Jennifer is owed $50 but owes $87. So, her net worth is $50 – $87 = −$37. The negative indicates that Jennifer owes more than she is owed. Multiplying and Dividing Integers Similar to addition and subtraction, the signs of the integers impact the results when multiplying and dividing integers. The rules are fairly straightforward, but again rely on the direction on the number line. There are only two rules. Rule 1: When multiplying or dividing two integers with the same sign, the result is positive. Rule 2: When multiplying or dividing two integers with opposite signs, the result is negative. Just as before, these rules can serve as a quick error check when using a calculator. Multiplying Positive Integers Use your calculator to calculate 4 × 8. Explain how the answer agrees with what was expected. Entering 4 × 8 into your calculator, the result is 32. This agrees with our expectation. The numbers have the same signs, so the result is positive. Multiplying Integers with Different Signs Use your calculator to calculate 9 × (−10). Explain how the answer agrees with what was expected. Entering 9 × (−10) into your calculator, the result is −90. This agrees with our expectation. The numbers have opposite signs, so the result is negative. Dividing Integers with Different Signs Use your calculator to calculate 400/(−25). Explain how the answer agrees with what was expected. Entering 400/(−25) into your calculator, the result is −16. This agrees with our expectation. The numbers have opposite signs, so the result is negative. Dividing Negative Integers Use your calculator to calculate −750/(−3). Explain how the answer agrees with what was expected. Entering −750/(−3) into your calculator, the result is 250. This agrees with our expectation. The numbers have the same signs, so the result is positive. At the end of a season, a team may wish to buy their coach an end-of season gift. It makes sense to share the cost equally among the members. To do so, the team would need to find the average (or mean) cost per member. The average (or mean) of a set of numbers is the sum of the numbers divided by the number values that are being averaged. Finding the Average of a Set of Numbers The daily low temperatures in Barrie, Ontario, for the week of February 14, 2021, were −20°, −12°, −15°, −23°, −17°, −13°, and −19° degrees Celsius. What was the average daily temperature for the week of February 14, 2021, in Barrie? Step 1: To find the average daily temperature, we first need to add the temperatures. (−20) + (−12) + (−15) + (−23) + (−17) + (−13) + (−19) = −119 Step 2: That sum will then be divided by 7 since we are averaging over seven days, giving −119/7 = −17. So, the average daily temperature in Barrie, Ontario the week of February 14, 2021, was −17° Celsius. Check Your Understanding Key Terms integer absolute value average of a set of numbers Key Concepts A set of numbers that can be built from the natural numbers are the integers, which consist of the natural numbers, zero (0), and the negatives of the natural numbers. Integers are often graphed on a number line, which helps display the relative positions and values of those numbers. The number line can be used to visualize when one integer is larger than or smaller than another integer. Arithmetic operations with integers are similar to the operations with natural numbers, except that the sign (positive or negative) of the numbers will determine the sign (positive or negative) of the result. Videos Graphing Integers on the Number Line Comparing Integers Using the Number Line Evaluating the Absolute Value of an Integer", "section": "The Integers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Order of Operations Calculators may automatically apply order of operations to calculations. (credit: “Precision” by Leonid Mamchenkov/Flickr, CC BY 2.0) After completing this module, you should be able to: Simplify expressions using order of operations. Simplify expressions using order of operations involving grouping symbols. Calculates else sure someone be rules expect explicit we what that needs to need make, we to that them what be calculate to calculated first. You probably read that sentence and couldn't make heads or tails of it. Seems like it might concern calculations, but maybe it concerns needs? You may even be attempting to unscramble the sentence as you read it, placing words in the order you might expect them to appear in. The reason that the sentence makes no sense is that the words don't follow the order you expect them to follow. Unscrambled, the sentence was intended to be “To be sure that someone else calculates when we expect them to calculate, we need rules that make explicit what needs to be calculated first.” Similarly, when working with math expressions and equations, if we don't follow the rules for order of operations , arithmetic expressions make no sense. Just a simple expression would be problematic if we didn't have some rules to tell us what to calculate first. For instance, 4 × 2 2 + 3 + 5 2 can be calculated in many ways. You could get 5,184. Or, you could get 80. Or, 96. The issue is that without following a set of rules for calculation, the same expression will give various results. In case you are curious, using the appropriate order of operations, we find 4 × 2 2 + 3 + 5 2 = 44 . Simplify Expressions Using Order of Operations The order in which mathematical operations is performed is a convention that makes it easier for anyone to correctly calculate. They follow the acronym EMDAS: E Exponents M/D Multiplication and division A/S Addition and Subtraction So, what does EMDAS tell us to do? In an equation, moving left to right, we begin by calculating all the exponents first. Once the exponents have been calculated, we again move left to right, calculating the multiplications and divisions, one at a time. Multiplication and division hold the same position in the ordering, so when you encounter one or the other at this step, do it. Once the multiplications and divisions have been calculated, we again move left to right, calculating the additions and subtractions, one at a time. Additions and subtractions hold the same position in the ordering, so when you encounter one or the other at this step, do it. (You may have previously learned the order of operations as PEMDAS, with parentheses first; we will add that aspect later on.) We’ll explore this as we work an example. Using Two Order of Operations Calculate 21 − 4 × 13 . There are no exponents in this expression, so the next operations to check are multiplication and division. Step 1: Moving left to right, the first multiplication encountered is 4 multiplied by 13. We perform that operation first. 21 − 4 × 13 = 21 − 52 Step 2: The only operation remaining is the subtraction. 21 − 52 = − 31 . So, 21 − 4 × 13 = − 31 . Using Two Order of Operations Calculate 4 × 8 3 . Step 1: Moving left to right, we see there is an exponent. We calculate the exponent first. 4 × 8 3 = 4 × ( 8 × 8 × 8 ) = 4 × 512 Step 2: The only operation remaining is the multiplication. 4 × 512 = 2,048 So, 4 × 8 3 = 2,048 . Using Three Order of Operations Calculate 2 + 3 2 × 4 . Step 1: To calculate this, move left to right, and compute all the exponents first. The only exponent we see is the squaring of the 3, so that is calculated first. 2 + 3 2 × 4 = 2 + ( 3 × 3 ) × 4 = 2 + 9 × 4 Step 2: Since the exponents are all calculated, now calculate all the multiplications and divisions moving left to right. The only multiplication or division present is 9 times 4. 2 + 9 × 4 = 2 + 36 Step 3: Moving left to right, perform the additions and subtractions. There is only one such operation, 2 plus 36. 2 + 36 = 38 So, 2 + 3 2 × 4 = 38 . Even if the expression being calculated gets more complicated, we perform the operations in the order: EMDAS. Order of Operations 1 Using Eight Order of Operations Correctly apply the order of operations to compute the following: 4 − 25 × 6 / 10 × 3 2 + 7 × 2 3 . Step 1: To do so, calculate the exponents first, moving left to right. There are two occurrences of exponents in the expression, 3 squared and 2 cubed. 4 − 25 × 6 / 10 × 3 2 + 7 × 2 3 = 4 − 25 × 6 / 10 × 9 + 7 × 8 Step 2: Now that the exponents are calculated, perform the multiplication and division, moving left to right. The first is the product of 25 and 6. 4 − 25 × 6 / 10 × 9 + 7 × 8 = 4 − 150 / 10 × 9 + 7 × 8 Step 3: Next is the 150 divided by 10. 4 − 150 / 10 × 9 + 7 × 8 = 4 − 15 × 9 + 7 × 8 Step 4: Next is 15 multiplied by 9. 4 − 15 × 9 + 7 × 8 = 4 − 135 + 7 × 8 Step 5: Finally, multiply the 7 and 8. 4 − 135 + 7 × 8 = 4 − 135 + 56 As all the multiplications and divisions have been calculated, the additions and subtractions are performed, moving left to right. 4 − 135 + 56 = − 131 + 56 = − 131 + 56 = − 75 The computed value is −75. Order of Operations 2 Using Six Order of Operations Correctly apply the rules for the order of operations to accurately compute the following: 10 − 3 × 5 3 / 15 + 56 / 4 . Step 1: Calculate exponents first, moving left to right: 10 − 3 × 5 3 / 15 + 56 / 4 = 10 − 3 × 125 / 15 + 56 / 4 Step 2: Multiply and divide, moving left to right: 10 − 3 × 125 / 15 + 56 / 4 = 10 − 375 / 15 + 56 / 4 = 10 − 25 + 56 / 4 = 10 − 25 + 14 Step 3: Add and subtract, moving left to right: 10 − 25 + 14 = − 15 + 14 = − 1 Using Order of Operations Correctly apply the rules for the order of operations to accurately compute the following: ( − 8 ) / 2 × 3 − 9 × 2 4 / 12 + 9 × ( − 4 ) 2 / 2 3 . Step 1: Calculate the exponents first, moving left to right: ( − 8 ) / 2 × 3 − 9 × 2 4 / 12 + 9 × ( − 4 ) 2 / 2 3 = ( − 8 ) / 2 × 3 − 9 × 16 / 12 + 9 × 12 2 / 2 3 = ( − 8 ) / 2 × 3 − 9 × 16 / 12 + 9 × 144 / 2 3 = ( − 8 ) / 2 × 3 − 9 × 16 / 12 + 9 × 144 / 8 Step 2: Multiply and divide, moving left to right: = ( − 8 ) / 2 × 3 − 9 × 16 / 12 + 9 × 144 / 8 = ( − 4 ) × 3 − 9 × 16 / 12 + 9 × 144 / 8 = ( − 12 ) − 9 × 16 / 12 + 9 × 144 / 8 = ( − 12 ) − 144 / 12 + 9 × 144 / 8 = ( − 12 ) − 12 + 9 × 144 / 8 = ( − 12 ) − 12 + 1,296 / 8 = ( − 12 ) − 12 + 162 Step 3: Add and subtract, moving left to right: = ( − 12 ) − 12 + 162 = ( − 24 ) + 162 = 138 Using the Order of Operations Involving Grouping Symbols We have examined how to use the order of operations, denoted by EMDAS, to correctly calculate expressions. However, there may be expressions where a multiplication should happen before an exponent, or a subtraction before a division. To indicate an operation should be performed out of order, the operation is placed inside parentheses. When parentheses are present, the operations inside the parentheses are performed first. Adding the parentheses to our list, we now have PEMDAS, as shown below. P Parentheses E Exponents M/D Multiplication and division (division is just the multiplication by the reciprocal) A/S Addition and subtraction (subtraction is just the addition of the negative) As said previously, parentheses indicate that some operation or operations will be performed outside the standard order of operation rules. For instance, perhaps you want to multiply 4 and 7 before squaring. To indicate that the multiplication happens before the exponent, the multiplication is placed inside parentheses: ( 4 × 7 ) 2 . This means operations inside the parentheses take precedence, or happen before other operations. Now, the first step in calculating arithmetic expressions using the order of operations is to perform operations inside parentheses first. Inside the parentheses, you follow the order of operation rules EMDAS. Prioritizing Parentheses in the Order of Operations Correctly apply the rules for the order of operations to accurately compute the following: ( 10 − 3 ) × 5 3 . Step 1: Perform all calculations within the parentheses before all other operations. ( 10 − 3 ) × 5 3 = 7 × 5 3 Step 2: Since all parentheses have been cleared, move left to right, and compute all the exponents next. 7 × 5 3 = 7 × 125 Step 3: Perform all multiplications and divisions moving left to right. 7 × 125 = 875 Be aware that there can be more than one set of parentheses, and parentheses within parentheses. When one set of parentheses is inside another set, do the innermost set first, and then work outward. Order of Operations 3 Working Innermost Parentheses in the Order of Operations Correctly apply the rules for order of operations to accurately compute the following: 4 + 2 × ( 3 2 − ( 2 + 5 ) 2 × 4 ) / ( 3 + 8 ) . Step 1: Perform all calculations within the parentheses before other operations. Evaluate the innermost parentheses first. We can work separate parentheses expressions at the same time. The innermost set of parentheses has the 2 + 5 inside. The 3 + 8 is in a separate set of parentheses, so that addition can occur at the same time as the 2 + 5. 4 + 2 × ( 3 2 − ( 2 + 5 ) 2 × 4 ) / ( 3 + 8 ) = 4 + 2 × ( 3 2 − ( 7 ) 2 × 4 ) / ( 11 ) Step 2: Now that those parentheses have been handled, move on to the next set of parentheses. Applying the order of operation rules inside that set of parentheses, the exponent is evaluated first, then the multiplication, and then the addition. 4 + 2 × ( 3 2 − 7 2 × 4 ) / 11 = 4 + 2 × ( 3 2 − 49 × 4 ) / 11 = 4 + 2 × ( − 187 ) / 11 Step 3: Since all parentheses have been cleared, apply the EMDAS rules to finish the calculation. 4 + 2 × ( − 187 ) / 11 = 4 − 374 / 11 = 4 − 34 = − 30 Order of Operations 4 Check Your Understanding Key Terms order of operations PEMDAS Key Concepts Establishing shared rules on which arithmetic operations are calculated first is necessary. Without them, different people may find different values for the same expression. The highest precedence is with expressions in parentheses. This allows parts of an expression to be calculated in an order different than the basic order of operations. The lowest precedence is addition and subtraction, as they are the basis for all other calculations. Multiplication and division have precedence over addition and subtraction, as they are representations of repeated addition or subtraction. Exponents have precedence over multiplication and division, as they represent repeated multiplication and division. Videos Order of Operations 1 Order of Operations 2 Order of Operations 3 Order of Operations 4", "section": "Order of Operations", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Rational Numbers Stock gains and losses are often represented as percentages.(credit: \"stock market quotes in newspaper\" by Andreas Poike/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Define and identify numbers that are rational. Simplify rational numbers and express in lowest terms. Add and subtract rational numbers. Convert between improper fractions and mixed numbers. Convert rational numbers between decimal and fraction form. Multiply and divide rational numbers. Apply the order of operations to rational numbers to simplify expressions. Apply density property of rational numbers. Solve problems involving rational numbers. Use fractions to convert between units. Define and apply percent. Solve problems using percent. We are often presented with percentages or fractions to explain how much of a population has a certain feature. For example, the 6-year graduation rate of college students at public institutions is 57.6%, or 72/125. That fraction may be unsettling. But without the context, the percentage is hard to judge. So how does that compare to private institutions? There, the 6-year graduation rate is 65.4%, or 327/500. Comparing the percentages is straightforward, but the fractions are harder to interpret due to different denominators. For more context, historical data could be found. One study reported that the 6-year graduation rate in 1995 was 56.4%. Comparing that historical number to the recent 6-year graduation rate at public institutions of 57.6% shows that there hasn't been much change in that rate. Defining and Identifying Numbers That Are Rational A rational number (called rational since it is a ratio) is just a fraction where the numerator is an integer and the denominator is a non-zero integer. As simple as that is, they can be represented in many ways. It should be noted here that any integer is a rational number. An integer, n , written as a fraction of two integers is n 1 . In its most basic representation, a rational number is an integer divided by a non-zero integer, such as 3 12 . Fractions may be used to represent parts of a whole. The denominator is the total number of parts to the object, and the numerator is how many of those parts are being used or selected. So, if a pizza is cut into 8 equal pieces, each piece is 1 8 of the pizza. If you take three slices, you have 3 8 of the pizza ( ). Similarly, if in a group of 20 people, 5 are wearing hats, then 5 20 of the people are wearing hats ( ). Pizza cut in 8 slices, with 3 slices highlighted Group of 20 people, with 5 people wearing hats Another representation of rational numbers is as a mixed number, such as 2 5 8 ( ). This represents a whole number (2 in this case), plus a fraction (the 5 8 ). Two whole pizzas and one partial pizza Rational numbers may also be expressed in decimal form; for instance, as 1.34. When 1.34 is written, the decimal part, 0.34, represents the fraction 34 100 , and the number 1.34 is equal to 1 34 100 . However, not all decimal representations are rational numbers. A number written in decimal form where there is a last decimal digit (after a given decimal digit, all following decimal digits are 0) is a terminating decimal , as in 1.34 above. Alternately, any decimal numeral that, after a finite number of decimal digits, has digits equal to 0 for all digits following the last non-zero digit. All numbers that can be expressed as a terminating decimal are rational. This comes from what the decimal represents. The decimal part is the fraction of the decimal part divided by the appropriate power of 10. That power of 10 is the number of decimal digits present, as for 0.34, with two decimal digits, being equal to 34 100 . Another form that is a rational number is a decimal that repeats a pattern, such as 67.1313… When a rational number is expressed in decimal form and the decimal is a repeated pattern, we use special notation to designate the part that repeats. For example, if we have the repeating decimal 4.3636…, we write this as 4. 36 ¯ . The bar over the 36 indicates that the 36 repeats forever. If the decimal representation of a number does not terminate or form a repeating decimal, that number is not a rational number. One class of numbers that is not rational is the square roots of integers or rational numbers that are not perfect squares , such as 10 and 25 6 . More generally, the number b is the square root of the number a if a = b 2 . The notation for this is b = a , where the symbol is the square root sign. An integer perfect square is any integer that can be written as the square of another integer. A rational perfect square is any rational number that can be written as a fraction of two integers that are perfect squares. Sometimes you may be able to identify a perfect square from memory. Another process that may be used is to factor the number into the product of an integer with itself. Or a calculator (such as Desmos) may be used to find the square root of the number. If the calculator yields an integer, the original number was a perfect square. Using Desmos to Find the Square Root of a Number When Desmos is used, there is a tab at the bottom of the screen that opens the keyboard for Desmos. The keyboard is shown below. On the keyboard ( ) is the square root symbol ( ) . To find the square root of a number, click the square root key, and then type the number. Desmos will automatically display the value of the square root as you enter the number. Desmos keyboard with square root key Identifying Perfect Squares Which of the following are perfect squares? 45 144 We could attempt to find the perfect square by factoring. Writing all the factor pairs of 45 results in 1 × 45 , 3 × 15 , and 5 × 9 . None of the pairs is a square, so 45 is not a perfect square. Using a calculator to find the square root of 45, we obtain 6.708 (rounded to three decimal places). Since this was not an integer, the original number was not a perfect square. We could attempt to find the perfect square by factoring. Writing all the factor pairs of 144 results in 1 × 144 , 2 × 72 , 3 × 48 , 6 × 24 , 8 × 18 , and 12 × 12 . Since the last pair is an integer multiplied by itself, 144 is a perfect square. Alternately, using Desmos to find the square root of 144, we obtain 12. Since the square root of 144 is an integer, 144 is a perfect square. Introduction to Fractions Identifying Rational Numbers Determine which of the following are rational numbers: 73 4.556 3 1 5 41 17 5 . 64 ¯ Since 73 is not a perfect square, its square root is not a rational number. This can also be seen when a calculator is used. Entering 73 into a calculator results in 8.544003745317 (and then more decimal values after that). There is no repeated pattern, so this is not a rational number. Since 4.556 is a decimal that terminates, this is a rational number. 3 1 5 is a mixed number, so it is a rational number. 41 17 is an integer divided by an integer, so it is a rational number. 5.646464... is a decimal that repeats a pattern, so it is a rational number. Simplifying Rational Numbers and Expressing in Lowest Terms A rational number is one way to express the division of two integers. As such, there may be multiple ways to express the same value with different rational numbers. For instance, 4 5 and 12 15 are the same value. If we enter them into a calculator, they both equal 0.8. Another way to understand this is to consider what it looks like in a figure when two fractions are equal. In , we see that 3 5 of the rectangle and 9 15 of the rectangle are equal areas. Two Rectangles with Equal Areas They are the same proportion of the area of the rectangle. The left rectangle has 5 pieces, three of which are shaded. The right rectangle has 15 pieces, 9 of which are shaded. Each of the pieces of the left rectangle was divided equally into three pieces. This was a multiplication. The numerator describing the left rectangle was 3 but it becomes 3 × 3 , or 9, as each piece was divided into three. Similarly, the denominator describing the left rectangle was 5, but became 5 × 3 , or 15, as each piece was divided into 3. The fractions 3 5 and 9 15 are equivalent because they represent the same portion (often loosely referred to as equal). This understanding of equivalent fractions is very useful for conceptualization, but it isn’t practical, in general, for determining when two fractions are equivalent. Generally, to determine if the two fractions a b and c d are equivalent, we check to see that a × d = b × c . If those two products are equal, then the fractions are equal also. Determining If Two Fractions Are Equivalent Determine if 12 30 and 14 35 are equivalent fractions. Applying the definition, a = 12 , b = 30 , c = 14 and d = 35 . So a × d = 12 × 35 = 420 . Also, b × c = 30 × 14 = 420 . Since these values are equal, the fractions are equivalent. That a × d = b × c indicates the fractions a b and c d are equivalent is due to some algebra. One property of natural numbers, integers, and rational numbers (also irrational numbers) is that for any three numbers a , b , and c with c ≠ 0 , if a = b , then a / c = b / c . In other words, when two numbers are equal, then dividing both numbers by the same non-zero number, the two newly obtained numbers are also equal. We can apply that to a × d and b × c , to show that a b and c d are equivalent if a × d = b × c . If a × d = b × c , and c ≠ 0 , d ≠ 0 , we can divide both sides by and obtain the following: a × d c = b × c c . We can divide out the c on the right-hand side of the equation, resulting in a × d c = b . Similarly, we can divide both sides of the equation by d and obtain the following: a × d c × d = b d . We can divide out d the on the left-hand side of the equation, resulting in a c = b d . So, the rational numbers a c and b d are equivalent when a × d = b × c . Equivalent Fractions Recall that a common divisor or common factor of a set of integers is one that divides all the numbers of the set of numbers being considered. In a fraction, when the numerator and denominator have a common divisor, that common divisor can be divided out . This is often called canceling the common factors or, more colloquially, as canceling . To show this, consider the fraction 36 63 . The numerator and denominator have the common factor 3. We can rewrite the fraction as 36 63 = 12 × 3 21 × 3 . The common divisor 3 is then divided out, or canceled, and we can write the fraction as 12 × 3 21 × 3 = 12 21 . The 3s have been crossed out to indicate they have been divided out. The process of dividing out two factors is also referred to as reducing the fraction . If the numerator and denominator have no common positive divisors other than 1, then the rational number is in lowest terms . The process of dividing out common divisors of the numerator and denominator of a fraction is called reducing the fraction . One way to reduce a fraction to lowest terms is to determine the GCD of the numerator and denominator and divide out the GCD. Another way is to divide out common divisors until the numerator and denominator have no more common factors. Reducing Fractions to Lowest Terms Express the following rational numbers in lowest terms: 36 48 100 250 51 136 One process to reduce 36 48 to lowest terms is to identify the GCD of 36 and 48 and divide out the GCD. The GCD of 36 and 48 is 12. Step 1: We can then rewrite the numerator and denominator by factoring 12 from both. 36 48 = 12 × 3 12 × 4 Step 2: We can now divide out the 12s from the numerator and denominator. 36 48 = 12 × 3 12 × 4 = 3 4 So, when 36 48 is reduced to lowest terms, the result is 3 4 . Alternately, you could identify a common factor, divide out that common factor, and repeat the process until the remaining fraction is in lowest terms. Step 1: You may notice that 4 is a common factor of 36 and 48. Step 2: Divide out the 4, as in 36 48 = 4 × 9 4 × 12 = 4 × 9 4 × 12 = 9 12 . Step 3: Examining the 9 and 12, you identify 3 as a common factor and divide out the 3, as in 9 12 = 3 × 3 3 × 4 = 3 4 . The 3 and 4 have no common positive factors other than 1, so it is in lowest terms. So, when 36 48 is reduced to lowest terms, the result is 3 4 . Step 1: To reduce 100 250 to lowest terms, identify the GCD of 100 and 250. This GCD is 50. Step 2: We can then rewrite the numerator and denominator by factoring 50 from both. 100 250 = 50 × 2 50 × 5 . Step 3: We can now divide out the 50s from the numerator and denominator. 100 250 = 50 × 2 50 × 5 = 2 5 So, when 100 250 is reduced to lowest terms, the result is 2 5 . Step 1: To reduce 51 136 to lowest terms, identify the GCD of 51 and 136. This GCD is 17. Step 2: We can then rewrite the numerator and denominator by factoring 17 from both. 51 136 = 17 × 3 17 × 8 Step 3: We can now divide out the 17s from the numerator and denominator. 51 136 = 17 × 3 17 × 8 = 3 8 So, when 51 136 is reduced to lowest terms, the result is 3 8 . Reducing Fractions to Lowest Terms Using Desmos to Find Lowest Terms Desmos is a free online calculator . Desmos supports reducing fractions to lowest terms. When a fraction is entered, Desmos immediately calculates the decimal representation of the fraction. However, to the left of the fraction, there is a button that, when clicked, shows the fraction in reduced form. Using Desmos to Reduce a Fraction Adding and Subtracting Rational Numbers Adding or subtracting rational numbers can be done with a calculator, which often returns a decimal representation, or by finding a common denominator for the rational numbers being added or subtracted. Using Desmos to Add Rational Numbers in Fractional Form To create a fraction in Desmos, enter the numerator, then use the division key (/) on your keyboard, and then enter the denominator. The fraction is then entered. Then click the right arrow key to exit the denominator of the fraction. Next, enter the arithmetic operation (+ or –). Then enter the next fraction. The answer is displayed dynamically (calculates as you enter). To change the Desmos result from decimal form to fractional form, use the fraction button ( ) on the left of the line that contains the calculation: Fraction button on the Desmos keyboard Adding Rational Numbers Using Desmos Calculate 23 42 + 9 56 using Desmos. Enter 23 42 + 9 56 in Desmos. The result is displayed as 0.70833333333 (which is 0.708 3 ¯ ). Clicking the fraction button to the left on the calculation line yields 17 24 . Performing addition and subtraction without a calculator may be more involved. When the two rational numbers have a common denominator, then adding or subtracting the two numbers is straightforward. Add or subtract the numerators, and then place that value in the numerator and the common denominator in the denominator. Symbolically, we write this as a c ± b c = a ± b c . This can be seen in the , which shows 3 20 + 4 20 = 7 20 . Partially Shaded Rectangle It is customary to then write the result in lowest terms. If c is a non-zero integer, then a c ± b c = a ± b c . Adding Rational Numbers with the Same Denominator Calculate 13 28 + 7 28 . Since the rational numbers have the same denominator, we perform the addition of the numerators, 13 + 7 , and then place the result in the numerator and the common denominator, 28, in the denominator. 13 28 + 7 28 = 13 + 7 28 = 20 28 Once we have that result, reduce to lowest terms, which gives 20 28 = 4 × 5 4 × 7 = 4 × 5 4 × 7 = 5 7 . Subtracting Rational Numbers with the Same Denominator Calculate 45 136 − 17 136 . Since the rational numbers have the same denominator, we perform the subtraction of the numerators, 45 − 17 , and then place the result in the numerator and the common denominator, 136, in the denominator. 45 136 − 17 136 − 45 − 17 136 = 28 136 Once we have that result, reduce to lowest terms, this gives 28 136 = 4 × 7 4 × 34 = 4 × 7 4 × 34 = 7 34 . When the rational numbers do not have common denominators, then we have to transform the rational numbers so that they do have common denominators. The common denominator that reduces work later in the problem is the LCM of the numerator and denominator. When adding or subtracting the rational numbers a b and c d , we perform the following steps. Step 1: Find LCM ( b , d ) . Step 2: Calculate n = LCM ( b , d ) b and m = L C M ( b , d ) d . Step 3: Multiply the numerator and denominator of a b by n , yielding a × n b × n . Step 4: Multiply the numerator and denominator of c d by m , yielding c × m d × m . Step 5: Add or subtract the rational numbers from Steps 3 and 4, since they now have the common denominators. You should be aware that the common denominator is LCM ( b , d ) . For the first denominator, we have b × n = b × L C M ( b , d ) b = L C M ( b , d ) , since we multiply and divide LCM ( b , d ) by the same number. For the same reason, d × m = d × L C M ( b , d ) b = L C M ( b , d ) . Adding Rational Numbers with Unequal Denominators Calculate 11 18 + 2 15 . The denominators of the fractions are 18 and 15, so we label b = 18 and d = 15 . Step 1: Find LCM(18,15). This is 90. Step 2 : Calculate n and m . n = 90 18 = 5 and m = 90 15 = 6 . Step 3: Multiplying the numerator and denominator of 11 18 by n = 5 yields 11 × 5 18 × 5 = 55 90 . Step 4: Multiply the numerator and denominator of 2 15 by m = 6 yields 2 × 6 15 × 6 = 12 90 . Step 5: Now we add the values from Steps 3 and 4: 55 90 + 12 90 = 67 90 . This is in lowest terms, so we have found that 11 18 + 2 15 = 67 90 . Subtracting Rational Numbers with Unequal Denominators Calculate 14 25 − 9 70 . The denominators of the fractions are 25 and 70, so we label b = 25 and d = 70 . Step 1: Find LCM(25,70). This is 350. Step 2: Calculate n and m : n = 350 25 = 14 and m = 350 70 = 5 . Step 3: Multiplying the numerator and denominator of 14 25 by n = 14 yields 14 × 14 25 × 14 = 196 350 . Step 4: Multiplying the numerator and denominator of 9 70 by m = 5 yields 9 × 5 70 × 5 = 45 350 . Step 5: Now we subtract the value from Step 4 from the value in Step 3: 196 350 − 45 350 = 151 350 . This is in lowest terms, so we have found that 14 25 − 9 70 = 151 350 . Adding and Subtracting Fractions with Different Denominators Converting Between Improper Fractions and Mixed Numbers One way to visualize a fraction is as parts of a whole, as in 5 12 of a pizza. But when the numerator is larger than the denominator, as in 23 12 , then the idea of parts of a whole seems not to make sense. Such a fraction is an improper fraction. That kind of fraction could be written as an integer plus a fraction, which is a mixed number . The fraction 23 12 rewritten as a mixed number would be 1 11 12 . Arithmetically, 1 11 12 is equivalent to 1 + 11 12 , which is read as “one and 11 twelfths.” Improper fractions can be rewritten as mixed numbers using division and remainders. To find the mixed number representation of an improper fraction, divide the numerator by the denominator. The quotient is the integer part, and the remainder becomes the numerator of the remaining fraction. Rewriting an Improper Fraction as a Mixed Number Rewrite 48 13 as a mixed number. When 48 is divided by 13, the result is 3 with a remainder of 9. So, we can rewrite 48 13 as 3 9 13 . Converting an Improper Fraction to a Mixed Number Using Desmos Similarly, we can convert a mixed number into an improper fraction. To do so, first convert the whole number part to a fraction by writing the whole number as itself divided by 1, and then add the two fractions. Alternately, we can multiply the whole number part and the denominator of the fractional part. Next, add that product to the numerator. Finally, express the number as that product divided by the denominator. Rewriting a Mixed Number as an Improper Fraction Rewrite 5 4 9 as an improper fraction. Step 1: Multiply the integer part, 5, by the denominator, 9, which gives 5 × 9 = 45 . Step 2: Add that product to the numerator, which gives 45 + 4 = 49 . Step 3: Write the number as the sum, 49, divided by the denominator, 9, which gives 49 9 . Using Desmos to Rewrite a Mixed Number as an Improper Fraction Desmos can be used to convert from a mixed number to an improper fraction. To do so, we use the idea that a mixed number, such as 5 6 11 , is another way to represent 5 + 6 11 . If 5 + 6 11 is entered in Desmos, the result is the decimal form of the number. However, clicking the fraction button to the left will convert the decimal to an improper fraction, 61 11 . As an added bonus, Desmos will automatically reduce the fraction to lowest terms. Converting Rational Numbers Between Decimal and Fraction Forms Understanding what decimals represent is needed before addressing conversions between the fractional form of a number and its decimal form , or writing a number in decimal notation . The decimal number 4.557 is equal to 4 557 1,000 . The decimal portion, .557, is 557 divided by 1,000. To write any decimal portion of a number expressed as a terminating decimal, divide the decimal number by 10 raised to the power equal to the number of decimal digits. Since there were three decimal digits in 4.557, we divided 557 by 10 3 = 1000 . Decimal representations may be very long. It is convenient to round off the decimal form of the number to a certain number of decimal digits. To round off the decimal form of a number to n (decimal) digits, examine the ( n + 1 )st decimal digit. If that digit is 0, 1, 2, 3, or 4, the number is rounded off by writing the number to the n th decimal digit and no further. If the ( n + 1 )st decimal digit is 5, 6, 7, 8, or 9, the number is rounded off by writing the number to the n th digit, then replacing the n th digit by one more than the n th digit. Rounding Off a Number in Decimal Form to Three Digits Round 5.67849 to three decimal digits. The third decimal digit is 8. The digit following the 8 is 4. When the digit is 4, we write the number only to the third digit. So, 5.67849 rounded off to three decimal places is 5.678. Rounding Off a Number in Decimal Form to Four Digits Round 45.11475 to four decimal digits. The fourth decimal digit is 7. The digit following the 7 is 5. When the digit is 5, we write the number only to the fourth decimal digit, 45.1147. We then replace the fourth decimal digit by one more than the fourth digit, which yields 45.1148. So, 45.11475 rounded off to four decimal places is 45.1148. To convert a rational number in fraction form to decimal form, use your calculator to perform the division. Converting a Rational Number in Fraction Form into Decimal Form Convert 47 25 into decimal form. Using a calculator to divide 47 by 25, the result is 1.88. Converting a terminating decimal to the fractional form may be done in the following way: Step 1: Count the number of digits in the decimal part of the number, labeled n . Step 2: Raise 10 to the n th power. Step 3: Rewrite the number without the decimal. Step 4: The fractional form is the number from Step 3 divided by the result from Step 2. This process works due to what decimals represent and how we work with mixed numbers. For example, we could convert the number 7.4536 to fractional from. The decimal part of the number, the .4536 part of 7.4536, has four digits. By the definition of decimal notation, the decimal portion represents 4,536 10 4 = 4,536 10,000 . The decimal number 7.4536 is equal to the improper fraction 7 4,536 10,000 . Adding those to fractions yields 74,536 10,000 . Converting from Decimal Form to Fraction Form with Terminating Decimals Convert 3.2117 to fraction form. Step 1: There are four digits after the decimal point, so n = 4 . Step 2: Raise 10 to the fourth power, 10 4 = 10,000 . Step 3: When we remove the decimal point, we have 32,117. Step 4: The fraction has as its numerator the result from Step 3 and as its denominator the result of Step 2, which is the fraction 32,117 10,000 . The process is different when converting from the decimal form of a rational number into fraction form when the decimal form is a repeating decimal. This process is not covered in this text. Multiplying and Dividing Rational Numbers Multiplying rational numbers is less complicated than adding or subtracting rational numbers, as there is no need to find common denominators. To multiply rational numbers, multiply the numerators, then multiply the denominators, and write the numerator product divided by the denominator product. Symbolically, a b × c d = a × c b × d . As always, rational numbers should be reduced to lowest terms. If b and d are non-zero integers, then a b × c d = a × c b × d . Multiplying Rational Numbers Calculate 12 25 × 10 21 . Multiply the numerators and place that in the numerator, and then multiply the denominators and place that in the denominator. 12 25 × 10 21 = 12 × 10 25 × 21 = 120 525 This is not in lowest terms, so this needs to be reduced. The GCD of 120 and 525 is 15. 120 525 = 15 × 8 15 × 35 = 8 35 Multiplying Fractions As with multiplication, division of rational numbers can be done using a calculator. Dividing Decimals with a Calculator Calculate 3.45 ÷ 2.341 using a calculator. Round to three decimal places if necessary. Using a calculator, we obtain 1.473729175565997. Rounding to three decimal places we have 1.474. Before discussing division of fractions without a calculator, we should look at the reciprocal of a number. The reciprocal of a number is 1 divided by the number. For a fraction, the reciprocal is the fraction formed by switching the numerator and denominator. For the fraction a b , the reciprocal is b a . An important feature for a number and its reciprocal is that their product is 1. When dividing two fractions by hand, find the reciprocal of the divisor (the number that is being divided into the other number). Next, replace the divisor by its reciprocal and change the division into multiplication. Then, perform the multiplication. Symbolically, b a ÷ c d = a b × d c = a × d b × c . As before, reduce to lowest terms. If b , c and d are non-zero integers, then b a ÷ c d = a b × d c = a × d b × c . Dividing Rational Numbers Calculate 4 21 ÷ 6 35 . Calculate 1 8 ÷ 5 28 . Step 1: Find the reciprocal of the number being divided by 6 35 . The reciprocal of that is 35 6 . Step 2: Multiply the first fraction by that reciprocal. 4 21 ÷ 6 35 = 4 21 × 35 6 = 140 126 The answer, 140 126 is not in lowest terms. The GCD of 140 and 126 is 14. Factoring and canceling gives 140 126 = 14 × 10 14 × 9 = 10 9 . Step 1: Find the reciprocal of the number being divided by, which is 5 28 . The reciprocal of that is 28 5 . Step 2: Multiply the first fraction by that reciprocal: 1 8 ÷ 5 28 = 1 8 × 28 5 = 28 40 The answer, 28 40 , is not in lowest reduced form. The GCD of 28 and 40 is 4. Factoring and canceling gives 28 40 = 4 × 7 4 × 10 = 7 10 . Dividing Fractions Applying the Order of Operations to Simplify Expressions The order of operations for rational numbers is the same as for integers, as discussed in Order of Operations . The order of operations makes it easier for anyone to correctly calculate and represent. The order follows the well-known acronym PEMDAS: P Parentheses E Exponents M/D Multiplication and division A/S Addition and subtraction The first step in calculating using the order of operations is to perform operations inside the parentheses. Moving down the list, next perform all exponent operations moving from left to right. Next (left to right once more), perform all multiplications and divisions. Finally, perform the additions and subtractions. Applying the Order of Operations with Rational Numbers Correctly apply the rules for the order of operations to accurately compute ( 5 7 − 2 7 ) × 2 3 . Step 1: To calculate this, perform all calculations within the parentheses before other operations. ( 5 7 − 2 7 ) × 2 3 = ( 3 7 ) × 2 3 Step 2: Since all parentheses have been cleared, we move left to right, and compute all the exponents next. ( 3 7 ) × 2 3 = ( 3 7 ) × 8 Step 3: Now, perform all multiplications and divisions, moving left to right. ( 3 7 ) × 8 = 24 7 Applying the Order of Operations with Rational Numbers Correctly apply the rules for the order of operations to accurately compute 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 2 3 + 5 ) ) 2 . To calculate this, perform all calculations within the parentheses before other operations. Evaluate the innermost parentheses first. We can work separate parentheses expressions at the same time. Step 1: The innermost parentheses contain 2 3 + 5 . Calculate that first, dividing after finding the common denominator. 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 2 3 + 5 ) ) 2 = 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 2 3 + 5 1 ) ) 2 = 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 2 3 + 15 3 ) ) 2 = 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 17 3 ) ) 2 Step 2: Calculate the exponent in the parentheses, ( 5 9 ) 2 . 4 + 2 3 ÷ ( ( 5 9 ) 2 − ( 17 3 ) ) 2 = 4 + 2 3 ÷ ( ( 25 81 ) − ( 17 3 ) ) 2 Step 3: Subtract inside the parentheses is done, using a common denominator. 4 + 2 3 ÷ ( ( 25 81 ) − ( 17 3 ) ) 2 4 + 2 3 ÷ ( ( 25 81 ) − ( 17 × 27 3 × 27 ) ) 2 4 + 2 3 ÷ ( ( 25 81 ) − ( 459 81 ) ) 2 4 + 2 3 ÷ ( ( − 434 81 ) ) 2 Step 4: At this point, evaluate the exponent and divide. 4 + 2 3 ÷ ( ( − 434 81 ) ) 2 4 + 2 3 ÷ ( 188,356 6,561 ) = 4 + 2 3 × ( 6,561 188,356 ) = 4 + 2,187 94,178 Step 5: Add. 4 + 2,187 94,178 = 378,899 94,178 Had this been done on a calculator, the decimal form of the answer would be 4.0232 (rounded to four decimal places). Order of Operations Using Fractions Applying the Density Property of Rational Numbers Between any two rational numbers, there is another rational number. This is called the density property of the rational numbers. Finding a rational number between any two rational numbers is very straightforward. Step 1: Add the two rational numbers. Step 2: Divide that result by 2. The result is always a rational number. This follows what we know about rational numbers. If two fractions are added, then the result is a fraction. Also, when a fraction is divided by a fraction (and 2 is a fraction), then we get another fraction. This two-step process will give a rational number, provided the first two numbers were rational. Applying the Density Property of Rational Numbers Demonstrate the density property of rational numbers by finding a rational number between 4 11 and 7 12 . To find a rational number between 4 11 and 7 12 : Step 1: Add the fractions. 4 11 + 7 12 = 4 × 12 11 × 12 + 7 × 11 12 × 11 = 48 132 + 77 132 = 125 132 Step 2: Divide the result by 2. Recall that to divide by 2, you multiply by the reciprocal of 2. The reciprocal of 2 is 1 2 , as seen below. 125 132 ÷ 2 = 125 132 × 1 2 = 125 264 So, one rational number between 4 11 and 7 12 is 125 264 . We could check that the number we found is between the other two by finding the decimal representation of the numbers. Using a calculator, the decimal representations of the rational numbers are 0.363636…, 0.473484848…, and 0.5833333…. Here it is clear that 125 264 is between 4 11 and 7 12 . Solving Problems Involving Rational Numbers Rational numbers are used in many situations, sometimes to express a portion of a whole, other times as an expression of a ratio between two quantities. For the sciences, converting between units is done using rational numbers, as when converting between gallons and cubic inches. In chemistry, mixing a solution with a given concentration of a chemical per unit volume can be solved with rational numbers. In demographics, rational numbers are used to describe the distribution of the population. In dietetics, rational numbers are used to express the appropriate amount of a given ingredient to include in a recipe. As discussed, the application of rational numbers crosses many disciplines. Mixing Soil for Vegetables James is mixing soil for a raised garden, in which he plans to grow a variety of vegetables. For the soil to be suitable, he determines that 2 5 of the soil can be topsoil, but 2 5 needs to be peat moss and 1 5 has to be compost. To fill the raised garden bed with 60 cubic feet of soil, how much of each component does James need to use? In this example, we know the proportion of each component to mix, and we know the total amount of the mix we need. In this kind of situation, we need to determine the appropriate amount of each component to include in the mixture. For each component of the mixture, multiply 60 cubic feet, which is the total volume of the mixture we want, by the fraction required of the component. Step 1: The required fraction of topsoil is 2 5 , so James needs 60 × 2 5 cubic feet of topsoil. Performing the multiplication, James needs 60 × 2 5 = 120 5 = 24 (found by treating the fraction as division, and 120 divided by 5 is 24) cubic feet of topsoil. Step 2: The required fraction of peat moss is also 2 5 , so he also needs 60 × 2 5 cubic feet, or 60 × 2 5 = 120 5 = 24 cubic feet of peat moss. Step 3: The required fraction of compost is 1 5 . For the compost, he needs 60 × 1 5 = 60 5 = 12 cubic feet. Determining the Number of Specialty Pizzas At Bella’s Pizza, one-third of the pizzas that are ordered are one of their specialty varieties. If there are 273 pizzas ordered, how many were specialty pizzas? One-third of the whole are specialty pizzas, so we need one-third of 273, which gives 1 3 × 273 = 273 3 = 91 , found by dividing 273 by 3. So, 91 of the pizzas that were ordered were specialty pizzas. Finding a Fraction of a Total Using Fractions to Convert Between Units A common application of fractions is called unit conversion , or converting units , which is the process of changing from the units used in making a measurement to different units of measurement. For instance, 1 inch is (approximately) equal to 2.54 cm. To convert between units, the two equivalent values are made into a fraction. To convert from the first type of unit to the second type, the fraction has the second unit as the numerator, and the first unit as the denominator. From the inches and centimeters example, to change from inches to centimeters, we use the fraction 2.54 cm 1 in . If, on the other hand, we wanted to convert from centimeters to inches, we’d use the fraction 1 in 2.54 cm . This fraction is multiplied by the number of units of the type you are converting from , which means the units of the denominator are the same as the units being multiplied. Converting Liters to Gallons It is known that 1 liter (L) is 0.264172 gallons (gal). Use this to convert 14 liters into gallons. We know that 1 liter = 0.264172 gal. Since we are converting from liters, when we create the fraction we use, make sure the liter part of the equivalence is in the denominator. So, to convert the 14 liters to gallons, we multiply 14 by 1 gal 0.264172 gal / 1 liter . Notice the gallon part is in the numerator since we’re converting to gallons, and the liter part is in the denominator since we are converting from liters. Performing this and rounding to three decimal places, we find that 14 liters is 14 liter × 0.264172 gal 1 liter = 3.69841 gal . Converting Centimeters to Inches It is known that 1 inch is 2.54 centimeters. Use this to convert 100 centimeters into inches. We know that 1 inch = 2.54 cm. Since we are converting from centimeters, when we create the fraction we use, make sure the centimeter part of the equivalence is in the denominator, 1 in 2.54 cm . To convert the 100 cm to inches, multiply 100 by 1 in 2.54 cm . Notice the inch part is in the numerator since we’re converting to inches, and the centimeter part is in the denominator since we are converting from centimeters. Performing this and rounding to three decimal places, we obtain 100 cm × 1 in 2.54 cm = 39.370 in . This means 100 cm equals 39.370 in. Converting Units Defining and Applying Percent A percent is a specific rational number and is literally per 100. n percent, denoted n %, is the fraction n 100 . Rewriting a Percentage as a Fraction Rewrite the following as fractions: 31% 93% Using the definition and n = 31 , 31% in fraction form is 31 100 . Using the definition and n = 93 , 93% in fraction form is 93 100 . Rewriting a Percentage as a Decimal Rewrite the following percentages in decimal form: 54% 83% Using the definition and n = 54 , 54% in fraction form is 54 100 . Dividing a number by 100 moves the decimal two places to the left; 54% in decimal form is then 0.54. Using the definition and n = 83 , 83% in fraction form is 83 100 . Dividing a number by 100 moves the decimal two places to the left; 83% in decimal form is then 0.83. You should notice that you can simply move the decimal two places to the left without using the fractional definition of percent. Percent is used to indicate a fraction of a total. If we want to find 30% of 90, we would perform a multiplication, with 30% written in either decimal form or fractional form. The 90 is the total , 30 is the percentage , and 27 (which is 0.30 × 90 ) is the percentage of the total . n % of x items is n 100 × x . The x is referred to as the total , the n is referred to as the percent or percentage , and the value obtained from n 100 × x is the part of the total and is also referred to as the percentage of the total . Finding a Percentage of a Total Determine 40% of 300. Determine 64% of 190. The total is 300, and the percentage is 40. Using the decimal form of 40% and multiplying we obtain 0.40 × 300 = 120 . The total is 190, and the percentage is 64. Using the decimal form of 64% and multiplying we obtain 0.64 × 190 = 121.6 . In the previous situation, we knew the total and we found the percentage of the total. It may be that we know the percentage of the total, and we know the percent, but we don't know the total. To find the total if we know the percentage of the total, use the following formula. If we know that n % of the total is x , then the total is 100 × x n . Finding the Total When the Percentage and Percentage of the Total Are Known What is the total if 28% of the total is 140? What is the total if 6% of the total is 91? 28 is the percentage, so n = 28 . 28% of the total is 140, so x = 140 . Using those we find that the total was 100 × 140 28 = 500 . 6 is the percentage, so n = 6 . 6% of the total is 91, so x = 91 . Using those we find that the total was 100 × 91 6 = 1,516.6 . The percentage can be found if the total and the percentage of the total is known. If you know the total, and the percentage of the total, first divide the part by the total. Move the decimal two places to the right and append the symbol %. The percentage may be found using the following formula. The percentage, n , of b that is a is a b × 100 % . Finding the Percentage When the Total and Percentage of the Total Are Known Find the percentage in the following: Total is 300, percentage of the total is 60. Total is 440, percentage of the total is 176. The total is 300; the percentage of the total is 60. Calculating yields 0.2. Moving the decimal two places to the right gives 20. Appending the percentage to this number results in 20%. So, 60 is 20% of 300. The total is 440; the percentage of the total is 176. Calculating yields 0.4. Moving the decimal two places to the right gives 40. Appending the percentage to this number results in 40%. So, 176 is 40% of 440. Solve Problems Using Percent In the media, in research, and in casual conversation percentages are used frequently to express proportions. Understanding how to use percent is vital to consuming media and understanding numbers. Solving problems using percentages comes down to identifying which of the three components of a percentage you are given, the total, the percentage, or the percentage of the total. If you have two of those components, you can find the third using the methods outlined previously. Percentage of Students Who Are Sleep Deprived A study revealed that 70% of students suffer from sleep deprivation, defined to be sleeping less than 8 hours per night. If the survey had 400 participants, how many of those participants had less than 8 hours of sleep per night? The percentage of interest is 70%. The total number of students is 400. With that, we can find how many were in the percentage of the total, or, how many were sleep deprived. Applying the formula from above, the number who were sleep deprived was 0.70 × 400 = 280 ; 280 students on the study were sleep deprived. Amazon Prime Subscribers There are 126 million users who are U.S. Amazon Prime subscribers. If there are 328.2 million residents in the United States, what percentage of U.S. residents are Amazon Prime subscribers? We are asked to find the percentage. To do so, we divide the percentage of the total, which is 126 million, by the total, which is 328.2 million. Performing this division and rounding to three decimal places yields 126 328.2 = 0.384 . The decimal is moved to the right by two places, and a % sign is appended to the end. Doing this shows us that 38.4% of U.S. residents are Amazon Prime subscribers. Finding the Percentage When the Total and Percentage of the Total Are Known Evander plays on the basketball team at their university and 73% of the athletes at their university receive some sort of scholarship for attending. If they know 219 of the student-athletes receive some sort of scholarship, how many student-athletes are at the university? We need to find the total number of student-athletes at Evander’s university. Step 1: Identify what we know. We know the percentage of students who receive some sort of scholarship, 73%. We also know the number of athletes that form the part of the whole, or 219 student-athletes. Step 2: To find the total number of student-athletes, use 100 × x n , with x = 219 and n = 73 . Calculating with those values yields 100 × 219 73 = 300 . So, there are 300 total student-athletes at Evander’s university Check Your Understanding Key Terms density property of rational numbers improper fraction lowest terms mixed number rational number repeating decimal terminating decimal Key Concepts Rational numbers are fractions of integers, and can always be written as an integer divided by an integer. The numerator and denominator of a fraction may have common factors. In such cases, the fraction can be reduced by canceling common factors. When the numerator and denominator of a fraction have no common factors, the fraction is said to be reduced. An improper fraction is one with a numerator larger than the denominator. Such a fraction can be rewritten as an integer plus a proper fraction. This is called a mixed number. Using division and remainder, an improper fraction may be written as a mixed number. A mixed number can be converted to an improper fraction by reversing the process for changing an improper fraction to a mixed number. The arithmetic operations or addition, subtraction, multiplication and division can all be performed on rational numbers. Addition and subtraction of rational numbers can be performed after a common denominator has been identified, and the fractions have been converted to forms having the common denominator. Multiplication and division of rational numbers can be performed without regard to common denominators. Between any two rational numbers, there is always another rational number. This is the density property of the rational numbers. Formulas a c ± b c = a ± b c a b × c d = a × c b × d a b × c d = a × c b × d a b ÷ c d = a d × d c = a × d b × c Videos Introduction to Fractions Equivalent Fractions Reducing Fractions to Lowest Terms Using Desmos to Reduce a Fraction Adding and Subtracting Fractions with Different Denominators Converting an Improper Fraction to a Mixed Number Using Desmos Multiplying Fractions Dividing Fractions Order of Operations Using Fractions Finding a Fraction of a Total Converting Units", "section": "Rational Numbers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Irrational Numbers The Pythagoreans were a philosophical sect of ancient Greece, often associated with mathematics. (credit: Fedor Andreevich Bronnikov (1827-1902) “Hymn of the Pythagoreans to the Rising Sun,” 1877, oil on canvas/Wikimedia, public domain) Learning Objectives After completing this section, you should be able to: Define and identify numbers that are irrational. Simplify irrational numbers and express in lowest terms. Add and subtract irrational numbers. Multiply and divide irrational numbers. Rationalize fractions with irrational denominators. The Pythagoreans were a philosophical sect in ancient Greece. Their philosophy included reincarnation and purifying the mind through the study and contemplation of mathematics and science. One of their principles was the cosmos is ruled by order, specifically mathematics and music. They even held mystic beliefs about specific numbers and figures. For example, the number 1 was associated with the mind and essence. Four represented justice, as it is the first product of two even numbers. Most famously, though, is the association with the Pythagorean Theorem, which states that in a right triangle, the sum of the squares of the shorter sides of the triangle (the legs) equals the square of the longer side (the hypotenuse). Even the ancient Egyptians used this relationship, as triangles with side measures 3, 4, and 5 were often used in surveying following the flooding of the Nile. There is a story of a Pythagorean, Hippasus, discovering that not all numbers could be expressed as fractions. In other words, not all numbers were rational numbers. The story ends with Hippasus, who shared this, or in some versions discovered it, put to death by drowning for sharing this fact, that not all quantities could be expressed as the ratio of two natural numbers. As colorful as that story may be, it is most likely false, as there are no contemporary sources to corroborate it. But it does seem to mark the discovery that not all quantities or measures were fractions of numbers. And so, irrational numbers were discovered. The Philosophy of the Pythagoreans Defining and Identifying Numbers That Are Irrational We defined rational numbers in the last section as numbers that could be expressed as a fraction of two integers. Irrational numbers are numbers that cannot be expressed as a fraction of two integers. Recall that rational numbers could be identified as those whose decimal representations either terminated (ended) or had a repeating pattern at some point. So irrational numbers must be those whose decimal representations do not terminate or become a repeating pattern. One collection of irrational numbers is square roots of numbers that aren’t perfect squares . x is the square root of the number a , denoted a , if x 2 = a . The number a is the perfect square of the integer n if a = n 2 . The rational number a b is a perfect square if both a and b are perfect squares. One method of determining if an integer is a perfect square is to examine its prime factorization. If, in that factorization, all the prime factors are raised to even powers, the integer is a perfect square. Another method is to attempt to factor the integer into an integer squared. It is possible that you recognize the number as a perfect square (such as 4 or 9). Or, if you have a calculator at hand, use the calculator to determine if the square root of the integer is an integer. Identifying Perfect Squares Determine which of the following are perfect squares. 45 81 9 28 144 400 The prime factorization of 45 is 45 = 3 2 × 5 . Since the 5 is not raised to an even power, 45 is not a perfect square. The prime factorization of 81 is 3 4 . All the prime factors are raised to even powers, so 81 is a perfect square. We must determine if both the numerator and denominator of 9 28 are perfect squares for the rational number to be a perfect square. The numerator is 9, and as mentioned above, 9 is a perfect square (it is 3 squared). Now we check the prime factorization of the denominator, 28, which is 28 = 2 2 × 7 . Since 7 is not raised to an even power, 28 is not a perfect square. Since the denominator is not a perfect square, 9 28 is not a perfect square. We must determine if both the numerator and denominator of 144 400 are perfect squares for the rational number to be a perfect square. The numerator is 144. The prime factorization of 144 is 144 = 2 4 × 3 2 . Since all the prime factors of 144 are raised to even powers, 144 is a perfect square. Now we check the prime factorization of the denominator, 400, which is 400 = 2 4 × 5 2 . Since all the prime factors of 400 are raised to even powers, 400 is a perfect square. Since the numerator and denominator of 144 400 are perfect squares, 144 400 is a perfect square. Using Desmos to Determine if a Number Is a Perfect Square Desmos may be used to determine if a number is a perfect square by using its square root function. When Desmos is opened, there is a tab in the lower left-hand corner of the Desmos screen. This tab opens the Desmos keypad, shown in . Desmos keyboard with square root key circles There you find the key for the square root, which is circled in . To find the square root of a number, click the square root key, which begins a calculation, and then enter the value for which you want a square root. If the result is an integer, then the number is a perfect square. Using Desmos to Find the Square Root of a Number Another collection of irrational numbers is based on the special number, pi , denoted by the Greek letter π , which is the ratio of the circumference of the diameter of the circle ( ). Circle with radius, diameter, and circumference labeled Any multiple or power of π is an irrational number. Any number expressed as a rational number times an irrational number is an irrational number also. When an irrational number takes that form, we call the rational number the rational part , and the irrational number the irrational part . It should be noted that a rational number plus, minus, multiplied by, or divided by any irrational number is an irrational number. Identifying Irrational Numbers Identify which of the following numbers are irrational. 35 0. 15 ¯ 121 4 π 35 can be factored as 5 × 7 , showing that 35 is not the square of an integer or a rational number. This mean its square root is an irrational number. Since 0. 15 ¯ is a decimal with a repeating pattern, it is rational, so it is not an irrational number. 121 = 11 2 . Since 121 is the square of an integer, its square root is a rational number. Since 4 π is a multiple of pi, it is irrational. In this case, the rational part of the number is 4, while the irrational part is π . Euler-Mascheroni Constant Determining if a number is rational or irrational is not trivial. There are numbers that defied such classification for quite a long time. One such is the Euler-Mascheroni constant. The Euler-Mascheroni constant is used in mathematics, and is primarily associated with the natural logarithm, which is a mathematical function. The constant has been around since around 1790. However, it was unknown if this constant was rational or irrational until 2013, at which point it was proven to be irrational. Simplifying Square Roots and Expressing Them in Lowest Terms To simplify a square root means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. The number inside the square root symbol is referred to as the radicand . So in the expression a the number a is referred to as the radicand. Before discussing how to simplify a square root, we need to introduce a rule about square roots. The square root of a product of numbers equals the product of the square roots of those number. Written symbolically, a × b = a × b . For any two numbers a and b , a × b = a × b . Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we’ve simplified the irrational number into lowest terms. To simplify the irrational number into lowest terms when n is an integer: Step 1: Determine the largest perfect square factor of n , which we denote a 2 . Step 2: Factor n into a 2 × b . Step 3: Apply a 2 × b = a 2 × b . Step 4: Write n in its simplified form, a b . When a square root has been simplified in this manner, a is referred to as the rational part of the number, and b is referred to as the irrational part. Simplifying a Square Root Simplify the irrational number 180 and express in lowest terms. Identify the rational and irrational parts. Begin by finding the largest perfect square that is a factor of 180. We can do this by writing out the factor pairs of 180: 1 × 180 2 × 90 3 × 60 4 × 45 5 × 36 6 × 30 9 × 20 10 × 18 12 × 15 Looking at the list of factors, the perfect squares are 4, 9, and 36. The largest is 36, so we factor the into 36 × 5 = 6 2 × 5 . In the formula, a = 6 and b = 5 . Apply a 2 × b = a 2 × b . 6 2 × 5 = 6 2 × 5 The simplified form of 180 is 6 5 . In this example, the 6 is the rational part, and the 5 is the irrational part. Simplifying Square Roots Simplifying a Square Root Simplify the irrational number 330 and express in lowest terms. Identify the rational and irrational parts. Begin by finding the largest perfect square that is a factor of 330. We can do this by writing out the factor pairs of 330: 1 × 330 2 × 165 3 × 110 5 × 66 6 × 55 10 × 33 11 × 30 15 × 22 Looking at the list of factors, there are no perfect squares other than 1, which means 330 is already expressed in lowest terms. In this case, 1 is the rational part, and 330 is the irrational part. Though we could write this as 1 330 , but the product of 1 and any other number is just the number. Simplifying a Square Root Simplify the irrational number 2,548 and express in lowest terms. Identify the rational and irrational parts. Begin by finding the largest perfect square that is a factor of 2,548. We can do this by writing out the factor pairs of 2,548: 1 × 2548 2 × 1274 4 × 637 7 × 364 13 × 196 14 × 182 26 × 98 28 × 91 49 × 52 Looking at the list of factors, the perfect squares are 4, 49, and 196. The largest is 196, so we factor the 2,548 into 196 × 13 = 14 2 × 13 . In the formula, a = 14 and b = 5 . Apply a 2 × b = a 2 × b . 14 2 × 13 = 14 2 × 13 The simplified form of 2,548 is 14 13 . In this example, 14 is the rational part, and 13 is the irrational part. Simplifying Square Roots Adding and Subtracting Irrational Numbers Just like any other number we’ve worked with, irrational numbers can be added or subtracted. When working with a calculator, enter the operation and a decimal representation will be given. However, there are times when two irrational numbers may be added or subtracted without the calculator. This can happen only when the irrational parts of the irrational numbers are the same. To add or subtract two irrational numbers that have the same irrational part, add or subtract the rational parts of the numbers, and then multiply that by the common irrational part. Let our first irrational number be a × x , where a is the rational and x the irrational parts. Let the other irrational number be b × x , where b is the rational and x the irrational parts. Then a × x ± b × x = ( a ± b ) × x . Subtracting Irrational Numbers with Similar Irrational Parts If possible, subtract the following irrational numbers without using a calculator. If this is not possible, state why. 3 7 − 8 7 Since these two irrational numbers have the same irrational part, 7 , we can subtract without using a calculator. The rational part of the first number is 3. The rational part of the second number is 8. Using the formula yields 3 7 − 8 7 = ( 3 − 8 ) × 7 = − 5 7 . Adding Irrational Numbers with Similar Irrational Parts If possible, add the following irrational numbers without using a calculator. If this is not possible, state why. 35 π + 17 π Since these two irrational numbers have the same irrational part, π , the addition can be performed without using a calculator. The rational part of the first number is 35. The rational part of the second number is 17. Using the formula yields 35 π + 17 π = ( 35 + 17 ) × π = 52 π . Subtracting Irrational Numbers with Different Irrational Parts If possible, subtract the following irrational numbers without using a calculator. If this is not possible, state why. 19 3 − 5.6 7 The two numbers being subtracted do not have the same irrational part, so the operation cannot be performed. Multiplying and Dividing Irrational Numbers Just like any other number that we’ve worked with, irrational numbers can be multiplied or divided. When working with a calculator, enter the operation and a decimal representation will be given. Sometimes, though, you may want to retain the form of the irrational number as a rational part times an irrational part. The process is similar to adding and subtracting irrational numbers when they are in this form. We do not need the irrational parts to match. Even though they need not match, they do need to be similar, such as both irrational parts are square roots, or both irrational parts are multiples of pi. Also, if the irrational parts are square roots, we may need to reduce the resulting square root to lowest terms. When multiplying two square roots, use the following formula. It is the same formula presented during the discussion of simplifying square roots. For any two positive numbers a and b , a × b = a × b . When dividing two square roots, use the following formula. For any two positive numbers a and b , with b not equal to 0, a ÷ b = a b = a b . To multiply or divide irrational numbers with similar irrational parts, do the following: Step 1: Multiply or divide the rational parts. Step 2: If necessary, reduce the result of Step 1 to lowest terms. This becomes the rational part of the answer. Step 3: Multiply or divide the irrational parts. Step 4: If necessary, reduce the result from Step 3 to lowest terms. This becomes the irrational part of the answer. Step 5: The result is the product of the rational and irrational parts. Dividing Irrational Numbers with Similar Irrational Parts Perform the following operations without a calculator. Simplify if possible. 3 15 ÷ ( 8 3 ) 14.7 135 ÷ ( 3 5 ) . In this division problem, 3 15 ÷ ( 8 3 ) , notice that the irrational parts of these numbers are similar. They are both square roots, so follow the steps given above. Step 1: Divide the rational parts. 3 ÷ 8 = 3 8 Step 2: If necessary, reduce the result of Step 1 to lowest terms. The 3 and 8 have no common factors, so 3 8 is already in lowest terms. Step 3: Divide the irrational parts. 15 ÷ 3 = 15 3 = 15 3 Step 4: If necessary, reduce the result from Step 3 to lowest terms. The radicand can be reduced, which yields 5 . Step 5: The result is the product of the rational and irrational parts, which is 3 8 5 . In this division problem, 14.7 135 ÷ ( 3 5 ) , notice that the irrational parts of these numbers are similar. They are both square roots, so follow the steps given above. Step 1: Divide the rational parts. 14.7 ÷ 3 = 4.9 Step 2: If necessary, reduce the result of Step 1 to lowest terms. This rational number is expressed as a decimal so will not be reduced. Step 3: Divide the irrational parts. 135 ÷ 5 = 135 5 = 135 5 Step 4: If necessary, reduce the result from Step 3 to lowest terms. The radicand can be reduced, which yields 135 5 = 27 = 9 × 3 = 3 3 . Step 5: The result is the product of the rational and irrational parts, which is 4.9 × 3 3 = 14.7 3 . Multiplying Irrational Numbers with Similar Irrational Parts Perform the following operations without a calculator. Simplify if possible. ( 19 3 ) × ( 5.6 12 ) 13 π × 8 π In this multiplication problem, ( 19 3 ) × ( 5.6 12 ) , notice that the irrational parts of these numbers are similar. They are both square roots. Follow the process above. Step 1: Multiply the rational parts. 19 × 5.6 = 106.4 Step 2: If necessary, reduce the result of Step 1 to lowest terms. This rational number is expressed as a decimal and will not be reduced. Step 3: Multiply the irrational parts. 3 × 12 = 3 × 12 = 36 Step 4: If necessary, reduce the result from Step 3 to lowest terms. The radicand is 36, which is the square of 6. The irrational part reduces to 36 = 6 . Step 5: The result is the product of the rational and irrational parts, which is 106.4 × 6 = 638.4 . Notice that sometimes multiplying or dividing irrational numbers can result in a rational number. In this multiplication problem, 13 π × 8 π , notice that the irrational parts of these numbers are the same, π . Follow the process above. Step 1: Multiply the rational parts. 13 × 8 = 104 Step 2: If necessary, reduce the result of Step 1 to lowest terms. That result is an integer. Step 3: Multiply the irrational parts. π × π = π 2 Step 4: If necessary, reduce the result from Step 3 to lowest terms. This cannot be reduced. Step 5: The result is the product of the rational and irrational parts, which is 104 π 2 . Rationalizing Fractions with Irrational Denominators Fractions often represent that some amount is being equally divided into some number of parts. But to conceptualize a fraction in that manner, the denominator needs to be an integer. An irrational number in the denominator interferes with that interpretation of a fraction. Fractions that have denominators that are just the square root of an integer can be altered into fractions with integer denominators using a process called rationalizing the denominator . The process relies on the following property of square roots: a × a = a and the following property of fractions: a b = a c b c for any non-zero number c . Using these two properties, when a fraction has a square root in the denominator, we can eliminate that square root. Multiply the numerator and denominator by that square root from the denominator, a b = a b b × b . Then apply a × a = a to the denominator, yielding a b b × b = a b b . Notice that there is no longer a square root in the denominator, which allows for interpreting the fraction as dividing a whole into equal parts. Rationalizing the Denominator Rationalizing the Denominator Rationalize the denominator of the following: 5 7 3 6 2 10 The square root in the denominator is 7 . In order to rationalize the denominator of 5 7 , we need to multiply the numerator and denominator by 7 and simplify. 5 7 = 5 7 7 × 7 = 5 7 7 The square root is in simplified form, so the final answer is 5 7 7 . The square root in the denominator is 10 . Step 1: In order to rationalize the denominator of 3 6 2 10 , we need to multiply the numerator and denominator by 10 and simplify. 3 6 2 10 = 3 6 × 10 2 10 × 10 = 3 60 2 × 10 = 3 60 20 Step 2: The 60 under the square root can be factored into the following factor pairs: 1 × 60 2 × 30 3 × 20 4 × 15 5 × 12 6 × 10 Step 3: The largest square factor of 60 is 4, so we simplify the 60 in the numerator into 2 15 . We also cancel any common factors. 3 60 20 = 3 × 2 15 20 = 6 15 20 = 3 15 10 This is completely simplified. There are occasions when the denominator is irrational but is the sum of two numbers where one or both involve square roots. For instance, 5 4 + 3 . The process used earlier required that the denominator was the square root of a number and would not work here. However, this type of denominator can be rationalized. In order to rationalize such a denominator, we will multiply the numerator and denominator of the fraction by the conjugate of the denominator. The conjugate of a + b is a – b . We say that a + b and a ‒ b are conjugate numbers . So, the conjugate of − 3 + 10 is just − 3 − 10 . But why is this of interest? The reason is because it leads to the difference of squares formula, which is used to factor the difference of two squares. Or, for our purposes, in reverse it allows us to eliminate a square root. For any two numbers, a and b , a 2 − b 2 = ( a − b ) ( a + b ) . Looking at that formula, you should see that the two factors on the right-hand side of the equals sign are conjugates of one another. So, for our purposes, we’re interested in ( a − b ) ( a + b ) = a 2 − b 2 . This tells us that when we multiply a + b by its conjugate, we get a squared minus b squared, or a 2 − b 2 . But how is this useful? Let’s return to the fraction above, 5 4 + 3 . The denominator is 4 + 3 . Its conjugate is 4 − 3 . According to the formula, and letting a = 4 and b = 3 , we see that ( 4 + 3 ) ( 4 − 3 ) = 4 2 − ( 3 ) 2 . But ( 3 ) 2 is just 3. That means the product is 16 − 3 or 13. This no longer has a square root. We use this to rationalize the denominator. We will also need the distributive property of numbers. For any three numbers a , b , and c , a × ( b ± c ) = a × b ± a × c . This is called the distributive property. Rationalizing the Denominator Using Conjugates Rationalize the denominator of 4 6 + 10 . Step 1: We recognize that the denominator is the sum of two numbers where one or both involve square roots. This means the conjugate can be used to remove the square root from the denominator. Step 2: To do so, we multiply the numerator and the denominator each by the conjugate of the denominator. Since the denominator is 6 + 10 , the conjugate we will use is 6 − 10 . Step 3: The conjugate is multiplied by the numerator and the denominator. 4 6 + 10 × 6 − 10 6 − 10 Step 4: Remembering how a number times its conjugate works, this becomes 4 6 + 10 × 6 − 10 6 − 10 = 4 × ( 6 − 10 ) 6 2 − ( 10 ) 2 . Step 5: In the numerator, we apply the distributive property. Using it yields 4 × ( 6 − 10 ) 6 2 − ( 10 ) 2 = 24 − 4 10 36 − 10 = 24 − 4 10 26 . Step 6: Notice that the denominator no longer contains a square root. It has been rationalized. If desired, this can then be written as a rational number minus an irrational number, by recalling that a − b c = a c − b c . Applying that to the answer, we have 4 6 + 10 = 24 − 4 10 26 = 24 26 − 4 10 26 . Step 7: With a bit of cancellation, this reduces to 4 6 + 10 = 24 26 − 4 10 26 = 12 13 − 2 10 13 . Rationalizing the Denominator Check Your Understanding Key Terms conjugate numbers difference of squares irrational numbers lowest terms rationalize the denominator Key Concepts Irrational numbers are numbers that cannot be written as an integer divided by another integer. One example is pi, denoted π . Another collection of irrational numbers are natural numbers that are not perfect squares. Some irrational numbers can be written as a rational part multiplied by an irrational part. If two irrational numbers have the same irrational parts, they can be added or subtracted. When irrational numbers are similar, on can multiply and divide the numbers without a calculator. Since a × b = a × b , and a ÷ b = a b = a b , products and quotients of square roots can be determined. Because a 2 = a and a × b = a × b , it is possible to simplify square root expressions so the radicand contains no perfect square factors. When a fraction has an irrational number as its denominator, it is possible to convert the denominator into a rational number using its conjugate. Doing so involves multiplying the numerator and denominator by the conjugate of the denominator, and then applying the difference of squares formula. With a single square root term Using conjugate numbers for two term denominators Formulas a × b = a × b a × x ± b × x = ( a ± b ) × x a ÷ b = a b = a b a 2 − b 2 = ( a − b ) ( a + b ) Videos The Philosophy of the Pythagoreans Using Desmos to Find the Square Root of a Number Simplifying Square Roots Rationalizing the Denominator", "section": "Irrational Numbers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Real Numbers Quick mental math involves using the known properties of real numbers. Learning Objectives After completing this section, you should be able to: Define and identify numbers that are real numbers. Identify subsets of the real numbers. Recognize properties of real numbers. Have you ever been impressed by the speed at which someone can do math in their head? Most of us at one time or another have witnessed a person speed through mental math, an impressive feat that often bests calculators. One such person is Neelkantha Bhanu Prakash. As of September 20, 2020, he is considered the world’s fastest human calculator. He currently holds four world records. How does someone do that, though? Have they memorized lots of arithmetic facts? Are they simply brilliant? The answer isn't simple so much as it is about knowledge. Real numbers behave in some very regular ways, following rules that can be learned. In this section, those rules are explored. Watch the video of Arthur Benjamin’s TED Talk to learn about another mathematician with remarkable mental abilities. Arthur Benjamin TED Talk, Faster Than a Calculator Defining and Identifying Real Numbers Real numbers are the rational and irrational numbers combined. The real numbers represent the collection of all physical distances that exist, along with 0 and the negatives of those physical distances. For example, if you take a measure of three units, and divide that distance into eight (8) equal lengths, the distance you have formed is 3 8 units. Also, if you draw a right triangle (a triangle with one angle equal to 90 degrees) with one side length of 1, and the other side length of 3, the long side of the triangle will have length 10 units, as shown in . Right triangle Of course, if we name something the real numbers, there must be numbers that aren't real. Otherwise, they’d just be called the numbers. One such not real number, one that cannot be a length, is − 1 . It is part of a collection of numbers called the complex numbers, it is denoted with the letter i . As an extension, the square root of any negative number is not a real number, but instead a complex number. To determine if a number is real, check to see if there are any negatives under a square root or any i ' s . If there are any present, the number is not real. Identifying Real Numbers Determine if each of the following are real numbers: 4 3 7 13.3381 17 − 8 4 3 7 is a real number, as there are no negatives under the square roots, nor is there any factor of i . 13.3381 is a rational number, and so it is a real number. 17 − 8 is not a real number, as there is a negative number under the square root. Identifying Subsets of Real Numbers The real numbers were built out of pieces, including integers, rational numbers, and irrational numbers. As such, the real numbers have named subsets, as shown in the table below. Set Name Set Symbol Set Description Natural Numbers ℕ The counting numbers Whole Numbers The counting numbers and 0 Integers ℤ The natural numbers, their negatives, and 0 Rational Numbers ℚ Fractions of integers Irrational Numbers ℙ Numbers that cannot be written as a fraction of integers Real Numbers ℝ The union of the rational and irrational numbers, all possible physical lengths, and their negatives When we categorize numbers using these sets, we use the smallest set that they belong to. For instance, −7 is an integer, and a rational number, and a real number. The smallest set to which −7 belongs is integer, so we’d say it belongs to the integers. We can also represent the relationships between the different sets of real numbers using set notation. All natural numbers are integers, but there are integers that are not natural numbers, so ℕ ⊂ ℤ . Similarly, every integer is a rational number, but there are rational numbers that are not integers, so ℤ ⊂ ℚ . The same is true of the rational numbers and the real numbers, so ℚ ⊂ ℝ . There is no agreed-upon symbol for the irrational numbers. If we represent the irrationals as the set A , we should note that the following are true: ℚ ∪ A = ℝ and ℚ ∩ A = ∅ . Recall that this means the irrationals are the complement of the rational numbers in the universal set of real numbers. Categorizing Numbers Identify all subsets of the real numbers to which the following real numbers belong: 14 − 14.223 17 14 is a natural number, integer, and rational number. − 14.223 is a rational number. 17 is an irrational number. Categorizing Numbers within a Venn Diagram Place the following numbers correctly in the Venn diagram ( ). − 4 2 − 10 37 150 41 1 20 4 π Since − 4 2 is irrational, it belongs in the real numbers, but outside the rational numbers ( ). Since −10 is an integer, it belongs in the integers but outside the natural numbers ( ). Since 37 150 is a rational number, it belongs in the rational numbers but not in the integers ( ). Since 41 is a natural number, it belongs in the natural numbers circle ( ). Since 1 20 is a rational number, it belongs in the rational numbers but not in the integers ( ). Since 4 π is irrational, it belongs in the real numbers, but outside the rational numbers ( ). Identifying Sets of Real Numbers Recognizing Properties of Real Numbers The real numbers behave in very regular ways. These behaviors are called the properties of the real numbers . Knowing these properties helps when evaluating formulas, working with equations, or performing algebra. Being familiar with these properties is helpful in all settings where numbers are used and manipulated. For example, when multiplying 4 × 13 × 25 , you could multiply the 4 and 25 first. If you know that product is 100, it makes the multiplication easier. The table below is a partial list of properties of real numbers. Property Example In Words Distributive property a × ( b + c ) = a × b + a × c 5 × ( 3 + 4 ) = 5 × 3 + 5 × 4 Multiplication distributes across addition Commutative property of addition a + b = b + a 3 + 7 = 7 + 3 Numbers can be added in any order Commutative property of multiplication a × b = b × a 10 × 4 = 4 × 10 Numbers can be multiplied in any order Associative property of addition a + ( b + c ) = ( a + b ) + c 4 + ( 3 + 8 ) = ( 4 + 3 ) + 8 Doesn't matter which pair of numbers is added first Associative property of multiplication a × ( b × c ) = ( a × b ) × c 2 × ( 5 × 7 ) = ( 2 × 5 ) × 7 Doesn't matter which pair of numbers is multiplied first Additive identity property a + 0 = a 17 + 0 = 17 Any number plus 0 is the number Multiplicative identity property a × 1 = a 21 × 1 = 21 Any number times one is the number Additive inverse property a + ( − a ) = 0 14 + ( − 14 ) = 0 Every number plus its negative is 0 Multiplicative inverse property a × ( 1 a ) = 1 , provided a ≠ 0 3 × ( 1 3 ) = 1 Every non-zero number times its reciprocal is 1 The names of the properties are suggestive. The commutative properties , for example, suggest commuting, or moving. Associative properties suggest which items are associated with others, or if order matters in the computation. The distributive property addresses how a number is distributed across parentheses. Identifying Properties of Real Numbers In each of the following, identify which property of the real numbers is being applied. 4 + ( 8 + 13 ) = ( 4 + 8 ) + 13 34 × ( 1 34 ) = 1 14 + 27 = 27 + 14 Here, the pair of numbers that is added first is switched. This is the associative property of addition. Here, a number is multiplied by its reciprocal, resulting in 1. This is the multiplicative inverse property. Here, the order in which numbers are added is switched. This is the commutative property of addition. Using these properties to perform arithmetic quickly relies on spotting easy numbers to work with. Look for numbers that add to a multiple of 10, or multiply to a multiple of 10 or 100. Using Properties of Real Numbers in Calculations Use properties of the real numbers and mental math to calculate the following: 2 × 13 × 50 13 + 84 + 27 9 × 16 × 11 Notice that 2 × 50 = 100 , so that becomes the multiplication to do first. Use the commutative property of multiplication to change the order of the numbers being multiplied. 2 × 13 × 50 = 2 × 50 × 13 = 100 × 13 = 1,300 Notice that 13 + 27 = 40 , so that becomes the addition to do first. Use the commutative property of addition to change the order in which the numbers are added. 13 + 84 + 27 = 13 + 27 + 84 = 40 + 84 = 124 Notice that 9 × 11 = 99 . Using that, the problem can be changed to 99 × 16 . That, however, doesn't look easy at all. But 99 = ( 100 − 1 ) . Using the distributive property, we rewrite and expand this as 99 × 16 = ( 100 − 1 ) × 16 = 100 × 16 − 1 × 16 = 1,600 − 16 . The last step is subtraction, so the final answer is 1,584. So, multiplying by 99 is the same as multiplying by 100, and then subtracting the other number once. Videos Properties of the Real Numbers 1 Properties of the Real Numbers 2 Properties of the Real Numbers 3 Check Your Understanding Key Terms complex number imaginary number real number Key Concepts Real numbers is the collection of all rational and irrational numbers. Conceptually, it is the collection of all values that can be represented on a number line, or, as a length along with sign. The subsets of the real numbers include the natural numbers, integers, rational numbers and irrational numbers. The natural numbers are a subset of the integers, which is a subset of the rational numbers. The rational and irrational numbers are disjoint sets. The real numbers, due to order of operation rules and that performing arithmetic operations on real number always results in a real number, have arithmetic properties that apply in all cases. There include the distributive property, the commutative property, and the associative property. Also, every real number has an additive inverse and, except for zero (0), have a multiplicative inverse. Formulas a × ( b + c ) = a × b + a × c a + b = b + a a × b = b × a a + ( b + c ) = ( a + b ) + c a × ( b × c ) = ( a × b ) × c a + 0 = a a × 1 = a a + ( − a ) = 0 a × ( 1 a ) = 1 Videos Arthur Benjamin TED talk, Faster than a Calculator Identifying Sets of Real Numbers Properties of the Real Numbers #1 Properties of the Real Numbers #2 Properties of the Real Numbers #3", "section": "Real Numbers", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Clock Arithmetic If a credit card number is entered incorrectly, error checking algorithms will often catch the mistake. (credit: modification of work “Senior couple at home checking finance on credit card from above” by Nenad Stojkovic/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Add, subtract, and multiply using clock arithmetic. Apply clock arithmetic to calculate real-world applications. Online shopping requires you to enter your credit card number, which is then sent electronically to the vendor. Using an ATM involves sliding your bank card into a reader, which then reads, sends, and verifies your card. Swiping or tapping for a purchase in a brick–and-mortar store is how your card sends its information to the machine, which is then communicated to the store’s computer and your credit card company. This information is read, recorded, and transferred many times. Each instance provides one more opportunity for error to creep into the process, a misrecorded digit, transposed digits, or missing digits. Fortunately, these card numbers have a built-in error checking system that relies on modular arithmetic, which is often referred to as clock arithmetic. In this section, we explore clock, or modular, arithmetic. Determining the Day of the Week for Any Date in History Adding, Subtracting, and Multiplying Using Clock Arithmetic When we do arithmetic, numbers can become larger and larger. But when we work with time, specifically with clocks, the numbers cycle back on themselves. It will never be 49 o’clock. Once 12 o’clock is reached, we go back to 1 and repeat the numbers. If it's 11 AM and someone says, “See you in four hours,” you know that 11 AM plus 4 hours is 3 PM, not 15 AM (ignoring military time for now). Math worked on the clock, where numbers restart after passing 12, is called clock arithmetic . Clock arithmetic hinges on the number 12. Each cycle of 12 hours returns to the original time ( ). Imagine going around the clock one full time. Twelve hours pass, but the time is the same. So, if it is 3:00, 14 hours later and two hours later both read the same on the clock, 5:00. Adding 14 hours and adding 2 hours are identical. As is adding 26 hours. And adding 38 hours. Clock showing 3:00 with arrow going around the clock one full time, or 12 hours What do 2, 14, 26, and 38 have in common in relation to 12? When they are divided by 12, they each have a remainder of 2. That's the key. When you add a number of hours to a specific time on the clock, first divide the number of hours being added by 12 and determine the remainder. Add that remainder to the time on the clock to know what time it will be. A good visualization is to wrap a number line around the clock, with the 0 at the starting time. Then each time 12 on the number line passes, the number line passes the starting spot on the clock. This is referred to as modulo 12 arithmetic. Even though the process says to divide the number being added by 12, first perform the addition; the result will be the same if you add the numbers first, and then divide by 12 and determine the remainder. In general terms, let n be a positive integer. Then n modulo 12, written ( n mod 12), is the remainder when n is divided by 12. If that remainder is x , we would write n = x (mod 12). Caution: 12 mod 12 is 0. So, if a mod 12 problem ends at 0, that would be 12 on the clock. Determining the Value of a Number modulo 12 Find the value of the following numbers modulo 12: 34 539 156 To determine the value of a number modulo 12, divide the number by 12 and record the remainder. To find the value 34 modulo 12: Step 1: Determine the remainder when 34 is divided by 12 using long division. The largest multiple of 12 that is less than or equal to 34 is 24, which is the product of 12 and 2. 12 34 2 24 _ Step 2: Performing the subtraction yields 10. 12 34 2 24 _ 10 Since that subtraction resulted in a number less than 12, that is the remainder, 10. The value of 34 modulo 12 is 10, or 34 = 10 (mod 12). To find the value 539 modulo 12: Step 1: Determine the remainder when 539 is divided by 12 using long division. We first look to the first two digits of 539, 53. The largest multiple of 12 that is less than or equal to 53 is 48, which is the product of 12 and 4. 12 539 4 48 _ Step 2: Performing the subtraction results in 5. 12 539 4 48 _ 5 Step 3: Now, the 9 is brought down. 12 539 4 48 _ 59 Step 4: The largest multiple of 12 that is less than or equal to 59 is once more 48 itself, which is 12 × 4 . 12 539 44 48 _ 59 48 _ Step 5: Finishing the process, the 48 is subtracted from the 59, yielding 11. 12 539 44 48 _ 59 48 _ 11 We've used all the digits of 539, and the last subtraction resulted in a number less than 12, so that number, 11, is the remainder. The value of 539 modulo 12 is 11, or, 539 = 11 (mod 12). To find the value 156 modulo 12: Step 1: Determine the remainder when 156 is divided by 12 using long division. We first look to the first two digits of 156, 15. The largest multiple of 12 that is less than or equal to 15 is 12 itself, which is the product of 12 and 1. 12 156 1 12 _ Step 2: Performing the subtraction results in 3. 12 156 1 12 _ 3 Step 3: Now, the 6 is brought down. 12 156 1 12 _ 36 Step 4: The largest multiple of 12 that is less than or equal to 36 is 36 itself, which is 12 × 3 . 12 156 13 12 _ 36 36 _ Step 5: Finishing the process, the 36 is subtracted from the 36, yielding 0. 12 156 13 12 _ 36 36 _ 0 We've used all the digits of 156, and the last subtraction resulted in a number less than 12, so that number, 0, is the remainder. The value of 156 modulo 12 is 0, or, 156 = 0 (mod 12). We should note here that, had we been speaking of time, the 0 would be interpreted as 12:00. Using Desmos to Determine the Value of a Number module 12 Desmos may be used to determine the value of a number modulo 12. It is flexible enough to find the value of a number modulo of any other integer you want. To determine the value of n modulo 12, type mod( n ,12) into Desmos. The result will be displayed immediately. This can be used to find 539 modulo 12, as shown in the . Display of 539 modulo 12 Clock arithmetic is modulo 12 arithmetic but applied to time. As time is divided into 12 hours that repeat a cycle, we use modulo 12 for clock arithmetic. Clock Arithmetic Adding with Clock Arithmetic If it's 3:00, what time will it be in 89 hours? To find that future time, we may determine the value of 89 (mod 12), either by long division or by using a calculator, such as Desmos. Then add the result to 3:00. Entering mod(89,12) in Desmos results in 5. Adding 5 hours, which was 89 (mod12), to 3:00 results in 8:00. Subtracting time on the clock works in much the same way as addition. Find the value of the number of hours being subtracted modulo 12, then subtract that from the original time. Subtracting with Clock Arithmetic If it is 4:00 now, what time was it 67 hours ago? To find that past time, we may determine the value of 67 (mod 12), either by long division or by using a calculator, such as Desmos. Then subtract the result to 4:00. Entering mod(67,12) in Desmos results in 7. Subtracting 7 hours from 4:00 results in ‒3:00. We know, though, that time is not represented with negative times. This value, ‒3:00, indicates three hours before 12:00, which is 9:00. So, 67 hours before 4:00 was 9:00. We see this in the . Clock showing 7 hours subtracted from 4:00 Recall that clock arithmetic was referred to as modulo 12 arithmetic. Multiplying in modulo 12 also relies on the remainder when dividing by 12. To multiply modulo 12 is just to multiply the two numbers, and then determine the remainder when divided by 12. Multiplying modulo 12 What is the product of 11 and 45 modulo 12? We begin by multiplying 11 and 45, which is 495. Next, we find 495 modulo 12, either by dividing the result by 12 to determine the remainder, or by using a calculator. Entering mod(495,12) in Desmos yields 3. Had long division been used, the remainder would be 3. So 11 × 45 = 3 modulo 12. Calculating Real-World Applications with Clock Arithmetic Applying Clock Arithmetic Suppose it is 3:00, and you decide to check your email every 5 hours. What time will it be when you check your email the ninth time? If you check your email every 5 hours nine times, that ninth check will occur 45 hours after 3:00, which is an addition of 45 hours to 3:00. So, we find 45 modulo 12, which is 9. Nine hours after 3:00 is 12:00. It will be 12:00 when you check your email the ninth time. Clock arithmetic processes can be applied to days of the week. Every 7 days the day of the week repeats, much like every 12 hours the time on the clock repeats. The only difference will be that we work with remainders after dividing by 7. In technical terms, this is referred to as modulo 7 . More generally, let n be a positive integer. Then n modulo 7, written n mod 7, is the remainder when n is divided by 7. If that value is x , we may write n = x (mod 7). Applying Clock Arithmetic to Days of the Week Your family has a cat, and no one wants to empty the litter box. However, it has to be done daily. The six of you agree to take turns, so everyone has to empty the litter box every 6 days. You empty the box on a Thursday. What day will you empty the box for the 10th time? The first time you emptied the litter box was on a Thursday. So,the 10th time you empty the litter box will be 9 times later (you've already had your first turn, so 9 turns left!). This will happen 54 (9 times 6) days later. Finding the value of 54 modulo 7, using division to determine the remainder or using a calculator to find the value of 54 modulo 7 gives the answer 5. Five days after a Thursday is Tuesday. Check Your Understanding Key Terms clock arithmetic modulo 7 modulo 12 Key Concepts Clock arithmetic uses the idea that after 12 o’clock comes 1 o’clock. For clock arithmetic, this means that every time 12 is passed in an arithmetic process, the next number is 1, not 13. To determine the clock result of an arithmetic operation, divide the final result by 12 and keep the remainder. If the remainder is 0, then the time is 12 o’clock. Clock arithmetic is technically called modulo 12 arithmetic. To perform modulo 12 arithmetic, calculate the expression, then divide the result by 12. The modulo 12 result is the remainder. Days, in our system, pass in groups of seven. To calculate in day arithmetic, modulo 7 is used. To perform modulo 7 arithmetic, calculate the expression, then divide the result by 7. The modulo 7 result is the remainder. Videos Determining the Day of the Week for Any Date in History Clock Arithmetic", "section": "Clock Arithmetic", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Exponents Astronomical distances are written using exponents. (credit: “Our Solar System (Artist's Concept)” by NASA/Jet Propulsion Laboratory-Caltech/Public Domain) Learning Objectives After completing this section, you should be able to: Apply the rules of exponents to simplifying expressions. Sometimes, we look for shorthand when writing or expressing something that simply takes too long. The use of LOL and tl;dr. This shorthand only works if everyone reading the shorthand knows what it stands for. Using exponents is a similar instance. Writing out a long string of a number times itself over and over takes too much time, and eventually one would forget how many of the value has been written or read. For example, 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 . There has to be a shorter and more efficient way to write 8 times itself 1, 2, 3….hmmmm, 19 times. And that’s the role that exponents play in mathematics. They are shorthand for multiplying a number by itself a number of times. Without it, calculations would become a mess and we’d have to write a lot more. Applying the Rules of Exponents to Simplify Expressions Squaring a number is multiplying it by itself, and has that name because it is the area of a square with that side length. Cubing a number is finding the volume of a cube with that length of sides. That’s why we refer to 5 2 as five squared, or 10 3 as ten cubed. Exponents represent that multiplication. Let’s remind ourselves of the terminology associated with exponents and what exponents represent. Suppose you want to multiply a number, let’s label that number a , by itself some number of times. Let’s label the number of times b . We denote that as a b . We say a raised to the b th power. When we write or see 7 5 , we call the 7 the base and we call 5 the exponent . What it represents is 7 multiplied by itself 5 times. This means exponents are used as a shorthand for repeated multiplications, where we write 7 5 = 7 × 7 × 7 × 7 × 7 . We would write 7 5 and say seven to the fifth power. Exponential Notation The definitions of base and exponent make it possible to understand the exponent rules. Product Rule for Exponents The first rule we examine is the product rule, a n a m = a n + m . This rule means that when we multiply a base raised to a power times the same base to another power, the result is the base raised to the sum of the powers. To demonstrate, consider 9 3 × 9 5 . If we apply the product rule to that we get 9 3 × 9 5 = 9 3 + 5 = 9 8 . This can be tested by looking at the multiplications that are represented. The 9 3 is 9 times itself 3 times, while 9 5 is 9 times itself 5 times. Substituting those into 9 3 × 9 5 we see 9 3 × 9 5 = ( 9 × 9 × 9 ) × ( 9 × 9 × 9 × 9 × 9 ) = 9 8 , which is what the formula told us would happen. Caution: The product rule only applies when the bases are the same. If the bases are different, we do not apply this rule. If a number, a , raised to a power, n , is then multiplied by a raised to another power, m , the result is a n a m = a n + m . Using the Product Rule for Exponents If possible, use the product rule to simplify the following: 21 9 × 21 15 5 9 × 8 4 We can apply the product rule to simplify the expression because the bases are the same and we are multiplying. 21 9 × 21 15 = 21 ( 9 + 15 ) = 21 24 Since the bases are not the same (one is 5, the other 8), this cannot be simplified using the product rule for exponents. These rules can be applied to unknowns too. Using the Product Rule for Exponents of Unknowns Use the product rule to simplify a 4 × a 10 . The bases are the same, and we are multiplying, so we apply the multiplication rule to simplify the expression. a 4 × a 10 = a ( 4 + 10 ) = a 14 Quotient Rule for Exponents The next rule we examine is the quotient, or division, rule. When a number, a , raised to a power, n , is divided by a raised to another power, m , then the result is a n a m = a ( n − m ) . This rule means that when we divide a base raised to a power by the same base to another power, the result is the base raised to the difference of the powers. To demonstrate, consider 14 13 14 6 . If we apply the quotient rule to that, we get 14 13 14 6 = 14 13 − 6 = 14 7 . This can be tested by looking at the division that is represented. Remember, 14 13 is 14 multiplied to itself 13 times, while 14 6 is 14 multiplied to itself 6 times. Substituting those into 14 13 14 6 gives the following: 4 13 4 6 = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 We see here that there are a LOT of fours to be divided out. = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 = 4 × 4 × 4 × 4 × 4 × 4 × 4 1 = 4 × 4 × 4 × 4 × 4 × 4 × 4 What remains is 4 to the 7th power, 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4 7 . All of the work above confirmed what the formula told us would be the result. Caution: The quotient rule only applies when the bases are the same. If the bases are different, we do not apply this rule. Using the Quotient Rule for Exponents Use the quotient rule to simplify 5 19 5 11 . We can apply the quotient rule to simplify the expression since the bases are the same and we are dividing. 5 19 5 11 = 5 ( 19 − 11 ) = 5 8 Product and Quotient Rule for Exponents A natural consequence of the quotient rule is what it means to raise a non-zero number to the zeroth power. Let’s look at the simplification when the exponents are equal. 3 6 3 6 = 3 ( 6 − 6 ) = 3 0 We know that a number divided by itself is 1, so 3 6 3 6 = 1 . From that is must be that 3 6 3 6 = 3 0 = 1 . This provides the rule for a number raised to the power 0: a ≠ 0 . If you have a non-zero number a , then a 0 = 1 . Distributive Rule for Exponents The next rule we look to is a distributive rule for exponents. If you have a product, ( a × b ) , and raise it to an exponent, n , then ( a × b ) n = a n × b n . This means that when we have two numbers multiplied together, and that is raised to a power, it is the same as raising each of the numbers to the same power first, then multiplying. For example, ( 3 × 7 ) 4 = 3 4 × 7 4 . This can be explained using the definition of exponents and multiplying all the factors. ( 3 × 7 ) 4 = ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 ) We may change the order in which numbers are multiplied. This is the commutative property of the real numbers. This can be written as 3 × 3 × 3 × 3 × 7 × 7 × 7 × 7 . Using exponents, that shortens to 3 4 × 7 4 . This also works in the other direction, a n × b n = ( a × b ) n . Read this way, if we have one base raised to an exponent, and another base raised to the same exponent, we can multiply the bases and raise that product to the shared exponent. For instance, 7 8 × 11 8 = ( 7 × 11 ) 8 = 77 8 . Caution: The exponent distributive rule , a n × b n = ( a × b ) n , only works if the exponents are the same . Using the Distributive Rule for Exponents Use the exponent distributive rule to expand ( 6 × 13 ) 7 . Applying the distributive rule to the product, we get ( 6 × 13 ) 7 = 6 7 × 13 7 . Using the Distributive Rule for Exponents Use the exponent distributive rule to expand ( c × d ) 10 . Applying the distributive rule to the product, we get ( c × d ) 10 = c 10 × d 10 . This distribution also works for quotients. A fraction raised to an exponent equals the numerator raised to the exponent divided by the denominator raised to the exponent. For example, ( 3 5 ) 7 = 3 7 5 7 . Demonstrating this is similar to the previous rule. When you have a fraction, a b , raised to an exponent, n , then ( a b ) n = a n b n . Using the Distributive Rule for Exponents with Fractions Use the exponent distributive rule to expand the following: ( 4 9 ) 6 ( 3 b ) 11 Applying the distributive rule to the quotient, we get ( 4 9 ) 6 = 4 6 9 6 . Applying the distributive rule to the quotient, we get ( 3 b ) 11 = 3 11 b 11 . Fraction Raised to a Power Power Rule In the previous two sets of rules, we’ve seen exponents applied to products and quotients. Now we look to exponents applied to other exponents. For example, ( 3 6 ) 4 = 3 ( 6 × 4 ) = 3 24 . This can be explained by examining what the outer exponent does. We raise 3 6 to the fourth power, so we multiply 3 6 by itself 4 times, ( 3 6 ) 4 = 3 6 × 3 6 × 3 6 × 3 6 . Now if we apply the product rule for exponents, this becomes 3 ( 6 + 6 + 6 + 6 ) = 3 24 . If you raise a non-zero base, say a , to an exponent n , and raise that to another exponent, m , you get the base raised to the product of the exponents, which is ( a n ) m = a ( n × m ) . Raising an Exponent to an Exponent Expand the following: ( 6 7 ) 3 ( b 12 ) 4 Using the power rule of exponents, ( 6 7 ) 3 = 6 ( 7 × 3 ) = 6 21 . Using the power rule of exponents, ( b 12 ) 4 = b ( 12 × 4 ) = b 48 . Negative Exponent Rule Up until now, we’ve only looked at positive exponents. The last exponent rule we look at is what negative exponents represent. Recall the quotient rule: a n a m = a ( n + m ) . What would happen if the exponent in the denominator was larger than that in the numerator? For example, 4 5 4 7 . If we apply the quotient rule, we obtain 4 5 4 7 = 4 5 − 7 = 4 − 2 . We need to make sense of that negative exponent. To do so, we can expand the quotient and see what happens: 4 5 4 7 = 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 × 4 . When we divide out common factors, only two factors of 4 are left in the denominator, as we see here: 1 4 × 4 . Using exponent notation, this is 1 4 2 . Since 4 − 2 and 1 4 2 represent the same number, 4 5 4 7 , they are equal. This demonstrates how negative exponents are defined. a − n = 1 a n provided that a ≠ 0 . Similarly, 1 a − n = a n . Eliminating Negative Exponents Convert the following to expressions with no negative exponent: 3 4 × 5 − 8 a − 9 × b 5 7 c − 2 Using the negative exponent rule on the 5 − 8 and multiplying, 3 4 × 5 − 8 = 3 4 × 1 5 8 = 3 4 5 8 . Using the negative exponent rule on the a − 9 and multiplying, a − 9 × b 5 = 1 a 9 × b 5 = b 5 a 9 . Begin by rewriting the expression as 7 c − 2 = 7 1 × 1 c − 2 . Apply the negative exponent rule to 1 c − 2 in the expression, which becomes 7 1 × 1 c − 2 = 7 × c 2 , which has no negative exponents. Eliminating Denominators by Using Negative Exponents Use negative exponents to rewrite the following expressions with no denominator: 7 3 13 9 c 4 d 8 Rewrite the expression 7 3 13 9 as 7 3 1 × 1 13 9 . Then use the definition of negative exponents to rewrite the 1 13 9 as 13 − 9 . Last, multiply, yielding 7 3 1 × 1 13 9 = 7 3 × 13 − 9 . Rewrite the expression c 4 d 8 as c 4 1 × 1 d 8 . Then use the definition of negative exponents to rewrite the 1 d 8 as d − 8 . Last, multiply, yielding c 4 1 × 1 d 8 = c 4 × d −8 . The table below shows a summary of the exponent rules from this section. Rule Example In Words Product Rule a n a m = a n + m 8 2 × 8 5 = 8 7 A base raised to a power, times the same based raised to another power, is the base raised to the sum of the powers. Quotient Rule a n a m = a ( n − m ) 11 14 11 12 = 11 12 A base raised to a power, divided by the same based raised to another power, is the base raised to the difference of the powers. Zero Power Rule a 0 = 1 provided that a ≠ 1 412 0 = 1 Any non-zero number raised to the zeroth power equals 1. Distributive Rule, Multiplication ( a × b ) n = a n × b n ( 14 × 31 ) 9 = 14 9 × 31 9 Exponents distribute across multiplication. Distributive Rule, Division ( a b ) n = a n b n ( 62 91 ) 8 = 62 8 91 8 Exponents distribute across division. Power Rule ( a n ) m = a ( n × m ) ( 5 9 ) 15 = 5 135 A base raised to a power, raised to another power, is the base raised to the first power times the second power. Negative Exponent Rule a − n = 1 a n provided that a ≠ 0 6 − 8 = 1 6 8 1 12 7 = 12 − 7 A base raised to a negative exponent is 1 divided by the base raised to the positive power, and vice versa. These rules often occur in tandem with each other, but it requires that you carefully apply the rules. Simplifying Expressions Using Exponent Rules Simplify the following: ( 4 2 × 7 9 3 ) 5 ( 5 a 4 b 9 ) 6 Step 1: To simplify this, we start by distributing the power 5 across the quotient: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) 5 ( 9 3 ) 5 Step 2: We distribute the power 5 in the numerator across that multiplication: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) ( 9 3 ) 5 5 = ( 4 2 ) 5 × 7 5 ( 9 3 ) 5 Step 3: We apply the power rule where indicated: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) ( 9 3 ) 5 5 = ( 4 2 ) 5 × 7 5 ( 9 3 ) 5 = 4 ( 2 × 5 ) × 7 5 9 ( 3 × 5 ) = 4 10 × 7 5 9 15 Step 1: To simplify this, we start by distributing the power 6 across the quotient: ( 5 a 4 b 9 ) 9 = ( 5 × a 4 ) 6 ( b 9 ) 6 Step 2: We distribute the power 5 in the numerator across that multiplication: ( 5 × a 4 ) 6 ( b 9 ) 6 = ( 5 ) 6 × ( a 4 ) 6 ( b 9 ) 6 Step 3: We apply the power rule where indicated: ( 5 ) 6 × ( a 4 ) 6 ( b 9 ) 6 = 5 6 a 24 b 54 Simplifying Expressions with Exponents Check Your Understanding Key Terms base exponent Key Concepts Exponents are used to express multiplying a number by itself a number of times. The number being multiplied by itself is the base. The number of times it is multiplied by itself is the exponent, which is often referred to as the power. Understanding that exponents represent repeated multiplication of a base makes it possible to establish some rules for combining exponential expressions, using the product rule, the quotient rule, and the power rule. Additionally, it allows us to formulate distributive rules for exponents. Any non-zero number raised to the 0th power is 1. This makes the definition of the 0th power consistent with the division rule for exponents. For consistency, negative exponents represent the reciprocal of the base raised to the power, so that a − n = 1 a n , provided that a ≠ 0 . Formulas a n a m = a n + m a n a m = a ( n − m ) a 0 = 1 , provided that a ≠ 0 ( a × b ) n = a n × b n ( a b ) n = a n b n ( a n ) m = a ( n × m ) a − n = 1 a n , provided that a ≠ 0 Videos Exponential Notation Product and Quotient Rule for Exponents Fraction Raised to a Power Simplifying Expressions with Exponents", "section": "Exponents", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Scientific Notation Calculations in the sciences often involve numbers in scientific notation form. Learning Objectives After completing this section, you should be able to: Write numbers in standard or scientific notation. Convert numbers between standard and scientific notation. Add and subtract numbers in scientific notation. Multiply and divide numbers in scientific notation. Use scientific notation in computing real-world applications. The amount of information available on the Internet is simply incomprehensible. One estimate for the amount of data that will be on the Internet by 2025 is 175 Zettabytes. A single zettabyte is one billion trillion. Written out, it is 1,000,000,000,000,000,000,000. One estimate is that we’re producing 2.5 quintillion bytes of data per day. A quintillion is a trillion trillion, or, written out, 1,000,000,000,000,000,000. To determine how many days it takes to increase the amount of information that is on the Internet by 1 zettabyte, divide these two numbers, a zettabyte being 1,000,000,000,000,000,000,000, and 2.5 quintillion, being 2,500,000,000,000,000,000, shows it takes 400 days to generate 1 zettabyte of information. But doing that calculation is awkward with a calculator. Keeping track of the zeros can be tedious, and a mistake can easily be made. On the other end of the scale, a human red blood cell has a diameter of 7.8 micrometers. One micrometer is one millionth of a meter. Written out, 7.8 micrometers is 0.0000078 meters. Smaller still is the diameter of a virus, which is about 100 nanometers in diameter, where a nanometer is a billionth of a meter. Written out, 100 nanometers is 0.0000001 meters. To compare that to engineered items, a single transistor in a computer chip can be 14 nanometers in size (0.000000014 meters). Smaller yet is the diameter of an atom, at between 0.1 and 0.5 nanometers. Sometimes we have numbers that are incredibly big, and so have an incredibly large number of digits, or sometimes numbers are incredibly small, where they have a large number of digits after the decimal. But using those representations of the names of the sizes makes comparing and computing with these numbers problematic. That’s where scientific notation comes in. Writing Numbers in Standard or Scientific Notation Form When we say that a number is in scientific notation , we are specifying the form in which that number is written. That form begins with an integer with an absolute value between 1 and 9, then perhaps followed the decimal point and then some more digits. This is then multiplied by 10 raised to some power. When the number only has one non-zero digit, the scientific notation form is the digit multiplied by 10 raised to an exponent. When the number has more than one non-zero digit, the scientific notation form is a single digit, followed by a decimal, which is then followed by the remaining digits, which is then multiplied by 10 to a power. The following numbers are written in scientific notation: 1.45 × 10 3 − 8.345 × 10 − 4 3 × 10 2 3.14159 × 10 0 The following numbers are not written in scientific notation: 1.45 because it isn't multiplied by 10 raised to a power − 50.053 × 10 7 because the absolute value of −50.053 is not at least 1 and less than 10 41.7 × 10 9 because 41.7 is not at least 1 and less than 10 0.036 × 10 − 3 because 0.036 is not at least one and less than 10 Identifying Numbers in Scientific Notation Which of the following numbers are in scientific notation? If the number is not in scientific notation, explain why it is not. − 9.67 × 10 20 145 × 10 − 8 1.45 The number − 9.67 × 10 20 is in scientific notation because the absolute value of −9.67 is at least 1 and less than 10. The number 145 × 10 − 8 is not in scientific notation because 145 is not at least 1 and less than 10. The number 1.45 is not in scientific notation form. Even though it is at least 1 but less than 10, it is not multiplied by 10 raised to a power. Some numbers are so large or so small that it is impractical to write them out fully. Avogadro’s number is important in chemistry. It represents the number of units in 1 mole of any substance. The substance many be electrons, atoms, molecules, or something else. Written out, the number is: 602,214,076,000,000,000,000,000. Another example of a number that is impractical to write out fully is the length of a light wave. The wavelength of the color blue is about 0.000000450 to 0.000000495 meters. Such numbers are awkward to work with, and so scientific notation is often used. We need to discuss how to convert numbers into scientific notation, and also out of scientific notation. Recall that multiplying a number by 10 adds a 0 to the end of the number or moves the decimal one place to the right, as in 43 × 10 = 430 or 3.89 × 10 = 38.9 . And if you multiply by 100, it adds two zeros to the end of the number or moves the decimal two places to the right, and so on. For example, 38 × 100,000 = 3,800,000 and 32.998 × 10,000 = 329,980 . Multiplying a number by 1 followed by some number of zeros just adds that many zeros to the end of the number or moves the decimal place that many places to the right. Numbers written as 1 followed by some zeros are just powers of 10, as in 10 1 = 10 , 10 2 = 100 , 10 3 = 1,000 , etc. Generally, 10 n = 1 0...0 ︸ n . We can use this to write very large numbers. For instance, Avogadro’s number is 602,214,076,000,000,000,000,000, which can be written as 6.02214076 × 10 23 . The multiplication moves the decimal 23 places to the right. Similarly, when we divide by 10, we move the decimal one place to the left, as in 46.7 10 = 4.67 . If we divide by 100, we move the decimal two places to the left, as in 3.456 100 = 0.03456 . In general, when you divide a number by a 1 followed by n zeros, you move the decimal n places to the left, as in 8,244.902 1,000,000 = 0.008244902 . This denominator could be written as 10 6 . If we use that in the expression and allow for negative exponents, rewrite the number as 8,244.902 1,000,000 = 8,244.902 10 6 = 8,244.902 × 10 − 6 . With this, we can write division by a 1 followed by n zeros as multiplication by 10 raised to ‒ n . Using that information, we can demonstrate how to convert from a number in standard form into scientific notation form. Case 1: The number is a single-digit integer. In this case, the scientific notation form of the number is d i g i t × 10 1 . Case 2: The absolute value of the number is less than 1. Follow the process below. Step 1: Count the number of zeros between the decimal and the first non-zero digit. Label this n . Step 2: Starting with the first non-zero digit of the number, write the digits. If the number was negative, include the negative sign. Step 3: If there is more than one digit, place the decimal after the first digit from Step 2. Step 4: Multiply the number from Step 3 by 10 n + 1 . Case 3: The absolute value of the number is 10 or larger. Follow the process below. Step 1: Count the number of digits that are to the left of the decimal point. Label this n . Step 2: Write the digits of the number without the decimal place, if one was present. If the number was negative, include the negative sign. Step 3: If there is more than one digit, place the decimal point after the first digit. Step 4: Multiply the number from Step 3 by 10 n − 1 . Writing a Number in Scientific Notation Write the following numbers in scientific notation form: 428.9 −0.00000981 8 Since the absolute value of 428.9 is 10 or larger, so we use the process from Case 3, above. Step 1: There are three digits to the left of the decimal point, so n = 3 . Step 2: Write the digits of the number without the decimal place, which is 4289. Step 3: Since there is more than one digit, place the decimal point after the first digit. We now have 4.289. Step 4: Since n = 3 , we multiply 4.289 by 10 raised to the second power, 4.289 × 10 2 . The scientific notation form of 428.9 is 4.289 × 10 2 . Since the absolute value of −0.00000981 is less than 1, we use the process from Case 2. Step 1: The number of zeros between the decimal and the first non-zero digit is 5, so n = 5 . Step 2: We write the non-zero digits, including the negative sign, yielding −981. Step 3: The decimal gets placed to the right of the first digit, resulting in −9.81. Step 4: Since n = 5 , we multiply −9.81 by 10 raised to the fourth power, − 9.81 × 10 − 6 . The scientific notation form of −0.00000981 is 9.81 × 10 − 6 . Since 8 is a single-digit integer, apply Case 1. The scientific notation form of 8 is 8 × 10 1 . When we write numbers in scientific notation form, we can manipulate the representation of the number by moving the decimal around, and making an appropriate change to the exponent of the 10. For instance, let’s look at 145.8141 × 10 8 . If we wanted to move the decimal one place to the left, we’d have to increase the power of 10, as shown here: 145.8141 × 10 8 = 14.58141 × 10 9 . Since we moved the decimal one to the left, we balance that with moving the exponent up by one. Similarly, if we move the decimal one place to the right, we have to balance that by moving the exponent one to the left, or subtracting one from the exponent, as shown here: 145.8141 × 10 8 = 1458.141 × 10 7 . Generally, for a number in the form n u m b e r × 10 n : If you move the decimal to the left by k digits, you increase the exponent by k . If you move the decimal to the right by k digits, you decrease the exponent by k digits. Increasing the Exponent Change 456.142 × 10 5 by moving the decimal two places to the left. Since we are moving the decimal to the left by two places, we increase the exponent of 10 by 2, so that the exponent is now 7. This gives us 456.142 × 10 5 = 4.56142 × 10 7 . Decreasing the Exponent Change 12.3 × 10 2 by moving the decimal five places to the right. Since we are moving the decimal to the right by five places, we decrease the exponent of 10 by 5, so that exponent is now −3. This give us 12.3 × 10 2 = 1230000.0 × 10 − 3 . Converting Numbers from Scientific Notation to Standard Form In the previous section, converting a number from standard form to scientific notation was explored. Now, we explore converting from scientific notation back into standard form. Doing so involves moving the decimal according to the power of the 10. The decimal is moved a number of steps equal to the exponent of the 10. As demonstrated previously, when the exponent of the 10 is negative, the decimal is moved to the left and when the exponent of the 10 is positive, the decimal is moved to the right. Converting from Scientific Notation to Standard Form Convert the following into standard form: 2.78 × 10 9 9.04 × 10 − 8 Since the exponent is positive, the decimal moves nine places to the right, so 2.78 × 10 9 is 2,780,000,000 . Since the exponent is negative, the decimal moves eight places to the left, so 9.04 × 10 − 8 is 0.0000000904 . Converting from Standard Form to Scientific Notation Form Converting from Scientific Notation Form to Standard Form Scientific Notation on a Calculator Most scientific and graphing calculators come with the ability to directly convert from standard form to scientific notation. On the TI-83, it is accessed through the MODE menus. For a commonly used, free phone scientific calculator, the calculator can be forced to work in scientific notation mode through its settings. Some calculators, such as the Desmos online calculator, display scientific notation as a number times 10 to a power as you’ve seen in this section. However, some calculators indicate scientific notation by replacing the × 10 n with an E (or EE) followed by the exponent. For example, shows what you may see on a TI-84. Calculator screens Adding and Subtracting Numbers in Scientific Notation To add or subtract numbers in scientific notation, the numbers first need to have the same exponent for the 10s. It is possible to add the following since the powers of 10 match: 4.5 × 10 4 + 3.15 × 10 4 = 7.65 × 10 4 Notice that the number parts were added, but the exponent part remained the same. This is due to the distributive property of the real numbers. The 10 4 is factored from the two terms, as shown: 4.5 × 10 4 + 3.15 × 10 4 = ( 4.5 + 3.15 ) × 10 4 = 7.65 × 10 4 Numbers in scientific notation can be added or subtracted directly using a calculator. Simply enter the values in scientific form and set your calculator to display scientific notation. Adding and Subtracting Numbers in Scientific Notation with the Same Powers of 10 Calculate the following: 3.8 × 10 − 3 + 1.006 × 10 − 3 9.61 × 10 8 − 3.85 × 10 8 Since the powers of 10 match, we use the distributive property of real numbers to factor 10 −3 from the numbers. We then add the number parts separately to get 4.806. 3.8 × 10 − 3 + 1.006 × 10 − 3 = ( 3.8 + 1.006 ) × 10 − 3 = 4.806 × 10 − 3 Since the powers of 10 match, we use the distributive property of real numbers to factor 10 8 from the numbers. We then subtract the number parts separately to get 5.76. 9.61 × 10 8 − 3.85 × 10 8 = ( 9.61 − 3.85 ) × 10 8 = 5.76 × 10 8 Adding and subtracting in scientific notation is straightforward when the exponents are the same. There are two issues that can arise. The first issue is what to do if after adding or subtracting the result is not in scientific notation. Correcting an Answer to Scientific Notation After Adding or Subtracting Calculate the following: 7.03 × 10 13 + 8.5 × 10 13 4.3 × 10 21 − 4.613 × 10 21 Since the powers of 10 match, we add the number parts and multiply that by 10 13 : 7.03 × 10 13 + 8.5 × 10 13 = ( 7.03 + 8.5 ) × 10 13 = 15.53 × 10 13 . However, 15.53 × 10 13 is not in scientific notation because the absolute value of 15.53 is more than 10. To put this number in scientific notation, the decimal needs to move one to the left. To balance that move, the power of 10 must be increased by 1. So, the answer in scientific notation is 1.553 × 10 14 . Since the powers of 10 match, we add the number parts: 4.3 × 10 21 − 4.613 × 10 21 = ( 4.3 − 4.613 ) × 10 21 = − 0.313 × 10 21 However, − 0.313 × 10 21 is not in scientific notation because it is less than 1. To put it in scientific notation, the decimal needs to move one to the right. To balance that move, the power of 10 must be decreased by 1. So, the answer in scientific notation is − 3.13 × 10 20 . The second issue that might be encountered when adding or subtracting is that the powers of 10 do not match. In that case, one of the numbers must be changed so that the powers of 10 match. It is easiest to make the smaller power of 10 larger to match the other power of 10. For example, to perform the following, 4.5 × 10 5 + 3.9 × 10 3 , we’d change the 3.9 × 10 3 so that the power of 10 is 5. To do so, we need to increase the power of 10 and move the decimal in the number part two places to the left. That would alter 3.9 × 10 3 into 0.039 × 10 5 . We would use 0.039 × 10 5 in the addition problem, so that the exponents match, allowing the addition to occur. 4.5 × 10 5 + 3.9 × 10 3 = 4.5 × 10 5 + 0.039 × 10 5 = ( 4.5 + 0.039 ) × 10 5 = 4.539 × 10 5 The steps to take when the exponents of the 10s are not equal are: Step 1: Increase the smaller exponent to equal the larger exponent. Label the amount increased as n . Step 2: For the number with the smaller power of 10, move the decimal point of the number part to the left n places. Step 3: Perform the addition or subtraction. Step 4: If the result is not in scientific notation, adjust the number to be in scientific notation. Adding Numbers in Scientific Notation with Different Powers of 10 Calculate the following: 6.1 × 10 4 + 4.8 × 10 5 Step 1: The lower exponent is 4. To make this equal to the larger exponent, we increased it by 1. Step 2: Since the smaller exponent was increased by 1, move the decimal one to the left, so the addition become 6.1 × 10 4 + 4.8 × 10 5 = 0.61 × 10 5 + 4.8 × 10 5 . Step 3: Now add the numbers, 0.61 × 10 5 + 4.8 × 10 5 = 5.41 × 10 5 Step 4: The result is in scientific notation, so no additional adjustment is necessary. 6.1 × 10 4 + 4.8 × 10 5 = 5.41 × 10 5 Subtracting Numbers in Scientific Notation with Different Powers of 10 Calculate the following: 7.9 × 10 − 15 − 6.8 × 10 − 13 Step 1: The lower exponent is −15 and the larger is −13. To make −15 equal to the larger exponent, we increased it by 2. Step 2: Since the smaller exponent increased by 2, move the decimal two to the left. The subtraction changes to 7.9 × 10 − 15 − 6.8 × 10 − 13 = 0.079 × 10 − 13 − 6.8 × 10 − 13 . Step 3: Subtract the numbers, 0.079 × 10 − 13 − 6.8 × 10 − 13 = − 6.721 × 10 − 13 . Step 4: The result is in scientific notation, so no additional adjustment is necessary. 7.9 × 10 − 15 − 6.8 × 10 − 13 = − 6.721 × 10 − 13 Multiplying and Dividing Numbers in Scientific Notation Multiplying and dividing numbers in scientific notation is somewhat easier than adding or subtracting, because the exponents of the 10s do not have to match. However, it is much more likely that the result will not be in scientific notation, and so that will have to be adjusted at the end. Generally, we multiply or divide the number parts of the two values, and then apply exponent rules to the 10 raised to the powers. To multiply two numbers in scientific notation: Step 1: Multiply the number parts. Step 2: Add the exponents of the 10s. Step 3: The result is the answer from Step 1 times 10 raised to the answer from Step 2. Step 4: If the number is not in scientific notation, adjust it appropriately. Multiplying Numbers in Scientific Notation Calculate the following: ( 4.3 × 10 3 ) × ( 1.8 × 10 7 ) ( 5 × 10 − 1 3 ) × ( 7.3 × 10 6 ) Step 1: Multiply the number parts to get 4.3 × 1.8 = 7.74 . Step 2: Add the exponents of the 10s to get 3 + 7 = 10 . Step 3: The result is then 7.74 × 10 10 . Step 4: This number is already in scientific notation, so no additional adjustment is necessary, ( 4.3 × 10 3 ) × ( 1.8 × 10 7 ) = 7.74 × 10 10 . Step 1: Multiply the number parts to get 5 × 7.3 = 36.5 . Step 2: Add the exponents of the 10s to get − 13 + 6 = − 7 . Step 3: The result then is 36.5 × 10 − 7 . Step 4: Since the number is not in scientific notation, it must be adjusted. To put 36.5 × 10 − 7 into scientific notation, the decimal moves one to the left, so the exponent would be increased by 1, giving 3.65 × 10 − 6 . ( 5 × 10 − 13 ) × ( 7.3 × 10 6 ) = 3.65 × 10 − 6 Multiplying Numbers in Scientific Notation Dividing Numbers in Scientific Notation To divide two numbers that are in scientific notation: Step 1: Divide the number parts. Step 2: Subtract the exponent of the denominator from the exponent of the numerator. Step 3: The answer is the result from Step 1 times 10 raised to the result from Step 2. Step 4: If the number is not in scientific notation, adjust it appropriately. Dividing Numbers in Scientific Notation Calculate the following: ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) Step 1: Divide the number parts to get 8.4 ÷ 2.1 = 4 . Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get 37 − 7 = 24 . Step 3: The result is then 4 × 10 24 . Step 4: This number is already in scientific notation, so no adjustment is necessary. ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) = 4 × 10 24 ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) Step 1: Divide the number parts to get 4.14 ÷ 8.28 = 0.5 . Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get − 13 − 9 = − 22 . Step 3: The result then is 0.5 × 10 − 22 . Step 4: Since this number is not in scientific notation, it must be adjusted. To put 0.5 × 10 − 22 into scientific notation, the decimal needs to move one to the right, so the exponent is decreased by 1, giving 5 × 10 − 23 . ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) = 5 × 10 − 23 Dividing Numbers in Scientific Notation Using Scientific Notation in Computing Real-World Applications As noted at the start of this section, scientific notation is useful when the standard representation of a number is awkward or impractical, which occurs when the numbers being used are extremely large or extremely small. For example, Venus is 67,667,000 miles from the sun. In scientific notation, this is 6.7667 × 10 7 . Planetary and galaxy distances is one set of numbers that is easier to express using scientific notation. Calculating Distances How much farther from the sun is Earth compared to Venus if Venus is 6.7667 × 10 7 miles from the sun and Earth is 9.1692 × 10 7 miles from the sun? To determine how much farther Earth is compared to Venus, we’d subtract the distances. 9.1692 × 10 7 − 6.7667 × 10 7 = 2.4025 × 10 7 . So, Earth is 2.4025 × 10 7 miles farther from the sun than Venus. Calculating Probability The probability of winning the Mega Millions lottery is published as 3.304693 × 10 − 9 . The probability of being hit by lightning is approximated to be 2 × 10 − 6 . How many times more likely are you to be hit by lightning than win the Mega Millions? To find out how many times more likely you are to be hit by lightning, divide the probability of being hit by lightning by the probability of winning the Mega Millions. ( 2 × 10 − 6 ) / ( 3.304693 × 10 − 9 ) = 6.609386 × 10 3 Step 1: Divide the number parts to get = 0.6052 (rounded to the fourth digit). Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get − 6 − ( − 9 ) = 3 . Step 3: The result then is 0.6052 × 10 3 . Step 4: Since this number is not in scientific notation, it must be adjusted. To put 0.6052 × 10 3 into scientific notation, the decimal needs to move one place to the right, so the exponent is decreased by 1, giving 6.052 × 10 2 . You are 6.052 × 10 2 , or 605.2, times more likely to be hit by lightning than you are to win the Mega Millions. Calculating Time and Length Sometimes it is entertaining to determine the time it takes for something to happen. Fingernails grow about 8.032 × 10 − 11 km per minute. How many kilometers long would fingernails be after 6 × 10 4 minutes? To find the length of the fingernails after the specified time, we multiply their rate of growth and the time they’ve grown. ( 8.032 × 10 − 11 ) × ( 6 × 10 4 ) = 48.192 × 10 − 7 = 4.8192 × 10 − 6 So, after 6 × 10 4 minutes, the fingernails would be 4.8192 × 10 − 6 km long. To put this in perspective, 1 × 10 − 6 km is a millimeter, and 6 × 10 4 minutes is about 4.16 days. So, after about 4.16 days, fingernails have grown about 4.8 millimeters. Calculating Data Generated As mentioned in the opening to this section, it is estimated that we’re producing 2.5 quintillion bytes of data per day. A good estimate is that there are 7.674 billion people on the planet. Convert both of those numbers to scientific notation, and then determine how much data is being generated per person each day. Written in standard form, 2.5 quintillion is 2,500,000,000,000,000,000. Changing that to scientific notation, move the decimal 18 places, so 2.5 quintillion bytes = 2.5 × 10 18 bytes. Writing 7.647 billion in scientific notation would be 7.647 × 10 9 because a billion is 1,000,000,000 = 10 9 . So, to find out how much data is being produced daily per person, we would divide these two numbers. 2.5 × 10 18 7.647 × 10 9 = 0.327 × 10 9 = 3.27 × 10 8 In standard form, that’s 327,000,000 bytes per person, so 327 million bytes of data daily are being produced per person. Application of Scientific Notation What Numbers Could Be Considered “Too Big” or “Too Small”? One wonders when the numbers we represent become too large or small for consideration. Perhaps the following examples put limits on what is meaningful. The number of particles in the known universe has been estimated at 4 × 10 80 particles. The smallest distance that has been measured is 1 × 10 − 18 m , though the theoretical smallest measurable value is 1 × 10 − 35 m . The distance across the universe is 4.4 × 10 26 m . Considering what those numbers represent, the extreme largest and extreme smallest, they might be numbers that constrain what we should reasonably be expected to deal with. Check Your Understanding Key Terms scientific notation standard notation Key Concepts Some numbers are so large or so small that writing the number out is clumsy and make it difficult to determine the true size of the number. Scientific notation makes the number more readable and make the relative size of the number immediately apparent. A number written in scientific notation is a number at least 1 and smaller than 10 multiplied by 10 raised to an exponent. Converting between scientific notation and standard notation involves correctly applying multiplication and division by powers of 10, which in practice equates to understanding how moving the decimal point of a number impacts the exponent of 10. Adding and subtracting numbers in base 10 requires the exponent of 10 in each number be the same. Once the numbers are converted to have the same exponent with the ten, then the numbers are added or subtracted as indicated, with the power of 10 remaining the same. If the result is not in scientific notation (for instance, the number has exceeded 10), then then number must be converted into scientific notation. Multiplying and dividing numbers in scientific notation is done by multiplying or dividing the number parts, then multiplying or dividing the 10 raised to the power parts, then multiplying those two results. If the new number is not in scientific notation, then the result must be converted into scientific notation. Videos Converting from Standard Form to Scientific Notation Form Converting from Scientific Notation Form to Standard Form Multiplying Numbers in Scientific Notation Dividing Numbers in Scientific Notation Application of Scientific Notation", "section": "Scientific Notation", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Arithmetic Sequences Learning Objectives After completing this section, you should be able to: Identify arithmetic sequences. Find a given term in an arithmetic sequence. Find the n th term of an arithmetic sequence. Find the sum of a finite arithmetic sequence. Use arithmetic sequences to solve real- world applications As we saw in the previous section, we are adding about 2.5 quintillion bytes of data per day to the Internet. If there are 550 quintillion bytes of data today, then there will be 552.5 quintillion bytes tomorrow, and 555 quintillion bytes in 2 days. This is an example of an arithmetic sequence. There are many situations where this concept of fixed increases comes into play, such as raises or table arrangements. Identifying Arithmetic Sequences A sequence of numbers is just that, a list of numbers in order. It can be a short list, such as the number of points earned on each assignment in a class, such as {10, 10, 8, 9, 10, 6, 10}. Or it can be a longer list, even infinitely long, such as the list of prime numbers. For example, here’s a sequence of numbers, specifically, the squares of the first 12 natural numbers. {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} Each value in the sequence is called a term . Terms in the list are often referred to by their location in the sequence, as in the n th term. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, the first term of the sequence is 1, the fourth term is 16, and so on. In the sequence of assignment scores {10, 10, 8, 9, 10, 6, 10}, the first term is 10 and the third term is 8 ( ). Sequence showing first, second, and fifth terms The notation we use with sequences is a letter, which represents a term in the sequence, and a subscript, which indicates what place the term is in the sequence. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, we will use the letter a as a value in the sequence, and so a 5 would be the term in the sequence at the fifth position. That number is 25, so we can write a 5 = 25 . In this section, we focus on a special kind of sequence, one referred to as an arithmetic sequence . Arithmetic sequences have terms that increase by a fixed number or decrease by a fixed number, called the constant difference (denoted by d ), provided that value is not 0. This means the next term is always the previous term plus or minus a specified, constant value. Another way to say this is that the difference between any consecutive terms of the sequence is always the same value. To see a constant difference, look at the following sequence: {7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87}. illustrates that each term of the sequence is the previous term plus 8. Eight is the constant difference here. Sequence of numbers with 8 added to each term Identifying Arithmetic Sequences Determine if the following sequences are arithmetic sequences. Explain your reasoning. { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } { 20 , 40 , 80 , 160 , 320 , 640 } { 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 } In the sequence { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } , every term is the previous term plus 3. The ellipsis indicates that the pattern continues, which means keep adding 3 to the previous term to get the new term. Therefore, this is an infinite arithmetic sequence. In the sequence { 20 , 40 , 80 , 160 , 320 , 640 } , terms increase by various amounts, for instance from term 1 to term 2, the sequence increases by 20, but from term 2 to term 3 the sequence increases by 40. So, this is not an arithmetic sequence. In the sequence { 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 } , every term is the previous term minus 6, so this is an arithmetic sequence. Arithmetic sequences can be expressed with a formula. When we know the first term of an arithmetic sequence, which we label a 1 , and we know the constant difference, which is denoted d , we can find any other term of the arithmetic sequence. The formula for the i th term of an arithmetic sequence is a i = a 1 + d × ( i − 1 ) . If we have an arithmetic sequence with first term a 1 and constant difference d , then the i th term of the arithmetic sequence is a i = a 1 + d × ( i − 1 ) . Let’s examine the formula with this arithmetic sequence: { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } . In this sequence a 1 = 4 and d = 3 . The table below shows the values calculated. i , Place in Sequence a i , i t h Term Value of Term Term Written as a 1 + 3 × ( i − 1 ) 1 a 1 4 4 + 3 × 0 2 a 2 7 4 + 3 × 1 3 a 3 10 4 + 3 × 2 4 a 4 13 4 + 3 × 3 5 a 5 16 4 + 3 × 4 i a i 4 + 3 × ( i − 1 ) We can see how the i th term can be directly calculated. In this sequence, the formula is a 1 + 3 × ( i − 1 ) where the first term, a 1 , is 4 and the constant difference d is 3. We can then determine the 47 th term of this sequence: a 47 = 4 + 3 × ( 47 − 1 ) = 4 + 3 × 46 = 4 + 138 = 142 . Calculating a Term in an Arithmetic Sequence Identify a 1 and d for the following arithmetic sequence. Use this information to determine the 60 th term. { 18 , 31 , 44 , 57 , 70 , 83 , ... } Inspecting the sequence shows that a 1 = 18 and d = 13 . We use those values in the formula, with i = 60 . a 60 = a 1 + d × ( i − 1 ) = 18 + 13 × ( 60 − 1 ) = 18 + 13 × 59 = 18 + 767 = 785 Arithmetic Sequences If we know two terms of the sequence, it is possible to determine the general form of an arithmetic sequence, a i = a 1 + d × ( i − 1 ) . If we have the i th term of an arithmetic sequence, a i , and the j th term of the sequence, a j , then the constant difference is d = a j − a i j − i and the first term of the sequence is a 1 = a i − d ( i − 1 ) . Determining First Term and Constant Difference Using Two Terms A sequence is known to be arithmetic. Two of its terms are a 7 = 56 and a 19 = 104 . Use that information to find the constant difference, the first term, and then the 50 th term of the sequence. To find the constant difference, use d = a j − a i j − i . The location of the terms is given by the subscript of the two a terms, i = 7 and j = 19 . So, the constant difference can be calculated as such: d = 104 − 56 19 − 7 = 48 12 = 4 . The constant difference of 4 is then used to find a 1 . a 1 = a i − d ( i − 1 ) = a 7 − 4 ( 7 − 1 ) = 56 − 4 × 6 = 32 . So d = 4 and a 1 = 32 . With this information, the 50 th term can be found. a 50 = a 1 + d × ( i − 1 ) = 32 + 4 × ( 50 − 1 ) = 32 + 4 × 49 = 32 + 196 = 228 . The 50 th term is a 50 = 228 . Finding the First Term and Constant Difference for an Arithmetic Sequence Finding the Sum of a Finite Arithmetic Sequence Sometimes we want to determine the sum of the numbers of a finite arithmetic sequence. The formula for this is fairly straightforward. The sum of the first n terms of a finite arithmetic sequence, written s n , with first and last term a 1 and a n , respectively, is s n = n ( a 1 + a n 2 ) . Finding the Sum of a Finite Arithmetic Sequence What is the sum of the first 60 terms of an arithmetic sequence with a 1 = 4.5 and d = 2.5 ? The formula requires the first and last terms of the sequence. The first term is given, a 1 = 4.5 . The 60 th term is needed. Using the formula a 1 = a i + d ( i − 1 ) provides the value for the 60 th term. a 60 = 4.5 + 2.5 ( 60 − 1 ) = 4.5 + 2.5 × 59 = 4.5 + 147.5 = 152 . Applying the formula s n = n ( a 1 + a n 2 ) provides the sum of the first 60 terms. s 60 = 60 ( 4.5 + 152 2 ) = 60 × 156.5 2 = 4,695 . The sum of the first 60 terms is 4,695. Finding the Sum of a Finite Arithmetic Sequence Using Arithmetic Sequences to Solve Real-World Applications Applications of arithmetic sequences occur any time some quantity increases by a fixed amount at each step. For instance, suppose someone practices chess each week and increases the amount of time they study each week. The first week the person practices for 3 hours, and vows to practice 30 more minutes each week. Since the amount of time practicing increases by a fixed number each week, this would qualify as an arithmetic sequence. Applying an Arithmetic Sequence Jordan has just watched The Queen’s Gambit and decided to hone their skills in chess. To really improve at the game, Jordan decides to practice for 3 hours the first week, and increase their time spent practicing by 30 minutes each week. How many hours will Jordan practice chess in week 20? Jordan’s practice scheme is an arithmetic sequence, as it increases by a fixed amount each week. The first week there are 3 hours of practice. This means a 1 = 3 . Jordan increases the time spent practicing by 30 minutes, or half an hour, each week. This means d = 0.5 . Using those values, and that we want to know the amount of time Jordan will study in week 20, we determine the time in week 20 using a i = a 1 + d × ( i − 1 ) . a 20 = 3 + 0.5 × ( 20 − 1 ) = 3 + 0.5 × 19 = 3 + 9.5 = 12.5 So, Jordan will practice 12.5 hours in week 20. Finding the Sum of a Finite Arithmetic Sequence Let’s check back in on Jordan. Recall, Jordan had just watched The Queen’s Gambit and decided to hone their skills, practicing for 3 hours the first week, and increasing the time spent practicing by 30 minutes each week. How many hours total will Jordan have practiced chess after 30 weeks of practice? To calculate the total amount of time that Jordan practiced, we need to use s n = n ( a 1 + a n 2 ) . The formula requires the first and last terms of the sequence. Since Jordan practiced 3 hours in the first week, the first term is a 1 = 3 . Because we want the total practice time after 30 weeks, we need the 30 th term. Because the constant difference is d = 0.5 , the 30 th term is a 30 = 3 + 0.5 ( 30 − 1 ) = 3 + 0.5 × 29 = 3 + 14.5 = 17.5 . Applying the formula s n = n ( a 1 + a n 2 ) provides the sum of the first 30 terms. s 30 = 30 ( 3 + 17.5 2 ) = 60 × 20.5 2 = 615 . This means that Jordan practiced a total of 615 hours after 30 weeks. The Fibonacci Sequence Not all sequences are arithmetic. One special sequence is the Fibonacci sequence , which is the sequence that has as its first two terms 1 and 1. Every term thereafter is the sum of the previous two terms. The first nine terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, and 34. This sequence is found in nature, architecture, and even music! In nature, the Fibonacci sequence describes the spirals of sunflower seeds, certain galaxy spirals, and flower petals. In music, the band Tool used the Fibonacci sequence in the song “Lateralus.” The Fibonacci sequence even relates to architecture, as it is closely related to the golden ratio. Fibonacci Sequence and “Lateralus” Check Your Understanding Key Terms sequence term of a sequence arithmetic sequence first term constant difference Key Concepts A sequence is a list of numbers. Any individual number in that list, or sequence, is a term of the sequence. A specific term of a sequence is denoted by the sequence symbol with a subscript indicating where the term in the sequence is. A special form of a sequence is an arithmetic sequence. Each arithmetic sequence is determined by its first term and its constant difference. Any term in an arithmetic sequence is determined by adding the constant difference to the preceding term. If the first term and the constant difference of an arithmetic sequence are known, then any term of the sequence can be found directly. Because arithmetic sequences follow such a strict pattern, the sum of the first n terms of an arithmetic sequence can be determined with the formula s n = n ( a 1 + a n 2 ) . Formulas a i = a 1 + d × ( i − 1 ) d = a j − a i j − i a 1 = a i − d ( i − 1 ) s n = n ( a 1 + a n 2 ) Videos Arithmetic Sequences Finding the First Term and Constant Difference for an Arithmetic Sequence Finding the Sum of a Finite Arithmetic Sequence Fibonacci Sequence and “Lateralus”", "section": "Arithmetic Sequences", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Geometric Sequences Savings grows in a geometric sequence. (credit: modification of “A big part of financial freedom is having your heart and mind free from worry about the what-ifs of life. – Suze Orman” by Morgan/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify geometric sequences. Find a given term in a geometric sequence. Find the n th term of a geometric sequence. Find the sum of a finite geometric sequence. Use geometric sequences to solve real-world applications. One of the concerns when investing is the doubling time , which is length of time it takes for the value of the investment to be twice, or double, that of its starting value. A shorter doubling times means the investment gets bigger, sooner. For example, if you invest $200 in an account with an 8-year doubling time, then in 8 years the value of the account will be double the starting amount, or 2 × $ 200 = $ 400 . After another 8 years (for a total of 16 years) the investment would be twice its value after the first 8 years, or 2 × ( 2 × $ 400 ) = 2 × ( $ 400 ) = $ 800 . Every 8 years, the investment would double again, so after the third 8-year period, the investment would be worth 2 × 2 × ( 2 × $ 400 ) = $ 1,600 . This process exhibits exponential growth, an application of geometric sequences, which is explored in this section. Identifying Geometric Sequences We know what a sequence is, but what makes a sequence a geometric sequence? In an arithmetic sequence, each term is the previous term plus the constant difference. So, you add a (possibly negative) number at each step. In a geometric sequence , though, each term is the previous term multiplied by the same specified value, called the common ratio . In the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , 1456 } the common ratio is 2. To see the difference between an arithmetic sequence and geometric sequence, examine these two sequences ( Figures 3.52 and 3.53 ). Arithmetic sequence Each term in this arithmetic sequence is the previous term plus 5. Geometric sequence Each term in this geometric sequence is the previous term times 2. In the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , 1456 } , the numbers get big fairly quickly, and stay positive. However, that’s not always the case with geometric sequences. Depending on the value of the common ratio, the terms could increase each time (like in the one shown in ), or the terms can get smaller each time, or the terms can alternate between positive and negative values. It all depends on the value of the common ratio, r . Consider this geometric sequence: { 5 , 15 , 45 , 135 , 405 , 2025... } Each term is the previous term times 5, which means the common ratio is 5. This common ratio is larger than 1, and so the terms increase each time. Now, look at this geometric sequence: { 2 , − 6 , 18 , − 54 , 162 , − 486 , 1458... } Each term is the previous term times −3, and the sign of the terms alternate from positive to negative. Then, there’s this geometric sequence: { 9 , 3 , 1 , 1 3 , 1 9 , 1 27 ... } Each term is the previous term times 1 3 , and the terms decrease each time. What we should take away from these three examples is if the common ratio is a positive number larger than 1, then the sequence increases. If the common ratio is a negative number, then the sign of the terms alternates between positive and negative. If the common ratio is between 0 and 1, then the terms decrease. Two special cases of geometric sequences are when the constant ratio is 1 and when the common ratio is 0. When the constant ratio is 1, every term of the sequence is the same, as in { 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 } . This is referred to as a constant sequence . When the constant ratio is 0, the first term can be any number, but every term after the first term is 0, as in { − 43.2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 } . Identifying Geometric Sequences For each sequence, determine if the sequence is a geometric sequence. If so, identify the common ratio. { 5 , 20 , 80 , 320 , 1,280 , 5,120 , 20,480 , ... } { − 3 , 6 , − 12 , 24 , 11 , 33 } { 4 , 2 , 1 , 1 2 , 1 4 , 1 8 , ... } In the sequence { 5 , 20 , 80 , 320 , 1,280 , 5,120 , 20,480 , ... } , the jump from 5 to 20 is a multiplication by 4, as is the next jump to 80, and the next to 320. Each term is 4 times the previous term. Since each term is 4 times the previous, this is a geometric sequence. The common ratio is 4. In the sequence { − 3 , 6 , − 12 , 24 , 11 , 33 } , notice that 6 is −3 times −2. The jump from 6 to −12 is another multiplication by negative. So, if this is a geometric sequence, each term should be the previous term times −2. But the change from 24 to 11 is not a multiplication by −2, This means the sequence is not a geometric sequence. In the sequence { 4 , 2 , 1 , 1 2 , 1 4 , 1 8 , ... } , the change from 4 to 2 is a multiplication by 1 2 , as is the next jump, from 2 to 1, as is the next from 1 to 1 2 . Each term is 1 2 times the previous term. Since each term is 1 2 times the previous, this is a geometric sequence. The common ratio is 1 2 . As with arithmetic sequences, the first term of a geometric sequence is labeled a 1 . The number that is multiplied by each term is called the common ratio and is denoted r . So, if the first term is known, a 1 , and the common ratio is known, r , then the n th term, a n , can be calculated with the formula a n = a 1 r n − 1 . The n th term of the geometric sequence, a n , with first term a 1 and common ratio r , is a n = a 1 r n − 1 . Return to the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , ... } . We observe that the first term is 3, so a 1 = 3 . We also found that the common ratio is 2, so r = 2 . The table below shows how any term can be calculated using just a 1 and r . i , Place in Sequence a i , i t h ,Term Value of Term Term Written as a 1 × r i − 1 1 a 1 3 3 × 2 0 2 a 2 6 3 × 2 1 3 a 3 12 3 × 2 2 4 a 4 24 3 × 2 3 5 a 5 48 3 × 2 4 i a i 3 × 2 i − 1 Determining the Value of a Specific Term in a Geometric Sequence In the following geometric sequences, determine the indicated term of the geometric sequence with a given first term and common ratio. Determine the 9 th term of the geometric sequence with a 1 × 6 and r = 3 . Determine the 11 th term of the geometric sequence with a 1 = 2 and r = − 5 . Using a n = a 1 r n − 1 with a 1 = 6 , r = 3 , and n = 9 , we calculate a 9 = a 1 r 9 − 1 = 6 × ( 3 ) 9 − 1 = 6 × ( 3 ) 8 = 6 × 6561 = 39366 . The 9 th term of the geometric sequence with a 1 = 6 and r = 3 is a 9 = 39366 . Using a n = a 1 r n − 1 with a 1 = 2 , r = − 5 , and n = 11 , we calculate a 11 = a 1 r 11 − 1 = 2 × ( − 5 ) 11 − 1 = 2 × ( − 5 ) 10 = 2 × 9,765,625 = 19,531,250 . Geometric Sequences Finding the Sum of a Finite Geometric Sequence As with arithmetic sequences, it is possible to add the terms of the geometric sequence. Like arithmetic sequences, the formula for the finite sum of the terms of a geometric sequence has a straightforward formula. The sum of the first n terms of a finite geometric sequence, written s n , with first term a 1 and common ratio r , is s n = a 1 ( 1 − r n 1 − r ) provided that r ≠ 1 . Calculating the Sum of a Finite Geometric Sequence What is the sum of the first 13 terms of the geometric sequence with first term a 1 = 5 and common ratio r = 3 ? What is the sum of the first 7 terms of the geometric sequence with first term a 1 = 16 and common ratio r = 1 8 ? Using a 1 = 5 , r = 3 , and n = 13 , we find that the sum is: S 13 = a 1 ( 1 − r 13 1 − r ) = 5 ( 1 − 3 13 1 − 3 ) = 5 ( 1 − 1,594,323 − 2 ) = 5 ( − 1,594,322 − 2 ) = 5 ( 797,161 ) = 3,985,805 The sum of the first 13 terms of this geometric sequence is 3,985,805. Using a 1 = 16 , r = 1 8 , and n = 7 , we find that the sum is: s 7 = a 1 ( 1 − r 7 1 − r ) = 16 ( 1 − ( 1 8 ) 7 1 − ( 1 8 ) ) = 16 ( 1 − ( 1 2,097,152 ) 7 8 ) = 16 ( 2,097,151 2,097,152 7 8 ) = 16 ( 299,593 262,144 ) = 299,593 16,384 = 18.2857 The sum of the first 7 terms of this geometric sequence is 18.2857 . Using Geometric Sequences to Solve Real-World Applications Geometric sequences have a multitude of applications, one of which is compound interest. Compound interest is something that happens to money deposited into an account, be it savings or an individual retirement account, or IRA. The interest on the account is calculated and added to the account at regular intervals. This means the interest that was earned later gains its own interest. This allows the money to grow faster. If that interest is added every month, we say it is compounded monthly. If the interest is added daily, then we say it is compounded daily. The amount of money that is deposited into the account is called the principal and is denoted P . The account earns money on that principal. The amount it earns is a percentage of the money in the account. The interest rate, expressed as a decimal, is denoted r . If you deposit P dollars in an account that earns interest compounded yearly, then the amount in the account, A , after t years is calculated with the formula: A = P ( 1 + r ) t . This is a geometric sequence, with constant ratio ( 1 + r ) and first term a 1 = P . Calculating Interest Compounded Yearly Daryl deposits $1,000 in an account earning 4 % interest compounded yearly. How much money is in the account after 25 years? Using A = P ( 1 + r ) t with P = 1000 , r = 0.04 , and t = 25 , we find that A = P ( 1 + r ) t = 1,000 × ( 1 + 0.04 ) 25 = 1,000 × ( 1.04 ) 25 = 1,000 × 2.66583633 = 2,665.85 . After 25 years, there is $ 2,665.84 in the account. Another application of geometric sequences is exponential growth. This arises in biology quite frequently, especially in relation to bacterial cultures, but also with other organism population models. In bacterial cultures, the time it takes the population to double is often recorded. This time to double is the same, regardless of how big the population gets. So, if the population doubles after 3 hours, it doubles again after another 3 hours, and again after another 3 hours, and so on. Put into geometric sequence language, it has a common ratio of 2. Doubling a Bacterial Culture When Escherichia coli ( E. coli ) is in a broth culture at 37°C, the population of E. coli doubles in number with 30 organisms, how many E. coli bacteria are present in the culture after 16 hours? Since the population is doubling every 20 minutes, this is a geometric sequence situation with common ratio r = 2 . The culture begins with 30 organisms, so a 1 = 30 . The time,16 hours, is 48 twenty-minute periods, so we’re looking for the 48th term in the sequence. Using these values in the geometric sequence formula gives a 48 = a 1 r n − 1 = 30 × 2 48 − 1 = 30 × 2 47 = 30 × ( 1.40737 × 10 14 ) = 4.22212 × 10 15 . So, after 16 hours, the culture contains 4.22212 × 10 15 E. coli organisms. That’s more than 4,000 trillion bacteria. Applying the Sum of a Finite Geometric Sequence A player places one grain of rice on the first square of a chess board. On the second square, the player places 2 grains of rice. On the third square, the player places 4 grains of rice. On each successive square of the board, the player doubles the number of grains of rice placed on the chess board. When the player places the last rice on the 64th square, how many total grains of rice have been placed on the board? Since the number of grains of rice is doubled at each step, this is a geometric sequence with first term a 1 = 1 and common ratio r = 2 . Rice is placed on 64 total squares, so we want the sum of the first 64 terms. Using this information and the formula, the total number of grains of rice on the board will be: s 64 = a 1 ( 1 − r n − 1 1 − r ) = 1 × ( 1 − 2 64 − 1 1 − 2 ) = ( 1 − 2 63 − 1 ) = − ( 1 − 2 63 ) = − ( − 9.2233720369 × 10 18 ) = 9.2233720369 × 10 18 That’s a 20-digit number! Sum of a Finite Geometric Sequence Check Your Understanding Key Terms geometric sequence common ratio Key Concepts A special form of a sequence is a geometric sequence. Each geometric sequence is determined by its first term and its constant ratio. Any term in a geometric sequence is determined by multiplying the constant ratio to the preceding term. If the first term and the constant ratio of a geometric sequence are known, then any term of the sequence can be found directly. Because geometric sequences follow such a strict pattern, the sum of the first n terms of a geometric sequence can be determined with the formula s n = a 1 ( 1 − r n − 1 1 − r ) . Finding the sum of a finite geometric sequence Applying arithmetic sequences Formulas a n = a 1 r n − 1 s n = a 1 ( 1 − r n − 1 1 − r ) Videos Geometric Sequences Sum of a Finite Geometric Sequence Encryption Throughout History Encryption began at least as far back as the Roman Empire. During the reign of Caesar, a particular cypher was used, fittingly named the Caesar Cypher. This encryption process granted the Romans a great tactical advantage. Even if a message was intercepted, it would not make sense to the person intercepting the message. Find four instances when encryption was used and cracked over the course of history. Projects The Golden Ratio in Art and Architecture The golden ratio has been used in art and architecture as far back as ancient Greece (possibly further). It also appears in South America (Incan architecture). Find five instances of the use of the golden ratio in art or architecture and describe its use in each of those instances. Your Budget Budgeting either is, or will shortly be, an important aspect of your life. Managing money well reduces stress in your life, and provides space for planning for future expenses, such as vacations or home improvements. Imagine your life 10 years from now. Estimate your monthly income. Identify expenses you will encounter monthly (mortgage or rent, car payment, insurance, entertainment, etc.). Decide on an amount you plan to save monthly (this is treated as an expense). Create a spreadsheet with those values. Record your monthly net income (your income minus your expenses). Determine how much money you will have saved over the course of 5 years (ignore interest). Write a reflection on your anticipated financial health. Estimating Pi The value of pi is the ratio of the circumference of a circle to the diameter of the circle. It is also equal to the ratio of the area of the circle to the square of the radius of the square. Research three ways to physically estimate pi. Estimate pi using all three processes you found. Present your process and solutions in class. Design Your Own Shift Cypher A cypher is a message written in such a way as to mask its contents. Changing a message into its cypher form is called encryption. Decryption or deciphering is the process of changing a cyphertext message back into the original (legible) message. One process of encryption is to scramble the letters, symbols, and punctuation of a message according to a mathematical rule. One rule that could be used for such a cypher is addition in a chosen modulus. In this project, you will create such a cypher, encrypt a message, and then decrypt the message. Step 1: Choose the letters, symbols, and punctuation marks you want to allow in your messages. This should include at least the uppercase letters and a space character. This is your character set. Step 2: Count the number of characters you will use. Label this number n . Step 3: Pair each character of your character set an integer from 0 to ( n − 1 ) . Do not assign more than one character to an integer. Step 4: Choose an integer between 1 and ( n − 1 ) . This will be the number used to create the cypher. Label this number s . Step 5: Write a message using your character set. Step 6: Replace every character in your message by the integer with which it was paired in Step 3. Step 7: For every number, x , from Step 7, perform the addition x + s (mod n ). Step 8: Replace every number found in Step 7 with the character with which it was paired in Step 3. This is your cyphertext. To decrypt your cyphertext, reverse the steps above. Step 1: Replace the cyphertext characters with the paired values. Step 2: For each value x , perform the subtraction x – s (mod n ). Step 3: Replace the numbers from Step 2 with their paired characters from the character set. The message is then deciphered. Design Your Own Cypher Using Multiplication A cypher is a message written in such a way as to mask its contents. Changing a message into its cypher form is called encryption. Decryption or deciphering is the process of changing a cyphertext message back into the original (legible) message. One process of encryption is to scramble the letters, symbols, and punctuation of a message according to a mathematical rule. One rule that could be used for such a cypher is multiplication in a chosen modulus. In this project, you will create such a cypher, encrypt a message, then decrypt the message. Step 1: Choose the letters, symbols, and punctuation marks you want to allow in your messages. This should include at least the uppercase letters and a space character. This is your character set. Step 2: Count the number of characters you will use. Label this number n . Step 3: Pair each character of your character set an integer from 0 to ( n − 1 ) . Do not assign more than one character to an integer. Step 4: Choose an integer, labeled s , between 1 and ( n − 1 ) so that GCF ( n , s ) = 1 . This will be the number used to create the cypher. Step 5: Write a message using your character set. Step 6: Replace every character in your message by the integer with which it was paired in Step 3. Step 7: For every number, x , from Step 6, perform the multiplication x − s (mod n ). Step 8: Replace every number found in Step 7 with the character with which it was paired in Step 3. This is your cyphertext. Before beginning to decrypt in this cypher, you need to know the multiplicative inverse of the value you chose as s . Step 1: The multiplicative inverse of s is the number that, when multiplied by s in your modulus, equals 1. To find this, you will have to multiply s and every number between 2 and ( n – 1 ) until the product is 1 (mod n ). Once this number is found, the message can be decrypted. Call this number r . Step 2: To decrypt your cyphertext, replace the cyphertext characters with the paired values. Step 3: For each of the value, x , perform the multiplication x × r ( mod n ) . Step 4: Replace the numbers from Step 3 with their paired characters from the character set. The message is then deciphered. Key Terms common ratio geometric sequence Key Concepts Geometric sequence. Finding an arbitrary term in a geometric sequence. Constant ratio. Finding the sum of a finite geometric sequence. Applying arithmetic sequences. Formulas a n = a 1 r n − 1 s n = a 1 ( 1 − r n − 1 1 − r ) Chapter Review Prime and Composite Numbers The Integers Order of Operations Rational Numbers Irrational Numbers Real Numbers Clock Arithmetic Exponents Scientific Notation Arithmetic Sequences Geometric Sequences Chapter Test", "section": "Geometric Sequences", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Different cultures developed different ways to record quantity. (credit: modification of work \"Tally sticks from the Swiss Alps\" by Sandstein, Swiss Alpine Museum permanent collection/Wikimedia Commons, CC BY 3.0) Right now, almost all cultures use the familiar Hindu-Arabic numbering system, which uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 along with place values based on powers of ten. This is a relatively recent development. The system didn’t develop until the 6th or 7th century C.E. and took some time to spread across the world, which means other cultures at other times had to develop their own methods of counting and recording quantity. Being different cultures and different times means there were significant differences in counting systems. Cultures needed to count and measure time for agriculture and for religious observations. It was needed for trade. Some languages had words only for one, two, and many. Other cultures developed more complex ways to represent quantity, with the Oksapmin people of New Guinea using an astonishing 27 words for their system. Representing these quantities in a recorded form likely began with a simple marking system, where one scratch on a stick or bone represented one of whatever was being counted. We still see this today with tally marks. These systems use repeated symbols to represent more than one. We also have systems where different symbols represent different quantities but still use some repetition, such as in Roman numerals. Other systems were devised that rely on place values, like the Hindu-Arabic system in use today. Place value systems needed a zero, though, and weren't immediately recognized and took time to develop. And within these positional systems there is variation. Some systems counted in twenties, others in tens, and some in a mix (adding another reason to visit Hawaii). Even now, though we all use and think using tens, computers are designed to work in groups of two, which requires a different perspective on numbering. In this chapter, we explore different numbering systems and grouping systems, eventually discussing base 2, the language of computers.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Hindu-Arabic Positional System This manuscript is an early example of Hindu numerals. (credit: modification of work “Bakshali manuscript”, Bodleian Libraries/ University of Oxford, public domain) Learning Objectives After completing this section, you should be able to: Evaluate an exponential expression. Convert a Hindu-Arabic numeral to expanded form. Convert a number in expanded form to a Hindu-Arabic numeral. The modern system of counting and computing isn’t necessarily natural. That different symbols are used to indicate different quantities or amounts is a relatively new invention. Simple marking by scratches or dots, one for each item being counted, was the norm long into human history. The modern system doesn’t use repeated symbols to indicate more than one of a thing. It uses the place of a digit in a numeral to determine what that digit represents. A numeral is a symbol used to represent a number. A number is an abstract idea that represents quantity or amount. Being clear about the difference between numeral and number is important. Just like a person can be called by various names, such as brother, father, husband, uncle, they are all representing the same person, John Smith. The person John Smith is the number, and the names brother, father, husband, and uncle are the numerals. Hindu-Arabic Numerals The numerals we currently use are referred to as Hindu-Arabic numerals, although they have changed as time has passed. Early forms of the numerals for 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 began in India, and passed through Persia to the Middle East. Place value was also employed in the early systems of India. Once this system was in north Africa and the Middle East, it spread to Europe, eventually replacing Roman numerals. Over time, the original symbols transformed into our modern ones. Read this article for another perspective on how the symbols began (based on the moon!) . The system we use for counting and computing uses place values based on powers of 10. In this section, we review exponents and our positional system. Evaluating Exponential Expressions Most modern numerical systems depend on place values, where the quantity represented depends not only on the digit, but also on where the digit is in the number. The place value is a power of some specific number, which means most numbering systems are actually exponential expressions. An exponential expression is any mathematical expression that includes exponents. So, evaluating such an expression means performing the calculation. In this chapter, we will be using exponents that are positive integer values. Before we do so, let’s remind ourselves about exponents and what they represent. Suppose you want to multiply a number. Let’s label that number a , by itself some number of times. Let’s label the number of times b . We denote that as a b . We say a , or the base , raised to the b th power, or the exponent . For example, if we are multiplying 13 by itself eight times, we write 13 8 and say 13 to the eighth power. When computing exponential expressions, we should be careful to remember the order of operations. Using the order of operation rules, calculations inside the parentheses are done first, then exponents are calculated, then multiplication and division calculations are performed, and then addition and subtraction. Exponential Notation Evaluating an Exponential Expression Evaluate the following exponential expressions. 4 × 5 2 + 2 × 6 3 6 × 8 2 + 3 × 8 1 + 4 × 8 0 3 × 10 2 + 0 × 10 1 + 6 × 10 0 To evaluate, or calculate, this expression, we use order of operations, which means the exponents are done first, then multiplications, and then additions. 4 × 5 2 + 2 × 6 3 = 4 × 5 × 5 + 2 × 6 × 6 × 6 = 4 × 25 + 2 × 216 = 100 + 432 = 532 To evaluate the expression, we use the order of operations, which means the exponents are done first, then the multiplications, then the additions. Remember that any base raised to the exponent 0 is 1. 6 × 8 2 + 3 × 8 1 + 4 × 8 0 = 6 × 8 × 8 + 3 × 8 + 4 × 1 = 6 × 64 + 3 × 8 + 4 × 1 = 384 + 24 + 4 = 412 To evaluate the expression, we use the order of operations, which means the exponents are done first, then the multiplications, and then the additions. Remember that any base raised to the exponent 0 is 1. 3 × 10 2 + 0 × 10 1 + 6 × 10 0 = 3 × 100 + 0 × 10 + 6 × 1 = 300 + 0 + 6 = 306 Converting Hindu-Arabic Numerals to Expanded Form When you see the number 738, and you speak the number out loud, what do you say? You probably said “seven hundred thirty-eight” while wondering what point could possibly be made by asking this. What you didn’t say was “seven, and three, and eight.” A pre-K student might say that. Which should make you wonder, why? The reason is that you’ve been taught place values , or the positions of digits in a number that determine the values of those digits. You know that in a three-digit number, the first digit is hundreds, the second digit is tens, and the last digit is ones. These place values rely on powers of 10, which makes this system a base 10 system . This sense of place value is what makes our system of numbers so useful. You’ve also been taught the Hindu-Arabic numeration system . This system, which uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, and also employs place value based on powers of 10, is in use today. Writing a number using these place values is writing them in expanded form . For a number with n digits, the expanded form is the first digit times 10 raised to one less than n , plus each following digit times 10 raised to one less than the previous power of 10. For example, the number 738 would be written as 7 × 10 2 + 3 × 10 1 + 8 × 10 0 . What about a four-digit number, like 5,825? Out loud, we’d say five thousand, seven hundred twenty-five. In expanded form, it would be 5 × 10 3 + 8 × 10 2 + 2 × 10 1 + 5 × 10 0 . Notice that the largest exponent is one less than the number of digits, and that the exponents go down by one as we move through the number. Aryabhata of Kusumapura and Brahmagupta The Hindu-Arabic numeral system developed in India, and Aryabhata of Kusumapura is credited with the place value notation in the 5th century. However, the system wasn’t as complete as it could be, until. roughly a century later, when Brahmagupta introduced the symbol for 0. The 0 is necessary to indicate that a given place value has been skipped, as in 4,098. In 4,098, the 10 2 power is skipped. Without such a symbol, 4,098 and 498 look similar. The value of both the place value notation and the introduction of the symbol 0 cannot be overstated, for math and the sciences. Writing a Number in Expanded Form Write the following in expanded form. 563 4,821 903,786 Step 1: Since there are three digits in 563, n is 3. So, this is the first digit times 10 raised to the power of 2, so we start with 5 × 10 2 . Step 2: Then we add the next digit, 6, multiplied by 10 to a power one less than the previous, at which point we have 5 × 10 2 + 6 × 10 1 . Step 3: Finally, the last digit is multiplied by 10 to the zeroth power and added to the previous. This results in 5 × 10 2 + 6 × 10 1 + 3 × 10 0 . Step 1: Since there are four digits in 4,821, n is 4. We multiply the first digit, 4, by 10 raised to the power of 3, which is 4 × 10 3 . Step 2: Then we add the next digit, 8, multiplied by 10 to a power one less than the previous, at which point we have 4 × 10 3 + 8 × 10 2 . Step 3: We continue to the next digit, lowering the exponent of 10 by one. Now we have 4 × 10 3 + 8 × 10 2 + 2 × 10 1 . Step 4: Finally, the last digit is multiplied by 10 to the zeroth power and added to the previous. This results in 4 × 10 3 + 8 × 10 2 + 2 × 10 1 + 1 × 10 0 . Since there are six digits in 903,786, n is 6. So, we begin the process with 9 times 10 raised to the 5th power and continue through the numbers, reducing the exponent of 10 by one each time. This results in 9 × 10 5 + 0 × 10 4 + 3 × 10 3 + 7 × 10 2 + 8 × 10 1 + 6 × 10 0 . Converting Numbers in Expanded Form to Hindu-Arabic Numerals Converting from expanded form back into a Hindu-Arabic numeral is the reverse process of expanding a number, and is equivalent to evaluating the exponential expression. Converting a Number from Expanded Form to a Hindu-Arabic Numeral Convert the following into Hindu-Arabic numerals. 3 × 10 2 + 4 × 10 1 + 8 × 10 0 5 × 10 3 + 0 × 10 2 + 9 × 10 1 + 9 × 10 0 6 × 10 6 + 2 × 10 5 + 0 × 10 4 + 9 × 10 3 + 1 × 10 2 + 1 × 10 1 + 7 × 10 0 Evaluating the expression results in: 3 × 10 2 + 4 × 10 1 + 8 × 10 0 = 3 × 100 + 4 × 10 + 8 × 1 = 300 + 40 + 8 = 348 Evaluating the expression results in: 5 × 10 3 + 0 × 10 2 + 9 × 10 1 + 9 × 10 0 = 5 × 1000 + 0 × 100 + 9 × 10 + 9 × 1 = 5000 + 0 + 90 + 9 = 5099 Evaluating the expression results in: 6 × 10 6 + 2 × 10 5 + 0 × 10 4 + 9 × 10 3 + 1 × 10 2 + 1 × 10 1 + 7 × 10 0 = 6 × 1,000,000 + 2 × 100,000 + 0 × 10,000 + 9 × 1,000 + 1 × 100 + 1 × 10 + 7 × 1 = 6,000,000 + 200,000 + 0 + 9,000 + 100 + 10 + 7 = 6,209,117 Check Your Understanding Key Terms numeral number exponential expression base exponent place value base 10 system Hindu-Arabic numeration system expanded form Key Concepts Exponents are used to represent repeated multiplication of a base. In arithmetic, exponents are computed before multiplication, division, addition, and subtraction. Computing an exponent is done by multiplying the base by itself the number of times equal to the exponent. The system of numbers currently used is the Hindu-Arabic system. Digits in this system take on values based on their place in the number. The place values are determined by multiplying the digit by 10 raised to the appropriate power. The expanded form of a Hindu-Arabic number is the sum of each digit times 10 raised to the exponent for that place value. Videos Exponential Notation", "section": "Hindu-Arabic Positional System", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Early Numeration Systems Babylonians used clay tablets for writing and record keeping. (credit: modification of work by Osama Shukir Muhammed Amin FRCP(Glasg), CC BY 4.0 International) Learning Objectives After completing this section, you should be able to: Understand and convert Babylonian numerals to Hindu-Arabic numerals. Understand and convert Mayan numerals to Hindu-Arabic numerals. Understand and convert between Roman numerals and Hindu-Arabic numerals. Each culture throughout history had to develop its own method of counting and recording quantity. The system used in Australia would necessarily differ from the system developed in Babylon that would, in turn, differ from the system developed in sub-Saharan Africa. These differences arose due to cultural differences. In nearly all societies, knowing the difference between one and two would be useful. But it might not be useful to know the difference between 145 and 167, as those quantities never had a practical use. For example, a shepherd likely didn't manage more than 100 sheep, so quantities larger than 100 might never have been encountered. This can even be seen in our use of the term few, which is an inexact quantity that most would agree means more than two. However, as societies became more complex, as commerce arose, as military bodies developed, so did the need for a system to handle large numbers. No matter the system, the issues of representing multiple values and how many symbols to use had to be addressed. In this section, we explore how the Babylonians, Mayans, and Romans addressed these issues. Understand and Convert Babylonian Numerals to Hindu-Arabic Numerals The Babylonians used a mix of an additive system of numbers and a positional system of numbers . An additive system is a number system where the value of repeated instances of a symbol is added the number of times the symbol appears. A positional system is a system of numbers that multiplies a “digit” by a number raised to a power, based on the position of the “digit.” The Babylonian place values didn’t use powers of 10, but instead powers of 60. They didn’t use 60 different symbols though. For the value 1, they used the following symbol: For values up to 9, that symbol would be repeated, so three would be written as To represent the quantity 10, they used For 20, 30, 40, and 50, they repeated the symbol for 10 however many times it was needed, so 40 would be written When they reached 60, they moved to the next place value. The complete list of the Babylonian numerals up to 59 is in . Babylonian Numerals You can see how Babylonians repeated the symbols to indicate multiples of a value. The number 6 is 6 of the symbol for 1 grouped together. The symbol for 30 is three of the symbols for 10 grouped together. However, their system doesn’t go past 59. To go past 59, they used place values. As opposed to the Hindu-Arabic system, which was based on powers of 10, the Babylonian positional system was based on powers of 60. You should also notice there is no symbol for 0, which has some impact on the number system. Since the Babylonian number system lacked a 0, they didn’t have a placeholder when a power of 60 was absent. Without a 0, 101, 110, and 11 all look the same. However, there is some evidence that the Babylonians left a small space between \"digits\" where we would use a 0, allowing them to represent the absence of that place value. To summarize, the Babylonian system of numbers used repeating a symbol to indicate more than one, used place values, and lacked a 0. Invention of 0 The idea of 0 is not a natural one. Most cultures failed to recognize the need for a 0. If someone asked a farmer in 300 B.C.E. how many cows they had, but they had none, they would not answer \"zero.\" They’d say “I don’t have any” and be done with it. It wasn’t until roughly 3 B.C.E. that 0 appeared in Mesopotamia. It was independently discovered (or invented!) in the Mayan culture around 4 C.E. it made its appearance in India in the 400s C.E., and began to spread at that point. It wasn’t developed earlier mostly because positional systems were not yet fully developed. Once positional systems arose, the need to represent a missing power had to be addressed. So how do we convert from Babylonian numbers to Hindu-Arabic numbers? To do so, we need to use the symbols from , and then place values based on powers of 60. If you have n digits in the Babylonian number, you multiply the first “digit” by 60 raised to one less than the number of “digits.” You then continue through the “digits,” multiplying each by 60 raised to a power that is one smaller. However, be careful of spaces, since they represent a zero in that place. Converting Two-Digit Babylonian Numbers to Hindu-Arabic Numbers Convert the Babylonian number into a Hindu-Arabic number. has two digits: and Step 1: So the first symbol, represents 4 in the Babylonian system. This is multiplied by 60 to the first power (just as would happen in a two digit number), which gives us 4 × 60 1 . Step 2: The next symbol is which represents 27 in the Babylonian system. This is multiplied by 60 raised to 0, which gives 4 × 60 1 + 27 × 60 0 . Step 3: Calculating that yields 4 × 60 1 + 27 × 60 0 = 240 + 27 = 267 . So the Babylonian number equals 267 in the Hindu-Arabic number system. Converting Three-Digit Babylonian Numbers to Hindu-Arabic Numbers Convert the Babylonian number into a Hindu-Arabic number. has three digits: and and Step 1: So the first symbol, represents 13 in the Babylonian system. This is multiplied by 60 to the second power (since there are 3 digits), which gives us 13 × 60 2 . Step 2: The next symbol is which represents 8 in the Babylonian system, is multiplied by 60 raised to the first power, which gives us 13 × 60 2 + 8 × 60 1 . Step 3: The last digit is representing 54, which is multiplied by 60 raised to 0, which gives 13 × 60 2 + 8 × 60 1 + 54 × 60 0 . Step 4: Calculating that yields 13 × 60 2 + 8 × 60 1 + 54 × 60 0 = 13 × 3,600 + 8 × 60 + 54 × 1 = 46,800 + 480 + 54 = 47,334 . So, the Babylonian number equals 47,334 in the Hindu-Arabic number system. Converting Four-Digit Babylonian Numbers to Hindu-Arabic Numbers Convert the Babylonian number into a Hindu-Arabic number. It appears that has three digits, but there is a space in between and Remember, the Babylonian system has no 0, it instead employs a space where we expect a zero. This means this is a four digit number. Step 1: The first symbol, represents 12 in the Babylonian system. This is multiplied by 60 to the third power since there are four digits, which gives us 12 × 60 3 . Step 2: The next symbol is a blank, which for us is a 0, representing 0 × 10 2 , giving us 12 × 60 3 + 0 × 10 2 . Step 3: The next symbol is which represents 42 in the Babylonian system, is multiplied by 60 raised to the first power, which gives us 12 × 60 3 + 0 × 10 2 + 42 × 60 1 . Step 4: The last Babylonian digit, represents 39 in the Babylonian system. This is multiplied by 60 raised to 0, which gives 12 × 60 3 + 0 × 10 2 + 42 × 60 1 + 39 × 60 0 . Step 5: Calculating that yields 12 × 60 3 + 0 × 10 2 + 42 × 60 1 + 39 × 60 0 = 12 × 216,000 + 0 × 10 2 + 42 × 60 + 39 × 1 = 2,592,000 + 0 + 2,520 + 39 = 2,594,559 So the Babylonian number equals 2,594,559 in the Hindu-Arabic number system. The Legacy of Babylonian System The Babylonian system can still be seen today. An hour is 60 minutes, and a minute is 60 seconds. Additionally, when measuring angles in degrees, each degree can be split into 60 minutes (1/60th of a degree) and 60 seconds (1/60th of a minute). Converting Between Babylonian and Hindu-Arabic Numbers Understand and Convert Mayan Numerals to Hindu-Arabic Numerals The Mayans employed a positional system just as we do and the Babylonians did, but they based their position values on powers of 20 and they had a dedicated symbol for zero. Similar to the Babylonians, the Mayans would repeat symbols to indicate certain values. A single dot was a 1, two dots were a 2, up to four dots. Then a five was a horizontal bar. The horizontal bars could be used three times, since the fourth horizontal bar would make a 20, which was a new position in the number. The 0 was a special picture, which appears like a turtle lying on its back. The shell would then be \"empty,\" so maybe that’s why the symbol was 0. The complete list is provided in . Another feature of Mayan numbers was that they were written vertically. The powers of 20 increased from bottom to top. Mayan Numerals To summarize, the Mayan system of numbers used repeating symbol to indicate more than one, used place values, and employed a 0. So how do we convert from Mayan numbers to Hindu-Arabic numbers? To do so, we need to use the symbols from and then place values based on powers of 20. If you have n digits in the Mayan number, you multiply the first “digit” by 20 raised to one less than the number of “digits.” You then continue through the “digits,” multiplying each by 20 raised to a power that is one smaller than the previous power. Fortunately, there is an explicit 0, so there is no ambiguity about numbers like 110, 101, and 11. Converting Two-Digit Mayan Numbers to Hindu-Arabic Numbers Convert the Mayan number into a Hindu-Arabic number. has two digits: and Step 1: So, the first symbol, represents 15 in the Mayan system. This is multiplied by 20 to the first power, which gives us 15 × 20 1 . Step 2: The next symbol is which represents 9 in the Mayan system. This is multiplied by 20 raised to 0, which gives 15 × 20 1 + 9 × 20 0 . Step 3: Calculating that yields 15 × 20 1 + 9 × 20 0 = 300 + 9 = 309 . So equals 309 in the Hindu-Arabic number system. Converting Three-Digit Mayan Numbers to Hindu-Arabic Numbers Convert the Mayan number into a Hindu-Arabic number. has three digits: and and Step 1: So the first symbol, represents 6 in the Mayan system. This is multiplied by 20 to the second power (since there are 3 digits), which gives us 6 × 20 2 . Step 2: The next symbol is which represents 8 in the Mayan system, is multiplied by 20 raised to the first power, which gives us 6 × 20 2 + 8 × 20 1 . Step 3: The last digit is representing 4, which is multiplied by 20 raised to 0, which gives 6 × 20 2 + 8 × 20 1 + 4 × 20 0 . Step 4: Calculating that yields 6 × 20 2 + 8 × 20 1 + 4 × 20 0 = 6 × 400 + 8 × 20 + 4 × 1 = 2,400 + 160 + 4 = 2,564 . So the Mayan number equals 2,564 in the Hindu-Arabic number system. Converting Four-Digit Mayan Numbers to Hindu-Arabic Numbers Convert the Mayan number into a Hindu-Arabic number. has four digits, so the first power of 20 that is used is 3. Step 1: The first symbol, represents 8 in the Mayan system. This is multiplied by 20 to the third power (since there are four digits), which gives us 8 × 20 3 . Step 2: The next symbol is which is a 0, representing 0 × 20 2 , giving us 8 × 20 3 + 0 × 20 2 . Step 3: The next symbol is which represents 16 in the Mayan system, is multiplied by 20 raised to the first power, which gives us 8 × 20 3 + 0 × 20 2 + 16 × 20 1 . Step 4: The last Mayan digit, represents 5 in the Mayan system. This is multiplied by 20 raised to 0, which gives 8 × 20 3 + 0 × 20 2 + 16 × 20 1 + 5 × 20 0 . Step 5: Calculating that yields 8 × 20 3 + 0 × 20 2 + 16 × 20 1 + 5 × 20 0 = 8 × 8000 + 0 × 400 + 16 × 20 + 5 × 1 = 64,000 + 0 + 320 + 5 = 64,325 So the Mayan number equals 64,325 in the Hindu-Arabic number system. The Mayan Calendar The Mayans used this base 20 system for everyday situations. But their culturally important, and extremely accurate, calendar system used a slightly different system. For their calendars, they used a system where the place values were 1, 20, then 20*18, then 20*18*18. The reason for this is 20*18 is 360, which is closer to the number of days in a year. Had they used a purely base 20 system for their calendar, they’d be very far off with 400 days in a year. Three hundred sixty days still left the Mayans a bit short, as there are 365 days in a year (ignoring leap years). The Mayan calendar also included 5 days, called Wayeb days, which brings their calendar to 365 days. As it happens, Wayeb is the Mayan god of misfortune, so these 5 days were considered the bad luck days. Converting Mayan Numbers to Hindu-Arabic Numbers Understand and Convert Between Roman Numerals and Hindu-Arabic Numerals The Mayan and Babylonian systems shared two features, one of which we are familiar with (place value) and one that we don’t use (repeated symbols). The Roman system of numbers used repeated symbols, but does not employ a place value. It also lacks a 0. The Roman system is built on the following symbols in . Roman Numeral Hindu-Arabic Value I 1 V 5 X 10 L 50 C 100 D 500 M 1,000 Roman Numerals As in the Mayan and Babylonian systems, a symbol may be repeated to indicate a larger value. However, at 4, they did not use IIII. They instead used IV. Since the I came before the V, the number stands for “one before five.” A similar process was used for 9, which was written IX, or “one before ten.” The value 40 was written XL, or “ten before fifty,” while 49 was written XLIX, or “forty plus nine.” The following are the rules for writing and reading Roman numerals. The representations for bigger values precede those for smaller values. Up to three symbols may be grouped together; for example, III for 3, or XXX for 30, or CC for 200. A larger value followed by a smaller value indicated addition; for example, VII for 7, XIII for 13, LV for 55, and MCC for 1200. I can be placed before V to indicate 4, or before X, to indicate 9. These are the only ways I is used as a subtraction. X can be placed before L to indicate 40, and before C to indicate 90. These are the only ways X is used as a subtraction. C can be placed before D to indicate 400, and before M to indicate 900. These are the only ways C is used as a subtraction. If multiple symbols are used, and a subtraction involving that symbol, the subtraction part comes after the multiple symbols. For example, XXIX for 29 and CCXC for 290. Legacy of Roman Numerals The Roman numbering system is still used today in some situations. Many cornerstones of buildings have the year written in Roman numerals. Movie titles often represent the year the movie was produced as Roman numerals. The most recognizable might be that the Super Bowl is numbered using Roman numerals. Converting Roman Numerals to Hindu-Arabic Numbers Convert the following Roman numerals into Hindu-Arabic numerals. XXVII XXXIV MMCMXLVIII The numeral XXVII begins with two X’s, which is then followed by a V. So, the two X’s combine to be 20. The V is followed by two I’s, so the V indicates the addition of 5. The two I’s that follow indicate addition of two. That ends the symbols, so the value is 20 plus 5 plus 2, or 27 in Hindu-Arabic numerals. The numeral XXXIV begins with three X’s, which is then followed by an I. So, the three X’s combine to be 30. The I is followed by a V, which indicates 4. That ends the symbols, so the value is 30 plus 4, or 34 in Hindu-Arabic numerals. The numeral MMCMXLVIII begins with two M’s, which is then followed by a C. So, the two M’s combine to make 2000. The C is followed by an M, which indicates 900. The CM is followed by XL, which indicates 40. The L is followed by V, which indicates 5. The V is followed by three I’s, indicating 3. Adding those values yields 2,948. Converting From Roman Numbers to Hindu-Arabic Numbers Of course, we can convert from Hindu-Arabic numerals, to Roman numerals, too. Converting Hindu-Arabic Numbers to Roman Numerals Convert the following Hindu-Arabic numerals into Roman numerals. 38 94 846 2,987 Thirty is represented as three X’s, and the 8 is represented with VIII, so 38 in Roman numerals is XXXVIII. Ninety is represented by XC, and four is represented by IV, so 94 in Roman numerals is XCIV. The number is less than 900 and more than 500, so the first symbol to be used is D, which is 500. To get to 800, we need 300 more, which is represented with three C’s. Forty is represented with XL, and the six. The Roman numerals are DCCCXLVI. The two thousand is represented by two M’s. The 900 is represented by CM. The 80 is represented by LXXX (50 plus 30). Finally, the 7 is represented by VII. We have that 2,987 in Roman numerals is MMCMLXXXVII. Converting From Hindu-Arabic Numbers to Roman Numbers Check Your Understanding Key Terms additive system of numbers positional system of numbers Babylonian system of numbers Mayan system of numbers Roman system of numbers Key Concepts Historically, there have been many systems for numbering. One system is an additive system, in which symbols are repeated to express larger numbers. Another system is a positional system, in which the digits and their positions determine the quantity being represented. The Babylonian system was a combination of a positional and additive system. It used 60 as its base. Using that in the positional system makes it possible to convert between Babylonian and Hindu-Arabic numbers. The Mayan system was a combination of a positional and additive system. It used 20 as its base. Using that in the positional system makes it possible to convert between Mayan and Hindu-Arabic numbers. The Roman system was an additive system. Knowing what each symbol represents makes it possible to convert between Roman and Hindu-Arabic numbers. Videos Converting Between Babylonian and Hindu-Arabic numbers Converting Mayan Numbers to Hindu-Arabic Numbers Converting From Roman Numbers to Hindu-Arabic Numbers Converting From Hindu-Arabic Numbers to Roman Numbers", "section": "Early Numeration Systems", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Converting with Base Systems Computers use Base 2, which only uses 0's and 1's, to represent quantity. (credit: modification of work “IMAG0933” by yvanhou/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Convert another base to base 10. Write numbers in different base systems. Convert base 10 to other bases. Determine errors in converting between bases. In our system of numbers, we use base 10, but using base 10 was not a given within other systems. There were other systems that used bases other than 10, as we saw with the Mayans and the Babylonians. The base 10 system comes down to grouping objects in sets of 10, but grouping in sets of 10 only happens if the culture values grouping by that many. We feel 10 is natural because we have 10 fingers. There are other systems using other grouping values, such as 4 or 20. One good reason for examining other bases is to remind ourselves how we had to learn arithmetic when we were young, memorizing rules for our base 10 system. We had to learn why those arithmetic rules made sense, such as why 1 + 1 = 2 and 1 + 2 = 3 . Another good reason for learning other base systems is due to computers; their circuitry instead uses base 2. In this section, we explore other base systems and how to convert between them. Conversion of Another Base into Base 10 and Other Bases We saw in Hindu-Arabic Positional System that our Hindu-Arabic system uses base 10 , which is a system using place values of digits that depend on powers of 10 (or, are based on powers of 10). We’ve already worked with bases other than base 10: The Babylonian system was base 60, while the Mayan system was base 20. To explore how our base 10 system is used, answer the following question: What’s the following quantity: 4,572? You probably said four thousand five hundred seventy-two (no, there is no “and” between hundred and seventy). But why do you think that 4 means four thousand? A very young person when learning their numbers might say that’s a four five seven and two. But you added the context of thousands to the four. Why? Place value, that’s why. You learned early on that where the numeral was gave it different meanings. Ten thousands, thousands, hundreds, tens, and ones. So, you translate that symbol string (4,572) into “four thousand five hundred seventy-two.” As we saw in Hindu-Arabic Positional System , expanding a Hindu-Arabic number involved writing the number using each digit times its appropriate power of 10. So, we could write 4,572 as 4 × 10 3 + 5 × 10 2 + 7 × 10 1 + 2 × 10 0 = 4 × 1000 + 5 × 100 + 7 × 10 + 2 × 1 . One possible reason we use base 10 is that we have 10 fingers, and in the cultures where the Hindu-Arabic system developed, that became the standard. Other cultures may have used other ways of organizing numbers, perhaps using 20 by including toes, or using 60 because 60 has many divisors. Mathematically though, base 10 is an awkward base to work in since 10 has limited divisors. But we think it is easy and simple because that’s what we’ve been taught to use. Using a base 10 system means we need 10 symbols to make our numbering system work: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Now imagine that we all only had 6 fingers instead of 10 and our counting system was based on those 6 fingers. We would be counting in groups of 6, not groups of 10. How would this change how we work with quantity? First, we’d need only six symbols. Let’s use 0, 1, 2, 3, 4, 5. Second, our place values would be based on powers of 6, not powers of 10. For instance, the number 3,024 in base 6 would be 3 × 6 3 + 0 × 6 2 + 2 × 6 1 + 4 × 6 0 . That is how you can translate a base 6 number into a base 10 number. When we calculate that expression we get 3 × 6 3 + 0 × 6 2 + 2 × 6 1 + 4 × 6 0 = 3 × 216 + 0 × 36 + 2 × 6 + 4 × 1 = 648 + 0 + 12 + 4 = 664 . This means the base 6 number 3,024 is equal to the base 10 number 664. From now on, if we are using a base 6 number, we will follow it with the subscript 6, like the following: 3,024 6 means the number is in base 6. A base 10 number gets no subscript (it’s the standard). So, 3,024 is a base 10 number. A base 13 number would be 4,672 13 . So, a base 6 system uses only the symbols 0, 1, 2, 3, 4, and 5. Also, the place values use powers of 6. However, we still don’t know how to count in base 6. In order to do so, we’d have to know how to represent the quantities larger than five in base 6. Let’s review how our base 10 system works by counting from 0 to 100, which shows how larger values are represented. In writing the base 10 numbers, you start with these first 10 values: 0 1 2 3 4 5 6 7 8 9 But you’ve run out of symbols. So, we use two digits: 10 11 12 13 14 15 16 17 18 19 The 1 out front means you’ve run out of digits one time. But now you’ve run out twice. Continuing with those numbers gives: 20 21 22 23 24 25 26 27 28 29 And so on, 30 31 32 33 34 etc.… Eventually, you hit the 90s, 90 91 92 93 94 95 96 97 98 99 And you’ve run out of the digits again! So, we say we’ve run out of digits in the tens place one time, hence: 100 101 102 103 104 105 106 107 108 109 That’s the pattern we use in base 10. We write out the symbols until we’ve used all the symbols, then add a digit in front that counts how many times we’ve used the digits. Knowing the numbers, or being able to count higher and higher, is necessary to understand how all the arithmetic works, as it all goes back to counting. The counting pattern is the same for any other base, including base 6. So, let’s start: 0 1 2 3 4 5 But we’ve run out of symbols! Just like in base 10, we use a second digit, where the first digit will tell us we’ve run out of symbols one time. 10 11 12 13 14 15 And we use the same pattern: 20 21 22 23 24 25 30 31 32 33 34 35 40 41 42 43 44 45 50 51 52 53 54 55 But we’ve run out of symbols for that front digit. So, we indicate it the same way as in base 10…by adding a third digit in front, indicating we’ve run out of symbols once in the second place: 100 101 102 103 104 105 110 111 etc. The symbol pattern is the same, but truncated. We only use the six symbols. So that is how we represent base 6. Being able to write out these numbers is important when working with addition in the base. When using a base larger than 10, though, we need more symbols. Instead of creating new symbols, we use capital letters, with A representing the digit for \"10,\" B representing the digit for \"11,\" and so on. Determining Digits of a Base with Less Than 10 Digits What are the digits used for base 7? Since this is base 7, we need only 7 symbols: 0, 1, 2, 3, 4, 5, 6. Determining Digits of a Base with More Than 10 Digits What are the digits used for base 14? Since this is base 14, we need 14 symbols. We don’t have single character numbers for 10, 11, 12, and 13, so, in a fit of inspired creativity, we use capital letters A, B, C, D to represent those quantities. So, the digits in base 14 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D. Using Base 12 As mentioned in the text, working in base 10 is mathematically awkward. Ten has only two natural number divisors: 2 and 5. This means dividing into groups is not easy. However, 12, or a dozen, has more divisors: 2, 3, 4, and 6. The Dozenal Society recognizes this more mathematically pleasant detail. It advocates for a switch to using base 12 for numbers. Their argument is based on the divisibility of the number 12. But has there ever been a society that used such a system? The answer is yes. A dialect of the Gwandara language in Nigeria uses the base 12 system. It is unlikely, though, that the Dozenal Society will achieve their goal, as the base 10 system is so entrenched in our society. Converting from One Base into Another Convert 3,601 7 into base 10. In base 7, the place values are powers of 7. Since there are four digits, the highest power of 7 that is used is 3. This yields 3,601 7 = 3 × 7 3 + 6 × 7 2 + 0 × 7 1 + 1 × 7 0 = 3 × 343 + 6 × 49 + 0 × 7 + 1 = 1,029 + 294 + 0 + 1 = 1,324 . Convert Base 7 to Base 10 Converting from Base 14 to Base 10 Convert 4B7 14 into base 10. In base 14, the place values are powers of 14. Since there are three digits, the highest power of 14 is 2. Also recall that in base 14, 10 is represented by A, 11 is represented by B, 12 is represented by C, and 13 is represented by D. Using that, we convert to base 10: 4 B 7 14 = 4 × 14 2 + B × 14 1 + 7 × 14 0 = 4 × 196 + 11 × 14 + 7 × 1 = 784 + 154 + 7 = 945 . Converting from Base 12 to Base 10 Convert A16 12 into base 10. In base 12, the place values are powers of 12. Since there are three digits, the highest power of 12 is 2. Also recall that in base 12, 10 is represented by A and 11 is represented by B. Using that, we convert to base 10: A 16 14 = 10 × 12 2 + 1 × 12 1 + 6 × 12 0 = 10 × 144 + 1 × 12 + 6 × 1 = 1,440 + 12 + 6 = 1,458 . Converting from Base 2 to Base 10 Convert 1011 2 into base 10. In base 2, the place values are powers of 2. Since there are four digits, the highest power of 2 is 3. Using that, we convert to base 10: 1011 2 = 1 × 2 3 + 0 × 2 2 + 1 × 2 1 + 1 × 2 0 = 1 × 8 + 0 × 4 + 1 × 2 + 1 × 1 = 8 + 0 + 2 + 1 = 11 . Before Napoleon Before Napoleon’s France, which adopted the base 10 system, a modified base 12 system was often used in Europe. Twelve is easily divisible into groups of 2, 3, 4, and 6, which makes it easier to work with. Even our numbering system retains a bit of this. You have likely noticed that we use the words thirteen , fourteen , fifteen , and so on to indicate 10 and 3, 10 and 4, 10 and 5, and so one. Even the 20s reinforce this idea, as in twenty-one, and twenty-two. However, two numbers don’t follow this pattern, namely 11 and 12. If they followed the same rules, they’d be one teen and two teen. We even have a special word for 12; that is, a dozen. However, etymologically speaking, the words eleven and twelve are likely derived by referencing the number 10. These two numbers may date back to the Old English words endleofan and twelf , which can be traced back further to ain lif and twa lif . The word lif here may be the base word for “to leave.” This would suggest ain lif is one left after 10, and twa lif is two left after 10, or, 11 and 12. Writing Numbers in Base Systems Other Than Base 10 Write the numbers in base 7 up to 100 7 . Step 1: Using the patterns we indicated earlier, we begin with the first seven digits. 0, 1, 2, 3, 4, 5, 6 Step 2: Since we’ve run out of digits, we start with 10, indicating we’ve run out of symbols once. 10, 11, 12, 13, 14, 15, 16 Step 3: Continuing in the same way, we get: 20, 21, 22, 23, 24, 25, 26 30, 31, 32, 33, 34, 35, 36 40, 41, 42, 43, 44, 45, 46 50, 51, 52, 53, 54, 55, 56 60, 61, 62, 63, 64, 65, 66 Now, all the digits have been used in the leading digits. Since the digits have all been used in that leading digit, we use 100, as in base 10. 100 Writing Numbers in Bases with More Than 10 Symbols Write the numbers in base 14 up to 100 14 . Step 1: Using the patterns we indicated earlier, we begin with the first 14 digits. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D Step 2: Since we’ve run out of digits, we start with 10, indicating we’ve run out of symbols once. 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D Step 3: Continuing in the same way, we get: 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 2A, 2B, 2C, 2D 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 3A, 3B, 3C, 3D 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 4A, 4B, 4C, 4D 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 5A, 5B, 5C, 5D 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 6A, 6B, 6C, 6D 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 7A, 7B, 7C, 7D 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 8A, 8B, 8C, 8D 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 9A, 9B, 9C, 9D A0, A1, A2, A3, A4, A5, A6, A7, A8, A9, AA, AB, AC, AD B0, B1, B2, B3, B4, B5, B6, B7, B8, B9, BA, BB, BC, BD C0, C1, C2, C3, C4, C5, C6, C7, C8, C9, CA, CB, CC, CD D0, D1, D2, D3, D4, D5, D6, D7, D8, D9, DA, DB, DC, DD 100 Base 2 is important in the digital age, as it is the system used by computers. It is the simplest base to work with, but has the drawback that the numbers in base 2 may use many, many digits. In Addition and Subtraction in Base Systems and Multiplication and Division in Base Systems , we will look at base 2 in each situation. Writing Numbers in Base 2 Write the numbers in base 2 up to 100 2 . Base 2 uses only two symbols: 0 and 1. Following the pattern established previously, the numbers in base 2 up to 100 2 are 0, 1, 10, 11, and 100. Early Hawaiian Numeration System Before the British arrived in Hawaii, people there used a system that combined two different bases. Objects were initially grouped into collections of four, and a collection of four was referred to as kauna . A person could have two kauna and three “ones” (in Hindu-Arabic, 11). Or they could have eight kauna and one “ones” (In Hindu-Arabic, 33). However, sets of four were grouped in collections of 10. A set of 10 kauna was ka’au . The collections of ka’au were grouped by 10 also. Which meant that 10 ka’au (this is 40 in Hindu-Arabic) would be lau (or 400 in Hindu-Arabic). What this shows is that the Hawaiian culture developed a system that used base 4 combined with base 10. Conversion of Base 10 into Another Base Converting from base 10 into another base uses repeated division, recording the remainder at each step. Then, the number in the new base is the remainder starting from the last remainder found. To be accurate in what we’re saying, we need to remind ourselves of some terminology associated with division. When integers are divided, the one being divided is the dividend , and the one that is dividing the dividend is the divisor . The quotient is the largest natural number that can be multiplied by the divisor where the product is less than the dividend. When the integer n is divided by the integer d , n is called the dividend and d is the divisor. To convert a base 10 number n into base d , we divide n by d , recording the remainder. Then we divide the quotient from that step by the base d , and record the remainder again. We continue this process until the quotient is 0. Then, the base d number has digits that start with the last remainder and use each remainder in reverse order. Converting from Base 10 into a Lower Base Convert 298 to base 6. We divide 298 by 6, and record the remainder. Then we divide the quotient from that step by 6, and record the remainder again. We continue this process until the quotient is 0. Then, the base 6 number has digits that start with the last remainder and use each remainder in reverse order. Step 1: When we divide 298 by 6, we get 4 9 r 4 6 298 . The quotient is 49 and the remainder is 4. Step 2: Now we divide the quotient, 49, by 6. This gives 8 r 1 6 49 . The quotient is 8 and the remainder is 1. Step 3: Repeating, we get 1 r 2 6 8 . The quotient is 1 and the remainder is 2. Step 4: Finally, we perform the operation on the quotient 1, 0 r 1 6 1 giving us a quotient of 0 and a remainder of 1. Step 5: The base 6 number has digits equal to the remainders in reverse order, 1214 6 . So, 298 in base 10 when converted to base 6 is 1214 6 . Converting from Base 10 to Another Base Converting from Base 10 into a Higher Base Convert 45,134 to base 13. We divide 45,134 by 13, and record the remainder. Then we divide the quotient from that step by 13, and record the remainder again. We continue this process until the quotient is 0. Then, the base 13 number has digits that start with the last remainder and use each remainder in reverse order. It is at this step that we’ll convert to the base 13 digits, which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C. Step 1: When we divide 45,134 by 13, we get 3,471 r 11 13 45,134 . The quotient is 3,471 and the remainder is 11. Step 2: Now we divide the quotient, 3,471, by 13. This gives 267 r 0 13 3,471 . The quotient is 267 and the remainder is 0. Step 3: Repeating, we get 20 r 7 13 267 . The quotient is 20 and the remainder is 7. Step 4: Again, and we get 1 r 7 13 20 . The quotient is 1 and the remainder is 7. Step 5: Finally, we get 0 r 1 13 1 , with quotient 0 and a remainder 1. Step 6: The base 13 number has digits equal to the remainders in reverse order, which were 1, 7, 7, 0, and 11. The 11 is written as B in base 13. So, 45,134 in base 10 when converted to base 13 is 1770B 13 . Converting from Base 10 into Base 2 Convert 100 to base 2. Following the pattern above: Step 1: We divide 100 by 2, and record the remainder. Step 2: Then we divide the quotient from that step by 2, and record the remainder again. Step 3: We continue this process until the quotient is 0. Step 4: Following this process, the remainders are, in order, 0, 0, 1, 0, 0, 1, 1. Writing those in reverse order gives the number in base 2, 1100100 2 . Notice that 100 in base 2 took seven digits. Converting from Hindu-Arabic Numbers to Mayan Numbers To convert from a Hindu-Arabic number to a Mayan number involves two distinct processes. First, the number must be converted to base 20, using the process described and demonstrated previously. Next, that base 20 number has to be written using Mayan numerals. For reference, the Mayan numerals and their values are below. Converting from Base 10 into the Mayan System Convert the following into Mayan numbers. 51 653 The Mayan system is base 20, so we must use 20 in the process from above. The first division has a quotient of 2 and remainder of 11. The 11 serves as the “ones” digit. Dividing that quotient, 2, by 20 has a quotient of 0 with a remainder of 2. The 2 becomes the “twenties” digit of the number. So, in base 20, the number would be 2 followed by 11. The Mayan symbols for 2 and 11 are and . Writing these vertically, with the “ones” digit on top, as appropriate for Mayan numbers, results in: The Mayan system is base 20, so we must use 20 in the process from above. The first division, 673 divided by 20, has a quotient of 32 and remainder of 13. Dividing that quotient, 32, by 20 has a quotient of 1 with a remainder of 12. Dividing that quotient, 1, by 20 has a quotient of 0 and a remainder of 1. Since there are three remainders here, this is a three-digit number. The 1 is the “20-squared” digit, the 12 is the “twenties” digit, and the 13 is the “ones” digit. So, in base 20, the number would be 1 followed by 12 followed by 13.. The Mayan symbols for 1, 12 and 13 are , , and . Writing these vertically, as appropriate for Mayan numbers, would result in: Other Languages, Other Bases There have been base systems that use bases other than 10. Some bases used were 20, 12, and 27! Visit this site to see more on the languages that used other bases . Errors in Converting Between Bases There are some common errors that are made when converting between bases. Often, it comes down to using an “illegal” symbol in the new base. Detecting an Illegal Symbol When Converting Between Bases A base 10 number is converted to base 7 and the result was 2081 7 . Was an error committed? How do you know? The result has the digit 8 in it. In base 7, 8 is an illegal symbol. Based on that, an error was committed. When converting from base 10 to another base, an illegal symbol will be used if a mistake was made in the division process used to find the number in the new base. Since the digits are based on the remainders, any remainder that is an illegal symbol would indicate an error. Detecting an Error in Division When Converting Between Bases When changing from base 10 to base 8, the division process resulted in the following remainders: 1, 0, 9, 2, 4. Was an error committed? How do you know? The remainders include 9, which in base 8 is an illegal symbol. Another possible way to detect an error in converting between bases is to count the number of digits. When converting from a higher base to a lower base, the number of digits cannot get smaller. Similarly, when converting from a lower base to a higher base, the number of digits cannot get bigger. So, if a base 10 number is converted to a base 3 number, the number of digits in the new base 3 numbers cannot be less than the number of digits in the base 10 number. Similarly, if a base 7 number is converted to base 10, the number of digits in the base 10 number cannot be more than the number of digits in the original base 7 number. Detecting an Error in Number of Digits When Converting Between Bases A five-digit base 10 number is converted to a base 5 number. The base 5 number has four digits. Was an error committed? How do you know? Since 10 is larger than 5, the base 5 number cannot have less digits than the base 10 number. Since it did, we know an error has been made. Check Your Understanding The Babylonian system used base 60. To convert from Hindu Arabic numbers into Babylonian numbers, the process for converting from base 10 to a different base would be done first. Then, the results found in the conversion process would be changed to Babylonian numerals. This process is similar to the one for Mayan numbers. Key Terms base 10 remainder dividend divisor quotient Key Concepts The system we use is the base 10 system. Base 10 is not the only base that can be used. To use another base, one could start with a list of numbers in that base. To indicate that a number is written in a base other than 10, a subscript is appended to the end of the number. That subscript indicates the base for the number. Numbers written in a base smaller than 10 use the same symbols as base 10. However, when using bases larger than 10, the symbols A, B, C, … are used to represent digits larger than 9. To convert from a number written in a base other than 10 into a base 10 number, the number is written in expanded form and then that expression is computed. To convert a number from base 10 into another base, the base 10 number is repeatedly divided by the new base. The remainders when performing these divisions become the digits for the number in the new base. Common errors can be detected when performing base conversions. Videos Convert Base 7 to Base 10 Converting from Base 10 to Another Base", "section": "Converting with Base Systems", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Addition and Subtraction in Base Systems All information in computers is represented by 0's and 1's, including quantity, which means computers use Base 2 for arithmetic. (credit: modification of work “Magnifying glass and binary code” by Marco Verch Professional Photographer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Add and subtract in bases 2–9 and 12. Identify errors in adding and subtracting in bases 2–9 and 12. Once we decide on a system for counting, we need to establish rules for combining the numbers we’re using. This begins with the rules for addition and subtraction. We are familiar with base 10 arithmetic, such as 2 + 5 = 7 or 3 × 5 = 15 . How does that change if we instead use a different base? A larger base? A smaller one? In particular, computers use base 2 for all number representation. When your calculator adds or subtracts, multiplies or divides, it uses base 2. This is because the circuitry recognizes only two things, high current and low current, which means the system is uses only has two symbols. Which is what base 2 is. In this section, we use addition and subtraction in bases other than 10 by referencing the processes of base 10, but applied to a new base system. Addition in Bases Other Than Base 10 Now that we understand what it means for numbers to be expressed in a base other than 10, we can look at arithmetic using other bases, starting with addition. When you think back to when you first learned addition, it is very likely you learned the addition table. Once you knew the addition table, you moved on to addition of numbers with more than one digit. The same process holds for addition in other bases. We begin with an addition table, and then move on to adding numbers with two or more digits. We worked with base 6 earlier, and have the numbers in base 6 up to 100 6 . Using that table of values, we can create the base 6 addition table. Here’s the beginning of the base 6 addition table: + 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 ? 2 2 3 4 5 ? ? 3 3 4 5 ? ? ? 4 4 5 ? ? ? ? 5 5 ? ? ? ? ? Many of the cells are not filled out. The ones filled in are values that never get past 5, which is the largest legal symbol in base 6, so they are acceptable symbols. But what do we do with 5 + 3 in base 6? We can’t represent the answer as “8” since “8” is not a symbol available to us. Let’s go back to the list of numbers we have for base 6. 0 1 2 3 4 5 10 11 12 13 14 15 20 21 22 23 24 25 30 31 32 33 34 35 40 41 42 43 44 45 50 51 52 53 54 55 So, what is 5 + 1 equal to in base 6? Well, start at the 5, and jump ahead one step. You land on 10. This means that, in base 6, 5 + 1 = 10. So, what is 5 + 2 in base 6? Well, 5 + 2 = 5 + 1 + 1, so 10 + 1…jump one more space and you land on 11. So, 5 + 2 = 11 in base 6. And so it goes. Using that process, stepping one more along the list, we can fill in the remainder of the base 6 addition table ( ). + 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 10 2 2 3 4 5 10 11 3 3 4 5 10 11 12 4 4 5 10 11 12 13 5 5 10 11 12 13 14 Base 6 Addition Table With this table, and with our understanding of “carrying the one,” we can then use the addition table to do addition in base 6 for numbers with two or more digits, using the same processes you learned for addition when you did it by hand. Adding in Base 6 Calculate 251 6 + 133 6 . Step 1: Let’s set up the addition using columns. 2 5 1 + 1 3 3 Step 2: Let’s do the one’s place first. According to the base 6 addition table ( ), 1 + 3 = 4. 2 5 1 + 1 3 3 4 Step 3: Now, we do the “tens” place (it’s really the sixes place). According to the base 6 addition table ( ), we have 5 + 3 = 12. So, like in base 10, we use the 2 and carry the 1. 1 2 5 1 + 1 3 3 2 4 Step 4: Now the “hundreds” place (really, thirty-sixes place). There, we have 1 + 2 + 1 = 3 + 1 = 4. 1 2 5 1 + 1 3 3 4 2 4 So, 251 6 + 133 6 = 424 6 . As you can see, the process is the same as when you learned base 10 addition, just a different symbol set. Creating an Addition Table for a Base Lower Than 10 Create the addition table for base 7. Create the addition table for base 2. We begin with the table below. + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 3 4 5 6 3 3 4 5 6 4 4 5 6 5 5 6 6 6 In base 7, the number that follows 6 is 10 (since we’ve run out of symbols!). So, 6 7 + 1 7 = 10 7 . Once that is established, 6 7 + 2 7 will be two numbers past 6, which is 11 in base 7. + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 10 2 2 3 4 5 6 11 3 3 4 5 6 4 4 5 6 5 5 6 6 6 10 11 Continuing, we can fill in the rows as we would in base 10, but being aware that we are working in base 7 ( ). + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 10 2 2 3 4 5 6 10 11 3 3 4 5 6 10 11 12 4 4 5 6 10 11 12 13 5 5 6 10 11 12 13 14 6 6 10 11 12 13 14 15 Base 7 Addition Table We revisit base 2 here. Begin with the table: + 0 1 0 0 1 1 1 Base 2 Addition Table In base 2, the number that follows 1 is 10 (since we’ve run out of symbols!). So, 1 2 + 1 2 = 10 2 . The complete table for base two then is below. + 0 1 0 0 1 1 1 10 This demonstrates that the rules necessary for base 2 addition are as small as possible: four rules. To summarize the creation of the addition tables for a given base, do the following. Step 1: Set up the table. Step 2: Fill in all the additions that use the “legal” symbols for the base. The diagonal that goes from upper left to lower right that is immediately next to the filled boxes all get the value 10, regardless of base. Step 3: Enter the values that are in the “teens.” This can all be done on one table without creating multiple copies of previously done work. Adding in Base 7 Calculate 536 7 + 433 7 . Step 1: Let’s set up the addition using columns. 5 3 6 + 4 3 3 Step 2: Let’s do the one’s place first. According to the base 7 addition table in the solution for , 6 + 3 = 12. We will carry the 1. 1 5 3 6 + 4 3 3 2 Step 3: Now, we do the “tens” place (it’s really the sevens place). According to the base 7 addition table in the solution for , we have 1 + 3 + 3 = 10. So, like in base 10, we use the 0 and carry the 1. 1 5 3 6 + 4 3 3 0 2 Step 4: Now the “hundreds” place (really, forty-ninths place). There, we have 1 + 5 + 4 = 6 + 4 = 13. 1 5 3 6 + 4 3 3 1 3 0 2 So, 536 7 + 333 7 = 1302 7 . As seen previously, when performing addition in another base, set up the problem exactly as you would for addition in base 10. At each step, check the addition table for the base. As in base 10 addition, move right to left, adding down the columns using the rules in the addition table. When necessary and just as in base 10, be sure to carry the 1. Creating an Addition Table for a Base Higher Than 10 Create the addition table for base 12. Step 1: Recall, in base 12, the symbol set is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, and B. So, the addition table begins as shown below. + 0 1 2 3 4 5 6 7 8 9 A B 0 0 1 2 3 4 5 6 7 8 9 A B 1 1 2 3 4 5 6 7 8 9 A B 2 2 3 4 5 6 7 8 9 A B 3 3 4 5 6 7 8 9 A B 4 4 5 6 7 8 9 A B 5 5 6 7 8 9 A B 6 6 7 8 9 A B 7 7 8 9 A B 8 8 9 A B 9 9 A B A A B B B Step 2: The diagonal immediately to the right of the filled in boxes is where the 10 goes for this base. + 0 1 2 3 4 5 6 7 8 9 A B 0 0 1 2 3 4 5 6 7 8 9 A B 1 1 2 3 4 5 6 7 8 9 A B 10 2 2 3 4 5 6 7 8 9 A B 10 3 3 4 5 6 7 8 9 A B 10 4 4 5 6 7 8 9 A B 10 5 5 6 7 8 9 A B 10 6 6 7 8 9 A B 10 7 7 8 9 A B 10 8 8 9 A B 10 9 9 A B 10 A A B 10 B B 10 Step 3: Using the pattern we’re familiar with, and counting in base 12, we can fill in the other cells. + 0 1 2 3 4 5 6 7 8 9 A B 0 0 1 2 3 4 5 6 7 8 9 A B 1 1 2 3 4 5 6 7 8 9 A B 10 2 2 3 4 5 6 7 8 9 A B 10 11 3 3 4 5 6 7 8 9 A B 10 11 12 4 4 5 6 7 8 9 A B 10 11 12 13 5 5 6 7 8 9 A B 10 11 12 13 14 6 6 7 8 9 A B 10 11 12 13 14 15 7 7 8 9 A B 10 11 12 13 14 15 16 8 8 9 A B 10 11 12 13 14 15 16 17 9 9 A B 10 11 12 13 14 15 16 17 18 A A B 10 11 12 13 14 15 16 17 18 19 B B 10 11 12 13 14 15 16 17 18 19 1A Base 12 addition table Notice that the lower-right entry is 1A 12 , as this is the number one past 19 12 . Adding in Base 12 Calculate 3A7 12 + 9BA 12 . Step 1: Using the process established in the earlier addition problem, set up the columns. 3 A 7 + 9 B A Step 2: Using the rules from the base 12 addition table in the solution for , and being careful to carry the 1 when necessary, we get the following: 1 1 3 A 7 + 9 B A 1 1 A 5 The ones that were carried are located over the columns. So, 3A7 12 + 9BA 12 = 11A5 12 . Adding in Base 2 We again return to base 2, the base used by computers. Calculate 1001 2 + 11011 2 . Step 1: Using the process established in the earlier addition problem, set up the columns. 1 0 0 1 + 1 1 0 1 1 Step 2: Using the rules from the base 2 addition table in the solution for , and being careful to carry the 1 when necessary (and shown at the top of the grid), we get the following: 1 1 1 1 0 0 1 + 1 1 0 1 1 1 0 0 1 0 0 Step 3: Calculate 1001 2 + 11011 2 = 100100 2 . So, 1001 2 + 11011 2 = 100100 2 . Subtraction in Bases Other Than Base 10 Subtraction in bases other than base 10 follow the same processes as base 10 subtraction, but, as with addition, using the addition table for the base. Subtracting in Base 6 Calculate 52 6 − 34 6 . Step 1: Let’s set up the subtraction using columns. 5 2 − 3 4 Step 2: Just as we might do in base 10, we borrow a 1 from the 5 for the ones digit. 5 4 12 − 3 4 Step 3: Referring to the base 6 addition table ( ), we see that 4 + 4 = 12, so 12 6 − 4 6 is 4 6 . 5 4 1 2 − 3 4 4 Step 4: Now we deal with the “tens” (really, sixes) digit, 4 6 − 3 6 , which equals 1 6 according to the base 6 addition table ( ). 5 4 1 2 − 3 4 1 4 So, 52 6 − 34 6 = 14 6 . Subtracting in Base 12 Calculate A17 12 − 4B3 12. Step 1: Let’s set up the subtraction using columns. A 1 7 − 4 B 3 Step 2: Even in base 12, 7 12 − 3 12 = 4 12 . A 1 7 − 4 B 3 4 Step 3: Moving to the “tens” digit, we have 1 12 − B 12 . Since 1 is less than B in base 12, we need to borrow a 1 from the A, just as we would for subtraction in base 10. A 9 11 7 − 4 B 3 4 Step 4: According to the base 12 addition table in the solution for , B 12 + 2 12 = 11 12 , so 11 12 − B 12 = 2 12 . A 9 1 1 7 − 4 B 3 2 4 Step 5: Finally, we deal with the “hundreds” digit. According to the base 12 addition table in the solution for , 4 12 + 5 12 = 9 12 , so 9 12 − 4 12 = 5 12 . A 9 11 7 − 4 B 3 5 2 4 So, A17 12 − 4B3 12 = 524 12 . Errors When Adding and Subtracting in Bases Other Than Base 10 Errors when computing in bases other than 10 often involve applying base 10 rules or symbols to an arithmetic problem in a base other than base 10. The first type of error is using a symbol that is not in the symbol set for the base. For instance, if a 9 shows up when working in base 7, you know an error has happened because 9 is not a legal symbol in base 7. Identifying an Illegal Symbol in Arithmetic in a Base Other Than Base 10 Explain the error in the following calculation: 15 6 + 34 6 = 49 6 Since the problem is in base 6, the symbol set available is 0, 1, 2, 3, 4 and 5. The 9 in the answer is clearly not a legal symbol for base 6. Looking back to the base 6 addition table ( ), we see that 5 6 + 4 6 = 13 6 . Correcting the error, we see the sum is 15 6 + 34 6 = 53 6 . The second type of error is using a base 10 rule when the numbers are not in base 10. For instance, if you are working in base 13, then 9 13 + 9 13 is not 18 13 , even though 18 is the correct answer in base 10. Identifying an Arithmetic Error in a Base Other Than Base 10 Explain the error in the following calculation, and correct the error: 89 12 + 76 12 = 165 12 If this problem was a base 10 problem, this would be the correct answer. However, in base 12, 9 + 6 is not 15, but is instead 13. To correct this error, carefully use the addition table for base 12 . If properly used, the correct answer would be 143 12 , as seen below: 8 9 + 7 6 1 4 3 Check Your Understanding Key Concepts Addition tables for bases other than 10 can be built using the same processes that are used in base 10, including using a number line. Addition in bases other than base 10 use the same processes as addition in base 10, but use the addition table for that base. Subtraction in bases other than base 10 use the same processes as subtraction in base 10, but use the addition table for that base.", "section": "Addition and Subtraction in Base Systems", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Multiplication and Division in Base Systems The processes for multiplication and division are the same for arithmetic in any bases. (credit: modification of work “NCTR Intern Claire Boyle” by Danny Tucker/U.S. Food and Drug Administration, Public Domain) Learning Objectives After completing this section, you should be able to: Multiply and divide in bases other than 10. Identify errors in multiplying and dividing in bases other than 10. Just as in Addition and Subtraction in Base Systems , once we decide on a system for counting, we need to establish rules for combining the numbers we’re using. This includes the rules for multiplication and division. We are familiar with those operations in base 10. How do they change if we instead use a different base? A larger base? A smaller one? In this section, we use multiplication and division in bases other than 10 by referencing the processes of base 10, but applied to a new base system. Multiplication in Bases Other Than 10 Multiplication is a way of representing repeated additions, regardless of what base is being used. However, different bases have different addition rules. In order to create the multiplication tables for a base other than 10, we need to rely on addition and the addition table for the base. So let’s look at multiplication in base 6. Multiplication still has the same meaning as it does in base 10, in that 4 × 6 is 4 added to itself six times, 4 × 6 = 4 + 4 + 4 + 4 + 4 + 4 . So, let’s apply that to base 6. It should be clear that 0 multiplied by anything, regardless of base, will give 0, and that 1 multiplied by anything, regardless of base, will be the value of “anything.” Step 1: So, we start with the table below: * 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 3 0 3 4 0 4 5 0 5 Step 2: Notice 2 × 2 = 4 is there. But we didn’t hit a problematic number there (4 works fine in both base 10 and base 6). But what is 2 × 3 ? If we use the repeated addition concept, 2 × 3 = 2 + 2 + 2 = 4 + 2 . According to the base 6 addition table ( ), 4 + 2 = 10 . So, we add that to our table: * 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 10 3 0 3 10 4 0 4 5 0 5 Step 3: Next, we need to fill in 2 × 4 . Using repeated addition, 2 × 4 = 2 + 2 + 2 + 2 = 10 + 2 = 12 (if we use our base 6 addition rules). So, we add that to our table: * 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 10 12 3 0 3 10 4 0 4 12 5 0 5 Step 4: Finally, 2 × 5 = 2 + 2 + 2 + 2 + 2 = 12 + 2 = 14 . And so we add that to our table: * 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 10 12 14 3 0 3 10 4 0 4 12 5 0 5 14 Step 5: A similar analysis will give us the remainder of the entries. Here is 4 × 5 demonstrated: 4 × 5 = 4 + 4 + 4 + 4 + 4 = 12 + 12 + 4 = 24 + 4 = 32 . This is done by using the addition rules from Addition and Subtraction in Base Systems , namely that 4 + 4 = 12 , and then applying the addition processes we’ve always known, but with the base 6 table ( ). In the end, our multiplication table is as follows: * 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 10 12 14 3 0 3 10 13 20 23 4 0 4 12 20 24 32 5 0 5 14 23 32 41 Base 6 Multiplication Table Notice anything about that bottom line? Is that similar to what happens in base 10? To summarize the creation of a multiplication in a base other than base 10, you need the addition table of the base with which you are working. Create the table, and calculate the entries of the multiplication table by performing repeated addition in that base. The table needs to be drawn only the one time. Creating a Multiplication Table for a Base Lower Than 10 Create the multiplication table for base 7. Step 1: Let’s apply the process demonstrated and outlined above to find the base 7 multiplication table. It should be clear that 0 multiplied by anything, regardless of base, will give 0, and that 1 multiplied by anything, regardless of base, will be the value of “anything.” So, we start with the table below: * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 3 0 3 6 4 0 4 5 0 5 6 0 6 Step 2: Notice 2 × 2 = 4 is there. But we didn’t hit a problematic number there (4 works fine in both base 10 and base 6). The same is true for 2 × 3 and 3 × 2 , which equal 6. But what is 2 × 4 ? If we use the repeated addition concept, 2 × 4 = 2 + 2 + 2 + 2 = 6 + 2 . According to the base 7 addition table in the solution for , 6 + 2 = 11 . So, we add that to our table: * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 3 0 3 6 4 0 4 11 5 0 5 6 0 6 Step 3: Next, we need to fill in 2 × 5 . Using repeated addition, 2 × 5 = 2 + 2 + 2 + 2 + 2 = 11 + 2 = 13 if we use our base 7 addition rules. So, we add that to our table: * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 3 0 3 6 4 0 4 11 5 0 5 13 6 0 6 Step 4: Finally, 2 × 6 = 2 + 2 + 2 + 2 + 2 + 2 + 2 = 13 + 2 = 15 . And so we add that to our table: * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 15 3 0 3 6 4 0 4 11 5 0 5 13 6 0 6 15 Step 5: A similar analysis will give us the remainder of the entries. Here is 4 7 × 5 7 demonstrated: 4 7 × 5 7 = 4 7 + 4 7 + 4 7 + 4 7 + 4 7 = 11 7 + 11 7 + 4 7 = 22 7 + 4 7 = 26 7 This is done by using the addition rules from Addition and Subtraction in Base Systems , namely that 4 7 + 4 7 = 11 7 and then applying the addition processes we’ve always known, but with the base 7 table in the solution for . Using those addition rules, the rest of the table is given below: * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 15 3 0 3 6 12 15 21 24 4 0 4 11 15 22 26 33 5 0 5 13 21 26 34 42 6 0 6 15 24 33 42 51 Creating a Multiplication Table for a Base Higher Than 10 Create the multiplication table for base 12. Let’s apply the repeated addition to base 12. Here is 7 12 × 9 12 demonstrated: 7 12 × 9 12 = 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 = 12 12 + 12 12 + 12 12 + 12 12 + 7 12 = 48 12 + 7 12 = 53 12 This is done by using the addition rules from Addition and Subtraction in Base Systems , namely that 7 12 + 7 12 = 12 12 and then applying the addition processes we’ve always known, but with the base 12 table in the solution for . Using those addition rules, the rest of the table is given below: * 0 1 2 3 4 5 6 7 8 9 A B 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 9 A B 2 0 2 4 6 8 A 10 12 14 16 18 1A 3 0 3 6 9 10 13 16 19 20 23 26 29 4 0 4 8 10 14 18 20 24 28 30 34 38 5 0 5 A 13 18 21 26 2B 34 39 42 47 6 0 6 10 16 20 26 30 36 40 46 50 56 7 0 7 12 19 24 2B 36 41 48 53 5A 65 8 0 8 14 20 28 34 40 48 54 60 68 74 9 0 9 16 23 30 39 46 53 60 69 76 83 A 0 A 18 26 34 42 50 5A 68 76 84 92 B 0 B 1A 29 38 47 56 65 74 83 92 A1 Base 12 Multiplication Table The multiplication table in base 2 below is as minimal as the addition table in the solution for . Since the product of 1 with anything is itself, the following multiplication table is found. * 0 1 0 0 0 1 0 1 Base 2 Multiplication Table As with the addition table, we can use the multiplication tables and the addition tables to perform multiplication of two numbers in bases other than base 10. The process is the same, with the same carry rules and placeholder rules. Multiplying in a Base Lower Than 10 Calculate 45 6 × 24 6 . Calculate 101 2 × 110 2 . Step 1: Use the base 6 multiplication table ( ) and, when necessary, the base 6 addition table ( ). Set up this calculation using columns: 4 5 x 2 4 Step 2: Multiply the 1s digits, 5 and 4, using the base 6 multiplication table ( ). There we see the result is 32 6 . So, we enter the 2 and carry the 3. 3 4 5 x 2 4 2 Step 3: So, now we multiply the 4 and the 4, then add the 3 (just as you would do if multiplying two base 10 numbers!). 4 6 × 4 6 = 24 6 (from the base 6 table [ ]), then 24 6 + 3 6 = 31 6 . So, we enter the 31. 3 4 5 x 2 4 3 1 2 Step 4: Now we move on to the 2 in the “tens” place in the bottom value. We multiply the 2 6 and the 5 6 , and we get 14 6 . So, we enter the 4 and carry the 1. Step 5: Next up, we multiply the 2 and the 4, and then add 1. This gives us 12 6 + 1 6 = 13 6 . We enter those on that second line. 1 4 5 x 2 4 3 1 2 1 3 4 0 Step 6: Now we add down the columns. 1 4 5 x 2 4 1 3 1 2 1 3 4 0 2 0 5 2 Step 7: The 3 and the 3 add to 10 in base 6, so we enter the 0 and carry the 1. We now have the result: 45 6 × 24 6 = 2052 6 . Step 1: Use the base 2 multiplication table ( ) and, when necessary, the base 2 addition table in the solution for . Set up this calculation using columns: 1 0 1 x 1 1 0 Step 2: Using the pattern established above, and the processes from multiplication from base 10, we find the following: 1 0 1 x 1 1 0 0 0 0 1 0 1 1 0 1 Step 3: Adding down the columns results in the following: 1 0 1 x 1 1 0 0 0 0 1 0 1 1 0 1 1 1 1 1 0 So, 101 2 × 110 2 = 11110 2 . Summarizing the process of multiplying two numbers in different bases, the multiplication table is referenced. Using that table, the multiplication is carried out in the same manner as it is in base 10. The addition rules for the base will also be referenced when carrying a 1 or when adding the results for each digit’s multiplication line. Multiplying in a Base Higher Than 10 Calculate 3 A 12 × 74 12 . Step 1: Use the base 12 multiplication table in the solution for and, when necessary, the base 12 addition table in the solution for . Set up this calculation using columns: 3 A × 7 4 Step 2: First, the 4 is multiplied by 3A, resulting in the first line. 3 A × 7 4 1 3 4 Step 3: Now we move on to the 7 in the “tens” place in the bottom value. 5 3 A x 7 4 1 3 4 2 2 A 0 Step 4: Now we add down the columns. 3 A x 7 4 1 3 4 2 2 A 0 2 4 1 4 Step 5: The 3 and the A add to 11 in base 12, so we enter the 1 and carry the 1. We now have the result: 3 A 12 × 74 12 = 2414 12 . Division in Bases Other Than 10 Just as with the other operations, division in a base other than 10, the process of division in a base other than 10 is the same as the process when working in base 10. For instance, 72 ÷ 9 = 8 because, we know that 9 × 8 = 72 . So, for many division problems, we are simply looking to the multiplication table to identify the appropriate multiplication rule. Dividing with a Base Other Than 10 Calculate 14 6 ÷ 5 6 . Calculate 5 A 12 ÷ 7 12 Looking at the multiplication table for base 6 ( ), we see that 5 6 × 2 6 = 14 6 . Using that, we know that 14 6 ÷ 5 6 = 2 6 . Looking at the multiplication table for base 12 in the solution for , we see that 7 12 × A 12 = 5 A 12 . Using that, we know that 5 A 12 ÷ 7 12 = A 12 . Errors in Multiplying and Dividing in Bases Other Than Base 10 The types of errors encountered when multiplying and dividing in bases other than base 10 are the same as when adding and subtracting. They often involve applying base 10 rules or symbols to an arithmetic problem in a base other than base 10. The first type of error is using a symbol that is not in the symbol set for the base. Identifying an Illegal Symbol in a Base Other Than Base 10 Explain the error in the following calculation, and determine the correct answer: 4 6 × 2 6 = 8 6 Since the problem is in base 6, the symbol set available is 0, 1, 2, 3, 4, and 5. The 8 in the answer is clearly not a legal symbol for base 6. Looking back to the base 6 multiplication table ( ), we see that 4 6 × 2 6 = 12 6 . The second type of error is using a base 10 rule when the numbers are not in base 10. For instance, in base 17, 6 17 × 9 17 = 54 17 would be incorrect, even though in base 10, 6 × 9 = 54 . That rule doesn’t apply in base 17. Identifying an Error in Arithmetic in a Base Other Than Base 10 Explain the error in the following calculation. Determine the correct answer: 18 12 × 7 12 = 126 12 If this problem was a base 10 problem, this would be the correct answer. However, in base 12, 8 12 × 7 12 is not 56, but is instead 48. To correct this error, carefully use the multiplication table for base 12 ( ). If properly used, the correct answer would be 18 12 × 7 12 = B 8 12 . Check Your Understanding Key Concepts Multiplication tables for bases other than 10 can be built using the same processes that are used in base 10, including using repeated addition and the addition table for the base. Multiplication in bases other than base 10 use the same processes as multiplication in base 10, but use the multiplication table for that base. Basic division in bases other than base 10 use the same processes as basic division in base 10, where the missing factor process is used. Projects Additive Systems Go online. Google “additive number systems.”What system comes up? Describe the additive system you found. Using Google, identify three more additive systems of numbers. Compare and contrast the systems you found. For instance, how many times can a symbol be used before a new symbol is used. Identify three situations where additive systems are still used. Computers and Bases Use Google to determine what base computers use. Were other bases attempted for use in computers? Determine why the base used in computers is appropriate. Determine how the base used in computers is related to the circuitry in computers. Determine how Boolean logic and the base used in computers are related, and might be identical. There is research into using quibits in computers. Find out what quibits are and how can they improve computing speed. Cultures Using Base Systems Other Than 10 Using Google, find three cultures, other than Babylonian or Mayan, that use base systems other than 10. Tell what base is used for each system. If possible, determine why the culture used that base system. Choose one of those systems. Explain that base system. Be sure to address whether the system is additive, place-value based, a blend of the two, and if it employs a zero. History of Zero Using any resources available to you, determine the history of 0 in at least three different numbering systems. Address at least when and why such a development occurred and why a 0 is vital to the use of a positional system. Numbering Systems from Other Global Regions Using any resources available to you, find at least three numbering systems from sub-Saharan Africa, Australia, China, or the Pacific Islands. Explore if they are positional or additive systems (or combinations!), the terminology of the system, if they used a 0, and what base they employed (if positional). Chapter Review Hindu-Arabic Positional System Early Numeration Systems Converting with Base Systems Addition and Subtraction in Base Systems Multiplication and Division in Base Systems Chapter Test", "section": "Multiplication and Division in Base Systems", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction In these algebraic equations, the x represents different numbers. (credit: Larissa Chu, CC BY 4.0) The jump from arithmetic to algebra can be a difficult one for many students. Many students struggle with the idea that mathematics can include situations that aren’t static and do change. In elementary arithmetic, a situation such as: 5 + 3 = ____ is a static situation and will yield the answer of 8 every time. However, a situation such as: 5 x + 3 = ____ can yield many different answers because the answer depends on what amount (number) that x represents. Since the value of x can vary (represent different values), it is known as a variable. Algebra is useful to better model real life situations. In the first equation shown, 5 + 3 = ____ can only model situations where you add those two numbers together. For example, if your uncle gives you five dollars and your aunt gives you three dollars, then you will always receive eight dollars. The second equation 5 x + 3 = ____ can model more complex situations. For example, you wish to buy a game that costs $38 but you only have three dollars. Your uncle will pay you five dollars an hour to work for him. If you’ve worked five hours, have you earned enough money? If not, how many hours will you have to work? Algebra and algebraic thinking open up a world of possibilities that arithmetic alone cannot do.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Algebraic Expressions Two college graduates! (credit: modification of work UC Davis College of Engineering/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Convert between written and symbolic algebraic expressions and equations. Simplify and evaluate algebraic expressions. Add and subtract algebraic expressions. Multiply and divide algebraic expressions. Algebraic expressions are the building blocks of algebra. While a numerical expression (also known as an arithmetic expression) like 5 + 3 can represent only a single number, an algebraic expression such as 5 x + 3 can represent many different numbers. This section will introduce you to algebraic expressions, how to create them, simplify them, and perform arithmetic operations on them. Algebraic Expressions and Equations Xavier and Yasenia have the same birthday, but they were born in different years. This year Xavier is 20 years old and Yasenia is 23, so Yasenia is three years older than Xavier. When Xavier was 15, Yasenia was 18. When Xavier will be 33, Yasenia will be 36. No matter what Xavier’s age is, Yasenia’s age will always be 3 years more. In the language of algebra, we say that Xavier's age and Yasenia's age are variable and the 3 is a constant. The ages change, or vary, so age is a variable . The 3 years between them always stays the same or has the same value, so the age difference is the constant . In algebra, letters of the alphabet are used to represent variables. The letters most often used for variables are x , y , z , a , b , and c . Suppose we call Xavier's age x . Then we could use x + 3 to represent Yasenia's age, as shown in the table below. Xavier’s Age Yasenia’s Age 15 18 20 23 33 36 x x + 3 To write algebraically, we need some symbols as well as numbers and variables. The symbols for the four basic arithmetic operations: addition, subtraction, multiplication, and division are summarized in , along with words we use for the operations and the result. Operation Notation Say: The result is… Addition a + b a plus b The sum of a and b Subtraction a − b a minus b The difference of a and b Multiplication a • b , ( a )( b ), ( a ) b , a ( b ), ab , ba a times b The product of a and b Division a ÷ b , a / b a divided by b The quotient of a and b Symbols for Operations In algebra, the cross symbol ( x ) is normally not used to show multiplication because that symbol could cause confusion. For example, does 3 x y mean 3 × y (three times y ) or 3 • x • y (three times x times y )? To make it clear, use • or parentheses for multiplication. We perform these operations on two numbers. When translating from symbolic form to words, or from words to symbolic form, pay attention to the words of or and to help you find the numbers. The sum of 5 and 3 means add 5 plus 3, which we write as 5 + 3 . The difference of 9 and 2 means subtract 9 minus 2, which we write as 9 − 2 . The product of 4 and 8 means multiply 4 times 8, which we can write as 4 • 8 . The quotient of 20 and 5 means divide 20 by 5, which we can write as 20 ÷ 5 . Translating from Algebra to Words Translate the following algebraic expressions from algebra into words. 12 + 14 ( 30 ) ( 5 ) 64 ÷ 8 x − y According to , this could be translated as 12 plus 14 OR the sum of 12 and 14. According to , this could be translated as 30 times 5 OR the product of 30 and 5. According to , this could be translated as 64 divided by 8 OR the quotient of 64 and 8. According to , this could be translated as x minus y OR the difference of x and y. Translating from Words to Algebra Translate the following phrases from words into algebraic expressions. The difference of 47 and 19 72 divided by 9 The sum of m and n 13 times 7 According to , these words could be translated as 47 − 19 . According to , these words could be translated as 72 ÷ 9 . According to , these words could be translated as m + n . According to , these words could be translated as ( 13 ) ( 7 ) . What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. “Running very fast” is a phrase, but “The football player was running very fast” is a sentence. A sentence has a subject and a verb. In algebra, we have expressions and equations. and used expressions. An expression is like an English phrase. Notice that the English phrases do not form a complete sentence because the phrase does not have a verb. The following table has examples of expressions, which are numbers, variables, or combinations of numbers and variables using operation symbols. Expression Words English Phrase 3 + 5 3 plus 5 The sum of three and five n − 1 n minus one The difference of n and one 6 • 7 6 times 7 The product of six and seven x ÷ y x divided by y The quotient of x and y Translating from an English Phrase to an Expression Translate the following phrases from words into algebraic expressions. Seven more than a number n . A number n times itself. Six times a number n , plus two more. The cost of postage is a flat rate of 10 cents for every parcel, plus 34 cents per ounce x . n + 7 n • n or n 2 6 n + 2 10 + 34 x An equation is two expressions linked with an equal sign (the symbol =). When two quantities have the same value, we say they are equal and connect them with an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb. So, a = b is read “ a is equal to b .” The following table has some examples of equations. Equation English Sentence 3 + 5 = 8 The sum of three and five is equal to eight. n − 1 = 14 n minus one equals fourteen. 6 • 7 = 42 The product of six and seven is equal to forty-two. x = 53 x is equal to fifty-three. y + 9 = 2 y − 3 y plus nine is equal to two times y minus three. Translating from an English Sentence to an Equation Translate the following sentences from words into algebraic equations. Two times x is 6. n plus 2 is equal to n times 3. The quotient of 35 and 7 is 5. Sixty-seven minus x is 56. 2 x = 6 n + 2 = 3 n 35 ÷ 7 = 5 67 − x = 56 The Use of Variables French philosopher and mathematician René Descartes (1596–1650) is usually given credit for the use of the letters x , y , and z to represent unknown quantities in algebra. He introduced these ideas in his publication of La Geometrie , which was printed in 1637. In this publication, he also used the letters a , b , and c to represent known quantities. There is a (possibly fictitious) story that, when the book was being printed for the first time, the printer began to run short of the last three letters of the alphabet. So the printer asked Descartes if it mattered which of x , y , or z were used for the mathematical equations in the book. Descartes decided it made no difference to him; so the printer decided to use x predominantly for the mathematics in the book, because the letters y and z would occur more often in the body of the text (written in French) than the letter x would! This might explain why the letter x is still used today as the most common variable to represent unknown quantities in algebra. Simplifying and Evaluating Algebraic Expressions To simplify an expression means to do all the math possible. For example, to simplify 4 • 2 + 1 we would first multiply 4 • 2 to get 8 and then add 1 to get 9. We have introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values. Consider 2 + 7 • 3 . Do you add first or multiply first? Do you get different answers? Add first: 9 • 3 = 27 Multiply first: 2 + 21 = 23 Which one is correct? Early on, mathematicians realized the need to establish some guidelines when performing arithmetic operations to ensure that everyone would get the same answer. Those guidelines are called the order of operations and are listed in the table below. Step 1: Parentheses and Other Grouping Symbols Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first. Step 2: Exponents Simplify all expressions with exponents. Step 3: Multiplication and Division Perform all multiplication and division in order from left to right. These operations have equal priority. Step 4: Addition and Subtraction Perform all addition and subtraction in order from left to right. These operations have equal priority. You may have heard about Please Excuse My Dear Aunt Sally or PEMDAS. Be careful to notice in Steps 3 and 4 in the table above that multiplication and division, as well as addition and subtraction, happen in order from LEFT to RIGHT. It is possible, for example, to have PEDMAS or PEMDSA. The PEMDAS trick can be misleading if not fully understood! Making a Numerical Equation True Using the Order of Operations Use parentheses to make the following statements true. 17 − 10 + 3 = 10 2 • 26 − 7 = 38 8 + 12 ÷ 5 − 3 = 14 5 + 2 3 • 7 = 91 Add the parentheses around the 17 − 10 . Then you have ( 17 − 10 ) + 3 = 7 + 3 = 10 . Add the parentheses around the 26 − 7 . Then you have 2 • ( 26 − 7 ) = 2 • 19 = 38 . Add the parentheses around the 5 − 3 . Then you have 8 + 12 ÷ ( 5 − 3 ) = 8 + 12 ÷ 2 = 8 + 6 = 14 . Add the parentheses around the 5 + 2 3 . Then you have ( 5 + 2 3 ) • 7 = ( 5 + 8 ) • 7 = 13 • 7 = 91 . In the last example, we simplified expressions using the order of operations. Now we'll evaluate some expressions—again following the order of operations. To evaluate an expression means to find the value of the expression when the variable is replaced by a given number. Evaluating and Simplifying an Expression Evaluate 3 x + 5 when x = 2 . Evaluate x 2 + 3 x + 1 when x = 2 . To evaluate, let x = 2 in the expression, and then simplify: 3 ( 2 ) + 5 = 6 + 5 = 11 . To evaluate, let x = 2 in the expression, and then simplify: 2 2 + 3 ( 2 ) + 1 = 4 + 6 + 1 = 11 . Operations of Algebraic Expressions Algebraic expressions are made up of terms . A term is a constant or the product of a constant and one or more variables. Examples of terms are 7, y , 5 x 2 , 9 a , and b 5 . The constant that multiplies the variable is called the coefficient . Think of the coefficient as the number in front of the variable. Consider the algebraic expressions 5 x 2 , which has a coefficient of 5, and 9 a , which has a coefficient of 9. If there is no number listed in front of the variable, then the coefficient is 1 since x = 1 • x . Some terms share common traits. When two terms are constants or have the same variable and exponent, we say they are like terms . If there are like terms in an expression, you can simplify the expression by combining the like terms. We add the coefficients and keep the same variable. Adding Algebraic Expressions Add ( x 2 + 4 x − 9 ) + ( 3 x 2 − x + 12 ) . Step 1: Add the terms in any order and get the same result (think: 2 + 3 = 3 + 2 ) and drop the parentheses: x 2 + 4 x − 9 + 3 x 2 − x + 12 Step 2: Group like terms together: x 2 + 3 x 2 + 4 x − x − 9 + 12 Step 3: Combine the like terms: 4 x 2 + 3 x + 3 Subtracting Algebraic Expressions Subtract ( 5 x 2 + 4 x − 9 ) − ( 3 x 2 − x + 12 ) . Step 1: Distribute the negative inside the parentheses (think: 2 − ( 3 − 4 ) = 2 − 3 + 4 = − 1 + 4 = 3 , which is the correct answer). You cannot just drop the parentheses (for example, 2 − 3 − 4 = − 1 − 4 = − 5 , which is not correct as we have already verified the answer is 3): 5 x 2 + 4 x − 9 − 3 x 2 + x − 12 Step 2: Group like terms together: 5 x 2 − 3 x 2 + 4 x + x − 9 − 12 Step 3: Combine the like terms: 2 x 2 + x − 21 Before looking at multiplying algebraic expressions we look at the Distributive Property , which says that to multiply a sum, first you multiply each term in the sum and then you add the products. For example, 5 ( 4 + 3 ) = 5 ( 4 ) + 5 ( 3 ) = 20 + 15 = 35 can also be solved as 5 ( 4 + 3 ) = 5 ( 7 ) = 35 . If we use a variable, then 5 ( x + 3 ) = 5 x + 15 . We can extended this example to ( 5 + 2 ) ( 4 + 3 ) = ( 5 ) ( 4 ) + ( 5 ) ( 3 ) + ( 2 ) ( 4 ) + ( 2 ) ( 3 ) = 20 + 15 + 8 + 6 = 49 , which can also be solved as ( 5 + 2 ) ( 4 + 3 ) = ( 7 ) ( 7 ) = 49 . If we use variables, then ( x + 5 ) ( x + 4 ) = ( x ) ( x ) + ( x ) ( 4 ) + ( 5 ) ( x ) + ( 5 ) ( 4 ) = x 2 + 4 x + 5 x + 20 = x 2 + 9 x + 20 . Distributive Property: a ( b + c ) = a b + a c Simplifying an Expression Using the Order of Operations Simplify each expression. ( x − 3 ) 5 ( − 3 ) ( x + y − 2 ) 5 2 ( 7 + 3 ) ( x ) 4 + x • 5 ( 4 + x ) • 5 5 x − 3 • 5 = 5 x − 15 ( − 3 ) • x + ( − 3 ) • y − ( − 3 ) • 2 = − 3 x − 3 y + 6 25 ( 7 + 3 ) ( x ) = 25 ( 10 ) ( x ) = 250 x 4 + 5 x ( 4 ) • ( 5 ) + ( x ) • ( 5 ) = 20 + 5 x Multiplying Algebraic Expressions Multiply ( 4 x − 9 ) ( x + 2 ) . Step 1: Use the Distributive Property: ( 4 x ) ( x ) + ( 4 x ) ( 2 ) − ( 9 ) ( x ) − ( 9 ) ( 2 ) Step 2: Multiply: 4 x 2 + 8 x − 9 x − 18 Step 3: Combine the like terms: 4 x 2 − x − 18 You may have heard the term FOIL which stands for: First, Outer, Inner, Last. FOIL essentially describes a way to use the Distributive Property if you multiply a two-term expression by another two-term expression, but FOIL only works in that specific situation. For example, suppose you have a two-term expression multiplied by a three-term expression, such as ( x + 2 ) ( x + y − 5 ) . What terms qualify as inner terms and what terms qualify as outer terms? In this particular situation, FOIL cannot possibly work; the multiplication of ( x + 2 ) ( x + y − 5 ) should yield six terms, where FOIL is designed to only give you four! The Distributive Property works regardless of how many terms there are. FOIL can be misleading and applied inappropriately if not fully understood! Dividing Algebraic Expressions Divide ( 8 x 2 + 4 x − 16 ) ÷ ( 4 x ) . Divide EACH term by 4 x : ( 8 x 2 ÷ 4 x ) + ( 4 x ÷ 4 x ) − ( 16 ÷ 4 x ) = 2 x + 1 − 4 x Be careful how you divide! Sometimes students incorrectly divide only one term on top by the bottom term. For example, 8 x 2 + 6 x − 3 2 x might turn into 4 x + 3 x − 3 = 7 x − 3 if done incorrectly. When we divide expressions, EACH term is divided by the divisor. So, 8 x 2 + 6 x − 3 2 x = 8 x 2 2 x + 6 x 2 x − 3 2 x = 4 x + 3 − 3 2 x . If you forget, it is always a good idea to check these rules by creating an example using numerical expressions. For example, 9 + 6 + 3 3 = 18 3 = 6 . Dividing each term on top by 3 would yield 9 + 6 + 3 3 = 9 3 + 6 3 + 3 3 = 3 + 2 + 1 = 6 , which is the correct answer. However, if you just divided the 9 on top by the 3 on the bottom, getting 9 + 6 + 3 3 = 3 + 6 + 3 = 12 , this does not result in the correct answer. Al-Khwarizmi Al-Khwarizmi Abu Ja’far Muhammad ibn Musa Al-Khwarizmi was born around 780 AD, probably in or around the region of Khwarizm, which is now part of modern-day Uzbekistan. For most of his adult life, he worked as a scholar at the House of Wisdom in Baghdad, Iraq. He wrote many mathematical works during his life, but is probably most famous for his book Al-kitab al-muhtasar fi hisab al-jabr w’al’muqabalah , which translates to The Condensed Book on the Calculation of al-Jabr ( completion ) and al’muqabalah ( balancing ). The word al-jabr would eventually become the word we use to describe the topic that he was writing about in this book: algebra . From another book of his, with the Latin title Algoritmi de numero Indorum ( Al-Khwarizmi on the Hindu Art of Reckoning ), our word algorithm is derived. In addition to writing on mathematics, Al-Khwarizmi wrote works on astronomy, geography, the sundial, and the calendar. In 2012, Andrew Hacker wrote an opinion piece in the New York Times Magazine suggesting that teaching algebra in high school was a waste of time. Keith Devlin, a British mathematician, was asked to comment on Hacker's article by his students in his Stanford University Continuing Studies course \"Mathematics: Making the Invisible Visible\" on iTunes University. Devlin concludes that Hacker was displaying his ignorance of what algebra is. Q&A: Why We Teach Algebra Check Your Understanding Key Terms variable constant expression equation equal sign term coefficient like terms Distributive Property Key Concepts Algebra is useful because it allows us to understand many situations in real life by modeling them with expressions. Algebraic expressions are the building blocks of algebra. From algebraic expressions we can create algebraic equations. Algebraic expressions are often simplified and evaluated using the four arithmetic operations. Videos Q&A: Why We Teach Algebra Formulas Distributive Property: a ( b + c ) = a b + a c", "section": "Algebraic Expressions", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Linear Equations in One Variable with Applications Most gyms have a monthly membership fee. (credit: modification of work \"Morning PT after the Holidays 2021\" by Fort Drum & 10th Mountain Division (LI)/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Solve linear equations in one variable using properties of equations. Construct a linear equation to solve applications. Determine equations with no solution or infinitely many solutions. Solve a formula for a given variable. In this section, we will study linear equations in one variable. There are several real-world scenarios that can be represented by linear equations: taxi rentals with a flat fee and a rate per mile; cell phone bills that charge a monthly fee plus a separate rate per text; gym memberships with a monthly fee plus a rate per class taken; etc. For example, if you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month? Linear Equations and Applications Solving any equation is like discovering the answer to a puzzle. The purpose of solving an equation is to find the value or values of the variable that makes the equation a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle! There are many types of equations that we will learn to solve. In this section, we will focus on a linear equation , which is an equation in one variable that can be written as a x + b = 0 where a and b are real numbers and a ≠ 0 , such that a is the coefficient of x and b is the constant. To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the Example 5.12, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier. Solving a Linear Equation Using a General Strategy Solve 7 ( n − 3 ) − 8 = − 15 Step 1: Simplify each side of the equation as much as possible. Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible. 7 ( n − 3 ) − 8 = − 15 7 n − 21 − 8 = − 15 7 n − 29 = − 15 Step 2: Collect all variable terms on one side of the equation. Nothing to do; all n -terms are on the left side. 7 n − 29 = − 15 Step 3: Collect constant terms on the other side of the equation. To get constants only on the right, add 29 to each side. Simplify. 7 n − 29 + 29 = − 15 + 29 7 n = 14 Step 4: Make the coefficient of the variable term equal to 1. Divide each side by 7. Simplify. 7 n 7 = 14 7 n = 2 Step 5: Check the solution. Let n = 2 Subtract. Check: 7 ( n − 3 ) − 8 = − 15 7 ( 2 − 3 ) − 8 = ? − 15 7 ( − 1 ) − 8 = ? − 15 − 7 − 8 = ? − 15 − 15 = − 15 ✓ In , we used both the addition and division property of equations. All the properties of equations are summarized in table below. Basically, what you do to one side of the equation, you must do to the other side of the equation to preserve equality. Operation Property Example Addition If a = b Then a + c = b + c 2 = 2 2 + 3 = 2 + 3 5 = 5 Subtraction If a = b Then a − c = b − c 5 = 5 5 − 2 = 5 − 2 3 = 3 Multiplication If a = b Then a • c = b • c 3 = 3 3 • 4 = 3 • 4 12 = 12 Division If a = b Then a ÷ c = b ÷ c for c ≠ 0 8 = 8 8 ÷ 2 = 8 ÷ 2 4 = 4 Be careful to multiply and divide every term on each side of the equation. For example, 2 + x = x 3 is solved by multiplying BOTH sides of the equation by 3 to get 3 ( 2 + x ) = 3 ( x 3 ) which gives 6 + 3 x = x . Using parentheses will help you remember to use the distributive property! A division example, such as 3 ( x + 2 ) = 6 x + 9 , can be solved by dividing BOTH sides of the equation by 3 to get 3 ( x + 2 ) 3 = 6 x + 9 3 , which then will lead to x + 2 = 2 x + 3 . Solving a Linear Equation Using Properties of Equations Solve 9 ( y − 2 ) − y = 16 + 7 y . Step 1: Simplify each side. 9 ( y − 2 ) − y = 16 + 7 y 9 y − 18 − y = 16 + 7 y 8 y − 18 = 16 + 7 y Step 2: Collect all variables on one side. 8 y − 18 − 7 y = 16 + 7 y − 7 y y − 18 = 16 Step 3: Collect constant terms on one side. y − 18 + 18 = 16 + 18 y = 34 Step 4: Make the coefficient of the variable 1. Already done! Step 5: Check. 9 ( 34 ) − 18 − ( 34 ) = ? 16 + 7 ( 34 ) 306 − 18 − 34 = ? 16 + 238 288 − 34 = ? 254 254 = 254 ✓ Who Invented the Symbol for Equals ? Before the creation of a symbol for equality, it was usually expressed with a word that meant equals, such as aequales (Latin), esgale (French), or gleich (German). Welsh mathematician and physician Robert Recorde is given credit for inventing the modern sign. It first appears in writing in The Whetstone of Witte , a book Recorde wrote about algebra, which was published in 1557. In this book, Recorde states, \"I will set as I do often in work use, a pair of parallels, or Gemowe (twin) lines of one length, thus: ===, because no two things can be more equal.\" Although his version of the sign was a bit longer than the one we use today, his idea stuck and \"=\" is used throughout the world to indicate equality in mathematics. In Algebraic Expressions , you translated an English sentence into an equation. In this section, we take that one step further and translate an English paragraph into an equation, and then we solve the equation. We can go back to the opening question in this section: If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month? We can create an equation for this scenario and then solve the equation (see ). Constructing a Linear Equation to Solve an Application The Beaudrie family has two cats, Basil and Max. Together, they weigh 23 pounds. Basil weighs 16 pounds. How much does Max weigh? Let b = Basil’s weight and m = Max’s weight. b + m = 23 We also know that Basil weighs 16 pounds so: Steps 1 and 2: 16 + m = 23 Since both sides are simplified, the variable is on one side of the equation, we start in Step 3 and collect the constants on one side: Step 3: 16 + m − 16 = 23 − 16 m = 7 Step 4: is already done so we go to Step 5: Step 5: 16 + 7 = ? 23 23 = 23 ✓ Basil weighs 16 pounds and Max weighs 7 pounds. Constructing a Linear Equation to Solve Another Application If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month? If we let x = number of classes, the expression 5 x + 10 would represent what you pay per month if each class is $5 and there’s a $10 monthly fee per class. $10 is your constant. If you want to know how many classes you can take if you have a $75 monthly gym budget, set the equation equal to 75. Then solve the equation 5 x + 10 = 75 for x . Steps 1 and 2: 5 x + 10 = 75 Step 3: 5 x + 10 − 10 = 75 − 10 5 x = 65 Step 4: 5 x 5 = 65 5 x = 13 Step 5: 5 ( 13 ) + 10 = ? 75 65 + 10 = ? 75 75 = 75 ✓ The solution is 13 classes. You can take 13 classes on a $75 monthly gym budget. Constructing an Application from a Linear Equation Write an application that can be solved using the equation 50 x + 35 = 185 . Then solve your application. Answers will vary. Let’s say you want to rent a snowblower for a huge winter storm coming up. If x = the number of days you rent a snowblower, then the expression 50 x + 35 represents what you pay if, for each day, it costs $50 to rent the snowblower and there is a $35 flat rental fee. $35 is the constant. To find out how many days you can rent a snowblower for $185, set the expression equal to 185. Then solve the equation 50 x + 35 = 185 for x . Steps 1 and 2: 50 x + 35 = 185 Step 3: 50 x + 35 − 35 = 185 − 35 50 x = 150 Step 4: 50 x 50 = 150 50 x = 3 Step 5: 50 ( 3 ) + 35 = ? 185 150 + 35 = ? 185 185 = 185 ✓ The equation is 50 x + 35 = 185 and the solution is 3 days. You can rent a snowblower for 3 days on a $185 budget. Linear Equations with No Solutions or Infinitely Many Solutions Every linear equation we have solved thus far has given us one numerical solution. Now we'll look at linear equations for which there are no solutions or infinitely many solutions. Solving a Linear Equation with No Solution Solve 3 ( x + 4 ) = 4 x + 8 − x . Step 1: Simplify each side. 3 ( x + 4 ) = 4 x + 8 − x 3 x + 12 − 3 x = 3 x + 8 − 3 x Step 2: Collect all variables to one side. 3 x + 12 − 3 x = 3 x + 8 − 3 x 12 = 8 The variable x disappeared! When this happens, you need to examine what remains. In this particular case, we have 12 = 8 , which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for x that can be put into the equation that will make it true. Solving a Linear Equation with Infinitely Many Solutions Solve 2 ( x + 5 ) = 4 ( x + 3 ) − 2 x − 2 . Step 1: 2 ( x + 5 ) = 4 ( x + 3 ) − 2 x − 2 2 x + 10 = 4 x + 12 − 2 x − 2 2 x + 10 = 2 x + 10 Step 2: 2 x + 10 − 2 x = 2 x + 10 − 2 x 10 = 10 As with the previous example, the variable disappeared. In this case, however, we have a true statement ( 10 = 10 ). When this occurs we say there are infinitely many solutions; any value for x will make this statement true. Solving a Formula for a Given Variable You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be able to manipulate formulas and solve for specific variables. To solve a formula for a specific variable means to isolate that variable on one side of the equal sign with a coefficient of 1. All other variables and constants are on the other side of the equal sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula. Solving for a Given Variable with Distance, Rate, and Time Solve the formula d = r t for t . This is the distance formula where d = distance, r = rate, and t = time. Divide both sides by r : d / r = r t / r d / r = t Solving for a Variable in an Equation Solving for a Given Variable in the Area Formula for a Triangle Solve the formula A = ½ bh for h . This is the area formula of a triangle where A = area, b = base, and h = height. Step 1: Multiply both sides by 2. 2 A = 2 ( ½ b h ) 2 A = b h Step 2: Divide both sides by b . 2 A b = b h b 2 A b = h h = 2 A b Using Algebra to Understand Card Tricks You will need to perform this card trick with another person. Before you begin, the two people must first decide which of the two will be the Dealer and which will be the Partner , as each will do something different. Once you have decided upon that, follow the steps here: Step 1: Dealer and Partner: Take a regular deck of 52 cards, and remove the face cards and the 10s. Step 2: Dealer and Partner: Shuffle the remaining cards Step 3: Dealer and Partner: Select one card each, but keep them face down and don’t look at them yet. Step 4: Dealer: Look at your card (just the Dealer!). Multiply its value by 2 (Aces = 1). Step 5: Dealer: Add 2 to this result. Step 6: Dealer: Multiply your answer by 5. Step 7: Partner: Look at your card. Step 8: Partner: Calculate: 10 - your card, and tell this information to the dealer. Step 9: Dealer: Subtract the value the Partner tells you from your total to get a final answer. Step 10: Dealer: verbally state the final answer. Step 11: Dealer and Partner: Turn over your cards. Now, answer the following questions Did the trick work? How do you know? Why did this occur? In other words, how does this trick work? Check Your Understanding Key Terms linear equation Key Concepts Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions. When solving a linear equation, most often you will have one solution; however, a linear equation may have no solutions or infinitely many solutions. Videos Solving for a Variable in an Equation", "section": "Linear Equations in One Variable with Applications", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Linear Inequalities in One Variable with Applications These poll results, showing a margin of error at 4 percent, are an example of a real-world scenario that can be represented by linear inequalities. Learning Objectives After completing this section, you should be able to: Graph inequalities in one variable. Solve linear inequalities in one variable. Construct a linear inequality to solve applications. In this section, we will study linear inequalities in one variable. Inequalities can be used when the possible values (answers) in a certain situation are numerous, not just a few, or when the exact value (answer) is not known but it is known to be within a range of possible values. There are many real-world scenarios that can be represented by linear inequalities. For example, consider the survey of the mayoral election in Surveys and polls are usually conducted with only a small group of people. The margin of error indicates a range of how the actual group of voters would vote given the results of the survey. This range can be expressed using inequalities. Another example involves college tuition. Say a local community college charges $113 per credit hour. You budget $1,500 for tuition this fall semester. What are the number of credit hours that you could take this fall? Since this answer could be many different values, it can be expressed as an inequality. Graphing Inequalities on the Number Line In Algebraic Expressions , we introduced equality and the = symbol. In this section, we look at inequality and the symbols < , > , ≤ , and ≥ . The table below summarizes the symbols and their meaning. Symbol Meaning < less than > greater than ≤ less than or equal to ≥ greater than or equal to Suppose you had the inequality statement x > 3 . What possible number or numbers would make the inequality x > 3 true? If you are thinking, \" x could be 4,\" that's correct, but x could also be 5, 6, 37, 1 million, or even 3.001. The number of solutions is infinite; any number greater than 3 is a solution to the inequality x > 3 . Rather than trying to list all possible solutions, we show all the solutions to the inequality x > 3 on the number line. All the numbers to the right of 3 on the number line are shaded, to show that all numbers greater than 3 are solutions. At the number 3 itself, an open parenthesis is drawn, since the number 3 is not part of the solutions of x > 3 . We can also represent inequalities using interval notation. There is no upper end to the solution to this inequality. In interval notation, we express x > 3 as ( 3 , ∞ ) . The symbol ∞ is read as \"infinity.\" Infinity is not an actual number. shows both the number line and the interval notation for x > 3 . The inequality x > 3 is graphed on this number line and written in interval notation. We used the left parenthesis symbol to show that the endpoint of the inequality is not included. Parentheses are used when the endpoints are not included as a possible answer to the inequality. The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals. The inequality x ≤ 1 means all numbers less than or equal to 1. To illustrate that solution on a number line, we first put a bracket at x = 1 ; brackets are used when the endpoint is included. We then shade in all the numbers to the left of 1, to show that all numbers less than one are solutions. There is no lower end to those numbers. We write x ≤ 1 in interval notation as ( − ∞ , 1 ] . The symbol − ∞ is read as \"negative infinity.\" shows both the number line and interval notation for x = 1 . The inequality x ≤ 1 is graphed on this number line and written in interval notation. summarizes the general representations in both number line form and interval notation of solutions for x > a , x < a , x ≥ a , and x ≤ a . Summary of representations in number line form and interval notation. Graphing an Inequality Graph the inequality x ≥ − 3 and write the solution in interval notation. Shade to the right of − 3 to show all the numbers greater than − 3 , and put a bracket at − 3 to show that the numbers are greater than or equal to − 3 ( ) Write in interval notation starting at − 3 with a bracket to show that − 3 is included in the solution and then infinity because the solution includes all the numbers greater than or equal to − 3 : [ − 3 , ∞ ) Graphing a Compound Inequality Graph the inequality x > − 3 and x < 4 and write the solution in interval notation. Step 1: Graph x > − 3 ( ). Step 2: Graph x < 4 ( ). Step 3: Graph both on the same number line and think of where the solutions are to BOTH inequalities . This will be where BOTH are shaded. Step 4: Write the solution in interval notation: ( − 3 , 4 ) Where Did the Inequality Symbols Come From? The first use of the < symbol to represent \"less than\" and > to represent \"greater than\" appeared in a mathematics book written by Englishman Thomas Harriot that was published in 1631. However, Harriot did not invent the symbols…the editor of the book did! Harriot used triangular symbols to represent less than and greater than; the editor, for reasons unknown, changed to symbols that are similar to the ones we use today. The symbols used to represent less than or equal to, and greater than or equal to ( ≤ and ≥ ) were first used in 1731 by French hydrologist and surveyor Pierre Bouguer. Interestingly, English mathematician John Wallis had used similar symbols as early as 1670, but he put the bar above the less than and greater than symbols instead of below them. Solving Linear Inequalities A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms a x + b < c , a x + b ≤ c , a x + b ≥ c , or a x + b > c , where a , b , and c are all real numbers. When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities. We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example, we know that 2 is less than 4, i.e., 2 < 4 . If we add 6 to both sides of this inequality, we still have a true statement: 2 + 6 < 4 + 6 8 < 10 The same would happen if we subtracted 6 from both sides of the inequality; the statement would stay true: 2 − 6 < 4 − 6 − 4 < − 2 Notice that the inequality signs stayed the same. This leads us to the Addition and Subtraction Properties of Inequality. For any numbers a , b , and c , if a < b , then a + c < b + c and a − c < b − c . For any numbers a , b , and c , if a > b , then a + c > b + c and a - c > b - c . We can add or subtract the same quantity from both sides of an inequality and still keep the inequality the same. But what happens to an inequality when we divide or multiply both sides by a number? Let's first multiply and divide both sides by a positive number, starting with an inequality we know is true, 10 < 15 . We will multiply and divide this inequality by 5: 10 < 15 10 < 15 10 ( 5 ) ? 15 ( 5 ) 10 5 ? 15 5 50 ? 75 2 ? 3 50 < 75 ( true ) 2 < 3 ( true ) The inequality signs stayed the same. Does the inequality stay the same when we divide or multiply by a negative number? Let's use our inequality 10 < 15 to find out, multiplying it and dividing it by − 5 : 10 < 15 10 < 15 10 ( − 5 ) ? 15 ( − 5 ) 10 − 5 ? 15 − 5 − 50 ? − 75 − 2 ? − 3 − 50 > − 75 ( true ) − 2 > − 3 ( true ) Notice that when we filled in the inequality signs, the inequality signs reversed their direction in order to make it true! To summarize, when we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses. This gives us the Multiplication and Division Property of Inequality. For any numbers a , b , and c , multiply or divide by a positive: if a < b and c > 0 , then a c < b c and a c < b c if a > b and c > 0 , then a c > b c and a c > b c multiply or divide by a negative: if a < b and c < 0 , then a c > b c and a c > b c if a > b and c < 0 , then a c < b c and a c < b c To summarize, when we divide or multiply an inequality by: a positive number, the inequality sign stays the same. a negative number, the inequality sign reverses. Be careful to only reverse the inequality sign when you are multiplying and dividing by a negative. You do NOT reverse the inequality sign when you add or subtract a negative. For example, 2 x < − 4 is solved by dividing both sides of the inequality by 2 to get x < − 2 . You do NOT reverse the inequality sign because there is a negative 4. As another example, − 2 x + 5 < 3 x is solved by adding − 2 x to both sides to get 5 < 5 x . This does not reverse the inequality sign because we were not multiplying or dividing by a negative. We then divide both sides by 5 and get 1 < x . Solving a Linear Inequality Using One Operation Solve 9 y < 54 , graph the solution on the number line, and write the solution in interval notation. 9 y < 54 9 y 9 < 54 9 y < 6 Solving a Linear Inequality Using Multiple Operations Solve the inequality 6 y ≤ 11 y + 17 , graph the solution on the number line, and write the solution in interval notation. 6 y ≤ 11 y + 17 6 y − 11 y ≤ 11 y + 17 − 11 y − 5 y ≤ 17 − 5 y − 5 ≥ − 17 5 y ≥ − 17 5 Solving Applications with Linear Inequalities Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality. Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down. Constructing a Linear Inequality to Solve an Application with Tablet Computers A teacher won a mini grant of $4,000 to buy tablet computers for their classroom. The tablets they would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets the teacher can buy? Let t = the number of tablets . t times $254.12 has to be less than $4,000, so 254.12 t ≤ 4 , 000 . Solve for t : 254.12 t 254.12 ≤ 4,000 254.12 t ≤ 15.74 The teacher can buy 15 tablets and stay under $4,000. Constructing a Linear Inequality to Solve a Tuition Application The local community college charges $113 per credit hour. Your budget is $1,500 for tuition this fall semester. What number of credit hours could you take this fall? Let c = the number of credit hours you could take. c times $113 has to be less than $1,500, so 113 c ≤ 1 , 500 . Solve for c : 113 c 113 ≤ 1500 113 c ≤ 13.27 You can take up to 13 credits and stay under $1,500. Constructing a Linear Inequality to Solve an Application with Travel Costs Brenda’s best friend is having a destination wedding and the event will last 3 days and 3 nights. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 for airfare, $375 for food and entertainment, and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip? Let b = number of babysitting hours. b times $15 plus $500 has to be more than $ 350 + $ 375 + $ 60 / night , so 15 b + 500 ≥ 350 + 375 + 60 ( 3 ) . Solve for b : 15 b + 500 − 500 ≥ 905 − 500 15 b ≥ 405 15 b 15 ≥ 405 15 b ≥ 27 Brenda must babysit at least 27 hours. The Desmos activities called \"Inequalities on a Number Line\" and \"Compound Inequalities on a Number Line\" are ways for students to develop and deepen their understanding of inequalities. Teachers will need a Desmos account to assign the activity for student use. Once they have assigned the activity to their students, teachers need to share the code for the activity with their students. Students will input the code to work on the activity. Check Your Understanding Key Terms linear inequality Addition and Subtraction Property of Linear Inequalities Multiplication and Division Property of Linear Inequalities Key Concepts Inequalities can be used when the possible values (answers) in a certain situation are numerous, or when the exact value (answer) is not known, but it is known to be within a range of possible values. Linear inequalities can be represented using a number line or using interval notation. Formulas For any numbers a , b , and , if a < b , then a + c < b + c and a − c < b − c . For any numbers a , b , and c , if a > b , then a + c > b + c and a − c > b − c . For any numbers a , b , and c , multiply or divide by a positive: if a < b and c > 0 , then a c < b c and a c < b c if a > b and c > 0 , then a c > b c and a c > b c multiply or divide by a negative: if a < b and c < 0 , then a c > b c and a c > b c if a > b and c < 0 , then a c < b c and a c < b c", "section": "Linear Inequalities in One Variable with Applications", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Ratios and Proportions This bar graph shows popular social media app usage. ( Source ) Learning Objectives After completing this section, you should be able to: Construct ratios to express comparison of two quantities. Use and apply proportional relationships to solve problems. Determine and apply a constant of proportionality. Use proportions to solve scaling problems. Ratios and proportions are used in a wide variety of situations to make comparisons. For example, using the information from , we can see that the number of Facebook users compared to the number of Twitter users is 2,006 M to 328 M. Note that the \"M\" stands for million, so 2,006 million is actually 2,006,000,000 and 328 million is 328,000,000. Similarly, the number of Qzone users compared to the number of Pinterest users is in a ratio of 632 million to 175 million. These types of comparisons are ratios. Constructing Ratios to Express Comparison of Two Quantities Note there are three different ways to write a ratio , which is a comparison of two numbers that can be written as: a to b OR a : b OR the fraction a / b . Which method you use often depends upon the situation. For the most part, we will want to write our ratios using the fraction notation. Note that, while all ratios are fractions, not all fractions are ratios. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only. Expressing the Relationship between Two Currencies as a Ratio The Euro (€) is the most common currency used in Europe. Twenty-two nations, including Italy, France, Germany, Spain, Portugal, and the Netherlands use it. On June 9, 2021, 1 U.S. dollar was worth 0.82 Euros. Write this comparison as a ratio. Using the definition of ratio, let a = 1 U.S. dollar and let b = 0.82 Euros. Then the ratio can be written as either 1 to 0.82; or 1:0.82; or 1 0.82 . Expressing the Relationship between Two Weights as a Ratio The gravitational pull on various planetary bodies in our solar system varies. Because weight is the force of gravity acting upon a mass, the weights of objects is different on various planetary bodies than they are on Earth. For example, a person who weighs 200 pounds on Earth would weigh only 33 pounds on the moon! Write this comparison as a ratio. Using the definition of ratio, let a = 200 pounds on Earth and let b = 33 pounds on the moon. Then the ratio can be written as either 200 to 33; or 200:33; or 200 33 . Using and Applying Proportional Relationships to Solve Problems Using proportions to solve problems is a very useful method. It is usually used when you know three parts of the proportion, and one part is unknown. Proportions are often solved by setting up like ratios. If a b and c d are two ratios such that a b = c d , then the fractions are said to be proportional . Also, two fractions a b and c d are proportional ( a b = c d ) if and only if a × d = b × c . Solving a Proportion Involving Two Currencies You are going to take a trip to France. You have $520 U.S. dollars that you wish to convert to Euros. You know that 1 U.S. dollar is worth 0.82 Euros. How much money in Euros can you get in exchange for $520? Step 1: Set up the two ratios into a proportion; let x be the variable that represents the unknown. Notice that U.S. dollar amounts are in both numerators and Euro amounts are in both denominators. 1 0.82 = 520 x Step 2: Cross multiply, since the ratios a b and c d are proportional, then a × d = b × c . 520 ( 0.82 ) = 1 ( x ) 426.4 = x You should receive 426.4 Euros ( 426.4 € ) . Solving a Proportion Involving Weights on Different Planets A person who weighs 170 pounds on Earth would weigh 64 pounds on Mars. How much would a typical racehorse (1,000 pounds) weigh on Mars? Round your answer to the nearest tenth. Step 1: Set up the two ratios into a proportion. Notice the Earth weights are both in the numerator and the Mars weights are both in the denominator. 170 64 = 1,000 x Step 2: Cross multiply, and then divide to solve. 170 x = 1,000 ( 64 ) 170 x = 64,000 170 x 170 = 64,000 170 x = 376.5 So the 1,000-pound horse would weigh about 376.5 pounds on Mars. Solving a Proportion Involving Baking A cookie recipe needs 2 1 4 cups of flour to make 60 cookies. Jackie is baking cookies for a large fundraiser; she is told she needs to bake 1,020 cookies! How many cups of flour will she need? Step 1: Set up the two ratios into a proportion. Notice that the cups of flour are both in the numerator and the amounts of cookies are both in the denominator. To make the calculations more efficient, the cups of flour ( 2 1 4 ) is converted to a decimal number (2.25). 2.25 60 = x 1020 Step 2: Cross multiply, and then simplify to solve. 2.25 ( 1,020 ) = 60 x 2,295 = 60 x 38.25 = x Jackie will need 38.25, or 38 1 4 , cups of flour to bake 1,020 cookies. Part of the definition of proportion states that two fractions a b and c d are proportional if a × d = b × c . This is the \"cross multiplication\" rule that students often use (and unfortunately, often use incorrectly). The only time cross multiplication can be used is if you have two ratios (and only two ratios) set up in a proportion. For example, you cannot use cross multiplication to solve for x in an equation such as 2 5 = x 8 + 3 x because you do not have just the two ratios. Of course, you could use the rules of algebra to change it to be just two ratios and then you could use cross multiplication, but in its present form, cross multiplication cannot be used. Eudoxus was born around 408 BCE in Cnidus (now known as Knidos) in modern-day Turkey. As a young man, he traveled to Italy to study under Archytas, one of the followers of Pythagoras. He also traveled to Athens to hear lectures by Plato and to Egypt to study astronomy. He eventually founded a school and had many students. Eudoxus made many contributions to the field of mathematics. In mathematics, he is probably best known for his work with the idea of proportions. He created a definition of proportions that allowed for the comparison of any numbers, even irrational ones. His definition concerning the equality of ratios was similar to the idea of cross multiplying that is used today. From his work on proportions, he devised what could be described as a method of integration, roughly 2000 years before calculus (which includes integration) would be fully developed by Isaac Newton and Gottfried Leibniz. Through this technique, Eudoxus became the first person to rigorously prove various theorems involving the volumes of certain objects. He also developed a planetary theory, made a sundial still usable today, and wrote a seven volume book on geography called Tour of the Earth , in which he wrote about all the civilizations on the Earth, and their political systems, that were known at the time. While this book has been lost to history, over 100 references to it by different ancient writers attest to its usefulness and popularity. Determining and Applying a Constant of Proportionality In the last example, we were given that 2 1 4 cups of flour could make 60 cookies; we then calculated that 38 1 4 cups of flour would make 1,020 cookies, and 720 cookies could be made from 27 cups of flour. Each of those three ratios is written as a fraction below (with the fractions converted to decimals). What happens if you divide the numerator by the denominator in each? 2.25 60 = 0.0375 38.25 1,020 = 0.0375 27 720 = 0.0375 . The quotients in each are exactly the same! This number, determined from the ratio of cups of flour to cookies, is called the constant of proportionality . If the values a and b are related by the equality a b = k , then k is the constant of proportionality between a and b . Note since a b = k , then b = a k . and b = a k . One piece of information that we can derive from the constant of proportionality is a unit rate. In our example (cups of flour divided by cookies), the constant of proportionality is telling us that it takes 0.0375 cups of flour to make one cookie. What if we had performed the calculation the other way (cookies divided by cups of flour)? 60 2.25 = 26.66666... 1,020 38.25 = 26.66666... 720 27 = 26.66666... In this case, the constant of proportionality ( 26.66666 … = 26 2 3 ) is telling us that 26 2 3 cookies can be made with one cup of flour. Notice in both cases, the \"one\" unit is associated with the denominator. The constant of proportionality is also useful in calculations if you only know one part of the ratio and wish to find the other. Finding a Constant of Proportionality Isabelle has a part-time job. She kept track of her pay and the number of hours she worked on four different days, and recorded it in the table below. What is the constant of proportionality, or pay divided by hours? What does the constant of proportionality tell you in this situation? Pay $87.50 $50.00 $37.50 $100.00 Hours 7 4 3 8 To find the constant of proportionality, divide the pay by hours using the information from any of the four columns. For example, 87.5 7 = 12.5 . The constant of proportionality is 12.5, or $12.50. This tells you Isabelle's hourly pay: For every hour she works, she gets paid $12.50. Applying a Constant of Proportionality: Running mph Zac runs at a constant speed: 4 miles per hour (mph). One day, Zac left his house at exactly noon (12:00 PM) to begin running; when he returned, his clock said 4:30 PM. How many miles did he run? The constant of proportionality in this problem is 4 miles per hour (or 4 miles in 1 hour). Since a b = k , where k is the constant of proportionality, we have a miles b hours = k a 4 .5 = 4 (30 minutes is ½ , or 0.5 , hours) a = 4 ( 4.5 ) , since from the definition we know a = k b a = 18 Zac ran 18 miles. Applying a Constant of Proportionality: Filling Buckets Joe had a job where every time he filled a bucket with dirt, he was paid $2.50. One day Joe was paid $337.50. How many buckets did he fill that day? The constant of proportionality in this situation is $2.50 per bucket (or $2.50 for 1 bucket). Since a b = k , where k is the constant of proportionality, we have a dollars b buckets = k 337.50 b = 2.50 Since we are solving for b , and we know from the definition that b = a k : b = 337.50 2.50 b = 135 Joe filled 135 buckets. Applying a Constant of Proportionality: Miles vs. Kilometers While driving in Canada, Mabel quickly noticed the distances on the road signs were in kilometers, not miles. She knew the constant of proportionality for converting kilometers to miles was about 0.62—that is, there are about 0.62 miles in 1 kilometer. If the last road sign she saw stated that Montreal is 104 kilometers away, about how many more miles does Mabel have to drive? Round your answer to the nearest tenth. The constant of proportionality in this situation is 0.62 miles per 1 kilometer. Since a b = k , where k is the constant of proportionality, we have a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48 Rounding the answer to the nearest tenth, Mabel has to drive 64.5 miles. Using Proportions to Solve Scaling Problems A map of the northeastern United States Ratio and proportions are used to solve problems involving scale . One common place you see a scale is on a map (as represented in ). In this image, 1 inch is equal to 200 miles. This is the scale. This means that 1 inch on the map corresponds to 200 miles on the surface of Earth. Another place where scales are used is with models: model cars, trucks, airplanes, trains, and so on. A common ratio given for model cars is 1:24—that means that 1 inch in length on the model car is equal to 24 inches (2 feet) on an actual automobile. Although these are two common places that scale is used, it is used in a variety of other ways as well. Solving a Scaling Problem Involving Maps is an outline map of the state of Colorado and its counties. If the distance of the southern border is 380 miles, determine the scale (i.e., 1 inch = how many miles). Then use that scale to determine the approximate lengths of the other borders of the state of Colorado. Outline Map of Colorado (credit: \"Map of Colorado Counties\" by David Benbennick/Wikimedia Commons, Public Domain) When the southern border is measured with a ruler, the length is 4 inches. Since the length of the border in real life is 380 miles, our scale is 1 inch = 95 miles. The eastern and western borders both measure 3 inches, so their lengths are about 285 miles. The northern border measures the same as the southern border, so it has a length of 380 miles. Solving a Scaling Problem Involving Model Cars Die-cast NASCAR model cars are said to be built on a scale of 1:24 when compared to the actual car. If a model car is 9 inches long, how long is a real NASCAR automobile? Write your answer in feet. The scale tells us that 1 inch of the model car is equal to 24 inches (2 feet) on the real automobile. So set up the two ratios into a proportion. Notice that the model lengths are both in the numerator and the NASCAR automobile lengths are both in the denominator. 1 24 = 9 x 24 ( 9 ) = x 216 = x This amount (216) is in inches. To convert to feet, divide by 12, because there are 12 inches in a foot (this conversion from inches to feet is really another proportion!). The final answer is: 216 12 = 18 The NASCAR automobile is 18 feet long. Check Your Understanding Key Terms ratio proportion constant of proportionality scale construct ratios solve proportions use proportions to solve scaling problems Key Concepts A ratio is a comparison of two numbers. The ratio of two numbers a and b can be written as: a to b OR a : b OR the fraction a / b . All fractions are ratios, but not all ratios are fractions. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only. When two ratios are equal, we say they are in proportion or are proportional. Setting up proportions allows us to solve many various situations where three of the four values of the proportion are known.", "section": "Ratios and Proportions", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Graphing Linear Equations and Inequalities How much would it cost to fill up your gas tank? (credit: \"Gas Under 4 Bucks\" by Mark Turnauckas, Flickr/CC BY 2.0) Learning Objectives After completing this section, you should be able to: Graph linear equations and inequalities in two variables. Solve applications of linear equations and inequalities. In this section, we will learn how to graph linear equations and inequalities. There are several real-world scenarios that can be represented by graphs of linear inequalities. Think of filling your car up with gasoline. If gasoline is $3.99 per gallon and you put 10 gallons in your car, you will pay $39.90. Your friend buys 15 gallons of gasoline and pays $59.85. You can plot these points on a coordinate system and connect the points with a line to create the graph of a line. You'll learn to do both in this section. Plotting Points on a Rectangular Coordinate System Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the x y -plane or the “coordinate plane.” The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the x -axis. The vertical number line is called the y -axis. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See . Quadrants on the Coordinate Plane In the rectangular coordinate system, every point is represented by an ordered pair ( ). The first number in the ordered pair is the x -coordinate of the point, and the second number is the y -coordinate of the point. The phrase \"ordered pair\" means that the order is important. At the point where the axes cross and where both coordinates are zero, the ordered pair is ( 0 , 0 ) . The point ( 0 , 0 ) has a special name. It is called the origin . Ordered Pair We use the coordinates to locate a point on the xy -plane. Let's plot the point ( 1 , 3 ) as an example. First, locate 1 on the x -axis and lightly sketch a vertical line through x = 1 . Then, locate 3 on the y -axis and sketch a horizontal line through y = 3 . Now, find the point where these two lines meet—that is the point with coordinates ( 1 , 3 ) . See . Point ( 1 , 3 ) Plotted on the Coordinate Plane Notice that the vertical line through x = 1 and the horizontal line through y = 3 are not part of the graph. The dotted lines are just used to help us locate the point ( 1 , 3 ) . When one of the coordinates is zero, the point lies on one of the axes. In , the point ( 0 , 4 ) is on the y -axis and the point (−2, 0) is on the x -axis. Points ( − 2 , 0 ) and ( 0 , 4 ) Plotted on the Coordinate Plane Plotting Points on a Coordinate System Plot the following points in the rectangular coordinate system and identify the quadrant in which the point is located: ( − 5 , 4 ) ( − 3 , − 4 ) ( 2 , − 3 ) ( 0 , − 1 ) ( 3 , 5 2 ) The first number of the coordinate pair is the x -coordinate, and the second number is the y -coordinate. To plot each point, sketch a vertical line through the x -coordinate and a horizontal line through the y -coordinate ( ). Their intersection is the point. Since x = − 5 , the point is to the left of the y -axis. Also, since y = 4 , the point is above the x -axis. The point ( − 5 , 4 ) is in quadrant II. Since x = − 3 , the point is to the left of the y -axis. Also, since y = − 4 , the point is below the x -axis. The point ( − 3 , − 4 ) is in quadrant III. Since x = 2 , the point is to the right of the y -axis. Since y = − 3 , the point is below the x -axis. The point ( 2 , − 3 ) is in quadrant IV. Since x = 0 , the point whose coordinates are ( 0 , − 1 ) is on the y -axis. Since x = 3 , the point is to the right of the y -axis. Since y = 5 2 , which is equal to 2.5, the point is above the x -axis. The point ( 3 , 5 2 ) is in quadrant I. Graphing Linear Equations in Two Variables Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation, you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form A x + B y = C . An equation of this form, where A and B are both not zero, is called a linear equation in two variables . Here is an example of a linear equation in two variables, x and y . A x + B y = C A + 4 y = 8 A = 1 , B = 4 , C = 8 The equation y = − 3 x + 5 is also a linear equation. But it does not appear to be in the form A x + B y = C . We can use the addition property of equality and rewrite it in A x + B y = C form. Step 1: Add 3 x to both sides. y + 3 x = − 3 x + 5 + 3 x Step 2: Simplify. y + 3 x = 5 Step 3: Put it in A x + B y = C form. 3 x + y = 5 By rewriting y = − 3 x + 5 as 3 x + y = 5 , we can easily see that it is a linear equation in two variables because it is of the form A x + B y = C . When an equation is in the form A x + B y = C , we say it is in standard form of a linear equation . Most people prefer to have A , B , and C be integers and A ≥ 0 when writing a linear equation in standard form, although it is not strictly necessary. Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair ( x , y ). When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the linear equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions. A graph is a visual representation of all the solutions of a linear equation. The line shows you all the solutions to that linear equation. Every point on the line is a solution of that linear equation. And every solution of the linear equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions! The graph of a linear equation A x + B y = C is a straight line. Every point on the line is a solution of the equation. Every solution of this equation is a point on this line. Determining Points on a Line is the graph of y = 2 x − 3 . Graph of y = 2 x – 3 For each ordered pair, decide: Is the ordered pair a solution to the equation? Is the point on the line? A: ( 0 , – 3 ) B: ( 3 , 3 ) C: ( 2 , – 3 ) D: ( – 1 , – 5 ) Substitute the x - and y -values into the equation to check if the ordered pair is a solution to the equation. A : ( 0 , − 3 ) y = 2 x − 3 − 3 = ? 2 ( 0 ) − 3 − 3 = − 3 ✓ ( 0 , − 3 ) is a solution. B : ( 3 , 3 ) y = 2 x − 3 3 = ? 2 ( 3 ) − 3 3 = 3 ✓ ( 3 , 3 ) is a solution. C : ( 2 , − 3 ) y = 2 x − 3 − 3 = ? 2 ( 2 ) − 3 − 3 ≠ 1 ( 2 , − 3 ) is not a solution. D : ( − 1 , − 5 ) y = 2 x − 3 − 5 = ? 2 ( − 1 ) − 3 − 5 = − 5 ✓ ( − 1 , − 5 ) is a solution. Plot the points ( 0 , - 3 ) , ( 3 , 3 ) , ( 2 , - 3 ) , and ( - 1 , - 5 ) . In , the points ( 0 , 3 ) , ( 3 , - 3 ) , and ( − 1 , − 5 ) are on the line y = 2 x − 3 , and the point ( 2 , − 3 ) is not on the line. The points that are solutions to y = 2 x − 3 are on the line, but the point that is not a solution is not on the line. The steps to take when graphing a linear equation by plotting points are: Step 1: Find three points whose coordinates are solutions to the equation. Organize them in a table. Step 2: Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work. Step 3: Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line, but it will not represent the solutions to the equation. It will be the wrong line. If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong, and you need to check your work. Graphing a Line by Plotting Points Graph the equation: y = 1 2 x + 3 . Find three points that are solutions to the equation. Since this equation has the fraction 1 2 as a coefficient of x , we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of x ? By choosing multiples of 2, the multiplication by 1 2 simplifies to a whole number. x = 0 x = 2 x = 4 y = 1 2 x + 3 y = 1 2 x + 3 y = 1 2 x + 3 y = 1 2 ( 0 ) + 3 y = 1 2 ( 2 ) + 3 y = 1 2 ( 4 ) + 3 y = 0 + 3 y = 1 + 3 y = 2 + 3 y = 3 y = 4 y = 5 x y ( x , y ) 0 3 ( 0 , 3 ) 2 4 ( 2 , 4 ) 4 5 ( 4 , 5 ) Plot the points, check that they line up, and draw the line ( ). Solving Applications Using Linear Equations in Two Variables Many fields use linear equalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used. Pumping Gas Gasoline costs $3.53 per gallon. You put 10 gallons of gasoline in your car, and pay $35.30. Your friend puts 15 gallons of gasoline in their car and pays $52.95. Your neighbor needs 5 gallons of gasoline, how much will they pay? Let x = the number of gallons of gasoline and let y = the total cost . If gas is $3.53 per gallon, then y = 3.53 x . The two points given are (10, 35.30) and (15, 52.95). Plot the points, check that they line up, and draw the line ( ). We can see the point at x = 5 . The y -value is found by multiplying 5 by $3.53 to get $17.65. Your neighbor will pay $17.65. René Descartes René Descartes (credit: Flickr, Public Domain) René Descartes was born in 1596 in La Haye, France. He was sickly as a child, so much so that he was allowed to stay in bed until 11:00 AM rather than get up at 5:00 AM like the other school children. He kept this habit of rising late for most of the rest of his life. After his primary schooling, Descartes attended the University of Poitiers, receiving a law degree in 1616. He then embarked on a myriad of journeys, joining two different militaries (one in the Netherlands, the other in Bavaria) and generally travelling around Europe until 1628, when he settled in the Netherlands. It was here that he began to delve deeply into his ideas of science, mathematics, and philosophy. In 1637, at the urging of his friends, Descartes published Discourse on the Method for Conducting One's Reason Well and Seeking the Truth in the Sciences . The book had three appendices: La Dioptrique , a work on optics; Les Météores , which pertained to meteorology; and La Géométrie , a work on mathematics. It was in this appendix that he proposed a geometric way of representing many different algebraic expressions and equations. It is this system of representation that almost all mathematical textbooks use today. These publications (along with several others) brought much fame to Descartes. So renowned was his reputation that late in 1649, Queen Christina of Sweden asked Descartes to come to Sweden to tutor her. However, she wished to do her studies at 5:00 in the morning; Descartes had to break his lifelong habit of sleeping in late. A few months later, in February 1650, Descartes died of pneumonia. Graphing Linear Inequalities Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables that can be written in one of the following forms: A x + B y ≥ C , A x + B y > C , A x + B y ≤ C , and A x + B y < C where A and B are not both zero. We will look at linear inequalities in two variables, which are very similar to linear equations in two variables. Like linear equations, linear inequalities in two variables have many solutions. Any ordered pair ( x , y ) that makes an inequality true when we substitute in the values is a solution to a linear inequality . Determining Solutions to an Inequality Determine whether each ordered pair is a solution to the inequality y > x + 4 : ( 0 , 0 ) ( 1 , 6 ) ( 2 , 6 ) ( – 5 , – 15 ) ( − 8 , 12 ) ( 0 , 0 ) y > x + 4 Substitute 0 for x and 0 for y 0 > ? 0 + 4 Simplify . 0 > 4 ( 0 , 0 ) is not a solution to y > x + 4. ( 1 , 6 ) y > x + 4 Substitute 1 for x and 6 for y 6 > ? 1 + 4 Simplify . 6 > 5 ( 1 , 6 ) is a solution to y > x + 4. ( 2 , 6 ) y > x + 4 Substitute 2 for x and 6 for y 6 > ? 2 + 4 Simplify . 6 > 6 ( 2 , 6 ) is not a solution to y > x + 4. ( − 5 , − 15 ) y > x + 4 Substitute − 5 for x and − 15 for y − 15 > ? − 5 + 4 Simplify . − 15 > − 1 So , ( − 5 , − 15 ) is not a solution to y > x + 4. ( − 8 , 12 ) y > x + 4 Substitute − 8 for x and 12 for y 12 > ? − 8 + 4 Simplify . 12 > − 4 So , ( − 8 , 12 ) is a solution to y > x + 4. Let us think about x > 3 . The point x = 3 separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See . Solution to x > 3 on a Number Line Similarly, the line y = x + 4 separates the plane into two regions. On one side of the line are points with y < x + 4 . On the other side of the line are the points with y > x + 4 . We call the line y = x + 4 a boundary line . For an inequality in one variable, the endpoint is shown with a parenthesis ( ) or a bracket ( ) depending on whether or not a is included in the solution: Endpoint with Parenthesis Endpoint with Bracket Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to show whether or not it the line is included in the solution. A x + B y < C A x + B y ≤ C A x + B y > C A x + B y ≥ C Boundary line is A x + B y = C Boundary line is A x + B y = C Boundary line is not included in solution. Boundary line is included in solution. Boundary line is dashed. Boundary line is solid. Now, let us take a look at what we found in Example 5.43. We will start by graphing the line y = x + 4 , and then we will plot the five points we tested, as graphed in . We found that some of the points were solutions to the inequality y > x + 4 and some were not. Which of the points we plotted are solutions to the inequality y > x + 4 ? The points ( 1 , 6 ) and ( − 8 , 12 ) are solutions to the inequality y > x + 4 . Notice that they are both on the same side of the boundary line y = x + 4 . The two points ( 0 , 0 ) and ( − 5 , − 15 ) are on the other side of the boundary line y = x + 4 , and they are not solutions to the inequality y > x + 4 . For those two points, y < x + 4 . What about the point ( 2 , 6 ) ? Because 6 = 2 + 4 , the point is a solution to the equation y = x + 4 , but not a solution to the inequality y > x + 4 . So, the point ( 2 , 6 ) is on the boundary line. Graph of y = x + 4 Let us take another point above the boundary line and test whether or not it is a solution to the inequality y > x + 4 . The point ( 0 , 10 ) clearly looks to be above the boundary line, doesn’t it? Is it a solution to the inequality? y > x + 4 10 > ? 0 + 4 10 > 4 ✓ Yes, ( 0 , 10 ) is a solution to y > x + 4 . Any point you choose above the boundary line is a solution to the inequality y > x + 4 . All points above the boundary line are solutions. Similarly, all points below the boundary line, the side with ( 0 , 0 ) and ( − 5 , − 15 ) , are not solutions to y > x + 4 , as shown in . Graph of y = x + 4 , with y > x + 4 Above the Boundary Line and y < x + 4 Below the Boundary Line The graph of the inequality y > x + 4 is shown in . The line y = x + 4 divides the plane into two regions. The shaded side shows the solutions to the inequality y > x + 4 . The points on the boundary line, those where y = x + 4 , are not solutions to the inequality y > x + 4 , so the line itself is not part of the solution. We show that by making the boundary line dashed, not solid. Graph of y > x + 4 Writing a Linear Inequality Shown by a Graph The boundary line shown in this graph is y = 2 x − 1 . Write the inequality shown in . The line y = 2 x − 1 is the boundary line. On one side of the line are the points with y > 2 x − 1 and on the other side of the line are the points with y < 2 x − 1 . Let us test the point ( 0 , 0 ) and see which inequality describes its position relative to the boundary line. At ( 0 , 0 ) , which inequality is true: y > 2 x − 1 or y < 2 x − 1 ? 0 > 2 ( 0 ) - 1 0 < 2 ( 0 ) - 1 0 > 0 - 1 0 < 0 - 1 0 > - 1 0 < − 1 True False Since y > 2 x − 1 is true, the side of the line with ( 0 , 0 ) , is the solution. The shaded region shows the solution of the inequality y > 2 x − 1 . Since the boundary line is graphed with a dashed line, the inequality does not include the equal sign. The graph shows the inequality y > 2 x − 1 . We could use an y point as a test point, provided it is not on the line. Why did we choose ( 0 , 0 ) ? Because it is the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that y < 2 x − 1 . Graphing a Linear Inequality Graph the linear inequality y ≥ 3 4 x − 2 . Step 1. Identify and graph the boundary line ( ). If the inequality is ≤ or ≥, the boundary line is solid. If the inequality is < or >, the boundary line is dashed. Replace the inequality sign with an equal sign to find the boundary line. Graph the boundary line y = 3 4 x − 2 . The inequality sign is ≥, so we draw a solid line. Step 2. Test a point that is not on the boundary line. Is it a solution of the inequality? We’ll test ( 0 , 0 ) . Is it a solution of the inequality? At ( 0 , 0 ) , is y ≥ 3 4 x − 2 ? 0 ≥ ? 3 4 ( 0 ) − 2 0 ≥ − 2 So, ( 0 , 0 ) is a solution. Step 3. Shade in one side of the boundary line ( ). If the test point is a solution, shade in the side that includes the point. If the test point is not a solution, shade in the opposite side. The test point ( 0 , 0 ) is a solution to y ≥ 3 4 x − 2 . So we shade in that side. All points in the shaded region and on the boundary line represent the solution to y ≥ 3 4 x − 2 . Graphing Linear Inequalities in Two Variables Solving Applications Using Linear Inequalities in Two Variables Many fields use linear inequalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used. Working Multiple Jobs Hilaria works two part time jobs to earn enough money to meet her obligations of at least $240 a week. Her job in food service pays $10 an hour and her tutoring job on campus pays $15 an hour. How many hours does Hilaria need to work at each job to earn at least $240? Let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. Write an inequality that would model this situation. Graph the inequality. Find three ordered pairs ( x,y ) that would be solutions to the inequality. Then, explain what that means for Hilaria. Let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will give the amount earned at that job. Amount earned at the food service job ︸ 10 x plus + the amount earned tutoring ︸ 15 y is at least ︸ ≥ 240 Graph the inequality: Step 1: Graph the boundary line 10 x + 15 y = 240 Create a table of values x y 0 10 ( 0 ) + 15 y = 240 − > y = 16 6 10 ( 6 ) + 15 y = 240 − > y = 12 12 10 ( 12 ) + 15 y = 240 − > y = 8 Step 2: Pick a test point. Let us pick ( 0 , 0 ) again: 10 ( 0 ) + 15 ( 0 ) ≥ 240 ? 0 ≥ 240 is false and not a solution so the shading happens on the other side of the boundary line ( ). From the graph, we see that the ordered pairs ( 15 , 10 ) , ( 0 , 16 ) , ( 24 , 0 ) represent three of infinitely many solutions. Check the values in the inequality. ( 15 , 10 ) 10 x + 15 y ≥ 240 10 ( 15 ) + 15 ( 10 ) ≥ ? 240 300 ≥ 240 True ( 0 , 16 ) 10 x + 15 y ≥ 240 10 ( 0 ) + 15 ( 16 ) ≥ ? 240 240 ≥ 240 True ( 24 , 0 ) 10 x + 15 y ≥ 240 10 ( 24 ) + 15 ( 0 ) ≥ ? 240 240 ≥ 240 True For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her food service job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service. Check Your Understanding Key Terms ordered pair origin points on the axes linear equation in two variables standards form of a linear equation solution linear inequality in two variables solution to a linear inequality boundary line Key Concepts Linear equations can be represented graphically on a rectangular coordinate system. Solving linear equations in two variables means finding the point where two lines intersect. There are three possibilities: The lines intersect at exactly one point; the lines do not intersect (they are parallel); or the lines intersect everywhere (they are the same line). Solving linear inequalities in two variables means finding a region of possible answers. Every point in this region will make both inequalities true statements. Plotting points is a standard way to help graph linear equations and linear inequalities. Videos Graphing Linear Inequalities in Two Variables", "section": "Graphing Linear Equations and Inequalities", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Quadratic Equations In One Variable with Applications The Gateway Arch in St. Louis, Missouri (credit: modification of work \"Gateway Arch - St. Louis - Missouri\" by Sam valadi/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Multiply binomials. Factor trinomials. Solve quadratic equations by graphing. Solve quadratic equations by factoring. Solve quadratic equations using square root method. Solve quadratic equations using the quadratic formula. Solve real world applications modeled by quadratic equations. In this section, we will discuss quadratic equations. There are several real-world scenarios that can be represented by the graph of a quadratic equation. Think of the Gateway Arch in St. Louis, Missouri. Both ends of the arch are 630 feet apart and the arch is 630 feet tall. You can plot these points on a coordinate system and create a parabola to graph the quadratic equation. Identify Polynomials, Monomials, Binomials and Trinomials You have learned that a term is a constant, or the product of a constant and one or more variables. When it is of the form a x m , where a is a constant and x m is a positive whole number, it is called a monomial . Some examples of monomial are 8, - 2 x 2 , 4 y 3 , and 11 z . A monomial or two or more monomials combined by addition or subtraction is a polynomial . Some examples include: b + 11 , 4 y 2 − 7 y + 2 , and 4 x 4 + x 3 + 8 x 2 − 9 x + 1 . Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term (examples: 14, 8 y 2 , - 9 x 3 y 5 , and − 13 ). A binomial has exactly two terms (examples: a + 7 , 4 b − 5 , y 2 − 16 , and 3 x 3 − 9 x 2 ), and a trinomial has exactly three terms (examples: x 2 − 7 x + 12 , 9 y 2 + 2 y − 8 , 6 m 4 − m 3 + 8 m , and x 4 + 3 x 2 − 1 ). Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials. Multiply Binomials Recall multiplying algebraic expressions from Algebraic Expressions . In this section, we will continue that work and multiply binomials as well. We can use an area model to do multiplication. Multiply Binomials Multiply ( x + 2 ) ( x + 3 ) . Step 1: Use the distributive property: ( x ) ( x ) + ( x ) ( 3 ) + ( 2 ) ( x ) + ( 2 ) ( 3 ) In the area model ( ) multiply each term on the side by each term on the top (think of it as a multiplication table). Step 2: After we multiply, we get the following equation: x 2 + 3 x + 2 x + 6 Step 3: Combine the like terms to arrive at: x 2 + 5 x + 6 Multiplying More Complex Binomials Multiply ( 2 x + 7 ) ( 3 x − 5 ) . Step 1: Use the Distributive Property: ( 2 x ) ( 3 x ) − ( 2 x ) ( 5 ) + ( 7 ) ( 3 x ) − ( 7 ) ( 5 ) Step 2: After multiplying, get the following equation: 6 x 2 − 10 x + 21 x - 35 Step 3: Combine the like terms to arrive at: 6 x 2 + 11 x - 35 They Are Teaching Multiplication of Binomials in Elementary School Manipulatives are often used in elementary school for students to experience a hands-on way to experience the mathematics they are learning. Base Ten Blocks, or Dienes Blocks, are often used to introduce place value and the operation of whole numbers. When multiplying two-digit numbers, students can make an array to visualize the Distributive Property. shows the value of each Base Ten Block and shows how to multiply 17 and 23 using an area model and Base Ten Blocks. You can see how this helps students visualize the multiplication using the Distributive Property. Consider how ( 10 + 7 ) ( 20 + 3 ) can easily extend to ( 10 + x ) ( 20 + y ) in algebra! The Value of Each Base Ten Block How to Multiply 17 and 23 Using an Area Model and Base Ten Blocks Factoring Trinomials We’ve just covered how to multiply binomials. Now you will need to “undo” this multiplication—to start with the product and end up with the factors. Let us review an example of multiplying binomials to refresh your memory. ( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6 To factor the trinomial means to start with the product, x 2 + 5 x + 6 , and end with the factors, ( x + 2 ) ( x + 3 ) . You need to think about where each of the terms in the trinomial came from. The first term came from multiplying the first term in each binomial. So, to get x 2 in the product, each binomial must start with an x . x 2 + 5 x + 6 ( x ) ( x ) The last term in the trinomial came from multiplying the last term in each binomial. So, the last terms must multiply to 6. What two numbers multiply to 6? The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use? Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in the following table, which will be very helpful when you work with numbers that can be factored in many different ways. Factors of 6 Sum of Factors 1, 6 1 + 6 = 7 2, 3 2 + 3 = 5 We see that 2 and 3 are the numbers that multiply to 6 and add to 5. We have the factors of x 2 + 5 x + 6 . They are ( x + 2 ) ( x + 3 ) . x 2 + 5 x + 6 product ( x + 2 ) ( x + 3 ) factors You can check if the factors are correct by multiplying. Looking back, we started with x 2 + 5 x + 6 , which is of the form x 2 + b x + c , where b = 5 and c = 6 . We factored it into two binomials of the form ( x + m ) and ( x + n ) . x 2 + 5 x + 6 x 2 + b x + c ( x + 2 ) ( x + 3 ) ( x + m ) ( x + n ) To get the correct factors, we found two number m and n whose product is c and sum is b . With the area model ( ), start with an empty box and then put in the x 2 term and c . Continue by putting in two terms that add up to 5 x : 2 x and 3 x ( ): Then you find the terms of the binomials on the top and side ( ): Factoring Trinomials Factor x 2 + 7 x + 12 . The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below. Factors of 12 Sum of Factors 1, 12 1 + 12 = 13 2, 6 2 + 6 = 8 3, 4 3 + 4 = 7 We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of x 2 + 7 x + 12 are ( x + 3 ) ( x + 4 ) . Factoring More Complex Trinomials Factor x 2 - 11 x + 28 . The numbers that must have a product of 28 will need a sum of − 11 . We will summarize the results in a table. Factors of 28 Sum of Factors 1 , 28 1 + 28 = 29 2 , 14 2 + 14 = 16 4 , 7 4 + 7 = 11 We see that 4 and 7 are the numbers that multiply to 28 and add to 11. But we needed − 11 , so we will need to use − 4 and − 7 because ( − 4 ) ( − 7 ) = 28 and ( − 4 ) + ( − 7 ) = − 11 . The factors of x 2 − 11 x + 28 are ( x − 4 ) ( x − 7 ) . Factoring with the Box Method (Area Model) Solving Quadratic Equations by Graphing We have already solved and graphed linear equations in Graphing Linear Equations and Inequalities , equations of the form A x + B y = C . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. The following are some examples of quadratic equations: x 2 + 5 x + 6 = 0 3 y 2 + 4 y = 106 4 u 2 − 81 = 0 n ( n + 1 ) = 42 The last equation does not appear to have the variable squared, but when we simplify the expression on the left, we will get n 2 + n . The general form of a quadratic equation is a x 2 + b x + c = 0 , where a , b , and c are real numbers, with a ≠ 0 . Remember that a solution of an equation is a value of a variable that makes a true statement when substituted into the equation. The solutions of quadratic equations are the values of the variables that make the quadratic equation a x 2 + b x + c = 0 true. To solve quadratic equations, we need methods different than the ones we used in solving linear equations. We will start by solving a quadratic equation from its graph. Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. Let us look first at graphing the quadratic equation y = x 2 . We will choose integer values of x between − 2 and 2 and find their y values, as shown in the table below. y = x 2 x y 0 0 1 1 − 1 1 2 4 − 2 4 Notice when we let x = 1 and x = − 1 , we got the same value for y . y = x 2 y = 1 2 y = 1 y = x 2 y = ( − 1 ) 2 y = 1 The same thing happened when we let x = 2 and x = − 2 . Now, we will plot the points to show the graph of y = x 2 . See . The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this. When y = 0 the solution to the quadratic y = x 2 is 0 because x 2 = 0 at x = 0 . Graphing a Quadratic Equation Graph y = x 2 - 1 and list the solutions to the quadratic equation. We will graph the equation by plotting points. Step 1: Choose integer values for x , substitute them into the equation, and solve for y . Step 2: Record the values of the ordered pairs in the chart. y = x 2 - 1 x y 0 − 1 1 0 − 1 0 2 3 − 2 3 Step 3: Plot the points and then connect them with a smooth curve. The result will be the graph of the equation y = x 2 - 1 ( ). The solutions are x = 1 and x = - 1 . Solving a Quadratic Equation From Its Graph Find the solutions of y = x 2 + 5 x + 4 from its graph ( ). The solutions of a quadratic equations are the values of x that make the equation a true statement when set equal to zero (i.e. when y = 0 ). x 2 + 5 x + 4 = 0 at x = - 4 and x = - 1 . Solving Quadratic Equations by Factoring Another way of solving quadratic equations is by factoring. We will use the Zero Product Property that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself. Solving a Quadratic Equation by Factoring Solve ( x + 1 ) ( x - 4 ) = 0 . Step 1. Set each factor equal to zero. The product equals zero, so at least one factor must equal zero. ( x + 1 ) ( x - 4 ) = 0 x + 1 = 0 or x - 4 = 0 Step 2. Solve the linear equations. Solve each equation. x = − 1 or x = 4 Step 3. Check. Substitute each solution separately into the original equation. x = - 1 ( x + 1 ) ( x - 4 ) = 0 ( − 1 + 1 ) ( − 1 − 4 ) = ? 0 ( 0 ) ( - 5 ) = ? 0 0 = 0 ✓ x = 4 ( x + 1 ) ( x - 4 ) = 0 ( 4 + 1 ) ( 4 − 4 ) = ? 0 ( 5 ) ( 0 ) = ? 0 0 = 0 ✓ Solve Another Quadratic Equation by Factoring Solve x 2 + 2 x − 8 = 0 . Step 1. Write the quadratic equation in standard form, ax 2 + bx + c = 0 . The equation is already in standard form. x 2 + 2 x - 8 = 0 Step 2. Factor the quadratic expression. Factor x 2 + 2 x - 8 ( x + 4 ) ( x - 2 ) ( x + 4 ) ( x - 2 ) = 0 Step 3. Use the Zero Product Property. Set each factor equal to zero. x + 4 = 0 or x - 2 = 0 Step 4. Solve the linear equations. We have two linear equations. x = - 4 or x = 2 Step 5. Check. Substitute each solution separately into the original equation. x 2 + 2 x - 8 = 0 x = - 4 ( − 4 ) 2 - 2 ( - 4 ) - 8 = ? 0 16 + ( - 8 ) - 8 = ? 0 0 = 0 ✓ x 2 + 2 x - 8 = 0 x = 2 2 2 - 2 ( 2 ) - 8 = ? 0 4 + 4 - 8 = ? 0 0 = 0 ✓ Solving Quadratics with the Zero Property Be careful to write the quadratic equation in standard form first. The equation must be set equal to zero in order for you to use the Zero Product Property! Often students start in Step 2 resulting in an incorrect solution. For example, x 2 + 2 x − 15 = - 7 cannot be factored to ( x − 3 ) ( x + 5 ) = − 7 and then solved by setting each factor equal to − 7 . The correct way to solve this quadratic equation is to set it equal to zero FIRST: x 2 + 2 x − 15 + 7 = − 7 + 7 which becomes x 2 + 2 x − 8 = 0 , then continue to factor. See the table below for the correct way to apply the Zero Product Property. x 2 + 2 x − 15 = − 7 x 2 + 2 x − 15 = − 7 Step 1 Skipped x 2 + 2 x − 15 + 7 = − 7 + 7 x 2 + 2 x − 8 = 0 Step 2 ( x − 3 ) ( x + 5 ) = − 7 ( x − 2 ) ( x + 4 ) = 0 Step 3 x − 3 = − 7 x + 5 = − 7 x − 2 = 0 x + 4 = 0 Step 4 x = − 4 x = − 12 x = 2 x = − 4 Step 5 ( − 4 ) 2 + 2 ( − 4 ) − 15 = 16 − 8 − 15 = − 7 ( − 12 ) 2 + 2 ( − 12 ) − 15 = 144 − 24 − 1 = 105 ≠ − 7 ( 2 ) 2 + 2 ( 2 ) − 8 = 4 + 4 − 8 = 0 ( − 4 ) 2 + 2 ( − 4 ) − 18 = 16 − 8 − 8 = 0 Solving Quadratic Equations Using the Square Root Property We just solved some quadratic equations by factoring. Let us use factoring to solve the quadratic equation x 2 = 9 . Step 1: Put the equation in standard form. x 2 − 9 = 0 Step 2: Factor the left side. ( x + 3 ) ( x - 3 ) = 0 Step 3: Use the Zero Product Property. x + 3 = 0 x − 3 = 0 Step 4: Solve each equation. x = − 3 x = 3 Step 5: Combine the two solutions into ± x = ± 3 The solution is read as “ x is equal to positive or negative 3.” What happens when we have an equation like x 2 = 7 ? Since 7 is not a perfect square, we cannot solve the equation by factoring. These equations are all of the form x 2 = k . We define the square root of a number in this way: If n 2 = m , then n is a square root of m . This leads to the Square Root Property . Using the Square Root Property to Solve a Quadratic Equation Solve using the square Root Property: x 2 = 169 . Step 1: Use the Square Root Property. x = ± 169 Step 2: Simplify the radical. x = ± 13 Step 3: Rewrite to show the two solutions. x = 13 x = - 13 Using the Square Root Property to Solve Another Quadratic Equation Solve using the Square Root Property: 144 q 2 = 25 . Step 1: Solve for q . q 2 = 25 144 Step 2: Use the Square Root Property. q = ± 25 144 Step 3: Simplify the radical. q = ± 5 12 Step 4: Rewrite to show the two solutions. q = 5 12 , q = - 5 12 Solving Quadratic Equations Using the Quadratic Formula This last method we will look at for solving quadratic equations is the quadratic formula . This method works for all quadratic equations, even the quadratic equations we could not factor! To use the quadratic formula, we substitute the values of a , b , and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation. Solving a Quadratic Equation Using the Quadratic Formula Solve using the quadratic formula: x 2 - 6 x + 5 = 0 . x 2 - 6 x + 5 = 0 This equation is in standard form. a x 2 + b x + c = 0 x 2 - 6 x + 5 = 0 Step 1: Identify the a , b , and c values. a = 1 , b = - 6 , c = 5 Step 2: Write the quadratic formula. x = - b ± b 2 - 4 a c 2 a Step 3: Substitute in the values of a , b , c . x = ( − 6 ) ± ( − 6 ) 2 − 4 ( 1 ) ( 5 ) 2.1 Step 4: Simplify. x = 6 ± 36 - 20 2 x = 6 ± 16 2 x = 6 ± 4 2 Step 5: Rewrite to show two solutions. x = 6 + 4 2 , x = 6 - 4 2 Step 6: Simplify. x = 10 2 , x = 2 2 x = 5 , x = 1 Step 7: Check. x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 5 2 − 6 ⋅ 5 + 5 = ? 0 1 2 − 6 ⋅ 1 + 5 = ? 0 25 − 30 + 5 = ? 0 1 − 6 + 5 = ? 0 0 = 0 ✓ 0 = 0 ✓ Solving Another Quadratic Equation Using the Quadratic Formula Solve using the quadratic formula: 2 x 2 + 9 x − 5 = 0 . Step 1. Write the quadratic equation in standard form. Identify the a , b , c values. This equation is in standard form. ax 2 + bx + c = 0 2 x 2 + 9 x - 5 = 0 a = 2 , b = 9 , c = - 5 Step 2. Write the quadratic formula. Then substitute in the values of a , b , c . Substitute in a = 2 , b = 9 , c = - 5 x = − b ± b 2 − 4 a c 2 a x = − 9 ± 9 2 − 4 ( 2 ) ( − 5 ) 2 ⋅ 2 Step 3. Simplify the fraction, and solve for x . x = − 9 ± 81 − ( − 40 ) 4 x = − 9 ± 121 4 x = − 9 ± 11 4 x = − 9 + 11 4 x = − 9 − 11 4 x = 2 4 x = − 20 4 x = 1 2 x = − 5 Step 4. Check the solutions. Put each answer in the original equation to check. Substitute x = 1 2 . 2 x 2 + 9 x - 5 = 0 2 ( 1 2 ) 2 + 9 ⋅ 1 2 - 5 = ? 0 2 ⋅ 1 4 + 9 ⋅ 1 2 - 5 = ? 0 1 2 + 9 2 - 5 = ? 0 10 2 - 5 = ? 0 5 - 5 = ? 0 0 = 0 ✓ Substitute x = - 5 . 2 x 2 + 9 x - 5 = 0 2 ( - 5 ) 2 + 9 ( - 5 ) - 5 = ? 0 2 ⋅ 25 - 45 - 5 = ? 0 50 - 45 - 5 = ? 0 0 = 0 ✓ Solving Quadratics with the Quadratic Formula Solving Real-World Applications Modeled by Quadratic Equations There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems: Step 1: Read the problem. Make sure all the words and ideas are understood. Step 2: Identify what we are looking for. Step 3: Name what we are looking for. Choose a variable to represent that quantity. Step 4: Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. Step 5: Solve the equation using good algebra techniques. Step 6: Check the answer in the problem and make sure it makes sense. Step 7: Answer the question with a complete sentence. Finding Consecutive Integers The product of two consecutive integers is 132. Find the integers. Step 1: Read the problem. Step 2: Identify what we are looking for. We are looking for two consecutive integers. Step 3: Name what we are looking for. Let n = the first integer Let n + 1 = the next consecutive integer. Step 4: Translate into an equation. Restate the problem in a sentence. n ( n + 1 ) = 132 The product of the two consecutive integers is 132. The first integer times the next integer is 132. Step 5: Solve the equation. n 2 + n = 132 Bring all the terms to one side. n 2 + n - 132 = 0 Factor the trinomial. ( n - 11 ) ( n + 12 ) = 0 Use the zero product property. n - 11 = 0 or n + 12 = 0 Solve the equations. n = 11 , n = - 12 There are two values for n that are solutions to this problem. So, there are two sets of consecutive integers that will work. If the first integer is n = 11 , then the next integer is 12. If the first integer is n = - 12 , then the next integer is − 11 . Step 6: Check the answer. The consecutive integers are 11, 12 and − 11 , − 12 . The product of 11 and 12 = 132 and the product of - 11 ( - 12 ) = 132 . Both pairs of consecutive integers are solutions. Step 7: Answer the question. The consecutive integers are 11, 12, and − 11 , − 12 . Were you surprised by the pair of negative integers that is one of the solutions? In some applications, negative solutions will result from the algebra, but will not be realistic for the situation. Finding Length and Width of a Garden A rectangular garden has an area 15 square feet. The length of the garden is 2 feet more than the width. Find the length and width of the garden. Step 1: Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation ( ). Step 2: Identify what you are looking for. We are looking for the length and width. Step 3: Name what you are looking for. The length is 2 feet more than width. Let W = the width of the garden. W + 2 = the length of the garden. Step 4: Translate into an equation. Restate the important information in a sentence. The area of the rectangular garden is 15 square feet. Use the formula for the area of a rectangle. A = L × W Substitute in the variables. 15 = ( W + 2 ) W Step 5: Solve the equation. Distribute first. 15 = W 2 + 2 W Get zero on one side. 0 = W 2 + 2 W - 15 Factor the trinomial. 0 = ( W + 5 ) ( W - 3 ) Use the Zero Product Property. 0 = W + 5 0 = W - 3 Solve each equation. - 5 = W 3 = W Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W . W = 3 Width is 3 feet. Find the value of the length. W + 2 = length. 3 + 2 5 Length is 5 feet. Step 6: Check the answer ( ). Does the answer make sense? Yes, this makes sense. Step 7: Answer the question. The width of the garden is 3 feet and the length is 5 feet. Completing the Square Recall the two methods used to solve quadratic equations of the form a x 2 + b x + c : by factoring and by using the quadratic formula. There are, however, many different methods for solving quadratic equations that were developed throughout history. Egyptian, Mesopotamian, Chinese, Indian, and Greek mathematicians all solved various types of quadratic equations, as did Arab mathematicians of the ninth through the twelfth centuries. It is one of these Arab mathematicians' methods that we wish to investigate with this activity. Muhammad ibn Musa al-Khwarizmi was employed as a scholar at the House of Wisdom in Baghdad, located in present day Iraq. One of the many accomplishments of Al-Khwarizmi was his book on the topic of algebra. In that book, he asks, “What must be the square which, when increased by ten of its own roots, amounts to 39?” Al-Khwarizmi, like many Arab mathematicians of his time, was well versed in Euclid's Elements. Like Euclid, he viewed algebra very geometrically, and thus had a geometric approach to solving a problem like the one above. In his approach, he used a method which today we refer to as completing the square . His description of the solution method for the above problem is: halve the number of roots, which in the present instance yields 5. This you multiply by itself; the product is 25. Add this to 39; the sum is 64. Now take the root of this which is 8, and subtract from it half the number of the roots, which is 5; the remainder is 3. This is the root of the square which you sought for. Thus the square is 9. So, what does all of this mean? Al-Khwarizmi would start with a square of unknown length of side (we will label the side length x ). See So, this square has area x 2 x He would then proceed to halve the number of roots (i.e., there are 10 roots by which the square is increasing) to get 5; this he would add to the first square. See The area of the two new pieces added into the original square are both 5 x . At this point, we have x 2 + 10 x = 39 . Now Al-Khwarizmi needed to “complete the square” by adding into the drawing a small square. See This square has an area of 25. x 2 + 10 x + 25 = 39 + 25 , or x 2 + 10 x + 25 = 64 . Notice that the completed square has side length x + 5 , so the large square has area ( x + 5 ) 2 . (Notice algebraically that the left half of the equation x 2 + 10 x + 25 = 64 factors to ( x + 5 ) 2 = 64. ) This means the area of large square equals 64. If ( x + 5 ) 2 = 64 , then x + 5 = 8 ; so x must be equal to 3 or − 13 to make this true. Note that Al-Kwarimi would not have considered the possibility of a negative solution, since he approached the solution geometrically, and negative distances do not exist. Check Your Understanding Key Terms monomial polynomial binomial trinomial quadratic equation Zero Product Property Key Concepts A quadratic equation is an algebraic equation where the highest power (degree) of the equation is two. To solve a quadratic equation is to find the value(s) that when substituted in for the variables, will make the equation equal to zero. There can be two, one, or no solutions to any quadratic equation. There are several methods to solve a quadratic equation. These methods include factoring quadratic equations, graphic quadratic equations, using the square root method, and using the quadratic formula. Videos Factoring with the Box Method (Area Model) Solving Quadratics with the Zero Property Solving Quadratics with the Quadratic Formula", "section": "Quadratic Equations In One Variable with Applications", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Functions A small group of elementary students learning from their teacher. (credit: modification of work \"Our school\" by Woodleywonderworks/Flickr, CC BY 2.0 ) Learning Objectives After completing this section, you should be able to: Use function notation. Determine if a relation is a function with different representations. Apply the vertical line test. Determine the domain and range of a function. In this section, we will learn about relations and functions. As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number, and our birthday are matched to our name. There is a relationship between our name and each of those items. When your teacher gets their class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation. (Student name, Student ID #) The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation. In general terms, a relation is any set of ordered pairs, ( x , y ). All the x -values in the ordered pairs together make up the domain . All the y -values in the ordered pairs together make up the range. There are many situations similar to the student's name and student ID # where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation. A special type of relation, called a function , occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x -value is matched with only one y -value. Let us look at the relation between your friends and their birthdays in . Every friend has a birthday, but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24 as their birthday and that June and Liz share August 2. Since each person has exactly one birthday, the relation is a function. Birthday Mapping Use Function Notation It is very convenient to name a function; most often functions are named f , g , h , F , G , or H . In any function, for each x -value from the domain, we get a corresponding y -value in the range. In the function f , we write this range value y as f ( x ). This notation f ( x ) is called function notation and is read \"f of x \" or \"the value of f at x .\" In this case the parentheses do not indicate multiplication. We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x . Much like when you first encountered the variable x , function notation may be rather unsettling. But the more you use the notation, the more familiar you become with the notation, and the more comfortable you will be with it. Let’s review the equation y = 4 x − 5 . To find the value of y when x = 2 , we know to substitute x = 2 into the equation and then simplify. y = 4 x − 5 Let x = 2 . y = 4 ⋅ 2 − 5 y = 3 The value of the function at x = 2 is 3. We do the same thing using function notation, the equation y = 4 x − 5 can be written as f ( x ) = 4 x − 5 . To find the value when x = 2 , we write: f ( x ) = 4 x − 5 Let x = 2 . f ( 2 ) = 4 ⋅ 2 − 5 f ( 2 ) = 3 The value of the function at x = 2 is 3. This process of finding the value of f ( x ) for a given value of x is called evaluating the function . Evaluating the Function For the function f ( x ) = 2 x 2 + 3 x − 1 , evaluate the function. f ( 3 ) f ( − 2 ) f ( a ) To evaluate f ( 3 ) , substitute 3, for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( 3 ) = 2 ( 3 ) 2 + 3 ⋅ 3 − 1 f ( 3 ) = 2 ⋅ 9 + 3 ⋅ 3 − 1 f ( 3 ) = 18 + 9 − 1 f ( 3 ) = 26 To evaluate f ( − 2 ) , substitute − 2 for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( − 2 ) = 2 ( − 2 ) 2 + 3 ( − 2 ) − 1 f ( − 2 ) = 2 ⋅ 4 + ( − 6 ) − 1 f ( − 2 ) = 8 + ( − 6 ) − 1 f ( − 2 ) = 1 To evaluate f ( a ) , substitute a for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( a ) = 2 ( a ) 2 + 3 ⋅ a − 1 f ( a ) = 2 a 2 + 3 a − 1 Evaluating the Function in an Application The number of unread emails in Sylvia’s inbox is 75. This number grows by 10 unread emails a day. The function N ( t ) = 75 + 10 t represents the relation between the number of emails, N , and the time, t , measured in days. Find N (5). Explain what this result means. Find N (5). Explain what this result means. Substitute in t = 5 . Simplify. N ( 5 ) = 75 + 10 ⋅ 5 N ( t ) = 75 + 10 t N ( 5 ) = 75 + 50 N ( 5 ) = 125 If 5 is the number of days, N ( 5 ) is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in Sylvia’s inbox. Determining If a Relation Is a Function with Different Representations We can determine whether a relation is a function by identifying the input and the output values. If each input value leads to only one output value, classify the relation as a function. If any input value leads to two or more outputs, do not classify the relation as a function. We will review three different representations of relations and determine if they are functions: ordered pairs, mapping, and equations. Determining If a Relation Is a Function with a Set of Ordered Pairs Use the set of ordered pairs to determine whether the relation is a function. { ( − 3 , 27 ) , ( − 2 , 8 ) , ( − 1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } { ( 9 , − 3 ) , ( 4 , − 2 ) , ( 1 , − 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) } Each x -value is matched with only one y -value. This relation is a function. The x -value 9 is matched with two y -values, both 3 and − 3 . This relation is not a function. A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range. Consider the example of the relation between your friends and their birthdays used in . In this particular example, the domain is the set of people’s names, and the range is the set of their birthdays. This mapping was a function because everybody’s name maps to exactly one birthday. Determining If a Relation Is a Function with Mapping Use the mapping in to determine whether the relation is a function. Both Lydia and Marty have two phone numbers. Each x -value is not matched with only one y -value. This relation is not a function. In algebra functions will usually be represented by an equation. It is easiest to see if the equation is a function when it is solved for y . If each value of x results in only one value of y , then the equation defines a function. Determining If a Relation Is a Function with an Equation Determine whether each equation is a function. Assume x is the independent variable. 2 x + y = 7 y = x 2 + 1 x + y 2 = 3 2 x + y = 7 For each value of x , we multiply it by − 2 and then add 7 to get the y -value. For example, if x = 3 : y = − 2 x + 7 y = − 2 ⋅ 3 + 7 y = 1 We have that when x = 3 , then y = 1 . It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function. y = x 2 + 1 For each value of x , we square it and then add 1 to get the y -value. For example, if x = 2 y = x 2 + 1 y = 2 2 + 1 y = 5 We have that when x = 2 , then y = 5 . It would work similarly for any value of x . Since each value of x corresponds to only one value of y , the equation defines a function. x + y 2 = 3 x + y 2 = 3 Isolate the y term. y 2 = − x + 3 Let us substitute x = 2 . y 2 = − 2 + 3 y 2 = 1 This gives us two values for y . y = 1 , y = − 1 We have shown that when x = 2 , then y = 1 and y = − 1 . It would work similarly for any value of x . Since each value of x does not corresponds to only one value of y the equation does not define a function. Relations and Functions Applying the Vertical Line Test We reviewed how to determine if a relation is a function. The relations we looked at were expressed as a set of ordered pairs, a mapping, or an equation. We will now cover how to tell if a graph is that of a function. An ordered pair ( x , y ) is a solution of a linear equation, if the equation is a true statement when the x -values and y -values of the ordered pair are substituted into the equation. The graph of a linear equation is a straight line where every point on the line is a solution of the equation, and every solution of this equation is a point on this line. we can see that in the graph of the equation y = 2 x − 3 , for every x -value there is only one y -value, as shown in the accompanying table. Graph of the Equation y = 2 x − 3 A relation is a function if every element of the domain has exactly one value in the range. The relation defined by the equation y = 2 x − 3 is a function. If we look at the graph, each vertical dashed line only intersects the solid line at one point. This makes sense as in a function, for every x -value there is only one y -value. If the vertical line hit the graph twice, the x -value would be mapped to two y -values, and so the graph would not represent a function. This leads us a graphical method of determining functions called the vertical line test , which states that a set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. If any vertical line intersects the graph in more than one point, the graph does not represent a function. Applying the Vertical Line Test Determine whether the graph ( ) is the graph of a function applying the vertical line test. On the graph ( ), only three vertical dashed lines are drawn. However, it can be determined that any vertical dashed line that is drawn will intersect the solid line at exactly one point. It is the graph of a function. Applying the Vertical Line Test to a Parabola Determine whether the graph is the graph of a function ( ). does not represent a function since the vertical dashed lines shown on the graph below intersect the solid line at two points. Determining the Domain and Range of a Function For the function y = f ( x ) , x is the independent variable as it can be any value in the domain, and y is the dependent variable since its value depends on x . For the function y = f ( x ) , the values of x make up the domain and the values of y make up the range. Finding the Domain and Range of Ordered Pairs For { ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) } : Find the domain of the relation. Find the range of the relation. The domain is the set of all x -values of the relation: { 1 , 2 , 3 , 4 , 5 } The range is the set of all y -values of the relation: { 1 , 4 , 9 , 16 , 25 } Finding the Domain and Range on a Graph Use to: List the ordered pairs of the relation. Find the domain of the relation. Find the range of the relation. The ordered pairs of the relation are: { ( 1 , 5 ) , ( − 3 , − 1 ) , ( 4 , − 2 ) , ( 0 , 3 ) , ( 2 , − 2 ) , ( − 3 , 4 ) } . The domain is the set of all x -values of the relation: { − 3 , 0 , 1 , 2 , 4 } . Notice that while − 3 repeats, it is only listed once. The range is the set of all y -values of the relation: { − 2 , − 1 , 3 , 4 , 5 } . Notice that while − 2 repeats, it is only listed once. Domain and Range on Graphs Function and Function Notation In 1673, Gottfried Leibniz, the German mathematician who co-invented calculus, seems to be the first person to use the word function in a mathematical sense, although his use of it does not exactly fit with the modern use and definition. The person who is credited with the modern definition of function is Swiss mathematician Johann Bernoulli, who wrote about it in a letter to Leibniz in 1698. Supposedly, Leibniz wrote Bernoulli back, approving of this use of the word. In 1734, the use of the notation f ( x ) for a function was first used by Swiss mathematician Leonhard Euler (pronounced “Oiler”). Euler had a knack for inventing notation. He also introduced the notation e for the base of natural logs (1727), i for the square root of − 1 (1777), ∑ for summation (1755), and many others. Euler also introduced many other ideas associated with functions. Euler defined exponential functions and defined logarithmic functions as their inverse; he also introduced the beta and gamma functions, and was the first person to consider the trigonometric identities (sine, cosine, etc.) as functions. Check Your Understanding Key Terms relation domain function mapping vertical line test Key Concepts A relation is any set of ordered pairs ( x , y ) . All of the x -values of the set are the domain, and all of the y -values of the set are the range. A relation is a function if each x -value in the domain is assigned to exactly one element in the range. A y -value in the range can have more than one x -value assigned to it; but each x -value can only be assigned to one y -value. For the function y = f ( x ) , f is the name of the function, x is the domain value variable, and y = f ( x ) is the range value variable. The vertical line test is a test that can be done on the graph of a relation to determine if it is a function. Videos Relations and Functions Domain and Range on Graphs", "section": "Functions", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Graphing Functions The ski lifts and the mountain both have a slope. (credit: modification of work \"colorado springs zoo tram\" by woodleywonderworks/Flickr, CC BY 2.0) Learning Objectives After completing this module, you should be able to: Graph functions using intercepts. Compute slope. Graph functions using slope and y -intercept. Graph horizontal and vertical lines. Interpret graphs of functions. Model applications using slope and y -intercept. In this section, we will expand our knowledge of graphing by graphing linear functions. There are many real-world scenarios that can be represented by graphs of linear functions. Imagine a chairlift going up at a ski resort. The journey a skier takes travelling up the chairlift could be represented as a linear function with a positive slope. The journey a skier takes down the slopes could be represented by a linear function with a negative slope. Graphing Functions Using Intercepts Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points. At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x -axis and the y -axis. These points are called the intercepts of a line. Let us review the graphs of the lines in . The table below lists where each of these lines crosses the x - and y -axis. Do you see a pattern? For each line, the y -coordinate of the point where the line crosses the x -axis is zero. The point where the line crosses the x -axis has the form ( a , 0 ) and is called the x -intercept of the line. The x -intercept occurs when y is zero. In each line, the x -coordinate of the point where the line crosses the y -axis is zero. The point where the line crosses the y -axis has the form ( 0 , b ) and is called the y -intercept of the line. The y -intercept occurs when x is zero. Figure The line crosses the x -axis at: Ordered Pair for this Point The line crosses the y -axis at: Ordered Pair for This Point Figure (a) 3 ( 3 , 0 ) 6 ( 0 , 6 ) Figure (b) 4 ( 4 , 0 ) − 3 ( 0 , − 3 ) Figure (c) 5 ( 5 , 0 ) − 5 ( 0 , − 5 ) Figure (d) 0 ( 0 , 0 ) 0 ( 0 , 0 ) General Figure a ( a , 0 ) b ( 0 , b ) Finding x - and y -Intercepts Find the x -intercept and y -intercept on the (a) and (b) graphs in . In , the graph crosses the x -axis at the point ( 4 , 0 ) . The x -intercept is ( 4 , 0 ) . The graph crosses the y -axis at the point ( 0 , 2 ) . The y -intercept is ( 0 , 2 ) . In , the graph crosses the x -axis at the point ( 2 , 0 ) . The x -intercept is ( 2 , 0 ) . The graph crosses the y -axis at the point ( 0 , − 6 ) . The y -intercept is ( 0 , − 6 ) . Graphing a Function Using Intercepts Find the intercepts of 2 x + y = 8 . Then graph the function using the intercepts. Let y = 0 to find the x -intercept, and let x = 0 to find the y -intercept. 2 x + y = 8 2 x + y = 8 To find the x -intercept, let y = 0 . 2 x + 0 = 8 To find the y -intercept, let x = 0 . 2 ( 0 ) + y = 8 Simplify. 2 x = 8 x = 4 Simplify. y = 8 The x -intercept is: ( 4 , 0 ) The y -intercept is: ( 0 , 8 ) Plot the intercepts to get the graph in . Computing Slope When graphing linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter. In mathematics, the measure of the steepness of a line is called the slope of the line. To find the slope of a line, we locate two points on the line whose coordinates are integers. Then we sketch a right triangle where the two points are vertices of the triangle and one side is horizontal and one side is vertical. Next, we measure or calculate the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run . We can assign a numerical value to the slope of a line by finding the ratio of the rise and run. The rise is the amount the vertical distance changes while the run measures the horizontal change, as shown in this illustration. Slope ( ) is a rate of change. To calculate slope ( m ) , use the formula m = rise run , where the rise measures the vertical change and the run measures the horizontal change. The concept of slope has many applications in the real world. In construction, the pitch of a roof, the slant of plumbing pipes, and the steepness of stairs are all applications of slope. As you ski or jog down a hill, you definitely experience slope. Finding the Slope from a Graph Find the slope of the line shown in . Step 1: Locate two points on the graph whose coordinates are integers, such as ( 0 , 5 ) and ( 3 , 3 ) . Starting at ( 0 , 5 ) , sketch a right triangle to ( 3 , 3 ) as shown in . Step 2: Count the rise; since it goes down, it is negative. The rise is −2. Step 3: Count the run. The run is 3. Step 4: Use the slope formula m = rise run substitute the values of the rise and run. m = − 2 3 The slope of the line is − 2 3 . The solution is y decreases by 2 units as x increases by 3 units. Sometimes we will need to find the slope of a line between two points when we don’t have a graph to measure the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing. First, we need to introduce some algebraic notation. We have seen that an ordered pair ( x , y ) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol ( x , y ) be used to represent two different points? Mathematicians use subscripts to distinguish such points. For example, ( x 1 , y 1 ) would be said aloud as “ x sub 1, y sub 1” and ( x 2 , y 2 ) read “ x sub 2, y sub 2.” The “sub” is a short way of saying “subscript.” We will use ( x 1 , y 1 ) to identify the first point and ( x 2 , y 2 ) to identify the second point in our slope equation. If we had more than two points, (if we were finding more than one slope), we could use ( x 3 , y 3 ), ( x 4 , y 4 ), and so on. Let’s review how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points ( 2 , 3 ) and ( 7 , 6 ) , as shown in . On the graph, we count the rise of 3 and the run of 5. Notice on the graph that that ( x 1 , y 1 ) is the point ( 2 , 3 ) and ( x 2 , y 2 ) is the point ( 7 , 6 ) . The rise can be found by subtracting the y -coordinates, 6 and 3, and the run can be found by subtracting the x -coordinates 7 and 2. m = rise run = 6 − 3 7 − 2 = 3 5 We have shown that m = y 2 − y 1 x 2 − x 1 is really another version of m = rise run . We can use this formula to find the slope of a line. To find the slope of the line between two points ( x 1 , y 1 ) and ( x 2 , y 2 ), use the formula m = y 2 − y 1 x 2 − x 1 Finding the Slope of the Line Using Points Use the slope formula to find the slope of the line through the points (−2, −3) and (−7, 4). We’ll call (−2, −3) point 1 and (−7, 4) point 2. Step 1: Use the slope formula: m = y 2 − y 1 x 2 − x 1 Step 2: Substitute the values: m = 4 − ( − 3 ) − 7 − ( − 2 ) Step 3: Simplify: m = 7 − 5 = − 7 5 Step 4: Verify the slope on the graph shown in . Graphing Functions Using Slope and y -Intercept We have graphed linear equations by plotting points and using intercepts. Once we see how an equation in slope-intercept form and its graph are related, we will have one more method we can use to graph lines. Review the graph of the equation y = 1 2 x + 3 in and find its slope and y -intercept. Graph of the equation y = 1 2 x + 3 . The vertical and horizontal lines in the graph show us the rise is 1 and the run is 2, respectively. Substituting into the slope formula: m = rise run = 1 2 The y -intercept is ( 0 , 3 ) . Look at the equation of this line. y = 1 2 x + 3 Look at the slope and y -intercept. slope m = 1 2 and y - intercept ( 0 , 3 ) When a linear equation is solved for y , the coefficient of the x term is the slope and the constant term is the y -coordinate of the y -intercept. We say that the equation y = 1 2 x + 3 is in slope-intercept form . Sometimes the slope-intercept form is called the y -form. Finding the Slope and y -Intercept of a Line Identify the slope and y -intercept of the line from the equation: y = − 4 7 x − 2 x + 3 y = 9 We compare our equation to the slope-intercept form of the equation. Step 1: Write the slope-intercept form of the equation of the line. y = m x + b Step 2: Write the equation of the line. y = − 4 7 x − 2 Step 3: Identify the slope. m = − 4 7 Step 4: Identify the y -intercept. y - intercept is ( 0 , − 2 ) When an equation of a line is not given in slope-intercept form, our first step will be to solve the equation for y . Step 1: Solve for y . x + 3 y = 9 Step 2: Subtract x from each side. 3 y = − x + 9 Step 3: Divide both sides by 3. 3 y 3 = − x + 9 3 Step 4: Simplify. y = − 1 3 x + 3 Step 5: Write the slope-intercept form of the equation of the line. y = m x + b Step 6: Write the equation of the line. y = − 1 3 x + 3 Step 7: Identify the slope. m = − 1 3 Step 8: Identify the y -intercept. y - intercept is ( 0 , 3 ) . Graphing the Slope and y -Intercept Graph the line of the equation y = − x + 4 using its slope and y -intercept. The equation is in slope-intercept form y = m x + b . y = − x + 4 . Step 1: Identify the slope and y -intercept. m = − 1 , y -intercept is ( 0 , 4 ) . Step 2: Plot the y -intercept on the coordinate system ( ). Identify the rise over the run. m = − 1 = − 1 1 Count out the rise and run to mark the second point. rise − 1 , run 1 Graphing Horizontal and Vertical Lines Some linear equations have only one variable. They may have just x without the y , or just y without an x . This changes how we make a table of values to get the points to plot. Let us consider the equation x = − 3 . This equation has only one variable, x . The equation says that x is always equal to − 3 , so its value does not depend on y . No matter what the value of y is, the value of x is always − 3 . To make a table of values, write − 3 in for all the x -values. Then choose any values for y . Since x does not depend on y , you can choose any numbers you like. But to fit the points on our coordinate graph, we will use 1, 2, and 3 for the y -coordinates in the table below. x = − 3 x y ( x , y ) −3 1 ( − 3 , 1 ) − 3 2 ( − 3 , 2 ) − 3 3 ( − 3 , 3 ) Plot the points from the table and connect them with a straight line ( ). Notice that we have graphed a vertical line. Graph of x = − 3 What is the slope? If we take the two points ( − 3 , 3 ) and ( − 3 , 1 ) then the rise is 2 and the run is 0. Using the slope formula we get: m = rise run = 2 0 The slope is undefined since division by zero is undefined. We say that the slope of the vertical line x = − 3 is undefined. The slope of any vertical line x = a (where a is any number) will be undefined. What if the equation has y but no x ? Let’s graph the equation y = 4 . This time the y -value is a constant, so in this equation, y does not depend on x . Fill in 4 for all the y values in the table below and then choose any values for x . We will use 0, 2, and 4 for the x -coordinates. y = 4 x y ( x , y ) 0 4 ( 0 , 4 ) 2 4 ( 2 , 4 ) 4 4 ( 4 , 4 ) In , we have graphed a horizontal line passing through the y -axis at 4. Graph of y = 4 What is the slope? If we take the two points ( 2 , 4 ) and ( 4 , 4 ) then the rise is 0 and the run is 2. Using the slope formula, we get m = rise run = 0 2 = 0 . The slope of the horizontal line y = 4 is 0. The slope of any horizontal line y = b (where b is any number) will be 0. When the y -coordinates are the same, the rise is 0. Graphing A Vertical Line Graph: x = 2 . The equation has only one variable, x , and x is always equal to 2. We create a table where x is always 2 and then put in any values for y . The graph is a vertical line passing through the x -axis at 2 ( ). x = 2 x y ( x , y ) 2 1 ( 2 , 1 ) 2 2 ( 2 , 2 ) 2 3 ( 2 , 3 ) Graphing A Horizontal Line Graph: y = − 1 . The equation y = − 1 has only one variable, y . The value of y is constant. All the ordered pairs in the next table have the same y -coordinate. The graph is a horizontal line passing through the y -axis at −1 ( ). y = − 1 x y ( x , y ) 0 − 1 ( 0 , − 1 ) 3 − 1 ( 3 , − 1 ) − 3 − 1 ( − 3 , − 1 ) The table below summarizes all the methods we have used to graph lines. Interpreting Graphs of Functions An important yet often overlooked area in algebra involves interpreting graphs. Oftentimes in math classes, students are given mathematical functions and can make graphs to represent them. But the interpretation of graphs is a more applicable skill to the real world. Being able to “read” a graph—understanding its domain and range, what the intercepts mean, and what the slope (or curve) means— that's a real-world skill. Interpreting a Graph In the x -axis on the graph represents the 120-minute bike ride Juan went on. The y -axis represents how far away he was from his home. Interpret the x - and y -intercept. For each segment, find the slope. Create an interpretation of this graph (i.e., make up a story that goes with it). ( 0 , 0 ) is the x - and y -intercept and represents Juan at home before his bike ride. The distance from home is 0 miles and 0 minutes have passed. In the first 30 minutes, the slope is 1 5 and indicates Juan is traveling 1 mile for every 5 minutes. Between 30 and 60 minutes, the slope is 0 and indicates that he’s not riding the bike (the distance is not increasing). Then between 60 and 90 minutes, the slope is 1 5 again. Finally, after 90 minutes the slope is − 4 15 , meaning Juan is getting 4 miles closer to home every 15 minutes. Answers will vary. Juan left his house for a bike ride. After 30 minutes, he was 6 miles from home and he stopped for ice cream at his local ice-cream truck. He enjoyed his ice cream for 30 minutes. He then jumped back on his bike and rode to his friend’s house. He arrived there 30 minutes later. His friend’s house was 12 miles from his home. His friend was not home so he immediately turned around and quickly rode home in 45 minutes. Modeling Applications Using Slope and y -Intercept Many real-world applications are modeled by linear equations. We will review a few applications here so you can understand how equations written in slope-intercept form relate to real-world situations. Usually when a linear equation model uses real-world data, different letters are used for the variables instead of using only x and y . The variable names often remind us of what quantities are being measured. Also, we often need to extend the axes in our rectangular coordinate system to bigger positive and negative numbers to accommodate the data in the application. Converting Temperature The equation F = 9 5 C + 32 is used to convert temperatures from degrees Celsius ( C ) to degrees Fahrenheit ( F ). Find the Fahrenheit temperature for a Celsius temperature of 0°. Find the Fahrenheit temperature for a Celsius temperature of 20°. Interpret the slope and F -intercept of the equation. Graph the equation. Find the Fahrenheit temperature for a Celsius temperature of 0°. Find F when C = 0 . F = 9 5 ( 0 ) + 32 Simplify. F = 32 Find the Fahrenheit temperature for a Celsius temperature of 20°. Find F when C = 20 . F = 9 5 ( 20 ) + 32 Simplify. F = 36 + 32 Simplify. F = 68 Interpret the slope and F -intercept of the equation. Even though this equation uses F and C , it is still in slope-intercept form. y = m x + b F = m C + b F = 9 5 C + 32 The slope 9 5 means that the temperature Fahrenheit ( F ) increases 9 degrees when the temperature Celsius ( C ) increases 5 degrees. The F -intercept means that when the temperature is 0° on the Celsius scale, it is 32° on the Fahrenheit scale. Graph the equation. We will need to use a larger scale than our usual. Start at the F -intercept ( 0 , 32 ) , and then count out the rise of 9 and the run of 5 to get a second point as shown in . Calculating Driving Costs Sam drives a delivery van. The equation C = 0.5 d + 60 models the relation between his weekly cost, C , in dollars and the number of miles, d , that he drives. Find Sam’s cost for a week when he drives 0 miles. Find the cost for a week when he drives 250 miles. Interpret the slope and C -intercept of the equation. Graph the equation. Find Sam’s cost for a week when he drives 0 miles. Find C when d = 0. C = 0.5 ( 0 ) + 60 Simplify. C = 60 Sam’s costs are $60 when he drives 0 miles. Find the cost for a week when he drives 250 miles. Find C when d = 250 . C = 0.5 ( 250 ) + 60 Simplify. C = 185 Sam’s costs are $185 when he drives 250 miles. Interpret the slope and C -intercept of the equation. y = m x + b C = 0.5 d + 60 The slope, 0.5, means that the weekly cost, C , increases by $0.50 when the number of miles driven, d , increases by 1. The C -intercept means that when the number of miles driven is 0, the weekly cost is $60. Graph the equation ( ). We’ll need to use a larger scale than usual. Start at the C -intercept (0, 60). To count out the slope m = 0.5 , we rewrite it as an equivalent fraction that will make our graphing easier. m = 0.5 = 5 10 = 50 100 So to graph the next point go up 50 from the intercept of 60 and then to the right 100. The second point will be (100, 110). Check Your Understanding Key Terms intercepts of a line slope slope-intercept form Key Concepts Every linear function can be graphically represented by a unique line that shows all the solutions of the equation. The points where the graph of a line intersects the x -axis and y -axis are called the intercepts of the line. Most lines will have one x -intercept and one y -intercept. Only if the line is straight vertical (no y -intercept) or straight horizontal (no x -intercept) will it not have both intercepts. Note that a line that is straight vertical is not a function, but a line that is straight horizontal is a function. Since any two points determine a straight line, any linear function can be graphed if both intercepts are known. The slope of a linear function is the ratio of the vertical change divided by the horizontal change. It is often referred to as r i s e r u n . A formula for finding the slope of linear functions is y 2 − y 1 x 2 − x 1 , for any two points of the linear function ( x 1 , y 1 ) and ( x 2 , y 2 ) . Formulas To calculate slope ( m ) , use the formula m = r i s e r u n , where the rise measures the vertical change and the run measures the horizontal change. To find the slope of the line between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) , use the formula m = y 2 − y 1 x 2 − x 1", "section": "Graphing Functions", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Systems of Linear Equations in Two Variables Fruits and vegetables at a farmer’s market. (credit: “California Ave. Farmers’ Market” by Jun Seita/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Determine and show whether an ordered pair is a solution to a system of equations. Solve systems of linear equations using graphical methods. Solve systems of linear equations using substitution. Solve systems of linear equations using elimination. Identify systems with no solution or infinitely many solutions. Solve applications of systems of linear equations. In this section, we will learn how to solve systems of linear equations in two variables. There are several real-world scenarios that can be represented by systems of linear equalities. Suppose two friends, Andrea and Bart, go shopping at a farmers market to buy some vegetables. Andrea buys 2 tomatoes and 4 cucumbers and spends $2.00. Bart buys 4 tomatoes and 5 cucumbers and spends $2.95. What is the price of each vegetable? Determining If an Ordered Pair Is a Solution to a System of Equations When we solved linear equations in Linear Equations in One Variable with Applications and Linear Inequalities in One Variable with Applications , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations . In this section, we will focus our work on systems of two linear equations in two unknowns (variables) and applications of systems of linear equations. An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. { 2 x + y = 7 x − 2 y = 6 A linear equation in two variables, such as 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line. To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) that make both equations true. These are called the solutions of a system of equations . To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. Determining Whether an Ordered Pair Is a Solution to the System Determine whether the ordered pair is a solution to the system. { x − y = − 1 2 x − y = − 5 ( − 2 , − 1 ) ( − 4 , − 3 ) { x − y = − 1 2 x − y = − 5 We substitute x = − 2 and y = − 1 into both equations. x − y = − 1 − 2 − ( − 1 ) = ? − 1 − 1 = − 1 ✓ 2 x − y = − 5 2 ( − 2 ) - ( − 1 ) = ? − 5 - 3 ≠ − 5 ( − 2 , − 1 ) does not make both equations true. ( − 2 , − 1 ) is not a solution. We substitute x = − 4 and y = − 3 into both equations. x − y = − 1 − 4 − ( − 3 ) = ? − 1 − 1 = − 1 ✓ 2 x − y = − 5 2 • ( − 4 ) − ( − 3 ) = ? − 5 − 5 = − 5 ✓ ( − 4 , − 3 ) makes both equations true. ( − 4 , − 3 ) is a solution. Determining Whether an Ordered Pair Is a Solution to the System Determine whether the ordered pair is a solution to the system { y = 3 2 x + 1 2 x − 3 y = 7 ( − 4 , − 5 ) ( − 4 , 5 ) { y = 3 2 x + 1 2 x − 3 y = 7 Substitute − 4 for x and − 5 for y into both equations. − 5 = ? 3 2 ( − 4 ) + 1 − 5 = ? 3 ( − 2 ) + 1 − 5 = ? − 6 + 1 − 5 = − 5 ✓ 2 ( − 4 ) − 3 ( − 5 ) = ? 7 ( − 8 ) − ( − 15 ) = ? 7 − 8 + 15 = ? 7 7 = 7 ✓ ( − 4 , − 5 ) is a solution. { y = 3 2 x + 1 2 x − 3 y = 7 Substitute − 4 for x and 5 for y into both equations. 5 = ? 3 2 ( − 4 ) + 1 5 = ? 3 ( − 2 ) + 1 5 = ? − 6 + 1 5 ≠ − 5 2 ( − 4 ) − 3 ( 5 ) = ? 7 ( − 8 ) − ( 15 ) = ? 7 − 8 − 15 = ? 7 − 23 ≠ 7 ( − 4 , 5 ) is not a solution. Solving Systems of Linear Equations Using Graphical Methods We will use three methods to solve a system of linear equations. The first method we will use is graphing. The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what points the lines have in common, we will find the solution to the system. Most linear equations in one variable have one solution; but for some equations called contradictions , there are no solutions, and for other equations called identities , all numbers are solutions. Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in . Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end you will decide which method was the most convenient way to solve this system. The steps to use to solve a system of linear equations by graphing are shown here. Step 1: Graph the first equation. Step 2: Graph the second equation on the same rectangular coordinate system. Step 3: Determine whether the lines intersect, are parallel, or are the same line. Step 4: Identify the solution to the system. If the lines intersect, identify the point of intersection. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions. Step 5: Check the solution in both equations. Solving a System of Linear Equations by Graphing Solve this system of linear equations by graphing. { 2 x + y = 7 x − 2 y = 6 Step 1: Graph the first equation. To graph the first line, write the equation in slope-intercept form. 2 x + y = 7 y = - 2 x + 7 m = - 2 b = 7 { 2 x + y = 7 x - 2 y = 6 Step 2: Graph the second equation on the same rectangular coordinate system. To graph the second line, use intercepts. x - 2 y = 6 ( 0 , - 3 ) ( 6 , 0 ) Step 3: Determine whether the lines intersect, are parallel, or are the same line. Look at the graph of the lines. The lines intersect. Step 4 and Step 5: Identify the solution to the system. If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions. Since the lines intersect, find the point of intersection. Check the point in both equations. The lines intersect at ( 4 , - 1 ) . 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 8 - 1 = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ The solution is ( 4 , - 1 ) . Solving Systems of Linear Equations Using Substitution We will now solve systems of linear equations by the substitution method. We will use the same system we used for graphing. { 2 x + y = 7 x − 2 y = 6 We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy. Then, we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those! After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true. This process is summarized here: Step 1: Solve one of the equations for either variable. Step 2: Substitute the expression from Step 1 into the other equation. Step 3: Solve the resulting equation. Step 4: Substitute the solution in Step 3 into either of the original equations to find the other variable. Step 5: Write the solution as an ordered pair. Step 6: Check that the ordered pair is a solution to both original equations. Solving a System of Linear Equations Using Substitution Solve this system of linear equations by substitution: { 2 x + y = 7 x − 2 y = 6 Step 1: Solve one of the equations for either variable. We’ll solve the first equation for y . { 2 x + y = 7 x - 2 y = 6 2 x + y = 7 y = 7 − 2 x Step 2: Substitute the expression from Step 1 into the other equation. We replace y in the second equation with the expression 7 - 2 x . x - 2 y = 6 x - 2 ( 7 - 2 x ) = 6 Step 3: Solve the resulting equation. Now we have an equation with just 1 variable. We know how to solve this! x - 2 ( 7 - 2 x ) = 6 x - 14 + 4 x = 6 5 x = 20 x = 4 Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable. We’ll use the first equation and replace x with 4. 2 x + y = 7 2 ( 4 ) + y = 7 8 + y = 7 y = - 1 Step 5: Write the solution as an ordered pair. The ordered pair is ( x , y ). ( 4 , - 1 ) Step 6: Check that the ordered pair is a solution to both original equations. Substitute x = 4 , y = - 1 into both equations and make sure they are both true. 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ Both equations are true. ( 4 , - 1 ) is the solution to the system. Solving Systems of Linear Equations Using Elimination We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small, and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression. The third method of solving systems of linear equations is called the elimination method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there. The elimination method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal. For any expressions a , b , c , and d : if a = b and c = d then a + c = b + d . To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable. Notice how that works when we add these two equations together: { 3 x + y = 5 2 x − y = 0 _ 5 x = 5 The y ’s add to zero and we have one equation with one variable. Let us try another one: { x + 4 y = − 2 2 x + 5 y = − 2 This time we do not see a variable that can be immediately eliminated if we add the equations. But if we multiply the first equation by − 2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by − 2 . { − 2 ( x + 4 y ) = − 2 ( –2 ) 2 x + 5 y = − 2 Then rewrite the system of equations. { − 2 x − 8 y = 4 2 x + 5 y = − 2 Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations. { − 2 x − 8 y = 4 2 x + 5 y = − 2 _ − 3 y = 2 Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations. Here’s a summary of using the elimination method: Step 1: Write both equations in standard form. If any coefficients are fractions, clear them. Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites. Step 3: Add the equations resulting from Step 2 to eliminate one variable. Step 4: Solve for the remaining variable. Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Step 6: Write the solution as an ordered pair. Step 7: Check that the ordered pair is a solution to both original equations. Solving a System of Linear Equations Using Elimination Solve this system of linear equations by elimination: { 2 x + y = 7 x − 2 y = 6 Step 1: Write both equations in standard form. If any coefficients are fractions, clear them. Both equations are in standard form, A x + B y = C . There are no fractions. { 2 x + y = 7 x - 2 y = 6 Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites. We can eliminate the y ’s by multiplying the first equation by 2. Multiply both sides of 2 x + y = 7 by 2. { 2 x + y = 7 x - 2 y = 6 { 2 ( 2 x + y ) = 2 ( 7 ) x - 2 y = 6 Step 3: Add the equations resulting from Step 2 to eliminate one variable. We add the x ’s, y ’s, and constants. { 4 x + 2 y = 14 x - 2 y = 6 _ 5 x = 20 Step 4: Solve for the remaining variable. Solve for x . x = 4 Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Substitute x = 4 into the second equation, x - 2 y = 6 . Then solve for y . x - 2 y = 6 4 - 2 y = 6 - 2 y = 2 y = - 1 Step 6: Write the solution as an ordered pair. Write it as ( x , y ) . ( 4 , - 1 ) Step 7: Check that the ordered pair is a solution to both original equations. Substitute x = 4 , y = - 1 into 2 x + y = 7 and x - 2 y = 6 . Do they make both equations true? Yes! 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ The solution is ( 4 , - 1 ) . Identifying Systems with No Solution or Infinitely Many Solutions In all the systems of linear equations so far, the lines intersected, and the solution was one point. In and , we will look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions. Solving a System of Linear Equations with No Solution Solve the system by a method of your choice: { y = 1 2 x − 3 x − 2 y = 4 Let us solve the system of linear equations by graphing. { y = 1 2 x − 3 x − 2 y = 4 To graph the first equation, we will use its slope and y -intercept. y = 1 2 x − 3 m = 1 2 b = − 3 To graph the second equation, we will use the intercepts. x − 2 y = 4 x y 0 −2 4 0 Graph the lines ( ). Determine the points of intersection. The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system. Solving a System of Linear Equations with Infinite Solutions Solve the system by a method of your choice: { y = 2 x − 3 − 6 x + 3 y = − 9 Let us solve the system of linear equations by graphing. { y = 2 x − 3 − 6 x + 3 y = − 9 Find the slope and y -intercept of the first equation. y = 2 x − 3 m = 2 b = − 3 Find the intercepts of the second equation. − 6 x + 3 y = − 9 x y 0 −3 3 2 0 Graph the lines ( ). The lines are the same! Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to this system. In the previous example, if you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system. We say the two lines are coincident. Coincident lines have the same slope and same y -intercept. A system of equations that has at least one solution is called a consistent system . A system with parallel lines has no solution. We call a system of equations like this an inconsistent system . It has no solution. We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent. If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines. Let us sum this up by looking at the graphs of the three types of systems. See and the table that follows Lines Intersecting Parallel Coincident Number of Solutions 1 point No solution Infinitely many Consistent/Inconsistent Consistent Inconsistent Consistent Dependent/Independent Independent Independent Dependent Using Matrices and Cramer’s Rule to Solve Systems of Linear Equations An m by n matrix is an array with m rows and n columns, where each item in the matrix is a number. Matrices are used for many things, but one thing they can be used for is to represent systems of linear equations. For example, the system of linear equations { 2 x + y = 7 x − 2 y = 6 can be represented by the following matrix: [ 2 1 7 1 − 2 6 ] To use Cramer’s Rule, you need to be able to take the determinant of a matrix. The determinant of a 2 by 2 matrix A , denoted | A | , is | A | = | a 11 a 12 a 21 a 22 | = ( a 11 × a 22 ) + ( a 21 × a 12 ) For example, the determinant of the matrix | 2 1 3 − 2 | = ( 2 × − 2 ) − ( 3 × 1 ) = − 4 − 3 = − 7. Cramer’s Rule involves taking three determinants: The determinant of the first two columns, denoted | D | ; The determinant of the first column and the third column, denoted | D y | ; The determinant of the third column and the first column, denoted | D x | . Going back to the original matrix [ 2 1 7 1 − 2 6 ] | D | = | 2 1 1 − 2 | = − 4 − 1 = − 5 | D y | = | 2 7 1 6 | = 12 − 7 = 5 | D x | = | 7 1 6 − 2 | = − 14 − 6 = − 20 Now Cramer’s Rule for the solution of the system will be: x = | D x | | D | , y = | D y | | D | Putting in the values for these determinants, we have x = − 20 − 5 = 4 ; y = 5 − 5 = − 1. The solution to the system is the ordered pair ( 4 , − 1 ) . Solving Applications of Systems of Linear Equations Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we will first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system. Step 1 : Read the problem. Make sure all the words and ideas are understood. Step 2: Identify what we are looking for. Step 3: Name what we are looking for. Choose variables to represent those quantities. Step 4: Translate into a system of equations. Step 5: Solve the system of equations using good algebra techniques. Step 6: Check the answer in the problem and make sure it makes sense. Step 7: Answer the question with a complete sentence. Applying System to a Real-World Application Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 a year plus $15 for each training session. Option B would pay her $10,000 a year plus $40 for each training session. How many training sessions would make the salary options equal? Step 1: Read the problem. Step 2: Identify what we are looking for. We are looking for the number of training sessions that would make the pay equal. Step 3: Name what we are looking for. Let s = Heather’s salary , and n = the number of training sessions Step 4: Translate into a system of equations. Option A would pay her $25,000 plus $15 for each training session. s = 25,000 + 15 n Option B would pay her $10,000 + $40 for each training session. s = 10,000 + 40 n The system is shown. { s = 25,000 + 15 n s = 10,000 + 40 n Step 5: Solve the system of equations. We will use substitution. Substitute 25,000 + 15 n for s in the second equation s = 25,000 + 15 n s = 10,000 + 40 n Solve for n . 25,000 + 15 n = 10,000 + 40 n 25,000 = 10,000 + 25 n 15,000 = 25 n 600 = n Step 6: Check the answer. Are 600 training sessions a year reasonable? Are the two options equal when n = 600 ? Substitute into each equation. s = 25 , 000 + 15 ( 600 ) = 34 , 000 s = 10 , 000 + 40 ( 600 ) = 34 , 000 Step 7: Answer the question. The salary options would be equal for 600 training sessions. Practice with Solving Applications of Systems of Equations Applications of Systems of Linear Equations Check Your Understanding Key Terms system of linear equations solutions of a system of equations contradictions identities coincident lines consistent system of linear equations inconsistent system of linear equations Key Concepts To solve a system of linear equations means finding the point or points where the two linear equations intersect. Two lines can intersect at one point, no points if they are parallel, or every point if they are the same equation. Systems of linear equations can be solved by graphing, by using substitution, or by using the elimination method. Videos Practice with Solving Applications of Systems of Equations Applications of Systems of Linear Equations", "section": "Systems of Linear Equations in Two Variables", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Systems of Linear Inequalities in Two Variables Many college students find part-time jobs at places such as coffee shops to help pay for college. (credit: modification of work “TULLY’s COFFEE” by MIKI Yoshihito/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Demonstrate whether an ordered pair is a solution to a system of linear inequalities. Solve systems of linear inequalities using graphical methods. Graph systems of linear inequalities. Interpret and solve applications of linear inequalities. In this section, we will learn how to solve systems of linear inequalities in two variables. In Systems of Linear Equations in Two Variables , we learned how to solve for systems of linear equations in two variables and found a solution that would work in both equations. We can solve systems of inequalities by graphing each inequality (as discussed in Graphing Linear Equations and Inequalities ) and putting these on the same coordinate system. The double-shaded part will be our solution to the system. There are many real-life examples for solving systems of linear inequalities. Consider Ming who has two jobs to help her pay for college. She works at a local coffee shop for $7.50 per hour and at a research lab on campus for $12 per hour. Due to her busy class schedule, she cannot work more than 15 hours per week. If she needs to make at least $150 per week, can she work seven hours at the coffee shop and eight hours in the lab? Determining If an Ordered Pair Is a Solution of a System of Linear Inequalities The definition of a system of linear inequalities is similar to the definition of a system of linear equations. A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here. { x + 4 y ≥ 10 3 x − 2 y < 12 To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) that make both inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the xy -coordinate system that includes all the points whose ordered pairs make the inequalities true. To determine if an ordered pair is a solution to a system of two inequalities, substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system. Determining Whether an Ordered Pair Is a Solution to a System Determine whether the ordered pair is a solution to the system: { x + 4 y ≥ 10 3 x − 2 y < 12 ( − 2 , 4 ) ( 3 , 1 ) Is the ordered pair ( − 2 , 4 ) a solution? We substitute x = − 2 and y = 4 into both inequalities. x + 4 y ≥ 10 - 2 + 4 ( 4 ) ≥ ? 10 14 ≥ 10 true 3 x − 2 y < 12 3 ( - 2 ) − 2 ( 4 ) < ? 12 − 14 < 12 true The ordered pair ( − 2 , 4 ) made both inequalities true. Therefore ( − 2 , 4 ) is a solution to this system. Is the ordered pair ( 3 , 1 ) a solution? We substitute x = 3 and y = 1 into both inequalities. x + 4 y ≥ 10 3 + 4 ( 1 ) ≥ ? 10 7 ≥ 10 false 3 x − 2 y < 12 3 ( 3 ) − 2 ( 1 ) < ? 12 7 < 12 true The ordered pair ( 3 , 1 ) made one inequality true, but the other one false. Therefore ( 3 , 1 ) is not a solution to this system. Solving Systems of Linear Inequalities Using Graphical Methods The solution to a single linear inequality was the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. We will review graphs of linear inequalities and solve the linear inequality from its graph. Solving a System of Linear Inequalities by Graphing Use to solve the system of linear inequalities: { y ≥ 2 x − 1 y < x + 1 To solve the system of linear inequalities we look at the graph and find the region that satisfies BOTH inequalities. To do this we pick a test point and check. Let's us pick ( − 1 , − 1 ) . Is ( - 1 , - 1 ) a solution to y ≥ 2 x − 1 ? - 1 ≥ ? 2 ( - 1 ) − 1 − 1 ≥ − 3 true Is ( - 1 , - 1 ) a solution to y < x + 1 ? - 1 < ? - 1 + 1 − 1 < 0 true The region containing ( − 1 , − 1 ) is the solution to the system of linear inequalities. Notice that the solution is all the points in the area shaded twice, which appears as the darkest shaded region. Graphing Systems of Linear Inequalities We learned that the solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region by graphing, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. Step 1: Graph the first inequality. Graph the boundary line. Shade in the side of the boundary line where the inequality is true. Step 2: On the same grid, graph the second inequality. Graph the boundary line. Shade in the side of that boundary line where the inequality is true. Step 3: The solution is the region where the shading overlaps. Step 4: Check by choosing a test point. Solving a System of Linear Inequalities by Graphing Solve the system by graphing: { x − y > 3 y < − 1 5 x + 4 Graph x − y > 3 by graphing x − y = 3 and testing a point ( ). The intercepts are x = 3 and y = − 3 and the boundary line will be dashed. Test ( 0 , 0 ) which makes the inequality false so shade the side that does not contain ( 0 , 0 ) . Graph y < − 1 5 x + 4 by graphing y = − 1 5 x + 4 using the slope m = − 1 5 and y -intercept b = 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Choose a test point in the solution and verify that it is a solution to both inequalities. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region. Graphing a System of Linear Inequalities Solve the system by graphing: { x − 2 y < 5 y > − 4 Graph x − 2 y < 5 by graphing x − 2 y = 5 ( ) and testing a point. The intercepts are x = 5 and y = − 2.5 and the boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Graph y > − 4 by graphing y = − 4 and recognizing that it is a horizontal line through y = − 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true so shade the side that contains ( 0 , 0 ) . The point ( 0 , 0 ) is in the solution, and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice, which appears as the darkest shaded region. Systems of linear inequalities where the boundary lines are parallel might have no solution. We will see this in the next example. Graphing Parallel Boundary Lines with No Solution Solve the system by graphing: { 4 x + 3 y ≥ 12 y < − 4 3 x + 1 Graph 4 x + 3 y ≥ 12 , by graphing 4 x + 3 y = 12 ( ) and testing a point. The intercepts are x = 3 and y = 4 and the boundary line will be solid. Test ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) . Graph y < − 4 3 x + 1 by graphing y = − 4 3 x + 1 using the slope m = − 4 3 and y -intercept b = 1 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . No shared point exists in both shaded regions, so the system has no solution. Some systems of linear inequalities where the boundary lines are parallel will have a solution. We will see this in the next example. Graphing Parallel Boundary Lines with a Solution Solve the system by graphing: { y > 1 2 x − 4 x − 2 y < − 4 Graph y > 1 2 x − 4 by graphing y = 1 2 x − 4 using the slope m = 1 2 and the y -intercept b = − 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Graph x − 2 y < − 4 by graphing x − 2 y = − 4 ( ) and testing a point. The intercepts are x = − 4 and y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalities. Test ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) . No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice which is also the solution to x − 2 y < − 4 . Interpreting and Solving Applications of Linear Inequalities When solving applications of systems of inequalities, first translate each condition into an inequality. Then graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so add inequalities to the system as additional requirements. Applying Linear Inequalities to Calculating Photo Costs A photographer sells their prints at a booth at a street fair. At the start of the day, they want to have at least 25 photos to display at their booth. Each small photo they display costs $4 and each large photo costs $10. They do not want to spend more than $200 on photos to display. Write a system of inequalities to model this situation. Graph the system. Could they display 10 small and 20 large photos? Could they display 20 large and 10 small photos? Let x = the number of small photos and y = the number of large photos . To find the system of equations translate the information. They want to have at least 25 photos. The number of small plus the number of large should be at least 25. x + y ≥ 25 $4 for each small and $10 for each large must be no more than $200 4 x + 10 y ≤ 200 The number of small photos must be greater than or equal to 0. x ≥ 0 The number of large photos must be greater than or equal to 0. y ≥ 0 We have our system of equations. { x + y ≥ 25 4 x + 10 y ≤ 200 x ≥ 0 y ≥ 0 Since x ≥ 0 and y ≥ 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only Quadrant I. To graph x + y ≥ 25 , graph x + y = 25 as a solid line. Choose ( 0 , 0 ) as a test point. Since it does not make the inequality true, shade the side that does not include the point ( 0 , 0 ) . To graph 4 x + 10 y ≤ 200 , graph 4 x + 10 y = 200 as a solid line. Choose ( 0 , 0 ) as a test point. Since it does make the inequality true, shade (bottom left) the side that include the point ( 0 , 0 ) . The solution of the system is the region of that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x -axis from ( 25 , 0 ) to ( 55 , 0 ) . To determine if 10 small and 20 large photos would work, we look at the graph to see if the point ( 10 , 20 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, so the photographer would not display 10 small and 20 large photos. To determine if 20 small and 10 large photos would work, we look at the graph to see if the point ( 20 , 10 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so the photographer could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality. Solving Systems of Linear Inequalities by Graphing Systems of Linear Inequalities Check Your Understanding Key Terms system of linear inequalities Key Concepts To solve a system of linear inequalities means to find the area(s) where the points in that area make all the linear inequalities true. Systems of linear inequalities can be solved by graphing the linear equations associated with the inequalities, then 'testing' points to see whether the values of the point make the equation true or not. Videos Solving Systems of Linear Inequalities by Graphing Systems of Linear Inequalities", "section": "Systems of Linear Inequalities in Two Variables", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Linear Programming The aftermath of an earthquake and tsunami. (credit: modification of work \"Earthquake and Tsunami Japan\" by Climate and Ecosystems Change Adaptation Research University Network/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compose an objective function to be minimized or maximized. Compose inequalities representing a system application. Apply linear programming to solve application problems. Imagine you hear about some natural disaster striking a far-away country; it could be an earthquake, a fire, a tsunami, a tornado, a hurricane, or any other type of natural disaster. The survivors of this disaster need help—they especially need food, water, and medical supplies. You work for a company that has these supplies, and your company has decided to help by flying the needed supplies into the disaster area. They want to maximize the number of people they can help. However, there are practical constraints that need to be taken into consideration; the size of the airplanes, how much weight each airplane can carry, and so on. How do you solve this dilemma? This is where linear programming comes into play. Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints. As a field, linear programming began in the late 1930s and early 1940s. It was used by many countries during World War II; countries used linear programming to solve problems such as maximizing troop effectiveness, minimizing their own casualties, and maximizing the damage they could inflict upon the enemy. Later, businesses began to realize they could use the concept of linear programming to maximize output, minimize expenses, and so on. In short, linear programming is a method to solve problems that involve finding a maximum or minimum where a linear function is constrained by various factors. A Mathematician Invents a “Tsunami Cannon” On December 26, 2004, a massive earthquake occurred in the Indian Ocean. This earthquake, which scientists estimate had a magnitude of 9.0 or 9.1 on the Richter Scale, set off a wave of tsunamis across the Indian Ocean. The waves of the tsunami averaged over 30 feet (10 meters) high, and caused massive damage and loss of life across the coastal regions bordering the Indian Ocean. Usama Kadri works as an applied mathematician at Cardiff University in Wales. His areas of research include fluid dynamics and non-linear phenomena. Lately, he has been focusing his research on the early detection and easing of the effects of tsunamis. One of his theories involves deploying a series of devices along coastlines which would fire acoustic-gravity waves (AGWs) into an oncoming tsunami, which in theory would lessen the force of the tsunami. Of course, this is all in theory, but Kadri believes it will work. There are issues with creating such a device: they would take a tremendous amount of electricity to generate an AGW, for instance, but if it would save lives, it may well be worth it. Compose an Objective Function to Be Minimized or Maximized An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized. Composing an Objective Function for Selling Two Products Miriam starts her own business, where she knits and sells scarves and sweaters out of high-quality wool. She can make a profit of $8 per scarf and $10 per sweater. Write an objective function that describes her profit. Let x represent the number of scarves sold, and let y represent the number of sweaters sold. Let P represent profit. Since each scarf has a profit of $8 and each sweater has a profit of $10, the objective function is P = 8 x + 10 y . Composing an Objective Function for Production William’s factory produces two products, widgets and wadgets. It takes 24 minutes for his factory to make 1 widget, and 32 minutes for his factory to make 1 wadget. Write an objective function that describes the time it takes to make the products. Let x equal the number of widgets made; let y equal the number of wadgets made; let T represent total time. The objective function is T = 24 x + 32 y . Composing Inequalities Representing a System Application For our two examples of profit and production, in an ideal world the profit a person makes and/or the number of products a company produces would have no restrictions. After all, who wouldn’t want to have an unrestricted profit? However in reality this is not the case; there are usually several variables that can restrict how much profit a person can make or how many products a company can produce. These restrictions are called constraints . Many different variables can be constraints. When making or selling a product, the time available, the cost of manufacturing and the amount of raw materials are all constraints. In the opening scenario with the tsunami, the maximum weight on an airplane and the volume of cargo it can carry would be constraints. Constraints are expressed as linear inequalities; the list of constraints defined by the problem forms a system of linear inequalities that, along with the objective function, represent a system application. Representing the Constraints for Selling Two Products Two friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for profit and the two constraints. Let x represent the number of scarves sold, and let y represent the number of sweaters sold. There are two constraints: the number of items the business can make in a day (a maximum of 8) and the number of bags of knitting wool they can use per day (a maximum of 27). The first constraint (total number of items in a day) is written as: x + y ≤ 8 Since each scarf takes 3 bags of knitting wool and each sweater takes 4 bags of knitting wool, the second constraint, total bags of knitting wool per day, is written as: 3 x + 4 y ≤ 27 In summary, here are the equations that represent the new business: P = 8 x + 10 y ; This is the profit equation: The business makes $8 per scarf and $10 per sweater. x + y ≤ 8 3 x + 4 y ≤ 27 Representing Constraints for Production A factory produces two products, widgets and wadgets. It takes 24 minutes for the factory to make 1 widget, and 32 minutes for the factory to make 1 wadget. Research indicates that long-term demand for products from the factory will result in average sales of 12 widgets per day and 10 wadgets per day. Because of limitations on storage at the factory, no more than 20 widgets or 17 wadgets can be made each day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for time and the two constraints. Let x equal the number of widgets made; let y equal the number of wadgets made. Based on the long-term demand, we know the factory must produce a minimum of 12 widgets and 10 wadgets per day. We also know because of storage limitations, the factory cannot produce more than 20 widgets per day or 17 wadgets per day. Writing those as inequalities, we have: x ≥ 12 y ≥ 10 x ≤ 20 y ≤ 17 The number of widgets made per day must be between 12 and 20, and the number of wadgets made per day must be between 10 and 17. Therefore, we have: 12 ≤ x ≤ 20 10 ≤ y ≤ 17 The system is: T = 24 x + 32 y 12 ≤ x ≤ 20 10 ≤ y ≤ 17 T is the variable for time; it takes 24 minutes to make a widget and 32 minutes to make a wadget. Applying Linear Programming to Solve Application Problems There are four steps that need to be completed when solving a problem using linear programming. They are as follows: Step 1: Compose an objective function to be minimized or maximized. Step 2: Compose inequalities representing the constraints of the system. Step 3: Graph the system of inequalities representing the constraints. Step 4: Find the value of the objective function at each corner point of the graphed region. The first two steps you have already learned. Let’s continue to use the same examples to illustrate Steps 3 and 4. Solving a Linear Programming Problem for Two Products Three friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Determine the number of scarves and sweaters they should make each day to maximize their profit. Step 1: Compose an objective function to be minimized or maximized. From , the objective function is P = 8 x + 10 y . Step 2: Compose inequalities representing the constraints of the system. From , the constraints are x + y ≤ 8 and 3 x + 4 y ≤ 27 . Step 3: Graph the system of inequalities representing the constraints. Using methods discussed in Graphing Linear Equations and Inequalities , the graphs of the constraints are shown below. Because the number of scarves ( x ) and the number of sweaters ( y ) both must be non-negative numbers (i.e., x ≥ 0 and y ≥ 0 ), we need to graph the system of inequalities in Quadrant I only. shows each constraint graphed on its own axes, while shows the graph of the system of inequalities (the two constraints graphed together). In , the large shaded region represents the area where the two constraints intersect. If you are unsure how to graph these regions, refer back to Graphing Linear Equations and Inequalities . Graphs of each constraint Graph of the System of Inequalities Step 4: Find the value of the objective function at each corner point of the graphed region. The “graphed region” is the area where both of the regions intersect; in , it is the large shaded area. The “corner points” refer to each vertex of the shaded area. Why the corner points? Because the maximum and minimum of every objective function will occur at one (or more) of the corner points. shows the location and coordinates of each corner point. Graph of Region with Corner Points Three of the four points are readily found, as we used them to graph the regions; the fourth point, the intersection point of the two constraint lines, will have to be found using methods discussed in Systems of Linear Equations in Two Variables , either using substitution or elimination. As a reminder, set up the two equations of the constraint lines: 3 x + 4 y = 27 x + y = 8 For this example, substitution will be used. x + y = 8 y = 8 - x . Substituting 8 - x into the first equation for y , we have 3 x + 4 ( 8 − x ) = 27 3 x + 32 − 4 x = 27 − x = − 5 x = 5 Now, substituting the 5 in for x in either equation to solve for y . Choosing the second equation, we have: 5 + y = 8 y = 3 Therefore, x = 5 , and y = 3 . To find the value of the objective function, P = 8 x + 10 y , put the coordinates for each corner point into the equation and solve. The largest solution found when doing this will be the maximum value, and thus will be the answer to the question originally posed: determining the number of scarves and sweaters the new business should make each day to maximize their profit. Corner ( x , y ) Objective Function P = 8 x + 1 0 y ( 0 , 0 ) P = 8 ( 0 ) + 10 ( 0 ) = 0 ( 0 , 6.75 ) P = 8 ( 0 ) + 10 ( 6.75 ) = 67.5 ( 5 , 3 ) P = 8 ( 5 ) + 10 ( 3 ) = 40 + 30 = 70 ( 8 , 0 ) P = 8 ( 8 ) + 10 ( 0 ) = 64 The maximum value for the profit P occurs when x = 5 and y = 3 . This means that to maximize their profit, the new business should make 5 scarves and 3 sweaters every day. Leonid Kantorovich Leonid Vitalyevich Kantorovich was born January 19, 1912, in St. Petersburg, Russia. Two major events affected young Leonid’s life: when he was five, the Russian Revolution began, making life in St. Petersburg very difficult; so much so that Leonid’s family fled to Belarus for a year. When Leonid was 10, his father died, leaving his mother to raise five children on her own. Despite the hardships, Leonid showed incredible mathematical ability at a young age. When he was only 14, he enrolled in Leningrad State University to study mathematics. Four years later, at age 18, he graduated with what would be equivalent to a Ph.D. in mathematics. Although his primary interests were in pure mathematics, in 1938 he began working on problems in economics. Supposedly, he was approached by a local plywood manufacturer with the following question: how to come up with a work schedule for eight lathes to maximize output, given the five different kinds of plywood they had at the factory. By July 1939, Leonid had come up with a solution, not only to the lathe scheduling problem but to other areas as well, such as an optimal crop rotation schedule for farmers, minimizing waste material in manufacturing, and finding optimal routes for transporting goods. The technique he discovered to solve these problems eventually became known as linear programming. He continued to use this technique for solving many other problems involving optimization, which resulted in the book The Best Use of Economic Resources , which was published in 1959. His continued work in linear programming would ultimately result in him winning the Nobel Prize of Economics in 1975. Check Your Understanding Key Terms linear programming objective function constraint Key Concepts Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints. An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized. In linear programming, a constraint is a restriction that affects the maximum or minimum values of an objective function. Through the creation of objective functions and restraints, a linear system can be developed and solved through linear programming. Projects Ratio and Proportion—Comparing Prices, Part 1 Go to your favorite coffee shops and find out what a same sized drink costs at each. You can do something similar for pizza as well. Find the unit rate (i.e., price per ounce or price per square inch). For example, go to your favorite coffee place and find the price per units on all their large coffee drinks. Or go to your favorite pizza place and compare prices of all their extra-large pizzas (by price per square inch). Write a report on the best deals. Ratio and Proportion—Comparing Prices, Part 2 Rather than comparing prices of different, but same sized drinks (or pizzas), compare unit prices of the same drinks but of different sizes. Find out what the best bargain is based on price per ounce, price per square inch, etc. For example, compare the prices of your favorite soft drink sold at a local store, but in various sizes (i.e., 12-ounce can, 16-ounce bottle, 20-ounce bottle, 1-liter bottle, and multipacks). Or go to a pizza place and find out what the best bargain is on their menu, based on price per square inch of pizza. Write a report on the best deals. Systems of Linear Inequalities—Comparing Cell Phone Plans Go to the websites of different cell phone companies and compare their plans. Write a report on “the best deals. \"Best Deals” doesn’t necessarily mean “cheapest.” You will need to look at what each company provides concerning restrictions (constraints) on minutes to talk. What are the constraints on the cell phone coverage for each company? Do they cover your area of the country well? Do they cover the entire United States well, or at least areas where you will be travelling? Is this coverage 5G, or is it less? Can you add a phone easily? Can you bring your previous phone number to this plan? The possibilities of constraints affecting each plan are several. So your task is to determine which plan is best, based on not only cost but also all constraints you deem important. Chapter Review Algebraic Expressions Linear Equations in One Variable with Applications Linear Inequalities in One Variable with Applications Ratio and Proportions Graphing Linear Equations and Inequalities Quadratic Equations In One Variable with Applications Functions Graphing Functions System of Linear Equations in Two Variables Systems of Linear Inequalities in Two Variables Linear Programming Chapter Test", "section": "Linear Programming", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Financial health helps you realize your goals. (credit: modification of work \"Budget and Bills\" by Alabama Extension/Flickr, Public Domain) The topic of money management is a broad and sometimes complex one. Ultimately, personal money management involves managing both our debt and also our savings and investments. In 2021, the average American had consumer debt balance of $96,371. Nearly $100,000 per person. And less than 25% of Americans are debt free. Consumer debt can include mortgages, credit cards, as well as student loans. A key question all consumers should consider is how to manage debt and not become overburdened by it. The first step is to create a budget, which puts earnings into perspective, indicating what we can, and cannot, afford. A budget also entails setting aside certain funds for savings and investment, which help us achieve our short- and long-term goals. Creating a budget requires an understanding of how money—debt and savings—works. Initially, percentages and interest need to be understood. They drive most of what happens with debt and savings. With that understanding, discussions of buying a house, a car, or incurring credit card debt can be addressed from a financial perspective. All the while, retirement is waiting. Preparing for retirement involves saving and saving earlier rather than later. The power of compound interest is on full display when saving early. This chapter covers some of the basics of money management: percentages, interest, budgeting, debt (student loans, mortgage, car, credit cards), savings, investments, and taxes.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Understanding Percent The federal budget describes how money is spent and how money is earned. (credit: \"Breakdown of revenues and outlays in 2021 US Federal budget\" Wikimedia Commons, Public Domain) Learning Objectives After completing this section, you should be able to: Define and calculate percent. Convert between percent, decimal, and fractional values. Calculate the total, percent, or part. Solve application problems involving percents. In 2020, the U.S. federal government budgeted $3.5 billion for the National Park Service , which appears to be a very large number (and is!) and a large portion of the total federal budget. However, the total outlays from the U.S. federal government in 2020 was $6.6 trillion . So, the amount budgeted for the National Park Service was less than one-tenth of 1 percent, or 1/10%, of the total outlays. This percent describes a specific number. Understanding that ratio puts the $3.5 billion budgeted to the National Park Service in perspective. This chapter focuses on percent as a primary tool for understanding money management. The interest paid on debt, the interest earned through investments, and even taxes are entirely determined using percent. This section introduces the basics of working with this invaluable tool. Define and Calculate Percent The word percent comes from the Latin phrase per centum, which means “by the hundred.” So any percent is a number divided by 100. Changing a percent to a fraction is to write the percent in its fractional form . To write n % in its fractional form is to write the percent as the fraction n 100 . A percent need not be an integer and does not have to be less than 100. Rewriting a Percent as a Fraction Rewrite the following as fractions: 18% 84% 38.7% 213% Using the definition and n = 18, 18% in fractional form is 18 100 . Using the definition and n = 84, 84% in fractional form is 84 100 . Using the definition and n = 38.7, 38.7% in fractional form is 38.7 100 . Using the definition and n = 213, 213% in fractional form is 213 100 . Convert Between Percent, Decimal, and Fractional Values When any calculation with a percent is to be performed, the form of the percent must be changed, either to its fractional form or its decimal form . We can change a percent into decimal form by dividing the percent by 100 and representing the result as a decimal. The decimal form of n % is found by calculating the decimal value of n ÷ 100 . Converting a Percent to Decimal Form Convert the following percents to decimal form: 17% 7% 18.45% To convert 17% to its decimal form, divide 17 by 100. This moves the decimal two places to the left, resulting in 0.17. The decimal form of 17% is 0.17. To convert 7% to its decimal form, divide 7 by 100. This moves the decimal two places to the left, resulting in 0.07. The decimal form of 7% is 0.07. To convert 18.45% to its decimal form, divide 18.45 by 100. This moves the decimal two places to the left, resulting in 0.1845. The decimal form of 18.45% is 0.1845. You should notice that, to convert from percent to decimal form, you can simply move the decimal two places to the left without performing the division. To convert the number x from decimal form to percent, multiply x by 100 and place a percent sign, %, after the number, ( x × 100 ) % . Converting the Decimal Form of a Percent to Percent Convert each of the following to percent: 0.34 4.15 0.0391 Using the formula and x = 0.34, we calculate ( 0.34 × 100 ) % , which gives us 34%. Using the formula and x = 4.15, we calculate ( 4.15 × 100 ) % , which gives us 415%. Using the formula and x = 0.0391, we calculate ( 0.0391 × 100 ) % , which gives us 3.91%. You should notice that, to convert from decimal form to percent form, you can simply move the decimal two places to the right without performing the multiplication. Calculate the Total, Percent, or Part The word “of” is used to indicate multiplication using fractions, as in “one-fourth of 56.” To find “one-fourth of 56” we would multiply 56 by one-fourth. We can think of percents as fractions with a specific denominator—100. So, to calculate “25% of 52,” we multiply 52 by 25%. But, first we need to convert the percent to either fractional form (25/100) or decimal form. Using the decimal form of 25% we have 0.25 × 52, which equals 13. In this problem, 52 is the total or base , 25 is the percentage , and 13 is the percentage of 52, or the part of 52. This is sometimes referred to as the amount . The mathematical formula relating the total (base), the percent in decimal form, and the part (amount) is part = percent × total , or, amount = percent × base . In all calculations, the percent is expressed in decimal form. Knowing any two of the values in our formula allows us to calculate the third value. In the following example, we know the total and the percent, and are asked to find the percentage of the total. Finding the Percent of a Total Determine 70% of 3,500 Determine 156% of 720 The total is x = 3,500, and the percent is n = 70. The decimal form of 70% is 0.70. To find the part, or percent of the total, substitute those values into the formula and calculate. part = percent × total = 0.70 × 3500 = 2450 From this, we say that 70% of 3,500 is 2,450. The total is x = 720, and the percent is n = 156. The decimal form of 156% is 1.56. To find the part, or percent of the total, substitute those values into the formula and calculate. part = percent × total = 1.56 × 720 = 1,123.2 From this, we say that 156% of 720 is 1,123.2. Finding Percent of a Total In the previous example, we knew the total and the percent and found the part using our formula. We may instead know the percent and the part, but not the total. We can use our formula again to solve for the total. Finding the Total from the Percent and the Part What is the total if 35% of the total is 70? What is the total if 10% of the total is 4,000? Step 1: The percent is 35, which in decimal form is 0.35. We were given that 35% of the total is 70, so the part is 70. We are to find the total. Substituting into the formula, we have part = percent × total 70 = 0.35 × total Step 2: To find the total, we solve the equation for the total. 70 = 0.35 × total 70 0.35 = 0.35 × total 0.35 200 = 0.35 × total 0.35 200 = total From this we see that 200 is the total, or, that 35% of 200 is 70. Step 1: The percent is 10, which in decimal form is 0.1. We were given that 10% of the total is 4,000, so the part is 4,000. Substituting into the formula, we have part = percent × total 4,000 = 0.1 × total Step 2: To find the total, we solve the equation for the total. 4,000 = 0.1 × total 4,000 0.1 = 0.1 × total 0.1 40,000 = 0.1 × total 0.1 40,000 = total From this we see that 40,000 is the total, or that 10% of 40,000 is 4,000. Finding the Total from the Percent and the Part Similarly, the percent can be found if the total and the percent of the total (the part) are known. This will result in the decimal form of the percent, so it must be converted to percent form. Finding the Percent from the Total and the Part What percent of 500 is 175? What percent of 228 is 155? Step 1: The total is 500, the percent of the total is 175. Substituting into the formula, we have part = percent × total 175 = percent × 500 Step 2: To find the percent, we solve the equation for the percent. 175 = percent × 500 175 500 = percent × 500 500 0.35 = percent × 500 500 0.35 = percent We see the percent in decimal form is 0.35. Converting from the decimal form yields 35%. We say that 175 is 35% of 500. Step 1: The total is 228, the percent of the total is 155. Substituting into the formula, we have part = percent × total 155 = percent × 228 Step 2: To find the percent, we solve the equation for the percent. 155 = percent × 228 155 228 = percent × 228 228 0.6798 = percent × 228 228 0.6798 = percent We see the percent is 0.6798 (rounded to four decimal places). Converting from the decimal form yields 67.98%. We say that 155 is 67.98% of 228. Finding the Percent When the Total and the Part Are Known Solve Application Problems Involving Percents Percents are frequently used in finance, research, science experiments, and even casual conversation. Understanding these types of values helps when consuming media or discussing finances, for instance. Effectively working with and interpreting numbers and percents will help you become an informed consumer of this information. In most cases, working through what is presented requires you to identify that you are indeed working with a question of percents, which two of the three values that are related through percents are known, and which of the three values you need to find. Retention Rate at College Justine applies to a medium size university outside her hometown and finds out that the retention rate (percent of students who return for their sophomore year) for the 2021 academic year at the university was 84%. During a visit to the registrar’s office, she finds out that 1,350 people had enrolled in academic year 2021. How many students from the academic year 2021 are returning for the 2022 academic year? The percent of students who will return for the 2022 academic year (the retention rate) is 84%. The total number of students who enrolled in the 2021 academic year was 1,350. This means the percent is known and the total is known. From this, we can determine the number of students who will return (percent of the total) for the 2022 academic year using the formula part = percent × total . Substituting into the formula and calculating, we find that the number of students that are returning is part = percent × total = 0.84 × 1,350 = 1,134 So 1,134 students will return for the 2022 academic year. Percent of Chemistry Majors Cameron enrolls in a calculus class. In this class of 45 students, there are 18 chemistry majors. What percent of the class are chemistry majors? In this situation, the percent is to be determined. We know the total number of students, 45, and the part of the students that are chemistry majors, 18. Using that information and the formula part = percent × total , the percent can be found. Substituting and solving, we have 18 = percent × 45 18 45 = percent × 45 45 0.4 = percent × 45 45 0.4 = percent Converting the 0.4 from decimal form, we find that 40% of the students in the calculus class are chemistry majors. Total Sales and Commission Mariel makes a 20% commission on every sale she makes. One week, her commission check is for $153.00. What were her total sales that week? In this problem, Mariel’s total sales is to be determined. We know the percent she earns is 20%. We also know that her sales commission was $153.00, which is the percent of the total. Using this information and the formula part = percent × total we can find Mariel’s total sales. The decimal form of 20% is 0.2. The part, or percent of the total, is 153. Substituting and solving, we obtain part = percent × total 153 = 0.2 × total 153 0.2 = 0.2 × total 0.2 765 = 0.2 × total 0.2 765 = total Mariel’s total sales were $765.00. LED Lightbulbs According to the energy website from the U.S. government , LED lightbulbs use at least 75% less energy than incandescent bulbs. They also last up to 25 times as long as an incandescent bulb. If lighting is a significant percent of your electrical use, replacing all incandescent bulbs with LED bulbs will significantly reduce your electric bill. Check Your Understanding For any answer, round to two decimal places, if necessary. Key Terms Percent Fractional form Decimal form Total Base Percent of the total Part Amount Key Concepts Know what a percent is as a fraction, a decimal, and as a part of the whole. Use the percent equation to find any of the three values that are related by the equation. Apply the percent equation in applications. Video Finding Percent of a Total Finding the Total from the Percent and the Part Finding the Percent When the Total and the Part Are Known Formulas part = percent x total", "section": "Understanding Percent", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Discounts, Markups, and Sales Tax Sale prices are often described as percent discounts. (credit: \"Close-up of a discount sign\" by Ivan Radic/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate discounts. Solve application problems involving discounts. Calculate markups. Solve application problems involving markups. Compute sales tax. Solve application problems involving sales tax. Many people first encounter percentages during a retail transaction such as a percent discount (SALE! 25% off!!), or through sales tax (\"Wait, I thought this was $1.99?\"), a report that something has increased by some percentage of the previous value (NOW! 20% more!!). These are examples of percent decreases and percent increases. In this section, we discuss decrease, increase, and then the case of sales tax. Calculating Discounts Retailers frequently hold sales to help move merchandise. The sale price is almost always expressed as some amount off the original price. These are discounts , a reduction in the price of something. The price after the discount is sometimes referred to as the reduced price or the sale price. When a reduction is a percent discount, it is an application of percent, which was introduced in Understanding Percent . The formula used was part = percentage × total . In a discount application, the discount plays the role of the part, the percent discount is the percentage, and the original price plays the role of the total. The formula for a discount based on a percentage is discount = percent discount × original price , with the percent discount expressed as a decimal. The price of the item after the discount is sale price = original price - discount . These are often combined into the following formula sale price = original price - percent discount × original price = original price × ( 1 - percent discount ) When the original price and the percent discount are known, the discount and the sale price can be directly computed. Calculating Discount for a Percent Discount Calculate the discount for the given price and discount percentage. Then calculate the sale price. Original price = $75.80; percent discount is 25% Original price = $168.90; percent discount is 30% Substituting the values into the formula discount = percent discount × original price , we find that the discount is discount = 0 .25 × 75.80 = 18.95 . The discount is $18.95. The sale price of the item is then sale price = original price - discount = 75.80 - 18.95 = 56.85 , or $56.85. Substituting the values into the formula discount = percent discount × original price , we find that the discount is discount = 0 .30 × 168.90 = 50.67 . The discount is $50.67. The sale price of the item is then sale price = original price - discount = 168.90 - 50.67 = 118.23 , or $118.23. Sometimes the original price and the sale price of an item is known. From this, the percent discount can be computed using the formula discount = percent discount × original price , by solving for the percent discount. Calculating the Percent Discount from the Original and Sale Prices Determine the percent discount based on the given original and sale prices. Original price = $1,200.00; sale price = $900.00 Original price = $36.70; sale price = $29.52 Step 1. Find the discount. Using the original price and the sale price, we can find the discount with the formula sale price = original price - discount . Substituting and calculating, we find the discount to be 900.00 = 1 , 200.00 - discount . Solving for the discount gives $300.00. Step 2. Find the percent discount. Substituting the discount of $300.00 and the original price of $1,200.00, into the formula discount = percent discount × original price , we can find the percent discount. 300 .00 = percent discount × 1,200 .00 300 .00 1 , 200.00 = percent discount 0.25 = percent discount Converting to percent form, the percent discount is 25%. Step 1. Find the discount. Using the original price and the sale price, we can find the discount with the formula sale price = original price - discount . Substituting and calculating, we find the discount to be 29.52 = 36.70 - discount . Solving for the discount gives $7.38. Step 2. Find the percent discount. Substituting the discount of $7.38 and the original price of $36.70, into the formula discount = percent discount × original price , we can find the percent discount. 29 .52 = percent discount × 36 .70 29 .52 36.70 = percent discount 0.2 = percent discount Converting to percent form, the percent discount is 20%. Sometimes the sale price and the percent discount of an item are known. From this, the original price can be found. To avoid multiple steps, though, the formula that we will use is sale price = original price × (1 - percent discount) . The original price can be found by solving this equation for the original price. Calculating the Original Price from the Percent Discount and Sale Price Determine the original price based on the percent discount and sale price. Percent discount 10%; sale price = $450.00 Percent discount 75%, sale price = $90.00 Using the percent discount and the sale price, we can find the original price with the formula sale price = original price × (1 - percent discount) . Substituting and solving for the original price, we find sale price = original price × (1 - percent discount) 450 .00 = original price × (1 - 0 .10) 450 .00 = original price × (0 .90) 500 .00 = original price The original price of the item was $500.00. Using the percent discount and the sale price, we can find the original price with the formula sale price = original price × (1 - percent discount) . Substituting and solving for the original price, we find sale price = original price × (1 - percent discount) 90 .00 = original price × (1 - 0 .75) 90 .00 = original price × (0 .25) 360 .00 = original price The original price of the item was $360.00. Solve Application Problems Involving Discounts In application problems, identify what is given and what is to be found, using the terms that have been learned, such as discount, original price, percent discount, and sale price. Once you have identified those, use the appropriate formula (or formulas) to find the solution(s). Determine Discount and New Price a Sale Rack Item The sale rack at a clothing store is marked “All Items 30% off.” Ian finds a shirt that had an original price of $80.00. What is the discount on the shirt? What is the sale price of the shirt? We are asked to find the discount, and the sale price. We know the percent discount is 30%, or 0.30 in decimal form. The original price was $80. Substituting into the percent discount formula, we find that the discount is discount = percent discount × original price = 0 .30 × 80 = 24 . The discount is $24 on that shirt. The sale price is the original price minus the discount, so the sale price is $80 – $24 = $56. Determine the Percent Discount of a Bus Pass An annual pass on the city bus is priced at $240. The student price, though, is $168. What is the percent discount for students for the bus pass? We know the original price of the item, $240. We also know the sale price of the item, $168. From this we know the discount is $ 240 - $ 168 = $ 72 . Substituting these values into the formula discount = percent discount × original price , we can find the percent discount. discount = percent discount × original price 72 = percent discount × 240 72 240 = percent discount × 240 240 0.3 = percent discount The student percent discount on the bus pass is 30%. Finding the Original Price of a New Pair of Tires Kendra’s car developed a flat, and the tire store told her that two tires had to be replaced. She got a 10% discount on the pair of tires, and the sale price came to $189.00. What was the original price of the tires? Using the percent discount and the sale price, we can find the original price with the formula sale price = original price × (1 - percent discount) . Substituting and solving for the original price, we find sales price = original price × (1 - percent discount) 189 .00 = original price × (1 - 0 .10) 189 .00 = original price × (0 .90) 210 .00 = original price The original price of the two tires Kendra bought was $210.00. Computing Price Based on a Percent Off Coupon There are cases where retailers allow multiple discounts to be applied. However, it is rare that the discount percentages are added together. For example, if you have a 15% coupon and qualify for a 20% price reduction, the retailer typically does not add those two percentages together to determine the new price. The retailer instead applies one discount, then applies the second discount to the price obtained after the first discount was deducted. Research the original prices of two different laptops offered by one retail outlet. Assume you will receive a student discount of 12% and your outlet of choice is having a 15% off sale on all laptops. For each laptop: List the original price and calculate the price after applying the student discount (12%) only. Then find the price after applying the sale discount (15% off) to the price found in Step 1. Determine the total saved on the laptop and what percent discount the total savings represents. Now, apply the discounts in reverse order (first the sale discount, then the student discount). Note anything interesting about your findings. Calculate Markups When retailers purchase goods to sell, they pay a certain price, called the cost . The retailer then charges more than that amount for the goods. This increase is called the markup . This selling price, or retail price , is what the retailer charges the consumer in order to pay their own costs and make a profit. Markup, then is very similar to discount, except we add the markup, while we subtract the discount. The formula for a markup based on a percentage is markup = percent markup × cost , with the percent markup expressed as a decimal. The price of the item after the markup is retail price = cost + markup . These are often combined into the following formula retail price = cost + percent markup × cost = cost × ( 1 + percent markup) It should be noted that the formulas used for a markup are very similar to those for a discount, with addition replacing the subtraction. Determining the Retail Price Based on the Cost and the Percent Markup Calculate the markup for the given cost and markup percentage. Then calculate the retail price. Cost = $62.00; percent markup is 15% Cost = $750.00; percent markup is 45% Substituting the values into the formula markup = percent markup × cost , we find that the markup is markup = 0.15 × 62.00 = 9.30 . The markup is $9.30. The retail price of the item is then retail price = cost + markup , or $62.00 + $9.30 = $71.30. Substituting the values into the formula markup = percent markup × cost , we find that the markup is markup = 0.45 × 750.00 = 337.50 . The markup is $337.50. The retail price of the item is then retail price = cost + markup , or $750.00 + $337.50 = $1,087.50. Sometimes the cost and the retail price of an item are known. From this, the percent markup can be computed using the formula markup = percent markup × cost , by solving for the percent markup. Calculating the Percent Markup from the Cost and Retail Price Determine the percent markup based on the given cost and retail price. Round percentages to two decimal places. Cost = $90.00; retail price = $103.50 Cost = $5.20; retail price = $9.90 Step 1: Using the cost and the retail price, we can find the markup with the formula retail price = cost + markup . Substituting and calculating, we find the markup to be 103.50 = 90.00 + markup . Solving for the markup gives $13.50. Step 2: After substituting the markup, $13.50, and the original price, $90.00, into the formula markup = percent markup × cost , we can find the percent markup. 13 .50 = percent markup × 90.00 13 .50 90.00 = percent markup 0.15 = percent markup Converting to percent form, the percent markup is 15%. Step 1: Using the cost and the retail price, we can find the markup with the formula retail price = cost + markup . Substituting and calculating, we find the markup to be 9.90 = 5.20 + markup . Solving for the markup gives $4.70. Step 2: After substituting the markup, $4.70, and the original price, $5.20, into the formula markup = percent markup × cost , we can find the percent markup. 4 .70 = percent markup × 5.20 4.70 5.20 = percent markup 0.9038 = percent markup Converting to percent form, the percent markup is 90.38%. Sometimes the retail price and the percent markup of an item are known. From this, the cost can be found. To avoid multiple steps, though, the formula that we will use is retail price = cost × (1 + percent markup) . The cost can be found by solving this equation for the cost. Calculating the Cost from the Percent Markup and Retail Price Determine the cost based on the percent markup and retail price. Percent markup 20%; retail price = $10.62 Percent markup 125%; retail price = $26.55 Using the percent markup and the retail price, we can find the cost with the formula retail price = cost × (1 + percent markup) . Substituting and solving for the cost, we find retail price = cost × (1 + percent markup) 10 .62 = cost × ( 1 + 0.2 ) 10 .62 = cost × (1 .2) 8 .85 = cost The cost of the item was $8.85. Using the percent markup and the retail price, we can find the cost with the formula retail price = cost × (1 + percent markup) . Substituting and solving for the original price, we find retail price = cost × (1 + percent markup) 26 .55 = cost × (1+2 .25) 26 .55 = cost × (3 .25) 11 .80 = cost The cost of the item was $11.80. Solve Application Problems Involving Markups As before when working with application problems, be sure to look for what is given and identify what you are to find. Once you have evaluated the problem, use the appropriate formula to find the solution(s). These application problems address markups. Determine Retail Price of a Power Bar Janice works at a convenience store near campus. It sells protein bars at a 60% markup. If a bar costs the store $1.30, how much is the retail price at the convenience store? We are asked to find the retail price. We know the percent markup is 60%. The cost of the bar was $1.30. Substituting into the percent markup formula, we find that the markup is markup = percent markup × cost = 0.60 × 1.30 = 0.78 . The markup is $0.78 on that protein bar. The retail price is the cost plus the markup, so the retail price is retail price = cost + markup = 1.30 + 0.78 = 2.08 . The retail price is $2.08. Determine the Percent Markup of a Phone Javi began working at a phone outlet. In a recent shipment, he noticed that the cost of the phone to the store was $480.00. The phone sells for $840.00 in the store. What is the percent markup on the phone? We know the cost of the phone, $480. We also know the retail price of the phone, $840.00. From this we know the markup is $ 840.00 - $ 480.00 = $ 360.00 . Substituting these values into the formula markup = percent markup × cost , we can find the percent markup. markup = percent markup × cost 360 = percent markup × 480 360 480 = percent markup × 480 480 0.75 = percent markup The markup on the phone is 75%. Finding the Cost of a T-Shirt Bob decided to order a t-shirt for his gaming friend online for $29.50. He knows the markup on such t-shirts is 18%. What was the t-shirt’s cost before the markup? Using the percent markup and the retail price, $29.50, we can find the cost with the formula retail price = original price × (1 + percent markup) . Substituting and solving for cost we find retail price = cost × (1 + percent markup) 29 .50 = cost × (1+0 .18) 29 .50 = cost × (1 .18) 25 .00 = cost The cost of the t-shirt was $25.00. Compute Sales Tax Sales tax is applied to the sale or lease of some goods and services in the United States but is not determined by the federal government. It is most often set, collected, and spent by individual states, counties, parishes, and municipalities. None of these sales tax revenues go to the federal government. For example, North Carolina has a state sales tax of 4.75% while New Mexico has a state sales tax of 5%. Additionally, many counties in North Carolina charge an additional 2% sales tax, bringing the total sales tax for most (72 of the 100) counties in North Carolinians to 6.75%. However, in Durham, the county sales tax is 2.25% plus an additional 0.5% tax used to fund public transportation, bringing Durham County’s sales tax to 7%. To find the sales tax in a particular place, then, add other locality sales taxes to the base state sales tax rate. How much we pay in sales tax depends on where we are, and what we are buying. To determine the amount of sales tax on taxable purchase, we need to find the product of the purchase price, or marked price, and the sales tax rate for that locality. To calculate the amount of sales tax paid on the purchase price in a locality with sales tax given in decimal form, calculate sales tax = purchase price × tax rate The total price is then Total price = purchase price + purchase price × tax rate = purchase price × ( 1 + tax rate) When the sales tax calculation results in a fraction of a penny, then normal rounding rules apply, round up for half a penny or more, but round down for less than half a penny. You should notice that this the same as markup, except using a different term. Sales tax plays the role of markup, the purchase price plays the role of cost, and the tax rate plays the role of percent markup. This means all the strategies developed for markups apply to this situation, with the changes indicated. Sales Tax in Kankakee Illinois The sales tax in Kankakee, Illinois, is 8.25%. Find the sales tax and total price of items based on the purchase price listed. Purchase price = $428.99 Purchase price = $34.88 The sales tax is found using sales tax = purchase price × tax rate . The purchase price is $428.99 and the tax rate is 8.25%. Substituting and calculating, the sales tax is sales tax = $ 428.99 × 0.0825 = $ 35.391675 . The sales tax needs to be rounded off. Since the third decimal place (fraction of a penny) is 1, we round down and the sales tax is $35.39. The total price is the sales tax plus the purchase price, so is $ 428.99 + $ 35.88 = $ 464.87 . The sales tax on the item is found using sales tax = purchase price × tax rate . The purchase price is $34.88 and the tax rate is 8.25%. Substituting and calculating, the sales tax is sales tax = $ 34.88 × 0.0825 = $ 2.8776 . The sales tax needs to be rounded off. Since the third decimal place (fraction of a penny) is 7, we round up and the sales tax is $2.88. The total price of the item is the sales tax plus the purchase price, so is $ 34.88 + $ 2.88 = $ 37.76 . As before, the information available might be different than only the purchase price and the sales tax rate. In these cases, use either sales tax = purchase price × tax rate or Total price = purchase price × (1 + tax rate) and solve for the indicated tax, price, or rate. These problems mirror those for percent markup. Be aware, almost all sales tax rates are structured as full percentages, or half percent, or one-quarter percent, or three-quarter percent. This means the decimal value of the sales tax rate, written as a percent, will be either 0, as in 5.0%, 5 as in 7.5%, 25 as in 3.25%, or 75 as in 4.75%. When rounding for the sales tax percentage, be sure to use this guideline. Calculating the Sales Tax from the Purchase Price and the Total Price Find the sales tax rate for the indicated purchase price and total price. Round using the guideline for sales tax percentages. Purchase price = $329.50; total price = $354.21 Purchase Price = $13.77; total price = $14.39 Step 1. Find the sales tax paid. First, the amount of sales tax must be found. Subtracting the purchase price from the total price, the amount of sales tax is $24.71. Step 2. Find the sales tax rate. Using the purchase price, the sales tax, and the formula sales tax = purchase price × tax rate , the sales tax rate can be found. Substituting and solving yields Sales Tax = purchase price × tax rate $ 24.71 = $ 329.50 × tax rate $ 24.71 $ 329.50 = tax rate 0 .07499 = tax rate Keeping in mind the guideline for rounding sales tax rate, the sales tax rate is 7.5%. Step 1. Find the sales tax paid. First, the amount of sales tax must be found. Subtracting the purchase price from the total price, the amount of sales tax is $0.62. Step 2. Find the sales tax rate. Using the purchase price, the sales tax, and the formula sales tax = purchase price × tax rate , the sales tax rate can be found. Substituting and solving yields sales tax = purchase price × tax rate $ 0.62 = $ 13.77 × tax rate $ 0.62 $ 13.77 = tax rate 0 .04503 = tax rate Keeping in mind the guideline for rounding sales tax rate, the sales tax rate is 4.5%. Calculating the Purchase Price from the Sales Tax and Total Price Find the purchase price for the indicated sales tax rate and total price. Sales tax rate = 5.75%; total price = $36.56 Sales tax rate = 4.25%; total price = $97.17 When the sales tax rate and the total price are known, the formula total price = purchase price × (1 + tax rate) can be used to find the purchase price. Substituting the tax rate and total price into the formula and solving, we find Total price = purchase price × (1 + tax rate) $ 36.56 = purchase price × (1 + 0 .0575) $ 36.56 1.0575 = purchase price $ 34.57 = purchase price The purchase price, the price before tax, was $34.57. When the sales tax rate and the total price are known, the formula total price = purchase price × (1 + tax rate) can be used to find the purchase price. Substituting the tax rate and total price into the formula and solving, we find Total price = purchase price × (1 + tax rate) $ 97.17 = purchase price × (1 + 0 .0425) $ 97.17 1.0425 = purchase price $ 93.21 = purchase price The purchase price, the price before tax, was $93.21. Solve Application Problems Involving Sales Tax Solving problems involving sales tax follows the same ideas and steps as solving problems for markups. But here we will use the following formula: total price = purchase price + sales tax We can also use the formula: total price = purchase price × (1 + sales tax rate) . This can be seen in the following examples. Compute Sales Tax for Denver, Colorado The sales tax rate in Denver Colorado is 8.81%. Keven buys a TV in Denver, and the purchase price (before taxes) is $499.00. How much will Keven pay in sales tax and what will be the total amount he spends when he buys the TV? The sales tax rate in Denver is 8.81%. To find the sales tax Keven will pay, find 8.81% of the purchase price. In decimal form, that sales tax rate is 0.0881. Using the formula and substituting 499.00 for purchase price, we find that Keven will pay purchase price × tax rate = $499 × 0.0881 = $ 43.96 in sales tax for the TV. The total price that Keven will pay is the purchase price plus the sales tax, or $ 499.00 + $ 43.96 = $ 542.96 . Compute Sales Tax for Austin, Texas Jillian visits Austin, Texas, and purchases a new set of weights for her home. She spends, including sales tax, $467.64. The sales tax rate in Austin Texas is 8.25%. How much of the total price is sales tax? The sales tax paid for this purchase is the difference in the total price and the purchase price. We know the total price is $467.64. We also know the sales tax rate, which is 8.25%. In decimal form, this is 0.0825. Using these values and the formula total price = purchase price × (1 + tax rate) to find the purchase price. total price = purchase price × (1 + tax rate) $ 467.64 = purchase price × ( 1 + 0.0825 ) $ 467.64 = purchase price × ( 1.0825 ) $ 432 = purchase price Knowing both the total price and the now the purchase price, we can find the difference, which is the sales tax. The total price was $467.64. The purchase price was $432. The difference of the total price and the purchase price, or the sales tax, is then $467.64 − $432.00, which is $35.64. Jillian pays $35.64 in sales tax. Finding Sales Tax Percentage West Virginia was the first state to impose a sales tax. This happened on May 3, 1921. Look up your locality on this website that lists standard state-level sales tax rates and compare the sales tax structure in your state to two nearby states (for the lower 48) and for any two states (Alaska and Hawaii). Check Your Understanding Key Terms Discount Cost Markup Retail price Key Concepts Discounts are markdowns from an original price. Mark-ups are increases to the price paid by a retailer to cover their costs. be able to calculate the markup based on a percentage of the cost Sales taxes vary from state to state and often county to county. Retail prices, sales prices and percent discounts can be calculated if the other two values are known. Original costs, retail prices, and percent markup can be calculated if the other two values are known. In calculations, sales tax acts like a markup. Videos Computing Price Based on a Percent Off Coupon Finding Sales Tax Percentage Formulas discount = percent discount × original price sale price = original price - discount sale price = original price - percent discount × original price = original price × (1 - percent discount) markup = percent markup × cost retail price = cost + markup retail price = cost + percent markup × cost = cost × (1 + percent markup) sales tax = purchase price × tax rate Total price = purchase price + purchase price × tax rate = purchase price × (1 + tax rate)", "section": "Discounts, Markups, and Sales Tax", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Simple Interest Interest is how savings earns money. (credit: “Interest Rates” by Mike Mozart/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute simple interest. Understand and compute future value. Compute simple interest loans with partial payments. Understand and compute present value. There is truth in the phrase “You need to have money to make money.” In essence, if you have money to lend, you can lend it at a cost to a borrower and make money on that transaction. When money is borrowed, the person borrowing the money (borrower) typically has to pay the person or entity that lent the money (the lender) more than the amount of money that was borrowed. This extra money is the interest that is to be paid. Interest is sometimes referred to as the cost to borrow, the cost of the loan, or the finance cost. This idea also applies when someone deposits money in a bank account or some other form of investment. That person is essentially lending the money to the bank or company. The money earned by the depositor is also called interest. The interest is typically based on the amount borrowed, or the principal . The pairing of borrower and lender can take various forms. The borrower may be a consumer using a credit card or taking out a loan from a bank, the lender. Companies also borrow from lending banks. Someone who invests in a company’s stock is the lender in this case; the company is essentially the borrower. In this section, we examine the basic building block of interest paid on loans and borrowed credit and also the returns on investments like bank accounts, simple interest. Compute Simple Interest Let’s get some terminology understood. Interest to be paid by a borrower is often expressed as an annual percentage rate , which is the percent of the principal that is paid as interest for each year the money is borrowed. This means that the more that is borrowed, the more that must be paid back. Sometimes, the interest to be paid back is simple interest , which means that the interest is calculated on the amount borrowed only. The length of time until the loan must be paid off is the term of the loan. The date when the loan must be paid off is when the loan is due . The day that the loan is issued is the origination date . We’ll put this terminology to use in the following examples. Note that in this section we will use letters, called variables, to represent the different parts of the formulas we’ll be using. This will help keep our formulas and calculations manageable. Simple Interest Loans with Integer Year Terms Calculating simple interest is similar to the percent calculations we made in Understanding Percent and Discounts, Markups, and Sales Tax , but must be multiplied by the term of the loan (in years, if dealing with an annual percentage rate). The simple interest, I , to be paid on a loan with annual interest rate r for a number of years (term of the loan) t , with principal P , is found using I = P × r × t , where the decimal form of the interest rate, r , is used. The total repaid, then is T = P + I or, more directly, T = P + P × r × t . This total is often referred to as the loan payoff amount , or more simply just the payoff. When the annual interest rate, the principal, and the number of years that the money is borrowed is known, the interest to be paid can be found and from there the total to be repaid can be calculated. Be aware, interest paid to a lender is almost uniformly rounded up to the next cent. Simple Interest on Loans with Integer Year Terms Calculate the simple interest to be paid on a loan with the given principal, annual percentage rate, and number of years. Then, calculate the loan payoff amount. Principal P = $4,000, annual interest rate r = 5.5%, and number of years t = 4 Principal P = $14,800, annual interest rate r = 7.9%, and number of years t = 7 Substitute the principal P = $4,000, the decimal form of the annual interest rate r = 0.055, and number of years t = 4 into the formula for simple interest, and calculate. I = P × r × t = 4,000 × 0.055 × 4 = 880 . The simple interest, or cost of the loan, to be paid on the loan is $880. The loan payoff amount, or the total to be repaid, is T = P + I = 4,000 + 880 = 4,880 , or $4,880.00. Substitute the principal P = $14,800, the decimal form of the annual interest rate r = 0.079, and number of years t = 7 into the formula for simple interest, and calculate. I = P × r × t = 14,800 × 0.079 × 7 = 8,184.4 . The simple interest, or cost to borrow, to be paid on the loan is $8,184.40. The loan payoff amount, or the total to be repaid, is T = P + I = 14,800 + 8,184.4 = 22,984.4 , or $22,984.40. Simple Interest Equipment Loan Riley runs an auto repair shop, and needs to purchase a new brake lathe, which costs $11,995. She takes out a two-year, simple interest loan at an annual interest rate of 14.9%. How much interest will she pay and how much total will she repay on the loan? Step 1. Determine the variables, or parts of the formula. The principal P is the cost of the brake lathe, so P = $11,995. The interest rate Riley pays is 14.9%, or r = 0.149 in decimal form. The length of the loan is two years, so t = 2. We are first asked to find I , the interest Riley will pay. Step 2. Substitute the known variables into the formula for simple interest I = P × r × t and solve for I . From Step 1 we have I = P × r × t = 11995 × 0.149 × 2 = 3,574.51 . This tells us that the simple interest, or cost to borrow, to be paid on the loan is $3,574.51. Step 3. Use the formula T = P + I to determine the total amount Riley will repay, T . The total to be repaid is T = P + I = 11,995 + 3,574.51 = 15,569.51 , or $15,569.51. Simple Interest Loans with Other Lengths of Terms In the previous example and Your Turn exercise, the loans were paid back in one payment after an integer number of years. However, there are also loans lasting a length of time not equal to an integer number of years (like 1, 2, or 3 years or more), but in a number of months (like 4 months, 18 months, and so on). What model would apply to these situations? When the loan is paid back after a term that is not an integer number of years but is instead a number of months, the term of the loan, or time, t , is expressed as a fraction of the year. So for a 2-month loan, the time, in years, is 2/12 = 1/6. For a 5-month loan, the time in years is 5/12. For an 18-month term, the term in years is 18/12 = 1.5. Loan to Purchase Equipment Abeje needs a loan to purchase equipment for the gym she is going to open. She visits the bank and secures a 4-month loan of $20,000. Her annual percentage rate is 6.75%. How much interest will Abeje pay and what is her loan payoff amount? Abeje’s loan is for $20,000, so her principal is P = 20,000. The interest rate Abeje will pay is 6.75%, or r = 0.0675 in decimal form. The length of the loan is 4 months, so t = 4 12 . Substituting these in the formula for simple interest, we find her interest to be I = P × r × t = 20,000 × 0.0675 × 4 / 12 = 450 The simple interest, or cost to borrow, to be paid on the loan is $450.00. The payoff is T = P + I = 20,000 + 450 = 20,450 , or $20,450.00. Those examples dealt in months. However, some loans are for days only (45 days, 60 days, 120 days). In such cases, we find the daily interest rate. The fraction we will use for the daily interest rate is the interest rate (as a decimal) divided by 365. This may be referred to as Actual/365. In order to find the term of the loan, divide the number of days in the term of the loan by 365. To determine the interest, I , on a loan with term t expressed in days, with principal of P , and interest rate in decimal form of r , calculate I = P × r 365 × t . Here, r 365 represents the daily interest rate. Alternately, the above formula is equivalent to I = P × r × t 365 , where the interest rate remains an annual rate, but the time is expressed as a fraction of the year. It seems reasonable to use 365 as the number of days in the year, since there are 365 days in most years. However, sometimes, banks have used (and continue to use) 360 as the number of days in a year. They may also treat all months as if they have 30 days. These differences lead to (sometimes small) differences in how much interest is paid. Since the number of days is in the denominator, a smaller denominator (360) will result in larger numbers (interest) that is 365 is used for the denominator. See this page from ACRE for a comparison. Loan for Moving Costs David plans to move his family from Raleigh, North Carolina to Tempe, Arizona. His company will reimburse (pay after the move) David for the move. David does research and determines that movers will cost $5,600 to move his family’s belongings to Tempe. He takes out a simple interest, 45-day loan at 11.75% interest to pay this cost. How much interest will be paid on this 45-day loan, and what is David’s loan payoff amount? This loan is in terms of days, so we will use the formula I = P × r 365 × t , where t is the number of days and r is the annual interest rate. The principal for the loan is the moving cost, or P = 5,600. The annual interest rate that David will pay is 11.75%, which in decimal is 0.1175. The length of time for the loan is 45 days, so t = 45. Substituting these values into the formula and calculating, we find that the interest to be paid is I = 5,600 × 0.1175 365 × 45 = 81.13 , or $81.13 (remember, interest is almost always rounded up to the next cent). The payoff for the loan is $5,681.13. Understand and Compute Future Value Money can be invested for a specific amount of time and earn simple interest while invested. The terminology and calculations are the same as we’ve already seen. However, instead of the total to be paid back, the investor is interested in the total value of the investment after the interest is added. This is called the future value of the investment. The future value, FV , of an investment that yields simple interest is F V = P + I = P + P × r × t , where P is the principal (amount invested at the start), r is the annual interest rate in decimal form, and t is the length of time the money is invested. The time t will be an integer if the term of the deposit is an integer number of years, will be number of months/12 if the term is in months, will be actual/365 if the deposit is for a number of days. Simple Interest on a Deposit In the following, determine how much interest was earned on the investment and the future value of the investment, if the investment yields simple interest. Principal is $1,000, annual interest rate is 2.01%, and time is 5 years Principal is $5,000, annual interest rate is 1.85%, and time is 30 years Principal is $10,000, annual interest rate is 1.25%, and time is 18 months Principal is $7,000, annual interest rate is 3.26%, and time is 100 days The principal is P = $1,000, the annual interest rate, in decimal form, is 0.0201, and the term is 5 years, or t = 5. Since the term is an integer number of years, the interest earned on the investment is I = P × r × t = 1,000 × 0.0201 × 5 = 100.5 , or the interest earned was $100.50. To find the future value, we use the formula F V = P + I . Substituting the values and calculating, we find the future value of the investment to be F V = P + I = 1,000 + 100.5 = 1100.5 . The future value of the investment at the end of 5 years is $1,100.50. Notice that the future value could have been calculated directly with F V = P + P × r × t The principal is P = $5,000, the annual interest rate, in decimal form, is 0.0185, and the term is 30 years, or t = 30. Since the term is an integer number of years, the interest earned on the investment is I = P × r × t = 5,000 × 0.0185 × 30 = 2,775 , or the interest earned was $2,775.00. To find the future value, we use the formula F V = P + I . Substituting the values and calculating, we find the future value of the investment to be F V = P + I = 5,000 + 2,775 = 7,775 . The future value of the investment at the end of 30 years is $7,775.00. The principal is P = $10,000, the annual interest rate, in decimal form, is 0.0125, and the term is 18 months. Since the term is in months, we have to write the months in terms of years. For 18 months, we use 18/12 as t . The interest earned on the investment is I = P × r × t = 10,000 × 0.0125 × 18 12 = 187.5 , or the interest earned was $187.50. To find the future value, we use the formula F V = P + I . Substituting the values and calculating, we find the future value of the investment to be F V = P + I = 10,000 + 187.5 = 10,187.5 . The future value of the investment at the end of 18 months is $10,187.50. Principal is $7,000, annual interest rate is 3.26%, and time is 100 days. The principal is P = $7,000, the annual interest rate, in decimal form, is 0.0326, and the term is 100 days. Since the term is in days, we have to write the time using actual/365, or t = 100/365. The interest earned on the investment is I = P × r × t = 7,000 × 0.0326 × 100 365 = 62.52 , or the interest earned was $62.52. To find the future value, we use the formula F V = P + I . Substituting the values and calculating, we find the future value of the investment to be F V = P + I = 7,000 + 62.52 = 7,062.52 . The future value of the investment at the end of 100 days is $7,062.52. You may have noticed that for these problems, the future value was rounded down. When the future value is paid, the amount is typically rounded down. A certificate of deposit (CD) is a savings account that holds a single deposit (the principal) for a fixed term at a fixed interest rate. Once the term of the CD is over, the CD may be redeemed (cashed in or withdrawn) and the owner of the CD receives the original principal plus the interest earned. The deposit often cannot be withdrawn until the term is up; if it can be withdrawn early, there is often a penalty imposed to do so. Certificate of Deposit Jonas deposits $2,500 in a CD bearing 3.25% simple interest for a term of 3 years. When he redeems his CD at the end of the 3 years, how much will he receive? This is a future value example. We know that P = $2,500 is the amount deposited. The annual simple interest rate in decimal form is r = 0.0325. The term of the investment is t = 3 years. Substituting those values into the future value formula, we have F V = P + P × r × t = 2,500 + 2,500 × 0.0325 × 3 = 2,500 + 243.75 = 2,743.75 . When the CD is redeemed, Jonas will receive $2,743.75. The reason CD (certificate of deposit) rates look so small is because they are extremely safe investments. Though overall interest rates for CDs change over time and individual returns vary with the terms of the CD, investors are offered predictable interest income for their investments. To investigate this yourself, search online to determine the strengths and weaknesses of CDs (investopedia.com offers good, basic information on investing). Then, online, identify five national banks and two local banks who offer CDs. Track the interest rates for the CDs at various terms (1 year, 3 years, 5 years) for each of the banks you found that offer CDs. Calculate the amount of interest earned for a $10,000 deposit for each CD at each of the terms. Compare the results from the various banks, CDs, and terms and decide which is the best investment. You may want to consider both the length of time that the money is locked up, and the return. Paying Simple Interest Loans with Partial Payments In every example above, there was one payment for the loan, or one withdrawal for the investment. However, for many loans (house, car, in-ground swimming pool), the loan will be paid back in two or more payments. Such a payment is called a partial payment , because they only pay off part of the loan. When a partial payment is made, some of the payment pays for the principal, but the rest of the payment pays for interest on the principal. When making the first partial payment, the interest is calculated on the principal for the time between the origination date of the loan and the date of the payment. If another partial payment is made, the interest is calculated based on the remaining principal and the time between the previous partial payment and the current partial payment date. Interest Paid in a Partial Payment on a Loan A simple interest loan for $6,500 is taken out at 12.6% annual percentage rate. A partial payment is made 45 days into the loan period. How much of the partial payment will be for interest? A simple interest loan for $13,700 is taken out at 6.55% annual interest rate. A partial payment is to be made after 60 days. How much of the partial payment will be for interest? To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 45 days. The principal is $6,500. The annual interest rate, in decimal form, is 0.126. The interest paid for 45 days is found by substituting the values for principal P , rate r , and time t into the formula I = P × r 365 × t . Calculating, we have I = P × r 365 × t = 6,500 × 0.126 365 × 45 = 100.9726 . Rounding up, the portion of the partial payment that will be paid for interest is $100.98. To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 60 days. The principal is $13,700. The annual interest rate, in decimal form, is 0.0655. The interest paid for those 60 days is found by substituting those values into the formula I = P × r 365 × t . Calculating, we have I = P × r 365 × t = 13,700 × 0.0655 365 × 60 = 147.5096 . Rounding up, the portion of the partial payment that will be paid for interest is $147.51. Remaining Balance The previous examples demonstrated how to determine the interest paid in a partial payment. Using this, we can determine the remaining balance after a partial payment. Step 1: determine the amount of the payment, P , that is applied to interest, I . Step 2: subtract the amount paid in interest from the payment, ( P − I ) . This is the amount applied to the balance. Step 3: subtract the amount applied to the balance (the value obtained in Step 2) from the balance of the loan, B − ( P − I ) . This is the remaining balance after the partial payment. Determining the Remaining Balance on a Loan After a Partial Payment A simple interest loan for $45,500 is taken out at 11.8% annual percentage rate. A partial payment of $20,000 is made 50 days into the loan period. After this payment, what will the remaining balance of the loan be? A simple interest loan for $150,000 is taken out at 5.85% annual percentage rate. A partial payment of $50,000 is made 70 days into the loan period. After this payment, what will the remaining balance of the loan be? The principal is $45,500, which will be treated as the balance, B , of the loan. The annual simple interest rate, in decimal form, is 0.118. The time is t = 50 days. Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formula I = P × r 365 × t and calculate. Calculating, the amount of the payment that is applied to interest is I = 45,500 × 0.118 365 × 50 = 735.4795 . Rounding up, we have I = $735.48. Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $2,000. The amount that is applied to the balance is P − I = $ 20,000 − $ 735.48 = $ 19,264.52 . Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or B − ( P − I ) = $ 45,500 − $ 19,264.52 = $ 26,235.48 . The remining balance after the partial payment is $26,235.48. The principal is $150,000, which will be treated as the balance, B , of the loan. The annual simple interest rate, in decimal form, is 0.0585. The time is t = 70 days. Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formula I = P × r 365 × t and calculate. Calculating, the amount of the payment that is applied to interest is I = 150,000 × 0.0585 365 × 70 = 1,682.8767 . Rounding up, we have I = $1,682.88. Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $50,000. The amount that is applied to the balance is P − I = $ 50,000 − $ 1,682.88 = $ 48,317.12 . Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or B − ( P − I ) = $ 150,000 − $ 48,317.12 = $ 101,682.88 . The remining balance after the partial payment is $101,682.88. Loan Payoff Finally, we will determine the amount to be paid at the end of the loan. To do so, we apply the formula for the loan payoff to the remaining balance. However, the length of time for that remaining balance is the time between the partial payment and the day the loan is paid off. Step 1: Determine the remaining balance after the partial payment. Step 2: Calculate the number of days between the partial payment and the date the loan is paid off. This will be the time t in the payment formula. Step 3: Calculate the amount to be paid at the end of the loan, or the payoff amount, using Payoff = P + P × r 365 × t , where P is the remaining balance and t is the time found in Step 2. Finding Loan Pay Off After a Partial Payment Laura takes out an $18,400 loan for 120 days at 17.9% simple interest. She makes a partial payment of $7,500 after 45 days. What is her payoff amount at the end of the loan? The initial balance, or principal, of her loan is $18,400. The interest rate in decimal form is 0.179. Her partial payment of $7,500 is made after 45 days. Using these values, we can determine how much of the partial payment is applied to the balance. From there, we can determine her final loan payoff after 120 days. Step 1: Determine the remaining balance after the partial payment. Using the partial payment process outlined in the previous example, we first find that the amount of the partial payment that is applied to the balance. Their interest paid in the partial payment is I = P × r 365 × t = 18,400 × 0.179 365 × 45 = 406.0603 , or $406.07 (remember to round up!). Using this and that the loan amount was for $18,400, the remaining balance on the loan after the partial payment is B - ( P - I ) = $ 18,400 - ( $ 7,500 - $ 406.07 ) = $ 11,306.07 . Step 2: The number of days between the partial payment and the date that the loan is to be paid off is 120 – 45 = 75. This means that the time between the partial payment and the final payment is 75 days. Step 3: To calculate the payoff amount, use payoff = P + P × r 365 × t , with P = $11,306.07 (the remaining balance), t = 75 (from Step 2) and r = 0.179. The payoff amount, then, is payoff = 11,306.07 + 11,306.07 × 0.179 365 × 75 = 11,721.9165 . Rounding up, the payoff amount is $11,721.92. Repeated Partial Payments Car loans and mortgages (loans for homes) are paid off through repeated partial payments, most often monthly payments. Since car loans are often 3 to 6 years, and mortgages 15 to 30 years, calculating each individual monthly payment one at a time is time consuming and tedious. Even a 3-year loan would involve applying the above steps 36 times! Fortunately, there is a formula for determining the amount of each partial payment for monthly payments on a simple interest loan. The amount of monthly payments, A , for a loan with principal P , monthly simple interest rate r (in decimal form), for t number of months is found using the formula A = P × r × ( 1 + r ) t ( 1 + r ) t − 1 . The monthly interest rate is the annual rate divided by 12. The number of months is the number of years times 12. Calculating Car Payments Desiree buys a new car, by taking a loan out from her credit union. The balance of her loan is $27,845.00. The annual interest rate that Desiree will pay is 7.3%. She plans to pay this off over 4 years. How much will Desiree’s monthly payment be? To use the formula for monthly payments, we need the principal, the interest rate, and the number of years. The principal is $27,845. The annual rate, in decimal form, is 0.073. Dividing 0.073 by 12 gives the monthly interest rate 0.073 / 12 = 0.00608 3 ¯ . She takes the loan out for 4 years, which is t = 12 × 4 = 48 months. Substituting these values into the formula, A = P × r × ( 1 + r ) t ( 1 + r ) t − 1 , we calculate: A = 27,845 × 0.00608 3 ¯ × ( 1 + 0.00608 3 ¯ ) 48 ( 1 + 0.00608 3 ¯ ) 48 - 1 = 27,845 × 0.00608 3 ¯ × ( 1.00608 3 ¯ ) 48 ( 1.00608 3 ¯ ) 48 - 1 = 27,845 × 0.00608 3 ¯ × 1.337918996 1.337918996 - 1 = 27,845 × 0.008139007 0.337918996 = 27,845 × 0.024085675 = 670.6656 Using the formula and rounding up to the next cent, we see that Desiree’s monthly payment will be $670.67. The calculation of payments is long, and involves many steps. However, most spreadsheet programs, including Google Sheets, have a payment function. In Google Sheets, that function is PMT. To find the payment for an installment loan (like for a car), you need to enter the interest rate per period, the number of payments, and the loan amount. From Your Turn 6.36 , the rate was 0.0476/12, the number of payments was 60, and the loan amount was $23,660. In Google Sheets, select any cell and enter the following: =PMT(0.0476/12,60,23660) And click the enter key. Immediately, in the cell you selected, the payment of $443.90 appears, though with a negative sign. The negative sign indicates it is a payment out of an account. Since we want to know the payment amount, we ignore the negative sign. The result, with the formula in the formula bar, is shown in . Google Sheets payment function In general, to use the PMT function in Google Sheets, enter =PMT( r /12, t *12, P ) where r is the annual interest rate, t is the number of years, and P is the principal of the loan. Understand and Compute Present Value for Simple Interest Investments When finding the future value of an investment, we know how much is deposited, but we have no idea how much that money will be worth in the future. If we set a goal for the future, it would be useful to know how much to deposit now so an account reaches the goal. The amount that needs to be deposited now to hit a goal in the future is called the present value . The present value, P V , of money deposited at an annual, simple interest rate of r (in decimal form) for time t (in years) with a specified future value of F V , is calculated with the formula P V = F V ( 1 + r t ) . Note: Present value, in this calculation, is always rounded up. Otherwise, future value may fall short of the target future value. Understanding what this tells you is important. When you find the present value, that is how much you need to invest now to reach the goal F V , under the conditions (time and rate) at which the money will be invested. Compute the present value of the investment described. Interpret the result. F V = $10,000, t = 15 years, annual simple interest rate of 5.5% F V = $150,000, t = 20 years, annual simple interest rate of 6.25% F V = $250,000, t = 486 months, annual simple interest rate of 4.75% The future value is F V = $10,000. The time of the investment is in years, so t = 15. The annual, simple interest rate is 5.5%, which in decimal form is 0.055. We substitute those values into the formula and calculate. P V = F V ( 1 + r t ) = 10,000 ( 1 + ( 0.055 ) × ( 15 ) ) = 10,000 ( 1 + 0.825 ) = 10,000 ( 1.825 ) = 5,479.452 . Rounding up, we see that the present value of $10,000 invested at a simple annual interest rate of 5.5% for 15 years is $5,479.46. This means that $5,479.46 needs to be invested so that, after 15 years at 5.5% interest, the investment will be worth $10,000. The future value is F V = $150,000. The time of the investment is in years, so t = 20. The annual, simple interest rate is 6.25%, which in decimal form is 0.0625. We substitute those values into the formula and calculate. P V = F V ( 1 + r t ) = 150,000 ( 1 + ( 0.0625 ) × ( 20 ) ) = 150,000 ( 1 + 1.25 ) = 150,000 ( 2.25 ) = 66,666. 6 ¯ . Rounding up, we see that the present value of $150,000 invested at a simple annual interest rate of 6.25% for 20 years is $66,666.67. This means that $66,666.67 needs to be invested so that, after 20 years at 6.25% interest, the investment will be worth $150,000. The future value is FV = $250,000. The time of the investment is 486 months. This needs to be converted to years. To do so, divide the number of months by 12, giving y e a r s = 486 12 = 40.5 , so t = 40.5 years. The annual, simple interest rate is 4.75%, which in decimal form is 0.0475. We substitute those values into the formula and calculate. P V = F V ( 1 + r t ) = 250,000 ( 1 + ( 0.0475 ) × ( 40.5 ) ) = 150,000 ( 1 + 1.92375 ) = 150,000 ( 2.92375 ) = 129,954.5159 . Rounding up, we see that the present value of $250,000 invested at a simple annual interest rate of 4.75% for 486 months is $129,954.52. This means that $129,954.52 needs to be invested so that, after 486 months at 4.75% interest, the investment will be worth $150,000. Present Value of a CD Beatriz will invest some money in a CD that yields 3.99% simple interest when invested for 30 years. How much must Beatriz invest so that after those 30 years, her CD is worth $300,000? Beatriz needs to know how much to deposit now so that her CD is worth $300,000 after 30 years. This means she needs to know the present value of that $300,000. The time is 30 years and the annual simple interest rate, in decimal form, is 0.0399. Using that information and the formula for present value, we calculate the present value of that $300,000. P V = F V ( 1 + r t ) = 300,000 ( 1 + ( 0.0399 ) × ( 30 ) ) = 300,000 ( 1 + 1.197 ) = 300,000 ( 2.197 ) = 136,549.8407 . Rounding up, Beatriz needs to invest $136,549.85 so that she has $300,000 in 30 years. Check Your Understanding Key Terms Interest Principal Annual percentage rate Simple interest Term Due Origination date Payoff amount Future value Partial payment Present value Key Concepts Interest is money that is paid by a borrower for the privilege of borrowing the money. Simple interest is computed by substituting the principal, interest rate, and number of years into the formula I = P × r × t The payoff for a loan is the amount of principal remaining on a loan plus the interest that accumulated on the loan since the last payment. The future value of an investment yielding simple interest is the original principal plus the interest earned on the investment. When making a partial payment, some of the payment pays off all the accumulated interest, while the remainder of the payment is applied to the principal of the loan. Finding the present value of an investment is used to determine how much should be invested now in order to achieve a specific goal. Formulas I = P × r × t T = P + I T = P + P × r × t I = P × r 365 × t F V = P + I = P + P × r × t A = P × r × ( 1 + r ) t ( 1 + r ) t − 1 P V = F V ( 1 + r t )", "section": "Simple Interest", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Compound Interest The impact of compound interest (credit: \"English Money\" by Images Money/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute compound interest. Determine the difference in interest between simple and compound calculations. Understand and compute future value. Compute present value. Compute and interpret effective annual yield. For a very long time in certain parts of the world, interest was not charged due to religious dictates. Once this restriction was relaxed, loans that earned interest became possible. Initially, such loans had short terms, so only simple interest was applied to the loan. However, when loans began to stretch out for years, it was natural to add the interest at the end of each year, and add the interest to the principal of the loan. After another year, the interest was calculated on the initial principal plus the interest from year 1, or, the interest earned interest. Each year, more interest was added to the money owed, and that interest continued to earn interest. Since the amount in the account grows each year, more money earns interest, increasing the account faster. This growth follows a geometric series ( Geometric Sequences ). It is this feature that gives compound interest its power. This module covers the mathematics of compound interest. Understand and Compute Compound Interest As we saw in Simple Interest , an account that pays simple interest only pays based on the original principal and the term of the loan. Accounts offering compound interest pay interest at regular intervals. After each interval, the interest is added to the original principal. Later, interest is calculated on the original principal plus the interest that has been added previously. After each period, the interest on the account is computed, then added to the account. Then, after the next period, when interest is computed, it is computed based on the original principal AND the interest that was added in the previous periods. The following example illustrates how compounded interest works. Interest Compounded Annually Abena invests $1,000 in a CD (certificate of deposit) earning 4% compounded annually. How much will Abena’s CD be worth after 3 years? Since the interest is compounded annually, the interest will be computed at the end of each year and added to the CD’s value. The interest at the end of the following year will be based on the value found form the previous year. Step 1: After the first year, the interest in Abena’s CD is computed using the interest formula I = P × r × t . The principal is P = 1,000, the rate, as a decimal, is 0.04, and the time is one year, so t = 1. Using that, the interest earned in the first year is I = P × r × t = 1,000 × 0.04 × 1 = 40 , so the interest earned in the first year was $40.00. This is added to the value of the CD, making the CD worth $ 1,000 + $ 40 = $ 1,040 . Step 2: At the end of the second year, interest is again computed, but is computed based on the CD’s new value, $1,040. Using this new value and the interest formula ( r and t are still 0.04 and 1, respectively), we see that the CD earned I = P × r × t = 1,040 × 0.04 × 1 = 41.6 , or $41.60. This is added to the value of the CD, making the CD now worth $ 1,040.00 + $ 41.60 = $ 1,081.60 . Step 3: At the end of the third year, interest is again computed, but is computed based on Abena’s CD’s new value, $1,081.60. Using this value and the interest formula ( r and t are still 0.04 and 1, respectively), we see that the CD earned I = P × r × t = 1,081.60 × 0.04 × 1 = 43.264 , or $43.26 (remember to round down). This is added to the value of the CD, making the CD now worth $ 1,081.60 + $ 43.26 = $ 1,124.86 . After 3 years, Abena’s CD is worth $1,124.86. Determine the Difference in Interest Between Simple and Compound Calculations It is natural to ask, does compound interest make much of a difference? To find out, we revisit Abena’s CD. Comparing Simple to Compound Interest on a 3-Year CD Abena invested $1,000 in a CD that earned 4% compounded annually, and the CD was worth $1,124.86 after 3 years. Had Abena invested in a CD with simple interest, how much would the CD have been worth after 3 years? How much more did Abena earn using compound interest? Had Abena invested $1,000 in a 4% simple interest CD for 3 years, her CD would have been worth P + P × r × t = 1,000 + 1,000 × 0.04 × 3 = 1,120 , or $1,120.00. With interest compounded annually, Abena’s CD was worth $1,124.86. The difference between compound and simple interest is $ 1,124.86 - $ 1,120.00 = $ 4.86 . So compound interest earned Abena $4.86 more than the simple interest did. Compound Interest Understand and Compute Future Value Imagine investing for 30 years and compounding the interest every month. Using the method above, there would be 360 periods to calculate interest for. This is not a reasonable approach. Fortunately, there is a formula for finding the future value of an investment that earns compound interest. The future value of an investment, A , when the principal P is invested at an annual interest rate of r (in decimal form), compounded n times per year, for t years, is found using the formula A = P ( 1 + r n ) n t . This is also referred to as the future value of the investment. Note, sometimes the formula is presented with the total number of periods, n , and the interest rate per period, r . In that case the formula becomes A = P ( 1 + r ) n . Computing Future Value for Compound Interest In the following, compute the future value of the investment with the given conditions. Principal is $5,000, annual interest rate is 3.8%, compounded monthly, for 5 years. Principal is $18,500, annual interest rate is 6.25%, compounded quarterly, for 17 years. The principal is P = $5,000, interest rate, in decimal form, r = 0.038, compounded monthly so n = 12, and for t = 5 years. Substituting these values into the formula, we find A = P ( 1 + r n ) n t = 5 , 000 ( 1 + 0.038 12 ) 12 × 5 = 5 , 000 ( 1 + 0.0031 6 ¯ ) 60 = 5 , 000 ( 1.0031 6 ¯ ) 40 = 5 , 000 × 1.20888663572 = 6 , 044.4332 The future value of the investment is $6,044.43. The principal is P = $18,500, interest rate, in decimal form, r = 0.0625, compounded quarterly so n = 4, and for t = 17 years. Substituting these values into the formula, we A = P ( 1 + r n ) n t = 18 , 500 ( 1 + 0.0625 4 ) 4 × 17 = 18 , 500 ( 1 + 0.015625 ) 68 = 18 , 500 ( 1.015625 ) 68 = 18 , 500 × 2.86992151999 = 53 , 093.5481 The future value of the investment is $53,093.54. Interest Compounded Quarterly Cody invests $7,500 in an account that earns 4.5% interest compounded quarterly (4 times per year). Determine the value of Cody’s investment after 10 years. Cody’s initial investment is $7,500, so P = $7,500. The annual interest rate is 4.5%, which is 0.045 in decimal form. Compounding quarterly means there are four periods in a year, so n = 4. He invests the money for 10 years. Substituting those values into the formula, we calculate A = P ( 1 + r n ) n t = 7 , 500 ( 1 + 0.045 4 ) 4 × 10 = 7 , 500 ( 1 + 0.01125 ) 40 = 7 , 500 ( 1.01125 ) 40 = 7 , 500 × 1.564376865391 = 11 , 732.8265 After 10 years, Cody’s initial investment of $7,500 is worth $11,732.82. Interest Compounded Daily Kathy invests $10,000 in an account that yields 5.6% compounded daily. How much money will be in her account after 20 years? Kathy’s initial investment is $10,000, so P = $10,000. The annual interest rate is 5.6%, which is 0.056 in decimal form. Compounding daily means there are 364 periods in a year, so n = 365. She invests the money for 20 years, so t = 20. Substituting those values into the formula, we calculate A = P ( 1 + r n ) n t = 10,000 ( 1 + 0.056 365 ) 365 × 20 = 10,000 ( 1 + 0.000153424657534 ) 7300 = 10,000 ( 1.000153424657534 ) 7300 = 10,000 × 3.06459091598 = 30,645.909 After 20 years, Kathy’s initial investment of $10,000 is worth $30,645.90. Compare Simple Interest to Interest Compounded Annually Compare Simple Interest and Compound Interest for Different Number of Periods Per Year To truly grasp how compound interest works over a long period of time, create a table comparing simple interest to compound interest, with different numbers of periods per year, for many years would be useful. In this situation, the principal is $10,000, and the annual interest rate is 6%. Create a table with five columns. Label the first column YEARS, the second column SIMPLE INTEREST, the third column COMPOUND ANNUALLY, the fourth column COMPOUND MONTHLY and the last column COMPOUND DAILY, as shown below. YEARS SIMPLE INTEREST COMPOUND ANNUALLY COMPOUND MONTHLY COMPOUND DAILY In the years column, enter 1, 2, 3, 5, 10, 20, and 30 for the rows. Calculate the account value for each column and each year. Compare the results from each of the values you find. How do the number of periods per year (compoundings per year) impact the account value? How does the number of years impact the account value? Redo the chart, with an interest rate you choose and a principal you choose. Are the patterns identified earlier still present? Understand and Compute Present Value When investing, there is often a goal to reach, such as “after 20 years, I’d like the account to be worth $100,000.” The question to be answered in this case is “How much money must be invested now to reach the goal?” As with simple interest, this is referred to as the present value. The money invested in an account bearing an annual interest rate of r (in decimal form), compounded n times per year for t years, is called the present value, P V , of the account (or of the money) and found using the formula P V = A ( 1 + r n ) n × t , where A is the value of the account at the investment’s end. Always round this value up to the nearest penny. Computing Present Value Find the present value of the accounts under the following conditions. A = $250,000, invested at 6.75 interest, compounded monthly, for 30 years. A = $500,000, invested at 7.1% interest, compounded quarterly, for 40 years. To reach a final account value of A = $250,000, invested at 6.75% interest, in decimal form r = 0.0675 (decimal form!), compounded monthly, so n = 12, for 30 years, substitute those values into the formula for present value. Calculating, we find the present value of the $250,000. P V = A ( 1 + r n ) n × t = 250,000 ( 1 + 0.0675 12 ) 12 × 30 = 250,000 ( 1 + 0.005625 ) 360 = 250,000 ( 1.005625 ) 360 = 250,000 7.5332454772 = 33,186.2277 In order for this account to reach $250,000 after 30 years, $33,186.23 needs to be invested. To reach a final account value of A = $500,000, invested at 7.1% interest, in decimal form r = 0.071, compounded quarterly, so n = 4, for 40 years, substitute those values into the formula for present value. Calculating, we find the present value of the $500,000. P V = A ( 1 + r n ) n × t = 500 , 000 ( 1 + 0.071 4 ) 4 × 40 = 500 , 000 ( 1 + 0.01775 ) 160 = 500 , 000 ( 1.01775 ) 160 = 500 , 000 16.6946672846 = 29 , 949.6834 In order for this account to reach $500,000 after 40 years, $29,949.69 needs to be invested. Investment Goal with Compound Interest Pilar plans early for retirement, believing she will need $1,500,000 to live comfortably after the age of 67. How much will she need to deposit at age 23 in an account bearing 6.35% annual interest compounded monthly? Knowing how much to deposit at age 23 to reach a certain value later is a present value question. The target value for Pilar is $1,500,000. The interest rate is 6.35%, which in decimal form is 0.0635. Compounded monthly means n = 12. She’s 23 and will leave the money in the account until the age of 67, which is 44 years, making t = 44. Using this information and substituting in the formula for present value, we calculate P V = A ( 1 + r n ) n × t = 1,500,000 ( 1 + 0.0635 12 ) 12 × 44 = 1,500,000 ( 1 + 0.005291 6 ¯ ) 528 = 1,500,000 ( 1.005291 6 ¯ ) 528 = 1,500,000 16.226302189 = 92,442.5037 Pilar will need to invest $92,442,51 in this account to have $1,500,000 at age 67. Compute and Interpret Effective Annual Yield As we’ve seen, quarterly compounding pays interest 4 times a year or every 3 months; monthly compounding pays 12 times a year; daily compounding pays interest every day, and so on. Effective annual yield allows direct comparisons between simple interest and compound interest by converting compound interest to its equivalent simple interest rate. We can even directly compare different compound interest situations. This gives information that can be used to identify the best investment from a yield perspective. Using a formula, we can interpret compound interest as simple interest. The effective annual yield formula stems from the compound interest formula and is based on an investment of $1 for 1 year. Effective annual yield is Y = ( 1 + r n ) n - 1 where Y = effective annual yield, r = interest rate in decimal form, and n = number of times the interest is compounded in a year. Y is interpreted as the equivalent annual simple interest rate. Determine and Interpret Effective Annual Yield for 6% Compounded Quarterly Suppose you have an investment paying a rate of 6% compounded quarterly. Determine and interpret that effective annual yield of the investment. Here, n = 4 (quarterly) and r = 0.06 (decimal form). Substituting into the formula we find that the effective annual yield is Y = ( 1 + 0.06 4 ) 4 - 1 = ( 1.015 ) 4 - 1 = 1.06136 - 1 = 0.0614 = 6.14 % Therefore, a rate of 6% compounded quarterly is equivalent to a simple interest rate of 6.14%. Determine and Interpret Effective Annual Yield for 5% Compounded Daily Calculate and interpret the effective annual yield on a deposit earning interest at a rate of 5% compounded daily. In this case, the rate is r = 0.05 and n = 365 (daily). Using the formula Y = ( 1 + r n ) n - 1 , we have Y = ( 1 + 0.05 365 ) 365 - 1 = ( 1.0001369863 ) 365 - 1 = 1.051267 - 1 = 0.0513 This tells us that an account earning 5% compounded daily is equivalent to earning 5.13% as simple interest. Choosing a Bank Minh has a choice of banks in which he will open a savings account. He will deposit $3,200 and he wants to get the best interest he can. The banks advertise as follows: Bank Interest Rate ABC Bank 2.08% compounded monthly 123 Bank 2.09% compounded annually XYZ Bank 2.05% compounded daily Which bank offers the best interest? To compare these directly, Minh could change each interest rate to its effective annual yield, which would allow direct comparison between the rates. Computing the effective annual yield for all three choices gives: ABC Bank: Y = ( 1 + 0.0208 12 ) 12 - 1 = 0.0210 = 2.10 % 123 Bank: Y = ( 1 + 0.0209 1 ) 1 - 1 = 0.0209 = 2.09 % XYZ Bank: Y = ( 1 + 0.0205 365 ) 365 - 1 = 0.0207 = 2.07 % ABC Bank has the highest effective annual yield, so Minh should choose ABC bank. Check Your Understanding Key Terms Compound interest Effective annual yield Key Concepts Compound interest means that the interest earned during one period will earn interest in later periods. Essentially, the amount of the principal grows from period to period. The important values in computing compound interest are the interest rate, the principal, the length of time the investment, and the number of times the investment is compounded. Compound interest has minimal impact early, but later has a very large impact. You can determine how much to invest today in order to reach a goal for some time later. Compound interest can be translated into an effective annual yield, which allows for comparison between investment options. Videos Compound Interest Compare Simple Interest to Interest Compounded Annually Compare Simple Interest and Compound Interest for Different Number of Periods Per Year Formulas A = P ( 1 + r n ) n t P V = A ( 1 + r n ) n × t Y = ( 1 + r n ) n - 1", "section": "Compound Interest", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Making a Personal Budget Calculating a budget is important to your financial health. (credit: “Budget planning concept on white desk” by Marco Verch Professional Photographer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Create a personal budget with the categories of expenses and income. Apply general guidelines for a budget. “That doesn’t fit in the budget.” “We didn’t budget for that.” “We need to figure out our budget and stick to it.” A budget is an outline of how money and resources should be spent. Companies have them, individuals have them, your college has one. But do you have one? Creating a realistic budget is an important step in careful stewardship of your financial health. Designing your budget will help understand the financial priorities you have, and the constraints on your life choices. You want to have enough income to pay not only for the necessities, but also for things that represent your wants, like trips or dinner out. You also may want to save money for large purchases or retirement. You do not want to just get by, and you do not want the problems associated with overdue balances, rising debt, and possibly losing something you have worked hard to obtain. While creating a budget may seem intimidating at first, coming up with your basic budget outline is the hardest part. Over time, you will adjust not only the numbers, but the categories. Creating a Budget You should view creating a budget as a financial tool that will help you achieve your long-term goals. A budget is an estimation of income and expenses over some period of time. You will be able to track your progress, which will help you to prepare for the future by making smart investment decisions. There are several budget-creating tools available, such as the apps Good budget and Mint, and Google Sheets. Getting started, though, begins well before you find an app. The following are steps that can be used to create your monthly budget. Track your income and expenses Review your income and expenses for the past 6 months to a year. This will give you an idea of your current habits. Set your income baseline Determine all the sources of income you will have. This income may from paychecks, investments, or freelance work. It even includes child support and gifts. Be sure to use income after taxes. This allows you to determine your maximum expenditures per month. For income that is not steady, such as gig work or freelance work, use the previous 6 to 12 months of income to find an average income from that gig or freelance work. Use this average in the budgeting process. Determine your expenses Review your bills from the past 6 months. You should include mortgage payments or rent, insurance, car payments, utilities, groceries, transportation expenses, personal care, entertainment, and savings. Using your credit card statements and bank statements will help you determine these amounts. Be aware that some of the expenses will not change over time. These are referred to as fixed expenses , like rent, car payments, insurance, internet service, and the like. Other expenses may vary widely from month to month and are appropriately called variable expenses , and include such expenses as gasoline, groceries. Some expenses are yearly, such as insurance or property taxes. Other expanses may be quarterly (four times per year) or semiannual (twice per year). To budget for such bills by month, divide the bill total by the number of months the bill covers. Categorize your expenses These categories may be housing, transportation, or food, for broad categories, or may get more specific, where you categorize car payments, car insurance, and gasoline separately. The categories are your choices. Be sure to account for the cost of maintaining a vehicle or home. The more specific you are, the better you’ll understand your spending needs and habits. Total your monthly income and monthly expenses and compare These values should be compared. If your expenses are higher than your income, then adjustments have to be made. Decisions of what to do with any extra income is part of the planning process also. Make plans for unplanned expenses Ask anyone, an unexpected car repair can ruin a carefully crafted budget. Have a plan for how you can be ready for these random expenses. This often means creating a cushion in your budget. Use your budget to make decisions and adjust for any changes Your budget is a changeable document. Add to it when you wish, refer to it when special purchases are to be made. Keeping your budget up to date helps accommodate changes in income and expenses. Creating a Budget In this section, we will focus on income and expenses. One of the easiest ways to manage a budget is to create a table, with one column containing income sources, another with income values, a third with expense categories, and a last containing expenses. An example is shown in . Income Source Amount Expense Amount Full-time job $3,565 Rent $975 Uber $185 Car Payment $355 TOTAL $3,750 Student Loan $418 Electric $76 Food $400 Gasoline $250 Car Insurance $165 Clothing $100 Entertainment $100 TOTAL $2,839 Table with Budget Gross Pay and Take Home Pay If you’ve ever had a paycheck, you know that taxes are taken out of your pay before you get your check. This amount of money varies from state to state, and sometimes even city to city. For a person making $50,000 per year gross salary in Salt Lake City, Utah, take home pay is about 75.6% of gross salary. In Detroit, Michigan, take home pay is about 74.5% of gross salary. Lakeland, Florida, take home pay is about 80.5% of gross salary. These also change based on how much a person earns! Before choosing a place to live, it makes sense to determine how much deductions from pay will impact your income. Creating a Budget Heather has graduated college and currently works as a nurse for a rural medical group. Her net monthly income from that job is $3,765.40. She also works part-time on the weekends, earning another $672.00 per month. Her monthly expenses are rent at $1,050, car payments at $489, student loan payments at $728, car insurance at $139, utilities at $130, clothing at $150, entertainment (going out with friends, Netflix, Amazon Prime, movies) at $300, credit card debt at $200, food at $360, and gasoline at $275. Create her budget in a table, compare the total income to total expenses, and determine how much excess income per month she has or how much she falls short by each month. Step 1: To begin, we create the table with appropriate headings. Income Source Amount Expense Amount Step 2: Her income categories are her nursing job, with $3,765.40 per month, and her part-time job, with $672.00 per month. Entering these into the table, we have the following. Income Source Amount Expense Amount Nursing $3,765.40 Part-time $672.00 Step 3: Her monthly expenses are listed above. Entering the categories and the amount for each of those expenses, the table is now Income Source Amount Expense Amount Nursing $3,765.40 Rent $1,050 Part-time $672.00 Car Payment $489 Student Loan $728 Car Insurance $139 Utilities $130 Clothing $150 Entertainment $300 Credit Card $200 Food $400 Gasoline $250 Step 4: Totaling the income and expenses, we see that her total income is $4,437.40 per month, and her total expenses are $3,836 per month. Comparing these, we see that Heather has $601.40 in excess income per month. This provides a cushion in her budget. Creating a Budget Carol is working in a dental lab, creating dentures and bridges. Monthly her take home pay is $2,816 (based on $22 per hour minus payroll taxes). She also receives $320 per month in child support for her one daughter. Her monthly expenses are rent at $700, car payments at $229, student loan payments at $250, car insurance at $119, health insurance at $225, utilities at $80, clothing at $75, entertainment at $200, food at $275, and gasoline at $275. Create Carol’s budget in a table, compare the total income to total expenses, and determine how much excess income per month she has or how much she falls short by each month. Step 1: To begin, we create the table with appropriate headings. Income Source Amount Expense Amount Step 2: Her income categories are from work, $2,816, and child support, $320, per month. Entering these into the table, we have the following. Income Source Amount Expense Amount Job $2,816.00 Child support $320.00 Step 3: Her monthly expenses are listed above. Entering the categories and the amount for each of those expenses, the table is now Income Source Amount Expense Amount Job $2,816.00 Rent $700 Child support $320.00 Car Payment $229 Student Loan $250 Car Insurance $119 Utilities $80 Health insurance $225 Clothing $75 Entertainment $200 Food $275 Gasoline $275 Step 4: Totaling the income and expenses, we see that her total income is $3,136.00 per month, and her total expenses are $2,428.00 per month. Comparing these, we see that Carol has $708.00 in excess income per month. This is the cushion in her budget. Using the budget process, we can make decisions on adding expenses to the budget. To do so, check the cushion of the budget to see if there is room in the budget for the new expense. Adding to an Existing Budget In the example above, Carol had excess income of $708.00. She looks up the cost of before-school care for her daughter. She finds that, monthly, the cost would be $252.00 per month. Is this an affordable program for Carol? Add this expense to her budget table. She can afford this, as the cost for the before school program is $252.00 and she had extra income of $708.00. Adding this to her budget, her budget table is now Income Source Amount Expense Amount Job $2,816.00 Rent $700 Child support $320.00 Car Payment $229 Student Loan $250 Car Insurance $119 Utilities $80 Health insurance $225 Clothing $75 Entertainment $200 Food $275 Gasoline $275 Before-school care $252 Now, she has $456.00 in excess income per month. The 50-30-20 Budget Philosophy It isn’t clear, obvious, or easy to decide how much of your income to allocate to various categories of expenses. Many people pay their bills and then consider all the leftover money to be spending money. However, when developing your own budget, you may want to follow the 50-30-20 budget philosophy , which provides a basic guideline for how your income could be allocated. Fifty percent of your budget is allotted to your needs, 30% of your budget is allotted to pay for your wants, and 20% of your budget is allotted for savings and debt service (paying off your debts). Knowing what expenses are necessary and what expenses are wants is important, since wants and needs are often confused. The following are necessary expenses that represent basic living requirements and debt services. This list isn’t complete: mortgage/rent, utilities, car, car insurance, health care, groceries, gasoline, child care (for working parents), and minimum debt payments. The 50-30-20 budget philosophy suggests that 50%, or half, your income go to these necessities. Wants, though, are things you could live without but still wish to have, such as Amazon Prime, restaurant dinners, coffee from Starbucks, vacation trips, and hobby costs. Even a gym membership or that new laptop are wants. Creating the room to afford these wants is important to our mental health. Not budgeting for things we want will negatively impact our quality of life. The remaining 20% should be set aside, either in retirement funds, stocks, other investments, an emergency fund (recommendations are that an emergency fund have 3 months of income), and perhaps extra spent to pay down debt. This 20% is very useful for addressing those unexpected costs, such as repairs or replacement of items that no longer work. Without budgeting this cushion, any expense that is a surprise can cause us to miss necessary payments. The list of necessary expenses was not complete. There are other expenses that could be included. Necessary Expenses and Expenses that are Wants For some people, an expense will be necessary while the same expense for someone else will be a want. A good example of this is internet service. Many people consider internet service as a need, especially those who work from home or who are not able to leave their homes. One could also call internet service a need if they have children in school. For others, internet service is a want. If a person’s job doesn’t require them to be online, if they are not in school, if they do not have kids, then internet service can be dropped. There are public options for internet service. One could even use their phone as a hot spot. Cars often fall into the category of need, but could also fall into the want category, depending on where and how you live. Bikes, public transportation, and walking are all options that could replace a car. This would then remove the cost of gasoline and car insurance. Another consideration when deciding if an item on your budget is a need or a want is about your choices and priorities. A car is a need for many. But the need for a car is not the same as the need for a specific car. If you choose to buy a car with payments that exceed your budgeted amount for the car, then that car is a want. The amount you exceed the budget now belongs in the want category. The same can be said for housing. If you want an apartment that costs $1,250 per month, but your budget only allows for an apartment that costs $900, then $350 of the rent is a want. The point of that is to carefully consider if an expense is a need as opposed to a want. When your expenses exceed your income, you may want to change how you budget your income to line up with these guidelines. This may mean cutting back, finding less-expensive living arrangements, finding a less-expensive (and more fuel-efficient) car, or sacrificing some specialty groceries. Using these guidelines keeps your financial life manageable. Better still, they can guide you as you begin your life after graduation. Evaluate a Budget Using 50-30-20 model In the example above, after Carol added before school care for her daughter to the budget, her budget was as shown below. Evaluate Carol’s budget using the 50-30-20 budget philosophy. Income Source Amount Expense Amount Job $2,816.00 Rent $700 Child support $320.00 Car Payment $229 Student Loan $250 Car Insurance $119 Utilities $80 Health insurance $225 Clothing $75 Entertainment $200 Food $275 Gasoline $275 Before-school care $252 Carol’s total income is $3,136.00. Applying the 50-30-20 budget philosophy to this income requires the calculation of each of those percentages. For the necessities, Carol should budget 50% of her income, or 0.5 × $ 3,136.00 = $ 1,568.00 . For her wants, she should budget 30% of her income, or 0.3 × $ 3,136.00 = $ 940.80 . For savings and extra debt service, she should budget 20% of her income, or 0.2 × $ 3,136.00 = $ 627.20 . In her budget, her necessities include all expenses except for entertainment. These expenses total $2,480, which exceeds the suggested budget amount of $1568.00. To follow the guidelines, Carol would have to cut back on these necessities. For her wants, she spends $200.00 on entertainment, which is well below the suggested budget amount of $940.80. If she modifies how much she spends on needs, she may be able to increase the spending on her wants. Her excess income is $456.00, which is below what she should be saving and using to pay down extra debt. If she does adjust how much she spends on needs, she could increase the amount for savings. Creating a Budget Based on the 50-30-20 Budget Philosophy Carmen is about to graduate and has been offered a job at a bank as a data scientist. She estimates her monthly take home pay to be $5,662.50. Apply the 50-30-20 philosophy to that monthly income. How should Carmen use this information? Step 1. To apply the 50-30-20 budget philosophy to Carmen’s income, she needs to calculate 50%, 30%, and 20% of her income. Fifty percent of her income is 0.5 × $ 5,662.50 = $ 2,831.25 . Thirty percent of her income is 0.3 × $ 5,662.50 = $ 1,698.75 . Twenty percent of her income is 0.2 × $ 5,662.50 = $ 1,132.50 . Step 2. She would then budget $2,831.25 for her needs, $1,698.75 for her wants, and $1,132.50 for savings and debt service. Step 3. When choosing where to live, what to eat, and what to drive, she should make choices that keep those costs, combined with her debt service costs, gasoline, and utilities, below $2,831.25. This means she will have to make decisions about what her priorities are. Step 4. She should then figure out what she wants to do with her money, and stay within the limits, that is, keep those costs below $1,698.75. Step 5. Finally, she can begin building her savings with the remaining $1,132.50. Using the 50-30-20 Budget Philosophy to Analyze Affordability Steve is thinking of moving out of his family’s home. He currently works at a full-time job making $18 per hour, which will give him, approximately, a net annual income of $29,180 (working 40 hours per week for 52 weeks per year). He has student debt that he pays off at $218.00 per month, and already owns a car that he pays $162.00 per month for. Apply the 50-30-20 budget philosophy to Steve’s income. If he follows the budget, how much does he have, after paying his car payment and student loan, to spend on necessities. If he follows the budget, how much will he set aside for wants? For savings? Discuss the affordability of moving out, based on Steve’s budget. Before the 50-30-20 philosophy can be applied, Steve’s monthly income needs to be determined. This is found by dividing his annual income by 12. This gives $ 29,180 / 12 = $ 2,431.67 . This will be used for his monthly budget. To apply the 50-30-20 philosophy to Steve’s income, find 50%, 30%, and 20% of his monthly income. Needs (50%): 50% of his income is 0.50 × $ 2,431.67 = $ 1,215.83 . Wants (30%): 0.30 × $ 2,431.67 = $ 729.50 Savings (20%): 0.20 × $ 2,431.67 = $ 486.33 The total for Steve’s needs is $1,215.83. From this, he already pays $218.00 for his student loans, and $162.00 for his car payment. Together that is $380.00. Subtracting from the amount he should budget for his needs, he can spend $835.83 on other needs. Steve budgeted $729.50 for wants, and $486.33 for savings and other debt servicing. Steve will have other needs to pay for, including rent, utilities, food, heath care, gasoline, and car insurance. It is difficult to imagine Steve being able to afford to move out, unless he reallocates money that he would want to save, or use for entertainment and other wants, or takes on another job. Even if Steve uses all the money that the 50-30-20 budget sets aside for savings, he still only has $1,322.16 to spend on those necessities. It does not appear he can afford to move out. 50-30–20 Budget Philosophy Check Your Understanding Key Terms Budget Necessary expenses Fixed expenses Variable expenses 50-30-20 budget philosophy Key Concepts A budget is a set of guidelines for how to allocate your income. Budgeting helps to plan for many of life’s expenses Budgets are used to compare income to expenses. When expenses exceed income, changes have to be made. Budgets can help evaluate the affordability of life changes. One guideline for setting a budget is the 50-30-20 budget philosophy. The guidelines suggest that 50% of income is allocated to necessary expenses, 30% to expenses that wants, and 20% to savings and other debt reduction. Videos Creating a Budget 50-30–20 Budget Philosophy", "section": "Making a Personal Budget", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Methods of Savings Money wisely invested grows over time. (credit: “Stack of Cash” by Janak Raja/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Distinguish various basic forms of savings plans. Compute return on investment for basic forms of savings plans. Compute payment to reach a financial goal. The stock market crash of 1929 led to the Great Depression, a decade-long global downturn in productivity and employment. A state of shock swept through the United States; the damage to people’s lives was immeasurable. Americans no longer trusted established financial institutions. By October 1931, the banking industry’s biggest challenge was restoring confidence to the American public. In the next 10 years, the federal government would impose strict regulations and guidelines on the financial industry. The Emergency Banking Act of 1933 created the Federal Deposit Insurance Corporation (FDIC), which insures bank deposits. The new federal guidelines helped ease suspicions among the general public about the banking industry. Gradually, things returned to normal, and today we have more investment instruments, many insured through the FDIC, than ever before. In this section, we will first look at the different types of savings accounts and proceed to discuss the various types of investments. There is some overlap, but we will try to differentiate among these financial instruments. Saving money should be a goal of every adult, but it can also be a difficult goal to attain. Distinguish Various Basic Forms of Savings Plans There are at least three types of savings accounts. Traditional savings accounts, certificates of deposit (CDs), and money market accounts are three main savings account vehicles. Savings Account A savings account is probably the most well-known type of investment, and for many people it is their first experience with a bank. A savings account is a deposit account, held at a bank or other financial institution, which bears some interest on the deposited money. Savings accounts are intended as a place to save money for emergencies or to achieve short-term goals. They typically pay a low interest rate, but there is virtually no risk involved, and they are insured by the FDIC for up to $250,000. Savings accounts have some strengths. They are highly flexible. Generally, there are no limitations on the number of withdrawals allowed and no limit on how much you can deposit. It is not unusual, however, that a savings account will have a minimum balance in order for the bank to pay maintenance costs. If your account should dip below the minimum, there are usually fees attached. Many banks are covered by FDIC insurance. The FDIC is the Federal Deposit Insurance Corporation and is an independent agency created by the U.S. Congress. One of its purposes is to provide insurance for deposits in banks, including savings accounts. Be aware, not all banks are FDIC insured. The FDIC insures up to $250,000 for a savings account, so you do not want your balance to exceed that federally insured limit. Having your savings account at the same bank as your checking account does offer a real advantage. For example, if your checking account is approaching its lower limit, you can transfer funds from your savings account and avoid any bank fees. Similarly, if you have an excess of funds in your checking account, you can transfer funds to your savings account and earn some interest. Checking accounts rarely pay interest. J.P. Morgan J.P. Morgan was a wealthy banker around the turn of the 20th century. His business interests included railroads and the steel industry. However, it was in 1907 that a financial crisis, caused by poor banking decisions and followed by such great distrust in the banking system that a frenzy of withdrawals from banks occurred, that J.P. Morgan and other wealthy bankers lent from their own funds to help stabilize and save the system. There are some weaknesses to savings accounts. Primarily, it is because savings accounts earn very low interest rates. This means they are not the best way to grow your money. Experts, though, recommend keeping a savings account balance to cover 3 to 6 months of living expenses in case you should lose your job, have a sudden medical expense, or other emergency. Around tax time, you will receive a 1099-INT form stating the amount of interest earned on your savings, which is the amount that must be reported when you file your tax return. A 1099 form is a tax form that reports earnings that do not come from your employer, including interest earned on savings accounts. These 1099 forms have the suffix INT to indicate that the income is interest income. Savings accounts earn interest, and those earnings can be found using the interest formulas from previous sections. The final value of these accounts is sometimes called the future value of the account. Single Deposit in a Savings Account Violet deposits $4,520.00 in a savings account bearing 1.45% interest compounded annually. If she does not add to or withdraw any of that money, how much will be in the account after 3 years? To find the compound interest, use the formula from Compound Interest , A = P ( 1 + r n ) n t , where A represents the amount in the account after t years, with initial deposit (or principal) of P , at an annual interest rate, in decimal form, of r , compounded n times per year. Violet has a principal of $4,520.00, which will earn an interest of r = 0.0145, compounded yearly (so n = 1), for t = 3 years. Substituting and calculating, we find that Violet’s account will be worth A = P ( 1 + r n ) n t = $ 4,520.00 ( 1 + 0.0145 1 ) 1 × 3 = $ 4,520.00 ( 1.0145 ) 3 = $ 4,719.48 Or, Violet will have $4,719.48 after 3 years. Banks have not always offered interest on savings accounts. An 1836 publication from Indiana noted that banks in other states allow small interest on deposits. It specifically says that in these other states, these deposits are what business transactions are based upon. And that giving interest would encourage deposits, and thus increase the business that banks can do. Journal of the House of Representatives of the Sate of Indiana Certificates of Deposit, or CDs We discussed certificates of deposit (CDs) in earlier sections. CDs differ from savings accounts in a few ways. First, the investment lasts for a fixed period of time, agreed to when the money is invested in the CD. These time periods often range from 6 months to 5 years. Money from the CD cannot be withdrawn (without penalty) until the investment period is up. Also, money cannot be added to an existing CD. Certificates of deposit have features similar to savings accounts. They are insured by the FDIC. They are entirely safe. They do, though, offer a better interest rate. The trade-off is that once the money is invested in a CD, that money is unavailable until the investment period ends. 5-Year CD Silvio deposits $10,000 in a CD that yields 2.17% compounded semiannually for 5 years. How much is the CD worth after 5 years? This also uses the compound interest formula from Compound Interest , A = P ( 1 + r n ) nt , Substituting the values P = $10,000, r = 0.0217, n = 2 (semiannually means twice per year), and t = 5, we find the account will be worth A = P ( 1 + r n ) nt = $ 10 , 000.00 ( 1 + 0.0217 2 ) 2 × 5 = $ 10,000.00 ( 1.01085 ) 10 = $ 11 , 239.53 The CD will be worth $11,219.53 after 5 years. Money Market Account A money market account is similar to a savings account, except the number of transactions (withdrawals and transfers) is generally limited to six each month. Money market accounts typically have a minimum balance that must be maintained. If the balance in the account drops below the minimum, there is likely to be a penalty. Money market accounts offer the flexibility of checks and ATM cards. Finally, the interest rate on a money market account is typically higher than the interest rate on a savings account. Single Deposit to a Money Market Account Marietta opens a money market account, and deposits $2,500.00 in the account. It bears 1.76% interest compounded monthly. If she makes no other transactions on the account, how much will be in the account after 4 years? This, once again, uses the compound interest formula from Compound Interest : A = P ( 1 + r n ) nt , Substituting the values P = $25,000, r = 0.0176, n = 12, and t = 4, we find the account will be worth A = P ( 1 + r n ) nt = $ 2,500.00 ( 1 + 0.0176 12 ) 12 × 4 = $ 2,500.00 ( 1.0014 6 ¯ ) 48 = $ 2,682.20 The money market account will be worth $2,682.20 after 4 years. Return on Investment If we want to compare the profitability of different investments, like savings accounts versus other investment tools, we need a measure that evens the playing field. Such a measure is return on investment . The return on investment, often denoted ROI, is the percent difference between the initial investment, P , and the final value of the investment, F V , or ROI = F V - P P , expressed as a percentage. The length of time of the investment is not considered in ROI. Calculating Return on Investment Determine the return on investment for the 5-year CD from . Round the percentage to two decimal places. Determine the return on investment for the money market account from . Round the percentage to two decimal places. The initial deposit in the CD was $10,000, so P = $10,000. The value at the end of 5 years was $11,239.53. so F V = $11,239.53. Substituting and computing we find the return on investment. ROI = F V - P P = $ 11,239.53 - $ 10,000 $ 10,000 = $ 1,239.53 $ 10 , 000 = 0.123953 The ROI is 12.40%. The initial deposit in the money market was $2,500, so P = $2,500. The value at the end of 4 years was $2,682.20. so F V = $2,682.20. Substituting and computing we find the return on investment. ROI = F V - P P = $ 2,682.20 - $ 2,500 $ 2,500 = $ 182.20 $ 2,500 = 0.07288 The ROI is 7.29%. Return on Investment, ROI Annuities as Savings In Compound Interest , we talked about the future value of a single deposit. In reality, people often open accounts that allow them to add deposits, or payments , to the account at regular intervals. This agrees with the 50-30-20 budget philosophy, where some income is saved every month. When a deposit is made at the end of each compounding period, such a savings account is called an ordinary annuity . The formula for the future value of an ordinary annuity is F V = pmt × ( 1 + r / n ) n × t - 1 r / n , where F V is the future value of the annuity, pmt is the payment, r is the annual interest rate (in decimal form), n is the number of compounding periods per year, and t is the number of years. It is important to note that the number of deposits per year and the number of periods per year are the same. Another form of annuity if the annuity due, which has deposits at the start of each compounding period. This other annuity type has different formulas and is not addressed in this text. Future Value of an Ordinary Annuity Jill has an account that bears 3.75% interest compounded monthly. She decides to deposit $250.00 each month, at the end of the compounding period, into this account. What is the future value of this account, after 8 years? These are regular payments into an account bearing compound interest. She is depositing them at the end of each compounding period. This makes this an ordinary annuity. Substituting the values pmt = 250, r = 0.0375, n = 12, and t = 8 into the formula, we find the future value of the account. F V = pmt × ( 1 + r / n ) n × t - 1 r / n = 250 × ( 1 + 0.0375 / 12 ) 12 × 8 - 1 0.0375 / 12 = 250 × ( 1.003125 ) 96 - 1 0.003125 = 250 × 1.34922752406 - 1 0.003125 = 250 × 0.34922752406 0.003125 = 250 × 111.752807699 = 27 , 938.202 The account, after 8 years, will contain $27,938.20. Setting Savings Account Interest Rates There are a number of factors that contribute to the amount a bank gives for savings accounts. The interest rate reflects how much the bank values deposits. It also reflects the money that the bank will earn when they lend out money. Finally, interest rates are impacted by the Federal Reserve Bank. When the Fed raises interest rates, so do banks. The Federal Reserve Chairperson The Federal Reserve Board monitors the risks in the financial system to help ensure a healthy economy for individuals, companies, and communities. The Board oversees the 12 regional reserve banks. The Chairperson of the Federal Reserve Board testifies to Congress twice per year, meets with the secretary of the Treasury, chairs the Federal Open Market Committee, and is the face of federal monetary policy. Currently, the Fed Chair is Jerome Powell, who has served since 2018. Saving for College When Yusef was born, Rita and George began to save for Yusef’s college years by investing $2,500 each year in a savings account bearing 3.4% interest compounded annually. How much will they have saved after 18 years? To find the future value of the account, we use the ordinary annuity formula F V = pmt × ( 1 + r / n ) n × t - 1 r / n . The payment is $2,500, rate is 0.034, the number of compounding periods is 1, and the number of years is 18. Substituting these values and computing, we have F V = pmt × ( 1 + r / n ) n × t - 1 r / n = $ 2,500 × ( 1 + 0.034 / 1 ) 1 × 18 - 1 0.034 / 1 = $ 2,500 × ( 1.034 ) 18 - 1 0.034 = $ 2,500 × 1.82544897331 - 1 0.034 = $ 2,500 × 24.2779109798 = $ 60,694.77 After saving for 18 years, Rita and George will have $60,694.77 for Yusef’s college. Google Sheets offers a function to calculate the future value of an ordinary annuity. To get Google Sheets to calculate the future value, you use the following: =fv(rate,number_of_periods, payment, present_value, end_or_beginning). To explain, the rate is the rate per compounding period. From our formula, that is r / n . Also, the number of periods must be entered. From our formula, that is n × t . The payment is the amount deposited each period. Present value is 0 if we begin with no money and rely only on the payments to be made. However, if some money is available to put in the account before the payments start, that amount, an initial deposit, would be the value of P V . Finally, for an ordinary annuity, enter 0 for end or beginning. Using the values for Jill, the payment amount is $250, r = 0.0375, n = 12, t = 8, and that there is no initial deposit, P V = 0, the Google Sheets formula is =fv(0.0375/12,12*8,250,0,0). shows the formula in Google Sheets. Google Sheets formula Hitting the enter key shows the payment value ( ). Payment value Notice that the future value is negative, since it is a payment leaving an account. Future Value Using Google Sheets Compute Payment to Reach a Financial Goal The formula used to get the future value of an ordinary annuity is useful, finding out what the final amount in the account will be. However, that isn’t how planning works. To plan, we need to know how much to put into the ordinary annuity each compounding period in order to reach a goal. Fortunately, that formula exists. The formula for the amount that needs to be deposited per period, pmt , of an ordinary annuity to reach a specified goal, FV , is pmt = F V × ( r / n ) ( 1 + r / n ) n × t - 1 , where r is the annual interest rate (in decimal form), n is the number of periods per year, and t is the number of years. With this formula, it is possible to plan the amount to be saved. Saving for a Car Yaroslava wants to save in order to buy a car, in 3 years, without taking out a loan. She determines that she’ll need $35,500 for the purchase. If she deposits money into an ordinary annuity that yields 4.25% interest compounded monthly, how much will she need to deposit each month? Yaroslava has a goal and needs to know the payments to make to reach the goal. Her goal is FV = $35,500, with an interest rate r = 0.0425, compounded per month so n = 12, and for 3 years, making t = 3. Substituting into the formula, Yaroslava finds the necessary payment. pmt = F V × ( r / n ) ( 1 + r / n ) n × t - 1 = 35,500 × ( 0.0425 / 12 ) ( 1 + 0.0425 / 12 ) 12 × 3 - 1 = 35,500 × ( 0.003541 6 ¯ ) ( 1.003541 6 ¯ ) 36 - 1 = 125.7291 6 ¯ 0.13572901696 = 926.325 To reach her goal, Yaroslava would need to deposit $926.33 in her account each month. This has been rounded up, so that the deposits don’t fall short of the goal. However, some round off using the standard rounding rules: if the last digit is 1, 2, 3, or 4, the number is rounded down; if the last digit is 5, 6, 7, 8, or 9 the number is rounded up. Google Sheets offers a function to calculate the payment necessary to reach a goal using ordinary annuities. To get Google Sheets to calculate the payment, you use the following: =pmt(rate,number_of_periods, present_value, future_value, end_or_beginning). To explain, the rate is per compounding period. From our formula, that is r / n . Also, the number of periods must be entered. From our formula, that is n × t . The present value is the amount of money that the account begins with. If we begin with no money and rely only on the payments to be made, then this number is 0. However, if some money is available to put in the account before the payments start, that amount, an initial deposit, would be the value of P V . Next, enter the future value, F V . Finally, for an ordinary annuity, enter 0 for end or beginning. Using the values for Yaroslava, r = 0.0425, n = 12, t = 3, and that there is no initial deposit, P V = 0, the Google Sheets formula is =pmt(0.0425/12,12*3,0,35500,0). shows the formula in Google Sheets. Google Sheets formula Hitting the enter key shows the payment value ( ). Payment value Notice that the payment is negative, since it is a payment leaving an account. Additionally, the payment is $926.32. We rounded that up, but Google Sheets rounded off. Check Your Understanding Key Terms Savings account 1099 form Certificate of deposit Money market account Return on investment Ordinary annuity Key Concepts There are three main types of savings accounts, saving accounts, certificates of deposit (CD), and money market accounts. Savings account are very risk free, and so yield low interest rates. The differences in the three types of savings accounts relate to their convenience. Savings account typically have a lower interest rate that money market accounts, which typically have lower interest rates than CDs. Ordinary annuities more accurately reflect how we save, in that money is deposited repeatedly over time. Spreadsheet software, such as Google Sheets, have built in functions that can be used to quickly calculate both the future value of an ordinary annuity account, but also the payment necessary to reach a goal using an ordinary annuity. Videos Return on Investment, ROI Future Value Using Google Sheets Formulas A = P ( 1 + r n ) n t ROI = F V - P P F V = p m t × ( 1 + r / n ) n × t - 1 r / n p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1", "section": "Methods of Savings", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Investments Stocks are bought and sold to improve investment values. (credit: modification of work \"FT ringing the Closing Bell at the NYSE\" by Financial Times/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Distinguish between basic forms of investments including stocks, bonds, and mutual funds. Understand what bonds are and how bond investments work. Understand how stocks are purchased and gain or lose value. Read and derive information from a stock table. Define a mutual fund and how to invest. Compute return on investment for basic forms of investments. Compute future value of investments. Compute payment to reach a financial goal. Identify and distinguish between retirement savings accounts. You can save your money in a safe or a vault (or worse, under the mattress!), but that money does not grow. It would be hard to save enough for retirement that way. What can be done to increase the value of the money you already have? The answer is to invest it. Use the money that you have to earn more money back. For instance, as we saw in Methods of Savings , you can save it in a bank. Or, to reach loftier goals, invest in something more likely to grow, such as stocks. A great example of this is Apple stock. Anyone who bought stock in Apple Inc. (formerly Apple Computer, Inc.) in 1997 and held onto the shares earned a lot of money. To be more specific, $100 worth of Apple shares bought in 1980, when it was first sold to the public, was valued at $67,564 in 2019, or 676 times more! Perhaps you have heard a story like that, of an investment opportunity taken that paid off, or the story of an investment opportunity missed. But such stories are the exceptions. In this section, we’ll investigate bonds, stocks, and mutual funds and their comparative strengths and weaknesses. We close the section with a discussion of retirement savings accounts. Distinguish Between Basic Forms of Investments Bonds, stocks, and mutual funds tend to offer higher returns, but to varying degrees, come with higher risks. Stocks and mutual funds also vary in how much they earn. Their predicted rates of return on investment are not guaranteed, but educated guesses based on market trends and historical performance. We will use the methods and formulas we learned earlier to evaluate these forms of investment. Bonds Bonds are issued from big companies and from governments. Selling bonds is an alternative to an institution taking a loan from a bank. The funds from the selling of bonds are often used for large projects, like funding the building of a new highway or hospital. Bonds are considered a conservative investment. They are bought for what is known as the issue price . The interest is fixed (does not change) at the time of purchase and is based on the issue price of the bond. The interest rate is often referred to as the coupon rate; the interest paid is often called the coupon yield. The interest paid is often higher than savings accounts and the risk is exceptionally low. The bond is for a fixed length of time. The end of this time is the maturity date of the bond. There are several types of bonds: Treasury bonds are issued by the federal government. Municipal bonds are issued by state and local governments. Corporate bonds are issued by major corporations. There are other types of bonds available, but they are beyond the scope of this section. Trading Bonds Bonds are often part of larger investment portfolios. These bonds may be traded. However, the interest paid is based on the price when the bond was bought (the issue price). These bonds can be bought and sold for more or less money than the issue price. If the bond is bought for more than the issue price, the interest is still paid on the issue price, not on the purchase price when the trade was made. This means the actual return on the bond decreases. If the bond is bought for less than the issue price, the return on the bond goes up. Bonds Bond Investment Muriel purchases a $3,000 bond with a maturity of 4 years at a fixed coupon rate of 5.5% paid annually. How much is Muriel paid each year, and how much does she receive on the maturity date? The coupon rate is 5.5%. 5.5% of her bond value is 0.055 × $ 3,000 = $ 165 . After year 1, Muriel receives $165. She receives $165 after years 2 and 3 also. In year 4, when the bond matures, Muriel receives $3,165, or the interest and the initial investment, or principal. Stocks Stocks are part ownership in a company. They come in units called shares . The performance and earnings of stocks is not guaranteed, which makes them riskier than any other investment discussed earlier. However, they can offer higher return on investment than the other investments. Their value grows in two ways. They offer dividends , which is a portion of the profit made by the company. And the price per share can increase based on how others see that value of the company changing. If the value of the company drops, or the company folds, the money invested in the stock also drops. Most stock transactions are executed through a broker. Brokers’ commissions can be a percentage of value of the trades made or a flat fee. There are full-service brokers who charge higher commission rates, but they also offer financial advice and perform the research that you may not have the time or the expertise to do on your own. A discount broker only executes the stock transactions, buying or selling, so they charge lower rates than full-service brokers. There are also brokers that offer commission-free trading. An important thing to remember is that stocks might provide a very large return on investment, but the trade-off is the risk associated with owning stocks. Chapter 11 Bankruptcy and Stocks In the fall of 2022, the parent company of Regal Theaters, named Cineworld, filed for Chapter 11 bankruptcy. According to news articles, the bankruptcy was necessitated due to its heavy debt load. Generally, a company can file for Chapter 11 bankruptcy to allow them time to reorganize and restructure debts. When this happens, the company, after the Chapter 11 process is over, offers new stock. This makes the previous stock worthless. However, the company may allow an exchange of old stock for a discounted amount of the new stock. This in effect reduces (maybe vastly) the wealth held by those who owned the original stock. Buying Stock in Company ABC Haniah buys stock in the ABC company, investing a total of $13,000. She expects the stock to grow, through stock price increase and reinvestment of dividends, by 12.3% per year and compounded annually. If she leaves that money invested, how much will the stocks be worth in 20 years? Calculating this is a compound interest calculation, if Haniah’s assumption about the stock’s performance is correct. If so, then the principal is $13,000, the rate is 0.123, the number of compounding periods per year is 1, and the time is 20 years. Substituting into the compound interest formula from Methods of Savings , and computing, we have A = P ( 1 + r n ) n t = $ 13,000 ( 1 + 0.123 1 ) 1 × 20 = $ 13,000 × 10.1764223996 = $ 132,293.49 . After 20 years, her stock is now worth $132,293.49. Risk and Volkswagen The question of risk hovers over every investment. How risky can it get? Volkswagen seems to be a rather safe investment. But in 2015, Volkswagen’s stock tumbled 30% over a few days when it was revealed that the company had installed software that altered the emission performance of some of their diesel engines. Volkswagen’s hope was that lower emissions would bolster US sales of some of their diesel models. This was a drastic drop, and many investors lost a lot of money. However, the stock has come back since then. This was mild compared to the 65% drop in the Martha Stewart Living Omnimedia stocks. Warren Buffett Warren Buffett is an investment legend. He began his career as an investment salesman in the 1950s. He formed Buffett associates in 1956. In 1965, he was in control of Berkshire Hathaway, which began as a merger between two textile companies. In his role there, he began to invest in a variety of companies. It is now a conglomerate holding company, and fully owns GEICO, Duracell, Diary Queen, and other large companies. His investment philosophy involves finding stocks and bonds from companies that have high intrinsic worth compared to their stock or bond prices. This means he focuses not on the supply and demand side of stock investing, but instead on the company’s worth in total. Using this philosophy, he has become one of the world’s most successful investors. Reading Stock Tables Information about particular stocks is contained in stock tables . This information includes how much the stock is selling for, and its high and low values form the past year (52 weeks). In a newspaper, the stock table may look like this: 52-Week High Low Stock SYM Div Yld % P/E Vol 100s High Low Close Net Chg 41.66 18.90 McDonald’s MCD .72 2.9 12 7588 25.73 23.87 25.42 +0.31 22.60 13.20 Monsanto MON .52 2.4 55 15474 21.86 21.48 21.64 -0.29 17.05 8.30 Motorola MOT .16 1.7 dd 16149 10.57 8.88 10.43 +0.14 31.75 22.99 Mueller MLI - - 16 1564 29.32 27.03 27.11 -0.02 excerpt from a stock table, 2008 The symbols and abbreviations are defined here: 52-week High 52-week Low The highest and lowest price of the stock over the past 52 weeks Stock SYM The name of the company and the symbol used for trading Annual DIV The current annual dividend per share Yld % Percent yield is = annual dividend share price × 100 P/E Price to earnings ratio, share price divided by earnings per share over past year (dd indicates loss) Vol 100s The number of shares traded yesterday in 100s High Low The highest and lowest prices at which stocks traded yesterday Close The price at which the stock traded at the close of the market yesterday Net Chg Net change; change in price from market close 2 days ago to yesterday’s close The formulas for yield and price to earnings is a good way to measure how much the stock returns per share. Their values are calculated in the stock table, but deserve attention here. The price to earnings ratio of a stock, P/E, is P / E = Share Price Dividend . The percent yield for a stock, Yld%, is Y l d % = Annual Dividend Share Price × 100 % . It should be noted that the price of a stock increases and decreases every moment, and so these value change as the share price changes. Computing Percent Yield Find the percent yield for a stock with a price of $30.69 and an annual dividend of $1.48. Find the percent yield for a stock with a price of $62.25 and an annual dividend of $1.76. Substituting the values for price, $30.69, and annual dividend, $1.48, we find the percent yield for the stock to be Y l d % = Annual Dividend Share Price × 100 % = $ 1.48 $ 30.69 × 100 % = 4.82 % Substituting the values for price, $62.25, and annual dividend, $1.76, we find the percent yield for the stock to be Yld% = Annual Dividend Share Price × 100 % = $ 1.76 $ 62.25 × 100 % = 2.83 % The stock table information is now, and has been, available online, from websites such as cnn.com/markets, markets.businessinsider.com/stocks, and marketwatch.com. The same information is available from these sites as from the newspaper listings, but are often accessed one stock at a time. shows the stock table for Lowe’s on September 7, 2022. Key data for Lowe's stock 9/7/2022 (data source: marketwatch.com) Other key data is further down on the website, and is shown in , below. Key data for Lowe's stock 9/7/2022 (data source: marketwatch.com) Notice that the 52-week high and low are now shown as the 52-week range. However, you get additional information, including the stock performance over the past 5 days, past month, past 3 months, the year to date (YTD), and over the past year. You can also read the number of shares outstanding, the expected date for the dividend (EX-DIVIDEND DATE), and importantly for the P/E ratio, the earning per share (EPS). Reading an Online Stock Table Consider the stock table ( ), and answer the questions based on the table. Key data for McDonald's stock 9/7/2022 (data source: marketwatch.com) What is the current price for McDonald’s Corp on this date? What is the 52-wk high? 52-wk low? When is the dividend expected? What is its yield? What is the earnings per share? Looking at the table, the current price of a share is $258.87. The high was $271.15, and the low was $217.68. August 31, 2022 2.13% The EPS value is $8.12. As mentioned, stocks earn money in two ways, through dividends and increase in share price. Dividends Paid Darma owns 150 shares of stock in the GDW company. This quarter, GDW is paying $0.87 per share in dividends. How much will Darma earn in dividends this quarter? Each share pays $0.87, so Darma earns 150 × $ 0.87 = $ 130.50 . Stock Price Increases Vincent buys 100 stocks in the REM company for $21.87 per share. One year later, he sells those 100 shares for $29.15 per share. How much money did Vincent make? What was his return on investment for that one year? Vincent spent $21.87 per share to buy the stock. The total he spent on the stock was $ 21.87 × 100 = $ 2,187.00 . When he sold the stock, the price was $29.15, so he received $ 29.15 × 100 = $ 2,915.00 . He made $ 2,915.00 - $ 2,187.00 = $ 728.00 . His return on investment was Earnings Original Price = $ 728 $ 2,187 = 33.29 % . Reading Stock Summary Online Mutual Funds A mutual fund is a collection of investments that are all bundled together. When you buy shares of a mutual fund, your money is pooled with the assets of other investors. This pooled money is invested in stocks, bonds, money market instruments, and other assets. Mutual funds are typically operated by professional money managers who allocate the fund's assets and attempt to produce capital gains or income for the fund's investors. A key benefit of mutual funds is that they allow small or individual investors to invest in professionally managed portfolios of equities, bonds, and other securities. This means each shareholder participates proportionally in the gains or losses of the fund. The performance of a mutual fund is usually stated as how much the mutual fund’s total value has increased or decreased. Since there are many different investments inside the mutual fund, the risk is reduced significantly, compared to direct ownership of stocks. Even so, mutual funds historically perform well and can earn more than 10% annually. The investments that make up a mutual fund are structured and maintained to match stated investment objectives, which are specified in its prospectus . A prospectus is a pamphlet or brochure that provides information about the mutual fund. Before buying shares of a mutual fund, consult its prospectus, consider its goals and strategies to see if they match your goals and values and also research any associated fees. Mutual Funds Investing in a Mutual Fund Kaitlyn has analyzed her $12,862.50 quarterly budget using the 50-30-20 budget philosophy, and sees she should be saving or paying down debt with $2,572.50 per quarter. She decides to invest $1,300 quarterly a mutual fund that reports an average return of 11.62% over the 18-year life of the mutual fund. Assuming that this interest rate continues, and is compounded quarterly, how much will her mutual fund account be worth after 5 years? Kaitlyn’s plan is an ordinary annuity, and so the future value of her account can be found using the formula F V = p m t × ( 1 + r / n ) n × t - 1 r / n , with a payment of $1,300, a rate of 0.1162, number of compounding periods 4, after 5 years. Substituting these values into the formula and calculating, we find F V = p m t × ( 1 + r / n ) n × t - 1 r / n = $ 1,300 × ( 1.02905 ) 20 - 1 0.02905 = $ 1,300 × 0.773084935178 0.032905 = $ 1,300 × 26.6122189784 = $ 34,595.88 Kaitlyn’s mutual fund will be worth $34,595.88 after 5 years. Investing in a Mutual Fund to Reach a Goal Kaitlyn wants to retire with $1,500,000 in her mutual fund account. She will invest for 35 years. The mutual fund reports an average return of 11.62% over the 18-year-long life of the mutual fund. Assuming that this interest rate continues, and is compounded quarterly, how much will she need to pay annually into her mutual fund to reach her goal? Kaitlyn’s plan is an ordinary annuity, and so the payment to reach her goal can be found using the formula p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1 , with a F V , or goal, of $1,500,000, a rate of 0.1162, for 35 years. Substituting these values into the formula and calculating, we find p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1 = $ 1,500,000 × ( 0.1162 / 1 ) ( 1 + 0.1162 / 1 ) 1 × 35 - 1 = $ 174,300 26.0558103113 = $ 6,689.49 Kaitlyn needs to invest $6,689.49 per year (or $557.46 per month) into the mutual fund to reach $1,500,000 in 35 years. Return on Investment As in Methods of Savings , the formula for return on investment is ROI = F V - P P . As indicated before, this formula does not take into account how long the investment took to reach its current value. It depends only on the initial value, P , and the value at the end of the investment, F V . Return on Investment for a Bond Recall , in which Muriel purchased a $3,000 bond with a maturity of 4 years at a fixed coupon rate of 5.5% paid annually. What was Muriel’s return on investment? Each year, Muriel received $165. She received this money four times, so earned a total of $660. This represents F V – P , or just the earnings. Using that we find that the ROI is ROI = 660 3000 = 0.22 , or 22%. As mentioned, the ROI does not address the length of time of the investment. A good way to do that is to equate the ROI to an account bearing interest that is compounded annually. The annual return is the average annual rate, or the annual percentage yield (APY) that would result in the same amount were the interest paid once a year. The formula for annual return is annual return = ( F V P ) ( 1 t ) - 1 , where t = the number of years, F V = new value, and P = starting principal. We apply this to the previous example. Annual Return on Investment for a Bond Recall , in which Muriel purchased a $3,000 bond with a maturity of 4 years at a fixed coupon rate of 5.5% paid annually. What was Muriel’s annual return on investment? Interpret this as compound interest. Muriel earned a total of $660. This represents F V – P , or just the earnings. The starting principal was $3,000. The value at the end of 4 years was $3,000 + $660 = $3,660. The time of the investment was 4 years. Using that we find that the annual return is annual return = ( F V P ) ( 1 t ) - 1 = ( $ 3,660 $ 3,000 ) ( 1 4 ) - 1 = 1.22 1 4 - 1 = 1.050969 = 0.050969 , or 5.10%. The 5.5% bond earned the equivalent of 5.10% compounded annually. In and Your Turn, the annual return was lower than the interest rate of the investment. This is because the interest from a bond is simple interest, but annual yield equates to compounded annually. Return on Investment for Stock in Company ABC Haniah buys stock in the ABC company, investing a total of $13,000. After 20 years, the stock is worth $132,293.49, including reinvestment of dividends. What is Haniah’s return on investment? What is Haniah’s annual return? To calculate Hanniah’s return on investment, substitute $13,000 for P and $132,293.49 for F V in the formula ROI = F V - P P and calculate. Doing so we find Haniah’s return on investment to be ROI = F V - P P = $ 132,293.49 - $ 13,000 $ 13,000 = 9.176422 , or 917.64% To calculate Haniah’s annual return, substitute $13,000 for P and $132,293.49 for F V in the formula annual return = ( F V P ) ( 1 t ) - 1 and calculate. Doing so we find her annual return to be annual return = ( F V P ) ( 1 t ) - 1 = ( $ 132,293.49 $ 13,000 ) ( 1 20 ) - 1 = 0.1230 , or 12.3% You should see that the annual return is equal to the annual compounded interest that was assumed for the stocks. Compute Payment to Reach a Financial Goal As in Methods of Savings , determining the payment necessary to reach a financial goal uses the payment formula for an ordinary annuity, p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1 . If dealing with mutual funds or stocks, an assumed annual interest rate, compounded, will be used. This value is often determined through research and informed speculation. Richard is saving for new siding for his home. He and his partner believe they will need $37,500 in 10 years to pay for the siding. How much should they invest yearly in a mutual fund they believe will have an annual interest rate of 12%, compounded annually, in order to reach their goal? The necessary annual payment is found using the function p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1 with F V = 37,500, r = 0.12, and n = 1. Substituting and calculating, we find the annual payment should be p m t = F V × ( r / n ) ( 1 + r / n ) n × t - 1 = $ 37,500 × ( 0.12 / 1 ) ( 1 + 0.12 / 1 ) 1 × 10 - 1 = $ 4,500 ( 1.12 ) 10 - 1 = $ 4,500 2.10584820834 = $ 2,136.91 Retirement Savings Plans We close this section by investigating the three main forms of retirement savings accounts: traditional individual retirement accounts (IRAs), Roth IRAs, and 401(k) accounts. Each has distinct characteristics that are suited to different investors’ needs. Individual Retirement Accounts A traditional IRA lets you contribute up to an amount set by the government, which may change from year to year. For example, the maximum contribution for 2022 is $6,000; $7,000 over age 50. Anyone is eligible to contribute to a traditional IRA, regardless of your income level. Your money grows tax-deferred, but withdrawals after age 59½ are taxed at current rates. Traditional IRAs also allow you to use the contribution itself as a deduction on a current year tax return. Roth IRAs allow contributions at the same levels as traditional IRAs, with a maximum $6,000 for 2022; $7,000 over age 50. However, to be eligible to make contributions, your earned income must be below a certain level. A Roth IRA allows after-tax contributions. In other words, the contribution itself is not tax-deductible, as it is with the traditional IRA. However, your money grows tax-free. If you make no withdrawals until you are age 59½, there are no penalties. IRAs pay a modest interest rate. In either case, IRA deposits have to be from earned income, which in effect means if your earned income is over $6,000 ($7,000) then you can deposit the maximum. Comparing Roth IRAs to Traditional IRAs Which type of IRA, Roth or traditional, has an income limit for its use? Roth IRAs require income to be below a certain limit. In 2022, the maximum that can be added to a Roth IRA was $6,000 for those under 50 years of age. For those over 50 years of age, the maximum that can be added to a Roth IRA is $7,000. However, to qualify for a Roth IRA in fall of 2022, a single person’s modified adjusted gross income (MAGI) must be below $129,000. Then, if a single person’s income is between $129,000 and $144,000, the maximum contribution is reduced from the limit for incomes below $129,000. For a married couples filing a joint tax return those values are $204,000 to $214,000. 401(k) Accounts Your employer may offer a retirement account to you. These are often in the form of a 401(k) account. There are traditional and Roth 401(k) accounts, which differ in how they are taxed, much as with other IRAs. In the traditional 401(k) plans, the money is deposited before tax is assessed, which means you do not pay taxes on this money. However, that means when money is withdrawn, it is taxed. These accounts are similar to mutual funds, in that the money is invested in a wide range of assets, spreading the risk. One of the perks some employers offer is to match some amount of your contributions to the 401(k) plan. For instance, they may match your deposits up to 5% of your income. This is an instant 100% return on the money that was matched. 401(k) Accounts Matching 401(k) Deposit Alice signs up for her employer-based 401(k). The employer matches any 401(k) contribution up to 6% of the employee salary. Alice’s annual salary is $51,600. What is the most money that Alice can deposit that will be fully matched by the company? How much total will be deposited into Alice’s account if she deposits the full 6%? How much return does Alice earn if she deposits exactly 6% in her 401(k)? The employer will match up to 6% of any employee’s salary. 6% of Alice’s salary is 0.06 × $ 51,600 = $ 3,096 . So Alice can deposit up to $3,096 and receive that amount in matching funds in her account. Alice’s contribution plus the company’s contribution is $ 3,096 + $ 3,096 = $ 6,192 , which is the total that is deposited into Alice’s account. She earns a 100% return on the day she deposits her $3,096. 401(k) plans with matching funds provide great value, as their rates of return are high compared to savings accounts, and are less risky that stocks since such funds invest across many investment vehicles. The next example demonstrates the power of constant deposits into a 401(k) plan that has some employer match. Constant Deposits into a 401(k) Plan DeJean begins depositing $300 per month from his paycheck each month in his employer-based 401(k) account. The employer matches this deposit as it falls below their matching threshold. DeJean expects the return to average 10% per year, compounded annually. How much will DeJean’s account be worth if he keeps making those payments for 30 years? What will his account be worth without the matching funds? This is a form of an ordinary annuity, so the formula F V = p m t × ( 1 + r / n ) n × t - 1 r / n will be used. The company matches DeJean’s full deposit, so each month $600 will be deposited. He is assuming the money will compound annually, so the amount deposited each year is needed as the value of pmt. For the year, he will deposit 12 × $ 600 = $ 7,200 . The rate is 0.1, the number of compounding periods is 1, and the number of years is 30. Substituting and calculating, the value of DeJean’s account after 30 years will be F V = p m t × ( 1 + r / n ) n × t - 1 r / n = $ 7,200 × ( 1 + 0.1 / 1 ) 1 × 30 - 1 0.1 / 1 = $ 7,200 × ( 1.1 ) 30 - 1 0.1 = $ 7,200 × 164.494022689 = $ 1,184,356.96 This is a form of an ordinary annuity, so the formula F V = p m t × ( 1 + r / n ) n × t - 1 r / n will be used but the deposit is now only $300 per month without the matching funds. For the year, he will deposit 12 × $ 300 = $ 3,600 . The rate is 0.1, the number of compounding periods is 1, and the number of years is 30. Substituting and calculating, the value of DeJean’s account after 30 years will be F V = p m t × ( 1 + r / n ) n × t - 1 r / n = $ 3,600 × ( 1 + 0.1 / 1 ) 1 × 30 - 1 0.1 / 1 = $ 3,600 × ( 1.1 ) 30 - 1 0.1 = $ 3,600 × 164.494022689 = $ 592,178.48 Check Your Understanding Key Terms Bonds Maturity date Stocks Dividend Mutual fund Prospectus Issue price Shares Stock table Individual retirement account Roth IRA 401(k) Key Concepts There are many different investments with different returns and risks. Bonds are loans form the purchaser to the entity selling the bond. Bonds have some tax benefits, low to no risk, and a low return. Stocks represent part ownership in a company. As such, stock holders share in the profits, and losses, of the company. Information, including price, P/E, yearly highs and lows, and dividend amount can be found in online stock tables available on many websites. Mutual funds represent collections of professionally administered investment vehicles. Have shares in a mutual fund has lower risk than ownership of stocks. Retirement accounts employ some of the same strategies as mutual funds, in that they spread the risk and are professionally managed. IRAs and Roth IRAs differ on when taxes are paid on the money, and who can use them. Roth IRAs have income limits while traditional IRAs do not. Videos Bonds Reading Stock Summary Online Mutual Funds 401(k) Accounts Formulas annual return = ( F V P ) ( 1 t ) - 1 P / E = Share Price Dividend Yld % = Annual Dividend Share Price × 100 %", "section": "Investments", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Basics of Loans Loans are contracts that allow people to buy now but require them to pay more. (credit: \"Closing\" by Tim Pierce/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe various reasons for loans. Describe the terminology associated with loans. Understand how credit scoring works. Calculate the payment necessary to pay off a loan. Read an amortization table. Determine the cost to finance for a loan. New car envy is real. Some people look at a new car and feel that they too should have a new car. The search begins. They find the model they want, in the color they want, with the features they want, and then they look at the price. That’s often the point where the new car fever breaks and the reality of borrowing money to purchase the car enters the picture. This borrowing takes the form of a loan. In this section, we look at the basics of loans, including terminology, credit scores, payments, and the cost of borrowing money. Reasons for Loans Even if you want a new car because you need one, or if you need a new computer since your current one no longer runs as fast or smoothly as you would like, or you need a new chimney because the one on your house is crumbling, it’s likely you do not have that cost in cash. Those are very large purchases. How do you buy that if you don ’t have the cash? You borrow the money. And for helping you with your purchase, the company or bank charges you interest. Loans are taken out to pay for goods or services when a person does not have the cash to pay for the goods or services. We are most familiar with loans for the big purchases in our lives, such as cars, homes, and a college education. Loans are also taken out to pay for repairs, smaller purchases, and home goods like furniture and computers. Loans can come from a bank, or from the company selling the goods or providing the service. The borrower agrees to pay back more than the amount borrowed. So there is a cost to borrowing that should be considered when deciding on a purchase bought with credit or a borrowed money. Even using a credit card is a form of a loan. Essentially, a loan can be obtained for just about any purchase, large or small, that has a cost beyond a person’s cash on hand. The Terminology of Loans There are many words and acronyms that get used in relation to loans. A few are below. APR is the annual percentage rate. It is the annual interest paid on the money that was borrowed. The principal is the total amount of the loan, or that has been financed. A fixed interest rate loan has an interest rate that does not change during the life of the loan. A variable interest rate loan has an interest rate that may change during the life of the loan. The term of the loan is how long the borrower has to pay the loan back. An installment loan is a loan with a fixed period, and the borrower pays a fixed amount per period until the loan is paid off. The periods are almost uniformly monthly. Loan amortization is the process used to calculate how much of each payment will be applied to principal and how much is applied to interest. Revolving credit , also known as open-end credit, is how most credit cards work but is also a kind of loan account. (We will learn about credit cards in Credit Cards ) You can use up to some specified value, called the limit, any way you want, and as long as you pay the issuer of the credit according to their terms, you can keep borrowing from this account. These and other terminologies can be researched further at Forbes . Credit Scores Not everyone pays the same interest for the same loan. One person might get an APR of 2.9% while another pays 6.9%. These rates are based on your credit score. Data about you and your credit is collected by three credit bureaus—Experian, Equifax, and TransUnion. They calculate your score using one of two main models: FICO and Vantage Score. The score they develop is based on the following categories: Payment History: Making your payments on time and not missing payments is by far the most important factor. All three credit types—revolving, installment, and open—contribute to this factor. Credit Utilization or Amount Owed: How much do you owe on your credit card accounts? This category is concerned with the ratio of how much you owe on revolving credit accounts relative to your available credit, also known as your credit utilization ratio. This is the only category that depends solely on your revolving credit accounts. Length of Credit History: This is the average age of your credit history, including the age of the oldest and newest accounts. All three types of credit accounts play a role in this category. Credit Mix: This number represents the different types of credit accounts you have, such as credit cards, car loan, mortgages, and whether you are successful managing both revolving and installment accounts. New Credit: Have you recently opened a new account or applied for new credit? Lenders want to know how much new credit you are taking on. So, if you are planning to buy a car and make another large purchase with a credit card, you may want to space these purchases out. If you have done well in these categories, your credit score will be high, and you will qualify for lower interest rates because you are not perceived as being a risky investment. However, if you do poorly in these categories, your score will be low and you will pay higher interest rates since you present a greater risk. Check out this nerdwallet article on credit scores to learn more! Credit Scores Explained Where Do Interest Rates Come From? There are many factors that impact your interest rate beyond your credit score. Banks have the authority to set their own rates, so competition between banks impacts interest rates. Bank A doesn’t want to charge interest rates that are too high, since borrowers will find banks with better rates. Banks also don’t want to charge too little interest. The too little interest is more involved than the too high. The bank needs to make a profit on its loans. Deposits at the bank are used by the bank to generate loans. The bank has to pay those depositors interest. The bank must charge more for loans they give than they pay to people with deposits in the bank. Banks also borrow money from each other. These loans have an interest rate, and once more, the bank making a loan must make a profit. More directly, they must charge more for loans they give than they pay for loans they take. In the United States, banks may also borrow from the Federal Reserve, which also charges an interest rate, which is also called the discount rate. This is where a bank’s prime rate comes from. A bank’s prime rate is the interest rate it will give to its very best customers, which means most customers will pay more than the bank’s prime rate. To confuse the issue, there is also the Wall Street Journal ’s prime rate. It is the average of the prime rates charged by individual banks. The Wall Street Journal surveys several banks to generate this value. Banks also increase the interest rate charged to customers based on both the credit risk presented by the customer, and the risk associate with what the loan will be used for. Calculating Loan Payments Loan payments are made up of two components. One component is the interest that accrued during the payment period. The other component is part of the principal. This should remind you of partial payments from Simple Interest . Over the course of the loan, the amount of principal remaining to be paid decreases. The interest you pay in a month is based on the remaining principal, just as in the partial payments of Simple Interest . The amount of interest, I , to be paid for one period of a loan with remaining principal P is I = P × r n , where r is the interest rate in decimal form and n is he number of payments in a year (most often n = 12). Since the interest is for the one period, the time is 1 and does not impact the calculation. Note, interest paid to lenders is always rounded up to the next penny. Interest for a Monthly Payment of a Loan Find the interest to be paid for the period on loans with the following remaining principal and given annual interest rate. Each period is a month. Remaining principal is $13,450, interest rate is 6.75% Remaining principal is $8,460, interest rate is 5.99% Substituting $13,450 for the remaining principal P , 0.0675 for r , and n = 12 since the period is a month into the formula, we find that the interest to be paid this period is I = P × r n = $ 13 , 450 × 0.0675 12 = $ 75.66 . Substituting $8,460 for the remaining principal P , 0.0599 for r , and n = 12 since the period is a month into the formula, we find that the interest to be paid this period is I = P × r n = $ 8 , 450 × 0.0599 12 = $ 42.18 . The payment of the loan has to be such that the principal of the loan is paid off with the last payment. In any period, the amount of interest is defined by the formula above, but changes from period to period since the principal is decreasing with each payment. The trick is knowing how much principal should be paid each payment so that the loan is paid off at the stated time. Fortunately, that is found using the following formula. The payment, pmt , per period to pay down a loan with beginning principal P is p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 , where r is the annual interest rate in decimal form, t is the number of years of the loan, and n is the number of payments per year (typically, loans are paid monthly making n = 12). Note, payment to lenders is always rounded up to the next penny. Often, the formula takes the form p m t = P × ( r ) × ( 1 + r ) n ( 1 + r ) n − 1 , where r is the interest rate per period (annual rate divided by the number of periods per year), and n is the total number of payments to be made. Calculating the Payment for a Loan In the following, calculate the payment necessary to pay off the loan with the given details. The payments are monthly. A car loan taken out for $28,500 at an annual interest rate of 3.99% for 5 years. A home loan taken out for $136,700 and an annual interest rate of 5.75% for 15 years. The loan is for $28,500, which is the principal. The rate is 3.99%, so r = 0.0399. The term of the loan is 5 years, so t =5. Monthly payments means n = 12. Substituting these values for P , r , n , and t into the formula p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating, we find the payment for the loan. p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 28 , 500 × ( 0.0399 / 12 ) × ( 1 + 0.399 / 12 ) 12 × 5 ( 1 + 0.0399 / 12 ) 12 × 5 − 1 = $ 28 , 500 × ( 0.003325 ) × ( 1.003325 ) 60 ( 1.003325 ) 60 − 1 = $ 115.6470437 0.220388273 = $ 524.75 The monthly payment needed is $524.75. The loan is for $136,000, which is the principal P . The rate is 5.75% so r = 0.0575. The term of the loan is 15 years, so n = 15. Monthly payments mean n =12. Substituting these values for P , r , n , and t into the formula p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating, we find the payment for the loan. p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 136 , 700 × ( 0.0575 / 12 ) × ( 1 + 0.0575 / 12 ) 12 × 15 ( 1 + 0.0575 / 12 ) 12 × 15 − 1 = $ 28 , 500 × ( 0.0047917 ) × ( 1.0047917 ) 180 ( 1.003325 ) 180 − 1 = $ 1 , 548.600986 1.364201118 = $ 1,135.18 The monthly payment needed is $1,135.18. Using Google Sheets to Calculate Loan Payments Google Sheets has a formula to calculate the monthly payment necessary to pay off a loan with a specified interest rate and term. The formula is the PMT formula. It uses the principal P , interest rate r , and t years. Follow these steps Open a Google worksheet and click on any cell. Type =-PMT(r/12,12*t,P). Hit the enter key. The cell displays the payment. Please note the negative sign. Since Google Sheets is a spreadsheet program, is sees the payment as funds leaving the account, and so they are, by default, negative. The negative sign in front of the formula makes the result positive. For example, for a $50,000 loan at 10.9% interest for 7 years, you would type =-PMT(0.109/12,12*7,50000). The formula and the result are shown in . Google Sheets formula Alternatively, you can also use an online calculator to find monthly payments for a loan, such as the one at Caluculator.net Reading Amortization Tables An amortization table or amortization schedule is a table that provides the details of the periodic payments for a loan where the payments are applied to both the principal and the interest. The principal of the loan is paid down over the life of the loan. Typically, the payments each period are equal. Importantly, one of the columns will show how much of each payment is used for interest, another column shows how much is applied to the outstanding principal, and another column shows the remaining principal or balance . Amortization table Reading from an Amortization Table Using the partial amortization table (Figure 6.23) , answer the following questions. Amortization table What is the loan amount (principal), the interest rate, and the term of the loan? How much is the monthly payment? How much remaining balance is there after the payment in month 15? How much was the interest in payment 10? What is the total of the interest paid after payment 18? What happens to the amount paid in interest each month? Reading the values at the top of the table, we see the principal is $10,000, the interest rate is 4.75%, and has a term of 20 years. The monthly payment is listed below the term of the loan, and is $64.62. $9,613.83 $38.68 $697.01 The amount paid to interest decreases each month. Reading an Amortization Table Cost of Finance There are often costs associated with a loan beyond the interest being paid. The cost of finance of a loan is the sum of all costs, fees, interest, and other charges paid over the life of the loan. Cost of Financing a Personal Loan Irena signed for a loan of $15,000 at 6.33% for 5 years. When she took out the loan, Irena paid a $750 origination fee. Over the course of the loan, she pays $2,537.96 in interest. What was her cost to finance the loan? The cost of finance is the sum is all interest and any fees paid for the loan. The fees paid were $750.00 and the interest was $2,537.96. Her cost of finance for this loan was $3,287.96. Check Your Understanding Key Terms Fixed interest rate Variable interest rate Installment loan Loan amortization Revolving credit Amortization table Cost of finance Key Concepts There are many reasons for a loan, but primarily it is taken out for a large expense when cash is not available. Each payment for an installment loan consists of an interest portion and a principal portion. There is a formula to calculate the payment necessary to pay off a loan in installments. Amortization schedules, or tables, show how each payment is applied to principal and interest. It also includes other details such as remaining balance and total interest paid. Loans often have other fees associated with them such as origination fees or application fees. The total of the interest paid and the fees is the cost of finance. Video Credit Scores Explained Reading an Amortization Table Formula I = P × r n p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1", "section": "The Basics of Loans", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Understanding Student Loans College is often paid for through loans. (credit: \"Sidewalk Scene in front of Science Building\" by IMCBerea College Wikimedia Commons, CC- BY 2.0) Learning Objectives After completing this section, you should be able to: Describe how to obtain a student loan. Distinguish between federal and private student loans and state distinctions. Understand the limits on student loans. Summarize the standard prepayment plan. Understand student loan consolidation. Summarize and describe benefits or drawbacks of other repayment plans. Summarize possible courses of action if a student loan defaults. Obtaining a Student Loan All college students are eligible to apply for a loan regardless of their financial situation or credit rating. Federal student loans do not require a co-signer or a credit check. Most students do not have a credit history when they begin college, and the federal government is aware of this. However, private loans will generally require a co-signer as well as a credit check. The co-signer will assume responsibility for paying off the loan if the student cannot make the payments. The first step in applying for student loans is to fill out the FAFSA (Free Application for Student Aid). FAFSA determines financial need and what type of loan the student is qualified to obtain. For students who are still dependents on their parent’s taxes, the parents also fill out the FAFSA, as their wealth and income impacts what the dependent student is eligible for. Students who cannot demonstrate financial need will also be helped by applying with FAFSA, as it will help guide them to the type of loan most appropriate. The FAFSA must be submitted each year. The FAFSA deadline is the spring of the student’s next academic year. The deadline is often in March. Do not allow this deadline to pass. As soon as an offer letter from the college is received, the student should start the application process. The college will determine the loan amount needed. Also, there are limits on the amount a student can borrow. There are both yearly limits and aggregate limits. See the table later in this section that outlines the loan limits per school year and in the aggregate. If the student receives a direct subsidized loan, there is a limit on the eligibility period. The time limit on eligibility depends on the college program into which the student enrolls. The school publishes how long a program is expected to take. The eligibility period is 150% of that published time. For example, if a student is enrolled in a 4-year program, such as a bachelor’s degree program, their eligibility period is 6 years, as 1.50(4) = 6. Therefore, the student may receive direct subsidized loans for a period of 6 years. Types and Features of Student Loans Once a tuition statement is received, and all the non-loan awards are analyzed that are applicable to the costs of college (such as scholarships and grants), there still may be quite of bit of an expense to attend college. This difference between what college will cost (including tuition, room and board, books, computers) and the non-loan awards received is the college funding gap. College Funding Gap Ishraq receives her award and tuition letter from the college she wants to attend. Her tuition, fees, books, and room and board all come to $24,845 for the year. Her non-loan awards include an instant scholarship from the school for $7,500, a scholarship she earned for enrolling in a STEM program for $3,750, and a $1,000 scholarship from her church. What is Ishraq’s college funding gap? Her awards total to $12,250. Her cost to attend is $24,845. Her college funding gap is then $ 24,845 − $ 12,250 = $ 12,595 . She will need to find $12,595 in funding. There are several loan types, which basically break down into four broad categories: subsidized loans, unsubsidized loans, PLUS loans, and private loans. These loans are meant to fill the college funding gap. Federal subsidized loans are backed by the U.S. Department of Education. These loans are intended for undergraduate students who can demonstrate financial need. Subsidized federal loans, including Stafford loans, defer payments until the student has graduated. During the deferment, the government pays the interest while the student is enrolled at least half-time. These loans are generally made directly to students. However, there are restrictions on how the money can be used. It can only be used for tuition, room and board, computers, books, fees, and college-related expenses. Interest rates are not based on the financial markets but determined by Congress. Federal loans are backed by the Department of Education. Federal unsubsidized loans , including unsubsidized Stafford loans, are available for undergraduate and graduate students who cannot demonstrate financial need. If the student meets the program requirements, they are automatically approved. The student is not required to pay these loans during their time in college (enrolled at least half-time). However, the interest rate is generally higher and there is no deferment period, as with subsidized loans. Interest begins accruing as soon as the money is disbursed. The immediate accrual of interest means the balance of the loan grows as the student attends school. A loan that was for $10,000 can grow past $13,000 over five years of college. Some advisors tell students to pay the interest portion of the loan while it is deferred to prevent this growth of debt. Parent Loans for Undergraduate Students (PLUS) are federal loans made directly to parents. They are available even if parents are not deemed financially needy. A credit check is performed and approval is not automatic. The limit to what parents can borrow from a PLUS loan each year is still the college funding gap, but the aggregate of the PLUS loans does not have a limit. This means the PLUS loan can cover whatever is left in the funding gap once all other aid and loans are applied. Payments do not begin until the student is out of school, but interest begins to accrue the moment funds are disbursed. Because the parents take out the loan, the parents are responsible for paying back the loan. Private student loans are backed by a bank or credit institution and require a credit check, and interest rates are variable. As private loans are not subsidized by the government, no one pays the interest but the borrower. The student does not have to start repaying the loan until after graduation, but interest starts to accrue immediately. This loan has fewer repayment options, more fees and penalties, and the loan cannot be discharged through bankruptcy. Many students need a co-signer to acquire a private loan. Like PLUS loans, private student loans can cover whatever is left in the funding gap once all other aid and loans are applied Student loans, in general, have a term of 10 years, that is, the loans are paid back over 10 years. This can vary, but 10 years is the standard. School-Channel Loans and Direct-to-Consumer Loans Private loans can fall into one of two categories: school-channel loans and direct to consumer loans. School-channel loans are disbursed directly to the school. The school verifies the loan does not exceed the cost to attend school. Direct–to-consumer loans do not have the verification process. Those proceeds are sent directly to the borrower. They are processed more quickly, but often have higher interest rates. Types of Student Loans Limits on Student Loans As mentioned earlier, there are limits to how much a student can borrow, per year and in total. The following table shows a general breakdown of the amounts the federal government and private lenders will lend. Amounts are based on level of need and whether the student is a dependent or an independent student. Independent students include those who are at least 24 years old, married, a professional, a graduate student, a veteran, a member of the armed forces, an emancipated minor, or an orphan. The amounts shown are as of this writing in 2022. Year Dependent Students Maximum Amounts Independent Students Maximum Amounts First-Year Undergraduate $5,500 but no more than $3,500 may be in subsidized loans $9,500 but no more than $3,500 may be in subsidized loans Second-Year Undergraduate $6,500 but no more than $4,500 may be subsidized loans $10,500 but no more than $4,500 in subsidized loans Third Year and Additional Years $7,500 but no more than $5,500 may be in subsidized loans $12,500 but no more than $5,500 may be in subsidized loans Graduate and Professional Not applicable $20,500 in unsubsidized loans Limits $31,000 but no more than $23,000 in subsidized loans $57,500 for undergraduates but no more than $23,000 in subsidized. $138,500 for graduate or professional but no more than $65,500 may be subsidized loans. Check out this Edvisors page on the limits of student borrowing to learn more! Loan for Year 5 of College Efraim is a dependent undergraduate student enrolled in a biology program. He’s about to attend for the fifth year. In year 1 he took out $5,000 in federal subsidized and unsubsidized loans, in year 2 he took out $6,400 in federal subsidized and unsubsidized loans, in years 3 and 4, he took out the maximum federal subsidized and unsubsidized loans amounts. He needs federal subsidized and unsubsidized loans for his fifth year of school. How much can he obtain in federal subsidized and unsubsidized student loans? The sum of his previous loans is $ 5,000 + $ 6,400 + $ 7,500 + $ 7,500 = $ 26,400 . The limit for federal subsidized and unsubsidized loans is $31,000, so in year 5 he can get student loans in the amount of $ 31,000 − $ 26,400 = $ 4,600 . Putting this all together, we have a way to determine the student loans needed for a student to attend college. First, determine the funding gap. If the student or family can cover the gap, then no loans are necessary. Second, determine how much in federal subsidized and unsubsidized student loans can be taken out. If the total federal loans available is more than the funding gap, no other loans are needed. Third, if the federal subsidized and unsubsidized loans do not cover the gap, PLUS and private student loans can be taken out to cover the remainder of the gap. At each step, if the student and family can cover some or all of the gap, they can do so without taking out a loan. College Funding Gap and PLUS and Private Student Loans Olivia receives her award and tuition letter from the college she wants to attend. Her tuition, fees, books, and room and board all come to $44,845 for her second year. Her non-loan awards include an instant scholarship from the school for $13,500, a scholarship she earned for enrolling in an engineering program for $5,750, and a $2,000 scholarship from her parent’s workplace. For her first year, what is Olivia’s college funding gap? How much can Olivia borrow in federal subsidized and unsubsidized student loans? Once Olivia takes out her maximum subsidized and unsubsidized federal student loans, how much will have to be paid for using PLUS and private student loans? Her awards total to $21,250. Her cost to attend is $44,845. Her college funding gap is then $ 44,845 − $ 21,250 = $ 23,595 . The maximum in federal student loans that Olivia can borrow is $6,500 in year 2. The remaining funding gap is $ 23,595 − $ 6,500 = $ 17,095 . Private student loans, PLUS loans, or other sources must be used to cover this gap. Student Loan Interest Rates Student loans are first and foremost loans. Students will pay them back and will pay interest. In the fall of 2022, the federal student loan interest rate was 4.99%. Private student loans rates ranged between 3.22% and 13.95%. Finding the lowest interest rate you can helps with the payments, and especially helps if the loan is not federally subsidized. Remember, if the loan is not federally subsidized, the student is on the hook for the interest that is accumulating with the loan. Interest Accrual The interest on student loans begins as soon as the loan is disbursed (paid to the borrower). When the loan is federally subsidized, the government pays that interest for the student. This means the loan for a subsidized loan of $3,000 is still a loan for $3,000 when the student graduates. However, if the loan is not federally subsidized, the student is responsible for the interest that accrues on the loan. The $3,000 loan from year 1 of college is now a loan for more due to that added interest. The interest on that loan grew while the student was in college. The formula for growth of the loan’s balance is the same as compound interest formula from Compound Interest , A = P ( 1 + r n ) n t . Denise takes out unsubsidized student loan, in August, in her first year of college for $2,000. She manages an interest rate of 8%. She graduates after her fifth year of college, in May. She does not pay the interest on the loan during her time in college. What is the balance of her first year loan in May of her graduation year? The principal of the loan is $2,000. Her interest rate is 8%. Since student loans are typically paid monthly, there are 12 periods per year. Since the time she has had the loan is not in years, we will use the number of months for the value of nt in the formula. She has had the loan for 4 years and 9 months, meaning 57 period have passed. Substituting those values into the formula and calculating, we find her balance in May of her graduating year is A = P ( 1 + r n ) n t = $ 2,000 ( 1 + 0.08 12 ) 57 = $ 2,000 ( 1.0 6 ¯ ) 57 = $ 2,920.89 . The Standard Repayment Plan There are various repayment plans available. The one most likely to apply to a student loan is the standard repayment plan, which is available to everyone. Borrowers pay a fixed amount monthly so the loan is paid in full within 10 years. Consolidated loans, discussed later in this section, also qualify for the standard repayment plan, and may allow the payoff period to range from 10 to 30 years. Direct subsidized and unsubsidized loans, PLUS loans, and federal Stafford loans are eligible. Since these are loans, they are paid back with interest. As with most installment loans, their payments are due monthly. The formula for paying back these loans is the same as the formula used for paying loans in The Basics of Loans : p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 . Using that formula, we can calculate how much the payment is for a student loan. Remember that all loan payments are rounded up to the next penny. Standard Repayment Plan Find the payment for the following student loans using the standard repayment plan: Loan is $3,500, interest is 4.99% Loan is for $6,200, interest is 6.75% The principal is P = $3,500 and the rate is r = 0.0499. Since this is the standard repayment plan, there are n = 12 payments per year for 10 years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 3,500 × ( 0.0499 / 12 ) × ( 1 + 0.0499 / 12 ) 12 × 10 ( 1 + 0.0499 / 12 ) 12 × 10 − 1 = $ 3,500 × ( 0.004158 3 ¯ ) × ( 1.004158 3 ¯ ) 120 ( 1.004158 3 ¯ ) 120 − 1 = $ 23.9469911276 0.645370131872 = $ 37.11 The principal is P = $6,200 and the rate is r = 0.0675. Since this is the standard repayment plan, there are n = 12 payments per year for 10 years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 6,200 × ( 0.0675 / 12 ) × ( 1 + 0.0675 / 12 ) 12 × 10 ( 1 + 0.0675 / 12 ) 12 × 10 − 1 = $ 6,200 × ( 0.005625 ) × ( 1.005625 ) 120 ( 1.005625 ) 120 − 1 = 68.3662233426 0.960321816275 = $ 71.20 Standard Repayment Plan for an Unsubsidized Loan Erson has a balance of $8,132.55 when he starts paying off the 8.6% unsubsidized student loan he took out in his third year. How much are his payments if the term for his loan is the standard 10 years? Using the payment formula, p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 , with P = $8,132.55, r = 0.086, t = 10 and n = 12, we calculate that his monthly payment will be p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 8,132.55 × ( 0.086 / 12 ) × ( 1 + 0.086 / 12 ) 12 × 10 ( 1 + 0.086 / 12 ) 12 × 10 − 1 = $ 8,132.55 × ( 0.0071 6 ¯ ) × ( 1.0071 6 ¯ ) 120 ( 1.0071 6 ¯ ) 120 − 1 = 137.310962384 1.35592393158 = $ 101.27 Student Loan Consolidation When a student graduates, they may have multiple different student loans. Keeping track of them and paying them off separately can be a burden. Instead, these loans can be consolidated into a single loan. If they are federal loans the combination is called federal consolidation. Combining private loans is often referred to as refinancing. Refinancing, or private consolidation, can be used to combine both private and federal student loans. Be aware that consolidated federal loans may still be subject to the rules and protections that govern subsidized loans. Refinancing loans, private or federal, are no longer subject to those rules and guidelines. Check out this Experian article about consolidation and refinancing for more deatil. In consolidation of federal direct student loans, the interest rate is the weighted average of the interest rates on the subsidized loans. This means the interest rate remains the same. However, if the term is extended, then the student will pay back more over time than if they did not extend the loan term. In refinancing, it is possible to obtain a lower interest rate on the student loans, which may lower how much is paid per month and lower the total paid back over time. These monthly payments are calculated using the same formula as for any other loan payment, p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 . The term of the refinanced loan may also be changed, which would also impact the payment per month. In either case, consolidating or refinancing, the monthly financial burden on the student can decrease. However, if the term is extended, the total amount repaid may increase. Federal Loan Consolidation and Interest Rates Ernest has four federal student loans that he wants to consolidate. He combines them into one loan. What is the maximum Ernest can reduce the interest rate by? Consolidating subsidized loans has no impact on the interest rate of the loans, so the maximum that the interest rate can be reduced is 0%. Payments for Consolidated Student Loans Brianna consolidates her student loans, some federal and some private, into a single refinanced student loan with a principal of $27,800. The interest rate that Brianna received was 8.375%. If Brianna’s new term is 15 years, how much are her payments per month? The principal is P = $27,800 and the rate is r = 0.08375. Since the payments are monthly, n =12. The loan term is for 15 years, so t = 15. years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 27,800 × ( 0.08375 / 12 ) × ( 1 + 0.08375 / 12 ) 12 × 15 ( 1 + 0.08375 / 12 ) 12 × 15 − 1 = $ 6,200 × ( 0.0069791 6 ¯ ) × ( 1.0069791 6 ¯ ) 180 ( 1.0069791 6 ¯ ) 180 − 1 = $ 678.477992464 2.49693370968 = $ 271.73 Other Repayment Plans There are various other repayment plans available to students. Plans other than the standard repayment plan typically require the student to meet certain criteria. The following plans are independent of student income, but may make early payments easier. Graduated repayment plans are plans where the amount of payments gradually increases so that the loan is paid off in 10 years, or within 10 to 30 years for consolidated loans. Payments start off small and increase approximately every 2 years. Almost all loan types are eligible, including direct subsidized and unsubsidized loans, Stafford loans, PLUS loans, and consolidated loans. Extended repayment plans are available to the direct loan borrower if the outstanding direct loans are over $30,000. The payments, fixed or graduated, are designed so that the loans are satisfied within 25 years. Eligible loans include both direct subsidized or unsubsidized loans, Stafford loans, PLUS loans, and consolidated loans. If student earnings are such that the standard, graduated, or extended repayment plans are unaffordable, then one can make payments that are based on their discretionary income. Discretionary income is federally defined to be the difference between (adjusted) gross income and 150% of the poverty guideline for location and family size. This discretionary income then depends on where one lives (contiguous United States or Hawaii or Alaska) and how many dependents one has. If married, a spouse’s income will be included in the adjusted gross income. Understanding discretionary income is necessary to understand how income driven payments plans work. Discretionary Income The poverty guideline for a single person living in Arkansas, is $12,000. If Harriet is a single person in Arkansas with a (adjusted) gross income of $23,500, what is her discretionary income? For California, the poverty guideline for a person with four people in the household is $27,750. If such a person has a (adjusted) gross income of $48,600, what is their discretionary income? The poverty guideline for a single person in Arkansas is $12,000. 150% of that guideline value is 1.5 × $ 12,000 = $ 18,000 . The gross income that Harriet makes over that $18,000 is her discretionary income. That gross income is $23,500, so her discretionary income is $ 23,500 - $ 18,000 = $ 5,500 . The poverty guideline for a household of four in California is $27,750. 150% of that guideline value is 1.5 × $ 27,750 = $ 41,625 . The gross income that Harriet makes over that $41,625 is her discretionary income. That gross income is $48,600, so her discretionary income is $ 48,600 − $ 41,625 = $ 6,975 . The following plans all depend on discretionary income. Pay As You Earn (PAYE) repayment plans have monthly payments that are 10% of discretionary income based on a student’s updated income and family size. If a borrower files a joint tax return, their spouse’s income and debt may also be considered. Eligible loans include direct subsidized and unsubsidized loans, PLUS loans made to students, and some consolidated loans. These loans are forgiven (student does not pay the remaining balance) after 20 years of monthly payments if they were direct federal student loans. Revised Pay As You Earn (REPAYE) repayment plans have payment amounts that are based on income and family size and calculated as 10% of discretionary income. Eligible loans include direct subsidized and unsubsidized loans, direct PLUS loans, and some consolidated loans. These loans are forgiven, that is, the student does not pay more, after 20 or 25 years, provided they were direct federal student loans. Income-Based Repayment (IBR) plans sound like a few of the others and there are similarities in all of them. The payments are either 10% or 15% of discretionary income, but this plan is meant for those with a relatively high debt. Every year, income and family size must be updated, and payments are calculated based on those figures. Eligible loans include direct subsidized and unsubsidized loans, Stafford loans, and PLUS loans made to students, but not PLUS loans made to parents. Income-Contingent Repayment (ICR) plans have payments that are either 20% of discretionary income or whatever would be paid if a student were on a fixed payment plan for more than 12 years, whichever is less. Eligible loans include direct subsidized and unsubsidized, PLUS loans to students, and consolidated loans. There are many similarities among these repayment plans, and it is easy to misunderstand the nuances of each. Therefore, be careful entering into any type of repayment contract until you fully understand all the details and repercussions of the plan you choose. For more detail, see this nerdwallet article \"Income-Driven Repayment: Is It Right for You?\" to learn more ! Payments for a REPAYE Program Warren qualifies for a REPAYE payment plan. His gross income is $32,700. He is single and live in Montana, so the federal poverty guideline for Warren is $12,000. What is Warren’s discretionary income? Under the REPAYE plan, he pays 10% of his discretionary income, but monthly. How much are Warren’s REPAYE payments? The poverty guideline for Warren is $12,000. 150% of that guideline value is 1.5 * $ 12,000 = $ 18,000 . The gross income that Warren makes over that $18,000 is his discretionary income. That gross income is $32,700, so his discretionary income is $ 32,700 − $ 18,000 = $ 14,700 . 10% of Warren’s discretionary income is 0.1 × $ 14,700 = $ 1,470 . He pays monthly, so his monthly payments are $1,470 divided by 12, or $122.50 per month. Using income to determine payments initially seems excellent. However, if there is no forgiveness at the end of the loan, then the income driven payment plans can cause problems. For one, your payment may not be sufficient to cover the interest rate of your loans. In that case, your loan balance actually increases as you make your payments. Eventually, you are paying for not only your original loan balance, but interest that has been growing and compounding over time. Also, if the loan term is extended, you may pay more, perhaps a lot more, money over time. You may find yourself in the position of paying these loans for decades. With those possible drawbacks, great care must be taken to avoid large problems down the line. Repayment Plans Student Loan Default and Consequences The first day a payment is late, the account becomes delinquent . After 90 days, this delinquency is reported to the credit bureaus, and goes into default . This is serious, as now a credit score is affected, meaning that it will be harder to buy a car, a home, get a credit card, or a cell phone. Even renting an apartment may be a task not easily overcome. The default rate for students who do not complete their degree is three times higher than for students who do. Further, defaulting on a student loan may mean that the borrower loses eligibility for repayment plans, as the balance and any unpaid interest may become due immediately, and any tax refunds may be withheld and applied to the loan, and wages may be garnished. One should immediately contact your loan servicer and try to make other arrangements for repayment if this situation becomes apparent, as different repayment plans are available, if actions are taken quickly. There are several options that may be open to avoid defaulting. One is called rehabilitation, or is the process in which a borrower may bring a student loan out of default by adhering to specified repayment requirements, and the other is consolidation. Certain criteria must be met to enter these programs. Both of these options are detailed, including the criteria required for eligibility, on the studentaid.gov loan management page . Professionals advise hiring an attorney if one of these paths is chosen. Check Your Understanding Key Terms FAFSA College funding gap Subsidized loan Unsubsidized loan Parent loan for undergraduate students Private student loan School-channel loan Direct-to-consumer loan Standard repayment plan Federal consolidation Refinancing Private consolidation Graduated repayment plan Extended repayment plan Discretionary income Pay as you earn (PAYE) repayment plan Revised pay as you earn (REPAYE) repayment plan Income-based (IBR) repayment plan Income-contingent (ICR) repayment plan Delinquent Default Rehabilitation Key Concepts The FAFSA must be filled out each year that a student wishes to borrow for. A student’s funding gap determines how much they need in loans to pay for college. Federal subsidized student loans defer payments until after graduation and interest does not accrue on these loans. Unsubsidized student loans defer payment until after graduation but interest begins accruing as soon as the loan finds are disbursed. There are both yearly and aggregate limits for student loans to prevent over-borrowing, among other reasons. Federal direct loans have a low interest rate set by the government, but other student loans have varying rates of interest set by the banks. The standard repayment plan lasts 10 years and is made up of monthly payments. Consolidating or refinancing student loans merges many student loans into one loan. If only federal loans are consolidated, the interest rate is the same as the individual loans, currently set at 4.99%. If other loans are refinanced together, the interest rate may be lower with the new loan. Other repayment plans are available. Such a plan may have payment that start small and grow as the loan is paid off, or it may have a longer term, or may be based on the discretionary income of the student. Being delinquent on a student loan is a precursor to being in default. Making payments in a timely fashion allows the student to avoid this situation. Video Types of Student Loans Repayment Plans Formula funding gap = total cost − all aid A = P ( 1 + r n ) n t p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 discretionary income = gross income − 1.5 × poverty guideline", "section": "Understanding Student Loans", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Credit Cards Credit cards are both a convenience and a danger. (credit: \"Credit Cards\" by Sean MacEntee/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Apply for a credit card armed with basic knowledge. Distinguish between three basic types of credit cards. Compare and contrast the benefits and drawbacks of credit cards. Read and understand the basic parts of a credit card statement. Compute interest, balance due, and minimum payment due for a credit card. It can be difficult to get along these days without at least one credit card. Most hotels and rental car agencies require that a credit card is used. There are even a number of retailers and restaurants that no longer accept cash. They make online purchasing easier. And nothing contributes more to a good credit rating than a solid history of making credit card payments on time. Being granted a credits card is a privilege. Used unwisely that privilege can become a curse and the privilege may be withdrawn. In this section, we will talk about the different types of credit cards and their advantages and disadvantages. The more knowledge a cardholder has about the credit card industry, the better able credit accounts can be managed, and that knowledge may cause major adjustments to a cardholder’s lifestyle. All credit cards are not equal, but they all represent consumers borrowing money, usually from a bank, to pay for needs and “wants.” As such, they are a type of loan, and your repayment may include interest. (You might want to review Section 6.8, which discusses loans and repayment plans.) There are many institutions and credit cards to choose from. Use caution as you shop around for a credit card that suits you. Your top concern is likely the interest rates on purchases and cash advances. But be careful to also read the small print regarding charges for late payments, and other fees such as an annual fee, where the credit card charges you (the cardholder) a fee each year for the privilege of using the cards. Many cards charge no such fee, but there are many that charge modest to heavy fees. Make sure to understand rules for reward programs , where the credit card issuer grants benefits based on one’s spending. Finally, once one applies for and is granted a credit card, pay attention to the credit limit the bank offers. Once a company is owed that much money, use of the card for purchases should be curtailed until some of the debt is paid off. The interest rate will not matter if the balance is paid every month. When the balance is paid every month, there is NO INTEREST charged. Types of Credit Cards There are basically three types of credit cards: bank-issued credit cards, store-issued credit cards, and travel/entertainment credit cards. We will look at all three and explain the good and the bad qualities of each. Bank-Issued Credit Cards Perhaps the most widely used credit card type is the bank-issued credit card , like Visa or MasterCard (and even American Express and Discover cards). These types of cards are an example of revolving credit, meaning that additional credit is extended before the previous balance is paid—but only up to the assigned credit limit. Bank-issued cards are considered the most convenient, as they can be used to purchase anything, including apparel, furniture, groceries, fuel for automobiles, meals, hotel bills, and so on, just as if paying with cash. The interest rates on bank-issued credit cards are usually lower than those for other credit cards we’ll discuss, and the credit limits are generally higher. Currently, bank-issued cards have an average 20.09% APR. Store-Issued Credit Cards Store-issued credit cards are issued by retailers. One can hardly walk into a store these days without being offered a discount on purchases if one applies for the store credit card. These cards can only be used in that store or family of stores that issues the card. However, if a store credit card is associated with Visa, MasterCard, or American Express, then the card might be used the same way that the bank-issued cards are used. This is called cobranding. The logo of the bank-issued card will be present on the store card. Many stores offer both types. Like other credit cards, they may come with an annual fee. Store credit cards usually charge higher interest rates than bank-issued cards. Currently, store credit cards have an APR (annual percentage rate) of 24.15%. Any rewards offered by store credit cards are usually limited to purchases made in their own store, and it typically takes longer to accumulate enough rewards or points to redeem them, whereas cobranded credit cards offer opportunities to earn rewards on all purchases, regardless of whether purchases are made in the issuing store or not. Store credit cards usually offer lower credit limits, at least in the beginning. After being proved to be a responsible credit card owner, credit limits can be raised. Nevertheless, store credit cards are a good choice for those new to the credit card industry. If on-time payments are consistently made, it is an excellent way to get started building a credit history. Travel/Entertainment Cards, or Charge Cards This is the third type of credit card. The travel and entertainment cards, also known as charge cards, first and foremost offer very high limits or unlimited credit, but they must be paid in full every month. They generally charge high annual fees and impose expensive penalties should a payment be late. On the other hand, they typically have longer grace periods and offer many and various kinds of rewards. Check out this nerdwallet article about the differences between a charge card and a credit card . An interest rate will greatly depend on credit score. Responsible use of credit cards will increase a credit score. See the WHO KNEW? from The Basics of Loans . Choosing a Credit Card Top Travel and Entertainment Cards At one time, Diner’s Club was the premier entertainment card. To be accepted into the Diner’s Club and be rewarded with a charge card meant one was special. Today, an American Express card is held with the same reverence as Diner’s Club was in the past. It was a privilege to own one of these cards. There is a lot of competition going on to supplant Diner’s Club, as shown by the Chase Sapphire Preferred Card’s travel rewards and benefits. Another thing worth mentioning here and something that appears to be an unusual offering is part of the American Express Gold and higher-level cards. Cardholders can actually open a savings account, buy a CD, or apply for a personal or business loan from American Express. The company boasts a higher interest rate than what is paid on traditional savings accounts and CDs. Comparing Credit Cards Which type of credit card is paid off every month so has no interest to be paid, but comes with high fees? Which type of credit cards are the most widely accepted? Which type of credit cards are the most limited? Charge cards are to be paid off completely each month. Bank-issued credit cards are the most flexible to use, because they are not limited to which retailers or service providers accept them. Store issues credit cards are the most limited, since they only work in that family of stores. Credit Card Statements Cardholders usually receive monthly statements and have 21 days to pay the minimum amount due. The statements itemize and summarize activity on the credit card for that statement’s billing period . The billing period for a credit card is generally a month long, but typically does not start and end on the first and last days of the month. The statement will include the current balance, interest rate, the minimum payment due, and the due date. Be aware, different companies produce statements that are laid out differently. The information will be clearly labeled though. The due date is a top concern. Missing a due date is one of the worst things a cardholder can do financially, and this is by far the biggest downfall of owning a credit card. Not only is the cardholder subject to late fees, but when a payment is late more than once there is a high probability that the cardholder will be negatively reported to the credit bureaus, which can quickly erode a credit score. shows an excerpt from an actual statement from a Chase Bank Visa card, based on the current $668.25 balance. Credit card statement Specifically pay attention to the late payment penalty and minimum payment warning statements. stating that if no other purchases are made and you continue making only the minimum payment, it will take 19 months to pay off the balance and you will pay $754.00. You can’t say you were not warned. It is critical that you examine your statement every month because it is always a possibility that your account may have been compromised. If you should notice fraudulent charges on your statement, notifying the credit card company is often enough to have those charges researched by the company and removed. The card with the fraudulent charges will be canceled and a new card with a new account number will be sent to you. Reading a Credit Card Statement On the credit card statement , identify The balance due The minimum required payment The length of time it takes to pay off the balance by paying the minimum payments and without charging more to the card The interest rate for purchases Credit card account statement The balance due is under the payment information heading and is $3,663.23. The minimum payment due is also under the payment information heading, and is $36.63. The time to pay off the balance using only minimum payments is below the payment information, and says it takes 2 years and 4 months to pay off the balance. The interest rate for purchases is toward the bottom of the statement. It is 19.99%. Reading Credit Card Statements Compute Interest, Balance Due, and Minimum Payment Due for a Credit Card Computing all of these values depends on understanding and computing the average daily balance on a credit card. Once that is known, the interest, balance due, and minimum payment can be found. Above all else, if you pay off the entire balance each month, interest is not charged. Average Daily Balance Most credit card companies compute interest using the average daily balance method. To find the average daily balance on your credit card, determine the balance on the card each day of the billing period (often that month), and take the average. One process to find that average daily balance follows these steps: Start with a list of transactions with their dates and amounts. For each day that had transactions, find the total of the transactions for the day. Expenditures are treated as positive values, payments are treated as negative values. Create a table containing each day with a different balance. The balance is the previous balance plus the day’s total transactions. Add a column for the number of days those balances until the balance changed. Add a column that contains the balances multiplied by the number of days until the balance changed. Find the sum of that last column. Divide the sum by the number of days in the billing period (often the number of days in the month). This is the average daily balance. Computing Average Daily Balance The billing cycle goes from May 1 to May 31. The balance at the start of the billing cycle is $450.21. The list of transactions on the card is below. Date Activity Amount 1-May Billing Date Balance $450.21 10-May Payment $120.00 15-May Groceries $83.43 26-May Auto Parts $45.12 26-May Restaurant $85.34 30-May Shoes $98.23 Find the average daily balance for the credit card during the month of May. To find the average daily balance, we use the following steps. Start with a list of transactions with their dates and amounts. This list is provided. For each day that had transactions, find the total of the transactions for the day. The only day with more than one transaction was May 26. The sum of those transactions is $130.46. Treating the payment as a negative value, the daily transaction amounts are Date Amount 1-May $450.21 10-May -$120.00 15-May $83.43 26-May $130.46 30-May $98.23 Create a table containing each day with a different balance. The new table with dates that had different balances is below. Date Balance 1-May $450.21 10-May $330.21 15-May $413.64 26-May $544.10 30-May $642.33 Now, add a column for the number of days those balances until the balance changed. The days until the balance changes is found by finding the difference in the dates. For instance, from May 15 to May 26 was 11. Adding that column to the table we have Date Balance Days Until Balance Changes 1-May $450.21 9 10-May $330.21 5 15-May $413.64 11 26-May $544.10 4 30-May $642.33 2 The last entry was 2 since there are 31 days in May. Add a column that contains the balances multiplied by the number of days until the balance changed. We create the column and multiply the values. Date Balance Days Until Balance Changes Balance Times Days 1-May $450.21 9 $4,051.89 10-May $330.21 5 $1,651.05 15-May $413.64 11 $4,550.04 26-May $544.10 4 $2,176.40 30-May $642.33 2 $1,284.66 Find the sum of that last column. Adding that last column we have a sum of $13,714.04. There are 31 days in May, so divide the sum by 31, which gives an average of $442.39, which is the average daily balance. Calculating the Interest for a Credit Card The interest charged for a credit card is based on the daily interest rate of the card, the number of days in the billing cycle, and the average daily balance on the card. The interest charge, I , for a credit card during a billing cycle is I = ADB × r × d 365 , where ADB is the average daily balance, r is the annual percentage rate, and d is the number of days in the billing cycle. As before, interest is rounded up to the next penny. Calculating Interest for a Credit Card Billing Cycle Compute the interest charged for the credit card based on the given average daily balance (ABD), annual interest rate, and number of days in the billing cycle. ADB = $2,765.00, annual interest rate 13.99%, billing cycle of 30 days ADB = $789.30, annual interest rate 17.99%, billing cycle of 31 days ADB = $1,037.85, annual interest rate 11.99%, billing cycle of 28 days Substituting $2,765.00 for ADB, 0.1399 for r and 30 for d and calculating, we find the interest charge to be I = ADB × r × d 365 = $ 2,765.00 × 0.1399 × 30 365 = $ 31.80 . Substituting $789.30 for ADB, 0.1799 for r and 31 for d and calculating, we find the interest charge to be I = ADB × r × d 365 = $ 789.30 × 0.1799 × 31 365 = $ 12.06 . Substituting $1,037.85 for ADB, 0.1199 for r and 28 for d and calculating, we find the interest charge to be I = ADB × r × d 365 = $ 1,037.85 × 0.1199 × 28 365 = $ 9.55 . Credit Cards Charge Stores Fees The interest you pay is not the only way a credit card company generates revenue. It also charges fees to the retailers, online stores, and service providers that allow the consumer, you, to use your credit card to pay them. These are called processing fees. Currently they typically range from 2.87% to 4.35% of each transaction. That means if you use your credit card at a store and spend $100.00, the store will have to pay the credit card company somewhere between $2.87 and $4.35. One type of processing fee is the interchange fee. Mastercard charges the vendor 1.35% of the sale, plus an additional percentage up to 3.25%, and a fixed $001 fee for each transaction. Added to that is an assessment fee. This fee is 0.14% for Visa cards. Calculating the Balance of a Credit Card The balance, or sometimes balance due, on a credit card is the previous balance, plus all expenses, minus all payments and credits, plus the interest on the card. As stated before, if the card was paid off, there is no interest to be paid. Calculating the Balance of a Credit Card Find the balance on the credit card with the given interest charge and balance before interest was charged. The cards were not paid off previously. Balance before interest is $708.50, interest charge is $8.15 Balance before interest is $1,395.10, interest charge is $21.32 Adding the balance before interest to the interest charge, we find the balance to be $716.65. Adding the balance before interest to the interest charge, we find the balance to be $1,416.42. The next example puts all those steps together. Find Balance Due from Transactions and Interest Rate Kaylen’s credit card charges 16.9% annual interest. His current billing period is from November 1 to November 30. The balance on November 1 was $1,845.23. Use Kaylen’s following transactions to determine his balance due at the end of the billing cycle. Date Activity Amount 1-Nov Billing Date Balance $1,845.23 3-Nov Groceries $78.50 4-Nov Tablet $159.00 4-Nov Online Game Purchase $39.99 4-Nov Restaurant $47.10 10-Nov Payment $300.00 13-Nov Gasoline $58.75 13-Nov Clothing $135.00 18-Nov Gift $30.00 18-Nov Restaurant $21.75 28-Nov Gasoline $43.79 The first step is to find Kaylen’s average daily balance. To find the average daily balance, we use the following steps. Start with a list of transactions with their dates and amounts. This list is provided. For each day that had transactions, find the total of the transactions for the day. The days with more than one transaction were Nov. 4, Nov. 13, and Nov. 18. Treating the payment on November 10 as a negative value, the daily transaction amounts are Date Amount 1-Nov $1,845.23 3-Nov $78.50 4-Nov $246.09 10-Nov -$300.00 13-Nov $193.75 18-Nov $51.75 28-Nov $43.79 Create a table containing each day with a different balance. The new table with dates that had different balances is below. Date Balance 1-Nov $1,845.23 3-Nov $1,923.73 4-Nov $2,169.82 10-Nov $1,869.82 13-Nov $2,063.57 18-Nov $2,115.32 28-Nov $2,159.11 Now, add a column for the number of days those balances until the balance changed. The days until the balance changes is found by finding the difference in the dates. For instance, from May 15 to May 26 was 11. Adding that column to the table we have The last entry was 3 since there are 30 days in November (the 28th, 29th, and 30th). Date Balance Days Until Balance Changes 1-Nov $1,845.23 2 3-Nov $1,923.73 1 4-Nov $2,169.82 6 10-Nov $1,869.82 3 13-Nov $2,063.57 5 18-Nov $2,115.32 10 28-Nov $2,159.11 3 Add a column that contains the balances multiplied by the number of days until the balance changed. We create the column and multiply the values. Date Balance Days Until Balance Changes Balance Times Days 1-Nov $1,845.23 2 $3,690.46 3-Nov $1,923.73 1 $1,923.73 4-Nov $2,169.82 6 $13,018.92 10-Nov $1,869.82 3 $5,609.46 13-Nov $2,063.57 5 $10,317.85 18-Nov $2,115.32 10 $21,153.20 28-Nov $2,159.11 3 $6,477.33 Find the sum of that last column. Adding that last column we have a sum of $62,190.95. There are 30 days in November, so divide the sum by 30, which gives an average of $2,073.03, which is the average daily balance. With the average daily balance, we can determine the interest that is charged for November. Substituting ADB = $2,073.03, r = 0.169, and d = 30 into the formula I = ADB × r × d 365 and calculating, we find the interest to be I = ADB × r × d 365 = $ 2,073.03 × 0.169 × 30 365 = $ 28.80 . This interest is added to the final balance from the table in step 3, $2,159.11, which yields a balance due of $2,101.83. Minimum Payment Due The minimum payment due is the smallest required amount to be paid on a credit card to avoid late fees and penalties, such as an increased interest rate. The calculations for this may differ from card to card. They also depend in the balance of the credit card. General guidelines for minimum payment due are: For larger balances (usually over $1,000), the minimum payment will be some percentage of the balance due. For moderate balances (between $25 and $1,000), the minimum would be a specified dollar amount. $25 seems to be a common value. If the balance is small (under $25 for instance), then the minimum payment is the balance. Those are just guidelines. Individual cards may vary in these values. Minimum payments should only be paid if money is short in a given month. The length of time to pay off a credit card using minimum payments is quite long, and results in paying a lot of interest. It is strongly discouraged. Calculate the Minimum Payment Due The FYA credit card company has the following minimum payment policy. For balances over $1,000, the minimum payment is 2.5% of the balance due plus fees, but not interest. For balances between $500.00 and $999.99, the minimum payment is $50.00. For balances $499.99 and under, the minimum payment is $25.00 or the balance due, whichever is smaller. In the following, calculate the minimum payment due given the credit card minimum payment policy, the balance due and fees charged. Balance due is $1,309.00, no fees Balance due is $265.50, $35 in fees Balance due is $784.90, no fees The balance is over $1,000, so the minimum payment is 2.5% of the balance due plus fees. 2.5% of the balance due is 0.025 × $ 1,309.00 = $ 32.73 . Since there are no fees, the minimum payment due is $32.73. The balance is under $499.99, so the minimum payment due is $25.00. The balance is between $500.00 and $999.00, so the minimum payment due is $50.00. Check out this nerdwallet article about minimum payments for more! Check Your Understanding Key Terms Reward program Annual fee Credit limit Bank-issued credit card Store-issued credit card Travel and entertainment cards Charge cards Billing period Balance Minimum payment Average daily balance Key Concepts Credit cards can be a flexible way to pay for almost anything, but can become a financial hazard if used unwisely. When deciding which credit card to apply for, evaluate the interest rate, fees (annual and late), reward programs and credit limit. Be sure they meet your criteria. Paying off the balance of your credit card every month will control your spending and will never result in paying interest. Credit card statements hold all important information about your credit card, including payment, balances, charges and billing cycle dates. Although the minimum payment is attractive precisely because it is so small, paying only the minimum results is a long payoff term and higher interest costs. Video Choosing a Credit Card Reading Credit Card Statements Formula I = ADB × r × d 365", "section": "Credit Cards", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Buying or Leasing a Car The choice to lease or buy is based on cost and other concerns. (credit: “Car Insurance” by Pictures of Money/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Evaluate basics of car purchasing. Compute purchase payments and identify the related interest cost. Evaluate the basics of leasing a car. Identify and contrast the pros and cons of purchasing versus leasing a car. Investigate the types of car insurance. Solve application problems involving owning and maintaining a car. There are people who don’t need a car and won’t purchase one. But for many people, whether or not to have a car is not a question. Having a car is a basic necessity for these people. Obtaining a car can be daunting. The models, the features, the additional costs, and finding funding are all steps that need to be taken. One of the big decisions is whether to buy the car or to lease the car. This section will address some of the issues associated with each option. The Basics of Car Purchasing The biggest questions you will answer before purchasing a car are, what do you want and what do you need? Does it have to be new? Does it have to be a make and model you are familiar with? Does it have to have assisted driving? What other details are important to you? For a new vehicle, every feature beyond standard features comes with additional cost, which leads to the question that constrains all of your decisions about a car. How much can you afford to spend on a car? What you can afford must include insurance costs (discussed later in this section) and maintenance and upkeep. Once you have this in mind, you can search for a car that matches, as closely as possible, what you want and can afford. Most, if not all, dealers have websites that you can search through to identify the car you want. If new cars are not affordable, used cars cost less but come with the wear and tear of use. The sticker price of the car, called the manufacturer’s suggested retail price (MSRP), or the negotiated price you arrive at, isn’t the end of the cost to buying a car. There are many fees that accompany the purchase of the car, and perhaps even sales tax. These include but aren’t necessarily limited to the following: the title and registration fee , which includes registering your car with the state, getting the license plate, and assigning the title of the car to the lender. This cannot be avoided. a destination fee , which covers the cost of delivering the vehicle to the dealer a documentation fee , sometimes referred to a processing fee of handling fee, is the cost of all the paperwork the dealer did to get you the car a dealer preparation fee , which is for washing the car and other preparation of that sort. You should try to negotiate that out of the cost of the dealer tries to charge for that extended warranties and maintenance plans, which help cover some of the costs of caring for the car. Sales tax. You could pay for these immediately, but they are often added to the financing of the car, meaning they become part of the principal of your loan. Total Cost to Purchase a Car Nichole negotiates with her car dealership so that the price is $21,800. She needs to pay the 6.75% sales tax on the car. Other fees are $31.00 for title and registration, $1,000 in destination fees, and a documentation fee of $175. What is the total cost of Nichole’s car? We add the car’s sales cost, sales tax and all other fees to arrive at this value. The sales tax is 6.75% on the price she negotiated, so is $ 21,800 × 0.0675 = $ 1,471.50 . Adding these up, we have a total cost of $ 21,800 + $ 1,471.50 + $ 31.00 + $ 1,000 + $ 175 = $ 24,477.50 . One way to bring down payments on a car is to provide a down payment or a trade in. This is money applied to the purchase price before financing happens. Be warned, the sales tax applies to the full purchase price! If you reduce the amount financed, the payments respond by going down. This often becomes part of the negotiating process. Total Cost to Purchase a Car with Down Payment Sophia negotiates a $19,800 price for her new car. The sales tax is 9.5% in her area, and the dealership charges her $300 in documentation fees. Her title, plates, and registration come to $321.50. The dealership adds to this a destination fee of $1,100. If she places a down payment of $5,000 on the car, what is the total she will finance for the car? The price was $19,800. The sales tax of 9.5% is based on this number. The sales tax comes to $ 19,800 × 0.095 = $ 1,881 . Adding all the fees to the price and the sales tax brings the total cost of the car to $ 19,800 + $ 1,881 + $ 300 + $ 321.50 + $ 1,100 = $ 23,402.50 . Her down payment is applied to this number, so the $5,000 is subtracted from $23,402.50. The subtraction yields the amount to be financed, which is $18,402.50. When purchasing a car, the total cost to obtain the car is not the only factor in your monthly price. You will also pay an interest rate for the loan you obtain. The interest rate you will get is dependent on your credit score (see The Basics of Loans ). But you can choose from different lenders. The dealership will likely offer to finance your car loan. Frequently, dealerships offer special financing with very low rates. This is to help move inventory, and may indicate their desire to make sales. This might make negotiating easier. Even if the dealership offers financing, check with your bank or credit union to determine the interest rates they are offering. To reduce your payments, choose the lowest rate you can find. Purchase Payments and Interest Whether or not you buy a new car or a used car, if you finance the purchase, you are taking out a loan. The interest rates available for used cards are frequently higher than those for new cars. These loan payments work exactly the same way as other loans do as far as payments are concerned. The payment function comes from The Basics of Loans . The difference between financing a new car or a used car is that financing a new car typically comes with a lower interest rate and a longer term that financing a used car. The payment, pmt , per period to pay off a loan with beginning principal P is p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 , where r is the annual interest rate in decimal form, t is the term in years, and n is the number of payments per year (typically, loans are paid monthly making n = 12). Note, payment to lenders is always rounded up to the next penny. Often, the formula takes the form p m t = P × ( r ) × ( 1 + r ) n ( 1 + r ) n − 1 , where r is the interest rate per period (annual rate divided by the number of periods per year), and n is the total number of payments to be made. New Car Payments In the following, calculate the monthly payment using the given total to be financed, the interest rate, and the term of the car loan. Total to be financed is $31,885, interest rate is 2.9%, for 5 years. Total to be financed is $22,778, interest rate is 4.5%, for 6 years. The amount to be financed is the principal, P , which is $31,885. The rate r is 0.029, and the term is t = 5 years. These are monthly payments, so n = 12. Substituting and calculating, we find the monthly payment to be p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 31,885 × ( 0.029 / 12 ) × ( 1 + 0.029 / 12 ) 12 × 5 ( 1 + 0.029 / 12 ) 12 × 5 − 1 = $ 31,885 × ( 0.00241 6 ¯ ) × ( 1.00241 6 ¯ ) 60 ( 1.00241 6 ¯ ) 60 − 1 = $ 31,885 × ( 0.00241 6 ¯ ) × ( 1.15583736592 ) 1.15583736592 − 1 = $ 89.0635298302 0.15583736592 = $ 571.52 The amount to be financed is the principal, P , which is $22,778. The rate r is 0.045, and the term is t = 6 years. These are monthly payments, so n = 12. Substituting and calculating, we find the monthly payment to be p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 22,778 × ( 0.045 / 12 ) × ( 1 + 0.045 / 12 ) 12 × 6 ( 1 + 0.045 / 12 ) 12 × 6 − 1 = $ 22,778 × ( 0.00375 ) × ( 1.00375 ) 72 ( 1.00375 ) 72 − 1 = $ 22,778 × ( 0.00375 ) × ( 1.3093031051 ) 1.3093031051 − 1 = $ 111.837397673 0.3093031051 = $ 361.58 Used Car Payments Calculate the monthly payment for the used car if the total to be financed is $16,990, the interest rate is 7.5%, and the loan term is 3 years. The amount to be financed is the principal, P , which is $16,990. The rate r is 0.075, and the term is t = 3 years. These are monthly payments, so n = 12. Substituting and calculating, we find the monthly payment to be p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 16,990 × ( 0.075 / 12 ) × ( 1 + 0.075 / 12 ) 12 × 3 ( 1 + 0.04775 / 12 ) 12 × 3 − 1 = $ 16,990 × ( 0.00625 ) × ( 1.00625 ) 36 ( 1.00625 ) 36 − 1 = $ 16990 × ( 0.00375 ) × ( 1.25144613551 ) 1.25144613551 − 1 = $ 132.887936515 0.25144613551 = $ 528.50 The Basics of Leasing a Car Leasing a car is an alternative to purchasing a car. It is still a loan, and acts like one in many respects. They typically last either 24 months or 36 months, though other terms are available. Leases also come with mileage limits, frequently 10,000, 12,000, or 15,000 miles per year. When the lease is over, the car is returned to the dealer. At that time there may be fees that have to be paid, such as for damage to the car or for extra miles driven over the limit. There are two components to lease costs. One is the monthly payment for the lease. The other is the fees for leasing, These often are paid before the lease is complete. These include: a down payment, which is your initial payment that is applied to the price of the car. It reduces the amount you finance, much the same as when you purchase a car. It is recommended that this be negotiated away. the acquisition fee , sometimes called the bank fee. This is the money charge for the company to set up the lease. It is essentially a paperwork fee. It is not likely that this can be negotiated. a security deposit , which might be required. It is about the same as 1 month’s payment for the lease. The deposit is returned to you if the car is in good shape at the end. This can be negotiated away. disposition fees , which cover the cost the company will incur when they take your car back and are typically between $200 and $450. the title, registration, and license fees, just as with the purchase of a car. sales tax, which will likely be applied. The sales tax only covers the depreciated portion of the car (more on depreciation later) in many states. Since this depends on the state in which the car is leased, you should determine the sales tax rules for where you lease the car. As you can imagine, this can come to a fairly high dollar amount. Cost to Obtain a Lease Donna wants to lease a Subaru Outback in Eden, New York. Find the total cost of obtaining her lease if there is no down payment, $175.00 in acquisition fees, a security deposit of $300.00, $350.00 in disposition fees, $102.50 in title and registration fees, and sales tax of $3,536.05. Adding these values together we find the total cost is $4,463.55. Zollie Frank Zollie Frank and Armund Shoen founded one of the original leasing companies, Four Wheels, in 1939. Their company leased automobiles to corporations. They began by leasing 5 cars to the Petrolager pharmaceutical company in year one. This saved Petrolager money and provided a steady cash flow to the Four Wheels business. In year two, the number of cars leased to Petrolager was 75. Their new idea was to lease cars directly to companies for one year. Previously, such companies might pay for mileage, gas, and a partial down payment. Sadly, the salesmen who were being so helped often left the company before the car was paid for, and so the company lost the down payment money. The lease was for $45 per month per car for one year. You have some obligations when you lease a car. You must keep the car in good condition, cleaned, maintained, and free of anything more than minor damage. If the car is in poor condition when the car is returned, you will be responsible for the cost to bring the car to an acceptable condition. You are also expected to keep the mileage under its limit. If you go over, you will pay 10 to 25 cents per mile over. Lease Payments Lease payments are similar to regular loan payments, but have some other details. Calculating a lease payment involves knowing the following values: The price of the car. This is the cost you would pay for the car after applying all discounts, incentives, and negotiations. Residual Value. This is the manufacturer's estimate of the car's value after a set period of time. The residual value is expressed as a percentage of the manufacturer’s suggested retail price (MSRP). Months. This is the length of the lease. Most leases are either 24- or 36-month leases, but other terms are available. Monthly Depreciation . The monthly depreciation is the difference between the price of the car and the residual value, divided by the number of months of the lease, and represents the monthly loss of value of the car while it’s being used. Money Factor (MF). This is the interest rate, but expressed in a different way for a lease. Converting from the money factor to the annual percentage rate (APR) is done by multiplying the MF by 2400. Naturally, converting an APR to a MF is done by dividing the APR by 2400. The monthly depreciation for a car, MD, is MD = P − R n , P is the price paid for the car, R is the residual value of the car, and n is the number of months of the lease. The annual percentage rate for a lease is APR = 2400 × MF , where MF is the money factor of the lease. The MF for a lease is MF = APR / 2,400 . Monthly Depreciation of a Car The purchase price of a car is $25,000. Its residual price is $14,500. What is its monthly depreciation for a 36-month lease? The purchase price of a car is $30,000. Its residual price is $18,600. What is its monthly depreciation for a 24-month lease? The monthly depreciation formula is MD = P − R n , Substituting $25,000 for P , $14,500 for R , and 36 for n , we find MD to be MD = P − R n = $ 25,000 − $ 14,500 36 = $ 10,500 36 = $ 291.67 . The monthly depreciation formula is MD = P − R n , Substituting $30,000 for P , $18,600 for R , and 24 for n , we find MD to be MD = P − R n = $ 30,000 − $ 18,600 24 = $ 11,400 24 = $ 475.00 . Converting Between APR and MF Find the annual percentage rate if the money factor is 0.00001875. Find the money factor if the APR is 6.25%. The APR is the money factor times 2400, so APR = 2400 × MF = 2400 × 0.00001875 = 0.045 . Expressed as a percentage, the APR is 4.5%. The MF is the APR divided by 2400, so MF = APR / 2400 = 0.0625 / 2400 = 0.000026041 6 ¯ . Once the values above are found, the payment for the lease can be calculated. The payment, PMT , for a lease is P M T = ( P − R ) n + ( P + R ) × MF , where P is the price paid for the car, R is the residual value of the car, n is the number of months of the lease, and MF is the money factor for the lease. Calculating Car Lease Payments Calculate the lease payments for car with the following price, residual price, length of lease, and money factor or APR. Price is $28,344, residual price is $18,140.16, 24-month lease, money factor is 0.000025. Price is $22,500, residual price is $13,050, 36-month lease, APR is 7.5%. Substituting the values P = $28,344, R = $18,140.16, n = 24 and MF = 0.000025 into the formula and calculating, the monthly lease payment is P M T = ( P − R ) n + ( P + R ) × MF = ( $ 28,344 − $ 18,140.16 ) 24 + ( $ 28,344 + $ 18,140.16 ) × 0.000025 = $ 10,203.84 24 + ( $ 46,484.16 ) × 0.000025 = $ 426.33 Given the APR, we find the MF which is MF = APR / 2400 = 0.075 / 2400 = 0.00003125 . Substituting the values P = $28,344, R = $18,140.16, n = 24 and the MF into the formula and calculating, the monthly lease payment is P M T = ( P − R ) n + ( P + R ) × MF = ( $ 22,500 − $ 13,050 ) 36 + ( $ 22,500 + $ 13,050 ) × 0.00003125 = $ 9,450 24 + ( $ 35,550 ) × 0.000025 = $ 263.62 Comparing Purchasing and Leasing When deciding to buy or lease a car, the differences between the two options should be carefully evaluated. The following is a list of points of comparison between the two. The payments for a lease are likely less than the payment for purchasing. When leasing, you get a new car after the lease term is over, typically 24 or 36 months. Buying the car means the same car is driven until it is re-sold and a new one bought. Essentially, leasing a car is equivalent to renting a car. The leased car is new, so all warrantees are in force and you drive the car during its best years. When the car is purchased, it may be kept past its warrantees and may be driven until it is quite old. Each time you lease a new car, all the fees and taxes must be paid again. When buying a car, these fees are only paid once. Leasing contracts carry restrictions on the mileage you can drive per year, and going over incurs more cost at the end of the lease. Buying the car means no such mileage limits. When leasing, you are obligated to keep the vehicle in good condition and maintained according to the dealer’s schedule. Some dealerships will even pay for oil changes over the life of the lease. When the car is purchased, the upkeep schedule is the choice of the owner. When a car is purchased and kept for long enough, the warranty expires and the owner is responsible for all maintenance items and repairs. The warrantee for a car won’t expire during the lease term. When a new car is purchased and the loan is paid off the car is still owned by the buyer and may be traded in when a new car is to be purchased. When leasing, the car is returned to the dealer when the lease term is over. When deciding between the two, you are choosing between these features. If you aren’t willing to drive an older car or deal with the upkeep that accompanies an older car, you may want to lease. This means you will need to pay those beginning costs each time the lease is up. If you want to own the car after the payments are over, then you may want to buy a car. This means you are paying for all the upkeep after the warrantees expire, but you have no limits on mileage and own the car at the end. It really depends on your preferences. Lease or Buy In the following, determine if a lease or purchase of a car is better. Joyce is concerned with large repairs and does not want to deal with them. Maurice prefers to drive newer cars. Since Joyce does not want to deal with repairs, so leasing would be a better choice. This way, the warranty covers most of the big repairs that could need to be done. Since Maurice likes to drive new cars, leasing is a better option, since he will lease a new car every 2 to 3 years. Car Insurance Car insurance is meant to cover costs associated with accidents involving cars. Most states (all except New Hampshire and Virginia) require some insurance. Without insurance, the state may not let you get a license for your car or register your car. Your state’s requirements can be hard to follow. Fortunately, insurance companies and brokers will make sure your insurance is sufficient for your state and will warn you if you try to not meet the requirements. Of course, they may offer more than what is sufficient, so it is your responsibility to determine how much coverage you want, as long as the minimum insurance requirements are met. The cost of insurance should be accounted for when evaluating the affordability of buying or leasing a car. Whether your car is leased or owned, you do need insurance. This contributes to the cost of having the vehicle. Leasing or owning makes no difference to the insurance company you choose, because they are insuring you based on what you are driving, your driving record, and other information about you including where you live and your age. These insurance policies have many components that address different costs that can come from auto accidents. This may make details confusing, and you may not realize what you are paying for until you must use it. Here is a brief outline of the different components of auto insurance, many of which are required by the state that issues your driver’s license. Liability insurance is mandatory coverage in most states. Liability insurance covers property damage and injuries to others should you be found legally responsible for an accident. You are required to have the minimum amount of coverage, as determined by your state, in both areas. Collision insurance is insurance covering the damage caused to your car if involved in an accident with another vehicle. Comprehensive insurance is an extra level of coverage if involved in an accident with another vehicle and covers other things like theft, vandalism, fire, or weather events as outlined in your policy. There is a deductible assigned to each type of insurance, an amount that you pay out of pocket before your comprehensive coverage takes effect. Comprehensive insurance is often required if you lease or finance the purchase of a vehicle. Uninsured or underinsured motorist insurance: If you are hit by an uninsured or underinsured motorist, this insurance will help pay medical bills and damage to your car. Medical payments insurance is mandatory in some states and helps pay for medical costs associated with an accident, regardless of who is at fault. Personal injury protection insurance is coverage for certain medical bills and other expenses due to a car accident. Other covered expenses may include loss of income or childcare, depending on your policy. Gap insurance is designed to cover the gap between what is owed on the car and what the car is worth in the event your car is a total loss. Rental reimbursement insurance is coverage for a rental car while your car is under repair resulting from an accident. You can also purchase other special insurance policies, such as classic car insurance, new car replacement insurance, and sound system replacement insurance, to name a few. It is important that you determine exactly what you need, as insurance policies can be expensive and vary according to your age, driving history, and where you live. Types of Insurance Which component of insurance pays if you are in an accident with a motorist without insurance? Which component of insurance pays for the remaining principal owed on your car in the case of a total loss? Uninsured motorist insurance covers accidents with those who have no insurance. Gap insurance will cover the gap between what is owed on the car and what it is worth if an accident results in a total loss. Monthly Cost of Owning a Car If your car payment is $287.50 per month and your car insurance is $930 every 6 months, what is the cost of the car per month when accounting for the insurance? The cost of the car including insurance is the monthly payment, $287.50, plus the monthly cost of the insurance. The insurance cost per month is $ 930 / 6 = $ 155 since the insurance cost is for every 6 months. Adding those the cost with insurance is $442.50. Maintaining a Car Cars are not a buy it and forget it item. They require upkeep, which adds to the cost of owning the car. Tires, brakes, and wipers need replacing. Oli changes, inspections, so many things other than gasoline. Below is a list of some maintenance requirements for cars, along with cost and roughly how often they should happen. Maintenance Frequency Cost Range New Tires Every 5 years $25–$300 per tire Oil Change Every 3,000–6,000 miles $35–$75 Wipers Every 6–12 months $20–$40 Inspection Annual $10–$50 Brake pads 10,000–20,000 miles $200–$300 Air Filter 15,000–30,000 miles $35–$80 When designing a budget, these expected costs should be accounted for. Extra money per month should be saved in addition to this budget category, to handle unanticipated, and perhaps very costly, repairs. Estella needs to budget for her car maintenance. She expects to buy new tires each 4 years, which will cost her $480 to replace them all. Oil changes near her cost $49.99, and she believes she will get one every 4 months. Her inspection costs $15 per year. Wipers for her car cost $95 for all three and she anticipates changing them every year. She drives less than 30,000 miles per year, so she plans to replace the air filter once per year. The air filter for her car costs $57.50. How much should she budget per month to cover these costs? Her yearly costs are the wipers, inspection, and tires, which total $167.50. Tires will be bought every 5 years, so per year she should budget $96. Her oil changes, which will happen three times per year, cost $49.99 each, so she’ll spend $149.97 for the year on oil changes. Adding these up, her yearly budget should include $413.47 for maintenance. Dividing by 12 gives the monthly budget for maintenance, which is $34.46 (rounded up to the next penny). Check Your Understanding Key Terms Title and registration fees Destination fee Documentation fee Dealer preparation fee Extended warranty Down payment Acquisition fee Security deposit Disposition fees Liability insurance Collision insurance Comprehensive insurance Uninsured or underinsured motorist insurance Medical payment insurance Personal injury insurance Gap insurance Rental reimbursement insurance Key Concepts There are many factors to consider when choosing to buy or lease a car. The cost of the car is increased by a number of fees and sales tax. There are advantages to buying a car and advantages to leasing a car. The decision between the two depends on the preference of the buyer. Insurance covers costs associate with accidents. It is made up of various components. The costs of owning a car, including insurance and maintenance, should be a part of the budgeting process. Budgeting for unexpected repairs can ease the stress of encountering large repair bill. Formula p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 MD = P − R n APR = 2400 × MF MF = APR / 2,400 P M T = ( P − R ) n + ( P + R ) × MF", "section": "Buying or Leasing a Car", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Renting and Homeownership Purchasing a home is a big investment, while renting is a lower cost alternative. (credit: \"New construction, new development house for sale\" by jongorey/houseandhammer.com, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Evaluate advantages and disadvantages of renting. Evaluate advantages and disadvantages of homeownership. Calculate the monthly payment for a mortgage and related interest cost. Read and interpret an amortization schedule. Solve application problems involving affordability of a mortgage. After renting an apartment for 10 years, you realize that it may be time to purchase a home. Your job is stable, and you could use more space. It is time to investigate becoming a homeowner. What are the things that you must consider, and what is the financial benefit of owning as opposed to renting? This section is about the advantages, disadvantages, and costs of homeownership as opposed to renting. Advantages and Disadvantages of Renting When renting, you will likely sign a lease , which is a contract between a renter and a landlord. A landlord is the person or company that owns property that is rented. The lease will detail your responsibilities, restrictions on activities, deposits, fees, maintenance, repairs, and rent during the term of the lease. It also defines what your landlord can, and cannot, do with the property while you occupy the property. Like leasing a car, there are advantages to renting but also some disadvantages. Some advantages are: Lower cost. Short-term commitment. Little to no maintenance cost. The landlord pays for or performs most maintenance. You need not stay at end of lease. Once the lease term is over (the lease is up), you are not obligated to stay. If renting in an apartment complex, there may be a pool, gym, or community room for renters to use. Of course, there are disadvantage too: No tax incentives. Housing cost is not fixed. When the lease is up, the rent can change. No equity. When you are done living in a rental, you have built no value. Restrictions on occupants. There may be a limit on how many can live in the apartment. Restrictions on decorating. The property is not yours, so any decorating or improvements need landlord permission. Limits on pets. Permission for pets, and their number and type, will be set forth in the lease. May not be able to remain when lease term is over. The landlord can, at the end of your lease, invite you to leave. The building may be sold, and the new landlord may institute changes to the lease when the previous lease expires. Renting has fees to be paid at the start of the lease. Typically, when you rent, you will pay first and last months’ rent and a security deposit. A security deposit is a sum of money that the landlord holds until the renter leaves the rental property. The deposit will cover repairs for damage to the apartment during the renter’s stay but may be returned if the apartment is in good condition. If your landlord runs a credit check on you, the landlord may charge you for that. Advantages of Buying a Home The advantages to buying a home mirror the disadvantages of renting, and the disadvantages of home ownership mirror the advantages of renting. Some advantages to buying a home are: There are tax incentives. The interest you pay for your mortgage (more on that later) is deductible on your federal income tax. There are no restrictions on pets or occupants, unless laws in your area specify limits for homes. You can redecorate any way you wish, limited only by the laws in your area. Once your mortgage is set with a fixed-interest rate, your housing cost is fixed. Your home grows equity, that is, the difference between what you owe and what the house is worth grows. You can use the equity to secure loans, and you recover the equity (and more if you’re fortunate) when you sell the house. As long as you pay your mortgage and maintain the home to the standards of your community, you can stay as long as you wish. Some disadvantages to home ownership are: The cost is higher than renting. Mortgages and associated costs are typically higher than rent for a similar living space. The owner is responsible for upkeep, maintenance, and repairs. These can be extremely costly. The owner cannot walk away from the property. It can be sold, but simply leaving the property, especially if not paid off yet, has serious consequences. The big question of affordability looms large over the decision to rent or buy. Renting, strictly from an affordability viewpoint, comes with much less initial outlay and smaller commitment. If you do not have sufficient income to regularly save for possibly expensive repairs, or your credit isn’t quite as good as it needs to be, then renting may be the best choice. Of course, even if you can afford to buy a home, you may choose to rent based on the comparative advantages. Rent or Buy Buying a home really involves two buyers. You and the mortgage company. The mortgage company has interest in the home, as they are providing the funds for the home. They want to protect their investment, and many fees are about the bank as much as the buyer. They fund a mortgage based on the value they assign the property. Not you. This means they will want some certainty that the home is sound, and you are a good investment. Closing Costs When a home is bought, there are many costs that need to be paid at the time of purchase, which are lumped under the term closing costs. At the start of 2022, the average closing costs for a single-family home exceeded $6,800. These costs include: The appraisal fee, which is what is paid to someone to establish the home’s worth. The value of the home to the bank may differ from what the home is listed for, or what an app tells you the home is worth. It may run approximately $350. The home inspection fee. The inspection should reveal any problems with the house that will need to be fixed either before or after you obtain the home. The title search. The is a records search to insure there are no issues with who actually owns the property. It can cost about 0.5% to 1% of the amount you are financing. Prepaid taxes. You will need to pay about 6 months of taxes at the time of purchase. The credit report fee. This is a fee for checking your credit. You might pay $25 or more for this. The origination fee. This is the price the mortgage company charges you to cover the costs of creating the mortgage. This could be 0.5% to 1% (or more) of the amount you are borrowing. The application fee. This is just a processing fee and could come to several hundred dollars. The underwriting fee. This covers the cost of verifying your financial qualifications. It could be a flat fee, or some small percentage of the amount financed. Such as 0.5% or 1%. Attorney fees. If you use an attorney, you will have to pay the attorney. State of local fees. This may include a filing fee charged by the county or municipality in which you reside. That’s a long list, and it is not even complete. When buying, be prepared to see these costs. It can be surprising. But in the end, you will have equity in the home, which means when you sell your home, you will get some of your money back. Closing Costs In the end, you must weigh your options and carefully consider your priorities in choosing to rent or buy a home. Mortgages Some people will purchase a home or condo with cash, but the majority of people will apply for a mortgage. A mortgage is a long-term loan and the property itself is the security. The bank decides the minimum down payment (with your input), the payment schedule, the duration of the loan, whether the loan can be assumed by another party, and the penalty for late payments. The title of the home belongs to the bank. Since a mortgage is a loan, everything about loans from The Basics of Loans holds true, including the formula for the payments. Monthly Mortgage Payments The formula to calculate your monthly payments of principal and interest uses APR as the annual interest rate. The payment, pmt , per month to pay down a mortgage with beginning principal P is p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 , where r is the annual interest rate in decimal form and t is the number of years of the payment. Note, payment to lenders is always rounded up to the next penny. To find the total amount of your payments over the life of the loan, multiply your monthly payments by the number of payments. 30-Year Mortgage at 4.8% Interest Evan buys a house. His 30-year mortgage comes to $132,650 with 4.8% interest. Find Evan’s monthly payments. Using the information above, P = $132,650, r = 0.048 and t = 30. Substituting those values into the formula p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 and calculating, we find the payment is p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 = $ 132,650 × ( 0.048 / 12 ) × ( 1 + 0.048 / 12 ) 12 × 30 ( 1 + 0.048 / 12 ) 12 × 30 − 1 = $ 132,650 × ( 0.004 ) × ( 1.004 ) 360 ( 1.004 ) 360 − 1 = $ 2,233.07781448 3.20858992551 = $ 695.97 His mortgage payment is $695.97. To find the total amount of your payments over the life of the loan, multiply your monthly payments by the number of payments. This can be useful information, but not too many people reach the end of their mortgage. They tend to move before the mortgage is paid off. The total paid, T , on an t year mortgage with monthly payments pmt is T = p m t × 12 × t . 30-Year Mortgage at 5.35% Interest Cassandra buys a house. Her 30-year mortgage comes to $99,596 with 5.35% interest. If Cassandra pays off the mortgage over those 30 years, how much will she have paid in total? To find the total paid over the life of the mortgage, use the formula T = p m t × 12 × t . To calculate this, the payment must be found. Using the information above, P = $99,596, r = 0.0535 and t = 30. Substituting those values into the formula p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 and calculating, we find the payment is p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 = $ 99,596 × ( 0.0535 / 12 ) × ( 1 + 0.0535 / 12 ) 12 × 30 ( 1 + 0.048 / 12 ) 12 × 30 − 1 = $ 99.596 × ( 0.004458 3 ¯ ) × ( 1.004458 3 ¯ ) 360 ( 1.004458 3 ¯ ) 360 − 1 = $ 2,202.45911222 3.96013414693 = $ 556.16 Using the mortgage payment of $556.16 and t = 30 years in the formula T = p m t × 12 × t , the total that Cassandra will pay for the mortgage is $200,217.60. With the principal of the mortgage and how much total is paid over the life of the mortgage, the cost of financing can be found by subtracting the principal of the mortgage from the total paid over the life of the mortgage. The cost of financing a mortgage, CoF, is CoF = T − P where P is the mortgage’s starting principal and T is the total paid over the life of the mortgage. 30-Year Mortgage at 5.35% Interest Cassandra buys a house. Her 30-year mortgage comes to $99,596 with 5.35% interest. What was Cassandra’s cost of financing? In , we found that the total Cassandra will pay for the $99,569 mortgage is $200,217.60. Subtracting those we find the cost of financing CoF = T − P = $ 200,217.60 − $ 99,596 = $ 100,621.60 . Private Mortgage Insurance (PMI) When you purchase a home, you will have to pay a down payment. This means you have money tied to the property, which lenders believe makes you less likely to walk away from a property. The amount of the down payment will be decided between you and the mortgage company. However, if your down payment is less than 20% of the property value, you will be required to pay private mortgage insurance (PMI). This is insurance you pay for so that the mortgage company is protected if you default on the loan. It often comes to between 0.5% and 2.25% of the original loan amount. It increases your monthly payment. Once you reach 20% of the loan value, you can request that the PMI be dropped. Even if you do not request cancelling the PMI, it will eventually and automatically be dropped. For more, see this article on ways to get rid of PMI . Reading and Interpreting Amortization Tables Amortization tables were addressed in The Basics of Loans . They are most frequently encountered when analyzing mortgages. The amortization table for a 30-year mortgage is quite long, containing 360 rows. A full table will not be reproduced here. We can, though, read information from a portion of an amortization table. Amortization Table for a 30-Year, $165,900 Mortgage shows a portion of an amortization table for a 30-year, $165,900 mortgage. Use that table to answer the following questions. What is the interest rate? How much are the payments? How much of payment 175 goes to principal? How much of payment 180 goes to interest? What’s the remaining balance on the mortgage after payment 170? Amortization table Reading at the top of the table, we see the interest rate is 5.61%. Reading from the top of the table or from the column labeled Payment, we see the payments are $953.44 per month. In the row for payment 175, we see that the amount that goes to principal is $400.44. In the row for payment 180, we see that the amount that goes to interest is $543.56. In the row for payment 170, we see the remaining balance is $119,873.35. Escrow Payments The last few examples have looked at mortgage payments, which cover the principal and interest. However, when you take out a mortgage, the payment is sometimes much higher than that. This is because your mortgage company also has you pay into an escrow account , which is a savings account maintained by the mortgage company. Your insurance payments will be set by your insurer and the mortgage company will pay them on time for you from your escrow account. Your property taxes are set by where you live and are typically a percentage of your property’s assessed value . The assessed value is the estimation of the value of your home and does not necessary reflect the purchase or resale value of the home. Your property taxes will also be paid on time by the mortgage company from your escrow account. For example, in Kalamazoo, Michigan, the effective tax rate for property is 1.69% of the assessed value of the home. These escrow payments, which cover bills for the home, can increase the monthly payments for your home well beyond the basic principal and interest payment. Adding Escrow Payments to Mortgage Payments Jenna decides to purchase a home, with mortgage of $108,450 at 6% interest for 30 years. The assessed value of her home is $75,600. Her property taxes come to 5.7% of her assessed value. Jenna also has to pay her home insurance every 6 months, which is $744 per six months. How much, including escrow, will Jenna pay per month? Using the payment function to find her mortgage payments, p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 , with P = $108,405, r = 0.06, and t = 30, her payments are p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 = $ 108,405 × ( 0.06 / 12 ) × ( 1 + 0.06 / 12 ) 12 × 30 ( 1 + 0.06 / 12 ) 12 × 30 − 1 = $ 649.95 Jenna also pays into escrow 1/12 of her property taxes per month. Her property taxes are 5.7% of the assessed value of $75,600, which comes to 0.057 × $ 75,600 = $ 4309.20 . This is an annual tax, so she pays 1/12 of that each month, or $359.10. Jenna’s home insurance is $744 per 6 months, so each month she pays $124.00 for insurance. Adding these together, her monthly payment is $ 649.95 + $ 359.10 + $ 124.00 = $ 1,133.05 . This is quite a bit more than the $649.95 for the principal and interest. Check Your Understanding Key Terms Lease Landlord Mortgage Escrow account Assessed value Key Concepts There are many points of comparison between renting and buying a house. Before deciding to buy a house, you should carefully consider all the responsibilities that come with home ownership. Renting comes with more restrictions on the renter, but with fewer costs and is easier to move from. Owning a house has more costs but has more freedom, plus the owner creates equity. Mortgages are loans, and payments are calculated in the same way as any other loan. Amortization tables help a homeowner understand the mortgage and how the payments are applied to the principal and interest. In addition to paying the amount financed for a mortgage, the monthly payment will include an escrow payment, which covers insurance and taxes. Video Rent or Buy Closing Costs Formula p m t = P × ( r / 12 ) × ( 1 + r / 12 ) 12 × t ( 1 + r / 12 ) 12 × t − 1 T = p m t × 12 × t CoF = T − P", "section": "Renting and Homeownership", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Income Tax Federal income tax is a concern for most US citizens. (credit: \"1040 US tax form\" by Marco Verch Professional Photographer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Determine gross, adjusted gross, and taxable income. Apply exemptions, deductions, and credits to basic income tax calculations. Compute FICA tax. Solve tax application problems for working students. Before the start of the American Civil War in 1861, most of the country’s revenue came from tariffs on trade and excise taxes. However, this fell far short of the high cost of the war. Because of this, the federal government enacted the nation’s first income tax with the Revenue Act of 1861, which created the Internal Revenue Service as we know it today. No one likes paying income tax, but it is a reality of life. In this section, we will learn about Form 1040, the U.S. Individual Income Tax Return, and ways to prepare for tax time. The U.S. tax code may change from year to year. Because of this, this section includes examples of how taxes, deductions, and exemptions might be computed. The types of income, deductions, and exemptions that are used in the examples are used in the current tax code. Gross, Adjusted Gross, and Taxable Income Your income drives how much you pay in taxes. The more you earn, the more you are likely to pay. But your income alone is not the full story. When you add all the money you earned from your job, freelance work, interest from savings, and other sources, you have your gross income . If you are an employee, your income from your job will be reported on a W-2, which is sent to you by your employer. Income from freelance work will be reported on a 1099-MISC form, and is sent by the company that paid you. Income from interest is reported on a 1099-INT form and comes from the entity that paid the interest. Before you determine how much you owe in taxes, you will make certain adjustments to that gross income. You will deduct, or subtract, some of income from the gross income. That’s your adjusted gross income , or AGI. That is still not what you are taxed on. Next, you need to apply exemptions to your income. These are pieces of income that the government does not tax. After that is done, you reach your taxable income . We will look at each of these parts of the taxable income. Gifts and Winning Money given as a gift my be taxed if the gift amount is high enough. If you win $50,000 in the lottery, that money is taxed as income. If you give a family member a large cash gift, that gift will be subject to a tax provided that the gift exceeds the federally set limits. You will notice that your paycheck already has taxes taken out of it. Your employer will withhold some of your income, sending it directly to the federal, state, and local governments. It is an estimate of how much you will owe in income tax. In the end, it reduces how much you will pay when your taxes are due. If they withhold too much income, you will receive the extra they withheld in the form of a refund. Computing Gross Income Roger is preparing to do his taxes. He worked two jobs, with reported income of $27,500 and $13,200. His two CDs yielded $327 together. He also won a split club raffle for $2,000. What was Roger’s gross earnings? This is the sum of his wages, winnings and interest earned. Add these Roger’s gross income was $43,027. Your adjusted gross income (AGI) is computed before your taxes are determined. It begins with the gross income, and then subtracts from that income any deduction. Deductions are expenditures on your part that the government won’t tax. These deductions include money deposited into tax-deferred investments, and mortgage interest that you paid, charitable contributions if you made any, medical bills over a threshold, medical insurance under certain circumstances, and property taxes. If you add all these up, and they are all legal deductions, the sum is subtracted from your gross income, leaving the AGI. Compute Adjusted Gross Income Chandra’s gross income is $58,400. She deposited $3,000 into her tax-deferred retirement account for the year. Her mortgage interest paid was $1,250 for the year, and her property taxes came to $4,200. What is her AGI? To get Chandra ’s AGI, total her deductions and subtract that total from her gross income. The total of her deductions is $ 3,000 + $ 1,250 + $ 4,200 = $ 8,450 . She subtracts that total from her gross income, making her AGI $ 58,400 − $ 8,450 = $ 57,950 . Remember that your AGI is not your taxable income . Exemptions need to be subtracted from the AGI to reach your taxable income. Exemptions are income that the government does not tax. Some examples of exempt income are disbursements from health savings accounts for qualified medical expenses, bond interest, some IRA distributions, and gifts given that are under $16,000. Note that exemptions are different from deductions: exemptions are excused incomes, whereas deductions are excused expenditures. Taxable Income Yelizaveta’s AGI is $75,490. However, she gave a $2,000 gift to her mother. What is her taxable income? Taxable income is AGI minus any exemptions. Gifts below $16,000 are exempted, so her taxable income is $ 75,490 − $ 2,000 = $ 75,490 . Tax Credits Another piece of the tax puzzle is tax credits . This is money subtracted from the tax you owe. Tax credits are very different from deductions or exemptions. Deductions and exemptions are taken away from your gross income before the tax you owe is calculated. A tax credit, is subtracted, dollar for dollar, from your tax bill. Once the tax you owe is calculated, subtract the any tax credits from that calculated tax. Some of the tax credits are refundable. This means that if subtracting them from your tax results in a negative number, you receive a tax refund. For more details, see this article about tax credits . The federal government has placed income limits and restrictions and on those eligible to receive tax credits because their value is so high. Here is a partial list of tax credits that you might qualify for: Earned income credit is a refundable tax credit for low- to moderate-income workers and ranges from $560 to $6,935 depending on dependents and income. This is refundable American opportunity credit is a credit taken by parents who have children enrolled in college at least half time and pursuing a degree. This credit is worth $2,500 per student for the first 4 years of undergraduate school, subject to income limits. This is a refundable tax credit. Lifetime learning credit is a credit is equivalent to 20% of educational expenses, up to $2,000 per year, subject to income limits. There is no cap to how many years you can apply for this credit. Child tax credit is worth $2,000 per child under the age of 17 if that child lives at home at least half the year, subject to income limits. This is a refundable tax credit. Child and dependent care tax credit was designed to help pay for child care while the parent works. The amount of the credit is dependent on your income. However, the maximum amount that can be received is, in 2022, $4,000 for one eligible person, or $8,000 for two or more qualifying people. A dependent qualifies if they are a child under 13 years old, a spouse who is unable to care for themselves, or some other qualifying person. This is a refundable tax credit. Premium tax credit was created by the Affordable Care Act, and it is one that is received by many people throughout the year. In essence it is a health insurance premium subsidy. The amount of the credit is based on your income and the price of health insurance in your area. This is a refundable tax credit. Apply a Tax Credit Kaitlyn has calculated the tax she owes to be $5,200. However, she receives an earned income tax credit of $1,715. How much does Kaitlyn owe after applying the earned income tax credit? The amount of taxes Katilyn owes is her calculated tax of $5,200 minus the credit she receives. The amount she owes in taxes is $ 5,200 − $ 1,715 = $ 3,485 . Tax Credits and a Refund Chanajah calculated their tax owed, which came to $4,300. They have an earned income tax credit of $2,190, a child tax credit of $2,000, and a child and dependent care tax credit of $4,000. How much tax does Chanajah owe, or how much will they get in a refund? Adding Chanajah’s tax credits together, we find their total to be $8,190. That is more than the tax they owe, which was $4,300. Each of those credits is refundable, which means they will receive a refund. Subtracting the credits from the tax owed yields $ 4,300 − $ 8,190 = − $ 3,890 . This is negative, so represents a refund of $3,890. Deductions Versus Credits Computing FICA Taxes FICA stands for the Federal Insurance Contributions Act of 1935. FICA taxes are used solely to fund Social Security and Medicare and are separate from federal income tax. It amounts to 7.65% of your gross pay, which is withheld from your paycheck automatically. Your employer is required to match the 7.65% amount. Of the 7.65%, 6.2% goes to Social Security (SSI), and 1.45% goes to Medicare. As of 2022, SSI tax only applies to the first $147,000 of earnings. Any gross income above that is not taxed for social security. This limit changes every year. Medicare tax, on the other hand, applies to the entirety of your gross income. Computed FICA Taxes McKenzi earned $2,700 in gross income, before taxes, in a given 2-week period. How much does she owe in FICA taxes, and how much of that is for SSI? The FICA tax is 7.65% of her gross earnings. 7.65% of her $2,700 is $206.55. Also, the SSI is 6.2% of her income, or $167.40 Social Security Tax with Higher Income Renard earned $195,000 in wages for the year. How much in SSI taxes does Renard owe for the year? Since Renard earned more than the taxable limit of $147,000 dollars, he only pays the 6.2% SSI tax on $147,000. This comes to $9,114. Calculating Your Income Tax Your income tax bill and your income tax rate are based on your taxable income. The tax system in the United States is progressive, meaning that the tax rates are marginal so the higher your taxable income the higher the tax rate you will pay. Taxable income is broken into brackets, or ranges of income. Each bracket has a different tax rate. The tax brackets and rates for single filers as of 2022 are given below: Bracket Lower Income Limit Upper Income Limit Tax Rate 1 0 $10,275 10% 2 $10,276 $41,775 12% 3 $41,776 $89,075 22% 4 $89,076 $170,050 24% 5 $170,051 $215,950 32% 6 $215,951 $539,900 35% 7 $539,901 37% Federal Income Tax Brackets for Single Filers, 2021–2022 (data source: https://www.irs.gov/newsroom/irs-provides-tax-inflation-adjustments-for-tax-year-2022) So if your taxable income is $76,500 and you are filing as a single filer, your tax bill will be 22% of that $76,500, right? Wrong. Your income is split among those brackets and the money in each bracket is taxed at that bracket’s tax rate. Seems confusing. Here is a list of steps to follow to find the tax owed. Step 1: Find the bracket for the income. Step 2: For each bracket below the income bracket, the tax from that bracket is: Step 2a: Find the difference between the upper limit of that bracket and upper limit of the next lower bracket. If this is bracket 1, use 0 as the upper limit of the previous bracket. Step 2b: The tax from that bracket is the bracket tax rate applied to the difference from Step 2a. Step 3: For the bracket that the income belongs to, find the income minus the lower limit for the bracket. Step 4: The tax for the bracket of the income is tax rate for that bracket applied to the difference found in Step 3. Step 5: Add these various tax values to get the total income tax. There are various tax brackets, and the rates may change in any given year. The income limits may also change. For all examples going forward, we will use the single filer tax brackets, even if those brackets are not appropriate (e.g., married or head of household filers). Income Tax on Taxable Income Faith has a taxable income of $103,650. How much income tax does Faith owe? Step 1: Faith’s income belongs to the 4th tax bracket, $89,076 to $170,050. So the taxes from the first three brackets follow steps 2a and 2b. For bracket 1: Step 2a: The upper limit is $10,275, the upper limit of the previous bracket was 0, so the difference is $10,275. Step 2b: For the first bracket, she owes 10% of $10,275, or $1,027.50. For bracket 2: Step 2a: The upper limit is $41,775, the upper limit from the previous bracket was $10,275, so the difference is $31,500 Step 2b: For the second bracket, she owes 12% of $31,500, or $3,780. For bracket 3: Step 2a: The upper limit is $89,075, the upper limit from the previous bracket is $41,175, so the difference was $47,300 Step 2b: For the third bracket, she owes 22% of $47,300, or $10,406. Step 3: For bracket 4, her income is $103,650, the upper limit of the previous bracket is $89,075. The difference of those is $ 103,650 − $ 89,075 = $ 14,575 . Step 4: The tax he owes for that bracket is 24% of the difference, which is $3,498. Step 5: The total in taxes that Emmanuel owes is the sum of the taxes found above, or $ 1,027.5 + $ 3,780 + $ 10,406 + $ 3,498 = $ 18,711.50 Remember that your employer will estimate how much tax you will owe and withholds it from paychecks. This means you may have already paid some, if not all and more, of the income tax you owe. Tax Brackets Finding Income Tax Owed Emmanuel is preparing his taxes. His W-2 from work shows gross income for the year of $95,250. He also has a 1099-MISC for some freelance art work he did, amounting to $7,500. Emmanuel also deposited $4,500 into a tax-deferred retirement plan. He paid $7,920 in mortgage interest for the year and $3,740 in property taxes. He also qualified for $4,000 in child tax credits. Based on this information, how much tax does Emmanuel owe or how much does he get in a refund? We need to know Emmanuel’s taxable income, based on gross income, deductions, and exemptions. His gross income is the amount from his W-2 and his 1099-MISC. Adding these gives a gross income of $ 95,250 + $ 7,500 = $ 102,750 . Subtracting his deductions will yield his AGI. His deductions are $7,920 for mortgage interest, $3,740 for property taxes, and $4,500 deposited into his retirement account. Adding these, his total deductions are $16,160. Subtracting the deductions from the gross income, we find his AGI to be $86,590. Emmanuel seems to have no exemptions, so his AGI and his taxable income are the same. We find the taxes Emmanuel owes using the process outlined above. Step 1: His income belongs to the third tax bracket, $41,776 to $89,0875. So the taxes from the first two brackets follow steps 2a and 2b. For bracket 1: Step 2a: The upper limit is $10,275, the upper limit of the previous bracket was 0, so the difference is $10,275. Step 2b: For the first bracket, he owes 10% of $10,275, or $1,027.50. For bracket 2: Step 2a: The upper limit is $41,775, the upper limit from the previous bracket was $10,275, so the difference is $31,500 Step 2b: For the second bracket, he owes 12% of $31,500, or $3,780. Step 3: For bracket 3, his income is $86,590, the upper limit of the previous bracket is $41,775. The difference of those is $ 86,590 − $ 41,775 = $ 44,815 . Step 4: The tax he owes for that bracket is 22% of the difference, which is $9,859.30. Step 5: The total in taxes that Emmanuel owes is the sum of the taxes found above, or $ 1027.5 + $ 3780 + $ 9,859.30 = $ 14,666.80 . Once his taxes are computed, he subtracts his tax credits. His only tax credit is $4,000. The total he owes in taxes is $ 14,666.80 − $ 4,000 = $ 10,666.80 . Check Your Understanding Key Terms Gross income Adjusted gross income Exemption Taxable income Deduction Tax credit Earned income credit American opportunity credit Lifetime learning credit Child tax credit Child and dependent care tax credit Premium tax credit Key Concepts Federal income tax is based on income after certain adjustments. Gross income is income from all sources, including gifts and winnings. Before taxes are calculated, the taxable income is found by subtracting deductions and exemptions from gross income. Income tax is progressive, increasing in rate as income increases. Being in the 32% tax bracket means some of your income is taxed at 10%, some at 12%, some at 22%, some at 24%, and the rest at 32%. Income in each tax bracket is taxed at that bracket’s rate, which means in 2022 the first $10,275 earned is taxed at 10% only. Tax credits are subtracted from the taxes that are owed. Some tax credits are refundable, which means they can make the amount you owe negative, which results in a refund. Video Deductions Versus Credits Tax Brackets Project Creating Your Future Budget In this project, you will create a budget based on a job you are likely to have after you graduate. Go online and research the average starting salary for the profession you are studying for. Use at least two sources. Be sure to record the web address from your search. Approximate your monthly take-home pay. You may use the SmartAsset website to estimate this. Use the 50-30-20 budget philosophy to determine how much you should budget for needs, wants and savings, or extra debt reduction. Create a list of likely expenses. This list must include rent/mortgage, utilities, food, and school loan repayment. You may also want to include car payments, gasoline, and other items. Categorize each expense as need, want, or savings. Using the amounts found in step 3, decide how much to allocate to each of your expenses. It may help to quickly research how much rent is where you want to live. Discuss the choices you had to make, and why you prioritized some expenses over others. Interest Rate and Time: What Is the Relationship? The interplay between interest rate and time for an annuity is not easily seen. How the amount that must be deposited per compounding period, pmt , changes based on the time and interest rate would be useful to understand. In this project, you will explore this relationship. We will use a fixed future value of FV = $1,000,000 and a fixed number of periods per year, 12 (monthly compounding). With those, we’ll find various annuity payments that must be made to reach the goal. The annual interest rate for the investment is in the top row. The number of years for the investment is in the left column. In each cell (or box), find the monthly payment necessary to reach the goal of $1,000,000. Annual Interest Rate 1.5% 2.0% 3.0% 5.0% 7.5% 10.0% Number of Years 10 15 20 30 40 45 Describe how the interest rates and number of years impact the payment necessary to reach the goal of $1,000,000. Finding a Home In this project you will identify a home you like, and then estimate the costs associated with that home. Find a home in your region that you would like to buy using an online search of listings in your area. Zillow is a good place to begin. Find the asking price for this home. Assume you would pay that price. Find an estimation for closing costs in your area. Assume you finance those costs also. Estimate the taxes to be paid on the home per year. It is likely that the online listing of the home has an estimate for the taxes for the house. Use Google to determine the average homeowner’s insurance cost in your region. Use the internet to determine the average interest rate for a 30-year mortgage. Find how much you would pay per month, based on the answers to the previous questions, including the escrow payments for taxes and insurance. Assume you will pay $50 per $100,000 borrowed in PMI. Add this to the monthly payment. Chapter Review Understanding Percent Discounts, Markups, and Sales Tax Simple Interest Compound Interest Making a Personal Budget Methods of Savings Investments The Basics of Loans Understanding Student Loans Credit Cards Buying or Leasing a Car Renting and Homeownership Income Tax Chapter Test", "section": "Income Tax", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Roulette is a game whose outcomes are based entirely on the concept of probability. (credit: modification of work \"Roulette wheel\" by Håkan Dahlström/Flickr, CC BY 2.0) Casinos are big business; according to the American Gaming Association, commercial casinos in the United States brought in over $43 billion in revenue in 2019. Casinos must walk a fine line in order to be profitable. Their customers must lose more money than they win, on average, in order to stay in business. But if the chances of a single customer winning more money than they lose is too small, people will stop coming in the door to play the games. In this chapter, we'll study the techniques a casino must use to determine how likely it is that a customer will win a particular game, and then how the casino decides how much money a winner will rake in so that the customers are happy, but the casino also turns a profit in the long run. In order to figure out those likelihoods, we have to be able to somehow consider every possible outcome of these games. For example, in a game that involves players receiving 5 cards from a deck of 52, there are 2,598,960 possibilities for each player. We'll start off this chapter by learning how to count those possible outcomes.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Multiplication Rule for Counting The Multiplication Rule for Counting allows us to compute more complicated probabilities, like drawing two aces from a deck. (credit: “Pair of Aces – Poker” by Poker Photos/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Apply the Multiplication Rule for Counting to solve problems. One of the first bits of mathematical knowledge children learn is how to count objects by pointing to them in turn and saying: “one, two, three, …” That’s a useful skill, but when the number of things that we need to count grows large, that method becomes onerous (or, for very large numbers, impossible for humans to accomplish in a typical human lifespan). So, mathematicians have developed short cuts to counting big numbers. These techniques fall under the mathematical discipline of combinatorics , which is devoted to counting. Multiplication as a Combinatorial Short Cut One of the first combinatorial short cuts to counting students learn in school has to do with areas of rectangles. If we have a set of objects to be counted that can be physically arranged into a rectangular shape, then we can use multiplication to do the counting for us. Consider this set of objects ( ): Certainly we can count them by pointing and running through the numbers, but it’s more efficient to group them ( ). If we group the balls by 4s, we see that we have 6 groups (or, we can see this arrangement as 4 groups of 6 balls). Since multiplication is repeated addition (i.e., 6 × 4 = 4 + 4 + 4 + 4 + 4 + 4 ), we can use this grouping to quickly see that there are 24 balls. Let’s generalize this idea a little bit. Let’s say that we’re visiting a bakery that offers customized cupcakes. For the cake, we have three choices: vanilla, chocolate, and strawberry. Each cupcake can be topped with one of four types of frosting: vanilla, chocolate, lemon, and strawberry. How many different cupcake combinations are possible? We can think of laying out all the possibilities in a grid, with cake choices defining the rows and frosting choices defining the columns ( Figure 7.5 ). Since there are 3 rows (cakes) and 4 columns (frostings), we have 3 × 4 = 12 possible combinations. This is the reasoning behind the Multiplication Rule for Counting , which is also known as the Fundamental Counting Principle. This rule says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are n × m ways to accomplish both tasks. We can tack on additional tasks by multiplying the number of ways to accomplish those tasks to our previous product. Using the Multiplication Rule for Counting Every card in a standard deck of cards has two identifying characteristics: a suit (clubs, diamonds, hearts, or spades; these are indicated by these symbols, respectively: ♣ , ♢ , ♡ , ♠ ) and a rank (ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king; the letters A, J, Q, and K are used to represent the words). Each possible pair of suit and rank appears exactly once in the deck. How many cards are in the standard deck? Since there are 4 suits and 13 ranks, the number of cards must be 4 × 13 = 52 ( ). Standard Deck of Cards, Sorted by Rank and Suit (credit: \"Playing Cards, USS Arkansas\" by Naval History & Heritage Command/Flickr, CC BY 2.0) Using the Multiplication Rule for Counting for 4 Groups The University Combinatorics Club has 31 members: 8 seniors, 7 juniors, 5 sophomores, and 11 first-years. How many possible 4-person committees can be formed by selecting 1 member from each class? Since we have 8 choices for the senior, 7 choices for the junior, 5 for the sophomore, and 11 for the first-year, there are 8 × 7 × 5 × 11 = 3,080 different ways to fill out the committee. Using the Multiplication Rule for Counting for More Groups The standard license plates for vehicles in a certain state consist of 6 characters: 3 letters followed by 3 digits. There are 26 letters in the alphabet and 10 digits (0 through 9) to choose from. How many license plates can be made using this format? Since there are 26 different letters and 10 different digits, the total number of possible license plates is 26 × 26 × 26 × 10 × 10 × 10 = 17 , 576 , 000 . Check Your Understanding Key Terms combinatorics Multiplication Rule for Counting (Fundamental Counting Principle) Key Concepts The Multiplication Rule for Counting is used to count large sets.", "section": "The Multiplication Rule for Counting", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Permutations We can use permutations to calculate the number of different orders of finish in an Olympic swimming heat. (credit: “London 2012 Olympics Park Stratford London” by Gary Bembridge/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Use the Multiplication Rule for Counting to determine the number of permutations. Compute expressions containing factorials. Compute permutations. Apply permutations to solve problems. Swimming events are some of the most popular events at the summer Olympic Games. In the finals of each event, 8 swimmers compete at the same time, making for some exciting finishes. How many different orders of finish are possible in these events? In this section, we’ll extend the Multiplication Rule for Counting to help answer questions like this one, which relate to permutations . A permutation is an ordered list of objects taken from a given population. The length of the list is given, and the list cannot contain any repeated items. Applying the Multiplication Rule for Counting to Permutations In the case of the swimming finals, one possible permutation of length 3 would be the list of medal winners (first, second, and third place finishers). A permutation of length 8 would be the full order of finish (first place through eighth place). Let’s use the Multiplication Rule for Counting to figure out how many of each of these permutations there are. Using the Multiplication Rule for Counting to Find the Number of Permutations The final heat of Olympic swimming events features 8 swimmers (or teams of swimmers). How many different podium placements (first place, second place, and third place) are possible? How many different complete orders of finish (first place through eighth place) are possible? Let’s start with the first place finisher. How many options are there? Since 8 swimmers are competing, there are 8 possibilities. Once that first swimmer completes the race, there are 7 swimmers left competing for second place. After the second finisher is decided, there are 6 swimmers remaining who could possibly finish in third place. Thus, there are 8 possibilities for first place, 7 for second place, and 6 for third place. The Multiplication Rule for Counting then tells us there are 8 × 7 × 6 = 336 different ways the winners’ podium can be filled out. To look at the complete order of finish, we can continue the pattern we can see in part 1 of this example: There are 5 possibilities for fourth place, 4 for fifth place, 3 for sixth place, 2 for seventh place, and then just 1 swimmer is left to finish in eighth place. Using the Multiplication Rule for Counting, we see that there are 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 , 320 possible orders of finish. Factorials The pattern we see in occurs commonly enough that we have a name for it: factorial . For any positive whole number n , we define the factorial of n (denoted n ! and read \" n factorial\") to be the product of every whole number less than or equal to n . We also define 0! to be equal to one. We will use factorials in a couple of different contexts, so let's get some practice doing computations with them. Computing Factorials Compute the following: 4 ! 8 ! 6 ! 9 ! 3 ! 4 ! 4 ! = 4 × 3 × 2 × 1 = 24 There are two ways to approach this calculation. The first way is to compute the factorials first, then divide: 8 ! 6 ! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 6 × 5 × 4 × 3 × 2 × 1 = 40,320 720 = 56 However, there is an easier way! You may notice in the second step that there are several terms that can be canceled; that’s always the case whenever we divide factorials. In this case, notice that we can rewrite the numerator like this: 8 ! = 8 × 7 × ( 6 × 5 × 4 × 3 × 2 × 1 ) = 8 × 7 × 6 ! With that in mind, we can proceed this way by canceling out the 6!: 8 ! 6 ! = 8 × 7 × 6 ! 6 ! = 8 × 7 = 56 That’s much easier! Let’s approach this one using our canceling technique. When we see two factorials in either the numerator or denominator, we should focus on the larger one first. So: 9 ! 3 ! 4 ! = 9 × 8 × 7 × 6 × 5 × 4 ! 3 ! 4 ! = 9 × 8 × 7 × 6 × 5 3 × 2 × 1 = 9 × 8 × 7 × 5 = 2,520 Permutations As we’ve seen, factorials can pop up when we’re computing permutations. In fact, there is a formula that we can use to make that connection explicit. Let’s define some notation first. If we have a collection of n objects and we wish to create an ordered list of r of the objects (where 1 ≤ r ≤ n ), we’ll call the number of those permutations n P r (read “the number of permutations of n objects taken r at a time”). We formalize the formula we'll use to compute permutations below. n P r = n ! ( n − r ) ! If you wondered why we defined 0 ! = 1 earlier, it was to make formulas like this one work; if we have n objects and want to order all of them (so, we want the number of permutations of n objects taken n at a time), we get n P n = n ! ( n − n ) ! = n ! 0 ! = n ! 1 = n ! . Next, we’ll get some practice computing these permutations. Computing Permutations Find the following numbers: The number of permutations of 12 objects taken 3 at a time The number of permutations of 8 objects taken 5 at a time The number of permutations of 32 objects taken 2 at a time 12 P 3 = 12 ! ( 12 − 3 ) ! = 12 ! 9 ! = 12 × 11 × 10 × 9 ! 9 ! = 12 × 11 × 10 = 1,320 8 P 5 = 8 ! ( 8 − 5 ) ! = 8 × 7 × 6 × 5 × 4 × 3 ! 3 ! = 8 × 7 × 6 × 5 × 4 = 6,720 32 P 2 = 32 ! ( 32 − 2 ) ! = 32 × 31 × 30 ! 30 ! = 32 × 31 = 992 Applying Permutations A high school graduating class has 312 students. The top student is declared valedictorian, and the second-best is named salutatorian. How many possible outcomes are there for the valedictorian and salutatorian? In the card game blackjack, the dealer’s hand of 2 cards is dealt with 1 card faceup and 1 card facedown. If the game is being played with a single deck of (52) cards, how many possible hands could the dealer get? The University Combinatorics Club has 3 officers: president, vice president, and treasurer. If there are 18 members of the club, how many ways are there to fill the officer positions? This is the number of permutations of 312 students taken 2 at a time, and 312 P 2 = 97,032 . We want the number of permutations of 52 cards taken 2 at a time, and 52 P 2 = 2,652 . Here we’re looking for the number of permutations of 18 members taken 3 at a time, and 18 P 3 = 4,896 . Very Big Permutations Permutations involving relatively small sets of objects can get very big, very quickly. A standard deck contains 52 cards. So, the number of different ways to shuffle the cards—in other words, the number of permutations of 52 objects taken 52 at a time—is 52 ! ≈ 8 × 10 67 (written out, that’s an 8 followed by 67 zeroes). The estimated age of the universe is only about 4 × 10 17 seconds. So, if a very bored all-powerful being started shuffling cards at the instant the universe began, it would have to have averaged at least 8 × 10 67 4 × 10 17 ≈ 2 × 10 50 shuffles per second since the beginning of time to have covered every possible arrangement of a deck of cards. That means the next time you pick up a deck of cards and give it a good shuffle, it’s almost certain that the particular arrangement you created has never been created before and likely never will be created again. Check Your Understanding Key Terms permutation factorial Key Concepts Using the Multiplication Rule for Counting to enumerate permutations. Simplifying and computing expressions involving factorials. Using factorials to count permutations. Formulas P n r = n ! ( n − r ) !", "section": "Permutations", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Combinations Combinations help us count things like the number of possible card hands, when the order in which the cards were drawn doesn’t matter. (credit: “IMG_3177” by Zanaca/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Distinguish between permutation and combination uses. Compute combinations. Apply combinations to solve applications. In Permutations , we studied permutations, which we use to count the number of ways to generate an ordered list of a given length from a group of objects. An important property of permutations is that the order of the list matters: The results of a race and the selection of club officers are examples of lists where the order is important. In other situations, the order is not important. For example, in most card games where a player receives a hand of cards, the order in which the cards are received is irrelevant; in fact, players often rearrange the cards in a way that helps them keep the cards organized. Combinations: When Order Doesn’t Matter In situations in which the order of a list of objects doesn’t matter, the lists are no longer permutations. Instead, we call them combinations . Distinguishing Between Permutations and Combinations For each of the following situations, decide whether the chosen subset is a permutation or a combination. A social club selects 3 members to form a committee. Each of the members has an equal share of responsibility. You are prompted to reset your email password; you select a password consisting of 10 characters without repeats. At a dog show, the judge must choose first-, second-, and third-place finishers from a group of 16 dogs. At a restaurant, the special of the day comes with the customer’s choice of 3 sides taken from a list of 6 possibilities. Since there is no distinction among the responsibilities of the 3 committee members, the order isn’t important. So, this is a combination. The order of the characters in a password matter, so this is a permutation. The order of finish matters in a dog show, so this is a permutation. A plate with mashed potatoes, peas, and broccoli is functionally the same as a plate with peas, broccoli, and mashed potatoes, so this is a combination. Counting Combinations Permutations and combinations are certainly related, because they both involve choosing a subset of a large group. Let’s explore that connection, so that we can figure out how to use what we know about permutations to help us count combinations. We’ll take a basic example. How many ways can we select 3 letters from the group A, B, C, D, and E? If order matters, that number is 5 P 3 = 60 . That’s small enough that we can list them all out in the table below. ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED BAC BAD BAE BCA BCD BCE BDA BDC BDE BEA BEC BED CAB CAD CAE CBA CBD CBE CDA CDB CDE CEA CEB CED DCA DAC DAE DBA DBC DBE DCA DCB DCE DEA DEB DEC EAB EAC EAD EBA EBC EBD ECA ECB ECD EDA EDB EDC Now, let’s look back at that list and color-code it so that groupings of the same 3 letters get the same color, as shown in : After color-coding, we see that the 60 cells can be seen as 10 groups (colors) of 6. That’s no coincidence! We’ve already seen how to compute the number of permutations using the formula To compute the number of combinations, let’s count them another way using the Multiplication Rule for Counting. We’ll do this in two steps: Step 1: Choose 3 letters (paying no attention to order). Step 2: Put those letters in order. The number of ways to choose 3 letters from this group of 5 (A, B, C, D, E) is the number of combinations we’re looking for; let’s call that number 5 C 3 (read “the number of combinations of 5 objects taken 3 at a time”). We can see from our chart that this is ten (the number of colors used). We can generalize our findings this way: remember that the number of permutations of n things taken r at a time is n P r = n ! ( n − r ) ! . That number is also equal to n C r × r ! , and so it must be the case that n ! ( n − r ) ! = n C r × r ! . Dividing both sides of that equation by r ! gives us the formula below. n C r = n ! r ! ( n − r ) ! Using the Combination Formula Compute the following: 8 C 3 12 C 5 15 C 9 8 C 3 = 8 ! 3 ! ( 8 − 3 ) ! = 8 × 7 × 6 × 5 ! 3 × 2 × 1 × 5 ! = 8 × 7 = 56 12 C 5 = 12 ! 5 ! ( 12 − 5 ) ! = 12 × 11 × 10 × 9 × 8 × 7 ! 5 × 4 × 3 × 2 × 1 × 7 ! = 11 × 9 × 8 = 792 15 C 9 = 15 ! 9 ! ( 15 − 9 ) ! = 15 × 14 × 13 × 12 × 11 × 10 × 9 ! 9 ! × 6 × 5 × 4 × 3 × 2 × 1 = 14 × 13 × 11 × 10 4 = 5,005 Applying the Combination Formula In the card game Texas Hold’em (a variation of poker), players are dealt 2 cards from a standard deck to form their hands. How many different hands are possible? The board game Clue uses a deck of 21 cards. If 3 people are playing, each person gets 6 cards for their hand. How many different 6-card Clue hands are possible? Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. How many different sets of winning numbers are possible? A standard deck has 52 cards, and a hand has 2 cards. Since the order doesn’t matter, we use the formula for counting combinations: 52 C 2 = 52 ! 2 ! ( 52 − 2 ) ! = 52 × 51 × 50 ! 2 × 1 × 50 ! = 52 × 51 2 = 1,326 . Again, the order doesn’t matter, so the number of combinations is: 21 C 6 = 21 ! 6 ! ( 21 − 6 ) ! = 21 × 20 × 19 × 18 × 17 × 16 × 15 ! 6 × 5 × 4 × 3 × 2 × 1 × 15 ! = 21 × 19 × 17 × 16 2 = 54,264 . There are 38 numbers to choose from, and we must pick 5. Since order doesn’t matter, the number of combinations is: 38 C 5 = 38 ! 5 ! ( 38 − 5 ) ! = 38 × 37 × 36 × 35 × 34 × 33 ! 5 × 4 × 3 × 2 × 1 × 33 ! = 501,492 . The notation and nomenclature used for the number of combinations is not standard across all sources. You’ll sometimes see ( n r ) instead of n C r . Sometimes you’ll hear that expression read as “ n choose r ” as shorthand for “the number of combinations of n objects taken r at a time.” Early Eastern Mathematicians Although combinations weren’t really studied in Europe until around the 13th century, mathematicians of the Middle and Far East had already been working on them for hundreds of years. The Indian mathematician known as Pingala had described them by the second century BCE; Varāhamihira (fl. sixth century) and Halayudha (fl. 10th century) extended Pingala’s work. In the ninth century, a Jain mathematician named Mahāvīra gave the formula for combinations that we use today. In 10th-century Baghdad, a mathematician named Al-Karaji also knew formulas for combinations; though his work is now lost, it was known to (and repeated by) Persian mathematician Omar Khayyam, whose work survives. Khayyam is probably best remembered as a poet, with his Rubaiyat being his most famous work. Meanwhile, in 11th-century China, Jia Xian also was working with combinations, as was his 13th-century successor Yang Hui. It is not known whether the discoveries of any of these men were known in the other regions, or if the Indians, Persians, and Chinese all came to their discoveries independently. We do know that mathematical knowledge and sometimes texts did get passed along trade routes, so it can’t be ruled out. Combining Combinations with the Multiplication Rule for Counting The student government at a university consists of 10 seniors, 8 juniors, 6 sophomores, and 4 first-years. How many ways are there to choose a committee of 8 people from this group? How many ways are to choose a committee of 8 people if the committee must consist of 2 people from each class? There are 28 people to choose from, and we need 8. So, the number of possible committees is 28 C 8 = 3,108,105 . Break the selection of the committee members down into a 4-step process: Choose the seniors, then choose the juniors, then the sophomores, and then the first-years, as shown in the table below: Class Number of Ways to Choose Committee Representatives senior 10 C 2 = 45 junior 8 C 2 = 28 sophomore 6 C 2 = 15 first-year 4 C 2 = 6 The Multiplication Rule for Counting tells us that we can get the total number of ways to complete this task by multiplying together the number of ways to do each of the four subtasks. So, there are 45 × 28 × 15 × 6 = 113,400 possible committees with these restrictions. Check Your Understanding Key Terms combination Key Concepts Permutations are used to count subsets when order matters; combinations work when order doesn't matter. Combinations can also be computed using factorials. Formulas The formula for counting combinations is: n C r = n ! r ! ( n − r ) !", "section": "Combinations", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Tree Diagrams, Tables, and Outcomes In genetics, the characteristics of an offspring organism depends on the characteristics of its parents. (credit: “Pea Plant” by Maria Keays/Flickr, CC BY 2.0)) Learning Objectives After completing this section, you should be able to: Determine the sample space of single stage experiment. Use tables to list possible outcomes of a multistage experiment. Use tree diagrams to list possible outcomes of a multistage experiment. In the 19th century, an Augustinian friar and scientist named Gregor Mendel used his observations of pea plants to set out his theory of genetic propagation. In his work, he looked at the offspring that resulted from breeding plants with different characteristics together. For applications like this, it is often insufficient to only know in how many ways a process might end; we need to be able to list all of the possibilities. As we’ve seen, the number of possible outcomes can be very large! Thus, it’s important to have a strategy that allows us to systematically list these possibilities to make sure we don’t leave any out. In this section, we’ll look at two of these strategies. Single Stage Experiments When we are talking about combinatorics or probability, the word “ experiment ” has a slightly different meaning than it does in the sciences. Experiments can range from very simple (“flip a coin”) to very complex (“count the number of uranium atoms that undergo nuclear fission in a sample of a given size over the course of an hour”). Experiments have unknown outcomes that generally rely on something random, so that if the experiment is repeated (or replicated ) the outcome might be different. No matter what the experiment, though, analysis of the experiment typically begins with identifying its sample space. The sample space of an experiment is the set of all of the possible outcomes of the experiment, so it’s often expressed as a set (i.e., as a list bound by braces; if the experiment is “randomly select a number between 1 and 4,” the sample space would be written { 1 , 2 , 3 , 4 } ). Finding the Sample Space of an Experiment For each of the following experiments, identify the sample space. Flip a coin (which has 2 faces, typically called “heads” and “tails”) and note which face is up. Flip a coin 10 times and count the number of heads. Roll a 6-sided die and note the number that is on top. Roll two 6-sided dice and note the sum of the numbers on top. If we use “H” to denote “heads is facing up” and “T” to denote “tails is facing up”, then the sample space is {H, T}. It’s possible (though unlikely) that there will be no heads flipped; the outcome in that case would be “0.” It’s also possible (more likely, but still quite unlikely) that only one flip will result in heads. Any other whole number is possible, up to the maximum: We’re flipping the coin 10 times, so we can’t get any more than 10 heads. So, the sample space is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. There are 6 numbers on the die: 1, 2, 3, 4, 5, and 6. So, the sample space for a single roll of the die is {1, 2, 3, 4, 5, 6}. If we roll 2 dice, the smallest possible sum we could get is 1 + 1 = 2 and the biggest is 6 + 6 = 12 . Every other whole number between those two is possible. So, the sample space is {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Multistage Experiments Some experiments have more complicated sample spaces because they occur in stages. These stages can occur in succession (like drawing cards one at a time) or simultaneously (rolling 2 dice). Sample spaces get more complicated as the complexity of the experiment increases, so it’s important to choose a systematic method for identifying all of the possible outcomes. The first method we’ll discuss is the table. Using Tables to Find Sample Spaces Tables are useful for finding the sample space for experiments that meet two criteria: (1) The experiment must have only two stages, and (2) the outcomes of each stage must have no effect on the outcomes of the other. When the stages do not affect each other, we say the stages are independent . Otherwise, the stages are dependent and so we can’t use tables; we’ll look at a method for analyzing dependent stages soon. Determining Independence Decide whether the two stages in these experiments are independent or dependent. You flip a coin and note the result, and then flip the coin again and note the result. You draw 2 cards from a standard deck (52 cards), one at a time. No matter what happens on the first flip, the second flip has the same sample space: {H, T} (You’ll sometimes hear the phrase “The coin has no memory”). So, these stages are independent. Let’s say that the first card you draw is A ♠ . The sample space for the second draw consists of all the cards except A ♠ (since that card is no longer in the deck, you can’t draw it again). If instead that first card was 2 ♡ , the sample space for the second draw is different: it’s every card except 2 ♡ . Since the sample space for the second card changes based on the result of the first draw, these stages are dependent. If you have a two-stage experiment with independent stages, a table is the most straightforward way to identify the sample space. To build a table, you list the outcomes of one stage of the experiment along the top of the table and the outcomes of the other stage down the side. The cells in the interior of the table are then filled using the outcomes associated with each cell’s row and column. Let’s look at an example. Using Tables to Identify Sample Spaces Identify the sample spaces of these experiments using tables. You roll two dice: one 4-sided and one 6-sided. You’re in an ice-cream shop, and you’re going to get a single scoop of ice cream with a topping. The flavors of ice cream you’re considering are vanilla, chocolate, and rocky road; the toppings are fudge, whipped cream, and sprinkles. The pea plants you’re breeding have two possible pod colors: green and yellow. These colors are decided by a particular gene, which comes in two types: “G” for green, and “g” for yellow (In genetics, capital letters usually denote dominant genes, while lower-case letters denote recessive genes). Each plant has two genes. If you breed a Gg pea plant with a gg plant, the offspring plant will get one gene from each parent. What are the possible outcomes? Step 1: Make the outline of a table, with the results of the 4-sided roll on one side and the results of the 6-sided roll on the other. In practice, it doesn’t matter which you choose; for this example, we’ll put the 4-sided results on top (labeling the columns of the chart) and the 6-sided results on the side (labeling the rows of the chart) as shown in the following table: 4-Sided Roll 1 2 3 4 6-Sided Roll 1 2 3 4 5 6 Step 2: Fill in the results in each cell of the table below. Use the notation of an ordered pair, with the row label first. 4-Sided Roll 1 2 3 4 6-Sided Roll 1 (1,1) (1,2) (1,3) (1,4) 2 (2,1) (2,2) (2,3) (2,4) 3 (3,1) (3,2) (3,3) (3,4) 4 (4,1) (4,2) (4,3) (4,4) 5 (5,1) (5,2) (5,3) (5,4) 6 (6,1) (6,2) (6,3) (6,4) So, for example, (3,2) represents the outcome where the 6-sided roll results in a 3, and the 4-sided roll gives us a 2. Thus, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4)}. Step 1: Let’s put the flavors on the rows and the toppings on the columns of the following table: Toppings fudge whipped cream sprinkles Flavors vanilla chocolate rocky road Step 2: We can fill in the cells of the table below with the resulting combinations. Toppings fudge whipped cream sprinkles Flavors vanilla vanilla with fudge vanilla with whipped cream vanilla with sprinkles chocolate chocolate with fudge chocolate with whipped cream chocolate with sprinkles rocky road rocky road with fudge rocky road with whipped cream rocky road with sprinkles So, the sample space is {vanilla with fudge, vanilla with whipped cream, vanilla with sprinkles, chocolate with fudge, chocolate with whipped cream, chocolate with sprinkles, rocky road with fudge, rocky road with whipped cream, rocky road with sprinkles}. Step 1: We’ll put the parents’ genes (P1 and P2) as labels on the rows and columns of the following table: P2 g g P1 G g Step 2: We’ll fill in the offspring’s gene composition, listing parent 1’s gene first in the table below. P2 g g P1 G Gg Gg g gg gg Thus, the sample space is {Gg, Gg, gg, gg}. (Diagrams like this, which allow us to identify the genotypes of offspring, are called Punnett squares in honor of Reginald Punnett (1875–1967), who first used them in the context of genetics.) Using Tree Diagrams to Identify Sample Spaces In experiments where there are more than two stages, or where the stages are dependent, a tree diagram is a helpful tool for systematically identifying the sample space. Tree diagrams are built by first drawing a single point (or node ), then from that node we draw one branch (a short line segment) for each outcome of the first stage. Each branch gets its own node at the other end (which we typically label with the corresponding outcome for that branch); from each of these, we draw another branch for each outcome of the second stage, assuming that the outcome of the first stage matches the branch we were on. If there are other stages, we can continue from there by continuing to add branches and nodes. This sounds really complicated, but it’s easier to understand through an example. Using a Tree Diagram to Identify a Sample Space Use a tree diagram to find the sample spaces of each of the following experiments: You flip a coin 3 times, noting the outcome of each flip in order. You flip a coin. If the result is heads, you roll a 4-sided die. If it’s tails, you roll a 6-sided die. You are planning to go on a hike with a group of friends. There are 3 trails to consider: Abel Trail, Borel Trail, and Condorcet Trail. One of your friends, Jess, requires a wheelchair; if she joins you, the group couldn’t handle the rocky Condorcet Trail. Step 1: Let’s start by placing our first node ( ). Step 2: We’ll add two branches, one for each outcome of the first coin flip, and label them ( ). Step 3: We’re ready for stage two of the experiment: another coin flip. At each node, we add in branches that represent those outcomes ( ). Finally, we can add another set of branches for the outcomes of the third stage ( ). (These final nodes are called leaves.) Step 4: We can write down the outcomes in the sample space by tracing the path out to each leaf, writing down the outcome at each node we pass through. For example, this leaf ( ): is reached via this path ( ): Step 5: We’ll label that leaf as \"HTH\" ( ), since the path passes through nodes labeled H, T, and H on its way out to our leaf. Step 6: We can label the remaining leaves using the same method ( ). The sample space is the labels on the leaves: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Step 1: We’ll start with our initial node ( ). Step 2: We’ll add in branches for the outcomes of the first stage ( ), which is the coin flip. Step 3: The second stage of the experiment depends on the outcome of the first stage. If the outcome of the first stage was H, then we roll a 4-sided die. So, only on the node for H, we’ll add in the outcomes of a 4-sided die roll ( ). If the outcome of the first stage was T, then we roll a 6-sided die. So, we’ll add those branches to the node for T ( ). Step 4: We can label the leaves to get the sample space ( ). The sample space is: {H1, H2, H3, H4, T1, T2, T3, T4, T5, T6}. Step 1: Let’s label the trails A, B, and C for ease of labeling. Even though the trails are listed first in the exercise, we can’t use the trail choice as our first stage: the trails available to us depend on whether Jess is able to join the trip. So, the first stage is whether Jess joins us (J) or not (N) ( ). Step 2: We list the appropriate trails on each branch ( ). So, the sample space is {JA, JB, NA, NB, NC}. Check Your Understanding Key Terms experiment replication sample space independent/dependent Key Concepts We identify the sample space of an experiment by identifying all of its possible outcomes. Tables can help us find a sample space by keeping the possible outcomes organized. Tree diagrams provide a visualization of the sample space of an experiment that involves multiple stages.", "section": "Tree Diagrams, Tables, and Outcomes", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Basic Concepts of Probability When you roll two dice, some outcomes (like rolling a sum of seven) are more likely than others (rolling a sum of twelve). (credit: “Dice Isn’t Just A Game; It's A Way of Life” by Leah Love/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Define probability including impossible and certain events. Calculate basic theoretical probabilities. Calculate basic empirical probabilities. Distinguish among theoretical, empirical, and subjective probability. Calculate the probability of the complement of an event. It all comes down to this. The game of Monopoly that started hours ago is in the home stretch. Your sister has the dice, and if she rolls a 4, 5, or 7 she’ll land on one of your best spaces and the game will be over. How likely is it that the game will end on the next turn? Is it more likely than not? How can we measure that likelihood? This section addresses this question by introducing a way to measure uncertainty. Introducing Probability Uncertainty is, almost by definition, a nebulous concept. In order to put enough constraints on it that we can mathematically study it, we will focus on uncertainty strictly in the context of experiments. Recall that experiments are processes whose outcomes are unknown; the sample space for the experiment is the collection of all those possible outcomes. When we want to talk about the likelihood of particular outcomes, we sometimes group outcomes together; for example, in the Monopoly example at the beginning of this section, we were interested in the roll of 2 dice that might fall as a 4, 5, or 7. A grouping of outcomes that we’re interested in is called an event . In other words, an event is a subset of the sample space of an experiment; it often consists of the outcomes of interest to the experimenter. Once we have defined the event that interests us, we can try to assess the likelihood of that event. We do that by assigning a number to each event ( E ) called the probability of that event ( P ( E ) ). The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur. Determining Certain and Impossible Events Consider an experiment that consists of rolling a single standard 6-sided die (with faces numbered 1-6). Decide if these probabilities are equal to zero, equal to one, or somewhere in between. P ( roll a 4 ) P ( roll a 7 ) P ( roll a positive number ) P ( roll a 1 3 ) P ( roll an even number ) P ( roll a single-digit number ) Let's start by identifying the sample space. For one roll of this die, the possible outcomes are {1, 2, 3, 4, 5,6}. We can use that to assess these probabilities: We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So, 0 < P ( roll a 4 ) < 1 . Notice that 7 is not in the sample space. So, P ( roll a 7 ) = 0 . Every outcome in the sample space is a positive number, so this event is certain. Thus, P ( roll a positive number ) = 1 . Since 1 3 is not in the sample space, P ( roll a 1 3 ) = 0 . Some outcomes in the sample space are even numbers (2, 4, and 6), but the others aren’t. So, 0 < P ( roll an even number ) < 1 . Every outcome in the sample space is a single-digit number, so P ( roll a single-digit number ) = 1 . Three Ways to Assign Probabilities The probabilities of events that are certain or impossible are easy to assign; they’re just 1 or 0, respectively. What do we do about those in-between cases, for events that might or might not occur? There are three methods to assign probabilities that we can choose from. We’ll discuss them here, in order of reliability. Method 1: Theoretical Probability The theoretical method gives the most reliable results, but it cannot always be used. If the sample space of an experiment consists of equally likely outcomes, then the theoretical probability of an event is defined to be the ratio of the number of outcomes in the event to the number of outcomes in the sample space. For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is the ratio P ( E ) = n ( E ) n ( S ) , where n ( E ) and n ( S ) denote the number of outcomes in the event and in the sample space, respectively. Computing Theoretical Probabilities Recall that a standard deck of cards consists of 52 unique cards which are labeled with a rank (the whole numbers from 2 to 10, plus J, Q, K, and A) and a suit ( ♣ , ♢ , ♡ , or ♠ ). A standard deck is thoroughly shuffled, and you draw one card at random (so every card has an equal chance of being drawn). Find the theoretical probability of each of these events: The card is 10 ♠ . The card is a ♡ . The card is a king (K). There are 52 cards in the deck, so the sample space for each of these experiments has 52 elements. That will be the denominator for each of our probabilities. There is only one 10 ♠ in the deck, so this event only has one outcome in it. Thus, P ( 10 ♠ ) = 1 52 . There are 13 ♡ s in the deck, so P ( ♡ ) = 13 52 = 1 4 . There are 4 cards of each rank in the deck, so P ( K ) = 4 52 = 1 13 . It is critical that you make sure that every outcome in a sample space is equally likely before you compute theoretical probabilities! Using Tables to Find Theoretical Probabilities In the Basic Concepts of Probability , we were considering a Monopoly game where, if your sister rolled a sum of 4, 5, or 7 with 2 standard dice, you would win the game. What is the probability of this event? Use tables to determine your answer. We should think of this experiment as occurring in two stages: (1) one die roll, then (2) another die roll. Even though these two stages will usually occur simultaneously in practice, since they’re independent, it’s okay to treat them separately. Step 1: Since we have two independent stages, let’s create a table ( ), which is probably the most efficient method for determining the sample space. Now, each of the 36 ordered pairs in the table represent an equally likely outcome. Step 2: To make our analysis easier, let’s replace each ordered pair with the sum ( ). Step 3: Since the event we’re interested in is the one consisting of rolls of 4, 5, or 7. Let’s shade those in ( ). Our event contains 13 outcomes, so the probability that your sister rolls a losing number is 13 36 . Using Tree Diagrams to Compute Theoretical Probability If you flip a fair coin 3 times, what is the probability of each event? Use a tree diagram to determine your answer You flip exactly 2 heads. You flip 2 consecutive heads at some point in the 3 flips. All 3 flips show the same result. Let’s build a tree to identify the sample space ( ). The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which has 8 elements. Flipping exactly 2 heads occurs three times (HHT, HTH, THH), so the probability is 3 8 . Flipping 2 consecutive heads at some point in the experiment happens 3 times: HHH, HHT, THH. So, the probability is 3 8 . There are 2 outcomes that all show the same result: HHH and TTT. So, the probability is 2 8 = 1 4 . Gerolamo Cardano The first known text that provided a systematic approach to probabilities was written in 1564 by Gerolamo Cardano (1501–1576). Cardano was a physician whose illegitimate birth closed many doors that would have otherwise been open to someone with a medical degree in 16th-century Italy. As a result, Cardano often turned to gambling to help ends meet. He was a remarkable mathematician, and he used his knowledge to gain an edge when playing at cards or dice. His 1564 work, titled Liber de ludo aleae (which translates as Book on Games of Chance ), summarized everything he knew about probability. Of course, if that book fell into the hands of those he played against, his advantage would disappear. That’s why he never allowed it to be published in his lifetime (it was eventually published in 1663). Cardano made other contributions to mathematics; he was the first person to publish the third degree analogue of the Quadratic Formula (though he didn’t discover it himself), and he popularized the use of negative numbers. Method 2: Empirical Probability Theoretical probabilities are precise, but they can’t be found in every situation. If the outcomes in the sample space are not equally likely, then we’re out of luck. Suppose you’re watching a baseball game, and your favorite player is about to step up to the plate. What is the probability that he will get a hit? In this case, the sample space is {hit, not a hit}. That doesn’t mean that the probability of a hit is 1 2 , since those outcomes aren’t equally likely. The theoretical method simply can’t be used in this situation. Instead, we might look at the player’s statistics up to this point in the season, and see that he has 122 hits in 531 opportunities. So, we might think that the probability of a hit in the next plate appearance would be about 122 531 ≈ 0.23 . When we use the outcomes of previous replications of an experiment to assign a probability to the next replication, we’re defining an empirical probability . Empirical probability is assigned using the outcomes of previous replications of an experiment by finding the ratio of the number of times in the previous replications the event occurred to the total number of previous replications. Empirical probabilities aren’t exact, but when the number of previous replications is large, we expect them to be close. Also, if the previous runs of the experiment are not conducted under the exact set of circumstances as the one we’re interested in, the empirical probability is less reliable. For instance, in the case of our favorite baseball player, we might try to get a better estimate of the probability of a hit by looking only at his history against left- or right-handed pitchers (depending on the handedness of the pitcher he’s about to face). Probability and Statistics One of the broad uses of statistics is called statistical inference, where statisticians use collected data to make a guess (or inference) about the population the data were collected from. Nearly every tool that statisticians use for inference is based on probability. Not only is the method we just described for finding empirical probabilities one type of statistical inference, but some more advanced techniques in the field will give us an idea of how close that empirical probability might be to the actual probability! Finding Empirical Probabilities Assign an empirical probability to the following events: Jose is on the basketball court practicing his shots from the free throw line. He made 47 out of his last 80 attempts. What is the probability he makes his next shot? Amy is about to begin her morning commute. Over her last 60 commutes, she arrived at work 12 times in under half an hour. What is the probability that she arrives at work in 30 minutes or less? Felix is playing Yahtzee with his sister. Felix won 14 of the last 20 games he played against her. How likely is he to win this game? Since Jose made 47 out of his last 80 attempts, assign this event an empirical probability of 47 80 ≈ 59 % . Amy completed the commute in under 30 minutes in 12 of the last 60 commutes, so we can estimate her probability of making it in under 30 minutes this time at 12 60 = 20 % . Since Felix has won 14 of the last 20 games, assign a probability for a win this time of 14 20 = 70 % . Buffon’s Needle A famous early question about probability (posed by Georges-Louis Leclerc, Comte de Buffon in the 18th century) had to do with the probability that a needle dropped on a floor finished with wooden slats would lay across one of the seams. If the distance between the slats is exactly the same length as the needle, then it can be shown using calculus that the probability that the needle crosses a seam is 2 π . Using toothpicks or matchsticks (or other uniformly long and narrow objects), assign an empirical probability to this experiment by drawing parallel lines on a large sheet of paper where the distance between the lines is equal to the length of your dropping object, then repeatedly dropping the objects and noting whether the object touches one of the lines. Once you have your empirical probability, take its reciprocal and multiply by 2. Is the result close to π ? Method 3: Subjective Probability In cases where theoretical probability can’t be used and we don’t have prior experience to inform an empirical probability, we’re left with one option: using our instincts to guess at a subjective probability . A subjective probability is an assignment of a probability to an event using only one’s instincts. Subjective probabilities are used in cases where an experiment can only be run once, or it hasn’t been run before. Because subjective probabilities may vary widely from person to person and they’re not based on any mathematical theory, we won’t give any examples. However, it’s important that we be able to identify a subjective probability when we see it; they will in general be far less accurate than empirical or theoretical probabilities. Distinguishing among Theoretical, Empirical, and Subjective Probabilities Classify each of the following probabilities as theoretical, empirical, or subjective. An eccentric billionaire is testing a brand new rocket system. He says there is a 15% chance of failure. With 4 seconds to go in a close basketball playoff game, the home team need 3 points to tie up the game and send it to overtime. A TV commentator says that team captain should take the final 3-point shot, because he has a 38% chance of making it (greater than every other player on the team). Felix is losing his Yahtzee game against his sister. He has one more chance to roll 2 dice; he’ll win the game if they both come up 4. The probability of this is about 2.8%. This experiment has never been run before, so the given probability is subjective. Presumably, the commentator has access to each player’s performance statistics over the entire season. So, the given probability is likely empirical. Rolling 2 dice results in a sample space with equally likely outcomes. This probability is theoretical. (We’ll learn how to calculate that probability later in this chapter.) Benford’s Law In 1938, Frank Benford published a paper (“The law of anomalous numbers,” in Proceedings of the American Philosophical Society ) with a surprising result about probabilities. If you have a list of numbers that spans at least a couple of orders of magnitude (meaning that if you divide the largest by the smallest, the result is at least 100), then the digits 1–9 are not equally likely to appear as the first digit of those numbers, as you might expect. Benford arrived at this conclusion using empirical probabilities; he found that 1 was about 6 times as likely to be the initial digit as 9 was! New Probabilities from Old: Complements One of the goals of the rest of this chapter is learning how to break down complicated probability calculations into easier probability calculations. We’ll look at the first of the tools we can use to accomplish this goal in this section; the rest will come later. Given an event E , the complement of E (denoted E ′ ) is the collection of all of the outcomes that are not in E . (This is language that is taken from set theory, which you can learn more about elsewhere in this text.) Since every outcome in the sample space either is or is not in E , it follows that n ( E ) + n ( E ′ ) = n ( S ) . So, if the outcomes in S are equally likely, we can compute theoretical probabilities P ( E ) = n ( E ) n ( S ) and P ( E ′ ) = n ( E ′ ) n ( S ) . Then, adding these last two equations, we get P ( E ) + P ( E ′ ) = n ( E ) n ( S ) + n ( E ′ ) n ( S ) = n ( E ) + n ( E ′ ) n ( S ) = n ( S ) n ( S ) = 1 Thus, if we subtract P ( E ′ ) from both sides, we can conclude that P ( E ) = 1 − P ( E ′ ) . Though we performed this calculation under the assumption that the outcomes in S are all equally likely, the last equation is true in every situation. P ( E ) = 1 − P ( E ′ ) How is this helpful? Sometimes it is easier to compute the probability that an event won’t happen than it is to compute the probability that it will . To apply this principle, it’s helpful to review some tricks for dealing with inequalities. If an event is defined in terms of an inequality, the complement will be defined in terms of the opposite inequality: Both the direction and the inclusivity will be reversed, as shown in the table below. If E is defined with: then E ′ is defined with: < ≥ ≤ > > ≤ ≥ < Using the Formula for Complements to Compute Probabilities If you roll a standard 6-sided die, what is the probability that the result will be a number greater than one? If you roll two standard 6-sided dice, what is the probability that the sum will be 10 or less? If you flip a fair coin 3 times, what is the probability that at least one flip will come up tails? Here, the sample space is {1, 2, 3, 4, 5, 6}. It’s easy enough to see that the probability in question is 5 6 , because there are 5 outcomes that fall into the event “roll a number greater than 1.” Let’s also apply our new formula to find that probability. Since E is defined using the inequality roll > 1 , then E ′ is defined using roll ≤ 1 . Since there’s only one outcome (1) in E ′ , we have P ( E ′ ) = 1 6 . Thus, P ( E ) = 1 − P ( E ′ ) = 5 6 . In , we found the following table of equally likely outcomes for rolling 2 dice ( ): Here, the event E is defined by the inequality sum ≤ 10 . Thus, E ′ is defined by sum > 10 . There are three outcomes in E ′ : two 11s and one 12. Thus, P ( E ) = 1 − P ( E ′ ) = 1 − 3 36 = 11 12 . In , we found the sample space for this experiment consisted of these equally likely outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Our event E is defined by T ≥ 1 , so E ′ is defined by T < 1 . The only outcome in E ′ is the first one on the list, where zero tails are flipped. So, P ( E ) = 1 − P ( E ′ ) = 1 − 1 8 = 7 8 . Check Your Understanding Key Terms event probability theoretical probability empirical probability subjective probability Key Concepts The theoretical probability of an event is the ratio of the number of equally likely outcomes in the event to the number of equally likely outcomes in the sample space. Empirical probabilities are computed by repeating the experiment many times, and then dividing the number of replications that result in the event of interest by the total number of replications. Subjective probabilities are assigned based on subjective criteria, usually because the experiment can’t be repeated and the outcomes in the sample space are not equally likely. The probability of the complement of an event is found by subtracting the probability of the event from one. Formulas For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is the ratio P ( E ) = n ( E ) n ( S ) where n ( E ) and n ( S ) denote the number of outcomes in the event and in the sample space, respectively. P ( E ) = 1 − P ( E ′ )", "section": "Basic Concepts of Probability", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Probability with Permutations and Combinations Bingo and many lottery games depend on selecting one or more numbers at random from a list; often this is done by drawing numbered balls from a bin. (credit: “Redundant Bingo Balls” by Greg Clarke/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate probabilities with permutations. Calculate probabilities with combinations. In our earlier discussion of theoretical probabilities, the first step we took was to write out the sample space for the experiment in question. For many experiments, that method just isn’t practical. For example, we might want to find the probability of drawing a particular 5-card poker hand. Since there are 52 cards in a deck and the order of cards doesn’t matter, the sample space for this experiment has 52 C 5 = 2,598,960 possible 5-card hands. Even if we had the patience and space to write them all out, sorting through the results to find the outcomes that fall in our event would be just as tedious. Luckily, the formula for theoretical probabilities doesn’t require us to know every outcome in the sample space; we just need to know how many outcomes there are. In this section, we’ll apply the techniques we learned earlier in the chapter ( The Multiplication Rule for Counting , permutations, and combinations) to compute probabilities. Using Permutations to Compute Probabilities Recall that we can use permutations to count how many ways there are to put a number of items from a list in order. If we’re looking at an experiment whose sample space looks like an ordered list, then permutations can help us to find the right probabilities. Using Permutations to Compute Probabilities In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win? You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president? A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD? Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are 9 P 2 = 72 possible outcomes. That’s the size of our sample space, so it will go in the denominator of the probability. Since only one of those outcomes is a winner, the numerator of the probability is 1. So, the probability of randomly selecting the winning exacta bet is 1 72 . There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are 10 P 2 = 90 different ways to select officers. Next, we must figure out how many outcomes are in our event. We’ll use the Multiplication Rule for Counting to find that number. There is only 1 choice for president in our event, and there are 3 choices for vice president. So, there are 1 × 3 = 3 outcomes in the event. Thus, the probability that you will serve as president with one of your friends as vice president is 3 90 = 1 30 . There are 12 slips of paper in the bag, and 3 will be drawn. So, there are 12 P 3 = 1320 possible outcomes. Now, we’ll compute the number of outcomes in our event. The first letter drawn must be a B, and there are 3 of those. Next must come an A (2 of those) and then a D (4 of those). Thus, there are 3 × 2 × 4 = 24 outcomes in our event. So, the probability that the letters drawn spell out the word BAD is 24 1320 = 1 55 . Combinations to Computer Probabilities If the sample space of our experiment is one in which order doesn’t matter, then we can use combinations to find the number of outcomes in that sample space. Using Combinations to Compute Probabilities Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. If you buy one ticket for this game, what is the probability that you win the top prize by choosing all 5 winning numbers? There’s a second prize in the Palmetto Cash 5 game that a player wins if 4 of the player's 5 numbers are among the 5 winning numbers. What’s the probability of winning the second prize? Scrabble is a word-building board game. Players make hands of 7 letters by selecting tiles with single letters printed on them blindly from a bag (2 tiles have nothing printed on them; these blanks can stand for any letter). Players use the letters in their hands to spell out words on the board. Initially, there are 100 tiles in the bag. Of those, 44 are (or could be) vowels (9 As, 12 Es, 9 Is, 8 Os, 4 Us, and 2 blanks; we’ll treat Y as a consonant). What is the probability that your initial hand has no vowels? There are 38 numbers to choose from, and the order of the 5 we pick doesn’t matter. So, there are 38 C 5 = 501 , 492 outcomes in the sample space. Only one outcome is in our winning event, so the probability of winning is 1 501 , 492 . As in part 1 of this example,, there are 501,492 outcomes in the sample space. The tricky part here is figuring out how many outcomes are in our event. To qualify, the outcome must contain 4 of the 5 winning numbers, plus one losing number. There are 5 C 4 = 5 ways to choose the 4 winning numbers, and there are 38 − 5 = 33 losing numbers. So, using the Multiplication Rule for Counting, there are 5 × 33 = 165 outcomes in our event. Thus, the probability of winning the second prize is 165 501 , 492 = 55 167 , 164 , which is about 0.00033. The number of possible starting hands is 100 C 7 = 16 , 007 , 560 , 800 . There are 100 − 44 = 56 consonants in the bag, so the number of all-consonant hands is 56 C 7 = 231 , 917 , 400 . Thus, the probability of drawing all consonants is 231 , 917 , 40 16 , 007 , 560 , 800 = 32 , 139 2 , 425 , 388 ≈ 0.0145 . Check Your Understanding Key Concepts We use permutations and combinations to count the number of equally likely outcomes in an event and in a sample space, which allows us to compute theoretical probabilities.", "section": "Probability with Permutations and Combinations", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "What Are the Odds? Scratch-off lottery tickets, as well as many other games, represent the likelihood of winning using odds. (credit: “My Scratch-off Winnings” by Shoshanah/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute odds. Determine odds from probabilities. Determine probabilities from odds. A particular lottery instant-win game has 2 million tickets available. Of those, 500,000 win a prize. If there are 500,000 winners, then it follows that there are 1,500,000 losing tickets. When we evaluate the risk associated with a game like this, it can be useful to compare the number of ways to win the game to the number of ways to lose. In the case of this game, we would compare the 500,000 wins to the 1,500,000 losses. In other words, there are 3 losing tickets for every winning ticket. Comparisons of this type are the focus of this section. Computing Odds The ratio of the number of equally likely outcomes in an event E to the number of equally likely outcomes not in the event E ′ is called the odds for (or odds in favor of) the event. The opposite ratio (the number of outcomes not in the event to the number in the event E ′ to the number in the event E is called the odds against the event. Both odds and probabilities are calculated as ratios. To avoid confusion, we will always use fractions, decimals, or percents for probabilities, and we’ll use colons to indicate odds. The rules for simplifying fractions apply to odds, too. Thus, the odds for winning a prize in the game described in the section opener are 500,000 : 1,500,000 = 1 : 3 and the odds against winning a prize are 3 : 1 . These would often be described in words as “the odds of winning are one to three in favor” or “the odds of winning are three to one against.” . Notice that, while probabilities must always be between zero and one inclusive, odds can be any (non-negative) number, as we’ll see in the next example. Computing Odds If you roll a fair 6-sided die, what are the odds for rolling a 5 or higher? If you roll two fair 6-sided dice, what are the odds against rolling a sum of 7? If you draw a card at random from a standard deck, what are the odds for drawing a ♡ ? If you draw 2 cards at random from a standard deck, what are the odds against them both being ♠ ? The sample space for this experiment is {1, 2, 3, 4, 5, 6}. Two of those outcomes are in the event “roll a five or higher,” while four are not. So, the odds for rolling a five or higher are 2 : 4 = 1 : 2 . In , we found the sample space for this experiment using the following table ( ): There are 6 outcomes in the event “roll a sum of 7,” and there are 30 outcomes not in the event. So, the odds against rolling a 7 are 30 : 6 = 5 : 1 . There are 13 ♡ in a standard deck, and 52 − 13 = 39 others. So, the odds in favor of drawing a ♡ are 13 : 39 = 1 : 3 . There are 13 C 2 = 78 ways to draw 2 ♠ , and 52 C 2 − 78 = 1,248 ways to draw 2 cards that are not both ♠ . So, the odds against drawing 2 ♠ are 1,248 : 78 = 16 : 1 . Odds as a Ratio of Probabilities We can also think of odds as a ratio of probabilities. Consider again the instant-win game from the section opener , with 500,000 winning tickets out of 2,000,000 total tickets. If a player buys one ticket, the probability of winning is 500 , 000 2 , 000 , 000 = 1 4 , and the probability of losing is 1 − 1 4 = 3 4 . Notice that the ratio of the probability of winning to the probability of losing is 1 4 : 3 4 = 1 : 3 , which matches the odds in favor of winning. For an event E , odds for E = n ( E ) : n ( E ′ ) = P ( E ) : P ( E ′ ) = P ( E ) : ( 1 − P ( E ) ) odds against E = n ( E ′ ) : n ( E ) = P ( E ′ ) : P ( E ) = ( 1 − P ( E ) ) : P ( E ) We can use these formulas to convert probabilities to odds, and vice versa. Converting Probabilities to Odds Given the following probabilities of an event, find the corresponding odds for and odds against that event. P ( E ) = 3 5 P ( E ) = 17 % Using the formula , we have: odds for E = P ( E ) : ( 1 − P ( E ) ) = 3 5 : ( 1 − 3 5 ) = 3 5 : 2 5 = 3 : 2. (Note that in the last step, we simplified by multiplying both terms in the ratio by 5.) Since the odds for E are 3 : 2 , the odds against E must be 2 : 3 . Again, we’ll use the formula: odds for E = P ( E ) : ( 1 − P ( E ) ) = 0.17 : ( 1 − 0.17 ) = 0.17 : 0.83 ≈ 1 : 4.88. (In the last step, we simplified by dividing both terms in the ratio by 0.17.) It follows that the odds against E are approximately 4.88 : 1 . Now, let’s convert odds to probabilities. Let’s say the odds for an event are A : B . Then, using the formula above , we have A : B = P ( E ) : ( 1 − P ( E ) ) . Converting to fractions and solving for P ( E ) , we get: A B = P ( E ) 1 − P ( E ) A ( 1 − P ( E ) ) = B × P ( E ) A − A × P ( E ) = B × P ( E ) A = A × P ( E ) + B × P ( E ) A = ( A + B ) × P ( E ) A A + B = P ( E ) . Let’s put this result in a formula we can use. If the odds in favor of E are A : B , then P ( E ) = A A + B . Converting Odds to Probabilities Find P ( E ) if E : The odds of E are 2 : 1 in favor The odds of E are 6 : 1 against Using the formula we just found, we have P ( E ) = 2 2 + 1 = 2 3 . If the odds against are 6 : 1 , then the odds for are 1 : 6 . Thus, using the formula, P ( E ) = 1 1 + 6 = 1 7 . Some places, particularly state lottery websites, will use the words “odds” and “probability” interchangeably. Never assume that the word “odds” is being used correctly! Compute one of the odds/probabilities yourself to make sure you know how the word is being used! Check Your Understanding Key Terms odds (for/against) Key Concepts Odds are computed as the ratio of the probability of an event to the probability of its compliment. Formulas For an event E , odds for E = n ( E ) : n ( E ′ ) = P ( E ) : P ( E ′ ) = P ( E ) : ( 1 − P ( E ) ) odds against E = n ( E ′ ) : n ( E ) = P ( E ′ ) : P ( E ) = ( 1 − P ( E ) ) : P ( E ) If the odds in favor of E are A : B , then P ( E ) = A A + B .", "section": "What Are the Odds?", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Addition Rule for Probability Students can be sorted using a variety of possible categories like class year, major, whether they are a varsity athlete, and so forth. (credit: “Multicultural Mashup Melds Languages, Cultures at COD 36” by COD Newsroom/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify mutually exclusive events. Apply the Addition Rule to compute probability. Use the Inclusion/Exclusion Principle to compute probability. Up to this point, we have looked at the probabilities of simple events. Simple events are those with a single, simple characterization. Sometimes, though, we want to investigate more complicated situations. For example, if we are choosing a college student at random, we might want to find the probability that the chosen student is a varsity athlete or in a Greek organization. This is a compound event : there are two possible criteria that might be met. We might instead try to identify the probability that the chosen student is both a varsity athlete and in a Greek organization. In this section and the next , we’ll cover probabilities of two types of compound events: those build using “or” and those built using “and.” We’ll deal with the former first. Mutual Exclusivity Before we get to the key techniques of this section, we must first introduce some new terminology. Let’s say you’re drawing a card from a standard deck. We’ll consider 3 events: H is the event “the card is a ♡ ,” T is the event “the card is a 10,” and S is the event “the card is a ♠ .” If the card drawn is J ♠ , then H and T didn’t occur, but S did. If the card drawn is instead 10 ♠ , then H didn’t occur, but both T and S did. We can see from these examples that, if we are interested in several possible events, more than one of them can occur simultaneously (both T and S , for example). But, if you think about all the possible outcomes, you can see that H and S can never occur simultaneously; there are no cards in the deck that are both ♡ and ♠ . Pairs of events that cannot both occur simultaneously are called mutually exclusive . Let’s go through an example to help us better understand this concept. Identifying Mutually Exclusive Events Decide whether the following events are mutually exclusive. If they are not mutually exclusive, identify an outcome that would result in both events occurring. You are about to roll a standard 6-sided die. E is the event “the die shows an even number” and F is the event “the die shows an odd number.” You are about to roll a standard 6-sided die. E is the event “the die shows an even number” and S is the event “the die shows a number less than 4.” You are about to flip a coin 4 times. J is the event “at least 2 heads are flipped” and K is the event “fewer than 3 tails are flipped.” Let’s look at the outcomes for each event: E = { 2 , 4 , 6 } and F = { 1 , 3 , 5 } . There are no outcomes in common, so E and F are mutually exclusive. Again, consider the outcomes in each event: E = { 2 , 4 , 6 } and S = { 1 , 2 , 3 } . Since the outcome 2 belongs to both events, these are not mutually exclusive. Suppose the results of the 4 flips are HTTH. Then at least 2 heads are flipped, and fewer than 3 tails are flipped. That means that both J and K occurred, and so these events are not mutually exclusive. The Addition Rule for Mutually Exclusive Events If two events are mutually exclusive, then we can use addition to find the probability that one or the other event occurs. If E and F are mutually exclusive events, then P ( E or F ) = P ( E ) + P ( F ) . Why does this formula work? Let’s consider a basic example. Suppose we’re about to draw a Scrabble tile from a bag containing A, A, B, E, E, E, R, S, S, U. What is the probability of drawing an E or an S? Since 3 of the tiles are marked with E and 2 are marked with S, there are 5 tiles that satisfy the criteria. There are ten tiles in the bag, so the probability is 5 10 = 1 2 . Notice that the probability of drawing an E is 3 10 and the probability of drawing an S is 2 10 ; adding those together, we get 3 10 + 2 10 = 5 10 . Look at the numerators in the fractions involved in the sum: the 3 represents the number of E tiles and the 2 is the number of S tiles. This is why the Addition Rule works: The total number of outcomes in one event or the other is the sum of the numbers of outcomes in each of the individual events. Using the Addition Rule For each of the given pairs of events, decide if the Addition Rule applies. If it does, use the Addition Rule to find the probability that one or the other occurs. You are rolling a standard 6-sided die. Event A is “roll an even number” and event B is “roll a 3.” You are drawing a card at random from a standard 52-card deck. Event R is “draw a ♡ ” and event S is “draw a king.” You are rolling a pair of standard 6-sided dice. Event E is “roll an odd sum” and event F is “roll a sum of 10.” The table we constructed in might help. Since 3 is not an even number, these events are mutually exclusive. So, we can use the Addition Rule: since P ( A ) = 3 6 and P ( B ) = 1 6 , we get P ( A ⁢ or ⁢ B ) = 3 6 + 1 6 = 2 3 . If the card drawn is K ♡ , then both R and S occur. So, they aren’t mutually exclusive, and the Addition Rule doesn’t apply. Since 10 is not odd, these events are mutually exclusive. Since P ( E ) = 18 36 and P ( F ) = 3 36 , the Addition Rule gives us P ( E ⁢ or ⁢ F ) = 18 36 + 3 36 = 7 12 . Finding Probabilities When Events Aren’t Mutually Exclusive Let’s return to the example we used to explore the Addition Rule: We’re about to draw a Scrabble tile from a bag containing A, A, B, E, E, E, R, S, S, U. Consider these events: J is “draw a vowel” and K is “draw a letter that comes after L in the alphabet.” Since there are 6 vowels, P ( J ) = 6 10 . There are 4 tiles with letters that come after L alphabetically, so P ( K ) = 4 10 . What is P ( J ⁢ or ⁢ K ) ? If we blindly apply the Addition Rule, we get 6 10 + 4 10 = 1 , which would mean that the compound event J or K is certain. However, it’s possible to draw a B, in which case neither J nor K happens. Where’s the error? The events are not mutually exclusive: the outcome U belongs to both events, and so the Addition Rule doesn’t apply. However, there’s a way to extend the Addition Rule to allow us to find this probability anyway; it’s called the Inclusion/Exclusion Principle . In this example, if we just add the two probabilities together, the outcome U is included in the sum twice: It’s one of the 6 outcomes represented in the numerator of 6 10 , and it’s one of the 4 outcomes represented in the numerator of 4 10 . So, that particular outcome has been “double counted.” Since it has been included twice, we can get a true accounting by excluding it once: 6 10 + 4 10 − 1 10 = 9 10 . We can generalize this idea to a formula that we can apply to find the probability of any compound event built using “or.” Inclusion/Exclusion Principle: If E and F are events that contain outcomes of a single experiment, then P ( E or F ) = P ( E ) + P ( F ) − P ( E and F ) . It’s worth noting that this formula is truly an extension of the Addition Rule. Remember that the Addition Rule requires that the events E and F are mutually exclusive. In that case, the compound event ( E ⁢ and ⁢ F ) is impossible, and so P ( E ⁢ and ⁢ F ) = 0 . So, in cases where the events in question are mutually exclusive, the Inclusion/Exclusion Principle reduces to the Addition Rule. Using the Inclusion/Exclusion Principle Suppose we have events E , F , and G , associated with these probabilities: P ( E ) = 0.45 P ( F ) = 0.6 P ( G ) = 0.55 P ( E and ⁢ F ) = 0.2 P ( E and ⁢ G ) = 0.2 P ( F and ⁢ G ) = 0.25 Compute the following: P ( E ⁢ or ⁢ F ) P ( E ⁢ or ⁢ G ) P ( F ⁢ or ⁢ G ) Using the Inclusion/Exclusion Principle, we get: P ( E ⁢ or ⁢ F ) = P ( E ) + P ( F ) − P ( E and ⁢ F ) = 0.45 + 0.6 − 0.2 = 0.85. Again, we’ll apply the Inclusion/Exclusion Principle: P ( E or ⁢ G ) = P ( E ) + P ( G ) − P ( E and ⁢ G ) = 0.45 + 0.55 − 0.2 = 0.8. Applying the Inclusion/Exclusion Principle one more time: P ( F ⁢ or ⁢ G ) = P ( F ) + P ( G ) − P ( F and ⁢ G ) = 0.6 + 0.55 − 0.25 = 0.9. Check Your Understanding Key Terms mutually exclusive Key Concepts The Addition Rule is used to find the probability that one event or another will occur when those events are mutually exclusive. The Inclusion/Exclusion Principle is used to find probabilities when events are not mutually exclusive. Formulas If E and F are mutually exclusive events, then P ( E or F ) = P ( E ) + P ( F ) . If E and F are events that contain outcomes of a single experiment, then P ( E or F ) = P ( E ) + P ( F ) − P ( E ⁢ and ⁢ F ) .", "section": "The Addition Rule for Probability", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Conditional Probability and the Multiplication Rule If you roll two dice by throwing them one at a time, the face showing on the first die will affect the possible outcomes for the sum of the two dice. (credit: “dice” by Ciarán Archer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate conditional probabilities. Apply the Multiplication Rule for Probability to compute probabilities. Back in , we constructed the following table ( ) to help us find the probabilities associated with rolling two standard 6-sided dice: For example, 3 of these 36 equally likely outcomes correspond to rolling a sum of 10, so the probability of rolling a 10 is 3 36 = 1 12 . However, if you choose to roll the dice one at a time, the probability of rolling a 10 will change after the first die comes to rest. For example, if the first die shows a 5, then the probability of rolling a sum of 10 has jumped to 1 6 —the event will occur if the second die also shows a 5, which is 1 of 6 equally likely outcomes for the second die. If instead the first die shows a 3, then the probability of rolling a sum of 10 drops to 0—there are no outcomes for the second die that will give us a sum of 10. Understanding how probabilities can shift as we learn new information is critical in the analysis of our second type of compound events: those built with “and.” This section will explain how to compute probabilities of those compound events. Conditional Probabilities When we analyze experiments with multiple stages, we often update the probabilities of the possible final outcomes or the later stages of the experiment based on the results of one or more of the initial stages. These updated probabilities are called conditional probabilities . In other words, if O is a possible outcome of the first stage in a multistage experiment, then the probability of an event E conditional on O (denoted P ( E | O ) , read “the probability of E given O ”) is the updated probability of E under the assumption that O occurred. In the example that opened this section, we might consider rolling two dice as a multistage experiment: rolling one, then the other. If we define E to be the event “roll a sum of 10,” O to be the event “first die shows 5,” and Q to be the event “first die shows 3,” then we computed P ( E ) = 1 12 , P ( E | O ) = 1 6 , and P ( E | Q ) = 0 . Computing Conditional Probabilities April is playing a coin-flipping game with Ben. She will flip a coin 3 times. If the coin lands on heads more than tails, April wins; if it lands on tails more than heads, Ben wins. Let A be the event “April wins,” H be “first flip is heads,” and T be “first flip is tails.” Compute P ( A ) , P ( A | H ) , and P ( A | T ) . You are about to draw 2 cards without replacement from a deck containing only these 10 cards: A ♡ , A ♠ , A ♣ , A ♢ , K ♠ , K ♣ , Q ♡ , Q ♠ , J ♡ , J ♠ . We’ll define the following events: F is “both cards are the same rank,” A is “first card is an ace,” and K is “first card is a king.” Compute P ( F | A ) and P ( F | K ) . Jim’s sock drawer contains 5 black socks and 3 blue socks. To avoid waking his partner, Jim doesn’t want to turn the lights on, so he puts on 2 socks at random. Let M be the event “Jim’s 2 socks match,” let K be the event “the sock on Jim’s left foot is black,” and let L be the event “the sock on Jim’s left foot is blue.” Compute P ( M ) , P ( M | K ) , and P ( M | L ) . Step 1. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The event A consists of the first 4 of those outcomes: HHH, HHT, HTH, and THH. Thus, P ( A ) = 4 8 = 1 2 . Step 2. Now, let’s compute P ( A | H ) . We are assuming the result of the first flip is heads. That leaves us with 4 possible outcomes: HHH, HHT, HTH, and HTT. Of those, April wins 3 (HHH, HHT, HTH) and loses one (HTT). So, P ( A | H ) = 3 4 . Step 3. If the result of the first flip is instead tails, the 4 possible outcomes are THH, THT, TTH, and TTT. Of those, April wins 1 (THH) and loses 3 (THT, TTH, TTT). So, P ( A | T ) = 1 4 . Step 1. If the event A happens, then 1 of the 4 aces is drawn first; the remaining cards in the deck are 3 aces, 2 kings, 2 queens, and 2 jacks. In order for the event F to occur, the second card drawn has to be an ace. Since there are 3 aces among the remaining 9 cards, P ( F | A ) = 3 9 = 1 3 . Step 2. If the event K happens instead, then the first card drawn is a king. That leaves 4 aces, 1 king, 2 queens, and 2 jacks in the deck. Under the assumption that the first card is a king, the event F will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have P ( F | K ) = 1 9 . Step 1. We can view the event M as a compound event using “or”: both socks are blue or both socks are black. Let’s compute the probability that both socks are blue using combinations. We’re choosing 2 socks from a group of 8; 3 of the 8 are blue. So, P ( both socks blue ) = 3 C 2 8 C 2 = 3 28 . Similarly, P ( both socks black ) = 5 C 2 8 C 2 = 10 28 . Therefore, since these events are mutually exclusive, we can use the Addition Rule: P ( M ) = P ( both socks blue ) + P ( both socks black ) = 3 28 + 10 28 = 13 28 . Step 2. If the sock on Jim’s left foot is black (i.e., K occurred), then there are 4 remaining black socks of the 7 in the drawer. So, P ( M | K ) = 4 7 . Step 3. If the sock on Jim’s left foot is blue ( L occurred), then there are 2 blue socks among the 7 remaining in the drawer. So, P ( M | L ) = 2 7 . In Tree Diagrams, Tables, and Outcomes , we introduced the concept of dependence between stages of a multistage experiment. We stated at the time that two stages were dependent if the result of one stage affects the other stage. We explained that dependence in terms of the sample space, but sometimes that dependence can be a little more subtle; it’s more properly understood in terms of conditional probabilities. Two stages of an experiment are dependent if P ( E | F ) ≠ P ( E | F ′ ) for some outcome of the second stage E and outcome of the first stage F . Protecting Bombers in World War II In his book How Not to Be Wrong , Jordan Ellenberg recounts this anecdote: During World War II, the American military wanted to add additional armor plating to bomber aircraft, in order to reduce the chances that they get shot down. So, they collected data on planes after returning from missions. The data showed that the fuselage, wings, and fuel system had many more bullet holes (per unit area) than the engine compartments, so the military brass wanted to add additional armor to the parts of the plane that were hit most often. Luckily, before they added the armor to the planes, they asked for a second opinion. Abraham Wald, a Jewish mathematician who had fled the rising Nazi regime, pointed out that it was far more important that the armor plating be added to areas where there were fewer bullet holes. Why? The planes they were studying had already completed their missions, so the military was essentially looking at conditional probabilities: the probability of suffering a bullet strike, given that the plane made it back safely. More bullet holes in an area on the plane indicated that was a region that wasn’t as important for the plane’s survival! Compound Events Using “And” and the Multiplication Rule For multistage experiments, the outcomes of the experiment as a whole are often stated in terms of the outcomes of the individual stages. Commonly, those statements are joined with “and.” For example, in the sock drawer example just above, one outcome might be “the left sock is black and the right sock is blue.” As with “or” compound events, these probabilities can be computed with basic arithmetic. Multiplication Rule for Probability: If E and F are events associated with the first and second stages of an experiment, then P ( E and ⁢ F ) = P ( E ) × P ( F | E ) . In The Addition Rule for Probability , we considered probabilities of events connected with “and” in the statement of the Inclusion/Exclusion Principle. These two scenarios are different; in the statement of the Inclusion/Exclusion Principle, the events connected with “and” are both events associated with the same single-stage experiment (or the same stage of a multistage experiment). In the Multiplication Rule, we’re looking at events associated with different stages of a multistage experiment. Using the Multiplication Rule for Probability You are president of a club with 10 members: 4 seniors, 3 juniors, 2 sophomores, and 1 first-year. You need to choose 2 members to represent the club on 2 college committees. The first person selected will be on the Club Awards Committee and the second will be on the New Club Orientation Committee. The same person cannot be selected for both. You decide to select these representatives at random. What is the probability that a senior is chosen for both positions? What is the probability that a junior is chosen first and a sophomore is chosen second? What is the probability that a sophomore is chosen first and a senior is chosen second? We need the probability that a senior is chosen first and a senior is chosen second. These are two stages of a multistage experiment, so we’ll apply the Multiplication Rule for Probability: P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . Since there are 4 seniors among the 10 members, P ( senior chosen first ) = 4 10 = 2 5 . Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 . Putting this all together, we get P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 . There are 3 juniors among the 10 members, so P ( junior chosen first ) = 3 10 . Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so P ( sophomore chosen second | junior chosen first ) = 2 9 . Thus, using the Multiplication Rule for Probability, we have P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 . The probability that a sophomore is chosen first is 2 10 = 1 5 , and the probability that a senior is chosen second given that a sophomore was chosen first is 4 9 . Thus, using the Multiplication Rule for Probability, we have: P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 . The Birthday Problem One of the most famous problems in probability theory is the Birthday Problem, which has to do with shared birthdays in a large group. To make the analysis easier, we’ll ignore leap days, and assume that the probability of being born on any given date is 1 365 . Now, if you have 366 people in a room, we’re guaranteed to have at least one pair of people who share a single birthday. Imagine filling the room by first admitting someone born on January 1, then someone born on January 2, and so on… The 365th person admitted would be born on December 31. If you add one more person to the room, that person’s birthday would have to match someone else’s. Let’s look at the other end of the spectrum. If you choose two people at random, what is the probability that they share a birthday? As with many probability questions, this is best addressed by find out the probability that they do not share a birthday. The first person’s birthday can be anything (probability 1), and the second person’s birthday can be anything other than the first person’s birthday (probability 364 365 ). The probability that they have different birthdays is 1 × 364 365 = 364 365 . So, the probability that they share a birthday is 1 − 364 365 = 1 365 . What if we have three people? The probability that they all have different birthdays can be obtained by extending our previous calculation: The probability that two people have different birthdays is 364 365 , so if we add a third to the mix, the probability that they have a different birthday from the other two is 363 365 . So, the probability that all three have different birthdays is 364 365 × 363 365 ≈ 0.9918 , and thus the probability that there’s a shared birthday in the group is 1 − 0.9918 ≈ 0.0082 . The big question is this: How many people do we need in the room to have the probability of a shared birthday greater than 1 2 ? Make a guess, then with a partner keep adding hypothetical people to the group and computing probabilities until you get there! It is often useful to combine the rules we’ve seen so far with the techniques we used for finding sample spaces. In particular, trees can be helpful when we want to identify the probabilities of every possible outcome in a multistage experiment. The next example will illustrate this. Using Tree Diagrams to Help Find Probabilities The board game Clue uses a deck of 21 cards: 6 suspects, 6 weapons, and 9 rooms. Suppose you are about to draw 2 cards from this deck. There are 6 possible outcomes for the draw: 2 suspects, 2 weapons, 2 rooms, 1 suspect and 1 weapon, 1 suspect and 1 room, or 1 weapon and 1 room. What are the probabilities for each of these outcomes? Step 1: Let’s start by building a tree diagram that illustrates both stages of this experiment. Let’s use S, W, and R to indicate drawing a suspect, weapon, and room, respectively ( ). Step 2: We want to start computing probabilities, starting with the first stage. The probability that the first card is a suspect is 6 21 = 2 7 . The probability that the first card is a weapon is the same: 2 7 . Finally, the probability that the first card is a room is 9 21 = 3 7 . Step 3: Let’s incorporate those probabilities into our tree: label the edges going into each of the nodes representing the first-stage outcomes with the corresponding probabilities ( ). Note that the sum of the probabilities coming out of the initial node is 1; this should always be the case for the probabilities coming out of any node! Step 4: Let’s look at the case where the first card is a suspect. There are 3 edges emanating from that node (leading to the outcomes SS, SW, and SR). We’ll label those edges with the appropriate conditional probabilities, under the assumption that the first card is a suspect. First, there are 5 remaining suspect cards among the 20 left in the deck, so P ( second is suspect | first is suspect ) = 5 20 = 1 4 . Using similar reasoning, we can compute P ( second is weapon | first is suspect ) = 6 20 = 3 10 and P ( second is room | first is suspect ) = 9 20 . Step 5: Checking our work, we see that the sum of these 3 probabilities is again equal to 1. Let’s add those to our tree ( ). Step 6: Let’s continue filling in the conditional probabilities at the other nodes, always checking to make sure the sum of the probabilities coming out of any node is equal to 1 ( ). Step 7: We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled “SW”) is 2 7 × 3 10 = 3 35 , as illustrated in . Step 8: Let’s fill in the rest of the probabilities ( ). Step 9: A helpful feature of tree diagrams is that the final outcomes are always mutually exclusive, so the Addition Rule can be directly applied. For example, the probability of drawing one suspect and one room (in any order) would be P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 . We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table: Outcome Probability 2 suspects 1 14 2 weapons 1 14 2 rooms 6 35 1 suspect and 1 weapon 3 35 + 3 35 = 6 35 1 suspect and 1 room 9 70 + 9 70 = 9 35 1 weapon and 1 room 9 70 + 9 70 = 9 35 Checking once again, the sum of these 6 probabilities is 1, as expected. The Monty Hall Problem On the original version of the game show Let’s Make a Deal , originally hosted by Monty Hall and now hosted by Wayne Brady, one contestant was chosen to play a game for the grand prize of the day (often a car). Here’s how it worked: On the stage were three areas concealed by numbered curtains. The car was hidden behind one of the curtains; the other two curtains hid worthless prizes (called “Zonks” on the show). The contestant would guess which curtain concealed the car. To build tension, Monty would then reveal what was behind one of the other curtains, which was always one of the Zonks (Since Monty knew where the car was hidden, he always had at least one Zonk curtain that hadn’t been chosen that he could reveal). Monty then turned to the contestant and asked: “Do you want to stick with your original choice, or do you want to switch your choice to the other curtain?” What should the contestant do? Does it matter? With a partner or in a small group, simulate this game. You can do that with a small candy (the prize) hidden under one of three cups, or with three playing cards (just decide ahead of time which card represents the “Grand Prize”). One person plays the host, who knows where the prize is hidden. Another person plays the contestant and tries to guess where the prize is hidden. After the guess is made, the host should reveal a losing option that wasn’t chosen by the contestant. The contestant then has the option to stick with the original choice or switch to the other, unrevealed option. Play about 20 rounds, taking turns in each role and making sure that both contestant strategies (stick or switch) are used equally often. After each round, make a note of whether the contestant chose “stick” or “switch” and whether the contestant won or lost. Find the empirical probability of winning under each strategy. Then, see if you can use tree diagrams to verify your findings. Check Your Understanding Key Terms conditional probability Key Concepts Conditional probabilities are computed under the assumption that the condition has already occurred. The Multiplication Rule for Probability is used to find the probability that two events occur in sequence. Formulas If E and F are events associated with the first and second stages of an experiment, then P ( E and ⁢ F ) = P ( E ) × P ( F | E ) .", "section": "Conditional Probability and the Multiplication Rule", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Binomial Distribution If one baseball team has a 65% chance of beating another in any single game, what’s the likelihood that they win a best-of-seven series? (credit: “baseball game” by Britt Reints/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify binomial experiments. Use the binomial distribution to analyze binomial experiments. It’s time for the World Series, which determines the champion for this season in Major League Baseball. The scrappy Los Angeles Angels are facing the powerhouse Cincinnati Reds. Computer models put the chances of the Reds winning any single game against the Angels at about 65%. The World Series, as its name implies, isn’t just one game, though: it’s what’s known as a “best-of-seven” contest: the teams play each other repeatedly until one team wins 4 games (which could take up to 7 games total, if each team wins three of the first 6 matchups). If the Reds truly have a 65% chance of winning a single game, then the probability that they win the series should be greater than 65%. Exactly how much bigger? If you have the patience for it, you could use a tree diagram like we used in to trace out all of the possible outcomes, find all the related probabilities, and add up the ones that result in the Reds winning the series. Such a tree diagram would have 2 7 = 128 final nodes, though, so the calculations would be very tedious. Fortunately, we have tools at our disposal that allow us to find these probabilities very quickly. This section will introduce those tools and explain their use. Binomial Experiments The tools of this section apply to multistage experiments that satisfy some pretty specific criteria. Before we move on to the analysis, we need to introduce and explain those criteria so that we can recognize experiments that fall into this category. Experiments that satisfy each of these criteria are called binomial experiments . A binomial experiment is an experiment with a fixed number of repeated independent binomial trials, where each trial has the same probability of success. Repeated Binomial Trials The first criterion involves the structure of the stages. Each stage of the experiment should be a replication of every other stage; we call these replications trials . An example of this is flipping a coin 10 times; each of the ten flips is a trial, and they all occur under the same conditions as every other. Further, each trial must have only two possible outcomes. These two outcomes are typically labeled “success” and “failure,” even if there is not a positive or negative connotation associated with those outcomes. Experiments with more than two outcomes in their sample spaces are sometimes reconsidered in a way that forces just two outcomes; all we need to do is completely divide the sample space into two parts that we can label “success” and “failure.” For example, your grade on an exam might be recorded as A, B, C, D, or F, but we could instead think of the grades A, B, C, and D as “success” and a grade of F as “failure.” Trials with only two outcomes are called binomial trials (the word binomial derives from Latin and Greek roots that mean “two parts”). Independent Trials The next criterion that we’ll be looking for is independence of trials. Back in Tree Diagrams, Tables, and Outcomes , we said that two stages of an experiment are independent if the outcome of one stage doesn’t affect the other stage. Independence is necessary for the experiments we want to analyze in this section. Fixed Number of Trials Next, we require that the number of trials in the experiment be decided before the experiment begins. For example, we might say “flip a coin 10 times.” The number of trials there is fixed at 10. However, if we say “flip a coin until you get 5 heads,” then the number of trials could be as low as 5, but theoretically it could be 50 or a 100 (or more)! We can’t apply the tools from this section in cases where the number of trials is indeterminate. Constant Probability The next criterion needed for binomial experiments is related to the independence of the trials. We must make sure that the probability of success in each trial is the same as the probability of success in every other trial. Identifying Binomial Experiments Decide whether each of the following is a binomial experiment. For those that aren’t, identify which criterion or criteria are violated. You roll a standard 6-sided die 10 times and write down the number that appears each time. You roll a standard 6-sided die 10 times and write down whether the die shows a 6 or not. You roll a standard 6-sided die until you get a 6. You roll a standard 6-sided die 10 times. On the first roll, we define “success” as rolling a 4 or greater. After the first roll, we define “success” as rolling a number greater than the result of the previous roll. Since we’re noting 1 of 6 possible outcomes, the trials are not binomial. So, this isn’t a binomial experiment. We have 2 possible outcomes (“6” and “not 6”), the trials are independent, the probability of success is the same every time, and the number of trials is fixed. This is a binomial experiment. Since the number of trials isn’t fixed (we don’t know if we’ll get our first 6 after 1 roll or 20 rolls or somewhere in between), this isn’t a binomial experiment. Here, the probability of success might change with every roll (on the first roll, that probability is 1 2 ; if the first roll is a 6, the probability of success on the next roll is zero). So, this is not a binomial experiment. The Binomial Formula If we flip a coin 100 times, you might expect the number of heads to be around 50, but you probably wouldn’t be surprised if the actual number of heads was 47 or 52. What is the probability that the number of heads is exactly 50? Or falls between 45 and 55? It seems unlikely that we would get more than 70 heads. Exactly how unlikely is that? Each of these questions is a question about the number of successes in a binomial experiment (flip a coin 100 times, “success” is flipping heads). We could theoretically use the techniques we’ve seen in earlier sections to answer each of these, but the number of calculations we’d have to do is astronomical; just building the tree diagram that represents this situation is more than we could complete in a lifetime; it would have 2 100 ≈ 1.3 × 10 30 final nodes! To put that number in perspective, if we could draw 1,000 dots every second, and we started at the moment of the Big Bang, we’d currently be about 0.00000003% of the way to drawing out those final nodes. Luckily, there’s a shortcut called the Binomial Formula that allows us to get around doing all those calculations! Binomial Formula: Suppose we have a binomial experiment with n trials and the probability of success in each trial is p . Then: P ( number of successes is a ) = C n a × p a × ( 1 − p ) n − a . We can use this formula to answer one of our questions about 100 coin flips. What is the probability of flipping exactly 50 heads? In this case, n = 100 , p = 1 2 , and a = 50 , so P ( flip 50 heads ) = C 100 50 × ( 1 2 ) 50 × ( 1 − 1 2 ) 100 − 50 . Unfortunately, many calculators will balk at this calculation; that first factor ( 100 C 50 ) is an enormous number, and the other two factors are very close to zero. Even if your calculator can handle numbers that large or small, the arithmetic can create serious errors in rounding off. Luckily, spreadsheet programs have alternate methods for doing this calculation. In Google Sheets, we can use the BINOMDIST function to do this calculation for us. Open up a new sheet, click in any empty cell, and type “=BINOMDIST(50,100,0.5,FALSE)” followed by the Enter key. The cell will display the probability we seek; it’s about 8%. Let’s break down the syntax of that function in Google Sheets: enter “=BINOMDIST( a , n , p , FALSE)” to find the probability of a successes in n trials with probability of success p . Using the Binomial Formula Find the probability of rolling a standard 6-sided die 4 times and getting exactly one 6 without using technology . Find the probability of rolling a standard 6-sided die 60 times and getting exactly ten 6s using technology. Find the probability of rolling a standard 6-sided die 60 times and getting exactly eight 6s using technology. We’ll apply the Binomial Formula, where n = 4 , a = 1 , and p = 1 6 : P ( rolling one 6 ) = C 1 4 × ( 1 6 ) 1 × ( 5 6 ) 4 − 1 = 4 ! 1 ! ( 4 − 1 ) ! × 1 6 × ( 5 6 ) 3 = 4 × 1 6 × 5 3 6 3 = 4 × 5 3 6 4 = 500 1,296 . Here, n = 60 , a = 10 , and p = 1 6 . In Google Sheets, we’ll enter “=BINOMDIST(10, 60, 1/6, FALSE)” to get our result: 0.137. This experiment is the same as in Exercise 2 of this example; we’re simply changing the number of successes from 10 to 8. Making that change in the formula in Google Sheets, we get the probability 0.116. The Binomial Distribution If we are interested in the probability of more than just a single outcome in a binomial experiment, it’s helpful to think of the Binomial Formula as a function, whose input is the number of successes and whose output is the probability of observing that many successes. Generally, for a small number of trials, we’ll give that function in table form, with a complete list of the possible outcomes in one column and the probability in the other. For example, suppose Kristen is practicing her basketball free throws. Assume Kristen always makes 82% of those shots. If she attempts 5 free throws, then the Binomial Formula gives us these probabilities: Shots Made Probability 0 0.000189 1 0.004304 2 0.0392144 3 0.1786432 4 0.4069096 5 0.3707398 A table that lists all possible outcomes of an experiment along with the probabilities of those outcomes is an example of a probability density function (PDF). A PDF may also be a formula that you can use to find the probability of any outcome of an experiment. Because they refer to the same thing, some sources will refer to the Binomial Formula as the Binomial PDF. If we want to know the probability of a range of outcomes, we could add up the corresponding probabilities. Going back to Kristen’s free throws, we can find the probability that she makes 3 or fewer of her 5 attempts by adding up the probabilities associated with the corresponding outcomes (in this case: 0, 1, 2, or 3): P ( makes 3 or fewer ) = P ( a = 0 ) + P ( a = 1 ) + P ( a = 2 ) + P ( a = 3 ) = 0.000189 + 0.004304 + 0.0392144 + 0.1786432 = 0.2223506 The probability that the outcome of an experiment is less than or equal to a given number is called a cumulative probability . A table of the cumulative probabilities of all possible outcomes of an experiment is an example of a cumulative distribution function (CDF). A CDF may also be a formula that you can use to find those cumulative probabilities. Cumulative probabilities are always associated with events that are defined using ≤ . If other inequalities are used to define the event, we must restate the definition so that it uses the correct inequality. Here are the PDF and CDF for Kristen’s free throws: Shots Made Probability Cumulative 0 0.000189 0.000189 1 0.004304 0.004493 2 0.0392144 0.0437073 3 0.1786432 0.2223506 4 0.4069096 0.6292602 5 0.3707398 1 Google Sheets can also compute cumulative probabilities for us; all we need to do is change the “FALSE” in the formulas we used before to \"TRUE.\" Using the Binomial CDF Suppose we are about to flip a fair coin 50 times. Let H represent the number of heads that result from those flips. Use technology to find the following: P ( H ≤ 22 ) P ( H < 26 ) P ( H > 28 ) P ( H ≥ 20 ) P ( 20 < H < 25 ) The event here is defined by H ≤ 22 , which is the inequality we need to have if we want to use the Binomial CDF. In Google Sheets, we’ll enter “=BINOMDIST(22, 50, 0.5, TRUE)” to get our answer: 0.2399. This event uses the wrong inequality, so we need to do some preliminary work. If H < 26 , that means H ≤ 25 (because H has to be a whole number). So, we’ll enter “=BINOMDIST(25, 50, 0.5, TRUE)” to find P ( H < 26 ) = P ( H ≤ 25 ) = 0.5561 . The inequality associated with this event is pointing in the wrong direction. If E is the event H > 28 , that means that E contains the outcomes {29, 30, 31, 32, 33, …}. Thus, E ′ must contain the outcomes {…, 25, 26, 27, 28}. In other words, E ′ is defined by H ≤ 28 . Since it uses ≤ , we can find P ( E ′ ) using “=BINOMDIST(28, 50, 0.5, TRUE)”: 0.8389 So, using the formula for probabilities of complements, we have P ( E ) = 1 − P ( E ′ ) = 1 − 0.8389 = 0.1611. As in part 3, this inequality is pointing in the wrong direction. If F is the event H ≥ 20 , then F contains the outcomes {20, 21, 22, 23, …}. That means F ′ contains the outcomes {…, 16, 17, 18, 19}, and so F ′ is defined by H ≤ 19 . So, we can find P ( F ′ ) using “=BINOMDIST(19, 50, 0.5, TRUE)”: 0.0595. Finally, using the formula for probabilities of complements, we get: P ( F ) = 1 − P ( F ′ ) = 1 − 0.0595 = 0.9405. If 20 < H < 25 , that means we are interested in the outcomes {21, 22, 23, 24}. This doesn’t look like any of the previous situations, but there is a way to find this probability using the Binomial CDF. We need to put everything in terms of “less than or equal to,” so we’ll first note that all of our outcomes are less than or equal to 24. But we don’t want to include values that are less than or equal to 20. So, we have three events: let I be the event defined by 20 < H < 25 (note that we’re trying to find P ( I ) ). Let J be defined by H ≤ 24 , and let K be defined by H ≤ 20 . Of these three events, J contains the most outcomes. If J occurs, then either K or I must have occurred. Moreover, K and I are mutually exclusive. Thus, P ( J ) = P ( K ) + P ( I ) , by the Addition Rule. Solving for the probability that we want, we get P ( I ) = P ( J ) − P ( K ) = P ( H ≤ 24 ) − P ( H ≤ 20 ) = 0.44386 − 0.10132 = 0.34254. Finally, we can answer the question posed at the beginning of this section . Remember that the Reds are facing the Angels in the World Series, which is won by the team who is first to win 4 games. The Reds have a 65% chance to win any game against the Angels. So, what is the probability that the Reds win the World Series? At first glance, this is not a binomial experiment: The number of games played is not fixed, since the series ends as soon as one team wins 4 games. However, we can extend this situation to a binomial experiment: Let’s assume that 7 games are always played in the World Series, and the winner is the team who wins more games. In a way, this is what happens in reality; it’s as though the first team to lose 4 games (and thus cannot win more than the other team) forfeits the rest of their games. So, we can treat the actual World Series as a binomial experiment with seven trials. If W is the number of games won by the Reds, the probability that the Reds win the World Series is P ( W ≥ 4 ) . Using the techniques from the last example, we get P ( Reds win the series ) = 0.8002 . Abraham de Moivre Abraham de Moivre was born in 1667 in France to a Protestant family. Though he was educated in Catholic schools, he remained true to his faith; in 1687, he fled with his brother to London to escape persecution under the reign of King Louis XIV. Once he arrived in England, he supported himself as a freelance math tutor while he conducted his own research. Among his interests was probability; in 1711, he published the first edition of The Doctrine of Chances: A Method of Calculating the Probabilities of Events in Play . This book was the second textbook on probability (after Cardano’s Liber de ludo aleae ). De Moivre discovered an important connection between the binomial distribution and the normal distribution (an important concept in statistics; we’ll explore that distribution and its connection to the binomial distribution in Chapter 8). De Moivre also discovered some properties of a new probability distribution that later became known as the Poisson distribution. Check Your Understanding Key Terms binomial experiment probability density function (PDF) cumulative distribution function (CDF) Key Concepts Binomial experiments result when we count the number of successful outcomes in a fixed number of repeated, independent trials with a constant probability of success. The binomial distribution is used to find probabilities associated with binomial experiments. Probability density functions (PDFs) describe the probabilities of individual outcomes in an experiment; cumulative distribution functions (CDFs) give the probabilities of ranges of outcomes. Formulas Suppose we have a binomial experiment with n trials and the probability of success in each trial is p . Then: P ( number of successes is a ) = C n a × p a × ( 1 − p ) n − a", "section": "The Binomial Distribution", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Expected Value The concept of expected value allows us to analyze games that involve randomness, like Roulette. (credit: “Roulette Table and Roulette Wheel in a Casino with People betting on numbers” by Marco Verch/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the expected value of an experiment. Interpret the expected value of an experiment. Use expected value to analyze applications. The casino game roulette has dozens of different bets that can be made. These bets have different probabilities of winning but also have different payouts. In general, the lower the probability of winning a bet is, the more money a player wins for that bet. With so many options, is there one bet that’s “smarter” than the rest? What’s the best play to make at a roulette table? In this section, we’ll develop the tools we need to answer these questions. Expected Value Many experiments have numbers associated with their outcomes. Some are easy to define; if you roll 2 dice, the sum of the numbers showing is a good example. In some card games, cards have different point values associated with them; for example, in some forms of the game rummy, aces are worth 15 points; 10s, jacks, queens, and kings are worth 10; and all other cards are worth 5. The outcomes of casino and lottery games are all associated with an amount of money won or lost. These outcome values are used to find the expected value of an experiment: the mean of the values associated with the outcomes that we would observe over a large number of repetitions of the experiment. (See Conditional Probability and the Multiplication Rule for more on means.) That definition is a little vague; How many is “a large number?” In practice, it depends on the experiment; the number has to be large enough that every outcome would be expected to appear at least a few times. For example, if we’re talking about rolling a standard 6-sided die and we note the number showing, a few dozen replications should be enough that the mean would be representative. Since the probability of each outcome is 1 6 , we would expect to see each outcome about 8 times over the course of 48 replications. However, if we’re talking about the Powerball lottery, where the probability of winning the jackpot is about 1 292 , 000 , 000 , we would need several billion replications to ensure that every outcome appears a few times. Luckily, we can find the theoretical expected value before we even run the experiment the first time. Expected Value: If O represents an outcome of an experiment and n ( O ) represents the value of that outcome, then the expected value of the experiment is: ∑ n ( O ) × P ( O ) , where Σ is the “sum,” meaning we add up the results of the formula that follows over all possible outcomes. Finding Expected Values Find the expected values of the following experiments. Roll a standard 6-sided die and note the number showing. Roll two standard 6-sided dice and note the sum of the numbers showing. Draw a card from a well-shuffled standard deck of cards and note its rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else). Step 1: Let’s start by writing out the PDF table for this experiment. Value Probability 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 Step 2: To find the expected value, we need to find n ( O ) × P ( O ) for each possible outcome in the table below. Value Probability n ( O ) × P ( O ) 1 1 6 1 × 1 6 = 1 6 2 1 6 2 × 1 6 = 1 3 3 1 6 3 × 1 6 = 1 2 4 1 6 4 × 1 6 = 2 3 5 1 6 5 × 1 6 = 5 6 6 1 6 6 × 1 6 = 1 Step 3: We add all of the values in that last column: 1 6 + 1 3 + 1 2 + 2 3 + 5 6 + 1 = 7 2 = 3.5 . So, the expected value of a single roll of a die is 3.5. Back in , we made this table of all of the equally likely outcomes ( ): Step 1: Let’s use to create the PDF for this experiment, as shown in the following table: Value Probability 2 1 36 3 1 18 4 1 12 5 1 9 6 5 36 7 1 6 8 5 36 9 1 9 10 1 12 11 1 18 12 1 36 Step 2: We can multiply each row to find n ( O ) × P ( O ) as shown in the following table: Value Probability n ( O ) × P ( O ) 2 1 36 1 18 3 1 18 1 6 4 1 12 1 3 5 1 9 5 9 6 5 36 5 6 7 1 6 7 6 8 5 36 10 9 9 1 9 1 10 1 12 5 6 11 1 18 11 18 12 1 36 1 3 Step 3: We can add the last column to get the expected value: 1 18 + 1 6 + 1 3 + 5 9 + 5 6 + 7 6 + 10 9 + 1 + 5 6 + 11 18 + 1 3 = 7 . So, the expected value is 7. Step 1: Let’s make a PDF table for this experiment. There are 3 events that we care about, so let’s use those events in the table below: Event Probability {A} 1 13 {10, J, Q, K} 4 13 {2, 3, 4, 5, 6, 7, 8, 9} 8 13 Step 2: Let’s add a column to the following table for the values of each event: Event Probability Value {A} 1 13 15 {10, J, Q, K} 4 13 10 {2, 3, 4, 5, 6, 7, 8, 9} 8 13 5 Step 3: We’ll add the column for the product of the values and probabilities to the table below: Event Probability Value n ( O ) × P ( O ) {A} 1 13 15 15 13 {10, J, Q, K} 4 13 10 40 13 {2, 3, 4, 5, 6, 7, 8, 9} 8 13 5 40 13 Step 4: We’ll find the sum of the last column: 15 13 + 40 13 + 40 13 = 95 13 ≈ 7.3 . Thus, the expected Rummy value of a randomly selected card is about 7.3. Let’s make note of some things we can learn from . First, as Exercises 1 and 3 demonstrate, the expected value of an experiment might not be a value that could come up in the experiment. Remember that the expected value is interpreted as a mean, and the mean of a collection of numbers doesn’t have to actually be one of those numbers. Second, looking at Exercise 1, the expected value (3.5) was just the mean of the numbers on the faces of the die: 1 + 2 + 3 + 4 + 5 + 6 6 = 3.5 . This is no accident! If we break that fraction up using the addition in the numerator, we get 1 6 + 2 6 + 3 6 + 4 6 + 5 6 + 6 6 , which we can rewrite as 1 × 1 6 + 2 × 1 6 + 3 × 1 6 + 4 × 1 6 + 5 × 1 6 + 6 × 1 6 . That’s exactly the computation we did to find the expected value! In fact, expected values can always be treated as a special kind of mean called a weighted mean , where the weights are the probabilities associated with each value. When the probabilities are all equal, the weighted mean is just the regular mean. Interpreting Expected Values As we noted, the expected value of an experiment is the mean of the values we would observe if we repeated the experiment a large number of times. (This interpretation is due to an important theorem in the theory of probability called the Law of Large Numbers.) Let’s use that to interpret the results of the previous example . Interpreting Expected Values Interpret the expected values of the following experiments. Roll a standard 6-sided die and note the number showing. Roll 2 standard 6-sided dice and note the sum of the numbers showing. Draw a card from a well-shuffled standard deck of cards and note its Rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else). If you roll a standard 6-sided die many times, the mean of the numbers you roll will be around 3.5. If you roll a pair of standard 6-sided dice many times, the mean of the sums of the numbers you roll will be about 7. If you draw a card from a well-shuffled deck many times, the mean of the Rummy values of the cards would be around 7.3. Pascal’s Wager The French scholar Blaise Pascal (1623–1662) was among the earliest mathematicians to study probabilities, and was the first to accurately describe and compute expected values. In his book Pensées ( Thoughts ), he turned the analysis of expected values to his belief in the Christian God. He said that there is no way for people to establish the probability that God exists, but since the “winnings” on a bet that God exists (and that you then lead your life accordingly) are essentially infinite, the expected value of taking that bet is always positive, no matter how unlikely it is that God exists. Using Expected Value Now that we know how to find and interpret expected values, we can turn our attention to using them. Suppose someone offers to play a game with you. If you roll a die and get a 6, you get $10. However, if you get a 5 or below, you lose $1. Is this a game you’d want to play? Let’s look at the expected value: The probability of winning is 1 6 and the probability of losing is 5 6 , so the expected value is $ 10 × 1 6 + ( − $ 1 ) × 5 6 = 5 6 ≈ $ 0.83 . That means, on average, you’ll come out ahead by about 83 cents every time you play this game. It’s a great deal! On the other hand, if the winnings for rolling a 6 drop to $3, the expected value becomes $ 3 × 1 6 + ( − $ 1 ) × 5 6 = − 1 3 ≈ − $ 0.33 , meaning you should expect to lose about 33 cents on average for every time you play. Playing that game is not a good idea! In general, this is how casinos and lottery corporations make money: Every game has a negative expected value for the player. Expected Values in Football In the 21st century, data analytics tools have revolutionized the way sports are coached and played. One tool in particular is used in football at crucial moments in the game. When a team faces a fourth down (the last possession in a series of four possessions, a fairly common occurrence), the coach faces a decision: Run one play to try to gain a certain number of yards, or kick the ball away to the other team. Here’s the interesting part of the decision: If the team “goes for it” and runs the play and they are successful, then they keep possession of the ball and can continue in their quest to score more points. If they are unsuccessful, then they lose possession of the ball, giving the other team an opportunity to score points. If, instead, the team punts, or kicks the ball away, then the other team gets possession of the ball, but in a worse position for them than if the original team goes for it and fails. To analyze this situation, data analysts have generated empirical probabilities for every fourth down situation, and computed the expected value (in terms of points) for each decision. Coaches frequently use those calculations when they decide which option to take! Pierre de Fermat and Blaise Pascal In 1654, a French writer and amateur mathematician named Antoine Gombaud (who called himself the Chevalier du Mére) reached out to his gambling buddy Blaise Pascal to answer a question that he’d read about called the “problem of points.” The question goes like this: Suppose you’re playing a game that is scored using points, and the first person to earn 5 points is the winner. The game is interrupted with the score 4 points to 2. If the winner stood to win $100, how should the prize money be divided between the players? Certainly the person who is 1 point away from victory should get more, but how much more? We have developed tools in this section to answer this question. At its heart, it’s a question about conditional probabilities and expected value. At the time that Pascal first started thinking about it, though, those ideas hadn’t yet been invented. Pascal reached out to a colleague named Pierre de Fermat, and over the course of a couple of months, their correspondence with each other would eventually solve the problem. In the process, they first described conditional probabilities and expected values! Apart from their work in probability, these men are famous for other work in mathematics (and, in Pascal’s case, philosophy and physics). Fermat is remembered for his work in geometry and in number theory. After his death, the statement of what came to be called “Fermat’s Last Theorem” was discovered scribbled in the margin of a book, with the note that Fermat had discovered a “marvelous proof that this margin is too small to contain.” The theorem says that any equation of the form a n + b n = c n has no positive integer solutions if n ≥ 3 . No proof of that theorem was discovered until 1994, when Andrew Wiles used computers and new branches of geometry to finally prove the theorem! Pascal is remembered for the “arithmetical triangle” that is named for him (though he wasn’t the first person to discover it; see the section on the binomial distribution for more), as well as work in geometry. In physics, Pascal worked on hydrodynamics and air pressure (the SI unit for pressure is named for him), and in philosophy, Pascal advocated for a mathematical approach to philosophical problems. Using Expected Values In the casino game keno, a machine chooses at random 20 numbers between 1 and 80 (inclusive) without replacement. Players try to predict which numbers will be chosen. Players don’t try to guess all 20, though; generally, they’ll try to predict between 1 and 10 of the chosen numbers. The amount won depends on the number of guesses they made and the number of guesses that were correct. At one casino, a player can try to guess just 1 number. If that number is among the 20 selected, the player wins $2; otherwise, the player loses $1. What is the expected value? At the same casino, if a player makes 2 guesses and they’re both correct, the player wins $14; otherwise, the player loses $1. What is the expected value? Players can also make 3 guesses. If 2 of the 3 guesses are correct, the player wins $1. If all 3 guesses are correct, the player wins $42. Otherwise, the player loses $1. What is the expected value? Which of these games is the best for the player? Which is the best for the casino? There are 20 winning numbers out of 80, so if we try to guess one of them, the probability of guessing correctly is 20 80 = 1 4 . The probability of losing is then 3 4 , and so the expected value is $ 2 × 1 4 + ( − $ 1 ) × 3 4 = − $ 0.25 . There are 20 C 2 = 190 winning choices out of 80 C 2 = 3 , 160 total ways to choose 2 numbers. So, the probability of winning is 190 3 , 160 and the probability of losing is 3 , 160 − 190 3 , 160 = 2 , 970 3 , 160 . So, the expected value of the game is $ 14 × 190 3 , 160 + ( − $ 1 ) × 2970 3160 ≈ − $ 0.10 . Step 1: Let’s start with the big prize. There are 20 C 3 = 1 , 140 ways to correctly guess 3 winning numbers out of 80 C 3 = 82 , 160 ways to guess three numbers total. That means the probability of winning the big prize is 1 , 140 82 , 160 ≈ 0.01388 . Step 2: Let’s find the probability of the second prize. The denominator is the same: 82,160. Let’s figure out the numerator. To win the second prize, the player must pick 2 of the 20 winning numbers and one of the 60 losing numbers. The number of ways to do that can be found using the Multiplication Rule for Counting: there are 20 C 2 = 190 ways to pick 2 winning numbers and 60 ways to pick 1 losing number, so there are 190 × 60 = 11 , 400 ways to win the second prize. So, the probability of winning that second prize is 11 , 400 82 , 160 ≈ 0.13875 . Step 3: Since the overall probability of winning is 1 , 140 82 , 160 + 11 , 400 82 , 160 = 12 , 540 82 , 160 ≈ 0.15263 , the probability of losing must be 1 − 0.15263 = 0.84737 . So, the expected value is $ 42 × 0.01388 + $ 1 × 0.13875 + ( − $ 1 ) × 0.84737 ≈ − $ 0.13 . The bet that’s the best for the player is the one with the highest expected value for the player, which is guessing two numbers. The best one for the casino is the one with the lowest expected value for the player, which is guessing one number. Make Your Own Lottery By yourself or with a partner, devise your own lottery scheme. Assume you would have access to one or more machines that choose numbers randomly. What will a lottery draw look like? How many numbers are players choosing from? How many will be drawn? Will they be drawn with replacement or without replacement? What conditions must be met for a player to win first or second (or more!) prize? Once you’ve decided that, decide the payoff structure for winners, and how much the game will cost to play. Try to make the game enticing enough that people will want to play it, but with enough negative expected value that the lottery will make money. Aim for the expected value to be about −0.25 times the cost of playing the game. Check Your Understanding Key Terms expected value Key Concepts The expected value of an experiment is the sum of the products of the numerical outcomes of an experiment with their corresponding probabilities. The expected value of an experiment is the most likely value of the average of a large number of replications of the experiment. Formulas If O represents an outcome of an experiment and n ( O ) represents the value of that outcome, then the expected value of the experiment is: ∑ n ( O ) × P ( O ) where Σ stands for the sum, meaning we add up the results of the formula that follows over all possible outcomes. Projects The Binomial Distribution is one of many examples of a discrete probability distribution . Other examples include the Geometric, Hypergeometric, Multinomial, Poisson, and Negative Binomial Distributions. Choose one of these distributions, and find out what makes it different from the Binomial Distribution. In what situations can it be applied? How is it used? Once you have an idea of how it’s used, write a series of five questions like the ones in this chapter that can be answered with that distribution, and find the answers. Binomial is a word that also comes up in algebra; the word describes polynomials with two terms. At first glance, there isn’t much to indicate that these two uses of the word are related, but it turns out there is a connection. Explore the connection between the Binomial Distribution and the algebraic concept of binomial expansion, (the process of multiplying out expressions like ( x + y ) n for a positive whole number n ). Search for a connection with the mathematical object known as Pascal’s Triangle. Hazard is a dice game that was mentioned in Chaucer’s Canterbury Tales . It was a popular game of chance played in taverns and coffee houses well into the 18th century; its popularity at the time of the foundation of probability theory means that it was a common example in early texts on finding expected values and probabilities. Find the rules of the game, and get some practice playing it. Then, analyze the choices that the caster gets to make, and decide which is most advantageous, using the language of expected values. Chapter Review The Multiplication Rule for Counting Permutations Combinations Tree Diagrams, Tables, and Outcomes Basic Concepts of Probability Probability with Permutations and Combinations What Are the Odds? The Addition Rule for Probability Conditional Probability and the Multiplication Rule The Binomial Distribution Expected Value Chapter Test", "section": "Expected Value", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Statistics can be used to decide on a fair salary for sports stars. (credit: “Jasmine Powell goes up for a shot in a game against Jacksonville” by Lorie Shaull/Flickr, CC BY 2.0) Before the 2021 WNBA season, professional basketball player Candace Parker signed a contract with the Chicago Sky, which entitled her to a salary of $190,000. This amount was the 23rd highest in the league at the time. How did the team’s management decide on her salary? They likely considered some intangible qualities, like her leadership skills. However, much of their deliberations probably took into account her performance on the court. For example, Parker led the league in rebounds in the 2020 season (214 of them) and scored 14.7 points per game (which ranked her 18th among all WNBA players). Further, Parker brought 13 seasons of experience to the team. All of these factors played a role in deciding the terms of her contract. Estimating the value of one variable (like salary) based on other, measurable variables (points per game, experience, rebounds, etc.) is among the most important applications of statistics, which is the mathematical field devoted to gathering, organizing, summarizing, and making decisions based on data.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Gathering and Organizing Data Surveys are commonly used to gather data. (credit: “survey” by Donnell King/Flickr, CC0 1.0 Public Domain) Learning Objectives After completing this section, you should be able to: Distinguish among sampling techniques. Organize data using an appropriate method. Create frequency distributions. When a polling organization wants to try to establish which candidate will win an upcoming election, the first steps are to write questions for the survey and to choose which people will be asked to respond to the survey. These can seem like simple steps, but they have far-reaching implications in the analysis the pollsters will later carry out. The process by which samples (or groups of units from which we collect data) are chosen can strongly affect the data that are collected. Units are anything that can be measured or surveyed (such as people, animals, objectives, or experiments) and data are observations made on units. One of the most famous failures of good sampling occurred in the first half of the twentieth century. The Literary Digest was among the most respected magazines of the early twentieth century. Despite the name, the Digest was a weekly newsmagazine. Starting in 1916, the Digest conducted a poll to try to predict the winner of each US Presidential election. For the most part, their results were good; they correctly predicted the outcome of all five elections between 1916 and 1932. In 1936, the incumbent President Franklin Delano Roosevelt faced Kansas governor Alf Landon, and once again the Digest ran their famous poll, with results published the week before the election. Their conclusion? Landon would win in a landslide, 57% to 43%. Once the actual votes had been counted, though, Roosevelt ended up with 61% of the popular vote, 18% more than the poll predicted. What went wrong? The short answer is that the people who were chosen to receive the survey (over ten million of them!) were not a good representation of the population of voting adults. The sample was chosen using the Digest's own base of subscribers as well as publicly available lists of people that were likely adults (and therefore eligible to vote), mostly phone books and vehicle registration records. The pollsters then mailed every single person on these lists a survey. Around a quarter of those surveys were returned; this constituted the sample that was used to make the Digest ’s disastrously incorrect prediction. However, the Digest made an error in failing to consider that the election was happening during the Great Depression, and only the wealthy had disposable income to spend on telephone lines, automobiles, and magazine subscriptions. Thus, only the wealthy were sent the Digest ’s survey. Since Roosevelt was extremely popular among poorer voters, many of Roosevelt’s supporters were excluded from the Digest ’s sample. Another more complicated factor was the low response rate; only around 25% of the surveys were returned. This created what’s called a non-response bias. Sampling and Gathering Data The Digest's failure highlights the need for what is now considered the most important criterion for sampling: randomness. This randomness can be achieved in several ways. Here we cover some of the most common. A simple random sample is chosen in a way that every unit in the population has an equal chance of being selected, and the chances of a unit being selected do not depend on the units already chosen. An example of this is choosing a group of people by drawing names out of a hat (assuming the names are well-mixed in the hat). A systematic random sample is selected from an ordered list of the population (for example, names sorted alphabetically or students listed by student ID). First, we decide what proportion of the population will be in our sample. We want to express that proportion as a fraction with 1 in the numerator. Let’s call that number D . Next, we’ll choose a random number between one and D . The unit at that position will go into our sample. We’ll find the rest of our sample by choosing every D th unit in the list, starting with our random number. To walk through an example, let’s say we want to sample 2% of the population: 2 % = 2 100 = 1 50 . (Note: If the number in the denominator isn’t a whole number, we can just round it off. This part of the process doesn’t have to be precise.) We can then use a random number generator to find a random number between 1 and 50; let's use 31. In our example, our sample would then be the units in the list at positions 31, 81 (31 + 50), 131 (81 + 50), and so forth. A stratified sample is one chosen so that particular groups in the population are certain to be represented. Let’s say you are studying the population of students in a large high school (where the grades run from 9th to 12th), and you want to choose a sample of 12 students. If you use a simple or systematic random sample, there’s a pretty good chance that you’ll miss one grade completely. In a stratified sample, you would first divide the population into groups (the strata ), then take a random sample within each stratum (that’s the singular form of “strata”). In the high school example, we could divide the population into grades, then take a random sample of three students within each grade. That would get us to the 12 students we need while ensuring coverage of each grade. A cluster sample is a sample where clusters of units are chosen at random, instead of choosing individual units. For example, if we need a sample of college students, we may take a list of all the course sections being offered at the college, choose three of them at random (the sections are the clusters), and then survey all the students in those sections. A sample like this one has the advantage of convenience: If the survey needs to be administered in person, many of your sample units will be located in one place at the same time. Random Sampling For each of the following situations, identify whether the sample is a simple random sample, a systematic random sample, a stratified random sample, a cluster random sample, or none of these. A postal inspector wants to check on the performance of a new mail carrier, so she chooses four streets at random among those that the carrier serves. Each household on the selected streets receives a survey. A hospital wants to survey past patients to see if they were satisfied with the care they received. The administrator sorts the patients into groups based on the department of the hospital where they were treated (ICU, pediatrics, or general), and selects patients at random from each of those groups. A quality control engineer at a factory that makes smartphones wants to figure out the proportion of devices that are faulty before they are shipped out. The phones are currently packed in boxes for shipping, each of which holds 20 devices. The engineer wants to sample 100 phones, so he selects five crates at random and tests every phone in those five crates. A newspaper reporter wants to write a story on public perceptions on a project that will widen a congested street. She stands on the side of the street in question and interviews the first five people she sees there. An executive at a streaming video service wants to know if her subscribers would support a second season of a new show. She gets a list of all the subscribers who have watched at least one episode of the show, and uses a random number generator to select a sample of 50 people from the list. An agent for a state’s Department of Revenue is in charge of selecting 100 tax returns for audit. He has a list of all of the returns eligible for audit (about 12,000 in all), sorted by the taxpayer’s ID number. He asks a computer to give him a random number between 1 and 120; it gives him 15. The agent chooses the 15th, 135th, 255th, 375th, and every 120th return after that to be audited. To decide which type of random sample is being used in each of these, we need to focus on how the randomization is being incorporated. The surveys are being given to households, so households are the units in this case. But households aren’t being chosen randomly; instead, streets are being chosen at random. These form clusters of units, so this is a cluster random sample. In this case, the administrator isn’t selecting patients at random from the entire list of patients. Instead, she is choosing at random from the patients who were in each of the departments (ICU, pediatrics, general) separately. The departments form strata , so this is a stratified random sample. The engineer is testing whether the phones are faulty, so those are the units. But the random process is being used to select the crates of phones. Those crates form clusters, so this is a cluster random sample. The reporter isn’t using a random process at all, so this sample doesn’t belong to any of the types we have been talking about. A sample like this one is sometimes described as a convenience sample , and shouldn’t be used in a statistical setting. The executive is choosing her sample completely at random from the full population, so this is a simple random sample. The agent is choosing from the full population, but is only choosing the first unit for the sample at random; the rest are chosen by skipping down the list systematically. Thus, this is a systematic random sample. George Gallup George Gallup was a founder of survey sampling techniques, and his legacy lives on to this day. (credit: \"George Gallup at the National Press Club, Washington, D.C., 1969\" by Bernard Gotfryd/Library of Congress Prints & Photographs Division, public domain) George Gallup (1901–1984) rose to fame in 1936 when his prediction of the percentage of the vote going to each candidate in that year’s U.S. Presidential election was more accurate than the one published in Literary Digest , and he did so using a sample that was much smaller than the Digest . He even took it one step farther, predicting with high accuracy the erroneous results of the poll that the Literary Digest would end up publishing! Gallup’s theories on public opinion polling essentially created that field. In 1948, Gallup’s reputation took a bit of a hit, when he famously, but incorrectly, predicted that Thomas Dewey would beat incumbent Harry Truman in that year’s Presidential election. Over the following decades, however, public trust in Gallup’s polls recovered and even steadily increased. The company Gallup founded continues to conduct daily public opinion polls, as well as provides consulting services for businesses. Organizing Data Once data have been collected, we turn our attention to analysis. Before we analyze, though, it’s useful to reorganize the data into a format that makes the analysis easier. For example, if our data were collected using a paper survey, our raw data are all broken down by respondent (represented by an individual response sheet). To perform an analysis on all the responses to an individual question, we need to first group all the responses to each question together. The way we organize the data depends on the type of data we’ve collected. There are two broad types of data: categorical and quantitative. Categorical data classifies the unit into a group (or category). Examples of categorical data include a response to a yes-or-no question, or the color of a person’s eyes. Quantitative data is a numerical measure of a property of a unit. Examples of quantitative data include the time it takes for a rat to run through a maze or a person’s daily calorie intake. We’ll look at each type of data in turn when considering how best to organize. Categorical Data Organization The best way to organize categorical data is using a categorical frequency distribution . A categorical frequency distribution is a table with two columns. The first contains all the categories present in the data, each listed once. The second contains the frequencies of each category, which are just a count of how often each category appears in the data. Creating a Categorical Frequency Distribution A teacher records the responses of the class (28 students) on the first question of a multiple choice quiz, with five possible responses (A, B, C, D, and E): A A C A B B A E A C A A A C E A B A A C A B E E A A C C Create a categorical frequency distribution that organizes the responses. Step 1: For each possible response, count the number of times that response appears in the data. In the responses for this class, “A” appears 14 times, “B” 4 times, “C” 6 times, “D” 0 times, and “E” 4 times. Step 2: Make a table with two columns. The first column should be labeled so that the reader knows what the responses mean, and the second should be labeled “Frequency.” Response to First Question Frequency A 14 B 4 C 6 D 0 E 4 Step 3: Check your work. If you add up your frequencies, you should get the same number as the total number of responses. Twenty-eight students answered that first question, and 14 + 4 + 6 + 0 + 4 = 28 . Quantitative Data We have a couple of options available for organizing quantitative data. If there are just a few possible responses, we can create a frequency distribution just like the ones we made for categorical data above. For example, if we’re surveying a group of high school students and we ask for each student’s age, we’ll likely only get whole-number responses between 13 and 19. Since there are only around seven (and likely fewer) possible responses, we can treat the data as if they’re categorical and create a frequency distribution as before. Creating a Quantitative Frequency Distribution Attendees of a conflict resolution workshop are asked how many siblings they have. The responses are as follows: 1 0 1 1 2 0 3 1 1 4 1 2 0 1 3 1 2 1 2 4 1 0 1 3 0 1 2 2 1 5 Create a frequency distribution to organize the responses. Step 1: Count the number of times you see each unique response: “0” appears 5 times, “1” appears 13 times, “2” appears 6 times, “3” appears 3 times, “4” appears twice, and “5” appears once. Step 2: Make a table with two columns. The first column should be labeled so that the reader knows what the responses mean, and the second should be labeled “Frequency.” Then fill in the results of our count. Number of Siblings Frequency Number of Siblings Frequency 0 5 3 3 1 13 4 2 2 6 5 1 Step 3: Check your work. If you add up your counts, you should get the same number as the total number of responses. Looking back at the raw data, there were 30 responses, and 5 + 13 + 6 + 3 + 2 + 1 = 30 . If there are many possible responses, a frequency distribution table like the ones we’ve seen so far isn’t really useful; there will likely be many responses with a frequency of one, which means the table will be no better than looking at the raw data. In these cases, we can create a binned frequency distribution . A binned frequency distribution groups the data into ranges of values called bins , then records the number of responses in each bin. For example, if we have height data for individuals measured in centimeters, we might create bins like 150–155 cm, 155–160 cm, and so forth (making sure that every data value falls into a bin). We must be careful, though; in this scenario, it’s not clear which bin would contain a response of 155 cm. Usually, responses on the edge of a bin are placed in the higher bin, but it’s good practice to make that clear. In cases where responses are rounded off, you can avoid this issue by leaving a gap between the bins that couldn’t contain any responses. In our example, if the measurements were all rounded off to the nearest centimeter, we could make bins like 150–154 cm, 155–159 cm, etc. (since a response like 154.2 isn’t possible). We’ll use this method going forward. How do we decide what the boundaries of our bins should be? There’s no one right way to do that, but there are some guidelines that can be helpful. Every data value should fall into exactly one bin. For example, if the lowest value in our data is 42, the lowest bin should not be 45–49. Every bin should have the same width. Note that if we shift the upper limits of our bins down a bit to avoid ambiguity (like described above), we can’t simply subtract the lower limit from the upper limit to get the bin width; instead, we subtract the lower limit of the bin from the lower limit of the next bin. For example, if we’re looking at GPAs rounded to the nearest hundredth, we might choose bins like 2.00–2.24, 2.25–2.49, 2.50–2.74, etc. These bins all have a width of 0.25. If the minimum or maximum value of the data falls right on the boundary between two bins, then it’s OK to bend the rule just a little in order to avoid having an additional bin containing just that one value. We’ll see an example of this in just a moment. If we have too many or too few bins, it can be difficult to get a good sense of the distribution. Seven or eight bins is ideal, but that’s not a firm rule; anything between five and twelve is fine. We often choose the number of bins so that the widths are round numbers. Creating a Binned Frequency Distribution The GPAs of students enrolled in an advanced sociology class are listed in the following table. At this institution, 4.00 is the maximum possible GPA. 3.93 3.43 2.87 2.51 2.70 1.91 2.32 2.85 3.06 3.03 3.49 1.84 3.72 2.56 1.99 3.40 3.74 3.23 1.98 3.05 1.43 2.90 1.20 3.72 3.56 3.07 2.58 4.00 2.79 3.81 2.60 3.69 2.88 3.34 1.51 3.63 3.45 1.89 2.30 2.98 3.04 2.70 Create a binned frequency distribution for the data. Step 1: Identify the max and min values in your bins. Looking at the dataset, you can see that the lowest value is 1.20, and the highest is 4.00. Step 2: Get a rough idea of bin widths. Aim for seven or eight bins, give or take a couple. For eight bins, the minimum width can be found by taking the difference between the largest and smallest data values and dividing by the number of bins: maximum - minimum # of bins = 4.00 - 1.20 8 = 0.35 . If we use 0.35 for our widths, starting at our minimum value of 1.20, we’ll get bins with these boundaries: 1.20, 1.55, 1.90, 2.25, 2.60, 2.95, 3.30, 3.65, 4.00. Step 3: Consider the context of the values. Because these are GPAs, there are natural breaks at 2.00 and 3.00 that are important. (People like whole numbers!) Since 0.35 is very close to 1 3 , let’s use that for our bin width instead, and make sure that whole numbers fall on the boundaries. That means our first bin needs to start at 1.00 and go up to 1.33 to make sure our minimum value is included. The next bin will run from 1.34 to 1.66, and so forth. Step 4: Create the distribution table. We start our distribution table by filling in the bins: GPA Range Frequency GPA Range Frequency GPA Range Frequency 1.00–1.33 2.00–2.33 3.00–3.33 1.34–1.66 2.34–2.66 3.34–3.66 1.67–1.99 2.67–2.99 3.67–4.00 Notice that the last bin doesn’t follow the pattern; since our maximum data value is right on the upper boundary of that last bin, this is a case where we can bend that rule just a little to avoid creating a bin for 4.00–4.33 (which wouldn’t really make sense in the context of these GPAs anyway, since 4.00 is the maximum possible GPA). Step 5: Complete the table with the frequencies. Finish the table by counting the number of data values that fall in each bin, and recording them in the frequency column: GPA Range Frequency GPA Range Frequency GPA Range Frequency 1.00–1.33 1 2.00–2.33 2 3.00–3.33 6 1.34–1.66 2 2.34–2.66 4 3.34–3.66 7 1.67–1.99 5 2.67–2.99 8 3.67–4.00 7 Step 6: Check your work. Add up the frequencies to make sure all the data values are included. We started with forty-two data values, and 1 + 2 + 5 + 2 + 4 + 8 + 6 + 7 + 7 = 42 . Check Your Understanding Key Terms sample units data population simple random sample systematic random sample stratified sample cluster sample quantitative data categorical data categorical frequency Distribution binned frequency distribution Key Concepts Categorical data places units into groups (categories), while quantitative data is a numerical measure of a property of a unit. The sampling method for a study depends on the way that randomization is used to select units for the sample. Frequency distributions help to summarize data by counting the number of units that fall into a particular category or range of quantitative values.", "section": "Gathering and Organizing Data", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Visualizing Data Data visualizations can help people quickly understand important features of a dataset. (credit: \"Group of diverse people having a business meeting\" by Rawpixel Ltd/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Create charts and graphs to appropriately represent data. Interpret visual representations of data. Determine misleading components in data displayed visually. Summarizing raw data is the first step we must take when we want to communicate the results of a study or experiment to a broad audience. However, even organized data can be difficult to read; for example, if a frequency table is large, it can be tough to compare the first row to the last row. As the old saying goes: a picture is worth a thousand words (or, in this case, summary statistics)! Just as our techniques for organizing data depended on the type of data we were looking at, the methods we’ll use for creating visualizations will vary. Let’s start by considering categorical data. Visualizing Categorical Data If the data we’re visualizing is categorical, then we want a quick way to represent graphically the relative numbers of units that fall in each category. When we created the frequency distributions in the last section, all we did was count the number of units in each category and record that number (this was the frequency of that category). Frequencies are nice when we’re organizing and summarizing data; they’re easy to compute, and they’re always whole numbers. But they can be difficult to understand for an outsider who’s being introduced to your data. Let’s consider a quick example. Suppose you surveyed some people and asked for their favorite color. You communicated your results using a frequency distribution. Jerry is interested in data on favorite colors, so he reads your frequency distribution. The first row shows that twelve people indicated green was their favorite color. However, Jerry has no way of knowing if that’s a lot of people without knowing how many people total took your survey. Twelve is a pretty significant number if only twenty-five people took the survey, but it’s next to nothing if you recorded a thousand responses. For that reason, we will often summarize categorical data not with frequencies, but with proportions . The proportion of data that fall into a particular category is computed by dividing the frequency for that category by the total number of units in the data. Proportion of a category = Category frequency Total number of data units Proportions can be expressed as fractions, decimals, or percentages. Finding Proportions Recall , in which a teacher recorded the responses on the first question of a multiple choice quiz, with five possible responses (A, B, C, D, and E). The raw data was as follows: A A C A B B A E A C A A A C E A B A A C A B E E A A C C We computed a frequency distribution that looked like this: Response to First Question Frequency A 14 B 4 C 6 D 0 E 4 Proportion of a category = Category frequency / Total number of data units Now, let's compute the proportions for each category. Step 1: In order to compute a proportion, we need the frequency (which we have in the table above) and the total number of units that are represented in our data. We can find that by adding up the frequencies from all the categories: 14 + 4 + 6 + 0 + 4 = 28 . Step 2: To find the proportions, we divide the frequency by the total. For the first category (“A”), the proportion is 14 28 = 1 2 = 0.5 = 50 % . We can compute the other proportions similarly, filling in the rest of the table: Response to First Question Frequency Proportion A 14 14 28 = 50 % B 4 4 28 = 14.3 % C 6 6 28 = 21.4 % D 0 0 28 = 0 % E 4 4 28 = 14.3 % Step 3: Check your work: If you add up your proportions, you should get 1 (if you’re using fractions or decimals) or 100% (if you’re using percentages). In this case, 50 % + 14.3 % + 21.4 % + 0 % + 14.3 % = 100 % . If you need to round off the results of the computations to get your percentages or decimals, then the sum might not be exactly equal to 1 or 100% in the end due to that rounding error. Now that we can compute proportions, let’s turn to visualizations. There are two primary visualizations that we’ll use for categorical data: bar charts and pie charts. Both of these data representations work on the same principle: If proportions are represented as areas, then it’s easy to compare two proportions by assessing the corresponding areas. Let’s look at bar charts first. Bar Charts A bar chart is a visualization of categorical data that consists of a series of rectangles arranged side-by-side (but not touching). Each rectangle corresponds to one of the categories. All of the rectangles have the same width. The height of each rectangle corresponds to either the number of units in the corresponding category or the proportion of the total units that fall into the category. Building a Bar Chart In , we computed the following proportions: Response to First Question Frequency Proportion A 14 50% B 4 14.3% C 6 21.4% D 0 0% E 4 14.3% Draw a bar chart to visualize this frequency distribution. Step 1: To start, we’ll draw axes with the origin (the point where the axes meet) at the bottom left: Step 2: Next, we’ll place our categories evenly spaced along the bottom of the horizontal axis. The order doesn’t really matter, but if the categories have some sort of natural order (like in this case, where the responses are labeled A to E), it’s best to maintain that order. We'll also label the horizontal axis: Step 3: Now, we have a decision to make: Will we use frequencies to define the height of our rectangles, or will we use proportions? Let’s try it both ways. First, let’s use frequencies. Notice that our frequencies run from zero to 14; this will correspond to the scale we put on the vertical axis. If we put a tick mark for every whole number between 0 and 14, the result will be pretty crowded; let’s instead put a mark on the multiples of 3 or 5: Step 4: Now, let’s draw in the first rectangle. The frequency associated with “A” is 14. So we’ll go to 14 on the vertical axis, and place a mark at that height above the “A” label: Step 5: Then, draw vertical lines straight down from the edges of your mark to make a rectangle: Step 6: Finally, we can build the rest of the rectangles, making sure that the bases all have the same length of the base = width of the rectangle, and the rectangles don’t touch. Notice that, since the frequency for “D” is zero, that category has no rectangle (but we’ll leave a space there so the reader can see that there is a category with frequency zero). Here’s the result: Step 7: That’s it! Now, let’s use proportions instead of frequencies. We'll label the vertical axis with evenly spaced numbers that run the full range of the percentages in our table: 0% to 50%. We can divide that into five equal parts (so that each has width 10%), and use that to label our vertical axis: Step 8: Then, we can fill in the rectangles just as we did before. The height of the “A” rectangle is 50%, the “B” rectangle goes up to 14.3%, “C” goes to 21.4%, there is no rectangle for “D” (since its proportion is 0%), and the “E” rectangle also goes up to 14.3%: Step 9: Notice that the rectangles are basically identical in our two final bar charts. That’s no coincidence! Bar charts that use proportions and those that use frequencies will always look identical (which is why it doesn’t really matter much which option you choose). Here’s why: look at the bars for “B” and “C”. The frequencies for these are 4 and 6 respectively. Notice that 6 is 50% bigger than 4 (since 6 = 1.5 × 4 ), which means that the “C” bar will be 50% higher than the “B” bar. Now look at the same bars using proportions: since 21.4 % = 1.5 × 14.3 % , the bar for “C” will be 50% higher than the bar for “B.” The same relationships hold for the other bars, too. In practice, most graphs are now made with computers. You can use Google Sheets , which is available for free from any web browser. Make a Simple Bar Graph in Google Sheets Now that we’ve explored how bar graphs are made, let’s get some practice reading bar graphs. Reading Bar Graphs The bar graph shown gives data on 2020 model year cars available in the United States. Analyze the graph to answer the following questions. (data source: consumerreports.org/cars) What proportion of available cars were sports cars? What proportion of available cars were sedans? Which categories of cars each made up less than 5% of the models available? The bar for sports cars goes up to 10%, so the proportion of models that are considered sports cars is 10%. The bar corresponding to sedan goes up past 30% but not quite to 35%. It looks like the proportion we want is between 33% and 34%. We’re looking for the bars that don’t make it all the way to the 5% line. Those categories are hatchback and wagon. Candy Color: Frequency and Distribution M&Ms, Skittles, and Reese’s Pieces are all candies that have pieces that are uniformly shaped, but which have different colors. Do the colors in each bag appear with the same frequency? Get a bag of one of these candies and make a bar chart to visualize the color distribution. Pie Charts A pie chart consists of a circle divided into wedges, with each wedge corresponding to a category. The proportion of the area of the entire circle that each wedge represents corresponds to the proportion of the data in that category. Pie charts are difficult to make without technology because they require careful measurements of angles and precise circles, both of which are tasks better left to computers. Create Pie Charts Using Google Sheets Pie charts are sometimes embellished with features like labels in the slices (which might be the categories, the frequencies in each category, or the proportions in each category) or a legend that explains which colors correspond to which categories. When making your own pie chart, you can decide which of those to include. The only rule is that there has to be some way to connect the slices to the categories (either through labels or a legend). Making Pie Charts Use the data that follows to generate a pie chart. Type Percent Type Percent SUV 43.6% Minivan 5.5% Sedan 33.6% Hatchback 3.6% Sports 10.0% Wagon 3.6% (data source: www.consumerreports.org/cars) First, enter the chart above into a new sheet in Google Sheets. Next, click and drag to select the full table (including the header row). Click on the “Insert” menu, then select “Chart.” The result may be a pie chart by default; if it isn’t, you can change it to a pie chart using the “Chart type” drop-down menu in the Chart Editor. (data source: consumerreports.org/cars) You can choose to use a legend to identify the categories, as well as label the slices with the relevant percentages. Florence Nightingale Florence Nightingale (1820–1910) is best remembered today for her contributions in the medical field; after witnessing the horrors of field hospitals that tended to the wounded during the Crimean War, she championed reforms that encouraged sanitary conditions in hospitals. For those efforts, she is today considered the founder of modern nursing. Florence Nightingale's significant contribution to the field of statistical graphics cannot be understated. (credit: \"Florence Nightingale\" by Library of Congress Prints and Photographs Division/http://hdl.loc.gov/loc.pnp/pp.print, public domain) Nightingale is also remembered for her contributions in statistics, especially in the ways we visualize data. She developed a version of the pie chart that is today known as a polar area diagram , which she used to visualize the causes of death among the soldiers in the war, highlighting the number of preventable deaths the British Army suffered in that conflict. In 1859, the Royal Statistical Society honored her for her contributions to the discipline by electing her to join the organization. She was the first woman to be so honored. She was later named an honorary member of the American Statistical Association. Nightingale's status as a revered pioneer in both nursing and statistics is a complex one, because some of her writings and opinions demonstrate a colonialist mindset and disregard for those who lost their lives and lands at the hands of the British. Her core statistical writings indicated that she felt superior to the Indigenous people she was treating. Members of both fields continue to debate her near-iconic role. Visualizing Quantitative Data There are several good ways to visualize quantitative data. In this section, we’ll talk about two types: stem-and-leaf plots and histograms. Stem-and-Leaf Plots Stem-and-leaf plots are visualization tools that fall somewhere between a list of all the raw data and a graph. A stem-and-leaf plot consists of a list of stems on the left and the corresponding leaves on the right, separated by a line. The stems are the numbers that make up the data only up to the next-to-last digit, and the leaves are the final digits. There is one leaf for every data value (which means that leaves may be repeated), and the leaves should be evenly spaced across all stems. These plots are really nothing more than a fancy way of listing out all the raw data; as a result, they shouldn’t be used to visualize large datasets. This concept can be difficult to understand without referencing an example, so let’s first look at how to read a stem-and-leaf plot. Reading a Stem-and-Leaf Plot A collector of trading cards records the sale prices (in dollars) of a particular card on an online auction site, and puts the results in a stem-and-leaf plot: 0 5 8 9 1 0 0 0 3 4 4 5 5 5 5 6 9 9 2 0 0 0 0 5 5 9 9 3 0 0 0 5 5 4 0 0 5 5 6 0 Answer the following questions about the data: How many prices are represented? What prices represent the five most expensive cards? The five least expensive? What is the full set of data? Each leaf (the numbers on the right side of the bar) represents one data value. So, on the first row (which looks like 0 | 5 8 9), there are three data values (one for each leaf: 5, 8, and 9). The next row has thirteen leaves, then eight, five, three, zero, and one. Adding those up, we get 3 + 13 + 8 + 5 + 3 + 0 + 1 = 33 data points or prices. The most expensive card is the last one listed. Its stem is 6 and its leaf is 0, so the price is $60. There are no leaves associated with the 5 stem, so there were no cards sold for $50 to $59. The next most expensive cards are then on the 4 stem: $45, $40, and $40 (remember, repeated leaves mean repeated values in the dataset). So, we have our four most expensive cards. The fifth would be on the next stem up. The biggest leaf on the 3 stem is a 5, so the fifth-most expensive card sold for $35. As for the five least-expensive cards, the smallest stem is 0, with leaves 5, 8, and 9. So, the three least expensive cards sold for $5, $8, and $9 (notice that we don’t write down that leading 0 from the stem in the tens place). The next two least-expensive cards will be the two smallest leaves on the next stem: $10 and $10. The full list of data is: 5, 8, 9, 10, 10, 10, 13, 14, 14, 15, 15, 15, 15, 16, 19, 19, 20, 20, 20, 24, 25, 25, 29, 29, 30, 30, 30, 35, 35, 40, 40, 45, 60. Stem-and-leaf plots are useful in that they give us a sense of the shape of the data. Are the data evenly spread out over the stems, or are some stems “heavier” with leaves? Are the heavy stems on the low side, the high side, or somewhere in the middle? These are questions about the distribution of the data, or how the data are spread out over the range of possible values. Some words we use to describe distributions are uniform (data are equally distributed across the range), symmetric (data are bunched up in the middle, then taper off in the same way above and below the middle), left-skewed (data are bunched up at the high end or larger values, and taper off toward the low end or smaller values), and right-skewed (data are bunched up at the low end, and taper off toward the high end). See below figures. Looking back at the stem-and-leaf plot in the previous example, we can see that the data are bunched up at the low end and taper off toward the high end; that set of data is right-skewed. Knowing the distribution of a set of data gives us useful information about the property that the data are measuring. Now that we have a better idea of how to read a stem-and-leaf plot, we’re ready to create our own. Constructing a Stem-and-Leaf Plot An entomologist studying crickets recorded the number of times different crickets (of differing species, genders, etc.) chirped in a one-minute span. The raw data are as follows: 89 97 82 102 84 99 93 103 120 91 115 105 89 109 107 89 104 82 106 92 101 109 116 103 100 91 85 104 104 106 Construct a stem-and-leaf plot to visualize these results. Step 1 : Before we can create the plot, we need to sort the data in order from smallest to largest: 82 82 84 85 89 89 89 91 91 92 93 97 99 100 101 102 103 103 104 104 104 105 106 106 107 109 109 115 116 120 Step 2: Next, we identify the stems. To do that, we cut off the final digit of each number, which leaves us with stems of 8, 9, 10, 11, and 12. Arrange the stems vertically, and add the bar to separate these from the leaves: 8 9 10 11 12 Step 3: Write down the leaves on the right side of the bar, giving just the final digit (that we cut off to make the stems) of each data value. List these in order, and make sure they line up vertically: 8 2 2 4 5 9 9 9 9 1 1 2 3 7 9 10 0 1 3 3 4 4 4 5 6 6 7 9 9 11 5 6 12 0 As we mentioned above, stem-and-leaf plots aren’t always going to be useful. For example, if all the data in your dataset are between 20 and 29, then you’ll just have one stem, which isn’t terribly useful. (Although there are methods like stem splitting for addressing that particular problem, we won’t go into those at this time.) On the other end of the spectrum, the data may be so spread out that every stem has only one leaf. (This problem can sometimes be addressed by rounding off the data values to the tens, hundreds, or some other place value, then using that place for the leaves.) Finally, if you have dozens or hundreds (or more) of data values, then a stem-and-leaf plot becomes too unwieldy to be useful. Fortunately, we have other tools we can use. Histograms Histograms are visualizations that can be used for any set of quantitative data, no matter how big or spread out. They differ from a categorical bar chart in that the horizontal axis is labeled with numbers (not ranges of numbers), and the bars are drawn so that they touch each other. The heights of the bars reflect the frequencies in each bin. Unlike with stem-and-leaf plots, we cannot recreate the original dataset from a histogram. However, histograms are easy to make with technology and are great for identifying the distribution of our data. Let’s first create one histogram without technology to help us better understand how histograms work. Constructing a Histogram In , we built a stem-and-leaf plot for the number of chirps made by crickets in one minute. Here are the raw data that we used then: 89 97 82 102 84 99 115 105 89 109 107 89 101 109 116 103 100 91 93 103 120 91 85 104 104 82 106 92 104 106 Construct a histogram to visualize these results. Step 1: Add data to bins. Histograms are built on binned frequency distributions, so we’ll make that first. Luckily, the stem-and-leaf plot we made earlier can help us do this much more quickly: 8 2 2 4 5 9 9 9 9 1 1 2 3 7 9 10 0 1 2 3 3 4 4 4 5 6 6 7 9 9 11 5 6 12 0 If we’re using bins of width 10, we can compute the frequencies by counting the numbers of leaves associated with the corresponding stem: Bin Frequency 80-89 7 90-99 6 100-109 14 110-119 2 120-129 1 (Note that, when we made binned frequency diagrams in the last module, we noted that if the biggest data value was right on the border between two bins, it was OK to lump it in with the lower bin. That’s not recommended when building histograms, so the data value 120 is all alone in the 120-129 bin.) Step 2: Create the axes. On the horizontal axis, start labeling with the lower end of the first bin (in this case, 80), and go up to the higher end of the last bin (120). Mark off the other bin boundaries, making sure they’re all evenly spaced. On the vertical axis, start with zero and go up at least to the greatest frequency you see in your bins (14 in this example), making sure that the labels you make are evenly spaced and that the difference between those numbers is the same. Let’s count off our vertical axis by threes: Step 3: Draw in the bars. Remember that the bars of a histogram touch, and that the heights are determined by the frequency. So, the first bar will cover 80 to 90 on the horizontal axis, and have a height of 7: Now, we can fill in the others: Step 4: Let’s compare the histogram we just created to the stem-and-leaf plot we made earlier: Notice that the leaves on the rotated stem-and-leaf plot match the bars on our histogram! We can view stem-and-leaf plots as sideways histograms. But, as we’ll see soon, we can do much more with histograms. Now that we’ve seen the connection between stem-and-leaf plots and histograms, we are ready to look at how we can use Google Sheets to build histograms. Make a Histogram Using Google Sheets Let’s use Google Sheets to create a histogram for a large dataset. Creating a Histogram in Google Sheets The data in “AvgSAT” contains the average SAT score for students attending every institution of higher learning in the US for which data is available. Create a histogram in Google Sheets of the average SAT scores. Use bins of width 50. Are the data uniformly distributed, symmetric, left-skewed, or right-skewed? Using the procedure described in the video above, we get this: (data source: https://data.ed.gov) The data are fairly symmetric, but slightly right-skewed. Bar Charts for Labeled Data Sometimes we have quantitative data where each value is labeled according to the source of the data. For example, in the Your Turn above, you looked at in-state tuition data. Every value you used to create that histogram was associated with a school; the schools are the labels. In YOUR TURN 8.11 , you found a histogram of the wins of every Major League Baseball team in 2019. Each of those win totals had a label: the team. If we’re interested in visualizing differences among the different teams, or schools, or whatever the labels are, we create a different version of the bar graph known as a bar chart for labeled data . These graphs are made in Google Sheets in exactly the same way as regular bar graphs. The only change is that the vertical axis will be labeled with the units for your quantitative data instead of just “Frequency.” Building a Bar Chart for Labeled Data The following table shows the gross domestic product (GDP) for the United States for the years 2010 to 2019: Year GDP (in $ trillions) Year GDP (in $ trillions) 2010 14.992 2015 18.225 2011 15.543 2016 18.715 2012 16.197 2017 19.519 2013 16.785 2018 20.580 2014 17.527 2019 21.433 (source: https://data.worldbank.org) Construct a histogram that represents these data. In this case, the years are the labels, and the data we are interested in are the GDP numbers. Once you have the table above (including the labels) entered into a spreadsheet, click and drag to select the full table. Then, in the “Insert” menu, click “Chart.” The result may not be a bar chart; if it’s not, select “Column chart” in the drop-down menu “Chart type” in the Chart Editor. If you want, you can edit things like the chart title in the “Customize” tab in the Chart Editor. (data source: https://data.worldbank.org) Misleading Graphs Graphical representations of data can be manipulated in ways that intentionally mislead the reader. There are two primary ways this can be done: by manipulating the scales on the axes and by manipulating or misrepresenting areas of bars. Let’s look at some examples of these. Misleading Graphs The table below shows the teams, and their payrolls, in the English Premier League, the top soccer organization in the United Kingdom. Team Salary (£1,000,000s) Team Salary (£1,000,000s) Manchester United F.C. 175.7 Newcastle United F.C. 56.9 Manchester City F.C. 136.5 Aston Villa F.C. 52.3 Chelsea F.C. 132.8 Fulham F.C. 52.1 Arsenal F.C. 130.7 Southampton F.C. 49.6 Tottenham Hotspur F.C. 129.2 Wolverhampton Wanderers F.C. 49.5 Liverpool F.C. 118.6 Brighton & Hove Albion 43.7 Crystal Palace 85.0 Burnley F.C. 35.5 Everton F.C. 82.5 West Bromwich Albion F.C. 23.8 Leicester City 73.7 Leeds United F.C. 22.5 West Ham United F.C. 69.2 Sheffield United F.C. 19.7 (source: www.spotrac.com) How might someone present this data in a misleading way? Step 1: Let’s focus on the top five teams. Here’s a bar chart of their payrolls: (data source: www.spotrac.com) Step 2: Now, here’s another bar chart visualizing exactly the same data: (data source: www.spotrac.com) Step 3: You should notice that despite using the same data, these two graphs look strikingly different. In the second graph, the gap between Manchester United and the other four teams looks significantly larger than in the first graph. The scale on the vertical axis has been manipulated here. The first graph's axis starts at zero, while the lowest value on the second graph's axis is 120. This trick has a strong impact on the viewer’s perception of the data. Beware of vertical axes that don’t start at zero! They overemphasize differences in heights. Step 4: To further emphasize the difference this creates in our perception, let's look at that data again, but this time using graphics instead of colored areas on our bar graph. (data source: www.spotrac.com) This graph uses an image of a £10 banknote in place of the bars. Using an image that evokes the context of the data in place of a standard, “boring” bar is a common tool that people use when creating infographics . However, this is generally not a good practice because it distorts the data. Notice that our “bars” (the banknotes) are just as tall here as they were in the previous figure. But, to maintain the right proportions, the widths had to be adjusted as well, which changes the area (height × width) of each bar. A key point is that when looking at rectangles, the human eye tends to process areas more easily than heights. Beware of infographics! Areas overemphasize a difference that should be measured with a height! Step 5: Now, let’s look at all 20 teams. This histogram indicates that the data are right-skewed, with the highest number of teams having a payroll between £40 million and £80 million: (data source: www.spotrac.com) Step 6: Now let's view this same data in another chart: (data source: www.spotrac.com) Step 7: Even though this chart uses the same data, the skew seems to be reversed. Why? Well, even though this graph looks like a histogram, it isn’t. Look closely at the labels on the horizontal axis; they don't correspond to spots on the axis, but instead provide a range, meaning this is a bar graph based on a binned frequency distribution. When we review these ranges, we can see that the last range is misleading as it consists of all data “over 80.” If the bins all had the same width, that last bin would run from 80 to 120. However, we can see from the histogram that the maximum value for this data is between 160 and 200. If the last bin in this bar graph were labeled honestly, it would read “80–200,” which would drive home the fact that the width of that bar is misleading. Always check the horizontal axis on histograms! The widths of all the bars should be equal. How to Spot a Misleading Graph Napoleon's Failed Invasion One of the most famous data visualizations ever created is the cartographic depiction by Charles Joseph Minard of Napoleon’s disastrous attempted invasion of Russia. Minard’s Napoleon Map (credit: Carte de Charles Minard/Wikimedia, public domain) Minard’s chart is remarkable in that it shows not just how the size of Napoleon’s army shrank drastically over time, but also the location on the map, the direction the army was traveling at the time, and the temperature during the retreat. Check Your Understanding Key Terms proportion bar chart pie chart stem-and-leaf plot distribution (of quantitative data) histogram bar chart for labeled data Key Concepts Categorical data can be visualized using pie charts or bar charts; quantitative data can be visualized using stem-and-leaf plots or histograms. Areas in pie charts and bar charts represent proportions of the data falling into a particular category, while areas in histograms represent proportions of the data that fall into a given range of data values (or “bins”). Stem-and-leaf plots are visual representations of entire datasets. By manipulating the axes, changing widths of bars, or making bad choices for bins, we can create data visualizations that misrepresent the distribution of data. Videos Make a Simple Bar Graph in Google Sheets Create Pie Charts Using Google Sheets Make a Histogram Using Google Sheets How to Spot a Misleading Graph", "section": "Visualizing Data", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Mean, Median and Mode What does it mean to say someone has average height? (credit: modification of work “I’m the tallest” by Jenn Durfey/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the mode of a dataset. Calculate the median of a dataset. Calculate the mean of a dataset. Contrast measures of central tendency to identify the most representative average. Solve application problems involving mean, median, and mode. What exactly do we mean when we describe something as \"average\"? Is the height of an average person the height that more people share than any other? What if we line up every person in the world, in order from shortest to tallest, and find the person right in the middle: Is that person’s height the average? Or maybe it’s something more complicated. Imagine a game where you and a friend are trying to guess the typical person’s height. Once the guesses are made, you bring in every person and measure their height. You and your friend figure out how far off each of your guesses were from the actual value, then square that number. The result is the number of points you earn for that person. After we check every height and award points accordingly, the person with the lower score wins (because a lower score means that person’s guess was, overall, closer to the actual values). Could we define the average height to be the number that you should guess to give you the smallest possible score? Each of these three methods of determining the “average” is commonly used. They are all methods of measuring centrality (or central tendency ). Centrality is just a word that describes the middle of a set of data. All give potentially different results, and all are useful for different reasons. In this section, we’ll explore each of these methods of finding the “average.” The Mode In our discussion of average heights, the first possible definition we offered was the height that more people share than any other. This is the mode , or the value that appears most often. If there are two modes, the data are bimodal . Let’s look at some examples. Finding the Mode Using a Stem-and-Leaf Plot In , we looked at a stem-and-leaf plot of the sale prices (in dollars) of a particular collectible trading card: 0 5 8 9 1 0 0 0 3 4 4 5 5 5 5 6 9 9 2 0 0 0 0 5 5 9 9 3 0 0 0 5 5 4 0 0 5 5 6 0 What is the mode price? The mode is the price that appears most often. Both 15 and 20 appear 4 times, more than any other values. So, they are the modes (and we can conclude that this set of data is bimodal). When we have a complete list of the data or a stem-and-leaf plot, it’s pretty straightforward to find the mode; we just need to find the number that appears most often. If we’re given a frequency distribution instead, the technique is different (but just as straightforward): we’re looking for the number with the highest frequency. Finding the Mode Using a Frequency Distribution In , we created a frequency distribution of the number of siblings of conflict resolution class attendees. Number of Siblings Frequency 0 5 1 13 2 6 3 3 4 2 5 1 What is the mode of the number of siblings? The mode is the value that appears the most often, which means it has the greatest frequency. Thirteen of the respondents have one sibling, more than any other number. So, the mode is 1. What happens if there is no number in the data that appears more than once? In that case, by our definition, every data value is a mode. But according to some other definitions, the data would have no mode. In practice, though, it doesn’t really matter; if no data value appears more than once, then the mode is not helpful at all as a measure of centrality. The Median Let's revisit our example of trying to identify the height of the “average” person. If we lined everyone up in order by height and found the person right in the middle, that person’s height is called the median , or the value that is greater than no more than half and less than no more than half of the values. Let’s look at a really simple example. Consider the following list of numbers: 11, 12, 13, 13, 14. Is the first number on the list, 11, the median? There are no values less than 11 (that’s 0%), and there are four values greater than 11 (that’s 80%). Since more than 50% of the data are greater than 11, the definition is violated; it’s not the median. Here’s a chart with the rest of the data, with red shading to show where the definition is violated: Data Value Number of Values Below Percentage of Values Below Number of Values Above Percentage of Values Above 11 0 0% 4 80% 12 1 20% 3 60% 13 2 40% 1 20% 14 4 80% 0 0% Only 13 has no violations, so it’s the median according to the definition. In practice, we find the median just like we described in the average height example: by lining up all the data values in order from smallest to largest and picking the value in the middle. For our easy example (with data values 11, 12, 13, 13, 14), that first 13 is right in the middle; there are two values to the left and two values to the right. If there’s not one value right in the middle, we pick the two closest, then choose the number exactly between them. For example, let’s say we have the data 41, 44, 46, 53. Since there are an even number of data values in our list, we can’t pick the one right in the middle. The two closest to the middle are 44 and 46, so we’ll choose the number halfway between those to be the median: 45. As this example shows, the median (unlike the mode) doesn’t have to be a number in our original set of data. In the examples we’ve looked at so far, it’s been pretty easy to identify which number is right in the middle. If we had a very large dataset, though, it might be harder. Fortunately, we have some formulas to help us with that. Suppose we have a set of data with n values, ordered from smallest to largest. If n is odd, then the median is the data value at position n + 1 2 . If n is even, then we find the values at positions n 2 and n 2 + 1 . If those values are named a and b , then the median is defined to be a + b 2 . Let’s put those formulas to work in an example. Finding the Median Using a Stem-and-Leaf Plot In , we looked at a stem-and-leaf plot that contained 33 sale prices (in dollars) of a particular collectible trading card: 0 5 8 9 1 0 0 0 3 4 4 5 5 5 5 6 9 9 2 0 0 0 0 5 5 9 9 3 0 0 0 5 5 4 0 0 5 5 6 0 What is the median price? Step 1: Since 33 is odd, the median is the data value at position n + 1 2 , where n is the number of values in the dataset. There are 33 total values, so our formula becomes 33 + 1 2 = 17 . That means we want to look for the 17th number in the dataset. Step 2: We'll want to count from the lowest value to the 17th number. We can use our stem and leaf plot to do this. 1 2 3 0 5 8 9 4 5 6 7 8 9 10 11 12 13 14 15 16 1 0 0 0 3 4 4 5 5 5 5 6 9 9 17 2 0 0 0 0 5 5 9 9 3 0 0 0 5 5 4 0 0 5 5 6 0 The seventeenth number is 20, so the median is 20. Now, let’s tackle an example with an even number of values. Finding the Median In , we looked at the number of times different crickets (of differing species, genders, etc.) chirped in a one-minute span. That data is again provided below: 89 97 82 102 84 99 115 105 89 109 107 89 101 109 116 103 100 91 93 103 120 91 85 104 104 82 106 92 104 106 Find the median. Step 1 : In order to find the median, we first need to sort the data so that they’re in order, smallest to largest: 82 82 84 85 89 89 89 91 91 92 93 97 99 100 101 102 103 103 104 104 104 105 106 106 107 109 109 115 116 120 Step 2 : Next, we figure out how many data values we have. Counting them up, we see there are 30, which is even. Step 3 : Since we have an even number of data values, we need to find the values in positions 30 2 = 15 and 30 2 + 1 = 16 . These are 101 and 102. Step 4 : We use the formula to compute the median: 101 + 102 2 = 101.5 . Finding the Median Using a Frequency Distribution In , we created a frequency distribution of the number of siblings of the people who attended a conflict resolution class. Let's review that data again: Number of Siblings Frequency 0 5 1 13 2 6 3 3 4 2 5 1 What is the median of the number of siblings? There are 30 data values total, so the median is between the 15th and 16th values in the ordered list. There are five 0s and thirteen 1s according to the frequency distribution, so items one through five are all 0s and items six through eighteen are all 1s. Since both items fifteen and sixteen are 1s, the median is 1. The Mean Recall our example of ways we could identify the “average” height of an individual. The last method we discussed was also the most complicated. It involved a game where the player guesses a height, then figures out how far off that guess is from every single person’s height. Those differences get squared and added together to get a score. Our next measure of centrality gives the lowest possible score: No other guess would beat it in the game. Given a dataset containing n total values, the mean of the dataset is the sum of all the data values, divided by n . This is a computation you have likely done before. In many places, including spreadsheet programs like Microsoft Excel and Google Sheets, this number is called the average . For statisticians, though, the word average has too many possible meanings, so they prefer the one we’ll use: mean. Finding the Mean Compute the mean of the numbers 12, 15, 17, 18, 18, and 19. The mean is the sum of the values, divided by the number of values on the list. So, we get: 12 + 15 + 17 + 18 + 18 + 19 6 = 99 6 = 16.5 . Finding the Mean Using a Frequency Distribution Refer again to the frequency distribution of the number of siblings people who attended a conflict resolution class reported: Number of Siblings Frequency 0 5 1 13 2 6 3 3 4 2 5 1 What is the mean of the number of siblings? Step 1: We compute the mean by adding up all the data values and then dividing by the number of data values on the list. Step 2: Adding up the frequencies, we get 5 + 13 + 6 + 3 + 2 + 1 = 30 data values in our list. Step 3: Now, to find the sum of all the data values, we could simply reconstruct the raw data and add up all the numbers there. But, there’s an easier way: Remember that repeatedly adding a number to itself is the definition of multiplication. So, for example, since there are six 2s in our data, the sum of all those 2s must be 6 × 2 = 12 . Step 4: Let’s add a column to our distribution for these products: Number of Siblings Frequency (Number of Siblings) × (Frequency) 0 5 0 1 13 13 2 6 12 3 3 9 4 2 8 5 1 5 Step 5: So, the sum of all our data values is 0 + 13 + 12 + 9 + 8 + 5 = 47 . The mean is 47 30 ≈ 1.567 . As the number of data values we are considering grows, the computation for the mean gets more and more complicated. That’s why people generally trust technology to perform that computation. Compute Measures of Centrality Using Google Sheets Note that a recent update to Google Sheets introduced a new function called “MODE.MULT,” which will find every mode (not just the first one on the list). Using Google Sheets to Compute Measures of Centrality The dataset \"InState\" contains the in-state tuitions of every college and university in the country that reported that data to the Department of Education. Find the mode, median, and mean of those in-state tuition values. Step 1: To find the mode, we select an empty cell type “=MODE(”, click on the header of the column to insert a reference to the column into our formula, and then close the parentheses. When we hit the enter key, our formula is replaced with the mode: $13,380. Step 2: We can find the median and mean using the same process, except using the functions “MEDIAN” and “AVERAGE” in place of “MODE”. We find that the median is $11,207 and the mean is $15,476.79. Which Is Better: Mode, Median, or Mean? If the mode, median, and mean all purport to measure the same thing (centrality), why do we need all three? The answer is complicated, as each measure has its own strengths and weaknesses. The mode is simple to compute, but there may be more than one. Further, if no data value appears more than once, the mode is entirely unhelpful. As for the mean and median, the main difference between these two measures is how each is affected by extreme values. Consider this example: the mean and median of 1, 2, 3, 4, 5 are both 3. But what if the dataset is instead 1, 2, 3, 4, 10? The median is still 3, but the mean is now 4. What this example shows is that the mean is sensitive to extreme values, while the median isn’t. This knowledge can help us decide which of the two is more relevant for a given dataset. If it is important that the really high or really low values are reflected in the measure of centrality, then the mean is the better option. If very high or low values are not important, however, then we should stick with the median. The decision between mean and median only really matters if the data are skewed. If the data are symmetric, then the mean and median are going to be approximately equal, and the distinction between them is irrelevant. If the data are skewed, the mean gets pulled in the direction of the skew (i.e., if the data are right-skewed, then the mean will be bigger than the median; if the data are left-skewed, the opposite relation is true). Choosing Which Measure of Centrality to Use For the following situations, decide which measure(s) of centrality would be best: You found a used car that you like, and you want to know if the price is too high or too low. You find a list of sale prices for that make and model, and you see that the distribution of those prices is skewed to the right. Some of the prices at the high end are close to the original sale price of the car, so you guess that those cars might have really low mileage, or have other enhancements added on that increased the value (but which don’t apply to the car you found). You are asked to analyze the responses to a survey. One of the questions asked, “How strongly do you agree with the statement, ‘I believe my elected representatives have my best interests in mind when they vote’?” Responses are a number between 1 and 5, with 1 representing “strongly disagree” and 5 representing “strongly agree.” You are asked to find the “average” household income for a zip code. Those values are skewed right. Thinking back to the situation at the beginning of this section: you want to find “average” height. The data you’ve collected seem to be distributed symmetrically. In this situation, the high values are not comparable to the value of the car you found and we don’t want them to affect the results. Also, we’re unlikely to find many repeated values, so the mode is probably not useful. Median is best. Here, we want to know what a typical result is. The mean doesn’t really make sense; it involves adding the numbers together, so it would treat two “strongly disagree” and two “strongly agree” responses (those add to 1 + 1 + 5 + 5 = 12 ) as exactly the same as four “neutral” responses ( 3 + 3 + 3 + 3 = 12 ). But those are really different situations; the first shows a strong polarization in the responses, while the second represents strong indifference. The mode is probably the best choice (because the data are actually categorical), but the median would be good too. The mode isn’t going to be useful in this situation because it’s unlikely you will find many households that have exactly the same income. The mean and median will be different because of the skew, so the choice comes down to the extreme values. Remember that the data are skewed right, so high values are prevalent. Because these high values are important for our analysis, we want them to be reflected in the results. Thus, the mean is best. That being said, the median is also useful; it allows us to say something like “50% of the households surveyed make more than” the median. Because we aren’t likely to find many people with exactly the same height, the mode won’t be useful. Since the data are symmetric, the mean and the median will be about the same. So, it doesn’t really matter which of those two we choose. Check Your Understanding Key Terms mode bimodal median mean Key Concepts The mode of a dataset is the value that appears the most frequently. The median is a value that is greater than or equal to no more than 50% of the data and less than or equal to no more than 50% of the data. The mean is the sum of all the data values, divided by the number of units in the dataset. The median of a dataset is not affected by outliers, but the mean will be biased toward outliers. This distinction might affect which measure of centrality is used to summarize a dataset. Formulas Suppose we have a set of data with n values, ordered from smallest to largest. If n is odd, then the median is the data value at position n + 1 2 . If n is even, then we find the values at positions n 2 and n 2 + 1 . If those values are named a and b , then the median is defined to be a + b 2 . Videos Compute Measures of Centrality Using Google Sheets", "section": "Mean, Median and Mode", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Range and Standard Deviation Measures of spread help us get a better understanding of test scores. (credit: \"Standardized test exams form with answers bubbled\" by Marco Verch Professional Photographer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the range of a dataset Calculate the standard deviation of a dataset Measures of centrality like the mean can give us only part of the picture that a dataset paints. For example, let’s say you’ve just gotten the results of a standardized test back, and your score was 138. The mean score on the test is 120. So, your score is above average! But how good is it really ? If all the scores were between 100 and 140, then you know your score must be among the best. But if the scores ranged from 0 to 200, then maybe 140 is good, but not great (though still above average). Knowing information about how the data are spread out can help us put a particular data value in better context. In this section, we’ll look at two numbers that help us describe the spread in the data: the range and the standard deviation. These numbers are called measures of dispersion. The Range Our first measure of dispersion is the range , or the difference between the maximum and minimum values in the set. It’s the measure we used in the standardized test example above. Let’s look at a couple of examples. Finding the Range You survey some of your friends to find out how many hours they work each week. Their responses are: 5, 20, 8, 10, 35, 12. What is the range? The maximum value in the set is 35 and the minimum is 5, so the range is 35 - 5 = 30 . For large datasets, finding the maximum and minimum values can be daunting. There are two ways to do it in a spreadsheet. First, you can ask the spreadsheet program to sort the data from smallest to largest, then find the first and last numbers on the sorted list. The second method uses built-in functions to find the minimum and maximum. Find the Minimum and Maximum Using Google Sheets In either method, once you’ve found the maximum and minimum, all you have to do is subtract to find the range. Finding the Range with Google Sheets The data in “AvgSAT” contains the average SAT score for students attending every institution of higher learning in the US for which data is available. What is the range of these average SAT scores? Step 1: To find the maximum, click on an empty cell in the spreadsheet, type “=MAX(”, and then click on the letter that marks the top of the column containing the AvgSAT data. That inserts a reference to the column into our function. Then we close the parentheses and hit the enter key. The formula is replaced with the maximum value in our data: 1566. Step 2: Using the same process (but with “MIN” instead of “MAX”), we find the minimum value is 785. Step 3: So, the range is 1566 - 785 = 781 . The range is very easy to compute, but it depends only on two of the data values in the entire set. If there happens to be just one unusually high or low data value, then the range might give a distorted measure of dispersion. Our next measure takes every single data value into account, making it more reliable. The Standard Deviation The standard deviation is a measure of dispersion that can be interpreted as approximately the average distance of every data value from the mean. (This distance from the mean is the “deviation” in “standard deviation.”) The standard deviation is computed as follows: s = ∑ ( x - x ¯ ) 2 n - 1 Here, x represents each data value, x ¯ is the mean of the data values, n is the number of data values, and the capital sigma ( Σ ) indicates that we take a sum. To compute the standard deviation using the formula, we follow the steps below: Compute the mean of all the data values. Subtract the mean from each data value. Square those differences. Add up the results in step 3. Divide the result in step 4 by n - 1 Take the square root of the result in step 5. Let’s see that process in action. Computing the Standard Deviation You surveyed some of your friends to find out how many hours they work each week. Their responses were: 5, 20, 8, 10, 35, 12. What is the standard deviation? Let’s follow the six steps mentioned previously to compute the standard deviation. Step 1 : Find the mean: x ¯ = 5 + 20 + 8 + 10 + 35 + 12 6 = 15 . Step 2 : Subtract the mean from each data value. To help keep track, let’s do this in a table. In the first row, we’ll list each of our data values (and we’ll label the row x ); in the second, we’ll subtract x ¯ = 15 from each data value. x 5 20 8 10 35 12 x − x ¯ −10 5 –7 –5 20 –3 Step 3: Square the differences. Let’s add a row to our table for those values: x 5 20 8 10 35 12 x − x ¯ −10 5 –7 –5 20 –3 ( x − x ¯ ) 2 100 25 49 25 400 9 Step 4: Add up those squares: 100 + 25 + 49 + 25 + 400 + 9 = 608 . Step 5: Divide the sum by n - 1 . Since we have 6 data values, that gives us 608 6 - 1 = 121.6 . Step 6: Take the square root of the result: 121.6 ≈ 11.027 . Thus, the standard deviation is s ≈ 11.027 . The computation for the standard deviation is complicated, even for just a small dataset. We’d never want to compute it without technology for a large dataset! Luckily, technology makes this calculation easy. Find the Standard Deviation Using Google Sheets Finding the Standard Deviation with Google Sheets The data in “AvgSAT” contains the average SAT score for students attending every institution of higher learning in the US for which data is available. What is the standard deviation of these average SAT scores? To find the standard deviation, we click in an empty cell in our spreadsheet and then type “=STDEV(”. Next, click on the letter at the top of the column containing our data; this will put a reference to that column into our formula. Then close the parentheses with and hit the enter key. The formula is replaced with the result: 125.517. Check Your Understanding Key Terms range standard deviation Key Concepts The range of a dataset is the difference between its largest and smallest values. The standard deviation is approximately the mean difference (in absolute value) that individual units fall from the mean of the dataset. Formulas s = ∑ ( x - x ¯ ) 2 n - 1 Here, s is the standard deviation, x represents each data value, x ¯ is the mean of the data values, n is the number of data values, and the capital sigma ( Σ ) indicates that we take a sum. Videos Find the Minimum and Maximum Using Google Sheets Find the Standard Deviation Using Google Sheets", "section": "Range and Standard Deviation", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Percentiles Two students graduating with the same class rank could be in different percentiles depending on the school population. (credit: \"graduation caps\" by John Walker/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute percentiles. Solve application problems involving percentiles. A college admissions officer is comparing two students. The first, Anna, finished 12th in her class of 235 people. The second, Brian, finished 10th in his class of 170 people. Which of these outcomes is better? Certainly 10 is less than 12, which favors Brian, but Anna’s class was much bigger. In fact, Anna beat out 223 of her classmates, which is 223 235 ≈ 95 % of her classmates, while Brian bested 160 out of 170 people, or 94%. Comparing the proportions of the data values that are below a given number can help us evaluate differences between individuals in separate populations. These proportions are called percentiles . If p % of the values in a dataset are less than a number n , then we say that n is at the p th percentile. Finding Percentiles There are some other terms that are related to \"percentile\" with meanings you may infer from their roots. Remember that the word percent means “per hundred.” This reflects that percentiles divide our data into 100 pieces. The word quartile has a root that means “four.” So, if a data value is at the first quantile of a dataset, that means that if you break the data into four parts (because of the quart- ), this data value comes after the first of those four parts. In other words, it’s greater than 25% of the data, placing it at the 25th percentile. Quintile has a root meaning “five,” so a data value at the third quintile is greater than three-fifths of the data in the set. That would put it at the 60th percentile. The general term for these is quantiles (the root quant – means “number”). In Mean, Median, and Mode , we defined the median as a number that is greater than no more than half of the data in a dataset and is less than no more than half of the data in the dataset. With our new term, we can more easily define it: The median is the value at the 50th percentile (or second quartile). Let’s look at some examples. Finding Percentiles Consider the dataset 5, 8, 12, 1, 2, 16, 2, 15, 20, 22. At what percentile is the value 5? What value is at the 60th percentile? Before we can answer these two questions, we must put the data in increasing order: 1, 2, 2, 5, 8, 12, 15, 16, 20, 22. There are three values (1, 2, and 2) in the set that are less than 5, and there are ten values in the set. Thus, 5 is at the 3 10 × 100 = 30 th percentile. To find the value at the 60th percentile, we note that there are ten data values, and 60% of ten is six. Thus, the number we want is greater than exactly six of the data values. Thus, the 60th percentile is 15. In each of the examples above, the computations were made easier by the fact that the we were looking for percentiles that “came out evenly” with respect to the number of values in our dataset. Things don’t always work out so cleanly. Further, different sources will define the term percentile in different ways. In fact, Google Sheets has three built-in functions for finding percentiles, none of which uses our definition. Some of the definitions you’ll see differ in the inequality that is used. Ours uses “less than or equal to,” while others use “less than” (these correspond roughly to Google Sheets’ ‘PERCENTILE.INC’ and ‘PERCENTILE.EXC’). Some of them use different methods for interpolating values. (This is what we did when we first computed medians without technology; if there were an even number of data values in our dataset, found the mean of the two values in the middle. This is an example of interpolation. Most computerized methods use this technique.) Other definitions don’t interpolate at all, but instead choose the closest actual data value to the theoretical value. Fortunately, for large datasets, the differences among the different techniques become very small. So, with all these different possible definitions in play, what will we use? For small datasets, if you’re asked to compute something involving percentiles without technology , use the technique we used in the previous example. In all other cases, we’ll keep things simple by using the built-in ‘PERCENTILE’ and ‘PERCENTRANK’ functions in Google Sheets (which do the same thing as the ‘PERCENTILE.INC’ and ‘PERCENTRANK.INC’ functions; they’re “inclusive, interpolating” definitions). Using RANK, PERCENTRANK, and PERCENTILE in Google Sheets Using Google Sheets to Compute Percentiles: Average SAT Scores The data in “AvgSAT” contains the average SAT score for students attending every institution of higher learning in the US for which data is available. What score is at the 3rd quartile? What score is at the 40th percentile? At what percentile is Albion College in Michigan (average SAT: 1132)? At what percentile is Oregon State University (average SAT: 1205)? The 3rd quartile is the 75th percentile, so we’ll use the PERCENTILE function. Click on an empty cell, and type “=PERCENTILE(“. Next, enter the data: click on the letter at the top of the column containing the average SAT scores. Then, tell the function which percentile we want; it needs to be entered as a decimal. So, type a comma (to separate the two pieces of information we’re giving this function), then type “0.75” and close the parentheses with “)”. The result should look like this (assuming the data are in column C): “=PERCENTILE(C:C, 0.75)”. When you hit the enter key, the formula will be replaced with the average SAT score at the 75th percentile: 1199. Using the PERCENTILE function, we find that an average SAT of 1100 is at the 40th percentile. Since we want to know the percentile for a particular score, we’ll use the PERCENTRANK function. Like the PERCENTILE function, we need to give PERCENTRANK two pieces of information: the data, and the value we care about. So, click on an empty cell, type “=PERCENTRANK(“, and then click on the letter at the top of the column containing the data. Next, type a comma and then the value we want to find the percentile for; in this case, we’ll type “, 1132”. Finally, close the parentheses with “)” and hit the enter key. The formula will be replaced with the information we want: an average SAT of 1132 is at the 54th percentile. Using the PERCENTRANK function, an average SAT of 1205 is at the 76.1th percentile. Using Google Sheets to Compute Percentiles: In-State Tuition The dataset \"InState\" contains the in-state tuitions of every college and university in the country that reported that data to the Department of Education. Use that dataset to answer these questions. What tuition is at the second quintile? What tuition is at the 95th percentile? At what percentile is Walla Walla University in Washington (in-state tuition: $28,035)? At what percentile is the College of Saint Mary in Nebraska (in-state tuition: $20,350)? The second quintile is the 40th percentile; using PERCENTILE in Google Sheets, we get $8,400. Using PERCENTILE again, we get $44,866. Using PERCENTRANK, we get the 81.6th percentile. Using PERCENTRANK, we get the 74.8th percentile. Check Your Understanding Key Terms percentile quartile quintile quantile Key Concepts The percentile rank of a data value is the percentage of all values in the dataset that are less than or equal to the given value. Videos Using RANK, PERCENTRANK, and PERCENTILE in Google Sheets", "section": "Percentiles", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Normal Distribution Symmetric, bell-shaped distributions arise naturally in many different situations, including coin flips. Learning Objectives After completing this section, you should be able to: Describe the characteristics of the normal distribution. Apply the 68-95-99.7 percent groups to normal distribution datasets. Use the normal distribution to calculate a z -score. Find and interpret percentiles and quartiles. Many datasets that result from natural phenomena tend to have histograms that are symmetric and bell-shaped. Imagine finding yourself with a whole lot of time on your hands, and nothing to keep you entertained but a coin, a pencil, and paper. You decide to flip that coin 100 times and record the number of heads. With nothing else to do, you repeat the experiment ten times total. Using a computer to simulate this series of experiments, here’s a sample for the number of heads in each trial: 54, 51, 40, 42, 53, 50, 52, 52, 47, 54 It makes sense that we’d get somewhere around 50 heads when we flip the coin 100 times, and it makes sense that the result won’t always be exactly 50 heads. In our results, we can see numbers that were generally near 50 and not always 50, like we thought. Moving Toward Normality Let’s take a look at a histogram for the dataset in our section opener: This is interesting, but the data seem pretty sparse. There were no trials where you saw between 43 and 47 heads, for example. Those results don’t seem impossible; we just didn’t flip enough times to give them a chance to pop up. So, let’s do it again, but this time we'll perform 100 coin flips 100 times. Rather than review all 100 results, which could be overwhelming, let's instead visualize the resulting histogram. From the histogram, we see that most of the trials resulted in between, say, 44 and 56 heads. There were some more unusual results: one trial resulted in 70 heads, which seems really unlikely (though still possible!). But we’re starting to maybe get a sense of the distribution. More data would help, though. Let’s simulate another 900 trials and add them to the histogram! We can still see that 70 is a really unusual observation, though we came close in another trial (one that had 68 heads). Now, the distribution is coming more into focus: It looks quite symmetric and bell-shaped. Let’s just go ahead and take this thought experiment to an extreme conclusion: 10,000 trials. The distribution is pretty clear now. Distributions that are symmetric and bell-shaped like this pop up in all sorts of natural phenomena, such as the heights of people in a population, the circumferences of eggs of a particular bird species, and the numbers of leaves on mature trees of a particular species. All of these have bell-shaped distributions. Additionally, the results of many types of repeated experiments generally follow this same pattern, as we saw with the coin-flipping example; this fact is the basis for much of the work done by statisticians. It’s a fact that’s important enough to have its own name: the Central Limit Theorem. John Kerrich Having enough time on your hands to actually perform this coin-flipping experiment may sound far-fetched, but the English mathematician John Kerrich found himself in just such a situation. While he was studying abroad in Denmark in 1940, that country was invaded by the Germans. Kerrich was captured and placed in an internment camp, where he remained for the duration of the war. Kerrich knew that he had all kinds of time on his hands. He also studied statistics, so he knew what should happen theoretically if he flipped a coin many, many times. He also knew of nobody who had ever tested that theory with an actual, large-scale experiment. So, he did it: While he was incarcerated, Kerrich flipped a regular coin 10,000 times and recorded the results. Sure enough, the theory held up! The Normal Distribution In the coin flipping example above, the distribution of the number of heads for 10,000 trials was close to perfectly symmetric and bell-shaped: Because distributions with this shape appear so often, we have a special name for them: normal distributions . Normal distributions can be completely described using two numbers we’ve seen before: the mean of the data and the standard deviation of the data. You may remember that we described the mean as a measure of centrality; for a normal distribution, the mean tells us exactly where the center of the distribution falls. The peak of the distribution happens at the mean (and, because the distribution is symmetric, it’s also the median). The standard deviation is a measure of dispersion; for a normal distribution, it tells us how spread out the histogram looks. To illustrate these points, let’s look at some examples. Identifying the Mean of a Normal Distribution This graph shows three normal distributions. What are their means? Step 1: Take a look at the three curves on the graph. Since the mean of a normal distribution occurs at the peak, we should look for the highest point on each distribution. Let’s draw a line from each curve's peak down to the axis, so we can see where these peaks occur: Step 2: The peak of the red (leftmost) distribution occurs over the number 1 on the horizontal axis. Thus, the mean of the red distribution is 1. Similarly, the mean of the blue (middle) distribution is 2, and the mean of the yellow (rightmost) distribution is 3. Identifying the Standard Deviation of a Normal Distribution This graph shows three distributions, all with mean 2. What are their standard deviations? Step 1: Identifying the standard deviation from a graph can be a little bit tricky. Let’s focus in on the yellow (lowest peaked) curve: Step 2: Notice that the graph curves downward in the middle, and curves upwards on the ends. Highlighted in red is the part that curves downward and in green, the part that curves upward: Step 3: The places where the graph changes from curving up to curving down (or vice versa) are called inflection points . Let’s identify where those occur by dropping a line straight down from each: Step 4: We can estimate that the inflection points occur at x = - 1 and x = 5 ; the mean is at x = 2 (as shown by the middle dotted line). The difference between the mean and the location of either inflection point is the standard deviation; since 5 - 2 = 2 - ( - 1 ) = 3 , we conclude that the standard deviation of the green distribution is 3. Step 5: Now, looking at the other two graphs, let's first identify the inflection points: Step 6: The red (tallest peaked) distribution has inflection points at 1 and 3, and the mean is 2. Thus, the standard deviation of the red distribution is 3 - 2 = 2 - 1 = 1 . The blue (lower peaked) distribution has inflection points at 0 and 4, and its mean is also 2. So, the standard deviation of the blue distribution is 2. Let’s put it all together to identify a completely unknown normal distribution. Identifying the Mean and Standard Deviation of a Normal Distribution Using the graph, identify the mean and standard deviation of the normal distribution. Step 1: Let’s start by putting dots on the graph at the peak and at the inflection points, then drop lines from those points straight down to the axis: Step 2: From the red (middle) line, we can see that the mean of this distribution is 55. The blue (outermost) lines are each 3 units away from the mean (at 52 and 58), so the standard deviation is 3. Properties of Normal Distributions: The 68-95-99.7 Rule The most important property of normal distributions is tied to its standard deviation. If a dataset is perfectly normally distributed, then 68% of the data values will fall within one standard deviation of the mean. For example, suppose we have a set of data that follows the normal distribution with mean 400 and standard deviation 100. This means 68% of the data would fall between the values of 300 (one standard deviation below the mean: 400 - 100 = 300 ) and 500 (one standard deviation above the mean: 400 + 100 = 500 ). Looking at the histogram below, the shaded area represents 68% of the total area under the graph and above the axis: Since 68% of the area is in the shaded region, this means that 100 % − 68 % = 32 % of the area is found in the unshaded regions. We know that the distribution is symmetric, so that 32% must be divided evenly into the two unshaded tails: 16% in each. Of course, datasets in the real world are never perfect; when dealing with actual data that seem to follow a symmetric, bell-shaped distribution, we’ll give ourselves a little bit of wiggle room and say that approximately 68% of the data fall within one standard deviation of the mean. The rule for one standard deviation can be extended to two standard deviations. Approximately 95% of a normally distributed dataset will fall within 2 standard deviations of the mean. If the mean is 400 and the standard deviation is 100, that means 95% calculation describes the way we compute standardized scores. (two standard deviations below the mean: 400 – 2 × 100 = 200 ) and 600 (two standard deviations above the mean: 400 + 2 × 100 = 600 ). We can visualize this in the following histogram: As before, since 95% of the data are in the shaded area, that leaves 5% of the data to go into the unshaded tails. Since the histogram is symmetric, half of the 5% (that’s 2.5%) is in each. We can even take this one step further: 99.7% of normally distributed data fall within 3 standard deviations of the mean. In this example, we’d see 99.7% of the data between 100 (calculated as 400 – 3 × 100 = 100 ) and 700 (calculated as 400 + 3 × 100 = 700 ). We can see this in the histogram below, although you may need to squint to find the unshaded bits in the tails! This observation is formally known as the 68-95-99.7 Rule . Using the 68-95-99.7 Rule to Find Percentages If data are normally distributed with mean 8 and standard deviation 2, what percent of the data falls between 4 and 12? If data are normally distributed with mean 25 and standard deviation 5, what percent of the data falls between 20 and 30? If data are normally distributed with mean 200 and standard deviation 15, what percent of the data falls between 155 and 245? Let’s look at a table that sets out the data values that are even multiples of the standard deviation (SD) above and below the mean: mean - 3 × SD mean - 2 × SD mean - 1 × SD Mean mean + 1 × SD mean + 2 × SD mean + 3 × SD 2 4 6 8 10 12 14 Since 4 and 12 represent two standard deviations above and below the mean, we conclude that 95% of the data will fall between them. Let’s build another table: mean - 3 × SD mean - 2 × SD mean - 1 × SD Mean mean + 1 × SD mean + 2 × SD mean + 3 × SD 10 15 20 25 30 35 40 We can see that 20 and 30 represent one standard deviation above and below the mean, so 68% of the data fall in that range. Let’s make one more table: mean - 3 × SD mean - 2 × SD mean − 1 × SD Mean mean + 1 × SD mean + 2 × SD mean + 3 × SD 155 170 185 200 215 230 245 Since 155 and 245 are three standard deviations above and below the mean, we know that 99.7% of the data will fall between them. Using the 68-95-99.7 Rule to Find Data Values If data are distributed normally with mean 100 and standard deviation 20, between what two values will 68% of the data fall? If data are distributed normally with mean 0 and standard deviation 15, between what two values will 95% of the data fall? If data are distributed normally with mean 14 and standard deviation 2, between what two values will 99.7% of the data fall? The 68-95-99.7 Rule tells us that 68% of the data will fall within one standard deviation of the mean. So, to find the values we seek, we’ll add and subtract one standard deviation from the mean: 100 - 1 × 20 = 80 and 100 + 1 × 20 = 120 . Thus, we know that 68% of the data fall between 80 and 120. Using the 68-95-99.7 Rule again, we know that 95% of the data will fall within 2 standard deviations of the mean. Let’s add and subtract two standard deviations from that mean: 0 - 2 × 15 = - 30 and 0 + 2 × 15 = 30 . So, 95% of the data will fall between -30 and 30. Once again, the 68-95-99.7 Rule tells us that 99.7% of the data will fall within three standard deviations of the mean. So, let’s add and subtract three standard deviations from the mean: and 14 + 3 × 2 = 20 . Thus, we conclude that 99.7% of the data will fall between 8 and 20. There are more problems we can solve using the 68-95-99.7 Rule. but first we must understand what the rule implies. Remember, the rule says that 68% of the data falls within one standard deviation of the mean. Thus, with normally distributed data with mean 100 and standard deviation 10, we have this distribution: Since we know that 68% of the data lie within one standard deviation of the mean, the implication is that 32% of the data must fall beyond one standard deviation away from the mean. Since the histogram is symmetric, we can conclude that half of the 32% (or 16%) is more than one standard deviation above the mean and the other half is more than one standard deviation below the mean: Further, we know that the middle 68% can be split in half at the peak of the histogram, leaving 34% on either side: So, just the “68” part of the 68-95-99.7 Rule gives us four other proportions in addition to the 68% in the rule. Similarly, the “95” and “99.7” parts each give us four more proportions: We can put all these together to find even more complicated proportions. For example, since the proportion between 100 and 120 is 47.5% and the proportion between 100 and 110 is 34%, we can subtract to find that the proportion between 110 and 120 is 47.5 - 34 = 13.5 % : Finding Other Proportions Using the 68-95-99.7 Rule Assume that we have data that are normally distributed with mean 80 and standard deviation 3. What proportion of the data will be greater than 86? What proportion of the data will be between 74 and 77? What proportion of the data will be between 74 and 83? Before we can answer these questions, we must mark off sections that are multiples of the standard deviation away from the mean: To figure out what proportion of the data will be greater than 86, let's start by shading in the area of data that are above 86 in our figure, or the data more than two standard deviations above the mean. We saw in that this proportion is 2.5%. To figure out what proportion of the data will be between 74 and 77, let's start by shading in that area of data. These are data that are more than one but less than two standard deviations below the mean. From , we know that the proportion of data less than two standard deviations below the mean is 47.5%. And, from YOUR TURN 8.33 , we know that 34% of the data is less than one standard deviation below the mean: Subtracting, we see that the proportion of data between 74 and 77 is 13.5%. To figure out what proportion of the data will be between 74 and 83, let's start by shading in that area of data in our figure. Next, we'll break this region into two pieces at the mean: From , we know the blue (leftmost) region represents 47.5% of the data. And, using YOUR TURN 8.33 , we get that the red (rightmost) region covers 34% of the data. Adding those together, the proportion we want is 81.5%. Standardized Scores When we want to apply the 68-95-99.7 Rule, we must first figure out how many standard deviations above or below the mean our data fall. This calculation is common enough that it has its own name: the standardized score . Values above the mean have positive standardized scores, while those below the mean have negative standardized scores. Since it's common to use the letter z to represent a standard score, this value is also often referred to as a z -score . So far, we’ve only really considered z -scores that are whole numbers, but in general they can be any number at all. For example, if we have data that are normally distributed with mean 80 and standard deviation 6, the value 85 is five units above the mean, which is less than one standard deviation. Dividing by the standard deviation, we get 5 6 . Since 85 is 5 6 of one standard deviation above the mean, we’d say that the standardized score for 85 is z = 5 6 (which is positive, since 85 > 80 ). This calculation describes the way we compute standardized scores. If x is a member of a normally distributed dataset with mean µ and standard deviation σ , then the standardized score for x is z = x - µ σ . If you know a z -score but not the original data value x , you can find it by solving the previous equation for x : x = µ + z × σ . The symbols μ and σ are the Greek letters mu and sigma . They are the analogues of the English letters m and s , which stand for mean and standard deviation . If you convert every data value in a dataset into its z -score, the resulting set of data will have mean 0 and standard deviation 1. This is why we call these standardized scores : the normal distribution with mean 0 and standard deviation 1 is often called the standard normal distribution . Standardizing Data Suppose we have data that are normally distributed with mean 50 and standard deviation 6. Compute the standardized scores (rounded to three decimal places) for these data values: 52 40 68 For each of these, we’ll plug the given values into the formula. Remember, the mean is μ = 50 and the standard deviation is σ = 6 : z = x - µ σ = 52 - 50 6 = 0.333 z = x - µ σ = 40 - 50 6 = - 1.667 z = x - µ σ = 68 - 50 6 = 3 Converting Standardized Scores to Original Values Suppose we have data that are normally distributed with mean 10 and standard deviation 2. Convert the following standardized scores into data values. 1.4 −0.9 3.5 We’ll use the formula previously introduced to convert z -scores into x -values. In this case, the mean is µ = 10 and the standard deviation is σ = 2 : x = µ + z × σ = 10 + 1.4 × 2 = 12.8 x = µ + z × σ = 10 + ( - 0.9 ) × 2 = 8.2 x = µ + z × σ = 10 + 3.5 × 2 = 17 Using Google Sheets to Find Normal Percentiles The 68-95-99.7 Rule is great when we’re dealing with whole-number z -scores. However, if the z -score is not a whole number, the Rule isn’t going to help us. Luckily, we can use technology to help us out. We’ll talk here about the built-in functions in Google Sheets, but other tools work similarly. Let’s say we’re working with normally distributed data with mean 40 and standard deviation 7, and we want to know at what percentile a data value of 50 would fall. That corresponds to finding the proportion of the data that are less than 50. If we create our histogram and mark off whole-number multiples of the standard deviation like we did before, we’ll see why the 68-95-99.7 Rule isn’t going to help: Since 50 doesn’t line up with one of our lines, the 68-95-99.7 Rule fails us. Looking back at and , the best we can say is that 50 is between the 84th and 99.5th percentiles, but that’s a pretty wide range. Google Sheets has a function that can help; it’s called NORM.DIST. Here’s how to use it: Click in an empty cell in your worksheet. Type “=NORM.DIST(“ Inside the parentheses, we must enter a list of four things, separated by commas: the data value, the mean, the standard deviation, and the word “TRUE”. These have to be entered in this order! Close the parentheses, and hit Enter. The result is then displayed in the cell; convert it to a percent to get the percentile. So, for our example, we should type “=NORM.DIST(50, 40, 7, TRUE)” into an empty cell, and hit Enter. The result is 0.9234362745; converting to a percent and rounding, we can conclude that 50 is at the 92nd percentile. Let’s walk through a few more examples. Using Google Sheets to Find Percentiles Suppose we have data that are normally distributed with mean 28 and standard deviation 4. At what percentile do each of the following data values fall? 30 23 35 By entering “=NORM.DIST(30, 28, 4, TRUE)” we find that 30 is at the 69th percentile. By entering “=NORM.DIST(23, 28, 4, TRUE)” we find that 23 is at the 11th percentile. By entering “=NORM.DIST(35, 28, 4, TRUE)” we find that 35 is at the 96th percentile. Google Sheets can also help us go the other direction: If we want to find the data value that corresponds to a given percentile, we can use the NORM.INV function. For example, if we have normally distributed data with mean 150 and standard deviation 25, we can find the data value at the 30th percentile as follows: Click on an empty cell in your worksheet. Type “=NORM.INV(“ Inside the parentheses, we’ll enter a list of three numbers, separated by commas: the percentile in question expressed as a decimal, the mean, and the standard deviation. These must be entered in this order! Close the parentheses and hit Enter. The desired data value will be in the cell! In our example, we want the 30th percentile; converting 30% to a decimal gives us 0.3. So, we’ll type “=NORM.INV(0.3, 150, 25)” to get 136.8899872; let’s round that off to 137. Using Google Sheets to Find the Data Value Corresponding to a Percentile Suppose we have data that are normally distributed with mean 47 and standard deviation 9. Find the data values (rounded to the nearest tenth) corresponding to these percentiles: 75th (that’s the third quartile) 12th 90th By entering “=NORM.INV(0.75, 47, 9)” we find that 53.1 is at the 75th percentile. By entering “=NORM.INV(0.12, 47, 9)” we find that 36.4 is at the 12th percentile. By entering “=NORM.INV(0.9, 47, 9)” we find that 58.5 is at the 90th percentile. Check Your Understanding normal distribution 68-95-99.7 Rule standardized score ( z -score) Key Concepts Normally distributed data follow a bell-shaped, symmetrical distribution. The mean of normally distributed data falls at the peak of the distribution. The standard deviation of normally-distributed data is the distance from the peak to either of the inflection points. Data that are normally distributed follow the 68-95-99.7 Rule, which says that approximately 68% of the data fall within one standard deviation of the mean, 95% fall within two standard deviations, and 99.7% fall within three standard deviations. The z -score for a data value is the number of standard deviations that value falls above (or below, if the z -score is negative) the mean. We can use the normal distribution to estimate percentiles. Formulas If x is a member of a normally distributed dataset with mean µ and standard deviation σ , then the standardized score for x is z = x - µ σ . If you know a z -score but not the original data value x , you can find it by solving the previous equation for x : x = µ + z × σ .", "section": "The Normal Distribution", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Applications of the Normal Distribution Standardized test results generally adhere to the normal distribution. (credit: “Taking a Test” by Marco Verch Professional Photographer/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Apply the normal distribution to real-world scenarios. As we saw in The Normal Distribution , the word “standardized” is closely associated with the normal distribution. This is why tests like college entrance exams, state achievement tests for K–12 students, and Advanced Placement tests are often called “standardized tests”: scores are assigned in a way that forces them to follow a normal distribution, with a mean and standard deviation that are consistent from year to year. Standardization also allows people like college admissions officers to directly compare an applicant who took the ACT (a college entrance exam) to an applicant who instead chose to take the SAT (a different college entrance exam). Standardization allows us to compare individuals from different groups; this is among the most important applications of the normal distribution. We’ll explore this and other real-world uses of the normal distribution in this section. College Entrance Exams There are two good ways to compare two data values from different groups: using z -scores and using percentiles. The two methods will always give consistent results (meaning that we won’t find, for example, that the first value is better using z -scores but the second value is better using percentiles), so use whichever method is more comfortable for you. Evaluating College Entrance Exam Scores According to the Digest of Education Statistics , composite scores on the SAT have mean 1060 and standard deviation 195, while composite scores on the ACT have mean 21 and standard deviation 5. At what percentile would an SAT score of 990 fall? What is the z -score of an ACT score of 27? Which is better: a score of 1450 on the SAT or 29 on the ACT? Using Google Sheets, we can answer this question with the formula “=NORM.DIST(990, 1060, 195, TRUE)”. A score of 990 would fall at the 36th percentile. Using the formula z = x - µ s , we get z = 27 - 21 5 = 1.2 . Let’s compare the values using both percentiles and z -values: Percentiles: Using “=NORM.DIST(1450, 1060, 195, TRUE)” we find that an SAT score of 1450 is at the 98th percentile. Meanwhile, by entering “=NORM.DIST(29, 21, 5, TRUE)” we see that an ACT score of 29 is around the 95th percentile. Since it’s at a higher percentile, we can conclude that an SAT score of 1450 is better than an ACT score of 29. z -scores: Using the formula, we see that the z -score for an SAT score of 1450 is z = 1450 - 1060 195 = 2 , while the z -score for an ACT score of 29 is z = 29 - 21 5 = 1.6 . Since it has a higher z -score, an SAT score of 1450 is better than an ACT score of 29. Coin flipping In the opening of The Normal Distribution , we saw that the number of heads we get when we flip a coin 100 times is distributed normally. It can be shown that if n is the number of flips, then the mean of that distribution is n 2 and the standard deviation is n 2 (as long as n ≥ 20 ). So, for 100 flips, the mean of the distribution is 50 and the standard deviation is 5. In that opening example, one of our early runs gave us 70 heads in 100 flips, which we noted seemed unusual. Using the normal distribution, we can identify exactly how unusual that really is. Using Google Sheets, the formula “=NORM.DIST(70, 50, 5, TRUE)” gives us 0.999968, which is the 99.997th percentile! How is that useful? Suppose you need to test whether a coin is fair, and so you flip it 100 times. While we might be suspicious if we get 70 heads out of the 100 flips, we now have a numerical measure for how unusual that is: If the coin were fair, we would expect to see 70 heads (or more) only 100 - 99.9968 = 0.0032 % of the time. That’s really unlikely! Analysis like this is related to hypothesis testing , an important application of statistics in the sciences and social sciences. Flipping a Coin Let’s say we flip a coin 64 times and count the number of heads. What would be the mean of the corresponding distribution? What would be the standard deviation of the corresponding distribution? Suppose we got 25 heads, which seems a little low. At what percentile would 25 heads fall? Since n = 64 , the mean is 64 2 = 32 . Again using n = 64 , we get a standard deviation of 64 2 = 4 . Using “=NORM.DIST(25, 32, 4, TRUE)”, we see that 25 heads is at the 4th percentile. Reading and Interpreting Scatter Plots Analyzing Data That Are Normally Distributed Whenever we’re working with a dataset that has a distribution that looks symmetric and bell-shaped, we can use techniques associated with the normal distribution to analyze the data. Using Normal Techniques to Analyze Data The data in “AvgSAT” contains the average SAT score for students attending every institution of higher learning in the United States for which data is available. In , we created a histogram for these data: (data source: https://data.ed.gov) This distribution is fairly symmetric (it’s just a little right-skewed) and bell-shaped, so we can use normal distribution techniques to analyze the data. What is the mean of these average SAT scores? What is the standard deviation of these SAT scores? Using the answers to the previous two questions, use NORM.DIST in Google Sheets to estimate at what percentile the University at Buffalo in New York (average SAT: 1250) falls. Use PERCENTRANK to find the actual percentile of the University at Buffalo, and see how close the estimate in the previous question came. Using the AVERAGE function in Google Sheets, we find that the mean is 1141.174. Using the STDEV function, we get that the standard deviation is 125.517. Entering “=NORM.DIST(1250, 1141, 125.517, TRUE)” into Google Sheets, we estimate that the University at Buffalo is at the 81st percentile. Using PERCENTRANK, we find that the actual percentile is the 84th. These are close! Political Meddling Exposed The normal distribution pops up in some unusual places. Recently, a team at Duke University has been using statistics to help identify partisan gerrymandering, where electoral districts have been carefully drawn in a way that benefits one political party over another. In their analysis, they found that hypothetical election results in randomly drawn districts are normally distributed. By using techniques similar to the ones we used above, they can quantify precisely how biased a particular electoral map is by finding the percentile rank of the actual election result on the normal distribution of the hypothetical results. You can find out more about their work at the \"Quantifying Gerrymandering\" site here. Check Your Understanding Key Concepts We can use z -scores to compare data values from different datasets.", "section": "Applications of the Normal Distribution", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Scatter Plots, Correlation, and Regression Lines A scatter plot is a visualization of the relationship between quantitative dataset. Learning Objectives After completing this section, you should be able to: Construct a scatter plot for a dataset. Interpret a scatter plot. Distinguish among positive, negative and no correlation. Compute the correlation coefficient. Estimate and interpret regression lines. One of the most powerful tools statistics gives us is the ability to explore relationships between two datasets containing quantitative values, and then use that relationship to make predictions. For example, a student who wants to know how well they can expect to score on an upcoming final exam may consider reviewing the data on midterm and final exam scores for students who have previously taken the class. It seems reasonable to expect that there is a relationship between those two datasets: If a student did well on the midterm, they were probably more likely to do well on the final than the average student. Similarly, if a student did poorly on the midterm, they probably also did poorly on the final exam. Of course, that relationship isn’t set in stone; a student’s performance on a midterm exam doesn’t cement their performance on the final! A student might use a poor result on the midterm as motivation to study more for the final. A student with a really good grade on the midterm might be overconfident going into the final, and as a result doesn’t prepare adequately. The statistical method of regression can find a formula that does the best job of predicting a score on the final exam based on the student’s score on the midterm, as well as give a measure of the confidence of that prediction! In this section, we’ll discover how to use regression to make these predictions. First, though, we need to lay some graphical groundwork. Relationships Between Quantitative Datasets Before we can evaluate a relationship between two datasets, we must first decide if we feel that one might depend on the other. In our exam example, it is appropriate to say that the score on the final depends on the score on the midterm, rather than the other way around: if the midterm depended on the final, then we’d need to know the final score first , which doesn’t make sense. Here’s another example: if we collected data on home purchases in a certain area, and noted both the sale price of the house and the annual household income of the purchaser, we might expect a relationship between those two. Which depends on the other? In this case, sale price depends on income: people who have a higher income can afford a more expensive house. If it were the other way around, people could buy a new, more expensive house and then expect a raise! (This is very bad advice.) It's worth noting that not every pair of related datasets has clear dependence. For example, consider the percent of a country’s budget devoted to the military and the percent earmarked for public health. These datasets are generally related: as one goes up, the other goes down. However, in this case, there’s not a preferred choice for dependence, as each could be seen as depending on the other. When exploring the relationship between two datasets, if one set seems to depend on the other, we’ll say that dataset contains values of the response variable (or dependent variable ). The dataset that the response variable depends on contains values of what we call the explanatory variable (or independent variable ). If no dependence relationship can be identified, then we can assign either dataset to either role. Identifying Explanatory and Response Variables For each of the following pairs of related datasets, identify which (if any) should be assigned the role of response variable and which should be assigned to be the explanatory variable. A person’s height and weight A professional basketball player’s salary and their average points scored per game (which is a measure of how good they are at basketball) The length and width of leaves on a tree As people get taller, their weight tends to increase. But if a person goes on a diet and loses weight, we don’t expect them to also get shorter. So, weight depends on height. That means we’ll say that the response variable is weight and the explanatory variable is height . The more points a basketball player scores, the more money they should make. But if a basketball player gets a raise, we wouldn’t expect them to get better at basketball as a result. So, the response is salary and the explanatory is points per game . The way that the length and width of leaves are connected isn’t clear. It seems reasonable that as the width goes up, so would the length. But the other direction is also plausible: as the length goes up, so does the width. Without a clear dependence relationship, we’re free to declare either to be the response and the other to be the explanatory. Once we’ve assigned roles to our two datasets, we can take the first step in visualizing the relationship between them: creating a scatter plot. Creating Scatter Plots A scatter plot is a visualization of the relationship between two quantitative sets of data. The scatter plot is created by turning the datasets into ordered pairs: the first coordinate contains data values from the explanatory dataset, and the second coordinate contains the corresponding data values from the response dataset. These ordered pairs are then plotted in the xy -plane. Let's return to our exam example to put this into practice. Creating Scatter Plots Without Technology Students are exploring the relationship between scores on the midterm exam and final exam in their math course. Here are some of the scores reported by their classmates: Name Midterm grade Final grade Student 1 88 84 Student 2 71 80 Student 3 75 77 Student 4 94 95 Student 5 68 73 Create a scatter plot to visualize the data. Step 1: Since it makes more sense to think of the final exam score as being dependent on the midterm exam score, we’ll let the final grade be the response. So, let’s think of these two datasets as a set of ordered pairs, midterm first, final second: Name (Midterm, Final) Student 1 (88, 84) Student 2 (71, 80) Student 3 (75, 77) Student 4 (94, 95) Student 5 (68, 73) Step 2: Next, let’s make the axes. On the horizontal axis, make sure the range of values is sufficient to cover all of the explanatory data. For our data, that’s 68 to 94. Similarly, the vertical axis should cover all of the response data (73 to 95): Step 3: Our first point is (88, 84). So, we’ll locate 88 on the horizontal axis, 84 on the vertical axis, and identify the point that’s directly above the first location and horizontally level with the second: Step 4: Repeat this process to place the other four points on the graph: Step 5: Finally, label the axes: For large datasets, it’s impractical to create scatter plots manually. Luckily, Google Sheets automates this process for us. Making Scatter Plots in Google Sheets Creating Scatter Plots in Google Sheets The dataset “NHL19” gives the results of the 2018–2019 National Hockey League season. The columns are team, wins (W), losses (L), overtime losses (OTL), total points (PTS), goals scored by the team (GF), goals scored against the team (GA), and goal differential (the difference in GF and GA). Use Google Sheets to create a scatter plot for GF vs. GA. When we talk about plotting one set versus another, the first is the response and the second is explanatory. Step 1: Open the dataset in Google Sheets, and click and drag to select the data we want to visualize (in this case, we want the columns for GF and GA; make sure you include those labels in the selection). Step 2: Next, click on the “Insert” menu, then click “Chart.” Sheets will automatically choose a chart format; if the result isn’t a scatter plot, click on the drop-down menu under “Chart type” in the Chart Editor on the right side of the window and select “Scatter chart.” Step 3: Next, check that the correct choices were made for the horizontal and vertical axes. In this case, we want to see GF on the vertical axis and GA on the horizontal axis. If Sheets made the wrong choice, we can fix it in the Chart Editor by clicking on the name of the dataset under “X-axis” to open up a dropdown menu, then selecting the variable that should go on the horizontal axis (GA in this case). Step 4: Now, click on the variable under “Series” and select the one that should go on the vertical axis (GF). If you had to make that change, the axis labels in the graph may also need changing; those labels can be fixed using the “Customize” tab in the Chart Editor under “Chart & axis titles.” Your result should look like this: (data source: www.nhl.com) Reading and Interpreting Scatter Plots Scatter plots give us information about the existence and strength of a relationship between two datasets. To break that information down, there are a series of questions we might ask to help us. First: Is there a curved pattern in the data? If the answer is “yes,” then we can stop; none of the linear regression techniques from here to the end of this section are appropriate. and show several examples of scatter plots that can help us identify these curved patterns. Curved pattern No curved pattern Curved pattern No curved pattern Once we have confirmed that there is no curved pattern in our data, we can move to the next question: Is there a linear relationship? To answer this, we must look at different values of the explanatory variable and determine whether the corresponding response values are different, on average . It's important to look at the values “on average” because, in general, our scatter plots won’t include just one corresponding response point for each value of the explanatory variable (i.e., there may be multiple response values for each explanatory value). So, we try to look for the center of those points. Let’s look again at , but consider some different values for the explanatory variable. Let’s highlight the points whose x -values are around 50 and those that are around 80: Now, we can estimate the middle of each group of points. Let's add our estimated averages to the plot as starred points: Since those two starred points occur at different heights, we can conclude that there’s likely a relationship worth exploring. Here’s another example using a different set of data: Let’s look again at the points near 50 and near 80, and estimate the middles of those clusters: Notice that there’s not much vertical distance between our two starred points. This tells us that there’s not a strong relationship between these two datasets. Positive and Negative Linear Relationships Another way to assess whether there is a relationship between two datasets in a scatter plot is to see if the points seem to be clustered around a line (specifically, a line that’s not horizontal). The stronger the clustering around that line is, the stronger the relationship. Once we’ve established that there’s a relationship worth exploring, it’s time to start quantifying that relationship. Two datasets have a positive linear relationship if the values of the response tend to increase, on average, as the values of the explanatory variable increase. If the values of the response decrease with increasing values of the explanatory variable, then there is a negative linear relationship between the two datasets. The strength of the relationship is determined by how closely the scatter plot follows a single straight line: the closer the points are to that line, the stronger the relationship. The scatter plots in to depict varying strengths and directions of linear relationships. Perfect negative relationship Strong negative relationship Weak negative relationship No relationship Weak positive relationship Strong positive relationship Perfect positive relationship The strength and direction (positive or negative) of a linear relationship can also be measured with a statistic called the correlation coefficient (denoted r ). Positive values of r indicate a positive relationship, while negative values of r indicate a negative relationship. Values of r close to 0 indicate a weak relationship, while values close to ± 1 correspond to a very strong relationship. Looking again at to , the correlation coefficients for each, in sequential order, are: ‒1, ‒0.97, ‒0.55, ‒0.03, 0.61, 0.97, and 1. There’s no firm rule that establishes a cutoff value of r to divide strong relationships from weak ones, but ± 0.7 is often given as the dividing line (i.e., if r > 0.7 or r < - 0.7 the relationship is strong, and if - 0.7 < r < 0.7 the relationship is weak). The formula for computing r is very complicated; it’s almost never done without technology. Google Sheets will do the computation for you using the CORREL function. The syntax works like this: if your explanatory values are in cells A2 to A50 and the corresponding response values are in B2 to B50, then you can find the correlation coefficient by entering “=CORREL(A2:A50, B2:B50)”. (Note that the order doesn’t matter for correlation coefficients; “=CORREL(B2:B50, A2:A50)” will give the same result.) Let’s put all of this together in an example. Interpreting Scatter Plots Consider the four scatter plots below: For each of these, answer the following questions: Is there a curved pattern in the data? If yes, stop here. If no, continue to part b. Classify the strength and direction of the relationship. Make a guess at the value of r . Yes, there is a curved pattern. No, there’s no curved pattern. Since the points tend upward as we move from left to right, this is a positive relationship. The points seem pretty closely grouped around a line, so it’s fairly strong. Comparing this scatter plot to those in to , we can see that the relationship is stronger than the one in ( r = 0.61 ) but not as strong as the one in ( r = 0.97 ). So, the value of the correlation coefficient is somewhere between the two. We might guess that r = 0.9 . No, there’s no curved pattern. Since the points tend downward as we move from left to right, this is a negative relationship. The points are not tightly grouped around a line, but the pattern is clear. It looks like it has approximately the same strength as the plot in , just with the opposite sign. So, we might guess that r = - 0.6 . No, there’s no curved pattern. Since the points don’t really tend upward or downward as we move from left to right, there is no real relationship here. Thus, r ≈ 0 . Finding the Correlation Coefficient The data that were plotted in the previous example can be found in the dataset “correlationcoefficient1” . All of them share the same values for the explanatory variable x . The four responses are labeled y 1 through y 4 . Compute the correlation coefficients for each, if appropriate, using Google Sheets. Round to the nearest hundredth. Step 1: There is a curved pattern in the data, so the correlation coefficient isn’t meaningful. Step 2: Using “=CORREL(A2:A101, C2:C101)” we get r = 0.89 . Step 3: Using “=CORREL(A2:A101, D2:D101)” we get r = - 0.66 . Step 4: Using “=CORREL(A2:A101, E2:E101)” we get r = - 0.04 . Winning with Statistics Billy Beane, the former general manager of the Oakland A’s baseball team, famously took his low budget team to unprecedented heights by using statistics to identify undervalued players; his story is recounted in the book Moneyball (which was later made into a movie, with Brad Pitt playing Beane). You can do the same thing: Take a look at team statistics in the sport of your choice and try to identify a statistic that’s most closely related to winning (meaning that it has the highest correlation coefficient with team wins). Linear Regression The final step in our analysis of the relationship between two datasets is to find and use the equation of the regression line. For a given set of explanatory and response data, the regression line (also called the least-squares line or line of best fit ) is the line that does the best job of approximating the data. What does it mean to say that a particular line does the “best job” of approximating the data? The way that statisticians characterize this “best line” is rather technical, but we’ll include it for the sake of satisfying your curiosity (and backing up the claim of \"best\"). Imagine drawing a line that looks like it does a pretty good job of approximating the data. Most of the points in the scatter plot will probably not fall exactly on the line; the distance above or below the line a given point falls is called that point’s residual . We could compute the residuals for every point in the scatter plot. If you take all those residuals and square them, then add the results together, you get a statistic called the sum of squared errors for the line (the name tells you what it is: “sum” because we’re adding, “squared” because we’re squaring, and “errors” is another word for “residuals”). The line that we choose to be the “best” is the one that has the smallest possible sum of squared errors. The implied minimization (“smallest”) is where the “least” in “least squares” comes from; the “squares” comes from the fact that we’re minimizing the sum of squared errors. This is very similar to the process we outlined in the \"game\" that we used to introduce the mean. Both the regression line and the mean are designed to minimize a sum of squared errors. Here ends the super technical part. Finding the Equation of the Regression Line So, how do we find the equation of the regression line? Recall the point-slope form of the equation of a line: If a line has slope m and passes through a point ( x 0 , y 0 ) , then the point-slope form of the equation of the line is: y = m ( x - x 0 ) + y 0 The regression line has two properties that we can use to find its equation. First, it always passes through the point of means . If x ¯ and y ¯ are the means of the explanatory and response datasets, respectively, then the point of means is ( x ¯ , y ¯ ) . We’ll use that as the point in the point-slope form of the equation. Second, if s x and s y are the standard deviations of the explanatory and response datasets, respectively, and if r is the correlation coefficient, then the slope is m = r × s y s x . Putting all that together with the point-slope formula gives us this: Suppose x and y are explanatory and response datasets that have a linear relationship. If their means are x ¯ and y ¯ respectively, their standard deviations are s x and s y respectively, and their correlation coefficient is r , then the equation of the regression line is: y = r ( s y s x ) ( x - x ¯ ) + y ¯ . Let's walk through an example. Finding the Equation of the Regression Line from Statistics Suppose you have datasets x and y with the following statistics: x has mean 21 and standard deviation 4, y has mean 8 and standard deviation 2, and their correlation coefficient is −0.4. What’s the equation of the regression line? Step 1: We’re given x ¯ = 21 , s x = 4 , y ¯ = 8 , s y = 2 , and r = - 0.4 . Let's start with the formula for the equation of the regression line: y = r ( s y s x ) ( x - x ¯ ) + y ¯ Step 2: Plugging in our values gives us: y = - 0.4 ( 2 4 ) ( x - 21 ) + 8 Step 3: Our final regression line equation is: y = - 0.2 x + 12.2 As you can see, finding the equation of the regression line involves a lot of steps if you have to find all of the values of the needed quantities yourself. But, as usual, technology comes to our rescue. This video (which you actually watched earlier when learning how to create scatter plots) covers the regression line at around the 3:30 mark. Note that Google Sheets calls it the \"trendline.\" Let's put this into practice. Finding the Equation of the Regression Line Using Google Sheets In , we considered the relationship between goals scored (GF) and goals against (GA) using the dataset “NHL19” . Recreate the scatter plot in Google Sheets, and use it to find the equation of the regression line. Once we have recreated the scatter plot, we find the equation of the regression line by clicking the three dots at the top right of the plot, selecting “Edit chart,” then clicking on “Customize” and “Series.” We add the regression line by checking the box next to “Trendline,” and then we show the equation by selecting “Use Equation” in the drop-down menu under “Label.” The equation of the tangent line is y = - 0.0554 x + 261 . Using the Equation of the Regression Line Once we’ve found the equation of the regression line, what do we do with it? We’ll look at two possible applications: making predictions and interpreting the slope. We can use the equation of the regression line to predict the response value y for a given explanatory value x . All we have to do is plug that explanatory value into the formula and see what response value results. This is useful in two ways: first, it can be used to make a guess about an unknown data value (like one that hasn’t been observed yet). Second, it can be used to evaluate performance (meaning, we can predict an outcome given a particular event). In , we created a scatter plot of final exam scores vs. midterm exam scores using this data: Name Midterm Grade Final Grade Allison 88 84 Benjamin 71 80 Carly 75 77 Daniel 94 95 Elmo 68 73 The equation of the regression line is y = 0.687 x + 27.4 , where y is the final exam score and x is the midterm exam score. If Frank scored 85 on the midterm, then our prediction for his final exam score is 0.687 × 85 + 27.4 = 85.795 . To use the regression line to evaluate performance, we use a data value we’ve already observed. For example, Allison scored 88 on the midterm. The regression line predicts that someone who scores an 88 on the midterm will get 0.687 × 88 + 27.4 = 87.856 on the final. Allison actually scored 84 on the final, meaning she underperformed expectations by almost 4 points ( 87.856 - 84 ) . The second application of the equation of the regression line is interpreting the slope of the line to describe the relationship between the explanatory and response datasets. For the exam data in the previous paragraph, the slope of the regression line is 0.687. Recall that the slope of a line can be computed by finding two points on the line and dividing the difference in the y -values of those points by the difference in the x -values. Keeping that in mind, we can interpret our slope as 0.687 = difference in final scores difference in midterm scores . Multiplying both sides of that equation by the denominator of the fraction, we get 0.687 × difference in midterm scores = difference in final scores . Thus, a one-point increase in the midterm score would result in a predicted increase in the final score of 0.687 points. A ten-point drop in the midterm score would give us a decrease in the predicted final score of 6.87 points. In general, the slope gives us the predicted change in the response that corresponds to a one unit increase in the explanatory variable. Applying the Equation of the Regression Line The data in “MLB2019Off” gives offensive team stats for the 2019 Major League Baseball season. Use that dataset to answer the following questions: What is the equation of the regression line for runs (R) vs. hits (H)? How many runs would we expect a team to score if the team got 1500 hits in a season? Did the Kansas City Royals (KCR) overperform or underperform in terms of runs scored, based on their hit total? By how much? Write a sentence to interpret the slope of the regression line. Using Google Sheets, we find that the regression line equation is y = 0.884 x – 456 , where y is the number of runs scored and x is the number of hits. Plugging 1500 into the equation of the regression line, we get 0.884 × 1500 - 456 = 870 . We would predict that a team with 1500 hits would score 870 runs. The Royals had 1356 hits, so we would predict their run total to be 0.884 × 1356 - 456 ≈ 743 . They actually scored 691 runs, so they underperformed expectations by 52 runs ( 743 – 691 ) . The slope gives us the predicted change in the response that corresponds to a one unit increase in the explanatory variable. So, we expect one additional hit to result in 0.884 more runs. Since 0.884 runs doesn’t really make sense, we can get a better interpretation by multiplying through by ten or one hundred: Ten additional hits will result in almost nine additional runs, or a hundred additional hits will yield on average just over 88 additional runs. Math and the Movies Statistics and regression are used by Hollywood movie producers to decide what movies to make, and to predict how much money they’ll earn at the box office. According to the American Statistical Association , not only do producers use statistics to identify the next potential blockbuster, but they’ve also pinned down how much money awards add to the bottom line. (An Academy Award is worth about $3 million!) In addition, studios use their streaming services to gather data about their customers and the types of movies they watch; this data helps them learn what kinds of entertainment their customers want more of. Collecting and Analyzing Your Own Data This section has demonstrated many pairs of related quantitative datasets. Think about some quantitative variables that you can ask your classmates about, which might be related. Once you have some ideas, collect the data from your classmates. Then analyze the data by creating a scatter plot, finding the equation of the regression line (if appropriate), and interpreting it. Extrapolation A very common misuse of regression techniques involves extrapolation , which involves making a prediction about something that doesn't belong in the dataset. More Applying the Equation of the Regression Line The data in “WNBA2019” gives team statistics from the 2019 WNBA season. Use that dataset to answer these questions about team wins (W) and the proportion of team field goals made (FG%, the number of shots made divided by the number of shots attempted. Even though this column is labeled using a percent sign, the values are not expressed as percentages): What is the equation of the regression line for wins vs. proportion of made field goals? How many wins would we expect for a team that makes 42% of its shots? Did the New York Liberty overperform or underperform in terms of wins, based on the team’s proportion of made field goals? Write a sentence to interpret the slope of the regression line. Using Google Sheets, we get the equation y = 178.097 x - 58.543 , where x is the proportion of field goals made and y is the number of wins. Since 42% corresponds to a proportion of 0.42, we’ll plug 0.42 into the regression equation for x , which gives us 178.097 × 0.42 - 58.543 ≈ 16.26 . We would predict that a team that makes 42% of its shots would win about 16 games. The New York Liberty made 41.4% of their shots, so we expect they would have 178.097 × 0.414 - 58.543 ≈ 15.19 wins. In fact, they had only 10 wins, so they underperformed expectations by over 5 wins. Step 1: The slope gives us the expected increase in the response that corresponds to a one unit increase in the explanatory variable. If we simply go with that interpretation, we would get a sentence like “We expect an increase in proportion of field goals made of 1 would result in an additional 178 wins.” However, that sentence doesn't make much sense. Let's consider why. Step 2: First, proportions must be between 0 and 1, the proportion of made field goals can’t be increased by 1 and still make sense. Second, the total number of games played is only 34, so no team could get an additional 178 wins! So, we’ll have to change the units. Step 3: Since the proportions of made field goals are often expressed as a percentage, we could try to use that. If we express the slope as a fraction with 1 in the denominator (remember, the denominator represents the proportion of field goals made), then convert the denominator to a percentage and simplify, we get 178.097 1 = 178.097 100 % ≈ 1.78 1 % . Step 4: So, an increase in field goal percentage of 1% would result in an expected increase of 1.78 wins. Correlation Does Not Imply Causation One of the most common fallacies about statistics has to do with the relationship between two datasets. In the dataset “Public” , we find that the correlation coefficient between the 75th percentile math SAT score and the 75th percentile verbal SAT score is 0.92, which is really strong. The slope of the regression line that predicts the verbal score from the math score is 0.729, which we might interpret as follows: “If the 75th percentile math SAT score goes up by 10 points, we’d expect the corresponding verbal SAT score to go up by just over 7 points.” Does the increasing math score cause the increase in the verbal score? Probably not. What’s really going on is that there’s a third variable that’s affecting them both: To raise the SAT math score by 10 points, a school will recruit students who do better on the SAT in general; these students will also naturally have higher SAT verbal scores. This third variable is sometimes called a lurking variable or a confounding variable . Unless all possible lurking variables are ruled out, we cannot conclude that one thing causes another. Dr. Talithia Williams A photo of Dr. Williams (credit: Used by permission of Talithia Williams) Dr. Talithia Williams is a statistician on the faculty of Harvey Mudd College, and the first Black woman to achieve tenure at this university. She advocates for more women to become involved in the fields of engineering and science, and is on the board of directors for the EDGE Foundation, an organization that helps women obtain advanced degrees in mathematics (EDGE standing for Enhancing Diversity in Graduate Education). In 2018, Dr. Williams published the book Power in Numbers: The Rebel Women of Mathematics, a retrospective look at historical female figures who have contributed to the development of the field of mathematics. Dr. Williams earned a Master’s degree in Mathematics from Howard University and a Master’s in Statistics from Rice University, and also went on to earn her Ph.D. in Statistics from Rice. She has held research appointments at the Jet Propulsion Laboratory, the National Security Agency, and NASA. Her research focuses on the environmental and medical applications of statistics. In 2014, she gave a popular TED talk titled “Own Your Body’s Data” that discussed the potential insights to be gained from collecting personal health data. She was even recently a host for the NOVA Wonders documentary series and a narrator for the NOVA Universe series on PBS. To stay up-to-date on Dr. Williams’s accomplishments, you can follow her on Twitter or her Facebook account . Statistics and Eugenics Some of the brightest minds in the history of statistics unfortunately decided to use their considerable intellects to further a pseudoscience known as eugenics. Eugenicists took Charles Darwin’s theories of evolution and ruthlessly applied them to the human race. Francis Galton (1822–1911), a cousin of Darwin and also the mathematician who invented the formula for standard deviation, claimed that people in the British upper classes possessed higher intelligence due to their superior breeding. Karl Pearson (1857–1936), who derived the formula for the correlation coefficient, argued in National Life from the Standpoint of Science , that, instead of providing social welfare programs, nations could better improve the fortunes of the poor by waging “war with inferior races.” Ronald Fisher (1890–1962) was possibly the most important statistician of the 20th century, having invented several new techniques (including the ubiquitous analysis of variance), and yet he also founded the Cambridge University Undergraduates Eugenics Society, whose self-prescribed goal was to evangelize “not by precept only, but by example, the doctrine of a new natural ability of worth and blood.” When eugenics took hold in the United States, it was used to justify terrible acts by the government, including the forced sterilization of individuals with mental illness, epilepsy, a physical impairment (like blindness), or a criminal history. The Nazi regime took these ideas to their ultimate, terrible conclusion: killing people who had mental or physical disabilities, or who were born into an “inferior” race. Over six million people died in this Holocaust, one of the darkest events in human history. To learn more, watch this video about Francis Galton and the legacy of eugenics Check Your Understanding Key Terms response variable (dependent variable) explanatory variable (independent variable) scatter plot positive linear relationship negative linear relationship correlation coefficient regression line (least-squares line, line of best fit) Key Concepts If one variable affects the value of another variable, we say the first is an explanatory variable and the second is a response variable. Scatter plots place a point in the xy -plane for each unit in the dataset. The x -value is the value of the explanatory variable, and the y -value is the value of the response variable. The correlation coefficient r gives us information about the strength and direction of the relationship between two variables. If r is positive, the relationship is positive: an increase in the value of the explanatory variable tends to correspond to an increase in the value of the response variable. If r is negative, the relationship is negative: an increase in the value of the explanatory variable tends to correspond to a decrease in the value of the response variable. Values of r that are close to 0 indicate weak relationships, while values close to –1 or indicate strong relationships. The regression line for a relationship between two variables is the line that best represents the data. It can be used to predict values of the response variable for a given value of the explanatory variable. Formulas If a line has slope m and passes through a point ( x 0 , y 0 ) , then the point-slope form of the equation of the line is: y = m ( x - x 0 ) + y 0 Suppose x and y are explanatory and response datasets that have a linear relationship. If their means are x ¯ and y ¯ respectively, their standard deviations are s x and s y respectively, and their correlation coefficient is r , then the equation of the regression line is: y = r ( s y s x ) ( x - x ¯ ) + y ¯ . Videos Making Scatter Plots in Google Sheets Projects Browse through some news websites to find five stories that report on data and include data visualizations. Can you tell from the report how the data were collected? Was randomization used? Are the visualizations appropriate for the data? Are the visualizations presented in a way that might bias the reader? We discussed three measures of centrality in this chapter: the mode, the median, and the mean. In a broader context, the mean as we discussed it is more properly called the arithmetic mean , to distinguish it from other types of means. Examples of these include the geometric mean, harmonic mean, truncated mean, and weighted mean. How are these computed? How do they compare to the arithmetic mean? In what situations would each of these be preferred to the arithmetic mean? Simpson’s Paradox is a statistical phenomenon that can sometimes appear when we observe a relationship within several subgroups of a population, but when the data for all thegroups are analyzed all together, the opposite relationship appears. Find some examples of Simpson’s Paradox in real-world situations, and write a paragraph or two that would explain the concept to someone who had never studied statistics before. Chapter Review Gathering and Organizing Data Visualizing Data Mean, Median and Mode Range and Standard Deviation Percentiles The Normal Distribution Applications of the Normal Distribution Scatter Plots, Correlation, and Regression Lines Chapter Test", "section": "Scatter Plots, Correlation, and Regression Lines", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction A road sign in Finland, a country that uses the metric system. (credit: modification of work by Anna Järvenpää/Flickr, Public Domain) You are planning a road trip from your home state to the sunny beaches of Mexico and need to prepare a budget. While in the United States, gasoline is sold in gallons and distances are measured in miles, but in almost any other country you will find that gasoline is sold in liters and distance is measured in kilometers. Whether you’re traveling, baking, watching an international sporting event, working with machine tools, or using scientific equipment, it's important to understand the metric system , or the International System of Units (SI). The metric system is a decimal measuring system that uses meters, liters, and grams to quantify length, capacity, and mass. It is used in all but three countries in the world, including the United States.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "The Metric System A scale that measures weight in both metric and customary units. (credit: “Weighing the homemade cheddar” by Ruth Hartnup/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify units of measurement in the metric system and their uses. Order the six common prefixes of the metric system. Convert between like unit values. Even if you don’t travel outside of the United States, many specialty grocery stores utilize the metric system. For example, if you want to make authentic tamales you might visit the nearest Hispanic grocery store. While shopping, you discover that there are two brands of masa for the same price, but one bag is marked 1,200 g and the other 1 kg. Which one is the better deal? Understanding the metric system allows you to understand that 1,200 grams is equivalent to 1.2 kg, so the 1,200 g bag is the better deal. Units of Measurement in the Metric System Units of measurement provide common standards so that regardless of where or when an object or substance is measured, the results are consistent. When measuring distance, the units of measure might be feet, meters , or miles. Weight might be expressed in terms of pounds or grams . Volume or capacity might be measured in gallons, or liters . Understanding how metric units of measure relate to each other is essential to understanding the metric system: The metric unit for distance is the meter (m). A person’s height might be written as 1.8 meters (1.8 m). A meter slightly longer than a yard (3 feet), while a centimeter is slightly less than half an inch. The metric unit for area is the square meter (m 2 ). The area of a professional soccer field is 7,140 square meters (7,140 m 2 ). The metric base unit for volume is the cubic meter (m 3 ). However, the liter (L), which is a metric unit of capacity, is used to describe the volume of liquids. Soda is often sold in 2-liter (2 L) bottles. The gram (g) is a metric unit of mass but is commonly used to express weight. The weight of a paper clip is approximately 1 gram (1 g). The metric unit for temperature is degrees Celsius (°C). The temperature on a warm summer day might be 24 °C. While the U.S. Customary System of Measurement uses ounces and pounds to distinguish between weight units of different sizes, in the metric system a base unit is combined with a prefix, such as kilo– in kilogram, to identify the relationship between smaller or larger units. When using abbreviations to represent metric measures, always separate the quantity and the units with a space, with no spaces between the letters or symbols in the units. For example, 7 millimeters is written as 7 mm, not 7mm. It is important to be able to convert between the U.S. Customar System of Measurement and the metric system. However, in this chapter we’ll focus on converting units within the metric system. Why? Typing “200 centimeters in inches” into any browser search bar will instantly convert those measures for you (Figure 9.3). You’ll have an opportunity in the Projects to work between measurement systems. (credit: Screenshot/Google) Let’s be honest. Most of us use computers or smartphones to perform many of the calculations and conversions we were taught in math class. But there is value in understanding the metric system since it exists all around us, and most importantly, knowing how the different metric units relate to each other allows you to compare prices, find the right tool in a workshop, or acclimate when in another country. No matter the circumstance, you cannot avoid the metric system. Determining the Correct Base Unit Which base unit would be used to express the following? amount of water in a swimming pool length of an electrical wire weight of one serving of peanuts Liquid volume is generally expressed in units of liters (L). Length is measured in units of meters (m). Weight is commonly expressed in units of grams (g). While there are other base units in the metric system, our discussions in this chapter will be limited to units used to express length, area, volume, weight, and temperature. Metric Prefixes Unlike the U.S. Customary System of Measurement in which 12 inches is equal to 1 foot and 3 feet are equal to 1 yard, the metric system is structured so that the units within the system get larger or smaller by a power of 10. For example, a centimeter is 10 − 2 , or 100 times smaller than a meter, while the kilometer is 10 3 , or 1,000 times larger than a meter. The metric system combines base units and unit prefixes reasonable to the size of a measured object or substance. The most used prefixes are shown in . An easy way to remember the order of the prefixes, from largest to smallest, is the mnemonic King Henry Died From Drinking Chocolate Milk. Prefix kilo– hecto– deca– base unit deci– cent– milli– Abbreviation k h da d c m Magnitude 10 3 10 2 10 1 10 0 ⁢ or ⁢ 1 10 − 1 10 − 2 10 − 3 Metric Prefixes Ordering the Magnitude of Units Order the measures from smallest unit to largest unit. centimeter, millimeter, decimeter Looking at the metric prefixes, we can see that the prefix order from smallest unit to largest unit is milli-, centi-, deci-, so the order of the units from smallest to largest is millimeter, centimeter, decimeter. Determining Reasonable Values for Length What is a reasonable value for the length of a person’s thumb: 5 meters, 5 centimeters, or 5 millimeters? Given that a meter is slightly longer than a yard, 5 meters is not a reasonable value for the length of a person’s thumb. Since a millimeter is 10 times smaller than a centimeter, which is approximately 1 2 inch, 5 millimeters is not a reasonable estimate for the length of a person’s thumb. The correct answer is 5 centimeters. Converting Metric Units of Measure Imagine you order a textbook online and the shipping detail indicates the weight of the book is 1 kg. By attaching the letter “k” to the base unit of gram (g), the unit used to express the measure is 10 3 or 1,000 times greater than a gram. One kilogram is equivalent to 1,000 grams. The tip of a highlighter measures approximately 1 cm. The letter “c” attached to the base unit of meter (m) means the unit used to express the measure is 10 − 2 or 1 100 of a meter. One meter is equivalent to 100 centimeters. A conversion factor is used to convert from smaller metric units to bigger metric units and vice versa. It is a number that when used with multiplication or division converts from one metric unit to another, both having the same base unit. In the metric system, these conversion factors are directly related to the powers of 10. The most common used conversion factors are shown in . Common Metric Conversion Factors for (a) Meters, (b) Liters, and (c) Grams Converting Metric Distances Using Multiplication The firehouse is 13.45 km from the library. How many meters is it from the firehouse to the library? When converting from a larger unit to a smaller unit, use multiplication. The conversion factor from kilometer to the base unit of meters is 1,000. 13.45 × 1,000 = 13,450 So, the firehouse is 13,450 meters away from the firehouse. Converting Metric Capacity Using Division How many liters is 3,565 milliliters? When converting from a smaller unit to a larger unit, use division. The conversion factor from milliliter to the base unit of liters is 1,000. 3,565 ÷ 1,000 = 3.565 So, 3,565 milliliters is 3.565 liters. Converting Metric Units of Mass to Solve Problems Caroline and Aiyana are working on a chemistry experiment together and must perform calculations using measurements taken during the experiment. Due to miscommunication, Caroline took measurements in centigrams and Aiyana used milligrams. Convert Caroline’s measurement of 125 centigrams to milligrams. When converting from a larger unit to a smaller unit, use multiplication. The conversion factor from centigrams to the milligrams is 10. 125 × 10 = 1,250 So, the 125 centigrams are 1,250 milligrams. Converting Metric Units of Volume to Solve Problems A bottle contains 500 mL of juice. If the juice is packaged in 24-bottle cases, how many liters of juice does the case contain? Step 1: Multiply the amount of juice in each bottle by the number of bottles. 500 ⁢ mL × 24 = 12,000 ⁢ mL Step 2: Divide by 1,000 to convert from milliliters to liters. 12,000 mL 1,000 = 12 L So, there are 12 liters of juice in each case. Valerie Antoine In the 1970s, people were told that they must learn the metric system because the United States was soon going to convert to using metric measurements. Children and young adults probably watched educational cartoons about the metric system on Saturday mornings. In 1975, President Gerald Ford signed the Metric Conversion Act and created a board of 17 people commissioned to coordinate the voluntary switch to the metric system in the United States. Among those 17 people was Valerie Antoine, an engineer who made it her life’s work to push for this change. Despite President Ronald Reagan dissolving the board in 1982, effectively killing the move to the metric system at the time, Antoine continued the movement out of her own home as the executive director of the U.S. Metric Association. Reagan’s decision followed intensive lobbying by American businesses whose factories used machinery designed to use customary measurements by workers trained in customary measurements. There was also intense public pressure from American citizens who didn’t want to go through the time consuming and expensive process of changing the country’s entire infrastructure. Fueled by a Congressional mandate in 1992 that required all federal agencies make the switch to the metric system, Antoine never gave up hope that the metric system would trickle down from the government and find its way into American schools, homes, and everyday life. U.S. Office of Education: Metric Education Converting Grams to Solve Problems The nutrition label on a jar of spaghetti sauce indicates that one serving contains 410 mg of sodium. You have poured two servings over your favorite pasta before recalling your doctor’s advice about keeping your sodium consumption below 1 g per meal. Have you followed your doctor’s recommendation? Step 1: Multiply the number of servings by the amount of sodium in each serving. 410 ⁢ mg × 2 = 820 ⁢ mg Step 2: Divide by 1,000 to convert from milligrams to grams. 820 ⁢ mg 1,000 = 0.82 ⁢ g You have followed doctor’s recommendation because 0.82 g is less than 1 gram. Comparing Different Units A student carefully measured 0.52 cg of copper for a science experiment, but their lab partner said they need 6 mg of copper total. How many more centigrams of copper does the student need to add? Step 1: Convert these two measurements into a common unit. Since the question asks for the number of centigrams, convert 6 mg to centigrams, which is 0.6 cg. Step 2: Find the difference by subtracting 0.6 − 0.52 which is 0.08 cg. This means the student must add another 0.08 cg of copper. The United States and the Metric System Did you know that the metric system pervades daily life in the United States already? While Americans still may purchase gallons of milk and measure house sizes in square feet, there are many instances of the metric system. Photographers buy 35 mm film and use 50 mm lenses. When you have a headache, you might take 600 mg of ibuprofen. And if you are eating a low-carb diet you probably restrict your carb intake to fewer than 20 g of carbs daily. Did you know even the dollar is metric? In the video, Neil DeGrasse Tyson and comedian co-host Chuck Nice provide an amusing perspective on the metric system. The International System of Units (SI) is the current international standard metric system and is the most widely used system around the world. In most English-speaking countries SI units such as meter, liter, and metric ton are spelled metre, litre, and tonne. Neil deGrasse Tyson Explains the Metric System Get to Know the Metric System Just how much is the metric system a part of your life now? Probably more than you think. For the next 24 hours, take notice as you move through your daily activities. When you are shopping, are the package sizes provided in metric units? Change the weather app on your phone to display the temperature in degrees Celsius. Are you able to tell what kind of day it will be now? While the United States is not officially using the metric system, you will still find the metric system all around you. Check Your Understanding Key Terms metric system meter (m) gram (g) liter (L) square meter (m 2 ) cubic meter (m 3 ) degrees Celsius (°C) conversion factor Key Concepts The metric system is decimal system of weights and measures based on base units of meter, liter, and gram. The system was first proposed in 1670 and has since been adopted as the International System of Units and used in nearly every country in the world. Each successive unit on the metric scale is 10 times larger than the previous one. To convert between units with the same base unit, you must either multiply or divide by a power of 10. The most used prefixes are listed in . An easy way to remember the order of the prefixes, from largest to smallest, is the mnemonic King Henry Died from Drinking Chocolate Milk. Videos U.S. Office of Education: Metric Education Neil deGrasse Tyson Explains the Metric System Formulas You can convert between unit sizes with the same base unit using the conversion factors shown in .", "section": "The Metric System", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Measuring Area A painter uses an extension roller to paint a wall. (credit: \"Paint Rollers are effective\" by WILLPOWER STUDIOS/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify reasonable values for area applications. Convert units of measures of area. Solve application problems involving area. Area is the size of a surface. It could be a piece of land, a rug, a wall, or any other two-dimensional surface with attributes that can be measured in the metric unit for distance-meters. Determining the area of a surface is important to many everyday activities. For example, when purchasing paint, you’ll need to know how many square units of surface area need to be painted to determine how much paint to buy. Square units indicate that two measures in the same units have been multiplied together. For example, to find the area of a rectangle, multiply the length units and the width units to determine the area in square units. 1 cm × 1 cm = 1 cm 2 Note that to accurately calculate area, each of the measures being multiplied must be of the same units. For example, to find an area in square centimeters, both length measures (length and width) must be in centimeters. The formula used to determine area depends on the shape of that surface. Here we will limit our discussions to the area of rectangular-shaped objects like the one in Figure 9.6. Given this limitation, the basic formula for area is: Area = length ( l ) × width ( w ) or A = l × w Rectangle with Length ( l ) and Width ( w ) Labeled Reasonable Values for Area Because area is determined by multiplying two lengths, the magnitude of difference between different square units is exponential. In other words, while a meter is 100 times greater in length than a centimeter, a square meter ( m 2 ) is 100 × 100 , or 10,000 times greater in area than a square centimeter ( cm 2 ) . The relationships between benchmark metric area units are shown in the following table. Units Relationship Conversion Rate km 2 to m 2 km × km = km 2 1 km = 1,000 m 1,000 m × 1,000 m = 1,000,000 m 2 1 ⁢ km 2 = 1,000,000 m 2 m 2 to cm 2 m × m = m 2 1 m = 100 cm 100 cm × 100 cm = 10,000 ⁢ cm 2 1 ⁢ m 2 = 10,000 ⁢ cm 2 cm 2 to mm 2 cm × cm = cm 2 1 cm = 10 mm 10 mm × 10 mm = 100 mm 2 1 cm 2 = 100 mm 2 An essential understanding of metric area is to identify reasonable values for area. When testing for reasonableness you should assess both the unit and the unit value. Only by examining both can you determine whether the given area is reasonable for the situation. Determining Reasonable Units for Area Which unit of measure is most reasonable to describe the area of a sheet of paper: km 2 , cm 2 , or mm 2 ? In the U.S. Customary System of Measurement, the length and width of paper is usually measured in inches. In the metric system centimeters are used for measures usually expressed in inches. Thus, the most reasonable unit of measure to describe the area of a sheet of paper is square centimeters. Square kilometers is too large a unit and square millimeters is too small a unit. Determining Reasonable Values for Area You want to paint your bedroom walls. Which represents a reasonable value for the area of the walls: 100 cm 2 , 100 m 2 , or 100 km 2 ? An area of 100 cm 2 is equivalent to a surface of 10 cm × 10 cm , which is much too small for the walls of a bedroom. An area of 100 km 2 is equivalent to a surface of 10 km × 10 km , which is much too large for the walls of a bedroom. So, a reasonable value for the area of the walls is 100 m 2 . Explaining Reasonable Values for Area A landscaper is hired to resod a school’s football field. After measuring the length and width of the field they determine that the area of the football field is 5,350 km 2 . Does their calculation make sense? Explain your answer. No, kilometers are used to determine longer distances, such as the distance between two points when driving. A football field is less than 1 kilometer long, so a more reasonable unit of value would be m 2 . An area of 5,350 km 2 can be calculated using the dimensions 53.5 by 100, which are reasonable dimensions for the length and width of a football field. So, a more reasonable value for the area of the football field is 5,350 m 2 . Converting Units of Measures for Area Just like converting units of measure for distance, you can convert units of measure for area. However, the conversion factor, or the number used to multiply or divide to convert from one area unit to another, is not the same as the conversion factor for metric distance units. Recall that the conversion factor for area is exponentially relative to the conversion factor for distance. The most frequently used conversion factors are shown in . Common Conversion Factors for Metric Area Units Converting Metric Units of Area Converting Units of Measure for Area Using Division A plot of land has an area of 237,500,000 m 2 . What is the area in square kilometers? Use division to convert from a smaller metric area unit to a larger metric area unit. To convert from m 2 to km 2 , divide the value of the area by 1,000,000. 237,500,000 1,000,000 = 237.5 The plot of land has an area of 237.5 km 2 . Converting Units of Measure for Area Using Multiplication A plot of land has an area of 0.004046 km 2 . What is the area of the land in square meters? Use multiplication to convert from a larger metric area unit to a smaller metric area unit. To convert from km 2 to m 2 , multiply the value of the area by 1,000,000. 0.004046 × 1,000,000 = 4,046 The plot of land has an area of 4,046 m 2 . Determining Area by Converting Units of Measure for Length First A computer chip measures 10 mm by 15 mm. How many square centimeters is the computer chip? Step 1: Convert the measures of the computer chip into centimeters 10 mm = 1 cm 15 mm = 1.5 cm Step 2: Use the area formula to determine the area of the chip. 1 × 1.5 = 1.5 The computer chip has an area of 1.5 cm 2 . Solving Application Problems Involving Area While it may seem that solving area problems is as simple as multiplying two numbers, often determining area requires more complex calculations. For example, when measuring the area of surfaces, you may need to account for portions of the surface that are not relevant to your calculation. Solving for the Area of Complex Surfaces One side of a commercial building is 12 meters long by 9 meters high. There is a rolling door on this side of the building that is 4 meters wide by 3 meters high. You want to refinish the side of the building, but not the door, with aluminum siding. How many square meters of aluminum siding are required to cover this side of the building? Step 1: Determine the area of the side of the building. 12 ⁢ m × 9 ⁢ m = 106 ⁢ m 2 Step 2: Determine the area of the door. 4 ⁢ m × 3 ⁢ m = 12 ⁢ m 2 Step 3: Subtract the area of the door from the area of the side of the building. 106 ⁢ m 2 − 12 ⁢ m 2 = 92 ⁢ m 2 So, you need to purchase 92 ⁢ m 2 of aluminum siding. When calculating area, you must ensure that both distance measurements are expressed in terms of the same distance units. Sometimes you must convert one measurement before using the area formula . Solving for Area with Distance Measurements of Different Units A national park has a land area in the shape of a rectangle. The park measures 2.2 kilometers long by 1,250 meters wide. What is the area of the park in square kilometers? Step 1: Use a conversion fraction to convert the information given in meters to kilometers. 1,250 ⁢ m × 1 ⁢ km 1,000 ⁢ m = 1.25 ⁢ km Step 2: Multiply to find the area. 2.2 ⁢ km × 1.25 ⁢ km = 2.75 ⁢ km 2 The park has an area of 2.75 km 2 . When calculating area, you may need to use multiple steps, such as converting units and subtracting areas that are not relevant. Solving for Area Using Multiple Steps A kitchen floor has an area of 15 m 2 . The floor in the kitchen pantry is 100 cm by 200 cm. You want to tile the kitchen and pantry floors using the same tile. How many square meters of tile do you need to buy? Step 1: Determine the area of the pantry floor in square centimeters. 100 ⁢ cm × 200 ⁢ cm = 20,000 ⁢ cm 2 Step 2: Divide the area in cm 2 by the conversion factor to determine the area in m 2 since the other measurement for the kitchen floor is in m 2 . 20,000 10,000 = 2 The area of the kitchen pantry floor is 2 m 2 . Step 3: Add the two areas of the pantry and the kitchen floors together. 15 ⁢ m 2 + 2 ⁢ m 2 = 17 ⁢ m 2 So, you need to buy 17 m 2 of tile. The Origin of the Metric System The metric system is the official measurement system for every country in the world except the United States, Liberia, and Myanmar. But did you know it originated in France during the French Revolution in the late 18th century? At the time there were over 250,000 different units of weights and measures in use, often determined by local customs and economies. For example, land was often measured in days, referring to the amount of land a person could work in a day. Why the Metric System Matters Check Your Understanding Key Terms area square units Key Concepts Area describes the size of a two-dimensional surface. It is the amount of space contained within the lines of a two-dimensional space. Area is measured in square meters units; in the metric system the base unit for area is square meters ( m 2 ). Videos Converting Metric Units of Area Why the Metric System Matters Formulas To determine the area of rectangular-shaped objects: Area = length ( l ) × width ( w ) or A = l × w Rectangle with Length ( l ) and Width ( w ) Labeled You can convert between metric area units using the conversion factors shown in .", "section": "Measuring Area", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Measuring Volume Packing cartons sit on a loading dock ready to be filled. (credit: “boxing day” by Erich Ferdinand/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify reasonable values for volume applications. Convert between like units of measures of volume. Convert between different unit values. Solve application problems involving volume. Volume is a measure of the space contained within or occupied by three-dimensional objects. It could be a box, a pool, a storage unit, or any other three-dimensional object with attributes that can be measured in the metric unit for distance–meters. For example, when purchasing an SUV, you may want to compare how many cubic units of cargo the SUV can hold. Cubic units indicate that three measures in the same units have been multiplied together. For example, to find the volume of a rectangular prism, you would multiply the length units by the width units and the height units to determine the volume in square units: 1 ⁢ cm × 1 ⁢ cm × 1 ⁢ cm = 1 ⁢ cm 3 Note that to accurately calculate volume, each of the measures being multiplied must be of the same units. For example, to find a volume in cubic centimeters, each of the measures must be in centimeters. The formula used to determine volume depends on the shape of the three-dimensional object. Here we will limit our discussions to the area to rectangular prisms like the one in Given this limitation, the basic formula for volume is: Volume = length ( l ) × width ( w ) × height ( h ) or V = l w h Rectangular Prism with Height ( h ) , Length ( l ) , and Width ( w ) Labeled. Reasonable Values for Volume Because volume is determined by multiplying three lengths, the magnitude of difference between different cubic units is exponential. In other words, while a meter is 100 times greater in length than a centimeter, a cubic meter, m 3 , is 100 × 100 × 100 , ⁢ or ⁢ 1,000,000 times greater in area than a cubic centimeter, cm 3 . This relationship between benchmark metric volume units is shown in the following table. Units Relationship Conversion Rate km 3 to m 3 km × km × km = km 3 1 km = 1,000 m 1,000 ⁢ m × 1,000 m × 1,000 m = 1,000,000,000 m 3 1 km 3 = 1,000,000,000 m 3 m 3 to dm 3 m × m × m = m 3 1 ⁢ m = 10 ⁢ dm 10 ⁢ dm × 10 ⁢ dm × 10 ⁢ dm = 1,000 ⁢ dm 3 1 m 3 = 1,000 dm 3 dm 3 to cm 3 dm × dm × dm = dm 3 1 ⁢ dm = 10 ⁢ cm 10 ⁢ cm × 10 ⁢ cm × 10 ⁢ cm = 1,000,000 ⁢ cm 3 1 cm 3 = 1,000 dm 3 cm 3 to mm 3 cm × cm × cm = cm 3 1 ⁢ cm = 10 ⁢ mm 10 ⁢ mm × 10 ⁢ mm × 10 ⁢ mm = 1,000 ⁢ mm 3 1 cm 3 = 1,000 mm 3 To have an essential understanding of metric volume, you must be able to identify reasonable values for volume. When testing for reasonableness you should assess both the unit and the unit value. Only by examining both can you determine whether the given volume is reasonable for the situation. Determining Reasonable Values for Volume A grandparent wants to send cookies to their grandchild away at college. Which represents a reasonable value for the volume of a box to ship the cookies: 3,375 km 3 , 3,375 m 3 , or 3,375 cm 3 ? A volume of 3,375 km 3 is equivalent to a rectangular prism with dimensions of 15 ⁢ km ⁢ × 15 ⁢ km × 15 ⁢ km , which is far too large for a shipping box. An area of 3,375 m 3 is equivalent to a surface of 15 ⁢ m ⁢ × 15 ⁢ m × 15 ⁢ m , which is also too large. A reasonable value for the volume of the box is 3,375 cm 3 . Identifying Reasonable Values for Volume A food manufacturer is prototyping new packaging for one of its most popular products. Which represents a reasonable value for the volume of the box: 2 dm 3 , 2 cm 3 , or 2 mm 3 ? A decimeter is equal to 10 centimeters. A box with a volume of 2 dm 3 might have the dimensions 1 ⁢ dm × 1 ⁢ dm × ⁢ 2 ⁢ dm , or 10 ⁢ cm × 10 ⁢ cm × 20 ⁢ cm , which is reasonable. A box with a volume of 2 cm 3 or 2 mm 3 would be too small. Explaining Reasonable Values for Volume A farmer has a hay loft. They calculate the volume of the hayloft as 64 cm 3 . Does the calculation make sense? Explain your answer. No. Centimeters are used to determine smaller distances, such as the length of a pencil. A hayloft is more than 64 centimeters long, so a more reasonable unit of value would be m 2 . A volume of 64 m 3 can be calculated using the dimensions 4 meters by 4 meters by 4 meters, which are reasonable dimensions for a hayloft. So, a more reasonable value for the volume of the hayloft is 64 m 3 . Converting Like Units of Measures for Volume Just like converting units of measure for distance, you can convert units of measure for volume. However, the conversion factor, the number used to multiply or divide to convert from one volume unit to another, is different from the conversion factor for metric distance units. Recall that the conversion factor for volume is exponentially relative to the conversion factor for distance. The most frequently used conversion factors are illustrated in . Common Conversion Factors for Metric Volume Units Converting Like Units of Measure for Volume Using Multiplication A pencil case has a volume of 1,700 cm 3 . What is the volume in cubic millimeters? Use multiplication to convert from a larger metric volume unit to a smaller metric volume unit. To convert from cm 3 to mm 3 , multiply the value of the volume by 1,000. 1,700 × 1,000 = 1,700,000 The pencil case has a volume of 1,700,000 mm 3 . How to Convert Cubic Centimeters to Cubic Meters Converting Like Units of Measure for Volume Using Multi-Step Multiplication A shipping container has a volume of 33.2 m 3 . What is the volume in cubic centimeters? Use multiplication to convert a larger metric volume unit to a smaller metric volume unit. To convert from m 3 to cm 3 , first multiply the value of the volume by 1,000 to convert from m 3 to dm 3 , and then multiply again by 1,000 to convert from dm 3 to cm 3 . 33.2 × 1,000 × 1,000 = 33,200,000 The shipping container has a volume of 32,200,000 cm 3 . Converting Like Units of Measure for Volume Using Multi-Step Division A holding tank at the local aquarium has a volume of 22,712,000,000 cm 3 . What is the volume in cubic meters? indicates that when converting from a smaller metric volume unit to a larger metric volume unit you divide using the given conversion factor. To convert from cm 3 to m 3 , divide the value of the volume by 1,000 to first convert from cm 3 to dm 3 , then divide again to convert from dm 3 to m 3 . cm 3 ⁢ to dm 3 : ⁢ 22,712,000,000 1,000 = 22,712,000 dm 3 ⁢ to m 3 : ⁢ 22,712,000 1,000 = 22,712 The holding tank has a volume of 22,712 m 3 . Understanding Other Metric Units of Volume When was the last time you purchased a bottle of soda? Was the volume of the bottle expressed in cubic centimeters or liters? The liter (L) is a metric unit of capacity often used to express the volume of liquids. A liter is equal in volume to 1 cubic decimeter. A milliliter is equal in volume to 1 cubic centimeter. So, when a doctor orders 10 cc (cubic centimeters) of saline to be administered to a patient, they are referring to 10 mL of saline. The most frequently used factors for converting from cubic meters to liters are listed in . Relationships Between Metric Volume and Metric Capacity Units m 3 to L m 3 to mL 1 ⁢ dm 3 = 1 ⁢ L 1 ⁢ dm 3 = 1,000 ⁢ mL 1,000 ⁢ cm 3 = 1 ⁢ L 1 ⁢ cm 3 = 1 ⁢ mL 1,000,000 ⁢ mm 3 = 1 ⁢ L 1 ⁢ mm 3 = 0.001 ⁢ mL Converting Metric Units of Volume Converting Different Units of Measure for Volume A holding tank at the local aquarium has a volume of 22,712,000,000 cm 3 ? What is the capacity of the holding tank in liters? Use division to convert from cubic centimeters to liters. To determine the equivalent volume in liters, convert from cm 3 to L by dividing the value of the volume in cm 3 by 1,000. 22,712,000,000 ⁢ cm 3 1,000 = 22,712,000 ⁢ L The holding tank holds 22,712,000 L of water. Converting Different Units of Measure for Volume Using Multiplication An airplane used 150 m 3 of fuel to fly from New York to Hawaii. How many liters of fuel did the airplane use? Because 1 liter is equivalent to 1 cubic decimeter, use multiplication to convert from m 3 to dm 3 . Multiply the value of the volume by 1,000 to convert from m 3 to dm 3 . Because 1 ⁢ dm 3 = 1 ⁢ L , the resulting value is equivalent to the number of liters used. 150 × 1,000 = 150,000 ⁢ dm 3 = 150,000 ⁢ L The airplane used 150,000 liters of fuel. Converting Different Units of Measure for Volume Using Multi-Step Division How many liters can a pitcher with a volume of 8,000,000 mm 3 hold? Use division to convert from a smaller metric volume unit to a larger metric volume unit. To convert from mm 3 to dm 3 , Step 1: Divide by 1,000 to convert from mm 3 to cm 3 . Step 2: Divide again by 1,000 to convert from cm 3 to dm 3 . Step 3: Use the unit value to express the volume in terms of liters. 8,000,000 mm 3 1,000 = 8,000 cm 3 8,000 cm 3 1,000 = 8 dm 3 = 8 L The pitcher can hold 8 liters of liquid. Solving Application Problems Involving Volume Knowing the volume of an object lets you know just how much that object can hold. When making a bowl of punch you might want to know the total amount of liquid a punch bowl can hold. Knowing how many liters of gasoline a car’s tank can hold helps determine how many miles a car can drive on a full tank. Regardless of the application, understanding volume is essential to many every day and professional tasks. Using Volume to Solve Problems A cubic shipping carton’s dimensions measure 2 m × 2 m × 2 m . A company wants to fill the carton with smaller cubic boxes that measure 10 ⁢ cm × 10 ⁢ cm × 10 ⁢ cm . How many of the smaller boxes will fit in each shipping carton? Step 1: Determine the volume of the shipping carton. 2 ⁢ m × 2 ⁢ m × 2 ⁢ m = 8 ⁢ m 3 Step 2: Use the appropriate conversion factor to convert the volume of the shipping carton from m 3 to cm 3 . 8 ⁢ m 3 × 1,000 = 8,000 ⁢ dm 3 8,000 ⁢ dm 3 × 1,000 = 8,000,000 ⁢ cm 3 Step 3: Determine the volume of the smaller boxes. 10 ⁢ cm × 10 ⁢ cm × 10 ⁢ cm = 1,000 ⁢ cm 3 Step 4: Divide the volume of the shipping carton, in cm 3 , by the volume of the smaller box, in cm 3 . 8,000,000 ⁢ cm 3 1,000 ⁢ cm 3 = 8,000 The shipping carton will hold 8,000 smaller boxes. Solving Volume Problems with Different Units A carton of juice measures 6 cm long, 6 cm wide and 20 cm tall. A factory produces 28,800 liters of orange juice each day. How many cartons of orange juice are produced each day? Step 1: Find the volume of the carton in cubic centimeters. 6 ⁢ cm × 6 ⁢ cm × 20 ⁢ cm = 720 ⁢ cm 3 Step 2: Convert the volume in cm 3 to liters. 1 cm 3 = 1 mL 720 cm 3 = 720 mL 720 mL 1,000 = 0.72 L Step 3: Divide the number of liters of orange juice produced each day by the volume of each carton. 28,800 ⁢ L 0.72 ⁢ L = 40,000 The factory produces 40,000 cartons of orange juice each day. Solving Complex Volume Problems A fish tank measures 60 cm long, 15 cm wide and 34 cm tall ( ). The tank is 25 percent full. How many liters of water are needed to completely fill the tank? Step 1: Determine the volume of the fish tank in cubic centimeters. 60 ⁢ cm × 15 ⁢ cm × 34 ⁢ cm = 30,600 ⁢ cm 3 Step 2: Convert the volume in cm 3 to volume in liters. 1 cm 3 = 1 mL 30,600 cm 3 = 30,600 mL 30,600 mL 1,000 = 30.6 L Step 3: Since the tank is 25 percent full, the tank is 75 percent empty. Convert 75 percent to its decimal equivalent. Multiply the total volume by 75 percent expressed in decimal form to determine how many liters of water are required to fill the tank. 75 % = 0.75 30.6 × 0.75 = 22.95 So, 22.95 liters of water are needed to fill the tank. How Does Shape Affect Volume? Take two large sheets of card stock. Roll one piece to tape the longer edges together to make a cylinder. Tape the cylinder to the other piece of card stock which serves as the base of the cylinder. Fill the cylinder to the top with cereal. Pour the cereal from the cylinder into a plastic storage or shopping bag. Remove the cylinder from the base and the tape from the cylinder. Re-roll the cylinder along the shorter edges a tape together. Attach the new cylinder to the base. Pour the cereal from the plastic bag into the cylinder. What do you observe? How does the shape of a container affect its volume? Check Your Understanding Key Terms volume cubic units Key Concepts Volume is a measure of the space contained within or occupied by three-dimensional objects. Volume is measured in cubic units; the base unit for volume in the metric system is cubic meters (m 3 ) The liter (L) is a metric unit of capacity but is often used to express the volume of liquids. One liter is equivalent to one cubic decimeter, which is the volume of a cube measuring 10 cm × 10 cm × 10 cm . Formulas To determine the volume of a rectangular prism: Volume = length ( l ) × width ( w ) × height ( h ) or V = l w h Rectangular Prism with Height ( h ) , Length ( l ) , and Width ( w ) Labeled You can convert between metric volume units and metric capacity units using the relationships shown in . Videos How to Convert Cubic Centimeters to Cubic Meters Converting Metric Units of Volume", "section": "Measuring Volume", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Measuring Weight Weight scale at the local Antigua market (credit: “Weight scale at the local Antigua market” by Lucía García González/Flickr, CC0 1.0 Public Domain Dedication) Learning Objectives After completing this section, you should be able to: Identify reasonable values for weight applications. Convert units of measures of weight. Solve application problems involving weight. In the metric system, weight is expressed in terms of grams or kilograms, with a kilogram being equal to 1,000 grams. A paper clip weighs about 1 gram. A liter of water weighs about 1 kilogram. In fact, in the same way that 1 liter is equal in volume to 1 cubic decimeter, the kilogram was originally defined as the mass of 1 liter of water. In some cases, particularly in scientific or medical settings where small amounts of materials are used, the milligram is used to express weight. At the other end of the scale is the metric ton (mt), which is equivalent to 1,000 kilograms. The average car weighs about 2 metric tons. Any discussion about metric weight must also include a conversation about mass . Scientifically, mass is the amount of matter in an object whereas weight is the force exerted on an object by gravity. The amount of mass of an object remains constant no matter where the object is. Identical objects located on Earth and on the moon will have the same mass, but the weight of the objects will differ because the moon has a weaker gravitational force than Earth. So, objects with the same mass will weigh less on the moon than on Earth. Since there is no easy way to measure mass, and since gravity is just about the same no matter where on Earth you go, people in countries that use the metric system often use the words mass and weight interchangeably. While scientifically the kilogram is only a unit of mass, in everyday life it is often used as a unit of weight as well. Reasonable Values for Weight To have an essential understanding of metric weight, you must be able to identify reasonable values for weight. When testing for reasonableness, you should assess both the unit and the unit value. Only by examining both can you determine whether the given weight is reasonable for the situation. Metric System: Units of Weight Identifying Reasonable Units for Weight Which is the more reasonable value for the weight of a newborn baby: 3.5 kg or 3.5 g? Using our reference weights, a baby weighs more than 3.5 paperclips, so 3.5 kilograms is a more reasonable value for the weight of a newborn baby. Determining Reasonable Values for Weight Which of the following represents a reasonable value for the weight of three lemons? 250 g, 2,500 g, or 250 kg? Because a kilogram is about 2.2 pounds, we can eliminate 250 kg as it is way too heavy. 2,500 grams is equivalent to 2.5 kilograms, or about five pounds, which is again, too heavy. So, a reasonable value for the weight of three lemons would be 250 grams. How Do You Measure the Weight of a Whale? It is impossible to weigh a living whale. Fredrik Christiansen from the Aarhus Institute of Advanced Studies in Denmark developed an innovative way to measure the weight of whales. Using images taken from a drone and computer modeling, the weight of a whale can be estimated with great accuracy. Using Drones to Weigh Whales? Explaining Reasonable Values for Weight The blue whale is the largest living mammal on Earth. Which of the following is a reasonable value for the weight of a blue whale: 149 g, 149 kg, or 149 mt? Explain your answer. A reasonable value for the weight of a blue whale is 149 metric tons. Both 149 g and 149 kg are much too small a value for the largest living mammal on Earth. Converting Like Units of Measures for Weight Just like converting units of measure for distance, you can convert units of measure for weight. The most frequently used conversion factors for metric weight are illustrated in . Common Conversion Factors for Metric Weight Units Converting Metric Units of Weight Using Multistep Division How many kilograms are in 24,300,000 milligrams? Use division to convert from a smaller metric weight unit to a larger metric weight unit. To convert from milligrams to kilograms, Step 1: Divide the value of the weight in milligrams by 1,000 to first convert from milligrams to grams. Step 2: Divide by 1,000 again to convert from grams to kilograms. 24,300,000 ⁢ mg 1,000 = 24,300 ⁢ g 24,300 ⁢ g 1,000 = 24.3 ⁢ kg So, 24,300,000 milligrams are equivalent to 24.3 kilograms. Converting Metric Units of Weight Using Multiplication The average ostrich weighs approximately 127 kilograms. How many grams does an ostrich weigh? Use multiplication to convert from a larger metric weight unit to a smaller metric weight unit. To convert from kilograms to grams, multiply the value of the weight by 1,000. 127 ⁢ kg × 1,000 = 127,000 ⁢ g The average ostrich weighs 127,000 grams. Converting Metric Units of Weight Using Multistep Multiplication How many milligrams are there in 0.025 kilograms? Use multiplication to convert from a larger metric weight unit to a smaller metric weight unit. To convert from kilograms to grams, Step 1: Multiply the value of the weight by 1,000. Step 2: Multiply the result by 1,000 to convert from grams to milligrams. 0.025 ⁢ kg × 1,000 = 25 ⁢ g 25 g × 1,000 = 25,000 mg So, 0.025 kilograms is equivalent to 25,000 milligrams. Metric Units of Mass: Convert mg, g, and kg Solving Application Problems Involving Weight From children’s safety to properly cooking a pie, knowing how to solve problems involving weight is vital to everyday life. Let’s review some ways that knowing how to work with metric weight can facilitate important decisions and delicious eating. Comparing Weights to Solve Problems The maximum weight for a child to safely use a car seat is 29 kilograms. If a child weighs 23,700 grams, can the child safely use the car seat? Step 1: Convert the child’s weight in grams to kilograms. 23,700 ⁢ g ÷ 1,000 = 23.7 ⁢ kg Step 2: Compare the two weights. 23.7 ⁢ kg < 29 ⁢ kg Yes, the child can safely use the car seat. Solving Multistep Weight Problems A recipe for scones calls for 350 grams of flour. How many kilograms of flour are required to make 4 batches of scones? Step 1: Multiply the grams of flour need by 4 to determine the total amount of flour needed. 350 ⁢ g × 4 = 1,400 ⁢ g Step 2: Convert from grams to kilograms. 1,400 ⁢ g 1,000 = 1.4 ⁢ kg So, 1.4 kilograms of flour are needed to make four batches of scones. Solving Complex Weight Problems The average tomato weighs 140 grams. A farmer needs to order boxes to pack and ship their tomatoes to local grocery stores. They estimate that this year’s harvest will yield 125,000 tomatoes. A box can hold 12 kilograms of tomatoes. How many boxes does the farmer need? Step 1: Determine the total estimated weight of the harvested tomatoes. 140 ⁢ g × 125,000 = 17,500,000 g Step 2: Convert the total weight from grams to kilograms. 17,500,000 ⁢ g ÷ 1,000 = 17,500 ⁢ kg Step 3: Divide the weight of the tomatoes by the weight each box can hold. 17,500 ⁢ kg ÷ 12 ⁢ kg ≈ 1,458 So, the farmer will need to order 1,458 boxes. Check Your Understanding Key Terms mass Key Concepts Mass is the amount of matter in an object whereas weight is the force exerted on an object by gravity. The mass of an object never changes; the weight of an object changes depending on the force of gravity. An object with the same mass would weigh less on the moon than on Earth because the moon’s gravity is less than that of Earth. In the metric system, weight and mass are often used interchangeably and are expressed in terms in grams or kilograms. Videos Metric System: Units of Weight Using Drones to Weigh Whales? Metric Units of Mass: Convert mg, g, and kg Formulas You can convert between metric weight units using the conversion factors shown in .", "section": "Measuring Weight", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Measuring Temperature A thermometer that measures temperature in both customary and metric units. (credit: “Thermometer” by Jeff Djevdet/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Convert between Fahrenheit and Celsius. Identify reasonable values for temperature applications. Solve application problems involving temperature. When you touch something and it feels warm or cold, what is that really telling you about that substance? Temperature is a measure of how fast atoms and molecules are moving in a substance, whether that be the air, a stove top, or an ice cube. The faster those atoms and molecules move, the higher the temperature. In the metric system, temperature is measured using the Celsius (°C) scale. Because temperature is a condition of the physical properties of a substance, the Celsius scale was created with 100 degrees separating the point at which water freezes, 0 °C, and the point at which water boils, 100 °C. Scientifically, these are the points at which water molecules change from one state of matter to another—from solid (ice) to liquid (water) to gas (water vapor). When reading temperatures, it’s important to look beyond the degree symbol to determine which temperature scale the units express. For example, 13 °C reads “13 degrees Celsius,” indicating that the temperature is expressed using the Celsius scale, while 13 °F reads “13 degrees Fahrenheit,” indicating that the temperature is expressed using the Fahrenheit scale. Misconceptions About Temperature How Many Temperature Scales Are There? Did you know that in addition to Fahrenheit and Celsius, there is a third temperature scale widely used throughout the world? The Kelvin scale starts at absolute zero, the lowest possible temperature at which there is no heat energy present at all. It is primarily used by scientists to measure very high or very low temperatures when water is not involved. Converting Between Fahrenheit and Celsius Temperatures Understanding how to convert between Fahrenheit and Celsius temperatures is an essential skill in understanding metric temperatures. You likely know that below 32 °F means freezing temperatures and perhaps that the same holds true for 0 °C. While it may be difficult to recall that water boils at 212 °F, knowing that it boils at 100 °C is a fairly easy thing to remember. But what about all the temperatures in between? What is the temperature in degrees Celsisus on a scorching summer day? What about a cool autumn afternoon? If a recipe instructs you to preheat the oven to 350 °F, what Celsius temperature do you set the oven at? lists common temperatures on both scales, because we don’t use Celsius temperatures daily it’s difficult to remember them. Fortunately, we don’t have to. Instead, we can convert temperatures from Fahrenheit to Celsius and from Celsius to Fahrenheit using a simple algebraic expression. Common Temperatures The formulas used to convert temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in . Fahrenheit to Celsius Celsius to Fahrenheit C = 5 9 ( F − 32 ) F = 9 5 C + 32 Temperature Conversion Formulas Converting Temperatures from Fahrenheit to Celsius A recipe calls for the oven to be set to 392 °F. What is the temperature in Celsius? Use the formula in to convert from Fahrenheit to Celsius. C = 5 9 ( F − 32 ) C = 5 9 ( 392 − 32 ) C = 5 9 ( 360 ) C = 200 So, 392 °F is equivalent to 200 °C. Converting Temperatures from Celsius to Fahrenheit On a sunny afternoon in May, the temperature in London was 20 °C. What was the temperature in degrees Fahrenheit? Use the formula in to convert from Celsius to Fahrenheit. F = 9 5 C + 32 F = 9 5 ( 20 ) + 32 F = 36 + 32 F = 68 The temperature was 68 °F. Comparing Temperatures in Celsius and Fahrenheit A manufacturer requires a vaccine to be stored in a refrigerator at temperatures between 36 °F and 46 °F. The refrigerator in the local pharmacy cools to 3 °C. Can the vaccine be stored safely in the pharmacy’s refrigerator? Use the formula in to convert from Celsius to Fahrenheit. F = 9 5 C + 32 F = 9 5 ( 3 ) + 32 F = 5.4 + 32 F = 37.4 Then, compare the temperatures. 36 F ° < 37.4 F ° < 46 F ° Yes. 37.4 °F falls within the acceptable range to store the vaccine, so it can be stored safely in the pharmacy’s refrigerator. Reasonable Values for Temperature While knowing the exact temperature is important in most cases, sometimes an approximation will do. When trying to assess the reasonableness of values for temperature, there is a quicker way to convert temperatures for an approximation using mental math. These simpler formulas are listed in . The formulas used to estimate temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in . Fahrenheit to Celsius Celsius to Fahrenheit C = F − 30 2 F = 2 C + 30 Estimate Temperature Conversion Temperature Conversion Trick Using Benchmark Temperatures to Determine Reasonable Values for Temperatures Which is the more reasonable value for the temperature of a freezer? 5 °C or –5 °C? We know that water freezes at 0 °C. So, the more reasonable value for the temperature of a freezer is −5 °C, which is below 0 °C. At temperature of 5 °C is above freezing. Using Estimation to Determine Reasonable Values for Temperatures The average body temperature is generally accepted as 98.6 °F. What is a reasonable value for the average body temperature in degrees Celsius: 98.6 °C, 64.3 °C, or 34.3 °C? To estimate the average body temperature in degrees Celsius, subtract 30 from the temperature in degrees Fahrenheit, and divide the result by 2. ( 98.6 − 30 ) 2 = 68.6 2 = 34.3 A reasonable value for average body temperature is 34.3 °C. Using Conversion to Determine Reasonable Values for Temperatures Which is a reasonable temperature for storing chocolate: 28 °C, 18 °C, or 2 °C? Use the formula in to determine the temperature in degrees Fahrenheit. F = 9 5 C + 32 F = 9 5 ( 28 ) + 32 = 82.4 F = 9 5 ( 18 ) + 32 = 64.4 F = 9 5 ( 2 ) + 32 = 35.6 A temperature of 82.4 °F would be too hot, causing the chocolate to melt. A temperature of 35.6 °F is very close to freezing, which would affect the look and feel of the chocolate. So, a reasonable temperature for storing chocolate is 18 °C, or 64.4 °F. Solving Application Problems Involving Temperature Whether traveling abroad or working in a clinical laboratory, knowing how to solve problems involving temperature is an important skill to have. Many food labels express sizes in both ounces and grams. Most rulers and tape measures are two-sided with one side marked in inches and feet and the other in centimeters and meters. And while many thermometers have both Fahrenheit and Celsius scales, it really isn’t practical to pull out a thermometer when cooking a recipe that uses metric units. Let’s review at few instances where knowing how to fluently use the Celsius scale helps solve problems. Using Subtraction to Solve Temperature Problems The temperature in the refrigerator is 4 °C. The temperature in the freezer is 21 °C lower. What is the temperature in the freezer? Use subtraction to find the difference. 4 − 21 = − 17 So, the temperature in the freezer is −17 °C. Using Addition to Solve Temperature Problems A scientist was using a liquid that was 35 °C. They needed to heat the liquid to raise the temperature by 6 °C. What was the temperature after the scientist heated it? Use addition to find the new temperature. 35 + 6 = 41 The temperature of the liquid was 41 °C after the scientist heated it. Solving Complex Temperature Problems The optimum temperature for a chemical compound to develop its unique properties is 392 °F. When the heating process begins, the temperature of the compound is 20 °C. For safety purposes the compound can only be heated 9 °C every 15 minutes. How long until the compound reaches its optimum temperature? Step 1: Determine the optimum temperature in degrees Celsius using the formula in . C = 5 9 ( F − 32 ) C = 5 9 ( 392 − 32 ) C = 5 9 ( 360 ) C = 200 Step 2: Subtract the starting temperature. 200 C ° − 20 C ° = 180 C ° Step 3: Determine the number of 15-minute cycles needed to heat the compound to its optimum temperature. 180 ÷ 9 = 20 Step 4: Multiply the number of cycles needed by 15 minutes and convert the product to hours and minutes. 15 ⁢ minutes × 9 = 135 ⁢ minutes 135 ⁢ minutes = 2 ⁢ hours ⁢ 15 ⁢ minutes So, it will take 2 hours and 15 minutes for the compound to reach its optimum temperature. Learn the Metric System in 5 Minutes Check Your Understanding Key Terms temperature Key Concepts Temperature is a measure of how fast atoms and molecules are moving in a substance, whether that be the air, a stove top, or an ice cube. The faster those atoms and molecules move, the higher the temperature. In the metric system, temperature is measured using the Celsius (°C) scale. The Celsius scale was created with 100 degrees separating the point at which water freezes, 0 °C, and the point at which water boils, 100 °C. Scientifically, these are the points at which water molecules change from one state of matter to another—from solid (ice) to liquid (water) to gas (water vapor). Videos Misconceptions About Temperature Temperature Conversion Trick Learn the Metric System in 5 Minutes Formulas To convert temperature from Fahrenheit to Celsius: C = 5 9 ( F − 32 ) To convert temperature from Celsius to Fahrenheit: F = 9 5 C + 32 To estimate temperature from Fahrenheit to Celsius: C = F − 30 2 To estimate temperature from Celsius to Fahrenheit: F = 2 C + 30 Projects Cooking Take a favorite recipe that uses customary measures and convert the measures and cooking temperature to the metric system. Find a recipe that uses metric measures and convert the measures and cooking temperature to the U.S. Customary System of Measurement, using cups, tablespoons, or teaspoons as required. What did you observe? Was it easier to convert from one system to another? Which system allows for more precise measurements? What kitchen tools would you need in your kitchen if you used the metric system? Shopping Compare the average gas price in California to the average gas price in Puerto Rico. What conversions did you need to make to do the comparison? Do you think that the price of the gasoline is affected by the units in which it is sold? Sports What system of measurement is used for track and field events? Why do you think this system is used? What system of measurement is used for football? Why do you think this system is used? Research various sports records. Which units of measurement is used? What do you think influenced the unit of measure used? Chapter Review The Metric System Measuring Area Measuring Volume Measuring Weight Measuring Temperature Chapter Test", "section": "Measuring Temperature", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction The School of Athens by the Renaissance artist Raphael, painted between 1509 and 1511, depicts some of the greatest minds of ancient times. (credit: modification of work “School of Athens” by Raphael (1483–1520), Vatican Museums/Wikimedia, Public Domain) The painting The School of Athens presents great figures in history such as Plato, Aristotle, Socrates, Euclid, Archimedes, and Pythagoras. Other scientists are also represented in the painting. To the ancient Greeks, the study of mathematics meant the study of geometry above all other subjects. The Greeks looked for the beauty in geometry and did not allow their geometrical constructions to be “polluted” by the use of anything as practical as a ruler. They permitted the use of only two tools—a compass for drawing circles and arcs, and an unmarked straightedge to draw line segments. They would mark off units as needed. However, they never could be sure of what the units meant. For instance, how long is an inch? These mathematicians defined many concepts. The Greeks absorbed much from the Egyptians and the Babylonians (around 3000 BCE), including knowledge about congruence and similarity, area and volume, angles and triangles, and made it their task to introduce proofs for everything they learned. All of this historical wisdom culminated with Euclid in 300 BCE. Euclid (325–265 BC) is known as the father of geometry, and his most famous work is the 13-volume collection known as The Elements , which are said to be “the most studied books apart from the Bible.” Euclid brought together everything offered by the Babylonians, the Egyptians, and the more refined contributions by the Greeks, and set out, successfully, to organize and prove these concepts as he methodically developed formal theorems. This chapter begins with a discussion of the most basic geometric tools: the point, the line, and the plane. All other topics flow from there. Throughout the eight sections, we will talk about how to determine angle measurement and learn how to recognize properties of special angles, such as right angles and supplementary angles. We will look at the relationship of angles formed by a transversal, a line running through a set of parallel lines. We will explore the concepts of area and perimeter, surface area and volume, and transformational geometry as used in the patterns and rigid motions of tessellations. Finally, we will introduce right-angle trigonometry and explore the Pythagorean Theorem.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Points, Lines, and Planes The lower right-hand corner of The School of Athens depicts a figure representing Euclid illustrating to students how to use a compass on a small chalkboard. (credit: modification of work “School of Athens” by Raphael (1483–1520), Vatican Museums/Wikimedia, Public Domain) Learning Objectives After completing this section, you should be able to: Identify and describe points, lines, and planes. Express points and lines using proper notation. Determine union and intersection of sets. In this section, we will begin our exploration of geometry by looking at the basic definitions as defined by Euclid. These definitions form the foundation of the geometric theories that are applied in everyday life. In The Elements , Euclid summarized the geometric principles discovered earlier and created an axiomatic system, a system composed of postulates. A postulate is another term for axiom, which is a statement that is accepted as truth without the need for proof or verification. There were no formal geometric definitions before Euclid, and when terms could not be defined, they could be described. In order to write his postulates, Euclid had to describe the terms he needed and he called the descriptions “definitions.” Ultimately, we will work with theorems, which are statements that have been proved and can be proved. Points and Lines The first definition Euclid wrote was that of a point. He defined a point as “that which has no part.” It was later expanded to “an indivisible location which has no width, length, or breadth.” Here are the first two of the five postulates, as they are applicable to this first topic: Postulate 1 : A straight line segment can be drawn joining any two points. Postulate 2 : Any straight line segment can be extended indefinitely in a straight line. Before we go further, we will define some of the symbols used in geometry in : Basic Geometric Symbols for Points and Lines From , we see the variations in lines, such as line segments, rays, or half-lines. What is consistent is that two collinear points (points that lie on the same line) are required to form a line. Notice that a line segment is defined by its two endpoints showing that there is a definite beginning and end to a line segment. A ray is defined by two points on the line; the first point is where the ray begins, and the second point gives the line direction. A half-line is defined by two points, one where the line starts and the other to give direction, but an open circle at the starting point indicates that the starting point is not part of the half-line. A regular line is defined by any two points on the line and extends infinitely in both directions. Regular lines are typically drawn with arrows on each end. Defining Lines For the following exercises, use this line ( ). Define D E ¯ . Define F . Define D F ↔ . Define E F ¯ . The symbol D E ¯ , two letters with a straight line above, refers to the line segment that starts at point D and ends at point E . The letter F alone refers to point F . The symbol D F ↔ , two letters with a line above containing arrows on both ends, refers to the line that extends infinitely in both directions and contains the points D and F . The symbol E F ¯ , two letters with a straight line above, refers to the line segment that starts at point E and ends at point F . There are numerous applications of line segments in daily life. For example, airlines working out routes between cities, where each city’s airport is a point, and the points are connected by line segments. Another example is a city map. Think about the intersection of roads, such that the center of each intersection is a point, and the points are connected by line segments representing the roads. See . Air Line Routes Determining the Best Route View the street map ( ) as a series of line segments from point to point. For example, we have vertical line segments A B ¯ , B C ¯ , and C D ¯ on the right. On the left side of the map, we have vertical line segments H I ¯ , F G . ¯ The horizontal line segments are H A ¯ , E I ¯ , I E ¯ , E B ¯ , F C ¯ , C F ¯ , and G B ¯ . There are two diagonal line segments, A E ¯ and E F ¯ . Assume that each location is on a corner and that you live next door to the library. Street Map Let’s say that you want to stop at the grocery store on your way home from school. Come up with three routes you might take to do your errand and then go home. In other words, name the three ways by the line segments in the order you would walk, and which way do you think would be the most efficient route? How about stopping at the library after school? Name four ways you might travel to the library and which way do you think is the most efficient? Suppose you need to go to the post office and the dry cleaners on your way home from school. Name three ways you might walk to do your errands and end up at home. Which way do you think is the most efficient way to walk, get your errands done, and go home? From school to the grocery store and home: first way A H ¯ , H I ¯ , I E ¯ , E F ¯ , F G ¯ ; second way A B ¯ , B E ¯ , E I ¯ , I E ¯ , E F ¯ , F G ¯ ; third way A E ¯ , E I ¯ , I E ¯ , E F ¯ , F G ¯ . It seems that the third way is the most efficient way. From school to the library: first way A E ¯ , E F ¯ , F G ¯ ; second way A B ¯ , B C ¯ , C D ¯ , D G ¯ ; third way A B ¯ , B E ¯ , E F ¯ , F G ¯ ; fourth way A B ¯ , B C ¯ , C F ¯ , F G ¯ . The first way should be the most efficient way. From school to the post office or dry cleaners to home: first way A E ¯ , E F ¯ , F C ¯ , C D ¯ , D G ¯ ; second way A B ¯ , B E ¯ , E F ¯ , F C ¯ , C D ¯ , D G ¯ ; third way A B ¯ , B C ¯ , C F ¯ , F G ¯ . The third way would be the most efficient way. Parallel Lines Parallel lines are lines that lie in the same plane and move in the same direction, but never intersect. To indicate that the line l 1 and the line l 2 are parallel we often use the symbol l 1 ∥ l 2 . The distance d between parallel lines remains constant as the lines extend infinitely in both directions. See . Parallel Lines Perpendicular Lines Two lines that intersect at a 90 ∘ angle are perpendicular lines and are symbolized by ⊥ . If l 1 and l 2 are perpendicular, we write l 1 ⊥ l 2 . When two lines form a right angle, a 90 ∘ angle, we symbolize it with a little square □ . See . Perpendicular Lines Identifying Parallel and Perpendicular Lines Identify the sets of parallel and perpendicular lines in . Drawing these lines on a grid is the best way to distinguish which pairs of lines are parallel and which are perpendicular. Because they are on a grid, we assume all lines are equally spaced across the grid horizontally and vertically. The grid also tells us that the vertical lines are parallel and the horizontal lines are parallel. Additionally, all intersections form a 90 ∘ angle. Therefore, we can safely say the following: A B ↔ ∥ C D ↔ , the line containing the points A and B is parallel to the line containing the points C and D . E F ↔ ∥ G H ↔ , the line containing the points E and F is parallel to the line containing the points G and H . A B ↔ ⊥ E F ↔ , the line containing the points A and B is perpendicular to the line containing the points E and F . We know this because both lines trace grid lines, and intersecting grid lines are perpendicular. We can also state that A B ↔ ⊥ G H ↔ ; the line containing the points A and B is perpendicular to the line containing the points G and H because both lines trace grid lines, which are perpendicular by definition. We also have C D ↔ ⊥ E F ↔ ; the line containing the points C and D is perpendicular to the line containing the points E and F because both lines trace grid lines, which are perpendicular by definition. Finally, we see that C D ↔ ⊥ G H ↔ ; the line containing the points C and D is perpendicular to the line containing the points G and H because both lines trace grid lines, which are perpendicular by definition. Defining Union and Intersection of Sets Union and intersection of sets is a topic from set theory that is often associated with points and lines. So, it seems appropriate to introduce a mini-version of set theory here. First, a set is a collection of objects joined by some common criteria. We usually name sets with capital letters. For example, the set of odd integers between 0 and 10 looks like this: A = { 1 , 3 , 5 , 7 , 9 } . When it involves sets of lines, line segments, or points, we are usually referring to the union or intersection of set. The union of two or more sets contains all the elements in either one of the sets or elements in all the sets referenced, and is written by placing this symbol ∪ in between each of the sets. For example, let set A = { 1 , 2 , 3 } , and let set B = { 4 , 5 , 6 } . Then, the union of sets A and B is A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 } . The intersection of two or more sets contains only the elements that are common to each set, and we place this symbol ∩ in between each of the sets referenced. For example, let’s say that set A = { 1 , 3 , 5 } , and let set B = { 5 , 7 , 9 } . Then, the intersection of sets A and B is A ∩ B = { 5 } . Defining Union and Intersection of Sets Use the line ( ) for the following exercises. Draw each answer over the main drawing. Find B D → ∩ C A ← . Find A B ¯ ∪ A D ¯ . Find A D ↔ ∪ B C ¯ . Find A D ↔ ∩ B C ¯ . Find B A ← ∩ C D ¯ . Find B D → ∩ C A ← . This is the intersection of the ray B D → and the ray C A ← . Intersection includes only the elements that are common to both lines. For this intersection, only the line segment B C ¯ is common to both rays. Thus, B D → ∩ C A ← = B C ¯ . Find A B ¯ ∪ A D ¯ . The problem is asking for the union of two line segments, A B ¯ and A D ¯ . Union includes all elements in the first line and all elements in the second line. Since A B ¯ is part of A D ¯ , A B ¯ ∪ A D ¯ = A D ¯ . Find A D ↔ ∪ B C ¯ . This is the union of the line A D ↔ with the line segment B C ¯ . As the line segment B C ¯ is included on the line A D ↔ , then the union of these two lines equals the line A D ↔ . Find A D ↔ ∩ B C ¯ . The intersection of the line A D ↔ with line segment B C ¯ is the set of elements common to both lines. In this case, the only element in common is the line segment B C ¯ . Find A B ← ∩ C D ¯ . This is the intersection of the ray A B ← and the line segment C D ¯ . Intersection includes elements common to both lines. There are no elements in common. The intersection yields the empty set, as shown in . Therefore, A B ← ∩ C D ¯ = ∅ . Planes A plane, as defined by Euclid, is a “surface which lies evenly with the straight lines on itself.” A plane is a two-dimensional surface with infinite length and width, and no thickness. We also identify a plane by three noncollinear points, or points that do not lie on the same line. Think of a piece of paper, but one that has infinite length, infinite width, and no thickness. However, not all planes must extend infinitely. Sometimes a plane has a limited area. We usually label planes with a single capital letter, such as Plane P , as shown in , or by all points that determine the edges of a plane. In the following figure, Plane P contains points A and B , which are on the same line, and point C , which is not on that line. By definition, P is a plane. We can move laterally in any direction on a plane. Plane P One way to think of a plane is the Cartesian coordinate system with the x -axis marked off in horizontal units, and y -axis marked off in vertical units. In the Cartesian plane, we can identify the different types of lines as they are positioned in the system, as well as their locations. See . Cartesian Coordinate Plane This plane contains points S , T , and R . Points T and R are colinear and form a line segment. Point S is not on that line segment. Therefore, this represents a plane. To give the location of a point on the Cartesian plane, remember that the first number in the ordered pair is the horizontal move and the second number is the vertical move. Point R is located at ( 4 , 2 ) ; point S is located at ( − 3 , 4 ) ; and point T is located at ( − 1 , − 1 ) . We can also identify the line segment as T R ¯ . Two other concepts to note: Parallel planes do not intersect and the intersection of two planes is a straight line. The equation of that line of intersection is left to a study of three-dimensional space. See . Parallel and Intersecting Planes To summarize, some of the properties of planes include: Three points including at least one noncollinear point determine a plane. A line and a point not on the line determine a plane. The intersection of two distinct planes is a straight line. Identifying a Plane For the following exercises, refer to Identify the location of points A , B , C , and D . Describe the line from point A to point D . Describe the line from point C containing point B . Does this figure represent a plane? Point A is located at ( 4 , 3 ) ; point B is located at ( − 4 , 1 ) ; point C is located at ( − 2 , − 3 ) ; point D is located at ( 1 , 1 ) . The line from point A to point D is a line segment A D . ¯ The line from point C containing point B is a ray C B → starting at point C in the direction of B . Yes, this figure represents a plane because it contains at least three points, points A and D form a line segment, and neither point B nor point C is on that line segment. Intersecting Planes Name two pairs of intersecting planes on the shower enclosure illustration ( ). The plane ABCD intersects plane CDEF , and plane CDEF intersects plane EFGH . Plato Part of a remarkable chain of Greek mathematicians, Plato (427–347 BC) is known as the teacher. He was responsible for shaping the development of Western thought perhaps more powerfully than anyone of his time. One of his greatest achievements was the founding of the Academy in Athens where he emphasized the study of geometry. Geometry was considered by the Greeks to be the “ultimate human endeavor.” Above the doorway to the Academy, an inscription read, “Let no one ignorant of geometry enter here.” The curriculum of the Academy was a 15-year program. The students studied the exact sciences for the first 10 years. Plato believed that this was the necessary foundation for preparing students’ minds to study relationships that require abstract thinking. The next 5 years were devoted to the study of the “dialectic.” The dialectic is the art of question and answer. In Plato’s view, this skill was critical to the investigation and demonstration of innate mathematical truths. By training young students how to prove propositions and test hypotheses, he created a culture in which the systematic process was guaranteed. The Academy was essentially the world’s first university and held the reputation as the ultimate center of learning for more than 900 years. Check Your Understanding Key Terms line segment plane union intersection parallel perpendicular Key Concepts Modern-day geometry began in approximately 300 BCE with Euclid’s Elements , where he defined the principles associated with the line, the point, and the plane. Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of each other. The union of two sets, A and B , contains all points that are in both sets and is symbolized as A ∪ B . The intersection of two sets A and B includes only the points common to both sets and is symbolized as A ∩ B .", "section": "Points, Lines, and Planes", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Angles This modern architectural design emphasizes sharp reflective angles as part of the aesthetic through the use of glass walls. (credit: “Société Générale @ La Défense @ Paris” by Images Guilhem Vellut/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify and express angles using proper notation. Classify angles by their measurement. Solve application problems involving angles. Compute angles formed by transversals to parallel lines. Solve application problems involving angles formed by parallel lines. Unusual perspectives on architecture can reveal some extremely creative images. For example, aerial views of cities reveal some exciting and unexpected angles. Add reflections on glass or steel, lighting, and impressive textures, and the structure is a work of art. Understanding angles is critical to many fields, including engineering, architecture, landscaping, space planning, and so on. This is the topic of this section. We begin our study of angles with a description of how angles are formed and how they are classified. An angle is the joining of two rays, which sweep out as the sides of the angle, with a common endpoint. The common endpoint is called the vertex . We will often need to refer to more than one vertex, so you will want to know the plural of vertex, which is vertices. In , let the ray A B → stay put. Rotate the second ray A C → in a counterclockwise direction to the size of the angle you want. The angle is formed by the amount of rotation of the second ray. When the ray A C → continues to rotate in a counterclockwise direction back to its original position coinciding with ray A B → , the ray will have swept out 360 ∘ . We call the rays the “sides” of the angle. Vertex and Sides of an Angle Classifying Angles Angles are measured in radians or degrees . For example, an angle that measures π radians, or 3.14159 radians, is equal to the angle measuring 180 ∘ . An angle measuring π 2 radians, or 1.570796 radians, measures 90 ∘ . To translate degrees to radians, we multiply the angle measure in degrees by π 180 . For example, to write 45 ∘ in radians, we have 45 ∘ ( π 180 ) = π 4 = 0.785398 radians. To translate radians to degrees, we multiply by 180 π . For example, to write 2 π radians in degrees, we have 2 π ( 180 π ) = 360 ∘ . Another example of translating radians to degrees and degrees to radians is 2 π 3 . To write in degrees, we have 2 π 3 ( 180 π ) = 120 ∘ . To write 30 ∘ in radians, we have 30 ∘ ( π 180 ) = π 6 . However, we will use degrees throughout this chapter. To translate an angle measured in degrees to radians, multiply by π 180 . To translate an angle measured in radians to degrees, multiply by 180 π . Several angles are referred to so often that they have been given special names. A straight angle measures 180 ∘ ; a right angle measures 90 ∘ ; an acute angle is any angle whose measure is less than 90 ∘ ; and an obtuse angle is any angle whose measure is between 90 ∘ and 180 ∘ . See . Classifying and Naming Angles An easy way to measure angles is with a protractor ( ). A protractor is a very handy little tool, usually made of transparent plastic, like the one shown here. Protractor (credit: modification of work “School drawing tools” by Marco Verch/Flickr, CC BY 2.0) With a protractor, you line up the straight bottom with the horizontal straight line of the angle. Be sure to have the center hole lined up with the vertex of the angle. Then, look for the mark on the protractor where the second ray lines up. As you can see from the image, the degrees are marked off. Where the second ray lines up is the measurement of the angle. Make sure you correctly match the center mark of the protractor with the vertex of the angle to be measured. Otherwise, you will not get the correct measurement. Also, keep the protractor in a vertical position. Notation Naming angles can be done in couple of ways. We can name the angle by three points, one point on each of the sides and the vertex point in the middle, or we can name it by the vertex point alone. Also, we can use the symbols ∠ or ∡ before the points. When we are referring to the measure of the angle, we use the symbol m ∡ . See . Naming an Angle We can name this angle ∡ B A C , or ∡ C A B , or ∡ A . Classifying Angles Determine which angles are acute, right, obtuse, or straight on the graph ( ). You may want to use a protractor for this one. Acute angles measure less than 90 ∘ . Obtuse angles measure between 90 ∘ and 180 ∘ . Right angles measure 90 ∘ . Straight angles measure 180 ∘ . ∠ E O F ∠ E O G ∠ F O G ∠ G O H ∠ F O H ∠ H O J ∠ H O K ∠ J O K ∠ K O L ∠ J O L ∠ E O J ∠ E O K ∠ F O J ∠ F O K ∠ F O L ∠ G O K ∠ G O L ∠ E O H ∠ H O L ∠ G O J ∠ E O L Most angles can be classified visually or by description. However, if you are unsure, use a protractor. Adjacent Angles Two angles with the same starting point or vertex and one common side are called adjacent angles. In , angle ∠ D B C is adjacent to ∠ C B A . Notice that the way we designate an angle is with a point on each of its two sides and the vertex in the middle. Adjacent Angles Supplementary Angles Two angles are supplementary if the sum of their measures equals 180 ∘ . In , we are given that m ∡ F B E = 35 ∘ , so what is m ∡ A B E ? These are supplementary angles. Therefore, because m ∡ A B F = 180 ∘ , and as 180 ∘ − 35 ∘ = 145 ∘ , we have m ∡ A B E = 145 ∘ . Supplementary Angles Solving for Angle Measurements and Supplementary Angles Solve for the angle measurements in . Step 1: These are supplementary angles. We can see this because the two angles are part of a horizontal line, and a horizontal line represents 180 ∘ . Therefore, the sum of the two angles equals 180 ∘ . Step 2: ( 32 x − 7 ) + ( 5 x + 2 ) = 180 37 x − 5 = 180 37 x = 185 x = 5 Step 3: Find the measure of each angle: 32 x − 7 = 32 ( 5 ) − 7 = 153 ∘ 5 x + 2 = 5 ( 5 ) + 2 = 27 ∘ Step 4: We check: 153 ∘ + 27 ∘ = 180 ∘ . Complementary Angles Two angles are complementary if the sum of their measures equals 90 ∘ . In , we have m ∡ A B C = 30 ∘ , and m ∡ A B D = 90 ∘ . What is the m ∡ C B D ? These are complementary angles. Therefore, because 90 ∘ − 30 ∘ = 60 ∘ , the ∡ C B D = 60 ° . Complementary Angles Solving for Angle Measurements and Complementary Angles Solve for the angle measurements in . We have that ( 9 x − 5 ) + 4 x + ( 7 x − 5 ) = 90 20 x = 100 x = 5 Then, m ∡ ( 9 x − 5 ) = 40 ∘ , m ∡ ( 4 x ) = 20 ∘ , and m ∡ ( 7 x − 5 ) = 30 ∘ . Vertical Angles When two lines intersect, the opposite angles are called vertical angles, and vertical angles have equal measure. For example, shows two straight lines intersecting each other. One set of opposite angles shows angle markers; those angles have the same measure. The other two opposite angles have the same measure as well. Vertical Angles Calculating Vertical Angles In , one angle measures 40 ∘ . Find the measures of the remaining angles. The 40-degree angle and ∠ 2 are vertical angles. Therefore, m ∡ 2 = 40 ∘ . Notice that ∡ 2 and ∡ 1 are supplementary angles, meaning that the sum of m ∡ 2 and m ∡ 1 equals 180 ∘ . Therefore, m ∡ 1 = 180 ∘ − 40 ∘ = 140 ∘ . Since ∡ 1 and ∡ 3 are vertical angles, then m ∡ 3 equals 140 ∘ . Transversals When two parallel lines are crossed by a straight line or transversal , eight angles are formed, including alternate interior angles, alternate exterior angles, corresponding angles, vertical angles, and supplementary angles. See . Angles 1, 2, 7, and 8 are called exterior angles, and angles 3, 4, 5, and 6 are called interior angles. Transversal Alternate Interior Angles Alternate interior angles are the interior angles on opposite sides of the transversal. These two angles have the same measure. For example, ∡ 3 and ∡ 6 are alternate interior angles and have equal measure; ∡ 4 and ∡ 5 are alternate interior angles and have equal measure as well. See . Alternate Interior Angles Alternate Exterior Angles Alternate exterior angles are exterior angles on opposite sides of the transversal and have the same measure. For example, in , ∡ 2 and ∡ 7 are alternate exterior angles and have equal measures; ∡ 1 and ∡ 8 are alternate exterior angles and have equal measures as well. Alternate Exterior Angles Corresponding Angles Corresponding angles refer to one exterior angle and one interior angle on the same side as the transversal, which have equal measures. In , ∡ 1 and ∡ 5 are corresponding angles and have equal measures; ∡ 3 and ∡ 7 are corresponding angles and have equal measures; ∡ 2 and ∡ 6 are corresponding angles and have equal measures; ∡ 4 and ∡ 8 are corresponding angles and have equal measures as well. Corresponding Angles Evaluating Space You live on the corner of First Avenue and Linton Street. You want to plant a garden in the far corner of your property ( ) and fence off the area. However, the corner of your property does not form the traditional right angle. You learned from the city that the streets cross at an angle equal to 150 ∘ . What is the measure of the angle that will border your garden? As the angle between Linton Street and First Avenue is 150 ∘ , the supplementary angle is 30 ∘ . Therefore, the garden will form a 30 ∘ angle at the corner of your property. Determining Angles Formed by a Transversal In given that angle 3 measures 40 ∘ , find the measures of the remaining angles and give a reason for your solution. m ∡ 2 = m ∡ 3 = 40 ∘ by vertical angles. ∡ 3 = m ∡ 7 by corresponding angles. m ∡ 7 = m ∡ 6 = 40 ∘ by vertical angles. m ∡ 1 = 180 − 40 = 140 ∘ by supplementary angles. m ∡ 4 = m ∡ 1 = 140 ∘ by vertical angles. m ∡ 8 = m ∡ 1 = 140 ∘ by alternate exterior angles. m ∡ 5 = m ∡ 8 = 140 ∘ by vertical angles. Measuring Angles Formed by a Transversal In given that angle 2 measures 23 ∘ , find the measure of the remaining angles and state the reason for your solution. m ∡ 2 = m ∡ 3 = 23 ∘ by vertical angles, because ∡ 2 and ∡ 3 are the opposite angles formed by two intersecting lines. m ∡ 1 = 157 ∘ by supplementary angles to m ∡ 2 or m ∡ 3. We see that ∡ 1 and ∡ 2 form a straight angle as does ∡ 1 and ∡ 3. A straight angle measures 180 ∘ , so 180 ∘ − 23 ∘ = 157 ∘ . m ∡ 4 = m ∡ 1 = 157 ∘ by vertical angles, because ∡ 4 and ∡ 1 are the two opposite angles formed by two intersecting lines. m ∡ 5 = m ∡ 1 = 157 ∘ by corresponding angles because they are the same angle formed by the transversal crossing two parallel lines, one exterior and one interior. m ∡ 8 = m ∡ 5 = 157 ∘ by vertical angles because ∡ 8 and ∡ 5 are the two opposite angles formed by two intersecting lines. m ∡ 7 = m ∡ 2 = 23 ∘ by alternate exterior angles because, like vertical angles, these angles are the opposite angles formed by the transversal intersecting two parallel lines. m ∡ 6 = m ∡ 7 = 23 ∘ by vertical angles because these are the opposite angles formed by two intersecting lines. Finding Missing Angles Find the measures of the angles 1, 2, 4, 11, 12, and 14 in and the reason for your answer given that l 1 and l 2 are parallel. m ∡ 12 = 118 ∘ , supplementary angles m ∡ 14 = 118 ∘ , vertical angles m ∡ 11 = 62 ∘ , vertical angles m ∡ 4 = 62 ∘ , corresponding angles m ∡ 1 = 62 ∘ , vertical angles m ∡ 2 = 56 ∘ , supplementary angles The Number 360 Did you ever wonder why there are 360 ∘ in a circle? Why not 100 ∘ or 500 ∘ ? The number 360 was chosen by Babylonian astronomers before the ancient Greeks as the number to represent how many degrees in one complete rotation around a circle. It is said that they chose 360 for a couple of reasons: It is close to the number of days in a year, and 360 is divisible by 2, 3, 4, 5, 6, 8, 9, 10, … Check Your Understanding Key Terms vertex right angle acute angle obtuse angle straight angle complementary supplementary Key Concepts Angles are classified as acute if they measure less than 90 ∘ , obtuse if they measure greater than 90 ∘ and less than 180 ∘ , right if they measure exactly 90 ∘ , and straight if they measure exactly 180 ∘ . If the sum of angles equals 90 ∘ , they are complimentary angles. If the sum of angles equals 180 ∘ , they are supplementary. A transversal crossing two parallel lines form a series of equal angles: alternate interior angles, alternate exterior angles, vertical angles, and corresponding angles Formula To translate an angle measured in degrees to radians, multiply by π 180 . To translate an angle measured in radians to degrees, multiply by 180 π .", "section": "Angles", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Triangles The appearance of triangles in buildings is part of modern-day architectural design. (credit: \"Inside Hallgrímskirkja church, Reykjavik, Iceland\" by O Palsson/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify triangles by their sides. Identify triangles by their angles. Determine if triangles are congruent. Determine if triangles are similar. Find the missing side of similar triangles. How were the ancient Greeks able to calculate the radius of Earth? How did soldiers gauge their target? How was it possible centuries ago to estimate the height of a sail at sea? Triangles have always played a significant role in how we find heights of objects too high to measure or distances between objects too far away to calculate. In particular, the concept of similar triangles has countless applications in the real world, and we shall explore some of those applications in this section. Technology has given us instruments that allow us to find measurements of distant objects with little effort. However, it is all based on the properties of triangles discovered centuries ago. In this section, we will explore the various types of triangles and their special properties, as well as how to measure interior and exterior angles. We will also explore congruence theorems and similarity. Identifying Triangles Joining any three noncollinear points with line segments produces a triangle. For example, given points A , B , and C , connected by the line segments A B ¯ , B C , ¯ and A C ¯ , we have a triangle, as shown in . Triangle Triangles are classified by their angles and their sides. All angles in an acute triangle measure < 90 ∘ . One of the angles in a right triangle measures 90 ∘ , symbolized by □. One angle in an obtuse triangle measures between 90 ∘ and 180 ∘ . Sides that have equal length are indicated by the same hash marks. illustrates the shapes of the basic triangles, their names, and their properties. A few other facts to remember as we move forward: The points where the line segments meet are called the vertices (plural for vertex ). We often refer to sides of a triangle by the angle they are opposite. In other words, side a is opposite angle A , side b is opposite angle B , and side c is opposite angle C . Types of Triangles We want to add a special note about right triangles here, as they are referred to more than any other triangle. The side opposite the right angle is its longest side and is called the hypotenuse , and the sides adjacent to the right angle are called the legs. One of the most important properties of triangles is that the sum of the interior angles equals 180 ∘ . Euclid discovered and proved this property using parallel lines. The completed sketch is shown in . Sum of Interior Angles This is how the proof goes: Step 1: Start with a straight line A B ↔ and a point C not on the line. Step 2: Draw a line through point C parallel to the line A B ↔ . Step 3: Construct two transversals (a line crossing the parallel lines), one angled to the right and one angled to the left, to intersect the parallel lines. Step 4: Because of the property that alternate interior angles inside parallel lines are equal, we have that m ∡ 2 = m ∢ 1 and m ∡ 3 = m ∡ 4. Step 5: Notice that m ∡ 2 + m ∡ 5 + m ∡ 4 = 180 ∘ by the straight angle property. Step 6: Therefore, by substitution, m ∡ 1 = m ∡ 2 , and m ∡ 3 = m ∡ 4 , we have that m ∡ 1 + m ∡ 3 + m ∢ 5 = 180 ∘ . Therefore, the sum of the interior angles of a triangle = 180 ∘ . Finding Measures of Angles Inside a Triangle Find the measure of each angle in the triangle shown ( ). We know that the sum of the angles must equal 180 ∘ . Step 1: As the sum of the interior angles equals 180 ∘ , we can use algebra to find the measures: x + ( x + 17 ) + ( 3 x − 62 ) = 180 5 x − 45 = 180 5 x = 225 x = 225 5 = 45 Step 2: Now that we have the value of x , we can substitute 45 into the other two expressions to find the measure of those angles: ( x + 17 ) = 45 + 17 = 62 ∘ ( 3 x − 62 ) = 3 ( 45 ) − 62 = 135 − 62 = 73 ∘ Step 3: Then, m ∡ x = 45 ∘ , m ∡ ( x + 17 ) = 62 ∘ , and m ∡ ( 3 x − 62 ) = 73 ∘ . Finding Angle Measures Find the measure of angles numbered 1–5 in . The m ∡ 1 = 119 ∘ because it is supplementary with the unknown angle of the adjacent triangle. The unknown angle measures 61 ∘ . The m ∡ 2 = 61 ∘ because of vertical angles. The m ∡ 5 = 59 ∘ because the angle that is supplementary to the 134 ∘ measures 46 ∘ , and angle 5 is the unknown angle in that triangle. The m ∡ 4 = 59 ∘ by vertical angles. Finally, m ∡ 3 = 60 ∘ , as it is the third angle in the triangle with angles measuring 59 ∘ and 61 ∘ . Congruence If two triangles have equal angles and their sides lengths are equal, the triangles are congruent. In other words, if you can pick up one triangle and place it on top of the other triangle and they coincide, even if you have to rotate one, they are congruent. Determining If Triangles Are Congruent In , is the triangle ABC congruent to triangle DEF ? Triangle ABC is congruent to triangle DEF . Angles A and C are congruent to angles D and F , which implies that angle B is congruent to angle E . Side AB is congruent to side DE , and side CB is congruent to side FE , which implies that side AC is congruent to side DF . The Congruence Theorems The following theorems are tools you can use to prove that two triangles are congruent. We use the symbol ≅ to define congruence. For example, Δ A B C ≅ Δ D E F . Side-Side-Side (SSS). If three sides of one triangle are equal to the corresponding sides of the second triangle, then the triangles are congruent. See . Side-Side-Side (SSS) We have that D F ¯ ≅ R T ¯ , E F ¯ ≅ S T ¯ , and DE ¯ ≅ RS ¯ , then Δ D E F ≅ Δ R S T . Side-Angle-Side (SAS). If two sides of a triangle and the angle between them are equal to the corresponding two sides and included angle of the second triangle, then the triangles are congruent. See . We see that A B ¯ ≅ A ′ B ′ ¯ and B C ¯ ≅ B ′ C ′ ¯ , m ∡ B = m ∡ B ′ , then Δ A B C ≅ Δ A ′ B ′ C ′ . Side-Angle-Side (SAS) Angle-Side-Angle (ASA). If two angles and the side between them in one triangle are congruent to the two corresponding angles and the side between them in a second triangle, then the two triangles are congruent. See . Notice that m ∡ A ≅ m ∡ F , and m ∡ C ≅ m ∡ D , AC ¯ ≅ DF ¯ , then Δ ABC ≅ Δ DEF . Angle-Side-Angle (ASA) Angle-Angle-Side (AAS). If two angles and a nonincluded side of one triangle are congruent to two angles and the nonincluded corresponding side of a second triangle, then the triangles are congruent. See . We see that m ∡ X ≅ m X ′ , m ∡ Z ≅ m ∡ Z ′ , and X Y ¯ ≅ X ′ Y ′ ¯ , then Δ X Y Z ≅ Δ X ′ Y ′ Z ′ . Angle-Angle-Side (AAS) Identifying Congruence Theorems What congruence theorem is illustrated in ? AAS: Two angles and a non-included side in one triangle are congruent to the corresponding angles and side in the second triangle. Determining the Congruence Theorem What congruence theorem is illustrated in ? The SSS theorem. Similarity If two triangles have the same angle measurements and are the same shape but differ in size, the two triangles are similar. The lengths of the sides of one triangle will be proportional to the corresponding sides of the second triangle. Note that a single fraction a b is called a ratio, but two fractions equal to each other is called a proportion, such as a b = c d . This rule of similarity applies to all shapes as well as triangles. Another way to view similarity is by applying a scaling factor , which is the ratio of corresponding measurements between an object or representation of the object, to an image that produces the second, similar image. For example, why are the two images in are similar? These two images have the same proportions between elements. Therefore, they are similar. Similarity Determining If Triangles Are Similar Are the two triangles shown in similar? Step 1: We will look at the proportions within each triangle. In triangle α (alpha), the side opposite the 57 ∘ angle measures 7, and the side opposite the 33 ∘ angle measures 4. Then, the measures of the corresponding sides in triangle β (beta) measures 3.5 and 2, respectively. We have 4 7 = 0.5714 2 3.5 = 0.5714. This is the proportion 4 7 = 2 3.5 . The scaling factor is 0.5714. Step 2: Let’s try another correspondence. In triangle α , the hypotenuse measures 8.06 and the side opposite the 57 ∘ angle measures 7. In triangle β , the hypotenuse measure 4.03 and the side opposite the 57 ∘ angle measures 3.5. We have 7 8.06 = 0.8685 3.5 4.03 = 0.8685. Step 3: Now, let’s look at the proportions between triangle α and triangle β . The side measuring 2 in triangle β corresponds to the side measuring 4 in triangle α , the side measuring 3.5 in triangle β corresponds to the side measuring 7 in triangle α , and the hypotenuse in triangle β corresponds to the hypotenuse in triangle α . We have 2 4 = 0.5 3.5 7 = 0.5 4.03 8.06 = 0.5 Thus, the corresponding angles are equal and the proportions between each pair of corresponding sides equals 0.5. In other words, the scaling factor is 0.5. Therefore, the triangles are similar. Proving Similarity In , is triangle δ (delta) similar to triangle ε (epsilon)? Find the lengths of sides x and y as part of your answer. We can see that all three angles in triangle δ are equal to the corresponding angles in triangle ε . That is enough to determine similarity. However, we want to find the values of x and y to prove similarity. Step 1: We have to do is set up the proportions between the corresponding sides. We have the side that measures 2.375 in triangle δ corresponding to the side measuring 1.069 in triangle ε . We have the hypotenuse/side in triangle δ measuring 6 corresponding to the hypotenuse/side labeled y in triangle ε . And, finally, the side labeled x in triangle δ corresponds to the side measuring 2.475 in triangle ε . Each proportion should be equal. We start with the proportion of the shorter sides. Thus 1.069 2.375 = 0.45 Step 2: We solve for y using the first proportion. Set the two ratios equal to each other, cross-multiply, and solve for y . We have: 1.069 2.375 = y 6 ( 6 ) ( 1.069 ) = ( 2.375 ) ( y ) 6.414 = 2.375 y 6.414 2.375 = 2.7 = y So, y = 2.7 . Step 3: Checking that length in the proportion factor of 0.45, we have: y 6 = 2.7 6 = 0.45 Step 4: Solving for x , we will use the same proportion we used to solve for y . We have: 1.069 2.375 = 2.475 x 1.069 ( x ) = 2.475 ( 2.375 ) 1.069 ( x ) = 5.878 x = 5.878 1.069 = 5.5 2.475 x = 2.475 5.5 = 0.45 Step 5: We test the proportions. We have the following: 2.7 6 = 2.475 5.5 = 1.069 2.375 = 0.45 The proportions are all equal. Therefore, we have proven the property of similarity between triangle δ and triangle ε . Applying Similar Triangles A person who is 5 feet tall is standing 50 feet away from the base of a tree ( ). The tree casts a 57-foot shadow. The person casts a 7-foot shadow. What is the height of the tree? The bigger triangle includes a tree at side x and the smaller triangle includes the person at the side labeled 5 ft. These two triangles are similar because the smaller triangle fits inside the larger triangle at the smallest angle. It would fit inside the larger triangle at either of the other two angles as well. That all angles are equal is one of the criteria for similar triangles, so we can solve using proportions: 5 7 = x 57 5 ( 57 ) = 7 x 285 7 = x = 40.7 The tree is 40.7 feet tall. Finding Missing Lengths At a certain time of day, a radio tower casts a shadow 180 feet long ( ). At the same time, a 9-foot truck casts a shadow 15 feet long. What is the height of the tower? These are similar triangles and the problem can be solved by using proportions: x 180 = 9 15 9 ( 180 ) = 15 x 1620 = 15 x 108 = x The height of the tower is 108 ft. Thales of Miletus Thales of Miletus, sixth century BC, is considered one of the greatest mathematicians and philosophers of all time. Thales is credited with being the first to discover that the two angles at the base of an isosceles triangle are equal, and that the two angles formed by intersecting lines are equal—that is, vertical or opposite angles, are equal. Thales is also known for devising a method for measuring the height of the pyramids by similar right triangles. shows his method. He measured the length of the shadow cast by the pyramid at the precise time when his own shadow ended at the same place. Thales and Similarity He equated the vertical height of the pyramid with his own height; the horizontal distance from the pyramid to the tip of its shadow with the distance from himself and the tip of his own shadow; and finally, the length of the shadow cast off the top of the pyramid with length of his own shadow cast off the top of his head. Using proportions, as shown in , he essentially discovered the properties of similarity for right triangles. That is, △ A B C is similar to △ A ′ B ′ C ′ . Note that to be similar, all corresponding angles between the two triangles must be equal, and the proportions from one side to another side within each triangle, as well as the proportions of the corresponding sides between the two triangles must be equal. Thales is also credited with discovering a method of determining the distance of a ship from the shoreline. Here is how he did it, as illustrated in . Thales and Similar Triangles Thales walked along the shoreline pointing a stick at the ship until it formed a 90 ∘ angle to the shore. Then he walked along the shot and placed the stick in the ground at point C . He continued walking until he reached point D . Then, he turned and walked away from the shore at a 90 ∘ angle until the stick he placed in the ground at point C lined up with the ship, point E . This is how he created similar triangles and estimated the distance of the ship to the shore by using proportions. Check Your Understanding Key Terms acute obtuse isosceles equilateral hypotenuse congruence similarity scaling factor Key Concepts The sum of the interior angles of a triangle equals 180 ∘ . Two triangles are congruent when the corresponding angles have the same measure and the corresponding side lengths are equal. The congruence theorems include the following: SAS, two sides and the included angle of one triangle are congruent to the same in a second triangle; ASA, two angles and the included side of one triangle are congruent to the same in a second triangle; SSS, all three side lengths of one triangle are congruent to the same in a second triangle; AAS, two angles and the non-included side of one triangle are congruent to the same in a second triangle. Two shapes are similar when the proportions between corresponding angles, sides or features of two shapes are equal, regardless of size.", "section": "Triangles", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Polygons, Perimeter, and Circumference Geometric patterns are often used in fabrics due to the interest the shapes create. (credit: \"Triangles\" by Brett Jordan/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify polygons by their sides. Identify polygons by their characteristics. Calculate the perimeter of a polygon. Calculate the sum of the measures of a polygon’s interior angles. Calculate the sum of the measures of a polygon’s exterior angles. Calculate the circumference of a circle. Solve application problems involving perimeter and circumference. In our homes, on the road, everywhere we go, polygonal shapes are so common that we cannot count the many uses. Traffic signs, furniture, lighting, clocks, books, computers, phones, and so on, the list is endless. Many applications of polygonal shapes are for practical use, because the shapes chosen are the best for the purpose. Modern geometric patterns in fabric design have become more popular with time, and they are used for the beauty they lend to the material, the window coverings, the dresses, or the upholstery. This art is not done for any practical reason, but only for the interest these shapes can create, for the pure aesthetics of design. When designing fabrics, one has to consider the perimeter of the shapes, the triangles, the hexagons, and all polygons used in the pattern, including the circumference of any circular shapes. Additionally, it is the relationship of one object to another and experimenting with different shapes, changing perimeters, or changing angle measurements that we find the best overall design for the intended use of the fabric. In this section, we will explore these properties of polygons, the perimeter, the calculation of interior and exterior angles of polygons, and the circumference of a circle. Identifying Polygons A polygon is a closed, two-dimensional shape classified by the number of straight-line sides. See for some examples. We show only up to eight-sided polygons, but there are many, many more. Types of Polygons If all the sides of a polygon have equal lengths and all the angles are equal, they are called regular polygons . However, any shape with sides that are line segments can classify as a polygon. For example, the first two shapes, shown in and , are both pentagons because they each have five sides and five vertices. The third shape is a hexagon because it has six sides and six vertices. We should note here that the hexagon in is a concave hexagon, as opposed to the first two shapes, which are convex pentagons. Technically, what makes a polygon concave is having an interior angle that measures greater than 180 ∘ . They are hollowed out, or cave in, so to speak. Convex refers to the opposite effect where the shape is rounded out or pushed out. Polygons While there are variations of all polygons, quadrilaterals contain an additional set of figures classified by angles and whether there are one or more pairs of parallel sides. See . Types of Quadrilaterals Identifying Polygons Identify each polygon. This shape has six sides. Therefore, it is a hexagon. This shape has four sides, so it is a quadrilateral. It has two pairs of parallel sides making it a parallelogram. This shape has eight sides making it an octagon. This is an equilateral triangle, as all three sides are equal. This is a rhombus; all four sides are equal. This is a regular octagon, eight sides of equal length and equal angles. Determining Multiple Polygons What polygons make up ? Shapes 1 and 5 are hexagons; shapes 2, 3, 4, 6, 7, 9, 10, 12, 13, 14, 15, and 16 are triangles; shapes 8 and 17 are parallelograms; and shape 11 is a trapezoid. Perimeter Perimeter refers to the outside measurements of some area or region given in linear units. For example, to find out how much fencing you would need to enclose your backyard, you will need the perimeter. The general definition of perimeter is the sum of the lengths of the sides of an enclosed region. For some geometric shapes, such as rectangles and circles, we have formulas. For other shapes, it is a matter of just adding up the side lengths. A rectangle is defined as part of the group known as quadrilaterals, or shapes with four sides. A rectangle has two sets of parallel sides with four angles. To find the perimeter of a rectangle, we use the following formula: The formula for the perimeter P of a rectangle is P = 2 L + 2 W , twice the length L plus twice the width W . For example, to find the length of a rectangle that has a perimeter of 24 inches and a width of 4 inches, we use the formula. Thus, 24 = 2 l + 2 ( 4 ) = 2 l + 8 24 − 8 = 2 l 16 = 2 l 8 = l The length is 8 units. The perimeter of a regular polygon with n sides is given as P = n ⋅ s . For example, the perimeter of an equilateral triangle, a triangle with three equal sides, and a side length of 7 cm is P = 3 ( 7 ) = 21 cm . Finding the Perimeter of a Pentagon Find the perimeter of a regular pentagon with a side length of 7 cm ( ). A regular pentagon has five equal sides. Therefore, the perimeter is equal to P = 5 ( 7 ) = 35 cm . Finding the Perimeter of an Octagon Find the perimeter of a regular octagon with a side length of 14 cm ( ). A regular octagon has eight sides of equal length. Therefore, the perimeter of a regular octagon with a side length of 14 cm is P = 8 ( 14 ) = 112 cm . Sum of Interior and Exterior Angles To find the sum of the measurements of interior angles of a regular polygon, we have the following formula. The sum of the interior angles of a polygon with n sides is given by S = ( n − 2 ) 180 ∘ . For example, if we want to find the sum of the interior angles in a parallelogram, we have S = ( 4 − 2 ) 180 ∘ = 2 ( 180 ) = 360 ∘ . Similarly, to find the sum of the interior angles inside a regular heptagon, we have S = ( 7 − 2 ) 180 ∘ = 5 ( 180 ) = 900 ∘ . To find the measure of each interior angle of a regular polygon with n sides, we have the following formula. The measure of each interior angle of a regular polygon with n sides is given by a = ( n − 2 ) 180 ∘ n . For example, find the measure of an interior angle of a regular heptagon, as shown in . We have a = ( 7 − 2 ) 180 ∘ 7 = 128.57 ∘ . Interior Angles Calculating the Sum of Interior Angles Find the measure of an interior angle in a regular octagon using the formula, and then find the sum of all the interior angles using the sum formula. An octagon has eight sides, so n = 8 . Step 1: Using the formula a = ( n − 2 ) 180 ∘ 8 : a = ( 8 − 2 ) 180 ∘ 8 = ( 6 ) 180 ∘ 8 = 135 ∘ . So, the measure of each interior angle in a regular octagon is 135 ∘ . Step 2: The sum of the angles inside an octagon, so using the formula: S = ( n − 2 ) 180 ∘ = ( 8 − 2 ) 180 ∘ = 6 ( 180 ) = 1,080 ∘ . Step 3: We can test this, as we already know the measure of each angle is 135 ∘ . Thus, 8 ( 135 ∘ ) = 1,080 ∘ . Calculating Interior Angles Use algebra to calculate the measure of each interior angle of the five-sided polygon ( ). Step 1: Let us find out what the total of the sum of the interior angles should be. Use the formula for the sum of the angles in a polygon with n sides: S = ( n − 2 ) 180 ∘ . So, S = ( 5 − 2 ) 180 ∘ = 540 ∘ . Step 2: We add up all the angles and solve for x : 5 ( x + 7 ) + 120 + ( 6 x + 25 ) + 5 ( 2 x + 5 ) + 5 ( 3 x − 5 ) = 540 5 x + 6 x + 10 x + 15 x + 180 = 540 36 x = 360 x = 10 Step 3: We can then find the measure of each interior angle: m ∡ A = 5 ( 10 + 7 ) = 85 ∘ m ∡ B = 120 ∘ m ∡ C = 6 ( 10 ) + 25 = 85 ∘ m ∡ D = 5 ( 2 * 10 + 5 ) = 125 ∘ m ∡ E = 5 ( 3 * 10 − 5 ) = 125 ∘ An exterior angle of a regular polygon is an angle formed by extending a side length beyond the closed figure. The measure of an exterior angle of a regular polygon with n sides is found using the following formula: To find the measure of an exterior angle of a regular polygon with n sides we use the formula b = 360 ∘ n . In , we have a regular hexagon ABCDEF . By extending the lines of each side, an angle is formed on the exterior of the hexagon at each vertex. The measure of each exterior angle is found using the formula, b = 360 ∘ 6 = 60 ∘ . Exterior Angles Now, an important point is that the sum of the exterior angles of a regular polygon with n sides equals 360 ∘ . This implies that when we multiply the measure of one exterior angle by the number of sides of the regular polygon, we should get 360 ∘ . For the example in , we multiply the measure of each exterior angle, 60 ∘ , by the number of sides, six. Thus, the sum of the exterior angles is 6 ( 60 ∘ ) = 360 ∘ . Calculating the Sum of Exterior Angles Find the sum of the measure of the exterior angles of the pentagon ( ). Each individual angle measures 360 5 = 72 ∘ . Then, the sum of the exterior angles is 5 ( 72 ∘ ) = 360 ∘ . Circles and Circumference The perimeter of a circle is called the circumference . To find the circumference, we use the formula C = π d , where d is the diameter, the distance across the center, or C = 2 π r , where r is the radius. The circumference of a circle is found using the formula C = π d , where d is the diameter of the circle, or C = 2 π r , where r is the radius. The radius is ½ of the diameter of a circle. The symbol π = 3.141592654 … is the ratio of the circumference to the diameter. Because this ratio is constant, our formula is accurate for any size circle. See . Circle Diameter and Radius Let the radius be equal to 3.5 inches. Then, the circumference is C = 2 π ( 3.5 ) = 21.99 in . Finding Circumference with Diameter Find the circumference of a circle with diameter 10 cm. If the diameter is 10 cm, the circumference is C = 10 π = 31.42 cm . Finding Circumference with Radius Find the radius of a circle with a circumference of 12 in. If the circumference is 12 in, then the radius is 12 = 2 π r 12 2 π = r = 1.91 in . Calculating Circumference for the Real World You decide to make a trim for the window in . How many feet of trim do you need to buy? The trim will cover the 6 feet along the bottom and the two 12-ft sides plus the half circle on top. The circumference of a semicircle is ½ the circumference of a circle. The diameter of the semicircle is 6 ft. Then, the circumference of the semicircle would be 1 2 π d = 1 2 π ( 6 ) = 3 π ft = 9.4 ft . Therefore, the total perimeter of the window is 6 + 12 + 12 + 9.4 = 39.4 ft . You need to buy 39.4 ft of trim. Archimedes Archimedes (credit: “Archimedes” Engraving from the book Les vrais pourtraits et vies des hommes illustres grecz, latins et payens (1586)/Wikimedia Commons, Public Domain) The overwhelming consensus is that Archimedes (287–212 BCE) was the greatest mathematician of classical antiquity, if not of all time. A Greek scientist, inventor, philosopher, astronomer, physicist, and mathematician, Archimedes flourished in Syracuse, Sicily. He is credited with the invention of various types of pulley systems and screw pumps based on the center of gravity. He advanced numerous mathematical concepts, including theorems for finding surface area and volume. Archimedes anticipated modern calculus and developed the idea of the “infinitely small” and the method of exhaustion. The method of exhaustion is a technique for finding the area of a shape inscribed within a sequence of polygons. The areas of the polygons converge to the area of the inscribed shape. This technique evolved to the concept of limits, which we use today. One of the more interesting achievements of Archimedes is the way he estimated the number pi, the ratio of the circumference of a circle to its diameter. He was the first to find a valid approximation. He started with a circle having a diameter of 1 inch. His method involved drawing a polygon inscribed inside this circle and a polygon circumscribed around this circle. He knew that the perimeter of the inscribed polygon was smaller than the circumference of the circle, and the perimeter of the circumscribed polygon was larger than the circumference of the circle. This is shown in the drawing of an eight-sided polygon. He increased the number of sides of the polygon each time as he got closer to the real value of pi. The following table is an example of how he did this. Sides Inscribed Perimeter Circumscribed Perimeter 4 2.8284 4.00 8 3.0615 3.3137 16 3.1214 3.1826 32 3.1365 3.1517 64 3.1403 3.1441 Archimedes settled on an approximation of π ≈ 3.1416 after an iteration of 96 sides. Because pi is an irrational number, it cannot be written exactly. However, the capability of the supercomputer can compute pi to billions of decimal digits. As of 2002, the most precise approximation of pi includes 1.2 trillion decimal digits. The Platonic Solids The Platonic solids ( ) have been known since antiquity. A polyhedron is a three-dimensional object constructed with congruent regular polygonal faces. Named for the philosopher, Plato believed that each one of the solids is associated with one of the four elements: Fire is associated with the tetrahedron or pyramid, earth with the cube, air with the octahedron, and water with the icosahedron. Of the fifth Platonic solid, the dodecahedron, Plato said, “… God used it for arranging the constellations on the whole heaven.” Platonic Solids Plato believed that the combination of these five polyhedra formed all matter in the universe. Later, Euclid proved that exactly five regular polyhedra exist and devoted the last book of the Elements to this theory. These ideas were resuscitated by Johannes Kepler about 2,000 years later. Kepler used the solids to explain the geometry of the universe. The beauty and symmetry of the Platonic solids have inspired architects and artists from antiquity to the present. Check Your Understanding Key Terms perimeter polygon pentagon hexagon heptagon octagon quadrilateral trapezoid parallelogram circumference Key Concepts Regular polygons are closed, two-dimensional figures that have equal side lengths. They are named for the number of their sides. The perimeter of a polygon is the measure of the outline of the shape. We determine a shape’s perimeter by calculating the sum of the lengths of its sides. The sum of the interior angles of a regular polygon with n sides is found using the formula S = ( n − 2 ) 180 ∘ . The measure of a single interior angle of a regular polygon with n sides is determined using the formula a = ( n − 2 ) 180 ∘ n . The sum of the exterior angles of a regular polygon is 360 ∘ . The measure of a single exterior angle of a regular polygon with n sides is found using the formula b = 360 ∘ n . The circumference of a circle is C = 2 π r , where r is the radius, or C = π d , and d is the diameter. Formulas The formula for the perimeter P of a rectangle is P = 2 L + 2 W , twice the length L plus twice the width W . The sum of the interior angles of a polygon with n sides is given by S = ( n − 2 ) 180 ∘ . The measure of each interior angle of a regular polygon with n sides is given by a = ( n − 2 ) 180 ∘ n . To find the measure of an exterior angle of a regular polygon with n sides we use the formula b = 360 ∘ n . The circumference of a circle is found using the formula C = π d , where d is the diameter of the circle, or C = 2 π r , where r is the radius.", "section": "Polygons, Perimeter, and Circumference", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Tessellations Penrose tiling represents one type of tessellation. (credit: \"Penrose Tiling\" by Inductiveload/Wikimedia Commons, Public Domain) Learning Objectives After completing this section, you should be able to: Apply translations, rotations, and reflections. Determine if a shape tessellates. The illustration shown above ( ) is an unusual pattern called a Penrose tiling. Notice that there are two types of shapes used throughout the pattern: smaller green parallelograms and larger blue parallelograms. What's interesting about this design is that although it uses only two shapes over and over, there is no repeating pattern. In this section, we will focus on patterns that do repeat. Repeated patterns are found in architecture, fabric, floor tiles, wall patterns, rug patterns, and many unexpected places as well. It may be a simple hexagon-shaped floor tile, or a complex pattern composed of several different motifs. These two-dimensional designs are called regular (or periodic) tessellations . There are countless designs that may be classified as regular tessellations, and they all have one thing in common—their patterns repeat and cover the plane. We will explore how tessellations are created and experiment with making some of our own as well. The topic of tessellations belongs to a field in mathematics called transformational geometry, which is a study of the ways objects can be moved while retaining the same shape and size. These movements are termed rigid motions and symmetries . M. C. Escher A good place to start the study of tessellations is with the work of M. C. Escher. The Dutch graphic artist was famous for the dimensional illusions he created in his woodcuts and lithographs, and that theme is carried out in many of his tessellations as well. Escher became obsessed with the idea of the “regular division of the plane.” He sought ways to divide the plane with shapes that would fit snugly next to each other with no gaps or overlaps, represent beautiful patterns, and could be repeated infinitely to fill the plane. He experimented with practically every geometric shape imaginable and found the ones that would produce a regular division of the plane. The idea is similar to dividing a number by one of its factors. When a number divides another number evenly, there are no remainders, like there are no gaps when a shape divides or fills the plane. Escher went far beyond geometric shapes, beyond triangles and polygons, beyond irregular polygons, and used other shapes like figures, faces, animals, fish, and practically any type of object to achieve his goal; and he did achieve it, beautifully, and left it for the ages to appreciate. M.C. Escher: How to Create a Tessellation The Mathematical Art of M.C. Escher Tessellation Properties and Transformations A regular tessellation means that the pattern is made up of congruent regular polygons, same size and shape, including some type of movement ; that is, some type of transformation or symmetry. Here we consider the rigid motions of translations, rotations, reflections, or glide reflections. A plane of tessellations has the following properties: Patterns are repeated and fill the plane. There are no gaps or overlaps. Shapes must fit together perfectly. (It was Escher who determined that a proper tessellation could have no gaps and no overlaps.) Shapes are combined using a transformation. All the shapes are joined at a vertex. In other words, if you were to draw a circle around a vertex, it would include a corner of each shape touching at that vertex. For a tessellation of regular congruent polygons, the sum of the measures of the interior angles that meet at a vertex equals 360 ∘ . In , the tessellation is made up of squares. There are four squares meeting at a vertex. An interior angle of a square is 90 ∘ and the sum of four interior angles is 360 ∘ . In , the tessellation is made up of regular hexagons. There are three hexagons meeting at each vertex. The interior angle of a hexagon is 120 ∘ , and the sum of three interior angles is 360 ∘ . Both tessellations will fill the plane, there are no gaps, the sum of the interior angle meeting at the vertex is 360 ∘ , and both are achieved by translation transformations. These tessellations work because all the properties of a tessellation are present. Tessellation – Squares Tessellation – Hexagons The movements or rigid motions of the shapes that define tessellations are classified as translations, rotations, reflections, or glide reflections. Let’s first define these movements and then look at some examples showing how these transformations are revealed. Translation A translation is a movement that shifts the shape vertically, horizontally, or on the diagonal. Consider the trapezoid A B C D in . We have translated it 3 units to the right and 3 units up. That means every corner is moved by the number of units and in the direction specified. Mathematicians will indicate this movement with a vector, an arrow that is drawn to illustrate the criteria and the magnitude of the translation. The location of the translated trapezoid is marked with the vertices, A ′ B ′ C ′ D ′ , but it is still the exact same shape and size as the original trapezoid A B C D . Translation Creating a Translation Suppose you have a hexagon on a grid as in . Translate the hexagon 5 units to the right and 3 units up. The best way to do this is by translating the individual points A , B , C , D , E , F . Once translated, the points become A ′ , B ′ , C ′ , D ′ , E ′ , F ′ . Rotation The rotation transformation occurs when you rotate a shape about a point and at a predetermined angle. In , the triangle is rotated around the rotation point by 90 ∘ , and then translated 7 units up and 4 units over to the right. That means that each corner is translated to the new location by the same number of units and in the same direction. Rotation We can see that Δ A is mapped to Δ A ′ by a rotation of 90 ∘ up and to the right. If rotated again by 90 ∘ , the triangle would be upside down. Applying a Rotation illustrates a tessellation begun with an equilateral triangle. Explain how this pattern is produced. A rotation to the right or to the left around the vertex by 60 ∘ , six times, produces the hexagonal shape. The sixth rotation brings the triangle back to its original position. Then, a reflection up and another one on the diagonal will reproduce the pattern. When a shape returns to its original position by a rotation, we say that it has rotational symmetry. Reflection A reflection is the third transformation. A shape is reflected about a line and the new shape becomes a mirror image. You can reflect the shape vertically, horizontally, or on the diagonal. There are two shapes in . The quadrilateral is reflected horizontally; the arrow shape is reflected vertically. Reflection Glide Reflection The glide reflection is the fourth transformation. It is a combination of a reflection and a translation. This can occur by first reflecting the shape and then gliding or translating it to its new location, or by translating first and then reflecting. The example in shows a trapezoid, which is reflected over the dashed line, so it appears upside down. Then, we shifted the shape horizontally by 6 units to the right. Whether we use the glide first or the reflection first, the end result is the same in most cases. However, the tessellation shown in the next example can only be achieved by a reflection first and then a translation. Glide Reflection Applying the Glide Reflection An obtuse triangle is reflected about the dashed line, and the two shapes are joined together. How does the tessellation shown in materialize? The new shape is reflected horizontally and joined with the original shape. It is then translated vertically and horizontally to make up the tessellation. Notice the blank spaces next to the vertical pattern. These areas are made up of the exact original shape rotated 180 ∘ , but with no line up the center. These rotated shapes are translated horizontally and vertically, and thus, the plane is tessellated with no gaps. This is an example of a glide reflection where the order of the transformations matters. Applying More Than One Tessellation Show how this tessellation ( ) can be achieved. This is a tessellation that has one color on the front of the trapezoid and a different color on the back. There is a translation on the diagonal, and a reflection vertically. These are two separate transformations resulting in two new placements of the trapezoid. We can call this a combination of two transformations or a glide reflection. Interior Angles The sum of the interior angles of a tessellation is 360 ∘ . In , the tessellation is made of six triangles formed into the shape of a hexagon. Each angle inside a triangle equals 60 ∘ , and the six vertices meet the sum of those interior angles, 6 ( 60 ° ) = 360 ° . Interior Angles at the Vertex of Triangles In , the tessellation is made up of trapezoids, such that two of the interior angles of each trapezoid equals 75 ° and the other two angles equal 105 ° . Thus, the sum of the interior angles where the vertices of four trapezoids meet equals 105 ° + 75 ° + 75 ° + 105 ° = 360 ° . Interior Angles at the Vertex of Trapezoids These tessellations illustrate the property that the shapes meet at a vertex where the interior angles sum to 360 ° . Tessellating Shapes We might think that all regular polygons will tessellate the plane by themselves. We have seen that squares do and hexagons do. The pattern of squares in is a translation of the shape horizontally and vertically. The hexagonal pattern in , is translated horizontally, and then on the diagonal, either to the right or to the left. This particular pattern can also be formed by rotations. Both tessellations are made up of congruent shapes and each shape fits in perfectly as the pattern repeats. Translation Horizontally and Vertically Translation Horizontally and Slide Diagonally We have also seen that equilateral triangles will tessellate the plane without gaps or overlaps, as shown in . The pattern is made by a reflection and a translation. The darker side is the face of the triangle and the lighter side is the back of the triangle, shown by the reflection. Each triangle is reflected and then translated on the diagonal. Reflection and Glide Translation Escher experimented with all regular polygons and found that only the ones mentioned, the equilateral triangle, the square, and the hexagon, will tessellate the plane by themselves. Let’s try a few other regular polygons to observe what Escher found. Tessellating the Plane Do regular pentagons tessellate the plane by themselves ( )? We can see that regular pentagons do not tessellate the plane by themselves. There is a gap, a gap in the shape of a parallelogram. We conclude that regular pentagons will not tessellate the plane by themselves. Tessellating Octagons Do regular octagons tessellate the plane by themselves ( )? Again, we see that regular octagons do not tessellate the plane by themselves. The gaps, however, are squares. So, two regular polygons, an octagon and a square, do tessellate the plane. Just because regular pentagons do not tessellate the plane by themselves does not mean that there are no pentagons that tessellate the plane, as we see in . Tessellation of Pentagons Another example of an irregular polygon that tessellates the plane is by using the obtuse irregular triangle from a previous example. What transformations should be performed to produce the tessellation shown in ? Tessellating with Obtuse Irregular Triangles First, the triangle is reflected over the tip at point A , and then translated to the right and joined with the original triangle to form a parallelogram. The parallelogram is then translated on the diagonal and to the right and to the left. Naming A tessellation of squares is named by choosing a vertex and then counting the number of sides of each shape touching the vertex. Each square in the tessellation shown in has four sides, so starting with square A , the first number is 4, moving around counterclockwise to the next square meeting the vertex, square B , we have another 4, square C adds another 4, and finally square D adds a fourth 4. So, we would name this tessellation a 4.4.4.4. The hexagon tessellation, shown in has six sides to the shape and three hexagons meet at the vertex. Thus, we would name this a 6.6.6. The triangle tessellation, shown in has six triangles meeting the vertex. Each triangle has three sides. Thus, we name this a 3.3.3.3.3.3. 4.4.4.4 6.6.6 3.3.3.3.3.3 Creating Your Own Tessellation Create a tessellation using two colors and two shapes. We used a parallelogram and an isosceles triangle. The parallelogram is reflected vertically and horizontally so that only every other corner touches. The triangles are reflected vertically and horizontally and then translated over the parallelogram. The result is alternating vertical columns of parallelograms and then triangles ( ). Check Your Understanding Key Terms tessellation translation reflection rotation glide reflection Key Concepts A tessellation is a particular pattern composed of shapes, usually polygons, that repeat and cover the plane with no gaps or overlaps. Properties of tessellations include rigid motions of the shapes called transformations. Transformations refer to translations, rotations, reflections, and glide reflections. Shapes are transformed in such a way to create a pattern. Videos M.C. Escher: How to Create a Tessellation The Mathematical Art of M.C. Escher", "section": "Tessellations", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Area The area of a regulation baseball diamond must adhere to specific measurements to be legal. (credit: \"Diagram of Regulation Diamond\" by Erica Fischer from \"Baseball\" The World Book, 1920/Flickr, Public Domain, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the area of triangles. Calculate the area of quadrilaterals. Calculate the area of other polygons. Calculate the area of circles. Some areas carry more importance than other areas. Did you know that in a baseball game, when the player hits the ball and runs to first base that he must run within a 6-foot wide path? If he veers off slightly to the right, he is out. In other words, a few inches can be the difference in winning or losing a game. Another example is real estate. On Manhattan Island, one square foot of real estate is worth far more than real estate in practically any other area of the country. In other words, we place a value on area. As the context changes, so does the value. Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. As opposed to linear measurements like perimeter, which in in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area. Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as in 2 or cm 2 . Note that linear measurements have no exponent above the units or we can say that the exponent is 1. Area of Triangles The formula for the area of a triangle is given as follows. The area of a triangle is given as A = 1 2 b h , where b represents the base and h represents the height. For example, consider the triangle in . Triangle 1 The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area: A = 1 2 ( 4 ) ( 5 ) = 1 2 ( 20 ) = 10 cm 2 In , the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then A = 1 2 ( 7 ) ( 3.5 ) = 12.25 cm 2 . Triangle 2 Finding the Area of a Triangle Find the area of this triangle that has a base of 4 cm and the height is 6 cm ( ). Using the formula, we have A = 1 2 ( 4 ) ( 6 ) = 24 2 = 12 cm 2 . Area of Quadrilaterals A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle. Square In , a 12 in × 12 in grid is represented with twelve 1 in × 1 in squares across each row, and twelve 1 in × 1 in squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where s = length of a side , is A = s ⋅ s . The area of the square in is A = 12 in × 12 in = 144 in 2 . Area of a Square The formula for the area of a square is A = s ⋅ s or A = s 2 . Rectangle Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in has width equal to 5 in and length equal to 12 in. The area is A = 5 ( 12 ) = 60 in 2 . Area of a Rectangle The area of a rectangle is given as A = l w . Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area? Finding the Area of a Rectangle You have a garden with an area of 196 square feet. How wide can the garden be if the length is 28 feet? The area of a rectangular region is A = l w . Letting the width equal w : 196 = 28 w 196 28 = w = 7 ft Determining the Cost of Floor Tile Jennifer is planning to install vinyl floor tile in her rectangular-shaped basement, which measures 29 ft by 16 ft. How many boxes of floor tile should she buy and how much will it cost if one box costs $50 and covers 20 ft 2 ? The area of the basement floor is A = 29 ( 16 ) = 464 ft 2 . We will divide this area by 20 ft . Thus, 464 20 = 23.2 . Therefore, Jennifer will have to buy 24 boxes of tile at a cost $1,200. Parallelogram The area of a parallelogram can be found using the formula for the area of a triangle. Notice in , if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram: A = 2 ( 1 2 b h ) A = b h Area of a Parallelogram The area of a parallelogram is A = b h . For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be A = ( 10 ) ( 5 ) = 50 in 2 . Finding the Area of a Parallelogram In the parallelogram ( ), if F B = 10 , A D = 15 , find the exact area of the parallelogram. Using the formula of A = b h , we have A = 10 ( 15 ) = 150 cm 2 . Finding the Area of a Parallelogram Park The boundaries of a city park form a parallelogram ( ). The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for $0.50 per square foot, installation is $1.50 per square foot, and the cost of the equipment for the day is $100. Step 1: As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft. Step 2: The park has the shape of a parallelogram, and the formula for the area is A = b h : A = 165 ( 117 ) = 19,305 ft 2 . Step 3: The city needs to purchase 19 , 305 ft 2 of sod. The cost will be $0.50 per square foot for the sod and $1.50 per square foot for installation, plus $100 for equipment: 19,305 ( $ 0.50 ) + 19,305 ( $ 1.50 ) = $ 9,652.5 + $ 28,957.5 + $ 100.00 = $ 38,710.00 Trapezoid Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases a and b and height h is given here. The formula for the area of a trapezoid is given as A = 1 2 h ( a + b ) . For example, find the area of the trapezoid in that has base a equal to 8 cm, base b equal to 6 cm, and height equal to 6 cm. Area of a Trapezoid The area is A = 1 2 ( 6 ) ( 6 + 8 ) = 42 cm 2 . Finding the Area of a Trapezoid A B C D ( ) is a regular trapezoid with A B ¯ ‖ C D ¯ . Find the exact perimeter of A B C D , and then find the area. The perimeter is the measure of the boundary of the shape, so we just add up the lengths of the sides. We have P = 31 + 13.5 + 11 + 13.5 = 69 in . Then, the area of the trapezoid using the formula is A = 1 2 ( 10 ) ( 11 + 31 ) = 210 in 2 . Rhombus The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals. The area of a rhombus is found using one of these formulas: A = d 1 d 2 2 , where d 1 and d 2 are the diagonals. A = 1 2 b h , where b is the base and h is the height. For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is 220 cm 2 , and the measure of d 2 = 11 , find the measure of d 1 . To solve this problem, we input the known values into the formula and solve for the unknown. See . Area of a Rhombus We have that 220 = 11 d 1 2 220 ( 2 ) = 11 d 1 440 11 = d 1 = 40 Finding the Area of a Rhombus Find the measurement of the diagonal d 1 if the area of the rhombus is 240 cm 2 , and the measure of d 2 = 24 cm . Use the formula with the known values: 240 = d 1 ( 24 ) 2 240 ( 2 ) = d 1 ( 24 ) 480 24 = d 1 = 20 Finding the Area of a Rhombus You notice a child flying a rhombus-shaped kite on the beach. When it falls to the ground, it falls on a beach towel measuring 36 in by 72 in . You notice that one of the diagonals of the kite is the same length as the 36 in width of the towel. The second diagonal appears to be 2 in longer. What is the area of the kite ( )? Using the formula, we have: A = d 1 d 2 2 = 36 ( 38 ) 2 = 684 in 2 Area of Polygons To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem a of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius r of a regular polygon is also a line segment that starts at the center but extends to a vertex. See . Apothem and Radius of a Polygon The area of a regular polygon is found with the formula A = 1 2 a p , where a is the apothem and p is the perimeter. For example, consider the regular hexagon shown in with a side length of 4 cm, and the apothem measures a = 2 3 . Area of a Hexagon We have the perimeter, p = 6 ( 4 ) = 24 cm . We have the apothem as a = 2 3 . Then, the area is: A = 1 2 ( 2 3 ) ( 24 ) = 24 3 = 41.57 cm 2 Finding the Area of a Regular Octagon Find the area of a regular octagon with the apothem equal to 18 cm and a side length equal to 13 cm ( ). Using the formula, we have the perimeter p = 8 ( 13 ) = 104 cm . Then, the area is A = 1 2 ( 10 ) ( 104 ) = 936 cm 2 . Changing Units Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency. Changing Units Carpeting comes in units of square yards. Your living room measures 21 ft wide by 24 ft long. How much carpeting do you buy? We must convert feet to yards. As there are 3 ft in 1 yd, we have 21 ft = 7 yds and 24 ft = 8 yds . Then, 7 ( 8 ) = 56 yd 2 . Area of Circles Just as the circumference of a circle includes the number π , so does the formula for the area of a circle. Recall that π is a non-terminating, non-repeating decimal number: π = 3.14159 … . It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area. The area of a circle is given as A = π r 2 , where r is the radius. For example, to find the of the circle with radius equal to 3 cm, as shown in , is found using the formula A = π r 2 . Circle with Radius 3 We have A = π r 2 = π ( 3 ) 2 = 9 π = 28.27 cm 2 Finding the Area of a Circle Find the area of a circle with diameter of 16 cm. The formula for the area of a circle is given in terms of the radius, so we cut the diameter in half. Then, the area is A = π ( 8 ) 2 = 201.1 cm 2 . Determining the Better Value for Pizza You decide to order a pizza to share with your friend for dinner. The price for an 8-inch diameter pizza is $7.99. The price for 16-inch diameter pizza is $13.99. Which one do you think is the better value? The area of the 8-inch diameter pizza is A = π ( 4 ) 2 = 50.3 in 2 . The area of the 16-inch diameter pizza is A = π ( 8 ) 2 = 201.1 in 2 . Next, we divide the cost of each pizza by its area in square inches. Thus, 13.99 201.1 = $ 0.07 per square inch and 7.99 50.3 = $ 0.16 per square inch. So clearly, the 16-inch pizza is the better value. Applying Area to the Real World You want to purchase a tinted film, sold by the square foot, for the window in . (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase. First, the rectangular portion has A = 5 ( 10 ) = 50 ft 2 . For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is A = 1 2 π ( 2.5 ) 2 = 9.8 ft 2 . Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is A = 50 + 9.8 = 59.8 ft 2 . Area within Area Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems. Finding the Area within an Area The patio in your backyard measures 20 ft by 10 ft ( ). On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over? The hot tub has a radius of 3 ft. That area is then A = π ( 3 ) 2 = 9 π = 28.27 ft 2 . The total square feet taken up with the table and chairs and the hot tub is 36 + 28.27 = 64.27 ft 2 . The area left over is equal to the total area of the patio, 200 ft 2 minus the area for the table and chairs and the hot tub. Thus, the area left over is 200 − 64.27 = 135.7 ft 2 . Finding the Cost of Fertilizing an Area A sod farmer wants to fertilize a rectangular plot of land 150 ft by 240 ft. A bag of fertilizer covers 5,000 ft 2 and costs $200. How much will it cost to fertilize the entire plot of land? The plot of land is 36 , 000 ft 2 . It will take 7.2 bags of fertilizer to cover the land area. Therefore, the farmer will have to purchase 8 bags of fertilizer at $200 a bag, which comes to $1,600. Heron of Alexandria Heron of Alexandria (credit: \"Heron of Alexandria\" from 1688 German translation of Hero's Pneumatics/Wikimedia Commons, Public Domain) Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, Mechanics and Optics . His contribution to the field of mathematics was enormous. Metrica , a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle A B C ( ), Heron’s formula is A = s ( s − a ) ( s − b ) ( s − c ) , where s is the semi-perimeter calculated as s = a + b + c 2 . Check Your Understanding Key Terms triangle square rectangle rhombus apothem radius circle Key Concepts The area A of a triangle is found with the formula A = 1 2 b h , where b is the base and h is the height. The area of a parallelogram is found using the formula A = b h , where b is the base and h is the height. The area of a rectangle is found using the formula A = l w , where l is the length and w is the width. The area of a trapezoid is found using the formula A = 1 2 h ( b 1 + b 2 ) , where h is the height, b 1 is the length of one base, and b 2 is the length of the other base. The area of a rhombus is found using the formula A = d 1 d 2 2 , where d 1 is the length of one diagonal and d 2 is the length of the other diagonal. The area of a regular polygon is found using the formula A = 1 2 a p , where a is the apothem and p is the perimeter. The area of a circle is found using the formula A = π r 2 , where r is the radius. Formulas The area of a triangle is given as A = 1 2 b h , where b represents the base and h represents the height. The formula for the area of a square is A = s ⋅ s or A = s 2 . The area of a rectangle is given as A = l w . The area of a parallelogram is A = b h . The formula for the area of a trapezoid is given as A = 1 2 h ( a + b ) . The area of a rhombus is found using one of these formulas: A = d 1 d 2 2 , where d 1 and d 2 are the diagonals. A = 1 2 b h , where b is the base and h is the height. The area of a regular polygon is found with the formula A = 1 2 a p , where a is the apothem and p is the perimeter. The area of a circle is given as A = π r 2 , where r is the radius.", "section": "Area", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Volume and Surface Area Volume is illustrated in this 3-dimensional view of an interior space. This gives a buyer a more realistic interpretation of space. (credit: \"beam render 10 with sun and cat tree\" by monkeywing/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the surface area of right prisms and cylinders. Calculate the volume of right prisms and cylinders. Solve application problems involving surface area and volume. Volume and surface area are two measurements that are part of our daily lives. We use volume every day, even though we do not focus on it. When you purchase groceries, volume is the key to pricing. Judging how much paint to buy or how many square feet of siding to purchase is based on surface area. The list goes on. An example is a three-dimensional rendering of a floor plan. These types of drawings make building layouts far easier to understand for the client. It allows the viewer a realistic idea of the product at completion; you can see the natural space, the volume of the rooms. This section gives you practical information you will use consistently. You may not remember every formula, but you will remember the concepts, and you will know where to go should you want to calculate volume or surface area in the future. We will concentrate on a few particular types of three-dimensional objects: right prisms and right cylinders. The adjective “right” refers to objects such that the sides form a right angle with the base. We will look at right rectangular prisms, right triangular prisms, right hexagonal prisms, right octagonal prisms, and right cylinders. Although, the principles learned here apply to all right prisms. Three-Dimensional Objects In geometry, three-dimensional objects are called geometric solids . Surface area refers to the flat surfaces that surround the solid and is measured in square units. Volume refers to the space inside the solid and is measured in cubic units. Imagine that you have a square flat surface with width and length. Adding the third dimension adds depth or height, depending on your viewpoint, and now you have a box. One way to view this concept is in the Cartesian coordinate three-dimensional space. The x -axis and the y -axis are, as you would expect, two dimensions and suitable for plotting two-dimensional graphs and shapes. Adding the z -axis, which shoots through the origin perpendicular to the x y -plane, and we have a third dimension. See . Three-Dimensional Space Here is another view taking the two-dimensional square to a third dimension. See . Going from Two Dimensions to Three Dimensions To study objects in three dimensions, we need to consider the formulas for surface area and volume. For example, suppose you have a box ( ) with a hinged lid that you want to use for keeping photos, and you want to cover the box with a decorative paper. You would need to find the surface area to calculate how much paper is needed. Suppose you need to know how much water will fill your swimming pool. In that case, you would need to calculate the volume of the pool. These are just a couple of examples of why these concepts should be understood, and familiarity with the formulas will allow you to make use of these ideas as related to right prisms and right cylinders. Right Prisms A right prism is a particular type of three-dimensional object. It has a polygon-shaped base and a congruent, regular polygon-shaped top, which are connected by the height of its lateral sides, as shown in . The lateral sides form a right angle with the base and the top. There are rectangular prisms, hexagonal prisms, octagonal prisms, triangular prisms, and so on. Pentagonal Prism Generally, to calculate surface area, we find the area of each side of the object and add the areas together. To calculate volume, we calculate the space inside the solid bounded by its sides. The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, S A = 2 B + p h , where B is equal to the area of the base and top, p is the perimeter of the base, and h is the height. The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, V = B ⋅ h , where B is the area of the base and h is the height. Calculating Surface Area and Volume of a Rectangular Prism Find the surface area and volume of the rectangular prism that has a width of 10 cm, a length of 5 cm, and a height of 3 cm ( ). The surface area is S A = 2 ( 10 ) ( 5 ) + 2 ( 5 ) ( 3 ) + 2 ( 10 ) ( 3 ) = 190 cm 2 . The volume is V = 10 ( 5 ) ( 3 ) = 150 cm 3 . In , we have three views of a right hexagonal prism. The regular hexagon is the base and top, and the lateral faces are the rectangular regions perpendicular to the base. We call it a right prism because the angle formed by the lateral sides to the base is 90 ∘ . See . Right Hexagonal Prism The first image is a view of the figure straight on with no rotation in any direction. The middle figure is the base or the top. The last figure shows you the solid in three dimensions. To calculate the surface area of the right prism shown in , we first determine the area of the hexagonal base and multiply that by 2, and then add the perimeter of the base times the height. Recall the area of a regular polygon is given as A = 1 2 a p , where a is the apothem and p is the perimeter. We have that A b a s e = 1 2 ( 5.2 ) ( 36 ) = 93.6 cm 2 Then, the surface area of the hexagonal prism is S A = 2 ( 93.6 ) + 36 ( 20 ) = 907.2 in 2 To find the volume of the right hexagonal prism, we multiply the area of the base by the height using the formula V = B h . The base is 93.6 cm 2 , and the height is 20 cm . Thus, V = 93.6 ( 20 ) = 1872 cm 3 . Calculating the Surface Area of a Right Triangular Prism Find the surface area of the triangular prism ( ). The area of the triangular base is A b a s e = 1 2 ( 12 ) ( 6 ) = 36 in 2 . The perimeter of the base is p = 12 + 8.49 + 8.49 = 28.98 in . Then, the surface area of the triangular prism is S A = 2 ( 36 ) + 28.98 ( 10 ) = 361.8 in 2 . Finding the Surface Area and Volume Find the surface area and the volume of the right triangular prism with an equilateral triangle as the base and height ( ). The area of the triangular base is A b a s e = 1 2 ( 6 ) ( 10.39 ) = 31.17 cm 2 . Then, the surface area is S A = 2 ( 31.17 ) + 36 ( 10 ) = 422.34 cm 2 . The volume formula is found by multiplying the area of the base by the height. We have that V = B ⋅ h = ( 31.17 ) ( 10 ) = 311.7 cm 3 . Determining Surface Area Application Katherine and Romano built a greenhouse made of glass with a metal roof ( ). In order to determine the heating and cooling requirements, the surface area must be known. Calculate the total surface area of the greenhouse. The area of the long side measures 95 ft × 6.5 ft = 1,235 ft 2 . Multiplying by 2 gives 2,470 ft 2 . The front (minus the triangular area) measures 22 ft × 6.5 ft = 286 ft 2 . Multiplying by 2 gives 572 ft 2 . The floor measures 95 ft × 22 ft = 2,090 ft 2 . Each triangular region measures A = 1 2 ( 22 ) ( 5 ) = 55 ft 2 . Multiplying by 2 gives 110 ft 2 . Finally, one side of the roof measures 12.1 ft × 95 ft = 1,149.5 ft 2 . Multiplying by 2 gives 2299 ft 2 . Add them up and we have S A = 2,470 + 572 + 2,090 + 110 + 2,299 = 7,541 ft 2 . Right Cylinders There are similarities between a prism and a cylinder. While a prism has parallel congruent polygons as the top and the base, a right cylinder is a three-dimensional object with congruent circles as the top and the base. The lateral sides of a right prism make a 90 ∘ angle with the polygonal base, and the side of a cylinder, which unwraps as a rectangle, makes a 90 ∘ angle with the circular base. Right cylinders are very common in everyday life. Think about soup cans, juice cans, soft drink cans, pipes, air hoses, and the list goes on. In , imagine that the cylinder is cut down the 12-inch side and rolled out. We can see that the cylinder side when flat forms a rectangle. The S A formula includes the area of the circular base, the circular top, and the area of the rectangular side. The length of the rectangular side is the circumference of the circular base. Thus, we have the formula for total surface area of a right cylinder. Right Cylinder The surface area of a right cylinder is given as S A = 2 π r 2 + 2 π r h . To find the volume of the cylinder, we multiply the area of the base with the height. The volume of a right cylinder is given as V = π r 2 h . Finding the Surface Area and Volume of a Cylinder Given the cylinder in , which has a radius of 5 inches and a height of 12 inches, find the surface area and the volume. Step 1: We begin with the areas of the base and the top. The area of the circular base is A b a s e = π ( 5 ) 2 = 25 π = 78.5 in 2 . Step 2: The base and the top are congruent parallel planes. Therefore, the area for the base and the top is A = 2 ( 78.5 ) = 157 in 2 . Step 3: The area of the rectangular side is equal to the circumference of the base times the height: A = 2 π ( 5 ) ( 12 ) = 377 in 2 Step 4: We add the area of the side to the areas of the base and the top to get the total surface area. We have S A = 157 + 377 = 534 in 2 . Step 5: The volume is equal to the area of the base times the height. Then, V = π ( 5 ) 2 ( 12 ) = 942.48 in 3 . Applications of Surface Area and Volume The following are just a small handful of the types of applications in which surface area and volume are critical factors. Give this a little thought and you will realize many more practical uses for these procedures. Applying a Calculation of Volume A can of apple pie filling has a radius of 4 cm and a height of 10 cm. How many cans are needed to fill a pie pan ( ) measuring 22 cm in diameter and 3 cm deep? The volume of the can of apple pie filling is V = π ( 4 ) 2 ( 10 ) = 502.7 cm 3 . The volume of the pan is V = π ( 11 ) 2 ( 3 ) = 1140.4 cm 3 . To find the number of cans of apple pie filling, we divide the volume of the pan by the volume of a can of apple pie filling. Thus, 1140 ÷ 502.7 = 2.3. We will need 2.3 cans of apple pie filling to fill the pan. Optimization Problems that involve optimization are ones that look for the best solution to a situation under some given conditions. Generally, one looks to calculus to solve these problems. However, many geometric applications can be solved with the tools learned in this section. Suppose you want to make some throw pillows for your sofa, but you have a limited amount of fabric. You want to make the largest pillows you can from the fabric you have, so you would need to figure out the dimensions of the pillows that will fit these criteria. Another situation might be that you want to fence off an area in your backyard for a garden. You have a limited amount of fencing available, but you would like the garden to be as large as possible. How would you determine the shape and size of the garden? Perhaps you are looking for maximum volume or minimum surface area. Minimum cost is also a popular application of optimization. Let’s explore a few examples. Maximizing Area Suppose you have 150 meters of fencing that you plan to use for the enclosure of a corral on a ranch. What shape would give the greatest possible area? So, how would we start? Let’s look at this on a smaller scale. Say that you have 30 inches of string and you experiment with different shapes. The rectangle in measures 12 in long by 3 in wide. We have a perimeter of P = 2 ( 12 ) + 2 ( 3 ) = 30 in and the area calculates as A = 3 ( 12 ) = 36 in 2 . The rectangle in , measures 8 in long and 7 in wide. The perimeter is P = 2 ( 8 ) + 2 ( 7 ) = 30 in and the area is A = 8 ( 7 ) = 56 in 2 . In , the square measures 7.5 in on each side. The perimeter is then P = 4 ( 7.5 ) = 30 in and the area is A = 7.5 ( 7.5 ) = 56.25 in 2 . If you want a circular corral as in , we would consider a circumference of 30 = 2 π r , which gives a radius of 30 ÷ 2 π = 4.77 in and an area of A = π ( 4.77 ) 2 = 71.5 in 2 . We see that the circular shape gives the maximum area relative to a circumference of 30 in. So, a circular corral with a circumference of 150 meters and a radius of 23.87 meters gives a maximum area of 1,790.5 m 2 . Designing for Cost Suppose you want to design a crate built out of wood in the shape of a rectangular prism ( ). It must have a volume of 3 cubic meters. The cost of wood is $15 per square meter. What dimensions give the most economical design while providing the necessary volume? To choose the optimal shape for this container, you can start by experimenting with different sizes of boxes that will hold 3 cubic meters. It turns out that, similar to the maximum rectangular area example where a square gives the maximum area, a cube gives the maximum volume and the minimum surface area. As all six sides are the same, we can use a simplified volume formula: V = s 3 , where s is the length of a side. Then, to find the length of a side, we take the cube root of the volume. We have 3 = s 3 3 3 = s = 1.4422 m The surface area is equal to the sum of the areas of the six sides. The area of one side is A = ( 1.4422 ) 2 = 2.08 m 2 . So, the surface area of the crate is S A = 6 ( 2.08 ) = 12.5 m 2 . At $15 a square meter, the cost comes to 12.5 ( $ 15 ) = $ 187.50. Checking the volume, we have V = ( 1.4422 ) 3 = 2.99 m 3 . Check Your Understanding Key Terms surface area volume right prism right cylinder Key Concepts A right prism is a three-dimensional object that has a regular polygonal face and congruent base such that that lateral sides form a 90 ∘ angle with the base and top. The surface area S A of a right prism is found using the formula S A = 2 B + p h , where B is the area of the base, p is the perimeter of the base, and h is the height. The volume V of a right prism is found using the formula V = B h , where B is the area of the base and h is the height. A right cylinder is a three-dimensional object with a circle as the top and a congruent circle is the base, and the side forms a 90 ∘ angle to the base and top. The surface area of a right cylinder is found using the formula S A = 2 π r 2 + 2 π r h , where r is the radius and h is the height. The volume is found using the formula V = π r 2 h , where r is the radius and h is the height. Formulas The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, S A = 2 B + p h , where B is equal to the area of the base and top, p is the perimeter of the base, and h is the height. The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, V = B ⋅ h , where B is the area of the base and h is the height. The surface area of a right cylinder is given as S A = 2 π r 2 + 2 π r h . The volume of a right cylinder is given as V = π r 2 h .", "section": "Volume and Surface Area", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Right Triangle Trigonometry In the lower left corner of the fresco The School of Athens by Raphael, the figure in white writing in the book represents Pythagoras. Alongside him, to the right, the figure with the long, light-brown hair is said to depict Archimedes. (credit: modification of work “School of Athens” by Raphael (1483–1520), Vatican Museums/Wikimedia, Public Domain) Learning Objectives After completing this section, you should be able to: Apply the Pythagorean Theorem to find the missing sides of a right triangle. Apply the 30 ∘ - 60 ∘ - 90 ∘ and 45 ∘ - 45 ∘ - 90 ∘ right triangle relationships to find the missing sides of a triangle. Apply trigonometric ratios to find missing parts of a right triangle. Solve application problems involving trigonometric ratios. This is another excerpt from Raphael’s The School of Athens. The man writing in the book represents Pythagoras, the namesake of one of the most widely used formulas in geometry, engineering, architecture, and many other fields, the Pythagorean Theorem. However, there is evidence that the theorem was known as early as 1900–1100 BC by the Babylonians. The Pythagorean Theorem is a formula used for finding the lengths of the sides of right triangles. Born in Greece, Pythagoras lived from 569–500 BC. He initiated a cult-like group called the Pythagoreans, which was a secret society composed of mathematicians, philosophers, and musicians. Pythagoras believed that everything in the world could be explained through numbers. Besides the Pythagorean Theorem, Pythagoras and his followers are credited with the discovery of irrational numbers, the musical scale, the relationship between music and mathematics, and many other concepts that left an immeasurable influence on future mathematicians and scientists. The focus of this section is on right triangles. We will look at how the Pythagorean Theorem is used to find the unknown sides of a right triangle, and we will also study the special triangles, those with set ratios between the lengths of sides. By ratios we mean the relationship of one side to another side. When you think about ratios, you should think about fractions. A fraction is a ratio, the ratio of the numerator to the denominator. Finally, we will preview trigonometry. We will learn about the basic trigonometric functions, sine, cosine and tangent, and how they are used to find not only unknown sides but unknown angles, as well, with little information. Pythagorean Theorem The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle). The Pythagorean Theorem states a 2 + b 2 = c 2 where a and b are two sides (legs) of a right triangle and c is the hypotenuse, as shown in . Pythagorean Right Triangle For example, given that side a = 6 , and side b = 8 , we can find the measure of side c using the Pythagorean Theorem. Thus, a 2 + b 2 = c 2 ( 6 ) 2 + ( 8 ) 2 = c 2 36 + 64 = c 2 100 = c 2 100 = c 2 10 = c Using the Pythagorean Theorem Find the length of the missing side of the triangle ( ). Using the Pythagorean Theorem , we have ( 6 ) 2 + b 2 = ( 14 ) 2 36 + b 2 = 196 b 2 = 196 − 36 b 2 = 160 b = ± 160 = 4 10 = 12.65 When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root. Distance The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this: In , the problem is to find the distance between the points ( − 3 , − 1 ) and ( 3 , 2 ) . We call the length from point ( − 3 , − 1 ) to point ( 3 , − 1 ) side a , and the length from point ( 3 , − 1 ) to point ( 3 , 2 ) side b . To find side c , we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 , such that ( x 2 − x 1 ) is a substitute for a in the Pythagorean Theorem and is equal to 3 − ( − 3 ) = 6 ; and ( y 2 − y 1 ) is a substitute for b in the Pythagorean Theorem and is equal to 2 − ( − 1 ) = 3. When we plug in these numbers to the distance formula, we have d = ( 3 − ( − 3 ) ) 2 + ( 2 − ( − 1 ) ) 2 = ( 6 ) 2 + ( 3 ) 2 = 36 + 9 = 45 = 3 5 = 6.7 Thus, d = c , the hypotenuse, in the Pythagorean Theorem. Distance Calculating Distance Using the Distance Formula You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive ( ). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft. You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then 9 ( 200 ) + 7 ( 200 ) = 1,800 ft + 1,400 ft = 3,200 ft . As the crow flies, use the distance formula. We have d = ( 1,800 − 0 ) 2 + ( 1,400 − 0 ) 2 = 3,240,000 + 1,960,000 = 5,200,000 = 2280.4 ft Calculating Distance with the Pythagorean Theorem The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches ( ). What should the vertical height be? The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a , is 130 in. The length of c , or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b . Then, by the Pythagorean Theorem , we have: a 2 + b 2 = c 2 ( 130 ) 2 + b 2 = ( 132.4 ) 2 16,900 + b 2 = 17,529.76 b 2 = 17,529.76 − 16,900 b 2 = 629.8 b = 629.8 = 25 If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change? The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be 12 ( 25 ) = 300 in . So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length: ( 300 ) 2 + ( 25 ) 2 = c 2 90,625 = c 2 90,625 = c = 301 in The new ramp will look like . 30 ∘ - 60 ∘ - 90 ∘ Triangles In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like ( 2 x + 5 y ) 2 , we do not have to expand it the long way: ( 2 x + 5 y ) 2 = ( 2 x + 5 y ) ( 2 x + 5 y ) = ( 2 x ) 2 + 10 x y + 10 x y + ( 5 y ) 2 = 4 x 2 + 20 x y + 25 y 2 If we know the perfect square formula, given as ( a + b ) 2 = a 2 + 2 a b + b 2 , we can skip the middle step and just start writing down the answer. This may seem trivial with problems like ( a + b ) 2 . However, what if you have a problem like ( 2 3 + 3 31.8 c 3 ) 2 ? That is a different story. Nevertheless, we use the same perfect square formula. The same idea applies in geometry. There are special formulas and procedures to apply in certain types of problems. What is needed is to remember the formula and remember the kind of problems that fit. Sometimes we believe that because a formula is labeled special, we will rarely have use for it. That assumption is incorrect. So, let us identify the 30 ∘ - 60 ∘ - 90 ∘ triangle and find out why it is special. See . The 3 0 ∘ - 6 0 ∘ - 9 0 ∘ We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the 30 ∘ - 60 ∘ - 90 ∘ triangle given as 1 : 3 : 2 , or x : x 3 : 2 x . Thus, you only need to know the length of one side to find the other two sides in a 30 ∘ - 60 ∘ - 90 ∘ triangle. Finding Missing Lengths in a 3 0 ∘ - 6 0 ∘ - 9 0 ∘ Triangle Find the measures of the missing lengths of the triangle ( ). We can see that this is a 30 ∘ - 60 ∘ - 90 ∘ triangle because we have a right angle and a 30 ∘ angle. The remaining angle, therefore, must equal 60 ∘ . Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side a is the shortest side, as it is opposite the smallest angle 30 ∘ , and we can substitute a = x . The ratios are x : x 3 : 2 x . We have the hypotenuse equaling 10, which corresponds to side c , and side c is equal to 2 x . Now, we must solve for x : 2 x = 10 x = 5 Side b is equal to x 3 or 5 3 . The lengths are 5 , 5 3 , 10. Applying 3 0 ∘ - 6 0 ∘ - 9 0 ∘ Triangle to the Real World A city worker leans a 40-foot ladder up against a building at a 30 ∘ angle to the ground ( ). How far up the building does the ladder reach? We have a 30 ∘ - 60 ∘ - 90 ∘ triangle, and the hypotenuse is 40 ft. This length is equal to 2 x , where x is the shortest side. If 2 x = 40 , then x = 20 . The ladder is leaning on the wall 20 ft up from the ground. 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangles The 45 ∘ - 45 ∘ - 90 ∘ triangle is another special triangle such that with the measure of one side we can find the measures of all the sides. The two angles adjacent to the 90 ∘ angle are equal, and each measures 45 ∘ . If two angles are equal, so are their opposite sides. The ratio among sides is 1 : 1 : 2 , or x : x : x 2 , as shown in . 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangles Finding Missing Lengths of a 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangle Find the measures of the unknown sides in the triangle ( ). Because we have a 45 ∘ - 45 ∘ - 90 ∘ triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and 2 . One leg measures 3, so the other leg, a , measures 3. Remember the ratio of x : x : x 2 . Then, the hypotenuse, c , equals 3 2 . Trigonometry Functions Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications. NOTE: You will need either a scientific calculator or a graphing calculator for this section. It must have the capability to calculate trigonometric functions and express angles in degrees. Trigonometry is based on three functions. We title these functions using the following abbreviations: sin = sine cos = cosine tan = tangent Letting r = x 2 + y 2 , which is the hypotenuse of a right triangle, we have . The functions are given in terms of x , y , and r , and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse. sin θ = y r = o p p h y p cos θ = x r = a d j h y p tan θ = y x = o p p a d j Trigonometric Ratios We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In , we have an illustration mainly to identify r and the sides labeled x and y . Angle θ An angle θ sweeps out in a counterclockwise direction from the positive x -axis and stops when the angle reaches the desired measurement. That ray extending from the origin that marks θ ∘ is called the terminal side because that is where the angle terminates. Regardless of the information given in the triangle, we can find all missing sides and angles using the trigonometric functions. For example, in , we will solve for the missing sides. Solving for Missing Sides Let’s use the trigonometric functions to find the sides x and y . As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for y . We have sin θ = y r , and θ = 60 ∘ . Then, sin 60 ∘ = y 2 2 sin 60 ∘ = y 1.732 = y 3 = y Next, let’s find x . This is the cosine function. We have cos θ = x r . Then, cos 60 ∘ = x 2 2 cos 60 ∘ = x = 1 Now we have all sides, 1 , 3 , 2. You can also check the sides using the 30 ∘ - 60 ∘ - 90 ∘ ratio of 1 : 3 : 2. is a list of common angles, which you should find helpful. sin 0 ∘ = 0 cos 0 ∘ = 1 sin 30 ∘ = 1 2 cos 30 ∘ = 3 2 sin 45 ∘ = 2 2 cos 45 ∘ = 2 2 sin 60 ∘ = 3 2 cos 60 ∘ = 1 2 sin 90 ∘ = 1 cos 90 ∘ = 0 Common Angles Using Trigonometric Functions Find the lengths of the missing sides for the triangle ( ). We have a 55 ∘ angle, and the length of the triangle on the x -axis is 6 units. Step 1: To find the length of r , we can use the cosine function, as cos θ = x r . We manipulate this equation a bit to solve for r : cos ( 55 ∘ ) = 6 r r cos ( 55 ∘ ) = 6 r = 6 cos ( 55 ∘ ) r = 6 0.5736 = 10.46 Step 2: We can use the Pythagorean Theorem to find the length of y . Prove that your answers are correct by using other trigonometric ratios: 6 2 + y 2 = 10.46 2 y 2 = 109.4 − 36 y = 8.57 Step 3: Now that we have y , we can use the sine function to prove that r is correct. We have sin θ = y r . sin ( 55 ∘ ) = 8.57 r r sin ( 55 ∘ ) = 8.57 r = 8.57 sin ( 55 ∘ ) = 8.57 0.819 = 10.46 To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as sin − 1 , cos − 1 , or tan − 1 . The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had x = 6 and r = 10.46. To find what angle has these values, enter the values for the inverse cosine function cos − 1 ( x r ) in your calculator: cos − 1 ( 6 10.46 ) = 55 ∘ . You can also use the inverse sine function and enter the values of sin − 1 ( y r ) in your calculator given y = 8.57 and r = 10.46. We have sin − 1 ( 8.57 10.46 ) = 55 ∘ . Finally, we can also use the inverse tangent function. Recall tan θ = y x . We have tan − 1 ( 8.57 6 ) = 55 ∘ . Solving for Lengths in a Right Triangle Solve for the lengths of a right triangle in which θ = 30 ∘ and r = 6 ( ). Step 1: To find side a , we use the sine function: sin 30 ∘ = a 6 6 sin 30 ∘ = a = 3 Step 2: To find b , we use the cosine function: cos 30 ∘ = b 6 6 cos 30 ∘ = b = 5.196 Step 3: Since this is a 30 ∘ - 60 ∘ - 90 ∘ triangle and side b should equal x 3 , if we input 3 for x , we have b = 3 3 . Put this in your calculator and you will get 3 3 = 5.196. Finding Altitude A small plane takes off from an airport at an angle of 31.3 ∘ to the ground. About two-thirds of a mile (3,520 ft) from the airport is an 1,100-ft peak in the flight path of the plane ( ). If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak. To solve this problem, we use the tangent function: tan 31.3 ∘ = x 3,520 3,520 tan 31.3 ∘ = 2,140 The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak. Finding Unknown Sides and Angles Suppose you have two known sides, but do not know the measure of any angles except for the right angle ( ). Find the measure of the unknown angles and the third side. Step 1: We can find the third side using the Pythagorean Theorem: 6 2 + 4 2 = c 2 52 = c 2 2 13 = c Now, we have all three sides. Step 2: To find θ , we will first find sin θ . sin θ = o p p h y p = 4 2 13 = 2 13 . The angle θ is the angle whose sine is 2 13 . Step 3: To find θ , we use the inverse sine function: θ = sin − 1 ( 2 13 ) = 33.7 ∘ Step 4: To find the last angle, we just subtract: 180 ∘ − 90 ∘ − 33.7 ∘ = 56.3 ∘ . Angle of Elevation and Angle of Depression Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression . These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level. Finding the Angle of Elevation A guy wire of length 110 meters runs from the top of an antenna to the ground ( ). If the angle of elevation of an observer to the top of the antenna is 43 ∘ , how high is the antenna? We are looking for the height of the tower. This corresponds to the y -value, so we will use the sine function: sin 43 ∘ = y 110 110 sin 43 ∘ = y 75 = y The tower is 75 m high. Finding Angle of Elevation You are sitting on the grass flying a kite on a 50-foot string ( ). The angle of elevation is 60 ∘ . How high above the ground is the kite? We can solve this using the sine function, sin θ = o p p h y p . sin 60 ∘ = x 50 50 sin 60 ∘ = x = 43.3 ft Pythagoras and the Pythagoreans The Pythagorean Theorem is so widely used that most people assume that Pythagoras (570–490 BC) discovered it. The philosopher and mathematician uncovered evidence of the right triangle concepts in the teachings of the Babylonians dating around 1900 BC. However, it was Pythagoras who found countless applications of the theorem leading to advances in geometry, architecture, astronomy, and engineering. Among his accolades, Pythagoras founded a school for the study of mathematics and music. Students were called the Pythagoreans, and the school’s teachings could be classified as a religious indoctrination just as much as an academic experience. Pythagoras believed that spirituality and science coexist, that the intellectual mind is superior to the senses, and that intuition should be honored over observation. Pythagoras was convinced that the universe could be defined by numbers, and that the natural world was based on mathematics. His primary belief was All is Number. He even attributed certain qualities to certain numbers, such as the number 8 represented justice and the number 7 represented wisdom. There was a quasi-mythology that surrounded Pythagoras. His followers thought that he was more of a spiritual being, a sort of mystic that was all-knowing and could travel through time and space. Some believed that Pythagoras had mystical powers, although these beliefs were never substantiated. Pythagoras and his followers contributed more ideas to the field of mathematics, music, and astronomy besides the Pythagorean Theorem. The Pythagoreans are credited with the discovery of irrational numbers and of proving that the morning star was the planet Venus and not a star at all. They are also credited with the discovery of the musical scale and that different strings made different sounds based on their length. Some other concepts attributed to the Pythagoreans include the properties relating to triangles other than the right triangle, one of which is that the sum of the interior angles of a triangle equals 180 ∘ . These geometric principles, proposed by the Pythagoreans, were proven 200 years later by Euclid. A Visualization of the Pythagorean Theorem In , which is one of the more popular visualizations of the Pythagorean Theorem, we see that square a is attached to side a ; square b is attached to side b ; and the largest square, square c , is attached to side c . Side a measures 3 cm in length, side b measures 4 cm in length, and side c measures 5 cm in length. By definition, the area of square a measures 9 square units, the area of square b measures 16 square units, and the area of square c measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have a 2 + b 2 = c 2 3 2 + 4 2 = 5 2 9 + 16 = 25 Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem. Check Your Understanding Key Terms right triangle sine cosine tangent Key Concepts The Pythagorean Theorem is applied to right triangles and is used to find the measure of the legs and the hypotenuse according the formula a 2 + b 2 = c 2 , where c is the hypotenuse. To find the measure of the sides of a special angle, such as a 30 ∘ - 60 ∘ - 90 ∘ triangle, use the ratio x : x 3 : 2 x , where each of the three sides is associated with the opposite angle and 2 x is associated with the hypotenuse, opposite the 90 ∘ angle. To find the measure of the sides of the second special triangle, the 45 ∘ - 45 ∘ - 90 ∘ triangle, use the ratio x : x : x 2 , where each of the three sides is associated with the opposite angle and x 2 is associated with the hypotenuse, opposite the 90 ∘ angle. The primary trigonometric functions are sin θ = o p p h y p , cos θ = a d j h y p , and tan θ = o p p a d j . Trigonometric functions can be used to find either the length of a side or the measure of an angle in a right triangle, and in applications such as the angle of elevation or the angle of depression formed using right triangles. Formula The Pythagorean Theorem states a 2 + b 2 = c 2 where a and b are two sides (legs) of a right triangle and c is the hypotenuse. Projects One of the reasons so many formulas in geometry were discovered was because of the importance in finding measurements of lengths, areas, perimeter, and angles. Find at least five examples of how geometry can be used in practical applications today. Who were the Pythagoreans? Why did this society exist? Explore what they did and discuss some of their beliefs. Chapter Review Points, Lines, and Planes Angles Triangles Polygons, Perimeter, and Circumference Tessellations Area Volume and Surface Area Right Triangle Trigonometry Chapter Test", "section": "Right Triangle Trigonometry", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Voters cast their ballots in one of the world’s many democracies. (credit: “Governor Votes Early” by Maryland GovPics/Flickr, CC BY 2.0) Suppose a friend asked you, “When did you last vote?” What would your answer be? Maybe you would tell your friend that the last time you voted was during the last presidential election, or perhaps you would tell your friend that you prefer not to vote. When thinking about voting, presidential campaigns or advertisements for reelections may come to mind, but you can cast your vote in many ways. Have you liked a post, followed a creator, friended a stranger, or clicked a heart online today? In the digital age, it's possible to vote several times a day. Voting systems are not only the machines that drive every democracy on Earth, but they are also the engines driving social media and many other aspects of life. A deeper understanding of these voting systems will enhance your ability to successfully engage with the world in which we live. In this chapter, you will become one of the founders of the new democratic country of Imaginaria. You have a great responsibility to the people of this fledgling democracy. You have been tasked with writing the portion of the constitution that lays out voting procedures. In preparation for this important task, you will explore the various types of voting systems, from school board elections to Twitter wars. You will see how these types are alike, how they differ, and how they might be applied in Imaginaria. Most importantly, you will learn about the mathematically inherent advantages and disadvantages of various voting systems so that you can make informed choices to better the lives of the Imaginarians.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Voting Methods President Barack Obama votes in the 2012 election. (credit: Pete Souza/White House, Public Domain) Learning Objectives After completing this section, you should be able to: Apply plurality voting to determine a winner. Apply runoff voting to determine a winner. Apply ranked-choice voting to determine a winner. Apply Borda count voting to determine a winner. Apply pairwise comparison and Condorcet voting to determine a winner. Apply approval voting to determine a winner. Compare and contrast voting methods to identify flaws. Today is the day that you begin your quest to collaborate on the constitution of Imaginaria! Let’s begin by thinking about the selection of a leader who can serve as president. It seems straightforward; if the majority of citizens prefer a particular candidate, that candidate should win. But not all votes are decided by a simple majority. Why not? What are the options? Majority versus Plurality Voting When an election involves only two options, a simple majority is a reasonable way to determine a winner. A majority is a number equaling more than half, or greater than 50 percent of the total. Let’s take a look at the outcomes of U.S. presidential elections to understand more. displays the results of the 2000 U.S. presidential election. Like most presidential elections, this election involved more than two options. If that is the case, is it possible that no single candidate will receive more than half of the votes cast? Candidate (Party Label) Popular Vote Total Al Gore (Democrat) 50,999,897 George W. Bush (Republican) 50,456,002 Ralph Nader (Green) 2,882,955 Patrick J. Buchanan (Reform/Independent) 448,895 Harry Browne (Libertarian) 384,431 Howard Phillips (Constitution) 98,020 Other 134,900 Total: 105,405,100 Results of the Popular Vote for the 2000 U.S. Presidential Election (source: https://www.fec.gov/introduction-campaign-finance/election-and-voting-information/federal-elections-2000/president2000/) Majority of Popular Vote in the 2000 U.S. Presidential Election Refer back to . Did any single candidate secure the majority of popular votes? Step 1: Calculate 50 percent of 105,405,100 by multiplying the decimal form of 50 percent, which is 0.50, by 158,394,605: 0.50 ( 105,405,100 ) = 52,702,550 Step 2: Determine the minimum number of votes needed to have a majority. The minimum number of votes required is the lowest counting number that is larger than 50 percent of the votes. To have a majority, an individual candidate must have more than 52,702,550; so, a majority candidate must have 52,702,551 votes or more. Step 3: Compare the number of votes each candidate received to 52,702,551. According to the data in , none of the candidates secured a majority. Unlike in the 2000 U.S. presidential election, a candidate won the majority of votes in the 2020 election (see ). It is a common occurrence for no single candidate to receive a majority of the votes in an election with more than two candidates. When this occurs, the candidate with the largest portion of the votes is said to have a plurality . Plurality of Popular Vote in the 2000 U.S. Presidential Election Refer again to . In the 2000 U.S. presidential election, which candidate had a plurality of popular votes? Al Gore secured 50,999,897 votes which was more than any other single candidate. Therefore, he had a plurality of the popular votes. Your plans for Imaginarian elections will likely include primary elections, or preliminary elections to select candidates for a principal or general election. displays the results of the 2018 U.S. Senate Republican primary for Maryland. Top Four Republican Candidates Votes Percentage of Party Votes Cambell, Tony 51,426 29.22% Chaffee, Chris 42,328 24.05% Grigorian, Christina J. 30,756 17.48% Graziani, John R. 15,435 8.77% Total Votes 175,981 100% (source: https://ballotpedia.org/United_States_Senate_election_in_Maryland_(June_26,_2018_Republican_primary)) Consider how election by plurality, not majority, is the most common method of selecting candidates for public office. The U.S. Electorial College: Winning the Presidency without a Plurality In the 2000 U.S. presidential election, Al Gore had a plurality of the popular votes, but he did not win the election. Why? This occurred because the U.S. president and vice president are elected by electors rather than a direct vote by the citizens. The electors are part of the Electoral College, a body of people representing the states. Why was the Electoral College created? The Electoral College was created as a compromise between those authors of the U.S. Constitution who believed Congress should elect the president, and those who believed the citizens should vote directly. The popular vote was not recorded until the presidential election of 1824. Since then, only five presidents have been elected without winning a plurality of the popular vote: John Quincy Adams in 1824, Rutherford B. Hayes in 1876, Benjamin Harrison in 1888, George Bush in 2000, and Donald Trump in 2016. Runoff Voting Has your family ever debated what to have for dinner? Suppose your family is deciding on a restaurant and exactly half of you want to have pizza but the other half want hamburgers. How do you decide when the result is a tie? You need a tiebreaker! Will the new democracy of Imaginaria need tiebreakers? When no candidate satisfies the requirements to win the election, a runoff election , or second election, is held to determine a winner. How would runoff voting work in Imaginaria? There are many types of runoff voting systems , which are voting systems that utilize a runoff election when the first round does not result in a winner. The method for implementing a runoff election can vary widely, particularly in the criteria used to determine whether a candidate will be on the ballot in the second election. For example, a two-round system is a runoff voting system in which only the top candidates advance to the runoff election. In some two-round systems, only the top two candidates are on the second ballot, or it may be any candidate who secures a certain percentage of the vote will advance. The Hare Method is another runoff voting system in which only the candidate(s) with the very least votes are eliminated. This can potentially result in several rounds of runoff elections. Runoff Election for Condominium Association President A condominium association elects a new president every two years by a two-round system of voting. If none of the candidates receive a majority, the association charter states that the top two candidates will be eligible to participate in a runoff election. In a particular year, five residents were nominated. The results of the first round are given in the table below. Candidate Votes in First Round Abou 18 Baiocchi 10 Campana 5 Dali 11 Eugene 4 Is there a winner based on the first round? Why or why not? If there is a winner, who won? If there should be a runoff, who will advance to the second round? A majority of 48 total votes is required to win. Begin by finding 50 percent of 48, which is calculated as follows: 0.50 ( 48 ) = 24 . A majority is 25 or more. No candidate has a majority, so there is no winner based on the first round. Abou and Dali advance to the second round. Steps to Determine Winner by Plurality or Majority Election with Runoff To determine the winner by plurality or when a majority election with runoff occurs, we take these three steps: Step 1: If a majority is required to win the election, determine the number of votes needed to achieve a majority. This is the least whole number greater than 50 percent of the total votes. If a majority is not required, move to Step 2. Step 2: Count the number of votes for each candidate in the current round of voting. If a single candidate has enough votes to win a plurality, or a majority as appropriate, then you are done! Otherwise, eliminate a predetermined number of candidates based on the rules of the election. Elimination conditions may vary. For example, the rules may state that the candidate(s) with the fewest votes will be eliminated (as in the Hare method), or that only the candidates meeting a certain threshold will move on (as in a two-round system). Once the appropriate candidates are eliminated, move on to Step 3. Step 3: Hold a runoff election. If the runoff is simulated using a list of voter’s preferences, renumber the preferences to reflect the remaining number of options in such a way that the original order of preference is retained. Then repeat Step 2. Note: The second and third steps may be repeated as many times as necessary for voting procedures that allow multiple runoffs. Family Dinner Night The five members of the Chionilis family—Annette, Rene, Seema, Titus, and Galen—have decided to get takeout for dinner. They are trying to decide on a restaurant. The options are Rainbow China, Dough Boys Pizza, Taco City, or Caribbean Flavor. They will use majority election with runoffs where the restaurant with the fewest votes is eliminated in each round. The preferences of each family member are listed by first initial in the table below. An entry of 1 represents the person’s first choice; 2, their second; and so on. For example, Annette’s second choice is Dough Boys Pizza. Options A R S T G Rainbow China 1 3 3 1 3 Dough Boys Pizza 2 2 1 2 1 Taco City 3 4 2 4 2 Caribbean Flavor 4 1 4 3 4 Use the information in the table to answer the following questions. Which common type of runoff voting method is this? List the results of each round of voting based on this information and determine which restaurant was ultimately chosen. the Hare Method Step 1: Determine the number of votes necessary to have a majority. There are five family members, so 50 percent of 5 is 0.50 ( 5 ) = 2.5 . A majority is three or more votes. Step 2: Count the number of votes for each restaurant in the first election. In a list of voter preferences, the 1s represent the top choice of each voter, which corresponds to their vote in the first round. Results of Round 1: Rainbow China — 2 votes Dough Boys — 2 votes Taco City — 0 votes Caribbean Flavor — 1 vote No restaurant received a majority. Eliminate Taco City, which has the fewest first place votes: Options A R S T G Rainbow China 1 3 3 1 3 Dough Boys Pizza 2 2 1 2 1 Caribbean Flavor 4 1 4 3 4 Step 3: Hold a runoff election. In other words, hold a second round. Since we have a list of the voters’ preferences with the eliminated option removed, we will renumber the preferences as first, second, and third so that we keep the original order of preference. The result is that we will count the second-place vote of any voter whose first choice was eliminated. Options A R S T G Rainbow China 1 3 2 1 2 Dough Boys Pizza 2 2 1 2 1 Caribbean Flavor 3 1 3 3 3 Step 4: Repeat the process from Step 2. Count the number votes for each restaurant in the first-round election. Since we are using a list of preferences, we need to count the number of 1s received by each restaurant. Results of Round 2: Rainbow China — 2 votes Dough Boys — 2 votes Caribbean Flavor — 1 vote No single restaurant has three votes. Eliminate Caribbean Flavor, which has the fewest first place votes. OPTIONS A R S T G Rainbow China 1 3 2 1 2 Dough Boys Pizza 2 2 1 2 1 Step 5: Repeat the process from Step 3. Hold another runoff election. This will be Round 3. Renumber the voters' preferences as first and second this time. Options A R S T G Rainbow China 1 2 2 1 2 Dough Boys Pizza 2 1 1 2 1 Step 6: Repeat the process from Step 2 one last time. Count the number of first place votes for each remaining restaurant. Results of Round 3: Rainbow China — 2 votes Dough Boys — 3 votes Determine whether any one choice has a majority. Yes! Dough Boys has three votes, so it is the winner! Ranked-Choice Voting In and Your Turn 11.4 , you were given a list that ordered each voter’s preferences. This ordering is called a preference ranking . A ballot in which a voter is required to give an ordering of their preferences is a ranked ballot , and any voting system in which a voter uses a ranked ballot is referred to as ranked voting . The vote for the Academy Awards uses a ranked ballot. The table below provides an example of a ranked ballot for the 2020 Academy Award nominees for Best Director. Candidate for Best Director Rank top choice as 1, next choice as 2, and so on. Martin Scorsese, The Irishman 1 2 3 4 5 Todd Phillips, Joker 1 2 3 4 5 Sam Mendes, 1917 1 2 3 4 5 Quentin Tarantino, Once Upon a Time in Hollywood 1 2 3 4 5 Bong Joon-ho, Parasite 1 2 3 4 5 As you decide on the voting methods that will be used in your new democracy, budget must be a consideration. You might consider a particular type of ranked voting called ranked-choice voting (RCV) , which simulates a series of runoff elections without the usual time and expense involved when voters must repeatedly return to the polls, like we did in . The method of ranked-choice voting (RCV), also called instant runoff voting (IRV) , is a version of the Hare Method, using preference ranking so that, if no single candidate receives a majority, the least popular selections can be eliminated and the results can be recounted, without the need for more elections. Ranked voting can be confused with ranked-choice voting, but ranked voting is a more general category which includes ranked-choice voting and several other voting methods. As we explore examples of ranked voting, we will summarize the voters’ preference rankings using a table in which the top row shows the number of ballots that ranked the options in the same order. Let’s practice interpreting the information in this type of table. Interpreting the Sample Preference Summary Refer to the table below containing voters’ preference rankings to answer the following questions. Number of Ballots 100 200 150 75 Option A 1 4 3 4 Option B 2 3 4 2 Option C 4 2 1 1 Option D 3 1 2 3 How many voters ranked the options in the following order: Option A in fourth place, Option B in second place, Option C in first place, and Option D in third place? How many ballots in total were collected? How many voters indicated that Option C was their first choice? The column farthest to the right displays this ordering. The top entry in this column is 75; so, there were 75 voters who ranked the options in this way. The sum of the top row gives the total number of ballots collected: 100 + 200 + 150 + 75 = 525 . So, there were 525 ballots collected. In the Option C row, there are two entries of 1 which indicate a first choice for that option. These occur in the last two columns. The sum of the top entries in these columns is 150 + 75 = 225 . So, 225 voters indicated Option C as their first choice. Now that we’ve covered how to read a summary of preference rankings, let’s practice using the ranked-choice method to determine the winner of an election. Recall that ranked-choice voting is still the Hare Method where the candidate with the very least number of votes is eliminated each round until a majority is attained. The difference here is that the voters have completed a ranked ballot, so they don't have to visit the polls multiple times. Here are the steps for ranked-choice voting. Steps to Determine Winner by Ranked-Choice Voting To determine the winner when ranked-choice voting occurs, we take these three steps: Step 1: Determine the number of votes needed to achieve a majority. This is the least whole number greater than 50 percent of the total votes. Step 2: Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election and we are done! Otherwise, eliminate the candidate(s) with the fewest votes and complete Step 3. Step 3: Reallocate the votes to the remaining candidates, and repeat Step 2. Be methodical to avoid arithmetic errors. Make sure that each time you count the number of first place votes they sum to the number of ballots. How Does Ranked-Choice Voting Work? Most Popular Color in Kindergarten Let’s review the kindergarten class color preferences again, and this time determine which color would be selected based on these results using the ranked-choice method. Number of Ballots 4 6 4 7 Red 2 6 2 5 Blue 1 2 1 4 Green 6 5 5 2 Yellow 5 4 6 3 Purple 4 3 4 1 Pink 3 1 3 6 Step 1: Determine whether any candidate received a majority. There were 21 ballots. Fifty percent of 21 is 0.50 ( 21 ) = 10.5 . A majority is 11. Step 2: Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes. Red: 0 Blue: 4 + 4 = 8 Green: 0 Yellow: 0 Purple: 7 Pink: 6 Notice that 8 + 7 + 6 = 21 , which is the total number of ballots. Confirming this helps to catch any arithmetic or counting errors. No candidate has a majority with 11 or more votes. We must eliminate red, green, and yellow which had the fewest votes with 0 each. The remaining votes that must be counted for Round 2 are given in the table below. Number of Ballots 4 6 4 7 Blue 1 2 1 4 Purple 4 3 4 1 Pink 3 1 3 6 Step 3: Reallocate the votes to the remaining candidates. We can do this by numbering the choices as 1, 2, and 3 in such a way that the order of preference is retained as seen in the table below: Number of Ballots 4 6 4 7 Blue 1 2 1 2 Purple 3 3 3 1 Pink 2 1 2 3 Step 4: Repeat the process from Step 2. Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes. Blue: 4 + 4 = 8 Purple: 7 Pink: 6 Confirm that 8 + 7 + 6 = 21 . Great! No candidate has 11 or more votes. We must eliminate pink which had the fewest votes with 6. The remaining votes that must be counted for Round 2 are shown in the table below. Number of Ballots 4 6 4 7 Blue 1 2 1 2 Purple 3 3 3 1 Step 5: Repeat the process from Step 3. Reallocate the votes to the remaining candidates. We can do this by numbering the choices as 1 and 2; Number of Ballots 4 6 4 7 Blue 1 1 1 2 Purple 2 2 2 1 Step 6: Repeat the process from Step 2 one last time. Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes. Blue: 4 + 6 + 4 = 14 Purple: 7 Blue has a majority and wins the election! Determine Winner of Election by Ranked-Choice Method (aka Instant Runoff) Borda Count Voting Ranked-choice voting is one type of ranked voting that simulates multiple runoffs based on ranked ballots. Another type of ranked voting is the Borda count method , which uses ranked ballots that award candidates points corresponding to the number of candidates ranked lower on each ballot. To understand how this works, let’s review the favorite colors of our kindergarten class from the table below. Let’s focus on the votes represented by the first column of the preference summary. Number of Ballots 4 6 4 7 Red 2 6 2 5 Blue 1 2 1 4 Green 6 5 5 2 Yellow 5 4 6 3 Purple 4 3 4 1 Pink 3 1 3 6 Each student had six options. This first column tells us that four students ranked blue as their first choice, red as their second choice, pink as their third choice, purple as their fourth choice, yellow as their fifth choice, and green as their sixth choice. Blue was ranked higher than 6 - 1 = 5 other colors. For each of the four students who completed their ballot in this way, blue would receive five points. Since there were four ballots with this ordering, blue would receive 5 ( 4 ) = 20 points from the first column. To determine the total points for each candidate, we have to find the sum of the points they received in each column. To determine the winner of a contest using the Borda count method, we must compare total number of points earned by each candidate. The candidate with the most points is the winner. Each row of the preference summary corresponds to a single candidate. To find the number of points received by a particular candidate in the preference summary, or their Borda score , we will need to focus on the row in which that candidate appears. Before we practice determining the winner of a Borda count election, let’s examine how to find the Borda score for a single candidate. Most Popular Color in Kindergarten Revisited Let’s review the ballots from the kindergarten class again, as shown in the table below. This time, let’s determine the Borda score received by the color purple. Number of Ballots 4 6 4 7 Red 2 6 2 5 Blue 1 2 1 4 Green 6 5 5 2 Yellow 5 4 6 3 Purple 4 3 4 1 Pink 3 1 3 6 Step 1: Find the number of points received by the candidate in each column. Column 1: 4 × ( 6 - 4 ) = 4 × 2 = 8 Column 2: 6 × ( 6 - 3 ) = 6 × 3 = 18 Column 3: 4 × ( 6 - 4 ) = 4 × 2 = 8 Column 4: 7 × ( 6 - 1 ) = 7 × 5 = 35 Step 2: Find the sum of the points received in each column. This is the total number of points received by this candidate: 8 + 18 + 8 + 35 = 69 This process can also be combined into one step as shown here. 4 × ( 6 - 4 ) + 6 × ( 6 - 3 ) + 4 × ( 6 - 4 ) + 7 × ( 6 - 1 ) = 4 × 2 + 6 × 3 + 4 × 2 + 7 × 5 = 8 + 18 + 8 + 35 = 69 Purple received 69 points in this election. When calculating a Borda score in one step, be careful to use the correct order of operations. Perform the subtraction inside each pair of parentheses first, then perform each multiplication, and then perform each addition. Now let’s determine the winner of an election by comparing the Borda scores for each of the candidates. Determine the Winner by Two Ranked Voting Methods Use the table below, which displays a sample preference summary, to answer the questions that follow. Number of Ballots 95 90 110 115 Option A 4 4 1 1 Option B 2 2 2 2 Option C 3 1 3 4 Option D 1 3 4 3 Use the ranked-choice voting method to determine the winner of the election. Use the Borda count method to determine the winner of the election. Step 1: Determine the number of votes needed to achieve a majority. The number of ballots is 95 + 90 + 110 + 115 = 410 . Fifty percent of 410 is 0.50 ( 410 ) = 205 . So, 206 votes or more is a majority. Step 2: Count the number of first place votes for each candidate. Option A has 110 + 115 = 225 Option B has 0 Option C has 90 Option D has 95 Since Option A has a majority, Option A is the winner by the ranked-choice method. The Borda scores would be: Option A: 95 ( 4 - 4 ) + 90 ( 4 - 4 ) + 110 ( 4 - 1 ) + 115 ( 4 - 1 ) = 675 Option B: 95 ( 4 - 2 ) + 90 ( 4 - 2 ) + 110 ( 4 - 2 ) + 115 ( 4 - 2 ) = 820 Option C: 95 ( 4 - 3 ) + 90 ( 4 - 1 ) + 110 ( 4 - 3 ) + 115 ( 4 - 4 ) = 475 Option D: 95 ( 4 - 1 ) + 90 ( 4 - 3 ) + 110 ( 4 - 4 ) + 115 ( 4 - 3 ) = 490 Since Option B has a Borda score of 820 points, Option B is the winner by the Borda count method. The Borda count method may seem too complicated to even consider using for Imaginaria, but each voting method has its own pros and cons. The Borda count method, for example, favors compromise candidates over divisive candidates. A compromise candidate is not the first choice of most of the voters, but is more acceptable to the population as a whole than the other candidates. A divisive candidate is simultaneously the first choice of a large portion of the voters and the last choice of another large portion of the voters. In , Candidate A was ranked first by 225 voters, but was ranked last by 185 voters. No voters ranked Candidate A as second or third. It appears that, although Candidate A had the majority of first place votes, there was a significant minority who strongly disliked them. Candidate A was a divisive candidate. Candidate B, on the other hand, was the second choice of every voter, making Candidate B a good compromise. The Borda count method chose Candidate B, a compromise candidate, that was more acceptable to the population as a whole. This scenario is cited by both opponents and proponents of the Borda count method. Determine Winner of Election by Borda Count Method Pairwise Comparison and Condorcet Voting We have discussed two kinds of ranked voting methods so far: ranked-choice and Borda count. A third type of ranked voting is the pairwise comparison method , in which the candidates receive a point for each candidate they would beat in a one-on-one election and half a point for each candidate they would tie. If one candidate earns more points than the others, then that candidate wins. This method is one of several Condorcet voting methods , which are methods in which candidates are ranked and then compared pairwise to each other, a candidate having to beat all others in order to win. These methods vary in the way candidates are scored, and there is not always a clear winner. A candidate who wins each possible pairing is known as a Condorcet candidate . These terms are named after the Marquis de Condorcet, a French philosopher and mathematician who preferred the pairwise comparison method to the Hare method and made public arguments in its favor. Marquis de Condorcet Condorcet voting methods are named for the Marquis de Condorcet, a French philosopher and mathematician known for, among other accomplishments, writing “ Sur l'admission des femmes au droit de Cité ” (“On the Admission of Women to the Rights of Citizenship”), in 1789, the first published essay on the political rights of women. For more details visit this Web site. If you include a Condorcet voting method in the constitution of Imaginaria, the election supervisors may want to use a pairwise comparison matrix like the one in . It’s a tool used to list the number of wins associated with each pairing of two candidates. Each candidate will receive a point for each win and a half a point for each tie. Each pairing is listed twice, once for the number of wins of a candidate over a particular challenger and once for the number of wins of the challenger over that candidate. Pairwise Comparison Matrix for Three Candidates Steps to Determine a Winner by Pairwise Comparison Method Using a Matrix To determine the winner when the pairwise comparison method is used, we take these three steps: Step 1: On the matrix, indicate a losing matchup by crossing out a box, × , and tie match ups by drawing a slash through the box, \\ . Step 2: Award each candidate 1 point for a win, half a point for a tie, and 0 points for a loss. Step 3: Identify the winner, which is the candidate with the most points. Determine Winner of Election by Using the Pairwise Comparison Method Before you decide on the pairwise comparison method for Imaginaria, review what’s involved in constructing a pairwise comparison matrix from a summary of ranked ballots. Then we can use the matrix to determine the winner of the election. Does the winner using the Borda method still win? Construct and Use a Pairwise Comparison Matrix Consider the summary of ranked ballots shown in the table below. Determine the winner of an election using the pairwise comparison method. Number of Ballots 95 90 110 115 Option A 4 4 1 1 Option B 2 2 2 2 Option C 3 1 3 4 Option D 1 3 4 3 Construct a pairwise comparison matrix for the sample summary of ranked ballots in the table above. Use the pairwise comparison method to determine a winner. Recall that in , Candidate A won by the ranked-ballot method, and Candidate B won by the Hare method. Did the same candidate win using the pairwise comparison method? Is the winner a Condorcet candidate? There are four candidates on the ballots. We will need a row and a column for each candidate in addition to the headings, so we will draw a five by five matrix. Pairwise Comparison Matrix for Four Candidates Step 1: Refer to to determine the values that belong in each cell. A over B: A is preferred to B in columns 3 and 4. So, A scores 110 + 115 = 225 points. A over C: A is preferred to C in columns 3 and 4. So, A scores 225 points again. A over D: Similarly, A scores 225 points. B over A: B is preferred to A in columns 1 and 2. So, B scores 95 + 90 = 185 points. Step 2: Continuing in this way, we complete the pairwise comparison matrix, as shown in . Pairwise Comparison Matrix for Four Candidates with Vote Counts Step 1: Losing pairings are crossed off with an × . In the event of a tie, we will draw a slash, \\ . Step 2: Determine the number of points for each candidate by analyzing their row of wins. Each win is 1 point, each loss, × , is 0 points, and each tie, \\ , is half a point. Construct an additional column for each candidate’s points. Pairwise Comparison Matrix for Four Candidates with Pairwise Winners and Points Column Added Step 3: The winner by the pairwise comparison method is Option A with 3 points. Option A was not the winner by the Hare method. The winner, Option A, is a Condorcet candidate because Option A won each pairwise comparison. Notice that the pairwise vote totals are not used to determine the points. Vote totals are only used to determine a win or a loss. Avoid the common error of adding the values in each row to get the points. Three Key Questions Before you decide if you want to use the pairwise comparison method for Imaginarian elections, let’s consider three questions that might affect your decision. Is there always a winner? If there is a winner, is the winner always a Condorcet candidate? If there is a Condorcet candidate, does that candidate always win? Let’s think about why these questions might be important to you if you chose the pairwise comparison method. First, if no candidate meets the criteria to win an election, you will need a backup plan such as a runoff election. Second, if the winner is not a Condorcet candidate, then there is at least one candidate who beat the winner in a pairwise matchup and the supporters of that candidate might question the validity of the election. Finally, if there is a Condorcet candidate who beat every other candidate in a pairwise matchup, it is reasonable to conclude that it would be unfair for anyone else to win. The rest of the examples in this section should illustrate these key concepts. Rock, Paper, Scissors by Pairwise Comparison Suppose that three people are playing the game Rock, Paper, Scissors. On the count of three, each person shows a hand signal for rock, paper, or scissors. Each hand signal beats another hand signal. The group keeps having a tie because Person A always picks rock, Person B always picks paper which beats rock, and Person C always picks scissors which beats paper, and is beaten by rock! This leads to a disagreement about which choice is best. They decide to use the pairwise comparison method determine the winner. Their preference rankings are given in the following table. Voters A B C Rock (R) 1 3 2 Paper (P) 2 1 3 Scissors (S) 3 2 1 Construct the comparison matrix: Rock, Paper, Scissors Pairwise Comparison Matrix There is a tie! There is no winner. illustrates the answer to the first key question. The pairwise comparison method does not always result in a winner. For example, much like the game of Rock, Paper, Scissors, it is possible for a cyclic pattern to emerge in which each candidate beats the next until the last candidate who beats the first. Rock, Paper, Scissors Cyclic Outcome Now, you have the answer to the second key question. The pairwise comparison matrix in YOUR TURN 11.10 is an example of a scenario where a winner is not a Condorcet candidate. The answer to the third question is not as clear. If there is a Condorcet candidate, does that candidate win? So far, we have not come across a contradictory example where the Condorcet candidate didn't win, but we cannot know with certainty that it is not possible by looking at examples. Instead, we will need to use some reasoning. Let’s review some particular cases of elections with a certain number of candidates, and then we will try to generalize the scenario to an election with n candidates. Does the Condorcet Candidate Win? Suppose there is an election with five candidates—A, B, C, D, and E—and that Candidate C is a Condorcet candidate. How many points did Candidate C win? What is the greatest number of points that any one of the other candidates could win? Is it possible for Candidate C to lose or tie? In any pairwise election with five candidates, each candidate must compete against four other candidates. It follows that the most points a single candidate can win is four points, which would occur if the candidate won every matchup. As a Condorcet candidate, Candidate C won all the pairwise matchups against Candidates A, B, D, and E, earning four points. The rest of the candidates lost to Candidate C. The most points a particular candidate could win if they won matchups with each of the other three candidates is three points. Since Candidate C has four points and the rest of the candidates have three points or less, Candidate C is the winner. Therefore, it is not possible for Candidate C to tie or lose. Let’s consider a general case where there are n candidates. One of the candidates is a Condorcet candidate. Since the Condorcet candidate wins all matchups, the Condorcet candidate wins n - 1 points. Since each of the other candidates lost to the Condorcet candidate, the most a single candidate could win is n - 2 . Since the Condorcet candidate won n - 1 points and each other candidate won n - 2 points or fewer, the Condorcet candidate is the winner. You have your answer to the third key question! If there is a Condorcet candidate, that candidate is always the winner. Approval Voting The last type of voting system you will consider for your budding democracy is an approval voting system . In this system, each voter may approve any number of candidates without rank or preference for one over another (among the approved candidates), and the candidate approved by the most voters wins. This voting system has aspects in common with plurality voting and Condorcet voting methods, but it has characteristics that distinguish it from both. An approval voting ballot lists the candidates and provides the option to approve or not approve each candidate. The term “approval voting” was not used until the 1970s Brams, Steven J.; Fishburn, Peter C. (2007), Approval Voting , Springer-Verlag, p. xv, ISBN 978-0-387-49895-9, although its use has been documented as early as the 13th century (Brams, Steven J. (April 1, 2006). The Normative Turn in Public Choice (PDF) (Speech). Presidential Address to Public Choice Society. New Orleans, Louisiana.) Approval voting has the appeal of being simpler than ranked voting methods. It also allows an individual voter to support more than one candidate equally. This has appeal for those who do not want a split vote among a few mainstream candidates to lead to the election of a fringe candidate. It also has appeal for those who want an underdog to have a chance of success because voters will not worry about wasting their vote on a candidate who is not believed likely to win. Rock, Paper, Scissors, Lizard, Spock Suppose that Person A/B/C were just about to give up on their game of Rock, Paper, Scissors when they were joined by Person D who reminded them that their updated version, Rock, Paper, Scissors, Lizard, Spock was a far superior game with the added rules that Lizard eats Paper, Paper disproves Spock, Spock vaporizes Rock, Rock crushes Lizard, Lizard poisons Spock, Spock smashes Scissors, and Scissors decapitates Lizard. Rock, Paper, Scissors, Lizard, Spock Dominance Person D encourages their friends to hold a new election. This time, for the sake of simplicity, the group decides to use approval voting to determine the best move in the game. The summary of approval ballots for Rock, Paper, Scissors, Lizard, Spock is given in the table below. Voters A B C D Rock Yes No No No Paper No Yes No No Scissors No No Yes No Lizard No No No Yes Spock Yes Yes Yes Yes Count the number of approval votes for each candidate by counting the number of “Yes” votes in each row of the table. Rock: 1 Paper: 1 Scissors: 1 Lizard: 1 Spock: 4 Spock is the winning candidate, approved by four voters! The Chionilis Family Is Hungry Again! The eight members of the Chionilis family—Annette, Rene, Seema, Titus, Galen, Elena, Max and Demitri—have another decision to make. Approval voting worked out nicely the last time. They are going to use it again, but this time, Annette, Rene, Seema, and Galen are feeling a little indecisive. They can't narrow their choice down to two. They will approve their three top choices, but the other family members will only approve two. These choices are reflected in the following table. Determine the restaurant that will be chosen. Options A R S T G E M D Rainbow China Yes Yes Yes Yes Yes No Yes Yes Dough Boys Pizza Yes Yes Yes Yes No Yes No Yes Taco City Yes No Yes No Yes Yes No No Caribbean Flavor No Yes No No Yes No Yes No Count the number of approval votes for each restaurant by counting the number of “Yes” votes in each row. Rainbow China: 7 Dough Boys Pizza: 6 Taco City: 4 Caribbean Flavor: 3 This time, Rainbow China won! Compare and Contrast Voting Methods to Identify Flaws Wow! We have covered a lot of options for the voting methods. Now, you need to decide which one is best for Imaginaria. Imaginarians might consider characteristics of certain voting systems desirable and others undesirable. In some cases, voters may consider these undesirable traits to be flaws in a voting system that are significant enough to motivate them to reject that system. If you are feeling a bit overwhelmed by this decision, maybe it would help to read about the experiences of others who have faced similar questions. Consider the 2000 U.S. presidential election in which Green Party candidate Ralph Nader and Reform Party candidate Pat Buchanan were on the ballet running against the mainstream candidates, Democrat Al Gore and Republican George W. Bush. The voting results for Florida are given in . Candidate Party Votes Percentage (G) George W. Bush Republican 2,912,790 48.85% (A) Al Gore Democrat 2,912,253 48.84% (R) Ralph Nader Green 97,488 1.63% (P) Pat Buchanan Reform 17,484 0.29% (H) Harry Brown Libertarian 16,415 0.28% (O) 7 Other Candidates Other 6,680 0.11% Total 5,963,110 Florida Results in the 2000 U.S. Presidential Election (source: https://www.fec.gov/resources/cms-content/documents/FederalElections2000_PresidentialGeneralElectionResultsbyState.pdf) In more than one state, Buchanan was able to split the Republican vote enough to allow Gore to win that state. Nader split the Democrat vote in Florida and New Hampshire by enough votes to prevent Gore from winning those states. Had Gore won either state, he would have had enough electoral votes to win the election. Instead, Bush won. This is an example of a flaw in the plurality system of voting: the spoiler. A spoiler is a less popular candidate who takes votes from a more popular candidate with similar positions, swinging the race to another candidate with vastly different views that they would not support. This encourages voters not to vote for the candidate that they perceive to be the best, but instead for the candidate they can live with who they perceive to have a better chance of winning. Some voters may prefer a method such as approval voting, which does not have this trait in common with plurality voting. The Spoiler Controversy Because the vote counts for George W. Bush and Al Gore differed by only 537 votes, many Democrats blamed Ralph Nader and the Green Party for their loss. Let’s consider how the election results might have differed if the approval voting method had been used. Use and the following assumptions to extrapolate the results of an approval method election: 100 percent of Pat Buchanan supporters would approve George W. Bush. 100 percent of Ralph Nader supporters would approve Al Gore. 72 percent of Libertarians would approve George W. Bush. 28 percent of Libertarians would approve Al Gore (as was roughly the known percentage at the time according to the Cato Institute). 50 percent of the supporters of other candidates would approve George Bush while 50 percent would approve Al Gore. Step 1: Create a summary of approval ballots based on the given assumptions. For the Libertarian candidate, 72 percent of 16,415 of the votes is 0.72 ( 16 , 415 ) = 11 , 819 and 28 percent is 0.28 ( 16 , 415 ) = 4 , 596 . For the other candidates, 50 percent of the votes is 0.50 ( 6 , 680 ) = 3 , 340 . Number of Votes 2,912,790 (G) 2,912,253 (A) 97,488 (R) 17,484 (B) 11,819 (72% H) 4,596 (28% H) 3,340 (50% O) 3,340 (50% O) (G) George W. Bush Yes No No Yes Yes No No Yes (A) Al Gore No Yes Yes No No Yes Yes No (R) Ralph Nader No No Yes No No No No No (P) Pat Buchanan No No No Yes No No No No (H) Harry Brown No No No No Yes Yes No No (O) 7 Other Candidates No No No No No No Yes Yes Step 2: Count the number of approval votes for each candidate. George W. Bush: 2 , 912 , 790 + 17 , 484 + 11 , 819 + 3340 = 2 , 945 , 433 Al Gore: 2 , 912 , 253 + 97 , 488 + 4 , 596 + 3 , 340 = 3 , 017 , 677 Ralph Nader: 97 , 488 Pat Buchanan: 17 , 484 Harry Brown: 11 , 819 + 4 , 596 = 16 , 415 Other Candidates: 6 , 680 In this scenario, Al Gore is the winner. The results in and Your Turn 11.14 highlight one of the characteristics of approval voting. Ralph Nader moved up from a distant third place finish to a close second place finish when Al Gore’s supporters approved him on their ballots. In this way, fringe candidates have a better chance of winning, which some voters consider a flaw but others consider a benefit. Another aspect of approval voting systems that is a concern to many voters is that candidates in approval elections might encourage their loyal supporters to approve them and only them to avoid giving support to any other candidate. If this occurred, the election in effect becomes a traditional plurality election. This is a flaw that cannot occur in an instant runoff system since all candidates are ranked. Three Habitable Planets In the future, humans have explored distant solar systems and found three habitable planets which could be colonized. Since it will take all available resources to colonize one planet, humans must agree on the planet. Planet A has the most comfortable climate and most plentiful resources, but it is the farthest from Earth making travel to the planet a challenge. Planet B is half the distance but will require more resources to make comfortable. Planet C is the least suitable of the three and terraforming will be required, but it is close enough to make travel between Earth and Planet C possible on a more regular basis. The table below provides the voter preferences for the colonization of each planet. Percentage of Voters 45% 15% 40% Planet A 1 3 3 Planet B 2 1 2 Planet C 3 2 1 If the entire population were able to vote, determine the winning planet using each of the methods listed below. Plurality Ranked-choice method Borda count The plurality method only considers the top choice of each voter. By this system, Planet A has 45 percent of the vote, Planet B has 15 percent of the vote, and Planet C has 40 percent of the vote. Planet A wins. Using either instant runoff or a two-round system, Planet B with only 15 percent of the vote will be eliminated in the first round. In Round 2, the 15 percent that voted for Planet B would vote for their second choice, Planet C. This leaves Planet A with 45 percent and Planet C with 55 percent. Planet C has a majority and wins the election. To find the Borda score for each candidate, imagine there are exactly 100 voters. Then the summary of ranked ballots looks like: Out of 100 Voters 45 15 40 Planet A 1 3 3 Planet B 2 1 2 Planet C 3 2 1 The Borda score for each candidate is as follows: Planet A: ( 3 - 1 ) ( 45 ) + ( 3 - 3 ) ( 15 ) + ( 3 - 3 ) ( 40 ) = 90 Planet B: ( 3 - 2 ) ( 45 ) + ( 3 - 1 ) ( 15 ) + ( 3 - 2 ) ( 40 ) = 115 Planet C: ( 3 - 3 ) ( 45 ) + ( 3 - 2 ) ( 15 ) + ( 3 - 1 ) ( 40 ) = 95 Planet B wins. The election in involves a scenario in which there are two extreme candidates, Planet A and Planet B, and a moderate candidate, Planet C. The supporters of the extreme candidates prefer the moderate candidate to the other extremist ones. This makes Planet C a compromise candidate. In this case, both the plurality method and ranked-choice voting resulted in the election of one of the extreme candidates, but the Borda count method elected the compromise candidate in this scenario. Depending on a person’s perspective, this may be perceived as a flaw in either ranked-choice and plurality systems, or the Borda count method. In Fairness in Voting Methods , we will analyze the fairness of each voting system in greater detail using objective measures of fairness. Voting Calculators It is possible to create Excel spreadsheets that complete the calculations necessary to determine the winner of an election by various voting methods. In some cases, this work has already been done and posted online. As you practice applying the various voting methods that could be used in Imaginaria, quick Internet search will lead to sites such as Ms. Hearn Math with free specialty calculators. These sites can be a great way to check your results! Check Your Understanding Key Terms majority plurality runoff election runoff voting system two-round system Hare Method preference ranking ranked ballot ranked-choice voting (RCV) instant runoff voting (IRV) Borda count method Borda score Compromise candidate divisive candidate pairwise comparison method Condorcet voting methods Condorcet candidate approval voting system approval voting ballot spoiler Key Concepts In plurality voting, the candidate with the most votes wins. When a voting method does not result in a winner, runoff voting can be used to do so. Ranked-choice voting, also known as instant runoff voting, is one type of ranked voting system. The Borda count method is a type of ranked voting system in which each candidate is given a Borda score based on the number of candidates ranked lower than them on each ballot. When pairwise comparison is used, the winner will be the Condorcet candidate if one exists. Approval voting allows voters to give equally weighted votes to multiple candidates. When a voter finds a characteristic of a particular voting method unappealing, they may consider that characteristic a flaw in the voting method and look for an alternative method that does not have that characteristic. Videos How Does Ranked-Choice Voting Work? Determine Winner of Election by Ranked-Choice Method (aka Instant Runoff) Determine Winner of Election by Borda Count Method Determine Winner of Election by Pairwise Comparison Method", "section": "Voting Methods", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Fairness in Voting Methods Citizens strive to ensure their voting system is fair. (credit: “Governor Votes Early” by Maryland GovPics/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compare and contrast fairness of voting using majority criterion. Compare and contrast fairness of voting using head-to-head criterion. Compare and contrast fairness of voting using monotonicity criterion. Compare and contrast fairness of voting using irrelevant alternatives criterion. Apply Arrow’s Impossibility Theorem when evaluating voting fairness. Now that we’ve covered a variety of voting methods and discussed their differences and similarities, you might be leaning toward one method over another. You will need to convince the other founders of Imaginaria that your preference will be the best for the country. Before your collaborators approve the inclusion of a voting method in the constitution, they will want to know that the voting method is a fair method. In this section, we will formally define the characteristics of a fair system. We will analyze each voting previously discussed to determine which characteristics of fairness they have, and which they do not. In order to guarantee one ideal, we must often sacrifice others. The Majority Criterion One of the most fundamental concepts in voting is the idea that most voters should be in favor of a candidate for a candidate to win, and that a candidate should not win without majority support. This concept is known as the majority criterion . With respect to the four main ranked voting methods we have discussed—plurality, ranked-choice, pairwise comparison, and the Borda count method—we will explore two important questions: Which of these voting systems satisfy the majority criterion and which do not? Is it always “fair” for a voting system to satisfy the majority criterion? Keep in mind that this criterion only applies when one of the candidates has a majority. So, the examples we will analyze will be based on scenarios in which a single candidate has more than 50 percent of the vote. Roommates Choose Fast Food It’s final exams week and seven college students are hungry. They must get food, but from which drive thru? Their preferences are listed in the table below. The majority have listed McDonald’s as their top choice. Let’s calculate what the results of the election will be using various voting methods. Voters A B C D E F G (M) McDonald’s 1 1 1 1 5 5 5 (B) Burger King 2 2 2 2 2 2 2 (T) Taco Bell 4 4 3 4 3 3 1 (O) Pollo Tropical 3 5 4 3 4 1 3 (I) Pizza Hut 5 3 5 5 1 4 4 Which restaurant is the winner using the plurality voting method? Which restaurant is the winner using the ranked-choice voting method? Does the majority criterion apply? If so, for which of voting method(s), if any, did the majority criterion fail? For plurality voting, we only need to count the first-place votes for each candidate. In this case, McDonald’s has four first place votes, which is a majority and wins the election automatically. For ranked-choice voting, McDonald’s also wins because it has a majority at the end of Round 1. The majority criterion does apply because one candidate had a majority of the first-place votes. The majority criterion did not fail by either method, because in each case the majority candidate won. From , it appears that the plurality and ranked-choice voting methods satisfy the majority criterion. In general, the majority candidate always wins in a plurality election because the candidate that has more than half of the votes has more votes than any other candidate. The same is true for ranked-choice voting; and there will never be a need for a second round when there is a majority candidate. Let’s examine how some of the other voting methods stand up to the majority criterion. Roommates Choose Fast Food Those seven college students are hungry again! Their preferences haven’t changed, as shown below. Let’s calculate if the results change when we use different voting methods. VOTERS A B C D E F G (M) McDonald’s 1 1 1 1 5 5 5 (B) Burger King 2 2 2 2 2 2 2 (T) Taco Bell 4 4 3 4 3 3 1 (O) Pollo Tropical 3 5 4 3 4 1 3 (I) Pizza Hut 5 3 5 5 1 4 4 Which restaurant is the winner using the pairwise comparison voting method? Which restaurant is the winner using the Borda count voting method? Does the majority criterion apply? If so, for which voting method(s), if any, did the majority criterion fail? For pairwise comparison, notice that McDonald’s is a Condorcet candidate because it wins every pairwise comparison. So, McDonald’s is the winner. For the Borda count, we must calculate the Borda score for each candidate: McDonald’s is 16, Burger King is 21, Taco Bell is 13, Pollo Tropical is 12, Pizza Hut is 8. The winner is Burger King! Yes, the majority criterion applies because McDonald’s has the majority of first place votes. The majority criterion only fails using the Borda method. demonstrates a concept that we also saw in Borda Count Voting —the Borda method frequently favors the compromise candidate over the divisive candidate. This can happen even when the divisive candidate has a majority, as it did in this example. Although a majority of the voters were in favor of McDonald’s, a significant minority was strongly opposed to McDonald’s, ranking it last. Since the Borda score includes all rankings, this strong opposition has an impact on the outcome of the election. Pairwise comparison will always satisfy the majority criterion because the candidate with the majority of first-place votes wins each pairwise matchup. While it is possible for the majority candidate to win by the Borda count method, it is not guaranteed. So, the Borda method fails the majority criterion. A summary of each voting method as it relates to the majority criterion is found in the following table. Voting Method Majority Criterion Plurality Satisfies Ranked-choice Satisfies Pairwise comparison Satisfies Borda count Violates If you prefer the Borda method, you might argue that its failure to satisfy the majority criterion is actually one of its strengths. As we saw in and , the majority have the power to vote for their own benefit at the expense of the minority. While four students were very enthusiastic about McDonald’s, three students were strongly opposed to McDonald’s. It is reasonable to say that the better option would be Burger King, the compromise candidate, which everyone ranked highly and no one strongly opposed. The Borda method is designed to favor a candidate that is acceptable to the population as a whole. In this way, the Borda method avoids a downfall of strict majority rule known as the tyranny of the majority , which occurs when a minority of a population is treated unfairly because their situation is different from the situation of the majority. The people of Imaginaria should know that the power of the majority to vote their will has serious implications for other groups. For example, according to the UCLA School of Law Williams Institute, the LGBTQ+ community in the United States makes up approximately 4.5 percent of the population. When elections occur that include issues that affect the LGBTQ+ community, members of the LGBTQ+ community depend on the 95.5 percent of the population who do not identify as LGBTQ+ to consider their perspectives when voting on issues such as same-sex marriage, the use of public restrooms by transgender people, and adoption by same-sex couples. Robert Dahl In his book, Democracy and Its Critics , Robert Dahl wrote, “If a majority is not entitled to do so, then it is thereby deprived of its rights; but if a majority is entitled to do so, then it can deprive the minority of its rights.” Dahl was a renowned political theorist, but he is also considered to be a mathematician since his work utilizes ideas from an area of mathematics known as Game Theory (Mathematics Genealogy Project, NDSU Department of Mathematics with the American Mathematical Society). Three Branches of Government Concerns about the consequences of majority rule are not new. In 1788, John Adams warned of the consequences of majority rule and he argued for three branches of government as a way to temper them. In the early 1800s, a young French aristocrat named Alexis de Tocqueville toured the United States and wrote Democracy in America , which focused on the impact of democracy on political and civil societies. He observed, even then, the dominance of the white majority over the Indigenous people and enslaved people, which was perpetuated by majority rule. Separation of Powers and Checks and Balances Head-to-Head Criterion Another fairness criterion you must consider as you select a voting method for Imaginaria is the Condorcet criterion , also known as the head-to-head criterion. An election method satisfies the Condorcet criterion provided that the Condorcet candidate wins the election whenever a Condorcet candidate exists. A Condorcet method is any voting method that satisfies the Condorcet criterion. Recall from Three Key Questions that not every election has a Condorcet candidate; the Condorcet criterion will not apply to every election. Also recall that a Condorcet candidate cannot lose an election by pairwise comparison. So, the pairwise comparison voting method is said to satisfy the Condorcet criterion. Spending Tax Refund A survey asked a random sample of 100 people in the United States to rank their priorities for spending their tax refund. The options were (V) go on vacation, (S) put into savings, (D) pay off debt, or (T) other. The pairwise comparison matrix for the results is in . Determine whether the Condorcet criterion applies. Pairwise Comparison Matrix for Tax Refund Spending The Condorcet criterion only applies when there is a Condorcet candidate. “Pay off debt” (D) is a Condorcet candidate because D wins every matchup. Yes, the Condorcet criterion applies to this election. Spending Tax Refund Let’s return to the survey about tax refund spending from . We know that the Condorcet criterion applies because Option D, “Pay off debt,” is a Condorcet candidate, which wins every pairwise match up. Use the information in the ballot summary from the table below to find the winner and determine whether the Condorcet criterion is satisfied in this election when each of the following voting methods are used. Votes 33 32 31 4 On a vacation (V) 1 3 3 2 Put into savings (S) 3 1 2 1 Pay off debt (D) 2 2 1 4 Other (T) 4 4 4 3 Plurality Ranked-choice voting Borda count V wins 33 first place votes; S, 36; D, 31; and T, 0. So candidate S, “Put into savings,” has a plurality and wins. Since the Condorcet candidate D didn’t win, the Condorcet criterion is violated. Use the steps outlined in Ranked-Choice Voting for determining the winner of an election by ranked-choice voting, the application of the Hare method in which instant runoffs are used. Step 1: The number of votes needed to achieve a majority is 51. Step 2: As illustrated in part 1, no candidate has a majority of first-place votes; so the candidate with the fewest votes, T, must be eliminated. Step 3: Reallocate votes to the remaining candidates for the second round: Votes 33 32 31 4 On a vacation (V) 1 3 3 2 Put into savings (S) 3 1 2 1 Pay off debt (D) 2 2 1 3 Step 4: Repeat the process from Step 2. Count the first-place votes for each candidate: V has 33 votes, S has 36 votes, D has 31. votes. Eliminate candidate D, “Pay off debt,” for the third round. Step 5: Repeat the process from Step 3. Reallocate votes to the remaining candidates for the third round: Votes 33 32 31 4 On a vacation (V) 1 2 2 2 Put into savings (S) 2 1 1 1 Step 6: Repeat the process from Step 2 one last time. Count the first-place votes for each candidate: V has 33, S has 67. Candidate S, “Put into savings,” has a majority and wins. Since the Condorcet candidate D candidate D, “Pay off debt,” didn’t win, the Condorcet criterion is violated. Calculate the Borda score for each candidate. V: 33 ( 4 - 1 ) + 32 ( 4 - 3 ) + 31 ( 4 - 3 ) + 4 ( 4 - 2 ) = 170 S: 33 ( 4 - 3 ) + 32 ( 4 - 1 ) + 31 ( 4 - 2 ) + 4 ( 4 - 1 ) = 203 D: 33 ( 4 - 2 ) + 32 ( 4 - 2 ) + 31 ( 4 - 1 ) + 4 ( 4 - 4 ) = 223 T: 33 ( 4 - 4 ) + 32 ( 4 - 4 ) + 31 ( 4 - 4 ) + 4 ( 4 - 3 ) = 4 Candidate D, “Pay off debt,” has the highest Borda score and wins. Since D was the Condorcet candidate, this election satisfies the Condorcet criterion. As we have seen, the plurality method, ranked-choice voting, and the Borda count method each fail the Condorcet criterion in some circumstances. Of the four main ranked voting methods we have discussed, only the pairwise comparison method satisfies the Condorcet criterion every time. A summary of each voting method as it relates to the Condorcet criterion is found in the following table. Voting Method Condorcet Criterion Plurality Violates Ranked-choice Violates Pairwise comparison Satisfies Borda count Violates Monotonicity Criterion The citizens of Imaginaria might be surprised to learn that it is possible for a voter to cause a candidate to lose by ranking that candidate higher on their ballot. Is that fair? Most voters would say, “Absolutely not!!” This is an example of a violation of the fairness criterion called the monotonicity criterion , which is satisfied when no candidate is harmed by up-ranking nor helped by down-ranking, provided all other votes remain the same. Consider a scenario in which voters are permitted a first round that is not binding, and then they may change their vote before the second round. Such a first round can be called a “straw poll.” Now, let’s suppose that a particular candidate won the straw poll. After that, several voters are convinced to increase their support, or up-rank , that winning candidate and no voters decrease that support. It is reasonable to expect that the winner of the first round will also win the second. Similarly, if some of the voters decide to decrease their support, or down-rank , a losing candidate, it is reasonable to expect that candidate will still lose in the second round. You might be wondering why it’s called the monotonicity criterion. In mathematics, the term monotonicity refers to the quality of always increasing or always decreasing. For example, a person’s age is monotonic because it always increases, whereas a person’s weight is not monotonic because it can increase or decrease. If the only changes to the votes for a particular candidate after a straw poll are in one direction, this change is considered monotonic. If you are going to make an informed decision about which voting method to use in Imaginaria, you need to know which of the four main ranked voting methods we have discussed—plurality, ranked-choice, pairwise comparison, and the Borda count method—satisfy the monotonicity criterion. Favorite Dog Breed by Plurality The local animal shelter is having a vote-by-donation charity event. For a $10 donation, an individual can complete a ranked ballot indicating their favorite large dog breed: standard poodle, golden retriever, Labrador retriever, or bulldog. Use the summary of ballots below to answer each question. Votes 42 53 61 24 (S) Standard Poodle 1 3 2 1 (G) Golden Retriever 3 1 4 4 (L) Labrador Retriever 4 2 1 2 (B) Bulldog 2 4 3 3 Determine the winner of the election by plurality. Suppose that the 53 voters in the second column increased their ranking of the winner by 1. Determine the winner by plurality with the new rankings. Does this election violate the monotonicity criterion? Do you think the result of part 3 is also true for plurality voting and the monotonicity criterion in general? Why or why not? The number of votes for each candidate are: S 66, G 53, L 61, and B 0. The winner is the standard poodle. If the 53 voters in the second column rank S as 2 and L as 3, then the number of votes for each candidate are: S with 66, G with 53, L with 61, and B with 0. The winner is still the standard poodle. This election does not violate the monotonicity criterion because the winner was not hurt by up-ranking. In general, increasing the ranking for a winner of a plurality election will either leave them with the same or more first place votes while leaving the other candidates with the same or fewer first place votes. So a plurality election will never violate the monotonicity criterion. Monotonicity Criterion in a Pairwise Comparison Election Earlier, we discovered that the summary of ranked ballots shown in the table below results in the pairwise comparison matrix in . Use this information to answer the questions. Number of Ballots 95 90 110 115 Option A 4 4 1 1 Option B 2 2 2 2 Option C 3 1 3 4 Option D 1 3 4 3 Analyzed Pairwise Comparison Matrix for Sample Summary of Ranked Ballots Determine the winner of the election by the pairwise comparison method. Suppose that the 95 voters in the first column increased their ranking of the winner by 1. Determine the winner by the pairwise comparison method with the new rankings. Does this election violate the monotonicity criterion? Do you think the result of part 3 is true for the pairwise comparison method and the monotonicity criterion in general? Why or why not? By the pairwise comparison method, Option A wins with three points. If the 95 voters in the first column of increased their ranking of the winner by 1, then C would fall into fourth place and A would move up to third place on those ballots. This would only affect the matchup between A and C, and the result would be that A would gain 95 votes while C would lose 95 votes. This means A would have 320 votes and C would have 90. Since A already was ahead of C, this just puts A further ahead and causes no change to the election results. Since the winner A is not hurt by an up-rank, and the loser C is not helped by a down-rank, this election is fair by the monotonicity criterion. Yes, the monotonicity criterion would be satisfied by the pairwise comparison method, because an up-rank of the winner can never decrease the number of pairwise wins. Similarly, a down-rank can never increase the number of pairwise wins. The last few examples illustrate that the plurality method, pairwise comparison voting, and the Borda count method each satisfy the monotonicity criterion. Of the four main ranked voting methods we have discussed, only the ranked-choice method violates the monotonicity criterion. A summary of each voting method as it relates to the Condorcet criterion is found in the table below. Voting Method Monotonicity Criterion Plurality Satisfies Ranked-choice Violates Pairwise comparison Satisfies Borda count Satisfies Irrelevant Alternatives Criterion We have covered a lot about voting fairness, but there is one more fairness criterion that you and the other Imaginarians should know. Consider this well-known anecdote that is sometimes attributed to the American philosopher Sidney Morgenbesser: A man is told by his waiter that the dessert options this evening are blueberry pie or apple pie. The man orders the apple pie. The waiter returns and tells him that there is also a third option, cherry pie. The man says, “In that case, I would like the blueberry pie.” ( Gaming the Vote: Why Elections Aren’t Fair (and What We Can Do About It) , William Pound stone, p. 50, ISBN 0-8090-4893-0) This story illustrates the concept of the Irrelevant Alternatives Criterion, also known as the Independence of Irrelevant Alternatives Criterion (IIA ), which means that the introduction or removal of a third candidate should not change or reverse the rankings of the original two candidates relative to one another. In particular, if a losing candidate is removed from the race or if a new candidate is added, the winner of the race should not change. Apple, Blueberry, or Cherry? Suppose that 30 students in a class are going to vote on whether to have apple, blueberry, or cherry pie. Use the summary of ranked ballots in below to answer each question. Number of Ballots 14 12 4 (A) Apple Pie 1 3 3 (B) Blueberry Pie 2 1 2 (C) Cherry Pie 3 2 1 Determine the winner of the election by plurality. Which candidate would win a plurality election if cherry pie were removed from the ballot? Does this election violate the IIA? The number of first place votes for each candidate is: A with 14, B with 12, and C with 4. Apple pie has the most first-place votes and wins the election. If cherry pie is removed from the ballot, then the four voters in the third column now rank blueberry pie as their first choice. So the four votes for C now belong to B. This means that blueberry pie has 16 votes compared to the 14 votes for apple pie. Blueberry pie now wins the plurality election. Yes, the election violates the IIA because the removal of a losing candidate from the ballot changed the winner of the election. Best Fourth Wall Breaking Stare on The Office The NBC sitcom The Office ran for nine years and has been one of the most popular streamed television shows of all time. One of the trademarks of the show was that characters would often break the fourth wall to communicate with the audience just by staring directly into the camera. In fact, there is a website dedicated to \" The Office stares\" where you can watch over 700 of these stares! Suppose that 36 fans were asked which character had the best \" The Office stare.\" Use the ballot summary in below to answer each question. Number of Ballots 9 11 7 6 3 (J) Jim Halpert (John Krasinski) 1 2 4 2 4 (P) Pam Beesly-Halpert (Jenna Fischer) 4 1 2 4 3 (D) Dwight Schrute (Rainn Wilson) 2 3 3 1 2 (M) Michael Scott (Steve Carell) 3 4 1 3 1 Determine the winner of the election by the pairwise comparison method. Determine the winner of the election by the pairwise comparison method if Michael Scott is removed from the ballot. Does this election violate IIA? Construct and analyze a pairwise comparison matrix: Pairwise Comparison Matrix for Jim, Pam, Dwight, and Michael Jim Halpert wins with 2 points. If Michael Scott is removed, the summary of ranked ballots becomes: Number of Ballots 9 11 7 6 3 (J) Jim Halpert (John Krasinski) 1 2 3 2 3 (P) Pam Beesly-Halpert (Jenna Fischer) 3 1 1 3 2 (D) Dwight Schrute (Rainn Wilson) 2 3 2 1 1 Construct and analyze a pairwise comparison matrix: Pairwise Comparison Matrix for Jim, Pam, and Dwight Pam wins with 1 1 2 points. Yes, this violates the IIA, because the winning candidate was hurt by the elimination of a losing candidate. We have seen that all four of the main voting systems we are working with fail the Irrelevant Alternatives Criterion (IIA). A summary of each voting method as it relates to the IIA criterion is found in the table below. Voting Method Irrelevant Alternatives Criterion Plurality Violates Ranked-choice Violates Pairwise comparison Violates Borda count Violates Electronic Voting: Does Your Vote Count? In order for an election to be fair, voting must be accessible to everyone and every vote must be counted. When hundreds of thousands to millions of votes must be collected and counted in a short period of time, deciding it can be challenging to be on counting procedures that are accurate and secure. Electronic voting machines or even Internet voting can speed up the process, but how reliable are these methods? This has been a subject for debate for years. In a press release on August 3, 2007, California Secretary of State Debra Bowen explained the results of an extensive review of electronic voting systems in her state. She said that transparency and auditability were key. She went on to say, “I think voters and counties are the victims of a federal certification process that hasn’t done an adequate job of ensuring that the systems made available to them are secure, accurate, reliable and accessible. Congress enacted the Help America Vote Act, which pushed many counties into buying electronic systems that—as we’ve seen for some time and we saw again in the independent UC review—were not properly reviewed or tested to ensure that they protected the integrity of the vote.” Secretary Bowden subsequently ordered that voting machines must have tighter security to be used in California. (DB07:042, Secretary of State Debra Bowen Moves to Strengthen Voter Confidence in Election Security Following Top-to-Bottom Review of Voting Systems, https://sos.ca.gov/elections.) In some instances, the use of electronic voting in parliamentary elections has been discontinued completely for security reasons. For example, according to the National Democratic Institute, the Netherlands returned to all paper ballots and hand counting in 2006. Will you use voting machines or Internet voting in Imaginaria? So far, every one of the voting methods we have analyzed has failed one or more of the fairness criteria in one election or another. Voting Method Majority Criterion Condorcet Criterion (Head-to-Head Criterion) Monotonicity Criterion (Independence of) Irrelevant Alternatives Criterion Plurality Satisfies Violates Satisfies Violates Ranked-choice Satisfies Violates Satisfies Violates Pairwise comparison Satisfies Satisfies Violates Violates Borda count Violates Violates Satisfies Violates You might be wondering if there is a voting system you could recommend for Imaginaria that satisfies all the fairness criteria. If there is one, it remains to be discovered and it is not a voting system that is based solely on preference rankings. In 1972, Harvard Professor of Economics Kenneth J. Arrow received the Nobel Prize in Economics for proving Arrow’s Impossibility Theorem , which states that any voting system, either existing or yet to be created, in which the only information available is the preference rankings of the candidates, will fail to satisfy at least one of the following fairness criteria: the majority criterion, the Condorcet criterion, the monotonicity criterion, and the independence of irrelevant alternatives criterion. This theorem only applies to a specific category of voting systems—those for which the preference ranking is the only information collected. There are other types of voting systems to which the Impossibility Theorem does not apply. For example, there is a class of voting systems called Cardinal voting systems that allow for rating the candidates in some way. “Rating” is different from “ranking” because a voter can give different candidates the same rating. Consider the five-star rating systems used by various industries, or the thumbs up/thumbs down rating system used on YouTube. Could there be a Cardinal voting system that does not violate any of the fairness criteria we have discussed? It’s possible, but more research must be done in order to prove it. Kenneth J. Arrow In 1972, Kenneth J. Arrow, a Harvard Professor of Economics, received the Nobel Prize in Economics jointly with Sir John Hicks, another world renowned economist, for their contributions to economic theory. In particular, Professor Arrow proved mathematically that no ranked voting system meets all four of the fairness criteria discussed in this section. The statement of this fact is known as Arrow’s Impossibility Theorem. Visit this site for more details on Professor Arrow Check Your Understanding Key Terms majority criterion tyranny of the majority Condorcet criterion Condorcet method monotonicity criterion up-rank down-rank independence of irrelevant alternatives criterion (IIA) Arrow’s Impossibility Theorem cardinal voting system Key Concepts There are several common measures of voting fairness, including the majority criterion, the head-to head criterion, the monotonicity criterion, and the irrelevant alternatives criterion. According to Arrow’s Impossibility Theorem, each voting method in which the only information is the order of preference of the voters will violate one of the fairness criteria. Videos Separation of Powers and Checks and Balances", "section": "Fairness in Voting Methods", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Standard Divisors, Standard Quotas, and the Apportionment Problem Every person at a party gets their fair slice of the cake. (credit: “apple spice cake” by Mark Bonica/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Analyze the apportionment problem and applications to representation. Evaluate applications of standard divisors. Evaluate applications of standard quotas. The Apportionment Problem In the new democracy of Imaginaria, there are four states: Fictionville, Pretendstead, Illusionham, and Mythbury. Each state will have representatives in the Imaginarian Legislature. You might now have an agreement on which voting method your citizens will use to elect representatives. However, before that process can even begin, you must decided on how many representatives each state will receive. This decision will present its own challenges. When sharing your birthday cake, it’s only fair that everyone gets the same portion size, right? You were portioning the cake by dividing it up equally and giving everyone a slice. A great thing about cake is that you can slice it any way you want, but how do you apportion , or divide and distribute, items that can't be sliced? Suppose that you have a box of 16 Ring Pops™, gem-shaped lollipops on a plastic ring. You are going to share the box with four other kids. Dividing the 16 Ring Pops™ among the group of five leads to a problem; after each person in the group gets three Ring Pops™, there is still one left! Who gets the last one? The apportionment problem is how to fairly divide, or apportion, available resources that must be distributed to the recipients in whole, not fractional, parts. The apportionment problem applies to many aspects of life, including the representatives in the Imaginarian legislature. The table below provides a short list of examples of resources that must be apportioned in whole parts, and the recipients of those resources. Resource Recipients Covid-19 Vaccines Nations around the world Airport Terminals Airlines Faculty Positions at a University Departments Public Schools Communities U.S. House of Representatives Seats States Parliamentary Seats Political Parties Fair division of a resource is not necessarily equal division of the resource like when distributing cake slices. When distributing airport terminals amongst airlines, there are many factors to consider such as the size of the airline, the number and types of aircraft they have, and the demand for the service. In most cases, fairness is defined as being proportional ; two quantities are proportional if they have the same relative size. In the case of the Covid-19 vaccine, the expectation would be that countries with larger populations get more doses of the vaccine. In the Imaginarian legislature, the expectation may be that the states with larger populations will receive the larger number of representatives. This concept is referred to as a part-to-part ratio . Suppose that a supermarket has a special on pies, two for $5. The first customer purchases four pies for $10, and the second customer purchases eight pies for $20. The dollar to pie ratio for the first customer is 10 dollars 4 pies = 2.5 dollars per pie and the dollar to pie ratio for the second customer is 20 dollars 5 pies = 2.5 dollars per pie . So, the dollar to pie ratio is constant. Although the customers do not spend the same amount of money, the amount each spent was proportional to the number of pies purchased. Now suppose that the supermarket changed the special to $5 for the first pie, and $2 for each additional pie. In that case, four pies would cost $ 5 + 3 ( $ 2 ) = $ 11 , while 8 pies would cost $ 5 + 7 ( $ 2 ) = $ 19 . The dollar to pie ratios would be 11 dollars 4 pies = 2.75 dollars per pie and 19 dollars 8 pies = 2.375 dollars per pie , respectively. This special does not result in a constant part to part ratio. The dollars spent are not proportional to the number of pies purchased. What Is a Ratio? What Are the Different Types of Ratios? Ratio of Faculty to Students at a College The following table provides a comparison of the number of faculty members in each department at a particular college to the student head count in that department and the number of class sections in that department in the Spring semester. Use this information to answer the questions. Department Mathematics English History Science (S) Student Head Count 4800 2376 1536 2880 (C) Class Sections 120 108 48 96 (T) Total Faculty 30 27 12 24 (F) Full-Time Faculty 10 9 4 8 (P) Part-Time Faculty 20 18 8 16 Determine the ratios for each department: S to C, C to T, S to T, F to P What are the units of the ratios that you found? Which of these pairs, if any, has a constant part to part ratio? State the ratio. Does it appear that the total number of faculty positions were allocated to each department based on student head count, the number of class sections, or neither? Justify your answer. Divide the first quantity by the second, for each department, as shown in the table below. Answers are provided in last column of the table. Department Mathematics English History Science Units of Ratios Found S to C 4800 120 = 40 2376 108 = 22 1536 48 = 32 2880 96 = 30 Students per class section C to T 120 30 = 4 108 27 = 4 48 12 = 4 96 24 = 4 Class sections per faculty member S to T 4800 30 = 160 2376 27 = 88 1536 12 = 128 2880 24 = 120 Students per faculty member F to P 10 20 = 1 2 9 18 = 1 2 4 8 = 1 2 8 16 = 1 2 Full-time faculty member per part-time faculty member The ratio of class sections to faculty members is a constant ratio of four. The ratio of full-time faculty to part-time faculty is a constant ratio of 1 2 . It appears that the faculty positions were allocated based on the number of class sections because there is a constant ratio of four class sections per faculty member. There are some useful relationships between quantities that are proportional to each other. When there is a constant ratio between two quantities, the one quantity can be found by multiplying the other by that ratio. Remember the supermarket special on pies, 2 pies for $5? The ratio of dollars to pies is 5 dollars 2 pies = 2.5 dollars per pie and the ratio of pies to dollars is 2 pies 5 dollars = 0.4 pies per dollar . These two values are reciprocals of each other, 1 2.5 = 0.4 and 1 0.4 = 2.5 . This means that multiplying by one has the same effect as dividing by the other. This also means that knowing either constant ratio allows us to calculate the price given the number of pies. To find the cost of 20 pies, multiply by the ratio of dollars to pies or divide by the ratio of pies to dollars. 20 pies × 2.50 dollars per pie = 50 dollars 20 pies ÷ 0.4 pies per dollar = 50 dollars These patterns are true in general. Let A be a particular item and B another such that there is a constant ratio of A to B ratio of B 's to A 's = 1 ratio of A 's to B 's and ratio of A 's to B 's = 1 ratio of B 's to A 's units of A = ( units of B ) × ( ratio of A 's to B 's ) = units of B ratio of B 's to A 's units of B = ( units of A ) × ( ratio of B 's to A 's ) = units of A ratio of A 's to B 's Ratio of Faculty to Students at a College Refer to the information given in . If there are 32 class sections each semester in the Fine Art department, and the same ratio is used to determine the number of faculty members, how many faculty members would you expect to see in the Fine Art department? If the Health Sciences department has 6 full-time faculty members, how many part-time faculty members are in the department? Multiply the number of class sections by the ratio of faculty members to class sections to find the number of faculty. Since there are 4 classess per faculty member, the ratio of faculty members to classs sections is 1 4 . This means that the number of faculty members for 32 classes should be 32 × 1 4 = 8 faculty members. Multiply the number of full-time faculty by the ratio of part-time to full-time to find the number of part-time. Since ratio of full-time faculty to part-time faculty at the college is 1 2 or 1 full-time per 2 part time, the ratio of part-time to full-time is 2 1 = 2 part-time to 1 full-time; so the number of part-time faculty in a department with 6 full-time faculty should be 6 . 4 = 24 part-time faculty. The apportionment application that will be important to the founders of Imaginaria occurs in representative democracies in which elected persons represent a group. The United Kingdom, France, and India each have a parliament, and the United States has a Congress, just as Imaginaria will have a legislature! The citizens of a country must decide what portion of the representatives each group, such as a state or province or even a political party, will have. A larger portion of representatives means greater influence over policy. Ratio of U.S. Representatives to State Population contains a list of the five U.S. states with the greatest number of representatives in the U.S. House of Representatives, along with the population of that state in 2021. Use the information in the table to answer the questions. State Representative Seats State Population (CA) California 53 39,613,000 (TX) Texas 36 29,730,300 (NY) New York 27 19,300,000 (FL) Florida 27 21,944,600 (PA) Pennsylvania 18 12,804,100 The First through Fifth Ranked States by Number of Representatives (sources: https://www.census.gov/popclock/ [state population], https://www.britannica.com/topic/United-States-House-of-Representatives-Seats-by-State-1787120 [representative seats]) What is the ratio of State Population to Representative Seats for each state to the nearest hundred thousand? What is the ratio of Representative Seats to State Population for each state rounded to seven decimal places? What is the ratio of Representative Seats to State Population for each state rounded to six decimal places? Does there appear to be a constant ratio? Justify your answer. CA 700,000; TX 800,000; NY 700,000; FL 800,000; PA 700,000 CA 0.0000013; TX 0.0000012; NY 0.0000014; FL 0.0000012; PA 0.0000014 CA 0.000001; TX 0.000001; NY 0.000001; FL 0.000001; PA 0.000001 The ratio of State Population to Representative Seats seems to be either 700,000 or 800,000. There does appear to be a constant ratio of about 0.000001 of Representative Seats to State Population if we round off to the sixth decimal place. Math Antics – Rounding You might be wondering why the ratio doesn't appear to be quite the same depending on the rounding of the values. We will see that the key to this variation lies in the fractions. Just like the five children sharing 16 Ring Pops™, there are going to be leftovers and there are many methods for deciding what to do with those leftovers. The Standard Divisor There are two houses of congress in the United States: the Senate and the House of Representatives. Each state has two senators, but the number of representatives depends on the population of the state. The number of representative seats in the U.S. House of Representatives is currently set by law to be 435. In order to distribute the seats fairly to each state, the ratio of the population of the U.S. to the number of representative seats must be calculated. The ratio of the total population to the house size is called the standard divisor , and it is the number of members of the total population represented by one seat. Although apportionment applies to many other scenarios, such as the pencil distribution during the SAT, the terminology of apportionment is based on the House of Representatives scenario. Thus, several government-related terms take on a more general meaning. The states are the recipients of the apportioned resource, the seats are the units of the resource being apportioned, the house size is the total number of seats to be apportioned, the state population is the measurement of the state's size, and the total population is the sum of the state populations. Standard Divisor = Total Population House Size The Standard Divisor of the U.S. House of Representatives 2021 As of this writing, the Census.gov website U.S. Population clock showed a population of 332,693,997. There are 435 seats in the U.S. House of Representatives. Find the standard divisor rounded to the nearest tenth. Dividing 330,147,881 people by 435 seats, there are 758,960.6 people per representative. Whether the standard divisor is less than, equal to, or greater than 1 depends on the ratio of the population to the number of seats. The standard divisor will be equal to 1 if the total population is equal to the number of seats. This would mean that each member of the population is allocated their own personal seat. The standard divisor will be a number between 0 and 1 when the total population is less than the number of seats. This means that each member of the population is allocated more than one seat. The standard divisor will be a number greater than 1when the total population is greater than the number of seats. This means that a certain number of members of the population will share 1 seat. If the population is five children and the house consists of five pieces of candy, the standard divisor is 5 children 5 candies = 1 child per candy meaning each child gets one candy. If the population is five children and ten pieces of candy, the standard divisor is 5 children 10 candies = 0.5 child per candy meaning that each child gets more than one candy. If the population is five children and four pieces of candy, the standard divisor is 5 children 4 candies = 1.25 child per candy meaning that each child gets less than one candy. If the seats in the Imaginarian legislature are distributed to the states based on population, then the house size will be less than the population and we should expect the standard divisor to be a number greater than 1. School Resource Officers in Brevard County, Florida The public schools in a certain county have been allotted 349 school resource officers to be distributed among 327 public schools attended by approximately 271,500 students. Identify the states, seats, house size, state population, and total population in this apportionment scenario. Describe the ratio the standard divisor represents in this scenario and calculate the standard divisor to the nearest tenth. The states are the schools in that county. The seats are the school resource officers. The house size is the number of school resource officers, which is 349. The state population is the number of students in a particular school, which was not given. The total population consists of the sum of the school populations, which is 271,500. The standard divisor is the ratio of the total population to the house size, which is the number of students served by each resource officer. Divide 271 , 500 students ÷ 349 officers = 777.9 students per officer. The Standard Quota Once the standard divisor for the Imaginarian legislature is calculated, the next task is to determine the number of seats that each state should receive, which is referred to as the state’s standard quota . Unless all the states have the same population, each state will receive a different number of seats because the quantities will be proportionate to the state populations. To determine those amounts, we will use an idea we learned earlier. Recall that, when the number of units of item A is proportionate to the number of units of item B , we have: units of A = units of B ratio of B 's to A 's In this case, we are trying to calculate the number of seats a state should be apportioned, the state’s standard quota. So A So A would refer to seats allocated to a particular state, while B would refer to the state population. This means that the ratio of B to A is the ratio of the total population to house size, which is the standard divisor. So in apportionment terms, we have the following formula. State's Standard Quota = State Population Standard Divisor seats The Standard Quota of the U.S. House of Representatives 2021 Example 27 outlined that the Census.gov website U.S. Population clock showed a population of 330,147,881, there are 435 seats in the U.S. House of Representatives, and the standard divisor was 758,960.6 people per representative. The state of California has a population of approximately 39,613,000. Use these values to determine the standard quota for California to two decimal places. California's standard Quota = California's population Standard Divisor = 39 , 613 , 000 758 , 960.6 = 52.19 representatives Apportionment of Laptops in a Science Department The science department of a high school has received a grant for 34 laptops. They plan to apportion them among their six classrooms based on each classroom’s student capacity. Use the values in the table below to find the standard quota for each classroom. Room Students A 30 B 25 C 28 D 32 E 24 F 27 Step 1: Identify the state population, total population, and the house size. The states are the classrooms, and the state populations are listed in the table. The total population is the sum of the state populations, which is 166. The house size is the number of seats, or laptops, to be allocated, which is 34. Step 2: Calculate the standard divisor by dividing the total population by the house size. Standard Divisor = Total of Room Capacities Number of Laptops = 166 34 ≈ 4.88 students per laptop. Step 3: Calculate the standard quota by dividing the state population by the standard divisors , as shown in the table below. Room's Standard Quota = Room Capacity Standard Divisor Room Room Capacity Room’s Standard Quota A 30 30 ÷ 4.88 ≈ 6.15 laptops B 25 25 ÷ 4.88 ≈ 5.12 laptops C 28 28 ÷ 4.88 ≈ 5.74 laptops D 32 32 ÷ 4.88 ≈ 6.56 laptops E 24 24 ÷ 4.88 ≈ 4.92 laptops F 27 27 ÷ 4.88 ≈ 5.53 laptops Step 4: Find the sum of the standard quotas. 6.15 + 5.12 + 5.74 + 6.56 + 4.91 + 5.53 = 34.01 . This is only slightly off from the number of laptops—34—which can be caused by rounding off in previous steps. This is a good indication that the calculations were correct. If you find that the value of the sum of the standard quotas is significantly different from the house size (number of seats), it is possible that the standard divisor was calculated using too few decimal places. Calculate the standard divisor and standard quotas again but round off to a greater number of decimal places. Check Your Understanding Key Terms apportion apportionment problem proportional part-to-part ratio representative democracies standard divisor states seats house size state population total population standard quota Key Concepts The apportionment problem is how to fairly divide and distribute available resources to recipients in whole, not fractional, parts. To distribute the seats in the U.S. House of Representatives fairly to each state, calculations are based on state population, total population, and house size, or the total number of seats to be apportioned. The standard divisor is the ratio of the total population to the house size, and the standard quota is the number of seats that each state should receive. Formulas Let A be a particular item and B another such that there is a constant ratio of A to B . ratio of B 's to A 's = 1 ratio of A 's to B 's and ratio of A 's to B 's = 1 ratio of B 's to A 's units of A = ( units of B ) × ( ratio of A 's to B 's ) = units of B ratio of B 's to A 's units of B = ( units of A ) × ( ratio of B 's to A 's ) = units of A ratio of A 's to B 's Standard Divisor = Total Population House Size State's Standard Quota = State Population Standard Divisor seats Videos What Is a Ratio? What Are the Different Types of Ratios? Math Antics – Rounding", "section": "Standard Divisors, Standard Quotas, and the Apportionment Problem", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Apportionment Methods Schoolchildren depend on apportionment of resources like laptops among schools and classroom. (credit: “Richmond Public Schools” by Virginia Department of Education/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and interpret the apportionment problem. Apply Hamilton’s Method. Describe and interpret the quota rule. Apply Jefferson’s Method. Apply Adams’s Method. Apply Webster’s Method. Compare and contrast apportionment methods. Identify and contrast flaws in various apportionment methods. A Closer Look at the Apportionment Problem In Standard Divisors, Standard Quotas, and the Apportionment Problem we calculated the standard divisor and the standard quotas in various apportionment scenarios. The results of those calculations routinely led to fractions and decimals of units. However, the seats in the House of Representatives, laptops in a classroom, or a variety of other resources, are indivisible, meaning they cannot be divided up into fractional parts. This leaves a decision to be made. For example, if the standard quota for the number of laptops to be distributed to a classroom is 12.44 units, how do we deal with the fractional part of 0.44? It is unclear if the classroom should receive 12 units, 13 units, or some other value. Let’s try traditional rounding to the nearest whole number value. Installing Emergency Lights The board of trustees of a college has recently approved the installation of 70 new emergency blue lights in three parking lots. The number of lights in each lot will be proportionate to the size of the parking lot, which is to be measured in acres. The total number of acres is 34; so the standard divisor is 34 70 ≈ 0.4857 . The standard quota for each lot is listed in the table below. Use this information to answer each question. Lot Acres Lot’s Standard Quota A 15 15 ÷ 0.4857 ≈ 30.88 emergency blue lights B 9 9 ÷ 0.4857 ≈ 18.53 emergency blue lights C 10 10 ÷ 0.4857 = 20.59 emergency blue lights Use traditional rounding to determine the number of lights assigned to each lot. Find the sum of the values from part 1. Does the sum found in part 2 equal the number of lights available? If traditional rounding is used, there will be 31, 19, and 21 lights distributed to each lot, respectively. The total of these values is 71. No, the total from part 2 is one more than the number of lights available. In other words, one of the parking lots must get 1 fewer light than apportioned. demonstrates that we cannot successfully apportion indivisible resources by rounding off each standard quota using traditional rounding. This leaves us with a problem. What is a fair way to distribute the fractional parts of the standard quotas? We will refer to this as the apportionment problem . Several methods for making this decision will be discussed. A-10C Thunderbolt II Aircraft In 2015, the U.S. Air Force had a fleet of approximately 281 A-10C Thunderbolt II aircraft. Suppose that the Air Force administration wanted to distribute 27 aircrafts across six bases based on the number of qualified pilots stationed at those bases. Use the information in the table below to answer each question. Base Pilots (A) Alpha 13 (B) Bravo 12 (C) Charlie 5 (D) Delta 16 (E) Echo 7 (F) Foxtrot 9 Identify the states, the seats, and the state population (the basis for the apportionment) in this scenario. Find the standard divisor for the apportionment of the aircraft. Round to four decimal places as needed. Include the units. Find each Air Force base’s standard quota for the apportionment of the aircraft. Round to the nearest hundredth as needed. What are the units? How does this example demonstrate the apportionment problem? Will traditional rounding solve the problem? The states are the bases, the seats are the aircraft, and the state populations are the pilots at a given base. Standard Divisor = Total Population House Size = 13 + 12 + 5 + 16 + 7 + 9 27 = 62 27 ≈ 2.2963 pilots per aircraft. A 13 2.2963 ≈ 5.66 , B 12 2.2963 ≈ 5.23 , C 5 2.2963 ≈ 2.18 , D 16 2.2963 ≈ 6.97 , E 7 2.2963 ≈ 3.05 , F 9 2.2963 ≈ 3.92 . The units are aircraft. This example demonstrates the apportionment problem because it is not possible to send a fractional number of aircraft to an Air Force base. On the other hand, if we use traditional rounding methods to get whole numbers, the results are 5 + 5 + 2 + 7 + 3 + 4 = 26 aircraft will be apportioned, which is one less than the number of aircraft that were supposed to be apportioned. Hamilton's Method of Apportionment One of the problems encountered when standard quotas are transformed into whole numbers using traditional rounding is that it is possible for the sum of the values to be greater than the number of seats available. A reasonable way to avoid this is to always round down, even when the first decimal place is five or greater. For example, a standard quota of 12.33 and a standard quota of 12.99 would both round down to 12. This is called the lower quota . Lower Quota for Apportionment of Aircraft The Air Force administration wants to distribute 27 aircrafts across six bases based on the number of qualified pilots stationed at those bases. The standard quotas for each base are listed in the table below. Use this information to answer the questions. Base Standard Quota (A) Alpha 13 2.2963 ≈ 5.66 aircraft (B) Bravo 12 2.2963 ≈ 5.23 aircraft (C) Charlie 5 2.2963 ≈ 2.18 aircraft (D) Delta 16 2.2963 ≈ 6.97 aircraft (E) Echo 7 2.2963 ≈ 3.05 aircraft (F) Foxtrot 9 2.2963 ≈ 3.92 aircraft Give the lower quota for each Air Force base. Find the sum of the lower quotas. By how much does this sum fall short of the actual number of aircraft? Round down. The lower quota for each Air Force base is 5, 5, 2, 6, 3, 3, respectively. The sum is 24. This is 3 fewer than the actual number of aircraft. If the standard quotas are all rounded down, their sum will always be less than or equal to the house size. Then, it would only remain to find a fair way to distribute any remaining seats. Alexander Hamilton, who was a general in the American Revolution, author of the Federalist Papers, and the first secretary of the treasury, took this approach to apportionment. Steps for Hamilton’s Method of Apportionment There are five steps we follow when applying Hamilton’s Method of apportionment: Find the standard divisor. Find each state’s standard quota. Give each state the state’s lower quota (with each state receiving at least 1 seat). Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain. Check the solution by confirming that the sum of the modified quotas equals the house size. Hamilton Method of Apportionment Hawaiian School Districts Suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of the five counties based on the residents under 19 years old, This data is provided in the table below. Using the Hamilton method, calculate how many schools would be funded in each state. Hawaii Honolulu Kalawao Kauai Maui Total Residents under age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: Calculate the standard divisor. Divide the total population, 325,570, by the house size, 616 seats. The standard divisor is 528.52. Step 2: Find each state’s standard quota: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Step 3: Find each state’s lower quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Lower Quota 87 424 1 31 72 615 Step 4: Compare the sum of the states’ lower quotas, 615, to the house size, 616. One seat remains to be apportioned and must be given to the state with the largest fractional part: Maui with 0.75. So, the final Hamilton quotas are as follows: Hawaii 87, Honolulu 424, Kalawao 1, Kauai 31, and Maui 73. Step 5: Find the total to confirm the sum of the quotas equals the house size, 616. Then 87 + 424 + 1 + 31 + 73 = 616 . The apportionment is complete. Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Hamilton apportionment calculator. This can be a useful tool to confirm your results! The Quota Rule A characteristic of an apportionment that is considered favorable is when the final quota values all either result from rounding down or rounding up from the standard quotas. The value that results from rounding down is called the lower quota, and the value that results from rounding up is called the upper quota . As we explore more methods of apportionment, we will consider whether they satisfy the quota rule. If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule. Which Apportionment Method Satisfies the Quota Rule? Several apportionment methods have been used to allocate 125 seats to ten states and the results are shown in the table below. Determine which apportionments do not satisfy the quota rule and justify your answer. State A State B State C State D State E State F State G Standard Quota 41.26 16.00 5.77 2.64 7.82 10.47 0.21 Lower Quota 41 16 5 2 7 10 0 Upper Quota 42 17 6 3 8 11 1 Method X 43 16 5 2 7 10 1 Method Y 41 16 6 2 8 10 1 Method Z 42 16 7 3 7 9 1 Look for states such that the number of seats allocated differs from the lower or upper quota. Method X violates the quota rule because State A receives 43 seats instead of 41 or 42. Method Z violates the quota rule because State C receives 7 seats instead of 5 or 6 and State F receives 9 instead of 10 or 11. It is possible for an apportionment method to satisfy the quota rule in some scenarios but violate it in others. However, because the Hamilton method always begins with the lower quota and either adds one to it or keeps it the same, the final Hamilton quota will always consist of values that are either lower quota values or upper quota values. When an apportionment method has this characteristic, it is said to satisfy the quota rule. So, we can say: The Hamilton method of apportionment satisfies the quota rule. Although the Hamilton method of apportionment satisfies the quota rule, it can result in some unexpected outcomes, which has caused it to pass in and out of favor of the U.S. government over the years. There are several apportionment methods that have been popular alternatives, such as Jefferson’s method of apportionment that the founders of Imaginaria should consider. Jefferson’s Method of Apportionment Another approach to dealing with the fractional parts of the standard quotas is to modify the standard divisor so that the total of the resulting modified lower quotas is the necessary number of seats. This is the approach used by Jefferson. In Jefferson’s method, the change to the standard divisor is made so that the total of the modified lower quotas equals the house size. The change in the standard divisor to get the modified divisor is relatively small. There is not a formula for this. The modified divisor is found by “guess and check.” It is important to remember that increasing the divisor decreases the quotas, but decreasing the divisor increases the quotas. So, if you need a larger quota, try reducing the divisor, and if you need a smaller quota, try increasing the divisor. Modifying a Standard Divisor Suppose the population of a state is 50 and the standard divisor is 12.5. Find the state’s standard quota. Increase the standard divisor by 2 units and use the modified divisor to determine the modified quota for the state. Decrease the modified divisor from part 2 by 1.5 units and use the new modified divisor to determine the modified quota for the state. Choose any value of divisor between the value of the modified divisor from part 2 and the value of the modified divisor from part 3 and use it to determine the modified quota for the state. Which modified quota was the largest, the modified quota from part 2, from part 3, or from part 4? Explain why. The state’s standard quota is 50 12.5 = 4 . The modified divisor is 14.5. The modified quota is 50 14.5 ≈ 3.45 . The modified divisor is 13. The modified quota is 50 13 ≈ 3.85 . One value between 13 and 14.5 is 13.5. With a modified divisor of 13.5, the modified quota is 50 13.5 ≈ 3.70 . The modified quota from part 3 was the largest because the divisor was the smallest of the three. Dividing the same number by a smaller value gives a larger result. When you use Jefferson’s method, you might have to adjust the divisor several times find modified lower quotas that sum to the house size. First, guess what the divisor should be based on the sum of the lower quotas and then increase or decrease it from there based on whether the sum needs to be smaller or larger respectively. If the result still does not produce lower quotas that sum to the house size, adjust again. Keep a record of the values that didn't work to help you narrow your search. Steps for Jefferson’s Method of Apportionment We take four steps to apply Jefferson’s Method of apportionment: Step 1: Find the standard divisor. Step 2: Find each state’s quota. This will be the standard quota the first time Step 2 is completed and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Step 3: Find the states’ lower quotas (with each state receiving at least one seat), and their sum. Step 4: If the sum from Step 3 equals the number of seats, the apportionment is complete. If the sum of the lower quotas is less than the number of seats, reduce the standard divisor. If the sum of the lower quotas is greater than the number of seats, increase the standard divisor. Return to Step 2 using the modified divisor. Hawaiian State Representative Districts Suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of five counties based on the residents under the age of 19. With the data in the table below, apply Jefferson’s method to apportion the schools to the counties. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: The process for finding the standard divisor, standard quotas, and lower quotas is the same in the Hamilton and Jefferson methods of apportionment. We walked through the Hamilton Method in , and following these steps resulted in lower quotas as shown in the table below. Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Lower Quota 87 424 1 31 72 615 Step 2: Compare the sum of the states’ lower quotas, 615, to the house size, 616. Since 615 is less than 616, use a modified divisor that is less than the standard divisor of 528.52. Try 526.00. Step 3: Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor of 526: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 526.00 ≈ 88.04 224 , 230 526.00 ≈ 426.29 20 526.00 ≈ 0.04 16 , 560 526.00 ≈ 31.48 38 , 450 526.00 ≈ 72.75 616 Lower Quota 88 426 1 31 72 618 Step 4: The new sum of the lower quotas is 2 units greater than 616. We have overshot the goal. So, increase the divisor to a value between 526.00 and 528.52. Try 527.00. Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor of 526.00: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 527.00 ≈ 87.87 224 , 230 527.00 ≈ 425.48 20 527.00 ≈ 0.04 16 , 560 527.00 ≈ 31.42 38 , 450 527.00 ≈ 72.96 616 Lower Quota 87 425 1 31 72 616 Step 6: The new sum of the lower quotas equals the house size. The apportionment is complete. The apportionment is: Hawaii County 87, Honolulu County 425, Kalawao County 1, Kauai 31, and Maui 72 schools. When using Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. Notice that, in this apportionment, Mythbury received more than the upper quota. Since this apportionment of representatives to Imaginarian states by Jefferson’s method does not satisfy the quota rule, we say that: Jefferson’s method violates the quota rule. We have discussed two apportionment methods: one that satisfies the quota rule and one that does not. Before you decide which method to use in Imaginaria, there are a couple more options to consider. Jefferson Apportionment Method Apportionment Calculators It is possible to create Excel spreadsheets that complete the calculations necessary to complete a Jefferson Apportionment. In some cases, this work has already been done and posted online. Check out websites such Ms. Hearn Math for a free Jefferson apportionment calculator. This can be a useful tool to confirm your results! Adams’s Method of Apportionment Adams’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did, Adams used the upper quotas. To apply Adams’s Method of apportionment, there are four steps we follow: Find the standard divisor. Find each state’s quota. This will be the standard quota the first time Step 2 is completed, and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Find the states’ upper quotas and their sum. If the sum from Step 3 equals the number of seats, the apportionment is complete. If the sum of the upper quotas is less than the number of seats, reduce the standard divisor. If the sum of the upper quotas is greater than the number of seats, increase the standard divisor. Return to Step 2 using the modified divisor. Hawaiian School Districts As in earlier examples, suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of the five counties based on the residents under the age of 19. Use the data in the following table and the Adams method to apportion the schools to the counties. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: The steps of finding the standard divisor and each state’s quota are the same in the Jefferson and Adams methods. As in , the standard divisor is 528.52. Step 2: Find each state’s upper quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Upper Quota 88 425 1 32 73 619 Step 3: Compare the sum of the states’ upper quotas, 619, to the house size, 616. Since 619 is greater than 616, we need to reduce the size of the quotas. Use a modified divisor that is greater than the standard divisor of 528.52. Try 534.00. Step 4: Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor of 534: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 534.00 ≈ 86.72 224 , 230 534.00 ≈ 419.91 20 534.00 ≈ 0.04 16 , 560 534.00 ≈ 31.01 38 , 450 534.00 ≈ 72.00 616 Upper Quota 88 420 1 32 72 613 Step 5: The new sum of the upper quotas is 3 units less than 616. Larger quotas are needed. So, decrease the divisor to a value between 534.00 and 528.52. Try 532.00. Step 6: Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor of 532.00: County Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 532.00 ≈ 87.05 224 , 230 532.00 ≈ 421.48 20 532.00 ≈ 0.04 16 , 560 532.00 ≈ 31.13 38 , 450 532.00 ≈ 72.27 616 Upper Quota 88 422 1 32 73 616 Step 7: The new sum of the upper quotas equals the house size. The apportionment is complete. The apportionment is Hawaii County 88, Honolulu County 422, Kalawao County 1, Kauai 32, and Maui 73 schools. When using Adams’s method, just as with Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. In this apportionment, Mythbury received less than the state’s lower quota. So, this apportionment is an example of a scenario in which the Adams’s method violates the quota rule. Adams’s method of apportionment violates the quota rule. So far, only Hamilton’s method satisfies the quota rule, but there is one more apportionment method you should consider for Imaginaria. Adams Method Apportionment Calculator Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Adams Method apportionment calculator. This can be a useful tool to confirm your results! Webster’s Method of Apportionment Webster’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did or the sum of the upper quotas as Adams did, Webster used traditional rounding. To apply Webster’s method of apportionment, there are four steps we take: Find the standard divisor. Find each state’s quota. This will be the standard quota the first time Step 2 is completed, and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Round each state’s quota to the nearest whole number and find the sum of these values. If the sum of the rounded quotas equals the number of seats, the apportionment is complete. If the sum of the rounded quotas is less than the number of seats, reduce the divisor. If the sum of the rounded quotas is greater than the number of seats, increase the divisor. Return to Step 2 using the modified divisor. When using Webster’s method, just as with Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. Hawaiian School Districts Use the data in the table below to apportion 616 schools to Hawaiian counties. This time, use Webster’s method. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 To apply Webster’s method of apportionment, there are four steps we take: Step 1: The processes of finding the standard divisor and standard quota are the same in the Jefferson, Adams, and Webster’s methods. As in the previous examples, the standard divisor is 528.52. Step 2: Find each state’s rounded quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Rounded Quota 88 424 1 31 73 617 Step 3: Compare the sum of the states’ rounded quotas, 617, to the house size, 616. Since 617 is greater than 616, we need to reduce the size of the quotas. Use a modified divisor that is greater than the standard divisor of 528.52. Try 534.00. Step 4: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 534: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 534.00 ≈ 86.72 224 , 230 534.00 ≈ 419.91 20 534.00 ≈ 0.04 16 , 560 534.00 ≈ 31.01 38 , 450 534.00 ≈ 72.00 616 Upper Quota 87 420 1 31 72 612 Step 5: The new sum of the rounded quotas is 4 units less than 616. Larger quotas are needed. So, decrease the divisor to a value between 534.00 and 528.52. Try 530.00. Step 6: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 530.00: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 530.00 ≈ 87.38 224 , 230 530.00 ≈ 423.08 20 530.00 ≈ 0.04 16 , 560 530.00 ≈ 31.25 38 , 450 530.00 ≈ 72.55 616 Upper Quota 87 423 1 31 73 615 Step 7: The new sum of the rounded quotas is 1 unit less than 616. Larger quotas are needed. So, decrease the divisor to a value between 528.52 and 530.00. Try 529.50. Step 8: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 529.50: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 529.50 ≈ 87.46 224 , 230 529.50 ≈ 423.48 20 529.50 ≈ 0.04 16 , 560 529.50 ≈ 31.27 38 , 450 529.50 ≈ 72.62 616 Upper Quota 87 423 1 31 73 615 Step 9: The new sum is still only 1 unit less than 616. Larger quotas are needed, but not much larger. So, decrease the divisor to a value between 528.52 and 529.50. Try 529.30. Step 10: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 529.30: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 529.30 ≈ 87.49 224 , 230 529.30 ≈ 423.63 20 529.30 ≈ 0.04 16 , 560 529.30 ≈ 31.29 38 , 450 529.30 ≈ 72.64 616 Upper Quota 87 424 1 31 73 616 Step 11: The new sum of the rounded quotas equals the house size. The apportionment is complete. The apportionment is Hawaii County 87, Honolulu County 424, Kalawao County 1, Kauai 31, and Maui 73 schools. So far, we know that the Hamilton method satisfies the quota rule, while the Jefferson and Adams methods do not. The apportionments in the Example and Your Turn above are both scenarios in which the Webster method satisfies the quota rule. Does it always? We have a little more work to do to find out. However, one thing is clear. Not all apportionment methods have the same results. Before you make such an important decision for Imaginaria, it’s important to think about the differences in the apportionments that result from these four methods. How will the differences affect the citizens of Imaginaria? Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Webster Method apportionment calculator. This can be a useful tool to confirm your results! Comparing Apportionment Methods Recall that the four apportionment methods discussed in this chapter differ in two main ways: Whether or not a modified divisor is used The type of rounding of the quotas that is used How might these differences affect Imaginarians? In the next two examples, we will compare the results when different apportionment methods are applied to the same scenario. Hawaiian School Districts with Different Apportionment Methods Let’s use the results from , , , and to compare the four apportionment methods we have discussed. The following table summarizes the results of the results of the Hamilton, Jefferson, Adams and Webster methods when applied to the apportionment of 616 schools to Hawaiian counties. Hawaii Honolulu Kalawao Kauai Maui Under 19 years old 46,310 224,230 20 16,560 38,450 Hamilton 87 424 1 31 73 Jefferson 87 425 1 31 72 Adams 88 422 1 32 73 Webster 87 424 1 31 73 Do any of the apportionment methods result in the same apportionment? If so, which ones? Which apportionment method would the citizens of the largest county likely favor most and least? Justify your answer. As a group, which apportionment method would the citizens of the other four counties likely favor most and least? Justify your answer. Yes, the Hamilton and Webster methods result in the same apportionment. The largest county is Honolulu. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams method of apportionment least because they received the least number of seats by that method. As a group, the other four counties received 192 seats by either the Hamilton or Webster method, 194 seats by the Adams method, and 191 seats by the Jefferson method. They would likely favor the Adams method the most and favor the Jefferson methods the least. The Adams method favored the smaller states and the Jefferson method favored the larger states in the previous example, but is this the case in general? Since the Jefferson method begins with the lower quotas, any adjustment to the quotas will be an increase. As you have seen, this is accomplished by using a modified divisor that is smaller than the standard divisor. The next example compares the impact of a decreasing divisor on the modified quotas of large states to the impact of the same size decrease on small states. Effect of Decreasing Divisors on Modified Quotas The following table displays the effect of reducing the size of the divisor. Observe the effect this has on the modified quotas of smaller states versus larger states and use the table answer each question. Modified Quotas State Population Divisor: 10,500 Divisor: 10,000 Divisor: 9,500 A 10,000 0.95 1 1.05 B 100,000 9.52 10 10.53 C 1,000,000 95.24 100 105.26 When the divisor decreases from 10,500 to 10,000, how many representatives are gained by each state based on the lower quota? When the divisor decreases from 10,000 to 9,500, how many representatives are gained by each state based on the lower quota? Which state gains the most representatives each time the divisor is decreased? Since a state must have at least one seat, State A begins with 1 seat and still has one seat. State B begins with 9 seats and increases to 10 seats. State C begins with 95 seats and increases to 100 seats. So, State A gains 0, B gains 1, and C gains 5 seats. State A begins with 1 and still has 1. State B begins with 10 and still has 10. State C begins with 100 and increases to 105. So, State A gains 0, State B gains 0, and State C gains 5. State C, the largest state, gains the most representatives each time the divisor is decreased. This example demonstrates that the Jefferson method is biased toward states with larger populations because the modified divisor is smaller than the standard divisor. On the other hand, the Adams’s method, which begins with the upper quotas, must increase the standard divisor in order to reduce the quotas. Once again, the effect on the number of seats is greater for the larger states, but this time they are decreased. This means that the Adams’s method favors states with smaller populations. Flaws in Apportionment Methods As we have seen, different apportionment methods can have the same results in some scenarios but different results in others. Citizens of states which receive fewer seats with a particular apportionment method will view the apportionment method as flawed and argue in favor of a different method. This inevitably creates debates regarding the use of one method over another. Methods that favor larger states are likely to be challenged by smaller states, methods that favor smaller states are likely to be challenged by larger states, and methods that violate the quota rule are likely to be challenged by states of any size depending on the circumstances. Suppose that the State of Hawaii House of Representatives had 51 representatives, each with their own district. Imagine that redistricting were underway, and the representative districts were to be apportioned to each of five counties based on population. The following table shows the apportionment that would result from the use of the Jefferson, Adams, and Webster methods of apportionment. Hawaii Honolulu Kalawao Kauai Maui Population 201,500 974,600 100 72,300 167,400 Lower Quota 7 35 0 2 6 Upper Quota 8 36 1 3 7 Jefferson 7 35 1 2 6 Adams 7 34 1 3 6 Webster 7 34 1 3 6 From the table, you can see that Hawaii, Kalawao, and Maui receive the same number of seats regardless of the method used. However, citizens of Honolulu would likely reject the Adams and Webster methods arguing that they violate the quota rule. Similarly, citizens of Kauai would probably reject the Jefferson method based on the argument that it unfairly favors the larger states. This scenario demonstrates that the Adams and Webster methods violate the quota rule, but the Jefferson method also violates the quota rule at times. The Hamilton method is the only method that satisfies the quota rule in all scenarios. It also consistently favors neither larger nor smaller states. Unfortunately, it can have some strange and results in certain circumstances, which you will see in the next section. Gerrymandering: A Subtle Way to Impact Apportionment In addition to your choice of voting method and your choice of apportionment method, there is another important decision to make which could potentially have a huge impact on the fairness of elections in Imaginaria—the creation of electoral districts. In example above, we imagined that there were 51 state legislators in Hawaii, each representing their own district. But how did the legislators decide on the boundaries for these districts? Typically, boundaries are drawn so that each district has approximately the same number of residents, but the percentage of residents in each district with a particular political affiliation can swing the power from one group to another. When the districts are drawn to impact the power of a political party, ethnic or racial group, or other group, this is called gerrymandering. For example, districts can be drawn so that a particular group is spread thinly across districts, increasing the likelihood that they will not have strong representation. This cartoon map conveys the idea that the drawing of the map may impact election outcomes. (credit: “The Gerry Mander”/Wikimedia Commons, Public Domain) \"The Gerry-mander\" first appeared in this cartoon-map in the Boston Gazette, March 26, 1812, and was soon reproduced in several other Massachusetts newspapers in response to election district changes initiated by governor Eldridge Gerry. Note that while the practice is named after him, Gerry was not the first to employ it. Jonathan Mattingly Jonathan Mattingly is a mathematician who was featured in a Nature article titled “The Mathematicians Who Want to Save Democracy” . Mattingly is a mathematician at Duke University in North Carolina and he runs election simulations based on alternate versions of electoral districts in order to analyze the effects of gerrymandering. He has even been asked to testify as an expert witness in court. Mattingly and other mathematicians who are working on the problem will potentially have an impact on the redistricting that will occur as a result of the 2020 census. (Carrie Arnold, “The Mathematicians Who Want to Save Democracy,” Nature 546 , 200–202, 2017.) Check Your Understanding Key Terms apportionment problem lower quota upper quota Key Concepts Hamilton’s method of apportionment uses the standard divisor and standard lower quotas, and it distributes any remaining seats based on the size of the fractional parts of the standard lower quota. Hamilton’s method satisfies the quota rule and favors neither larger nor smaller states. Jefferson’s method of apportionment uses a modified divisor that is adjusted so that the modified lower quotas sum to the house size. Jefferson’s method violates the quota rule and favors larger states. Adams’s method of apportionment uses a modified divisor that is adjusted so that the modified upper quotas, sum to the house size. Adams’s method violates the quota rule and favors smaller states. Webster’s method of apportionment uses a modified divisor that is adjusted so that the modified state quotas, rounded using traditional rounding, sum to the house size. Webster’s method violates the quota rule but favors neither larger nor smaller states. Videos Hamilton Method of Apportionment Jefferson Apportionment Method Adams Method Apportionment Calculator", "section": "Apportionment Methods", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Fairness in Apportionment Methods In this seating chart for the House of Representatives, each color indicates representatives from a particular state. Learning Objectives After completing this section, you should be able to: Describe and illustrate the Alabama paradox. Describe and illustrate the population paradox. Describe and illustrate the new-states paradox. Identify ways to promote fairness in apportionment methods. Apportionment Paradoxes The citizens of Imaginaria will want the apportionment method to be as fair as possible. There are certain characteristics that they would reasonably expect from a fair apportionment. If the house size is increased, the state quotas should all increase or remain the same but never decrease. If one state’s population is growing more rapidly than another state’s population, the faster growing state should not lose a seat while a slower growing state maintains or gains a seat. If there is a fixed number of seats, adding a new state should not cause an existing state to gain seats while others lose them. However, apportionment methods are known to contradict these expectations. Before you decide on the right apportionment for Imaginarians, let’s explore the apportionment paradox , a situation that occurs when an apportionment method produces results that seem to contradict reasonable expectations of fairness. There is a lot that the founders of Imaginaria can learn from U.S. history. The constitution of the United States requires that the seats in the House of Representatives be apportioned according to the results of the census that occurs every decade, but the number of seats and the apportionment method is not stipulated. Over the years, several different apportionment methods and house sizes have been used and scrutinized for fairness. This scrutiny has led to the discovery of several of these apportionment paradoxes. The Alabama Paradox At the time of the 1880 U.S. Census, the Hamilton method of apportionment had replaced the Jefferson method. The number of seats in the House of Representatives was not fixed. To achieve the fairest apportionment possible, the house sizes were chosen so that the Hamilton and Webster methods would result in the same apportionment. The chief clerk of the Census Bureau calculated the apportionments for house sizes between 275 and 350. There was a surprising result that became known as the Alabama paradox , which is said to occur when an increase in house size reduces a state’s quota. Alabama would receive eight seats with a house size of 299, but only receive seven seats if the house size increased to 300. (Michael J. Caulfield (Gannon University), \"Apportioning Representatives in the United States Congress - Paradoxes of Apportionment,\" Convergence (November 2010), DOI:10.4169/loci003163) The 1880 Alabama Quota The 1880 census recorded the population of Alabama as 1,513,401 and that of the U.S. as 62,979,766. Calculate the standard divisor and standard quota for the State of Alabama based on a house size of 299. Calculate the standard divisor and standard quota for the State of Alabama based on a house size of 300. Did the standard quota increase or decrease when the house size increased? Consider the Hamilton method of apportionment. Explain how Alabama’s final quota could be smaller with a larger standard quota. The standard divisor is 62,979,766 ÷ 299 = 210 , resulting in 634.6689 citizens per seat. The standard quota for Alabama is 1,513,401 ÷ 210,634.7 = 7.1850 seats. The standard divisor is 62,979,766 ÷ 300 = 209 , resulting in 932.5533 citizens per seat. The standard quota for Alabama is 1,513,401 ÷ 209,932.6 = 7.2090 seats. The standard quota increased. In each case, the state would receive the lower quota of 7 and then be awarded one more seat if the fractional part of the standard quota were high enough relative to the fractional parts of the other states’ standard quotas. When the house size was 299, Alabama received one of the remaining seats after the lower quotas were distributed. When the house size was 300, Alabama did not receive one of the remaining seats after the lower quotas were distributed. It must have been the case that either the fractional part 0.2090 ranked lower amongst the other fractional parts of the state quotas than the fractional part 0.1850 did, or there were fewer remaining seats, or both. After the 1900 census, the Census Bureau again calculated the apportionment based on various house sizes. It was determined that Colorado would receive three seats with a house size of 356, but only two seats with a house size of 357. Hamilton’s Method and the Alabama Paradox Suppose that States A and B each have a population of 6, while State C has a population of 2. Use the Hamilton method to apportion 10 seats. Use the Hamilton method to apportion 11 seats. Does this example demonstrate the Alabama paradox? If so, how? Step 1: The total population is 14. The standard divisor is 14 ÷ 10 = 1.4 individuals per seat. Step 2: The states’ standard quotas are as follows: A 6 ÷ 1.4 ≈ 4.29 , B 6 ÷ 1.4 ≈ 4.29 , and C 2 ÷ 1.4 ≈ 1.43. Step 3: The states’ lower quotas are as follows: A 4, B 4, and C 1. Step 4: The sum of the lower quotas is 9, which means there is one seat remaining to be apportioned. State C has the highest fractional part and receives the additional seat. Step 5: The final apportionment is as follows: A 4, B 4, and C 2, which sums to 10. Step 1: The total population is 14. The standard divisor is 14 ÷ 11 ≈ 1.2727 individuals per seat. Step 2: The states’ standard quotas are: A 6 ÷ 1.2727 ≈ 4.71 , B 6 ÷ 1.2727 ≈ 4.71 , and C 2 ÷ 1.2727 ≈ 1.57 . Step 3: The states’ lower quotas are: A 4, B 4, and C 1. Step 4: The sum of the lower quotas is 9, which means there are two seats remaining to be apportioned. A and B have the highest fractional parts and receive the additional seats. Step 5: The final apportionment is: A 5, B 5, and C 1. Yes, this demonstrates the Alabama paradox because State C receives two seats if the house size is 10, but only one seat if the house size is 11. The Population Paradox It is important for the founders of Imaginaria to keep in mind that the populations of states change as time passes. Some populations grow and some shrink. Some populations increase by a large amount while others increase by a small amount. These changes may necessitate a reapportionment of seats, or the recalculation of state quotas due to a change in population. It would be reasonable for Imaginarians to expect that the state with a population that has grown more than others will gain a seat before the other states. Once again, this is not always the case with the Hamilton method of apportionment. The population growth rate of a state is the ratio of the change in the population to the original size of the population, often expressed as a percentage. This value is positive if the population is increasing and negative if the population is decreasing. The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats. Notice that the population paradox definitions has two parts. If either part applies, then the population paradox has occurred. The first part of the definition only applies when one state has a decreasing population. The second part of the definition applies in all situations, whether there is a state with a decreasing population or not. It will be easier to identify situations that involve a decreasing population. The other situations requires the calculation of a growth rate. The reason that we don't have to calculate a growth rate when one state has a decreasing population and the other has an increasing population is that increasing population has a positive growth rate which is always greater than the negative growth rate of a decreasing population. A state must lose a seat in order for the population paradox to apply. It is not enough for a state with a lower growth rate to gain a seat while a state with a higher growth rate retains the same number of seats. Apportionment of Respirators to Hospitals Suppose that 18 respirators are to be apportioned to three hospitals based on their capacities. The Hamilton method is used to allocate the respirators in 2020, then to reallocate based on new capacities in 2021. The results are shown in the table below. How does this demonstrate the population paradox? Hospital Capacity in 2020 Respirators in 2020 Change in Capacity Growth Rate = Change in Capacity Capacity in 2020 Capacity in 2021 Respirators in 2021 A 825 9 57 57 825 ≈ 0.0691 = 6.91 % 882 9 B 613 7 13 13 613 ≈ 0.0212 = 2.12 % 626 6 C 239 2 3 3 239 = 0.0126 = 1.26 % 242 3 Hospital B lost a respirator while hospital C gained one, even though hospital B had a higher growth rate than hospital C. The growth rate of a population can be calculated by subtracting the previous population size from the current population size, and then dividing the difference by the previous population size. population growth rate = current population size − previous population size previous population size Make sure to calculate the subtraction before the division. If you are entering the values in a calculator, it is helpful to put parentheses around the subtracted terms. The Congress of Costaguana The country of Costaguana has three states: Azuera with a population of 894,000; Punta Mala with a population of 696,000; and Esmeralda with a population of 215,000. There are 38 seats in the Congress of Costaguana. The apportionment of the seats is determined by Hamilton’s method to be: 19 for Azuera, 15 for Punta Mala, and 4 for Esmerelda. A census reveals that the population has grown and the seats must be reapportioned. If Azuera now has 953,000 residents, Punta Mala now has 706,000 residents, and Esmerelda now has 218,000 residents, how many seats will each state receive upon reapportionment? How is this an example of the population paradox? The Hamilton reapportionment is: 19 for Azuera, 14 for Punta Mala, and 5 for Esmerelda. This is an example of the population paradox because Punta Mala lost a seat to Esmerelda, even though Punta Mala’s population grew by 1.44 percent while Esmerelda’s only grew by 1.40 percent. The New-States Paradox As a founder of Imaginaria, you might also consider the possibility that Imaginaria could annex nearby lands and increase the number of states. This occurred several times in the United States such as when Oklahoma became a state in 1907. The House size was increased from 386 to 391 to accommodate Oklahoma’s quota of five seats. When the seats were reapportioned using Hamilton’s method, New York lost a seat to Maine despite the fact that their populations had not changed. This is an example of the new-state paradox , which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment. New State of Oscuridad The country of San Lorenzo has grown from two states to three. The house size of the congress has been increased by eight and the seats have been reapportioned to accommodate the new state of Oscuridad. The constitution mandates the use of the Hamilton method of apportionment. Use this information and the following table to answer the questions. State Population (in hundreds) Original Apportionment Reapportionment Clara 7,100 39 40 Velasco 9,080 51 50 Oscuridad 1,500 Not Applicable 8 What was the original house size? What is the new house size? How is this reapportionment an example of the new-states paradox? 39 + 51 = 90 40 + 50 + 8 = 98 The original state of Velasco lost a seat to the original state of Clara when the new state of Oscuridad was added. The Growing Country of Gulliversia The country of Gulliversia has two states: Lilliput with a population of 700,000 and Brobdingnag with a population of 937,000. The constitution of Gulliversia requires that the 90 congressional seats be apportioned by Hamilton’s method. Lilliput has received 38 seats while Brobdingnag has received 52 seats. Recently, the island of Houyhnhnmsland with a population of 170,000 has joined the union, becoming a state of Gulliversia. When Houyhnhnmsland is included, nine additional seats must be apportioned to maintain the same ratio of seats to citizens. Use Hamilton’s method to reapportion the 99 seats to the three states. How is the resulting apportionment an example of the new-states paradox? The reapportionment gives 39 seats to Lilliput, 51 seats to Brobdingnag, and 9 seats to Houyhnhnmsland. This is an example of the new-states paradox because the original state of Brobdingnag lost a seat to the original state of Lilliput when the new state was added to the union. When a new state is added, it is necessary to determine the amount that the house size must be increased to retain the original ratio of population to seats, in other words to keep the original standard divisor. To calculate the new house size, divide the new population by the original standard divisor, and round to the nearest whole number. New House Size = New Population Original Standard Divisor rounded to the nearest whole number. Oklahoma Joins the Union Oklahoma was admitted as the 46th state on November 16, 1907. Before Oklahoma joined the union, the U.S. population was approximately 75,030,000 and the House of Representatives had 386 seats. The new state had a population of approximately 970,000. Use this information to estimate the original standard divisor to the nearest hundred, the new population, the new house size, and the number of seats Oklahoma should receive. Step 1: Original Standard Divisor ≈ 75 , 030 , 000 386 ≈ 194 , 400 Step 2: New Population ≈ 75 , 030 , 000 + 970 , 000 = 76 , 000 , 000 Step 3: New House Size ≈ 76,000,000 194 , 400 ≈ 391 Step 4: There are 391 - 386 = 5 new seats to be apportioned to Oklahoma. The Search for the Perfect Apportionment Method The ideal apportionment method would simultaneously satisfy the following four fairness criteria. Satisfy the quota rule Avoid the Alabama paradox Avoid the population paradox Avoid the new-states paradox We have seen that the Hamilton method allows the Alabama paradox, the population paradox, and the new-states paradox in some apportionment scenarios. Let’s explore the results of the other methods of apportionment we have discussed in some of the same scenarios. Orange Grove and the New-States Paradox The incorporated town of Orange Grove consists of two subdivisions: The Oaks with 1,254 residents and The Villages with 10,746 residents. A council with 100 members supervises the municipality’s operations. The council votes to annex an unincorporated subdivision called The Lakes with a population of 630. They plan to increase the size of the council to maintain the ratio of seats to residents such that the new council will have 100 seats plus the number of seats given to The Lakes. Use each of the following apportionment methods and indicated number of additional seats to find the original and new apportionment and determine whether the new-state paradox occurs. Jefferson’s method with five additional seats. Adams’s method with six additional seats. Webster’s method with five additional seats Using a modified divisor of 119, the original apportionment would have been: The Oaks 10 and The Villages 90. Using a modified divisor of 119, the new apportionment would be: The Oaks 10, The Villages 90, and The Lakes 5. The new-state paradox does not occur. Using a modified divisor of 121, the original apportionment would have been: The Oaks 11 and The Villages 89. Using a modified divisor of 121, the new apportionment would be: The Oaks 11, The Villages 89, and The Lakes 6. The new-state paradox does not occur. Using the standard divisor of 120, the original apportionment would have been: The Oaks 10 and The Villages 90. Using a modified divisor of 119.5, the new apportionment would be: The Oaks 10, The Villages 90, and The Lakes 5. The new-state paradox does not occur. We have seen in our examples that neither the population paradox nor the new-states paradox occurred when using the Jefferson, Adams, and Webster methods. It turns out that, although all three of these divisor methods violate the quota rule, none of them ever causes the population paradox, new-states paradox, or even the Alabama paradox. On the other hand, the Hamilton method satisfies the quota rule, but will cause the population paradox, the new-states paradox, and the Alabama paradox in some scenarios. In 1983, mathematicians Michel Balinski and Peyton Young proved that no method of apportionment can simultaneously satisfy all four fairness criteria. There are other apportionment methods that satisfy different subsets of these fairness criteria. For example, the mathematicians, Balinski and Young who proved the Balinski-Young Impossibility Theorem created a method that both satisfies the quota rule and is free of the Alabama paradox. (Balinski, Michel L.; Young, H. Peyton (November 1974). “A New Method for Congressional Apportionment.” Proceedings of the National Academy of Sciences . 71 (11): 4602–4606.) However, no method may always follow the quota rule and simultaneously be free of the population paradox. (Balinski, Michel L.; Young, H. Peyton (September 1980). \"The Theory of Apportionment\" (PDF). Working Papers. International Institute for Applied Systems Analysis. WP-80-131.) So, as you and your fellow founders of Imaginaria make the important decision about the right apportionment method for Imaginaria, do not look for a perfect apportionment method. Instead, look for an apportionment method that best meets the needs and concerns of Imaginarians. Check Your Understanding Key Terms apportionment paradox Alabama paradox reapportionment population growth rate population paradox new-state paradox Key Concepts Several surprising outcomes can occur when apportioning seats that voters may find unfair: Alabama paradox, population paradox, and new-state paradox. Apportionment methods are susceptible to apportionment paradoxes and may violate the quota rule. The Balinsky-Young Impossibility Theorem indicates that no apportionment can satisfy all fairness criteria. Formulas population growth rate = current population size - previous population size previous population size New House Size = New Population Original Standard Divisor rounded to the nearest whole number. Projects The First Census: Challenges of Collecting Population Data As you and your fellow founders write the Imaginarian constitution, you must also design systems to collect accurate information about the population of Imaginaria. Why is this important and what are the challenges you will face? Let’s find out! Research and answer each question. (Questions adapted from “Authorizing the First Census–The Significance of Population Data,” Census.gov, United States Census Bureau) The First Congress laid out a plan for collecting this data in Chapter II, An Act Providing for the Enumeration of the Inhabitants of the United States , which was approved March 1, 1790. What group did Congress select to carry out the first enumeration? Why did they choose this group? What might be the advantages and disadvantages to this approach? Choose at least two challenges faced by the U.S. government during the first enumeration and explain how the information gathered might have helped address them. President James Madison was a U.S. Representative who participated in the census debate in the first congress. What were the main points of his remarks and how were they relevant to the overall debate about the first enumeration? What issues do you think the U.S. Census Bureau encounters today as it continues to collect and process data about the U.S. population that might be significant to you and the other founders of Imaginaria? The Party System: How Many Political Parties Are Enough? The electoral system of Imaginaria will likely involve multiple political parties. The way these parties interact with the system may be determined by the founders of the new democracy. Let’s explore the ways in which political parties interact with the governments of various democracies by researching and answering each of the following questions. What concerns did the founders of the United States have about political parties? Have any of their concerns become a reality? Were political parties addressed in the U.S. Constitution? How did they become such a large part of politics in the U.S.? As a founder of Imaginaria, would you address political parties in your constitution? What is frontloading? Does our current system of frontloading impact fair representation? Why do two small, racially homogeneous states hold their primaries first? Do you think this impacts the final results? How does the interaction political parties and the electoral system in the United States differ from that of other countries? Give at least three specific examples. Of these three, which would you be most likely to use as a model for Imaginaria? Technology and Voting: How Has the Digital Age Impacted Elections? Unlike most other countries, Imaginaria will be founded in the digital age. Let’s explore the impact this might have on how you choose to set up your electoral system. Research and answer each question. Participation in an electoral system if very important. In what ways does the Internet positively affect participation? In what ways does it negatively affect participation. What roll, if any, should government of Imaginaria play in tempering or promoting the Impact of the Internet? Find at least three examples of governments that utilize Internet voting around the world. What concerns have slowed the spread of this technology? What are the advantages and disadvantages of Internet voting? Would you be in favor of Internet voting for Imaginaria? Why or why not? Chapter Review Voting Methods Fairness in Voting Methods Standard Divisors, Standard Quotas, and the Apportionment Problem Apportionment Methods Fairness in Apportionment Methods Chapter Test", "section": "Fairness in Apportionment Methods", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Networks connect cities around the globe. (credit: \"Globalization\" by faith.e.murphy/Flickr, Public Domain) In this chapter, you will learn the fundamental skills needed to work with graphs used in an area of mathematics known as graph theory. You can think of these graphs as a kind of map. Maps have served many purposes over the course of history. You probably use GPS maps to navigate to various destinations. A scientist from ancient Greece named Ptolemy wanted an accurate map of the world to make more accurate astrological predications. In recent years, neurobiologists have mapped the cerebral cortex to better understand the human brain. Social network analysts map online interactions to assist advertisers in reaching target audiences. Like other maps, the graphs you will study in this chapter can serve many purposes, but they do not have a lot of the details you might expect in a map such as size, shape, and distance between objects. All of that is stripped away so that we can focus on one element of maps, the connections between objects.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Graph Basics Cell phone networks connect individuals. (credit: \"Business people using their phones\" by Rawpixel Ltd./Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify parts of a graph. Model applications of graph basics. When you hear the word, graph , what comes to mind? You might think of the xy -coordinate system you learned about earlier in this course, or you might think of the line graphs and bar charts that are used to display data in news reports. The graphs we discuss in this chapter are probably very different from what you think of as a graph. They look like a bunch of dots connected by short line segments. The dots represent a group of objects and the line segments represent the connections, or relationships, between them. The objects might be bus stops, computers, Snapchat accounts, family members, or any other objects that have direct connections to each other. The connections can be physical or virtual, formal or casual, scientific or social. Regardless of the kind of connections, the purpose of the graph is to help us visualize the structure of the entire network to better understand the interactions of the objects within it. Parts of a Graph In a graph , the objects are represented with dots and their connections are represented with lines like those in . displays a simple graph labeled G and a multigraph labeled H . The dots are called vertices ; an individual dot is a vertex , which is one object of a set of objects, some of which may be connected. We often label vertices with letters. For example, Graph G has vertices a, b, c , and d , and Multigraph H has vertices, e, f, g , and h . Each line segment or connection joining two vertices is referred to as an edge . H is considered a multigraph because it has a double edge between f and h , and a double edge between h and g . Another reason H is called a multigraph is that it has a loop connecting vertex e to itself; a loop is an edge that joins a vertex to itself. Loops and double edges are not allowed in a simple graph. To sum up, a simple graph is a collection of vertices and any edges that may connect them, such that every edge connects two vertices with no loops and no two vertices are joined by more than one edge. A multigraph is a graph in which there may be loops or pairs of vertices that are joined by more than one edge. In this chapter, most of our work will be with simple graphs, which we will call graphs for convenience. A Graph and a Multigraph It is not necessary for the edges in a graph to be straight. In fact, you can draw an edge any way you want. In graph theory, the focus is on which vertices are connected, not how the connections are drawn (see ). In a graph, each edge can be named by the two letters of the associated vertices. The four edges in Graph X in are ab, ac, ad , and ae . The order of the letters is not important when you name the edge of a graph. For example, ab refers to the same edge as ba . A graph may have vertices that are not joined to other vertices by edges, such as vertex f in Graph X in , but any edge must have a vertex at each end. Different Representations of the Same Graph Identifying Edges and Vertices Name all the vertices and edges of graph F in . Graph F The vertices are v, w, x, y , and z . The edges are vw, vx, wx, wz, xy , and xz . When listing the vertices and edges in a graph, work in alphabetical order to avoid accidentally listing the same item twice. When you are finished, count the number of vertices or edges you listed and compare that to the number of vertices or edges on the graph to ensure you didn’t miss any. Since the purpose of a graph is to represent the connections between objects, it is very important to know if two vertices share a common edge. The two vertices at either end of a given edge are referred to as neighboring , or adjacent . For example, in , vertices x and w are adjacent, but vertices y and w are not. Identifying Vertices That Are Not Adjacent Name all the pairs of vertices of graph F in that are not adjacent. The pairs of vertices that are not adjacent in graph F are v and y, v and z, w and y , and y and z . Sergey Brin and Laurence Page The “Google boys,” Sergey Brin and Laurence Page, transformed the World Wide Web in 1998 when they used the mathematics of graph theory to create an algorithm called Page Rank, which is known as the Google Search Engine today. The two computer scientists identified webpages as vertices and hyperlinks on those pages as edges because hyperlinks connect one website to the next. The number of edges influences the ranking of a website on the Google Search Engine because the websites with more links to “credible sources” are ranked higher. (\"Page Rank: The Graph Theory-based Backbone of Google,\" September 20, 2011, Cornell University, Networks Blog. Analyzing Geographical Maps with Graphs Commercial airlines' route systems create a global network. When graphs are used to model and analyze real-world applications, the number of edges that meet at a particular vertex is important. For example, a graph may represent the direct flight connections for a particular airport as in . Representing the connections with a graph rather than a map shifts the focus away from the relative positions and toward which airports are connected. In , the vertices are the airports, and the edges are the direct flight paths. The number of flight connections between a particular airport and other South Florida airports is the number of edges meeting at a particular vertex. For example, Key West has direct flights to three of the five airports on the graph. In graph theory terms, we would say that vertex FYW has degree 3. The degree of a vertex is the number of edges that connect to that vertex. Direct Flights between South Florida Airports Determining the Degree of a Vertex Determine the degree of each vertex of Graph J in . If graph J represents direct flights between a set of airports, do any of the airports have direct flights to two or more of the other cities on the graph? Graph J For each vertex, count the number of edges that meet at that vertex. This value is the degree of the vertex. In , the dashed edges indicate the edges that meet at the marked vertex. Degrees of Vertices of Graph J Vertex a has degree 3, vertex b has degree 1, vertices c and d each have degree 2, and vertex e has degree 0. Airports a, c, and d have direct flights to two or more of the other airports. Graphs are also used to analyze regional boundaries. The states of Utah, Colorado, Arizona, and New Mexico all meet at a single point known as the “Four Corners,” which is shown in the map in . Map of the Four Corners In , each vertex represents one of these states, and each edge represents a shared border. States like Utah and New Mexico that meet at only a single point are not considered to have a shared border. By representing this map as a graph, where the connections are shared borders, we shift our perspective from physical attributes such as shape, size and distance, toward the existence of the relationship of having a shared boundary. Graph of the Shared Boundaries of Four Corners States Graphing the Midwestern States A map of the Midwest is given in . Create a graph of the region in which each vertex represents a state and each edge represents a shared border. Map of Midwestern States Step 1: For each state, draw and label a vertex as in . Vertex Assigned to Each Midwestern State Step 2: Draw edges between any two states that share a common land border as in . Edge Assigned to Each Pair of Midwestern States with Common Border The graph is given in . Final Graph Representing Common Boundaries between Midwestern States Graph Theory: Create a Graph to Represent Common Boundaries on a Map Graphs of Social Interactions Geographical maps are just one of many real-world scenarios which graphs can depict. Any scenario in which objects are connected to each other can be represented with a graph, and the connections don’t have to be physical. Just think about all the connections you have to people around the world through social media! Who is in your network of Twitter followers? Whose Snapchat network are you connected to? Graphing Chloe’s Roblox Friends Roblox is an online gaming platform. Chloe is interested to know how many people in her network of Roblox friends are also friends with each other so she polls them. Explain how a graph or multigraph might be drawn to model this scenario by identifying the objects that could be represented by vertices and the connections that could be represented by edges. Indicate whether a graph or a multigraph would be a better model. The objects that are represented with vertices are Roblox friends. A Roblox friendship between two friends will be represented as an edge between a pair of vertices. There will be no double edges because it is not possible for two friends to be linked twice in Roblox ; they are either friends or they are not. Also, a player cannot be a friend to themself, so there is no need for a loop. Since there are no double edges or loops, this is best represented as a graph. Using Graph Theory to Reduce Internet Fraud Could graphs be used to reduce Internet fraud? At least one researcher thinks so. Graph theory is used every day to analyze our behavior, particularly on social network sites. Alex Buetel, a computer scientist from Carnegie Mellon University in Pittsburgh, Pennsylvania, published a research paper in 2016 that discussed the possibilities of distinguishing the normal interactions from those that might be fraudulent using graph theory. Buetel wrote, “To more effectively model and detect abnormal behavior, we model how fraudsters work, catching previously undetected fraud on Facebook, Twitter, and Tencent Weibo and improving classification accuracy by up to 68%.” In the same paper, the researcher discusses how similar techniques can be used to model many other applications and even, “predict why you like a particular movie.” (Alex Beutel, \"User Behavior Modeling with Large-Scale Graph Analysis,\" http://reports-archive.adm.cs.cmu.edu/anon/2016/CMU-CS-16-105.pdf, May 2016, CMU-CS-16-105, Computer Science Department, School of Computer Science, Carnegie Mellon University, Pittsburgh, PA) Check Your Understanding Key Terms vertex edge loop graph (simple graph) multigraph adjacent (neighboring) degree Key Concepts Graphs and multigraphs represent objects as vertices and the relationships between the objects as edges. The degree of a vertex is the number of edges that meet it and the degree can be zero. An edge must have a vertex at each end. Multigraphs may contain loops and double edges, but simple graphs may not. Videos Graph Theory: Create a Graph to Represent Common Boundaries on a Map", "section": "Graph Basics", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Graph Structures Neuroimaging shows brain activity. (credit: \"MRI Scan\" by NIH Image Gallery/Flickr, Public Domain) Learning Objectives After completing this section, you should be able to: Describe and interpret relationships in graphs. Model relationships with graphs. Graph theory is used in neuroscience to study how different parts of the brain connect. Neurobiologists use functional magnetic resonance imaging (fMRI) to measure levels of blood in different parts of the brain, called nodes. When nodes are active at the same time, it suggests there is a functional connection between them so they form a network. This network can be represented as a graph where the vertices are the nodes and the functional connections are the edges between them. (Mikey Taylor, \"Graph Theory & Machine Learning in Neuroscience,\" Medium.com, June 24, 2020. Importance of the Degrees of Vertices One reason scientists study these networks is to determine how successful the communication within a network continues to be when it experiences failures in nodes and connections. Graphs can be used to study the resilience of these networks. (Mikey Taylor, \"Graph Theory & Machine Learning in Neuroscience,\" Medium.com, June 24, 2020) Using Graphs to Understand Relationships The graphs in represent neural networks, where the vertices are the nodes, and the edges represent functional connections between them. Which graph do you think would represent a network with the highest resistance to failure? Which graph would probably be the most vulnerable to failure? How might this relate to the degrees of the vertices? Models of Neural Networks Graph B represents a network that would have the highest resistance to failure because there are more edges connecting the nodes indicating there are more connections between nodes than on either of the other graphs. This would make the network more resistant to failure because there are more options to work around a faulty edge or node. Graph A would probably be the most susceptible to failure because the network depends on a few particular edges and vertices for connections. There are more vertices of higher degree in Graph B than in Graph A because there are more edges connecting the nodes. Relating the Number of Edges to the Degrees of Vertices In the applications of graph theory to neuroscience and sociology in and YOUR TURN 12.6 , there was a correlation between the degrees of vertices and the resilience of a network. Researchers in many fields have also observed a direct relationship between the number of edges in a graph and the degrees of the vertices. To begin to understand this relationship, consider a graph with five vertices and zero edges as in . Graph with Five Vertices and zero Edges Instead of being marked with a name, each vertex in is marked with its degree. In this case, all of the degrees are 0 so the sum of the degrees is also zero. Suppose that we add an edge between any two existing vertices and indicate the degrees of the vertices. This gives us a graph with five vertices and one edge like the graph in . Graph with Five Vertices and One Edge Note that the degrees of two vertices increased, each by 1. So, the sum of the degrees is now 2. Suppose that we continue in this way, adding one edge at a time and making note of the number of edges and the sum of the degrees of the vertices as in . Comparing Number of Edges to Sum of Degrees demonstrates a characteristic that is true of all graphs of any shape or size. When the number of edges is increased by one, the sum of the degrees increases by two. This happens because each edge has two ends and each end increases the degree of one vertex by one unit. As a result, the sum of the degrees of the vertices on any graph is always twice the number of edges. This relationship is known as the Sum of Degrees Theorem . For the Sum of Degrees Theorem, sum of the degrees = 2 × number of edges or number of edges = sum of degrees 2 Finding the Sum of Degrees Suppose that a graph has five edges. Find the sum of the degrees of the vertices. Draw two different graphs that demonstrate this conclusion. The sum of the degrees is twice the number of edges: 2 × 5=10. The sum of the degrees is 10. See . Completeness Suppose that there were five strangers in a room, A , B , C , D , and E , and each one would be introduced to each of the others. How many introductions are necessary? One way to begin to answer this question is to draw a graph with each vertex representing an individual in the room and each edge representing an introduction as in . Model of Introductions between Five Strangers Let’s approach the problem by thinking about how many new people Person A would meet, then Person B , and so on, making sure not to repeat any introductions. The first graph in shows Person A meeting Persons B , C , D , and E , for a total of 4 introductions. The next graph shows that Person B still has to meet Persons C , D , and E , for a total of 3 more introductions. The next graph shows that Person C still has to meet Persons D and E , which is 2 more introductions. The next graph shows that Person D only remains to meet Person E , which is 1 more introduction. The final graph has 4 + 3 + 2 + 1 = 10 edges representing 10 introductions. The last graph is an example of a complete graph because each pair of vertices is joined by an edge. Another way of saying this is that the graph is complete because each vertex is adjacent to every other vertex. shows complete graphs with three, four, five, and six vertices. Complete Graphs with Up to Six Vertices Suppose we want to know the number of introductions necessary in a room with six people. This would be represented by a complete graph with six vertices, and the total number of introductions would be 5 + 4 + 3 + 2 + 1 = 15 , the number of edges in the graph. In fact, you can always find the number of introductions in a room with n people by adding all the whole numbers from 1 to n − 1 . The number of edges in a complete graph with n vertices is the sum of the whole numbers from 1 to n − 1 , 1 + 2 + 3 + ⋯ + ( n − 1 ) . Suppose that we want to determine how many introductions are necessary in a room with 500 strangers. In other words, suppose that we want to determine the number of edges in a complete graph with 500 vertices. Adding up all the numbers from 1 to 499 could take a long time! In the next example, we use the Sum of Degrees Theorem to make the problem more manageable. Using the Sum of Degrees Theorem Use the Sum of Degrees Theorem to determine the number of introductions required in a room with 6 strangers. 10 strangers. n strangers. Since there are 6 strangers, there are 6 vertices. Since each individual must meet 5 other individuals, there are 5 edges meeting at each vertex which means each vertex has degree 5. Since there are 6 vertices of degree 5, the sum of degrees is 6 ⋅ 5 = 30 . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is 30 2 = 15 . So, the total number of introductions is 15. Since there are 10 strangers, there are 10 vertices. Since each individual must meet 9 other individuals, there are 9 edges meeting at each vertex which means each vertex has degree 9. Since there are 10 vertices of degree 9, the sum of degrees is 10 ⋅ 9 = 90 . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is 90 2 = 45 . So, the total number of introductions is 45. Since there are n strangers, there are n vertices. Since each individual must meet n − 1 other individuals, there are n − 1 edges meeting at each vertex which means each vertex has degree n − 1 . Since there are n vertices of degree n − 1 , the sum of degrees is n ( n − 1 ) . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is n ( n − 1 ) 2 . So, the total number of introductions is n ( n − 1 ) 2 . Now we have a shorter way to calculate the number of introductions in a room with n strangers, and the number of edges on a complete graph with n vertices. Let’s update our formula. The number of edges in a complete graph with n vertices is 1 + 2 + 3 + ⋯ + ( n − 1 ) = n ( n − 1 ) 2 . Subgraphs Sometimes a graph is a part of a larger graph. For example, the graph of South Florida Airports from is part of a larger graph that includes Orlando International Airport in Central Florida, which is shown in Orlando and South Florida Airports The graph in includes an additional vertex, MCO, and additional edges shown with dashed lines. The graph of direct flights between South Florida airports from is called a subgraph of the graph that also includes direct flights between Orlando and the same South Florida airports in . In general terms, if Graph B consists entirely of a set of edges and vertices from a larger Graph A , then B is called a subgraph of A . Identifying a Subgraph In , Graph G is given, along with four diagrams. Determine whether each diagram is or is not a subgraph of Graph G and explain why. Graph G and Potential Subgraphs Diagram J is not a subgraph of Graph G because edge ec is not in Graph G . Diagram K is a subgraph of Graph G because all of its vertices and edges were part of Graph G . Diagram L is not a graph at all, because there is a line extending from vertex a that does not connect a to another vertex. So, Diagram L cannot be a subgraph. Diagram M is not a subgraph of Graph G because it contains a vertex f , which is not part of G . Identifying and Naming Cycles The water cycle begins and ends with water. (credit: \"Diagram of the water cycle\" by NASA, Public Domain) When you think of a cycle in everyday life, you probably think of something that begins and ends the same way. For example, the water cycle ( ) begins with water in a lake or ocean, which evaporates into water vapor, condenses into clouds, and then returns to earth as rain or some other form of precipitation that settles into lakes or oceans. A cycle in graph theory is similar in that it begins and ends in the same way: It is a series of connected edges that begin and end at the same vertex but otherwise never repeat any vertices. In a cycle, there are always the same number of vertices as edges, and all vertices must be of degree 2. Cycles are often referred to by the number of vertices. For example, a cycle with 5 vertices can be called a 5-cycle. Cycles can also be named after polygons based on the number of edges. For example a 5-cycle is also called a pentagon. lists these names for cycles of size 3 through size 10. Cycle Category Number of Edges Example triangle 3 quadrilateral 4 pentagon 5 hexagon 6 heptagon 7 octagon 8 nonagon 9 decagon 10 Categories of Cycles There are many more polygon names, including a megagon that has a million edges and a googolgon that has 10 100 edges, but usually we just say n -gon when the number n is past 10. For example, a cycle with 11 edges could be called an 11-gon. Notice that the 10-cycle, or decagon, appears to cross over itself in . Remember, graphs can be drawn differently but represent the same connections. In , the same decagon is transformed into a graph that does not appear to overlap itself. We have done this without changing any of the connections so both diagrams represent the same relationships, and both diagrams are considered decagons. Transformation of a Decagon Googol to Google Did the name \"googolgon\" ring a bell for you? If so, there is a good reason for it. The creators of Google.com renamed their search engine Google, a misspelled reference to the number \"googol\" alluding to the enormous number of calculations their algorithm can tackle. A googol is a 1 followed by 100 zeroes, or 10 100 . (William L. Hosch, \"Google, American Company,\" https://www.britannica.com/topic/Google-Inc, Encyclopedia Britannica online) Cyclic Subgraphs and Cliques When cycles appear as subgraphs within a larger graph, they are called cyclic subgraphs . Cyclic subgraphs are named by listing their vertices sequentially. The vertex where you begin is not important. Graph K in has two triangle cycles ( g , h , j ) and ( h , i , j ), and one quadrilateral cycle ( g , h , i , j ). Cycles in Graph K The same cycle can be named in many ways, but we must keep in mind that listing two vertices consecutively indicates an edge exists between them. For example, (g, h, i, j) can also be named (h, i, j, g) or (j, i, h, g). However, you cannot name it (g, i, h, j,) which doesn't reflect the correct order of the vertices. We can see that the order (g, i, h, j) is incorrect because the cycle has no edges gi or hj. Identifying Cyclic Subgraphs Identify the types of cyclic subgraphs in Graph H in and name them. Graph H In order to avoid missing a cycle, begin by analyzing the vertex a , then proceed in alphabetical order. Vertex a is part of two cycles, the quadrilateral cycle ( a , b , f , g ,) and the hexagonal cycle ( a , b , c , d , e , f ). Next look at vertex b . It is part of the hexagonal cycle ( b , c , d , e , f , g ). After checking each of the remaining vertices, we see there are no other cycles. In order to avoid naming the same cycle twice, consider naming cycles beginning with the letter closest to the beginning of the alphabet and then progressing clockwise. We have seen that sociologists use graphs to study the structures of social networks. In sociology, there is a principle known as Triadic Closure. It says that if two individuals in a social network have a friend in common, then it is more likely those two individuals will become friends too. Sociologists refer to this as an open triad becoming a closed triad. This concept can be visualized as graphs in . (Chakraborty, Dutta, Mondal, and Nath, \"Application of Graph Theory in Social Media, International Journal of Computer Sciences and Engineering , 6(10):722-729) In the open triad in , person a and person b each has a friendship with person c . In the closed triad, person a and person b have also developed a friendship. Notice that the graph representing the closed triad is a three-cycle, or a triangle, in graph theory terms. Graphs of Open Triad and Closed Triad Another topic of interest to sociologists, as well as computer scientists and scientists in many fields, is the concept of a clique . In a social group, a clique is a subgroup who are all friends. A triad is an example of a clique with three people, but there can be cliques of any size. In graph theory, a clique is a complete subgraph. Identifying Triads and Cliques The graphs in YOUR TURN 12.6 represent social communities. The vertices are individuals and the edges are social connections. Use the graphs to answer the questions. Which graph has the most triads? Name the triangles. Which graph has the fewest triads? Name the triangles. How might an increase in the number of triads in a graph affect the resilience of a community? Explain your reasoning. Identify a clique with more than three vertices in Graph E by naming its vertices. Graph E has the most triads. The triangles are ( a, b, c ), ( a, c, e ), ( a, c, f ), ( a, c, g ), ( a, e, g ), ( c, e, g ), ( e, g, h ), ( e, h, i ), and ( f, g, i ). Graph D has the fewest triads. There are no triangles in the graph. More triads means vertices of greater degree, which indicates greater resilience. Specifically, if an edge is removed from a closed triad, there the two individuals who are no longer adjacent are still connected by way of the third member of the triad. The vertices a, e, c, and g form a clique with four vertices. Three-Cycles in Complete Graphs Just as complete graphs have a predictable number of edges, complete graphs have a predictable number of cyclic subgraphs. Let’s look at the three-cycles within complete graphs with up to six vertices, which are shown in . Complete Graphs with Up to Six Vertices Let's list the names of all the triangles in each graph. Since every pair of vertices is adjacent, any three vertices on a complete graph form a triangle . There is only one triangle in the complete graph with three vertices, ( a , b , c ). For the rest of the graphs, it is important that we take an organized approach. Start with the vertex that is first alphabetically, listing any triangles that include that vertex also in alphabetical order. Then, proceed to the next vertex in the alphabet, and list any triangles that include that vertex, except those that are already listed. Keep going in this way as shown in . Complete Graph With: 3 Vertices 4 Vertices 5 Vertices 6 Vertices All a triangles ( a , b , c ) ( a, b, c ), ( a, b, d ), ( a, c, d ) ( a, b, c ), ( a, b, d ), ( a, b, e ) ( a, c, d ), ( a, c, e ) ( a, d, e ) ( a, b, c ), ( a, b, d ), ( a, b, e ), ( a, b, f ) ( a, c, d ), ( a, c, e ), ( a, c, f ) ( a, d, e ), ( a, d, f ) ( a, e, f ) Other b triangles None ( b, c, d ) ( b, c, d ), ( b, c, e ), ( b, d, e ) ( b, c, d ), ( b, c, e ), ( b, c, f ) ( b, d, e ), ( b, d, f ) ( b, e, f ) Other c triangles None None ( c, d, e ) ( c, d, e ), ( c, d, f ) ( c, e, f ) Other d triangles None None None ( d, e, f ) Total 1 ( 2 + 1 ) + 1 = 3 + 1 = 4 ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 6 + 3 + 1 = 10 ( 4 + 3 + 2 + 1 ) + ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 10 + 6 + 3 + 1 = 20 Listing Triangles in Complete Graphs Look at the last row in . Do you see a pattern emerge for counting triangles in a complete graph? Without drawing a complete graph with 7 vertices, we can predict that it will have ( 5 + 4 + 3 + 2 + 1 ) + ( 4 + 3 + 2 + 1 ) + ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 35 triangles inside it. This pattern also appears in a famous diagram known to Western mathematicians as \"Pascal’s Triangle.\" displays the first 11 rows of Pascal’s Triangle. Row 0 of Pascal’s Triangle only has the number 1 in it. The first and last entries in each of the other rows are also 1s. Otherwise, all the entries are formed by adding two entries from the previous row. For example, in row 6, entry 1 is 6, which was found by adding 1 and the 5 from the previous row, and entry 2 is 15, which was found by adding the 5 and the 10 from the previous row, as shown in . Pascal’s Triangle IMPORTANT! When you count the rows and entries of Pascal’s Triangle begin with row 0 and entry 0, not row 1 and entry 1. The Mathematical Secrets of Pascal's Triangle by Wajdi Mohamed Ratemi If we begin to list the third entries in each row of Pascal's triangle from the top down, we see 1, 4, 10, 20, 35, and so on. Notice that these values are exactly the number of triangles in a complete graph of sizes 3, 4, 5, 6, and 7, respectively. Specifically, entry 3 in row 7 is 35, the number of triangles in a complete graph of size 7. Let's practice using Pascal’s triangle to find the number of triangles in a complete graph of a particular size. STEPS TO FIND THE NUMBER OF TRIANGLES IN A COMPLETE GRAPH OF SIZE n Step 1 Calculate the entries in row n of Pascal's Triangle using sums of pairs of entries in previous rows as needed. Step 2 The number of triangles is entry 3 in row n . Remember to count the entries 0, 1, 2, and 3, from left to right. Using Pascal’s Triangle to Find the Number of Triangles in a Complete Graph Use Pascal’s Triangle in to find the number of triangles in a complete graph with 11 vertices. Step 1: Identify the row number and calculate its entries. Since the complete graph has 11 vertices, we will need to look in row 11 of Pascal’s Triangle. Use row 10 in to find the entries in row 11 as shown in . Rows 10 and 11 of Pascal’s Triangle Step 2: Find entry number 3. The number of triangles is in entry 3. Remember to count the entries beginning with entry 0 as shown in . Identifying an Entry in Pascal’s Triangle Entry 3 in row 11 is 165. So, there are 165 triangles in a complete graph with 11 vertices. Since the entries in Pascal’s Triangle are useful in many applications, many resources such as online and traditional calculator functions have been developed to assist in calculating the values. The website dCode.xyz has a tool that automatically calculates entries in Pascal’s Triangle using these steps: Step 1: Make sure to select Index 0. Step 2: Select your preference, display the triangle until a particular row (line), display only one row (line), or calculate the value at coordinates (which means give the value of a single entry). Step 3: Enter row number (line number) you are interested, or enter the row number and entry number in parentheses: (row number, entry number). Step 4: Select Calculate. Step 5: Retrieve the answer from the results box. Pascal's Triangle Tool on dCode.xyz Check Your Understanding Key Terms complete subgraph cycle cyclic subgraph clique Key Concepts The sum of the degrees of the vertices in a graph is twice the number of edges. In a complete graph every pair of vertices is adjacent. A subgraph is part of a larger graph. Cycles are a sequence of connected vertices that begin and end at the same vertex but never visit any vertex twice. Formulas For the Sum of Degrees Theorem, sum of the degrees = 2 × number of edges or number of edges = sum of degrees 2 The number of edges in a complete graph with n vertices is the sum of the whole numbers from 1 to n − 1 , 1 + 2 + 3 + ⋯ + ( n − 1 ) . The number of edges in a complete graph with n vertices is 1 + 2 + 3 + ⋯ + ( n − 1 ) = n ( n − 1 ) 2 . Videos The Mathematical Secrets of Pascal's Triangle by Wajdi Mohamed Ratemi", "section": "Graph Structures", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Comparing Graphs A flat map represents the surface of Earth in two dimensions. (credit: modification of work \"World Map\" by Scratchinghead/Wikimedia Commons, Public Domain) A globe represents the surface of Earth in three dimensions. (credit: \"Globe\" by Wendy Cope/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify the characteristics used to compare graphs. Determine when two graphs represent the same relationships Explore real-world examples of graph isomorphisms Find the complement of a graph Maps of the same region may not always look the same. For example, a map of Earth on a flat surface looks distorted at the poles. When the same regions are mapped on a spherical globe, the countries that are closer to the polls appear smaller without the distortion. Despite these differences, the two maps still communicate the same relationships between nations such as shared boundaries, and relative position on the earth. In essence, they are the same map. This means that for every point on one map, there is a corresponding point on the other map in the same relative location. In this section, we will determine exactly what characteristics need to be the shared by two graphs in order for us to consider them “the same.” When Are Two Graphs Really the Same Graph? In arithmetic, when two numbers have the same value, we say they are equal, like ½ = 0.5. Although ½ and 0.5 look different, they have the same value, because they are assigned the same position on the real number line. When do we say that two graphs are equal? shows Graphs A and F , which are identical except for the labels. Graphs are visual representations of connections. As long as two graphs indicate the same pattern of connections, like Graph A and Graph F , they are considered to be equal, or in graph theory terms, isomorphic . Identical Graphs with Different Labels Two graphs are isomorphic if either one of these conditions holds: One graph can be transformed into the other without breaking existing connections or adding new ones. There is a correspondence between their vertices in such a way that any adjacent pair in one graph corresponds to an adjacent pair in the other graph. It is important to note that if either one of the isomorphic conditions holds, then both of them do. When we need to decide if two graphs are isomorphic, we will need to make sure that one of them holds. For example, shows how Graph T can be bent and flipped to look like Graph Z , which means that Graphs T and Z satisfy condition 1 and are isomorphic. Graph T Transformed Into Graph Z Also, notice that the vertices that were adjacent in the first graph are still adjacent in the transformed graph as shown in . For example, vertex 3 is still adjacent to vertex 4, which means they are still neighboring vertices joined by a single edge. Adjacent Vertices Are Still Adjacent When Are Two Graphs Really Different? Verifying that two graphs are isomorphic can be a challenging process, especially for larger graphs. It makes sense to check for any obvious ways in which the graphs might differ so that we don’t spend time trying to verify that graphs are isomorphic when they are not. If two graphs have any of the differences shown in , then they cannot be isomorphic. Unequal number of vertices Unequal number of edges Unequal number of vertices of a particular degree Different cyclic subgraphs Characteristics of Graphs That Are Not Isomorphic Recognizing Isomorphic Graphs Isomorphic graphs that represent the same pattern of connections can look very different despite having the same underlying structure. The edges can be stretched and twisted. The graph can be rotated or flipped. For example, in , each of the diagrams represents the same pattern of connections. Four Representations of the Same Graph Looking at , how can we know that these graphs are isomorphic? We will start by checking for any obvious differences. Each of the graphs in has four vertices and five edges; so, there are no differences there. Next, we will focus on the degrees of the vertices, which have been labeled in . Graphs with Vertices of the Same Degrees As shown in , each graph has two vertices of degree 2 and two vertices of degree 3; so, there are no differences there. Now, let’s check for cyclic subgraphs. These are highlighted in . Graphs with the Same Cyclic Subgraphs As shown in , each graph has two triangles and one quadrilateral; so, no differences there either. It is beginning to look likely that these graphs are isomorphic, but we will have to look further to be sure. To know with certainty that these graphs are isomorphic, we need to confirm one of the two conditions from the definition of isomorphic graphs. With smaller graphs, you may be able to visualize how to stretch and twist one graph to get the other to see if condition 1 holds. Imagine the edges are stretchy and picture how to pull and twist one graph to form the other. If you can do this without breaking or adding any connections, then the graphs are really the same. demonstrates how to change graph A 4 to get A 3 , graph A 3 to get A 2 , and graph A 2 to get A 1 . Transforming Graphs Now that we have used visual analysis to see that condition 1 holds for graphs A 1 , A 2 , A 3 , and A 4 in , we know that they are isomorphic. In , one of the edges of graph A 4 crossed another edge of the graph. By transforming it into graph A 3 , we have “untangled” it. Graphs that can be untangled are called planar graphs. The complete graph with five vertices is an example of a nonplanar graph-that means that, no matter how hard you try, you can’t untangle it. But, when you try to figure out if two graphs are the same, it can be helpful to untangle them as much as possible to make the similarities and differences more obvious. Identifying Isomorphic Graphs Which of the three graphs in are isomorphic, if any? Justify your answer. Three Similar Graphs Step 1: Check for differences in number of vertices, number of edges, degrees of vertices, and types of cycles to see if an isomorphism is possible. Vertices: They all have the same number of vertices, 4. Edges: They all have the same number of edges, 4. Degrees: Graph B 1 and Graph B 2 each have a vertex of degree 3, while Graph B 3 does not. So, Graph B 3 is not isomorphic to either of the other two graphs, but Graph B 1 and Graph B 2 could possibly be isomorphic. Cycles: Focus on any cycles in Graph B 1 and Graph B 2 . Each graph has a triangle as shown in . Cycles of Graph B 1 and Graph B 2 Step 2: If no differences were found and an isomorphism is possible, verify one of the conditions in the definition of isomorphic. Since we were able to determine that Graph B 1 and Graph B 2 have no obvious differences, and they are relatively small graphs, we will attempt to transform one graph into the other, which would verify condition 1. Graph B 1 can be transformed into Graph B 2 without breaking or adding connections as shown in . Begin by untangling graph B 1 . Then rotate or flip as needed to see that the graphs match. Transformation of Graph B 1 So, Graph B 1 and Graph B 2 have the same structure and are isomorphic. Have you ever noticed that many popular board games may look different but are really the same game? A good example is the many variations of the board game Monopoly ®, which was submitted to the U.S. Patent Office in 1935. Although the rules have been revised a bit, a very similar game board is still in use today. There have been many versions of Monopoly over the years. Many have been stylized to reflect a popular theme, such as a show or sports team, while retaining the same game board structure. If we were to represent these different versions of the game using a graph, we would find that the graphs are isomorphic. . Let's analyze some game boards using graph theory to determine if they have the same structure despite having different appearances. Two Game Boards Deciding If Graphs are Isomorphic A teacher uses games to teach her students about colors and numbers as shown in . In the Colors Game, shown in Figure A , each player begins in the space marked START and proceeds in a counterclockwise direction. On each turn, the player spins a spinner marked 1 and 2 and moves forward the number of spaces shown on the spinner. If the player lands on a space marked with any color other than white, the player must move forward or back to the other space of the same color. The first player to land in or pass the space marked END wins. In the Clock Game, shown in Figure B , each player begins in the space marked START and proceeds in a clockwise direction. On each turn, the player spins a spinner marked 1 and 2 and moves forward the number of spaces shown on the spinner. In the same turn, the player must read the number in the space and move forward the number of spaces indicated by a positive value or backward the number of spaces indicated by a negative value. Then the turn ends. The first player to land in or pass the space marked END wins. Draw a graph or multigraph to represent each game in which the vertices are the spaces and the edges represent the ability of a player to move between the spaces either by a spin or as dictated by a marked color or number. (We will ignore the direction of motion for simplicity.) Transform one of the graphs to show that it is isomorphic to the other and explain what this tells you about the games. The graph representing the Clock Game can be transformed as shown in so that we can see that the graphs are isomorphic. There is a correspondence between their vertices in such a way that any adjacent pair in one graph corresponds to an adjacent pair in the other graph. Graphs for Clock Game and Colors Game This tells us that the games are essentially the same game with the same moves even though the games appear to be different. Elizabeth Magie Game designer, engineer, comedian, and political activist, Elizabeth Magie designed a game called “Landlord’s Game” to educate fellow citizens about the dangers of monopolies and the benefits of wealth redistribution through a land tax. Magie, whose father had campaigned with Abraham Lincoln, was a proponent of a land tax, an idea popularized by Henry George’s 1879 book, Progress and Poverty . She designed the game to be played by two sets of rules for comparison. In one version, the goal was to dominate opponents by creating monopolies, leaving one wealthy player standing in the end. In the other version, all the players were rewarded when a monopoly was created through a simulated land tax. She patented this game for the first time in 1904. Does the game board in look familiar? Landlord’s Game Patent (credit: \"Landlord's Game Patent\" Wikimedia Commons, Public Domain) That’s right! A modified version of the game was obtained from friends by a man named Charles Darrow who renamed it Monopoly and sold to Parker Brothers. Parker Brothers later paid Magie $500 for the right to her patent on it. The property tax version was left behind, and the modern game of Monopoly was born. (Mary Pilon, Monopoly’s Lost Female Inventor , September 1, 2018, National Women’s History Museum, “Monopoly’s Lost Female Inventor,” Identifying and Naming Isomorphisms When two graphs are isomorphic, meaning they have the same structure, there is a correspondence between their vertices, which can be named by listing corresponding pairs of vertices. This list of corresponding pairs of vertices in such a way that any adjacent pair in one graph corresponds to an adjacent pair in the other graph is called an isomorphism . Consider the isomorphic graphs in . In , we could replace the labels Graph F with the labels from Graph A and have an identical graph, as in . Identical Graphs with Different Labels Corresponding Vertices So, we can identify an isomorphism between Graph A and Graph F by listing the corresponding pairs of vertices: b-g , c-h , d-i , and e-j . Notice that b is adjacent to c and g is adjacent to h . This must be the case since b corresponds to g and c corresponds to h . The same is true for other pairs of adjacent vertices. An isomorphism between graphs is not necessarily unique. There can be more than one isomorphism between two graphs. We can see how to form a different isomorphism between Graph A and Graph F from by rotating Graph F clockwise and comparing the rotated version of F to Graph A as in YOUR TURN 12.13 . Now, we can see that a second isomorphism exists, which has the correspondence: b-j , d-h , c-i , and e-g as shown in . Transforming Graph F into Graph A When you name isomorphisms, one way to check that your answer is reasonable is to make sure that the degrees of corresponding vertices are equal. Determine If Two Graphs Are Isomorphic and Identify the Isomorphism Identifying Isomorphisms In , we showed that the Graphs B 1 and B 2 in are isomorphic. In , labels have been assigned to the vertices of Graphs B 1 and B 2 . Identify an isomorphism between them by listing corresponding pairs of vertices. Two Isomorphic Graphs showed how to transform Graph B 1 to get Graph B 2 . In , we will do the same, but this time we will include the labels. Transform Graph B 1 into Graph B 2 From YOUR TURN 12.13 , we can see the corresponding vertices: a-q , d-p , c-r , and b-s , which is an isomorphism of the two graphs. Recognizing an Isomorphism Determine whether Graphs G and S in are isomorphic. If not, explain how they are different. If so, name the isomorphism. Graph G and Graph S Step 1: Check for any differences. Vertices: Graphs G and S each have six vertices. Edges: Graphs G and S each have six edges. Degrees: Each graph also has two vertices of degree 1, two vertices of degree 2, and two vertices of degree 3. Cycles: From , we can see that Graph G contains a quadrilateral cycle ( b , f , d , c ) but Graph S has no quadrilaterals. Also, Graph S contains a triangle cycle ( m , r , o ) but Graph G has no triangles. This means that the graphs are not isomorphic. Comparing Subgraphs Step 2: This step is not necessary because we now know Graphs G and S are not isomorphic. Complementary Graphs Suppose that you are a camp counselor at Camp Woebegone and you are holding a camp Olympics with four events. The campers have signed up for the events. You drew a graph in to help you visualize which events have campers in common. Graph of Events with Campers in Common Graph E in shows that some of the same campers will be in events a and b , as well as b and d , c and d , and a and c . What do you think the graph would look like that represented the events that do not have campers in common? It would have the same vertices, but any pair of adjacent edges in Graph E , would not be adjacent in the new graph, and vice versa. This is called a complementary graph, as shown in . Two graphs are complementary if they have the same set of vertices, but any vertices that are adjacent in one, are not adjacent in the other. In this case, we can say that one graph is the complement of the other. Graphs of Camp Olympics One way to find the complement of a graph is to draw the complete graph with the same number of vertices and remove all the edges that were in the original graph. Let’s say you wanted to find the complement of Graph E from , and you didn’t already know it was Graph F . You could start with the complete graph with four vertices and remove the edges that are in Graph E as shown in . Use Complete Graph to Find Complement Finding a Complement A particular high school has end-of-course exams in (E3) English 3, (E4) English 4, (M) Advanced Math, (C) Calculus, (W) World History, (U) U.S. History, (B) Biology, and (P) Physics. No English 3 students are taking English 4, World History, or Biology; no English 4 students are also in Calculus, Advanced Math, U.S. History, or Physics; no Physics students are also taking Advanced Math; No World History students are also taking U.S. History; and no Advanced Math students are also taking Calculus. Create a graph in which the vertices represent the exams, and an edge between a pair of vertices indicates that there are no students taking both exams. Find the complement of the graph in part 1. Explain what the graph in part 2 represents. In , we have drawn a vertex for each exam and edges between any vertices that have no students in common. Graph of Exams with No Students in Common One way to get the complement of the graph in is to draw a complete graph with the same number of vertices and remove the edges they have in common as shown in . Remove Unwanted Edges from Complete Graph The final graph of the complement is in . Graph of Exams with Students in Common In the graph in , the vertices are still the exams, and a pair of adjacent vertices represents a pair of exams that have students in common. When two graphs are really the same graph, they have the same missing edges. So, when two graphs have a lot of edges, it may actually be easier to determine if they are isomorphic by looking at which edges are missing rather than which edges are included. In other words, we can determine if two graphs are isomorphic by checking if their complements are isomorphic. Using a Complement to Find an Isomorphism Use to answer each question. Graph K Find the complement of Graph K . Identify an isomorphism between the complement of Graph K from part 1, and the complement of Graph H in YOUR TURN 12.17 . Confirm that the correspondence between the vertices you found in part 2 also gives an isomorphism between Graph H from YOUR TURN 12.17 , and Graph K from . The complement of Graph K can be found by removing the edges of Graph K from a complete graph with the same vertices as shown in . Find Complement of Graph K An isomorphism between the complement of Graph K and the complement of Graph H is A-M , C-L , E-N , B-O , and D-P , which is confirmed by transforming the complement of Graph K in . Isomorphism between Complements of K and H shows how Graph K can be transformed into Graph H to confirm that the correspondence is A-M , C-L , E-N , B-O , and D-P also gives an isomorphism between Graph K and Graph H . Isomorphism between K and H This means we now have three conditions that guarantee two graphs are isomorphic. First Way: One graph can be transformed into the other without breaking existing connections or adding new ones. Second Way: There is a correspondence between their vertices in such a way that any adjacent pair in one graph corresponds to an adjacent pair in the other graph. Third Way: Their complements are isomorphic. If any one of these statements is true, then they are all true. If any one of these statements is false, then they are all false. Here is an activity you can do with a few of your classmates that will build your graph comparison skills. Step 1: Draw a planar graph with the following characteristics: exactly five vertices, one vertex of degree four, at least two vertices of degree three, and exactly eight edges. Give names to the vertices. Make sure you do not use the same letters or numbers to label your vertices as your classmates do. Step 2: Analyze your graph. What is the degree of each vertex? Does your graph have any cyclic subgraphs? If so, list them and indicate their sizes. Step 3: Draw and analyze the complement of your graph. How many edges and vertices does it have? What is the degree of each vertex? Does the complementary graph have any cyclic subgraphs? If so, list them and indicate their sizes. Step 4: Compare your graphs to each of your classmates’ graphs. Does your graph have the same number of edges and vertices as the graph of your classmate? Does your graph have the same size cyclic subgraphs as the graph of your classmate? How does the complement of your graph compare to the complement of the graph of your classmate? Determine if your graph is isomorphic to your classmates’ graph. If so, give a correspondence that demonstrates the isomorphism. If not, explain how you know. Check Your Understanding Key Terms isomorphic isomorphism planar nonplanar complement complementary Key Concepts Two graphs are isomorphic if they have the same structure. When graphs are relatively small, we can use visual inspection to identify an isomorphism by transforming one graph into another without breaking connections or adding new ones. An isomorphism between two graphs preserves adjacency. If two graphs differ in number of vertices, number of edges, degrees of vertices, or types of subgraphs, they cannot be isomorphic. When the complements of two graphs are isomorphic, so are the graphs themselves. Videos Determine If Two Graphs Are Isomorphic and Identify the Isomorphism", "section": "Comparing Graphs", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Navigating Graphs Visitors navigate a garden maze. (credit: “Longleat Maze” by Niki Odolphie/Wikimedia, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and identify walks, trails, paths, and circuits. Solve application problems using walks, trails, paths, and circuits. Identify the chromatic number of a graph. Describe the Four-Color Problem. Solve applications using graph colorings. Now that we know the basic parts of graphs and we can distinguish one graph from another, it is time to really put our graphs to work for us. Many applications of graph theory involve navigating through a graph very much like you would navigate through a maze. Imagine that you are at the entrance to a maze. Your goal is to get from one point to another as efficiently as possible. Maybe there are treasures hidden along the way that make straying from the shortest path worthwhile, or maybe you just need to get to the end fast. Either way, you definitely want to avoid any wrong turns that would cause unnecessary backtracking. Luckily, graph theory is here to help! Walks Suppose is a maze you want to solve. You want to get from the start to the end. Your Maze You can approach this task any way you want. The only rule is that you can’t climb over the wall. To put this in the context of graph theory, let’s imagine that at every intersection and every turn, there is a vertex. The edges that join the vertices must stay within the walls. The graph within the maze would look like . The Graph in Your Maze One approach to solving a maze is to just start walking. It is not the most efficient approach. You might cross through the same intersection twice. You might backtrack a bit. It’s okay. We are just out for a walk. It might look something like the black sequence of vertices and edges in . The Walk Through the Graph in Your Maze This type of sequence of adjacent vertices and edges is actually called a walk (or directed walk ) in graph theory too! A walk can be identified by naming the sequence of its vertices (or by naming the sequence of its edges if those are labeled). Let’s take the graph out of the context of the maze and give each vertex a name and each edge of the walk a direction as in . The Graph without Your Maze The name of this walk from p to r is p → q → o → n → i → j → c → d → c → j → k → s → r . When a particular edge on our graph was traveled in both directions, it had arrows in both directions and the letters of vertices that were visited more than once had to be repeated in the name of the walk. Not every list of vertices or list of edges makes a path. The order must take into account the way in which the edges and vertices are connected. The list must be a sequence, which means they are in order. No edges or vertices can be skipped and you cannot go off the graph. A Walk or Not a Walk? The highlighted edges Graph Y in represent a walk between f and b . The highlighted edges in Graph X do not represent a walk between f and b , because there is a turn at a point that is not a vertex. This is like climbing over a wall when you are walking through a maze. Another way of saying this is that b → d → f is not a walk, because there is no edge between b and d . A point on a graph where two edges cross is not a vertex. Naming a Walk Through A House shows the floor plan of a house. Use the floor plan to answer each question. Floor Plan of a House Draw a graph to represent the floor plan in which each vertex represents a different room (or hallway) and edges represent doorways between rooms. Name a walk through the house that begins in the living room, ends in the garage and visits each room (or hallway) at least once. Step 1: We will need a vertex for each room and it is convenient to label them according to the names of the rooms. Visualize the scenario in your head as shown in . You don’t have to write this step on your paper. Assigning Vertices to Rooms Step 2: Draw a graph to represent the scenario. Start with the vertices. Then connect those vertices that share a doorway in the floorplan as shown in . Graph of the Floor plan Step 1: Draw a path that begins at vertex L , representing the living room, and ending at vertex G , representing the garage making sure to visit every room at least once. There are many ways this can be done. You may want to number the edges to keep track of their order. One example is shown in . Draw the Path from L to G Step 2: Name the path that you followed by listing the vertices in the order you visited them. L → R → L → K → L → H → B → H → M → H → G Paths and Trails A walk is the most basic way of navigating a graph because it has no restrictions except staying on the graph. When there are restrictions on which vertices or edges we can visit, we will call the walk by a different name. For example, if we want to find a walk that avoids travelling the same edge twice, we will say we want to find a trail (or directed trail ). If we want to find a walk that avoids visiting the same vertex twice, we will say, we want to find a path (or directed path ). Walks, trails, and paths are all related. All paths are trails, but trails that visit the same vertex twice are not paths. All trails are walks, but walks in which an edge is visited twice would not be trails. We can visualize the relationship as in . Walks, Trails, and Paths Let’s practice identifying walks, trails, and paths using the graphs in . Graphs A and K Identifying Walks, Paths, and Trails Consider each sequence of vertices from Graph A in . Determine if it is only a walk, both a walk and a path, both a walk and a trail, all three, or none of these. b → c → d → e → f c → b → d → b → e c → f → e → d → b → c b → e → f → c → b → d First, check to see if the sequence of vertices is a walk by making sure that the vertices are consecutive. As you can see in , there is no edge between vertex c and vertex d . This means that the sequence is not a walk. If it is not a walk, then it can’t be a path and it cannot be a trail, so, it is none of these. First, check to see if the sequence is a walk. As you can see in , the vertices are consecutive. This means that the sequence is a walk. Since the vertex b is visited twice, this walk is not a path. Since edge bd is traveled twice, this walk is not a trail. So, the sequence is only a walk. First, check to see if the sequence is a walk. We can see in that the vertices are consecutive, which means it is a walk. Next, check to see if any vertex is visited twice. Remember, we do not consider beginning and ending at the same vertex to be visiting a vertex twice. So, no vertex was visited twice. This means we have a walk that is also a path. Next check to see if any edge was visited twice; none were. So, the sequence is a walk, a path, and a trail. First, check to see if the sequence is a walk. We can see in that the vertices are consecutive, which means it is a walk. Next check to see if any vertex is visited twice. Since vertex b is visited twice, this is not a path. Finally, check to see if any edges are traveled twice. Since no edges are traveled twice, this is a trail. So, the sequence of vertices is a walk and a trail. Walks, Trails, and Paths in Graph Theory Circuits In many applications of graph theory, such as creating efficient delivery routes, beginning and ending at the same location is a requirement. When a walk, path, or trail end at the same location or vertex they began, we call it closed . Otherwise, we call it open (does not begin and end at the same location or vertex). Some examples of closed walks, closed trails, and closed paths are given in in the following table. DESCRIPTION EXAMPLE CHARACTERISTICS A closed walk is a walk that begins and ends at the same vertex. d → f → b → c → f → d Alternating sequence of vertices and edges Begins and ends at the same vertex A closed trail is a trail that begins and ends at the same vertex. It is commonly called a circuit . d → f → b → c → f → e → d No repeated edges Begins and ends at the same vertex A closed path is a path that begins and ends at the same vertex. It is also referred to as a directed cycle because it travels through a cyclic subgraph. d → f → b → c → d No repeated edges or vertices Begins and ends at the same vertex Closed Walks, Trails, and Paths Since walks, trails, and paths are all related, closed walks, circuits, and directed cycles are related too. All circuits are closed walks, but closed walks that visit the same edge twice are not circuits. All directed cycles are circuits, but circuits in which a vertex is visited twice are not directed cycles. We can visualize the relationship as in . Closed Walks, Circuits, and Directed Cycles The same circuit can be named using any of its vertices as a starting point. For example, the circuit d → f → b → c → d can also be referred to in the following ways. a → b → c → d → a is the same as { b → c → d → a → b c → d → a → b → c d → a → b → c → d Let’s practice working with closed walks, circuits (closed trails), and directed cycles (closed paths). In the graph in , the vertices are major central and south Florida airports. The edges are direct flights between them. Major Central and South Florida Airports Determining a Closed Walk, Circuit, or Directed Cycle Suppose that you need to travel by air from Miami (MIA) to Orlando (MCO) and you were restricted to flights represented on the graph. For the trip to Orlando, you decide to purchase tickets with a layover in Key West (EYW) as shown in , but you still have to decide on the return trip. Determine if your roundtrip itinerary is a closed walk, a circuit, and/or a directed cycle, based on the return trip described in each part. MIA to EYW to MCO You returned to Miami (MIA) by reversing your route. Your direct flight back left Orlando (MCO) but was diverted to Fort Lauderdale (FLL)! From there you flew to Tampa (TPA) before returning to Miami (MIA). The whole trip was MIA → EYW → MCO → EYW → MIA . This is a closed walk, because it is a walk that begins and ends at the same vertex. It is not a circuit, because it repeats edges. If it is not a circuit, then it cannot be a directed cycle. The whole trip was MIA → EYW → MCO → FLL → TPA → MIA . This is a closed walk, because it is a walk that begins and ends at the same vertex. It is a circuit because no edges were repeated. It is also a directed cycle because no vertices were repeated either. So, it is all three! Closed Walks, Closed Trails (Circuits), and Closed Paths (Directed Cycles) in Graph Theory Graph Colorings In this section so far, we have looked at how to navigate graphs by proceeding from one vertex to another in a sequence that does not skip any vertices, but in some applications we may want to skip vertices. Remember the camp Olympics at Camp Woebegone in Comparing Graphs ? You were planning a camp Olympics with four events. The campers signed up for the events. You drew a graph to help you visualize which events have campers in common. The vertices of Graph E in represent the events and adjacent vertices indicate that there are campers who are participating in both. Graphs of Camp Olympics In this case, we do NOT want events represented by two adjacent vertices to occur in the same timeslot, because that would prevent the campers who wanted to participate in both from doing so. We can use the graph in to count the timeslots we need so there are no conflicts. Let’s assign each timeslot a different color. We could categorize events that happen at 1 pm as Red; 2 pm, Purple; 3 pm, Blue; and 4 pm, Green. Then assign different colors to any pair of adjacent vertices to ensure that the events they represent do not end up in the same timeslot. shows several of the ways to do this while obeying the rule that no pair of adjacent vertices can be the same color. Graph Colorings In , the graphs with vertices colored so that no adjacent vertices are the same color are called graph colorings . Notice that Graph 3 has the fewest colors, which means it shows us how to have the fewest number of timeslots. The events marked in red, a and d , can be held at the same time because they are not adjacent and do not have conflicts. Also, the events marked in purple, b and c , can be held at the same time. We would not need green or blue timeslots at all! A graph that uses n colors is called an n -coloring . The smallest number of colors needed to color a particular graph is called its chromatic number . Graph colorings can be used in many applications like the scheduling scenario at camp Woebegone. Let's look at how they work in more detail. shows two different colorings of a particular graph. Coloring A is called a four-coloring, because it uses four colors, red (R), green (G), blue (B), and purple (P). Coloring B is called a three-coloring because it uses three. The colors allow us to visually subdivide the graphs into groups. The only rule is that adjacent vertices are different colors so that they are in different groups. Two Colorings of the Same Graph It turns out that a three-coloring is the best we can do with the graph in . No matter how many different patterns you try with only two colors, you will never find one in which the adjacent vertices are always different colors. In other words, the graph has a chromatic number of three. For large graphs, computer assistance is usually required to find the chromatic numbers. There is no formula for finding the chromatic number of a graph, but there are some facts that are helpful in . Fact Example Recall that planar graphs are untangled; that is, they can be drawn on a flat surface so that no two edges are crossing. If a graph is planar, it can be colored with four colors or possibly fewer. Each vertex in a complete graph is adjacent to every other vertex forcing us to color them all different colors. The chromatic number of a complete graph is the same as the number of vertices. If a graph has a clique, a complete subgraph, then each vertex in the clique must have a different color and the vertices outside the clique may or may not have even more colors. The chromatic number of a graph is at least the number of vertices in its biggest clique. Facts about Graph Colorings Creating Colorings to Solve Problems Let’s see how these facts can help us color the graph in . Since the graph is planar, the chromatic number is no more than four. The graph is not complete, but it has complete subgraphs of three vertices. In other words, it has triangles like the one shown with blue vertices in . This means that the chromatic number is at least three. A Graph with Triangles We know we can color this graph in three or four colors. It is usually best to start by coloring the vertex of highest degree as shown in . In this case, we used red (R). The color is not important. Color Highest Degree Vertex First We want to color as many of the vertices the same color as possible; so, we look at all the vertices that are not adjacent to the red vertex and begin to color them, red starting with the one among them of highest degree. Since the only vertices that are not adjacent are both degree 2, choose either one and color it red as shown in . Color More Red from Highest to Lowest Degree Now, there is only one vertex left that is not adjacent to a red; so, color it red. Of the remaining vertices, the highest degree is four; so, color one of the vertices of degree four in a different color. These two steps are shown in . Complete the Reds and Start Another Color Repeat the same procedure. There are three remaining vertices that are not adjacent to a blue. Color as many blue as possible, with priority going to vertices of higher degree as shown in . Repeat Procedure for Color Blue All the remaining vertices are adjacent to blue. So, it is time to repeat the procedure with another color as shown in . Repeat Procedure for Color Purple All the vertices are now colored with a three-coloring so we know the chromatic number is at most three, but we knew the chromatic number was at least three because the graph has a triangle. So, we are now certain it is exactly three. Coloring Graphs Part 1: Coloring and Identifying Chromatic Number Suppose the vertices of the graph in represented the nine events in the Camp Woebegone Olympics, and edges join any events that have campers in common, but nonadjacent vertices do not. Coloring for Nine Event Camp Woebegone Olympics Any vertices of the same color are nonadjacent; so, they have no conflicts. Since this graph is a three-coloring, all nine events could be scheduled in just three timeslots! Understanding Chromatic Numbers In , we discussed a high school, which holds end-of-course exams in (E3) English 3, (E4) English 4, (M) Advanced Math, (C) Calculus, (W) World History, (U) U.S. History, (B) Biology, and (P) Physics. We were given a list of courses that had no students in common. We used that information to find the graph in , which shows edges between exams with students in common. Use the graph we found in to answer each question. Graph of Exams with Students in Common The graph contains a clique of size 4 formed by the vertices P, E3, C, and U. What does this tell you about the chromatic number? The graph is not planar, meaning that you cannot untangle it. What does this tell you about the chromatic number? Create a coloring by coloring vertex of highest degree first, coloring as many other vertices as possible each color from highest to lowest degree, then repeating this process for the remaining vertices. Do you know what the minimum number of timeslots is? If so, what is it and how do you know? If not, what are the possibilities? We would need four different colors just for the clique with four vertices; so, the chromatic number is at least four. It is possible for the chromatic number to be greater than four. The process is shown in . This is the original graph. Vertex B has highest degree. Color vertex B. Vertex E3 is the only remaining vertex that is NOT adjacent to B. Color vertex E3 the same color. Vertices P, U, W, and C are the remaining vertices with highest degree. Pick one to color. Color vertex P a new color. Vertices E4 and M are NOT adjacent to P, and M has higher degree. Color vertex M the same color. Vertex E4 is the only remaining vertex NOT adjacent to P or M. Color vertex E4 the same color. Vertices U, W, and C are the remaining vertices with highest degree. Pick one to color. Color U a new color. Vertex W is the only remaining vertex NOT adjacent to U. Color vertex W the same color. Vertex C is the only remaining vertex that has NOT been colored. Color vertex C a new color. The coloring is final. We used four colors. Coloring the Graph The last graph in is the final coloring. Yes, the minimum number of times slots is the chromatic number. We knew the chromatic number had to be at least four because there was a clique with four vertices. Now we have found a four-coloring of the graph which tells us that the chromatic number is at most four. So, we know four must be the chromatic number. The procedure used in Example 22 to color the graph is not guaranteed to result in a graph that has the minimum number of colors possible, but it is usually results in a coloring that is close to the chromatic number. The Four Color Problem Using only four colors, no two adjacent regions have the same color. The idea of coloring graphs to solve problems was inspired by one of the most famous problems in mathematics, the “four color problem.” The idea was that, no matter how complicated a map might be, only four colors were needed to color the map so that no two regions that shared a boundary would be the same color. For many years, everyone suspected this to be true, because no one could create a map that needed more than four colors, but they couldn’t prove it was true in general. Finally, graphs were used to solve the problem! The Four Color Map Theorem – Numberphile We saw how maps can be represented as graphs in Graph Basics . from shows a map of the midwestern region of the United States. Map of Midwestern States shows how this map can be associated with a graph in which each vertex represents a state and each edge indicates the states that share a common boundary. shows the final graph. Edge Assigned to Each Pair of Midwestern States with Common Border Final Graph Representing Common Boundaries between Midwestern States Notice that the graph representing the common boundaries between midwestern states is planar, meaning that it can be drawn on a flat surface without edges crossing. As we have seen, any planar graph has a chromatic number of four or less. This very well-known fact is called the Four-Color Theorem , or Four-Color Map Theorem . Coloring Graphs Part 2: Coloring Maps and the Four Color Problem Four-Color Theorem In the 19th century, Francis Guthrie was coloring a map of British counties and noticed that he could color so that no two adjacent counties were the same color using only four colors. This seemed to be the case for other maps as well, but he could not figure out why. He shared this with his brother, Frederick, who subsequently took the problem to his professor, August De Morgan, one of the most famous mathematicians of all time. De Morgan never was able to solve the problem, but he shared it with his colleagues. The problem continued to intrigue and perplex mathematicians for over a hundred years, inspiring discoveries in several areas of mathematics. It was finally proven by Kenneth Appel and Wolfgang Haken from the University of Illinois in 1976 using a combination of computer-aided calculations and graph theory. This was the first time in history that a mathematical theorem was proven using a computer! (Jesus Najera, “The Four-Color Theorem,” Cantor’s Paradise , a Medium publication, October 22, 2019) Using Four Colors or Fewer to Color a Graph Find a coloring of the graph in , which uses four colors or fewer. Use the resulting coloring as a guide to recolor the map in . How many colors did you use? Does this support the conclusion of the Four-Color Theorem? If so, how? The steps to color the graph are shown in . Step 1: Graph with degrees of vertices labeled. Vertex IA has highest degree. Step 2: Color vertex IA any color. Vertices ND, KS, MI, OH, and IN are NOT adjacent to IA. MI and IN have highest degree, 3. Step 3: Color either MI or IN the same color as IA. Vertex ND and KS are the only remaining vertices not adjacent to a red vertex. They both have degree 2. Step 4: Since KS and ND are not adjacent to each other, we can save a step and color both red. The highest degree of remaining vertices is four. Step 5: Choose one of the vertices of degree 4, SD, to color with a new color, blue. The vertices WI, IL, MO, IN,and OH are NOT adjacent to blue. WI, IL, and MO have the highest degree. Step 6: Choose one of WI, IL, or MO to color. We color WI blue. MO, IN and OH are NOT adjacent to blue. MO has the highest degree of these. Step 7: Color MO blue. All remaining vertices are adjacent to blue. Choose a new color. Four vertices remain, MN, NE, IL, and OH. MN, NE, and IL have the highest degree. Step 8: Since MN, IL, and NE are not adjacent, save steps and color all three the new color, purple. Vertex OH is the only remaining vertex that has NOT been colored. Step 9: Since vertex OH is not adjacent to purple, color it purple. This is the final graph. We used three colors. Steps to Color Graph of Midwestern States The final graph in shows how we would color the map. In we have colored the map to correspond to the colors on the graph. Midwestern States in Three Colors We used three colors to color the graph. This supports the Four-Color Theorem, because the graph is planar and its chromatic number is less than four. Coloring A Möbius Strip Have you ever heard of a möbius strip? It is a flat object with only one side. This might sound theoretical, but you can make one for yourself. Take a narrow strip of paper, make a half twist in it, and tape the ends together. To prove to yourself that it has only one side, pick a point anywhere on the paper and start drawing a line. You can draw the line continuously without lifting your pen from the paper and eventually, you will get back to the point at which you started. Since you never had to flip over the paper, you have just proved it has one side! Now, you might be wondering what this has to do with Graph Theory. It turns out that a map drawn on a möbius strip cannot necessarily be drawn using four colors. This doesn’t contradict the Four-Color Theorem, which only applies to graphs that are planar. A möbius strip does not live on a “plane,” or flat surface, which means the maps are not always planar like those drawn on a flat piece of paper. In the case of a möbius strip, it is the Six-Color Map Theorem! is an example of a möbius strip with a map that requires five colors. A Five-Coloring on a Möbius Strip Neil deGrasse Tyson Explains the Möbius Strip Check Your Understanding Key Terms walk (directed walk) trail (directed trail) path (directed path) closed open closed walk circuit (closed trail) directed cycle (closed path) coloring (graph coloring) n -coloring chromatic number Key Concepts Walks, trails, and paths are ways to navigate through a graph using a sequence of connected vertices and edges. Closed walks, circuits, and directed cycles are ways to navigate from a vertex on a graph and return to the same vertex. Colorings are a way to organize the vertices of a graph into groups so that no two members of a group are adjacent. Maps can be represented with planar graphs, which can always be colored using four colors or fewer. Videos Walks, Trails, and Paths in Graph Theory Closed Walks, Closed Trails (Circuits), and Closed Paths (Directed Cycles) in Graph Theory Coloring Graphs Part 1: Coloring and Identifying Chromatic Number The Four Color Map Theorem – Numberphile Coloring Graphs Part 2: Coloring Maps and the Four Color Problem Neil deGrasse Tyson Explains the Möbius Strip", "section": "Navigating Graphs", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Euler Circuits Delivery trucks move goods from place to place. (credit: “Mack Midliner” by Jason Lawrence/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Determine if a graph is connected. State the Chinese postman problem. Describe and identify Euler Circuits. Apply the Euler Circuits Theorem. Eulerize graphs in real-world applications. The delivery of goods is a huge part of our daily lives. From the factory to the distribution center, to the local vendor, or to your front door, nearly every product that you buy has been shipped multiple times to get to you. If the cost and time of that delivery is too great, you will not be able to afford the product. Delivery personnel have to leave from one location, deliver the goods to various places, and then return to their original location and do all of this in an organized way without losing money. How do delivery services find the most efficient delivery route? The answer lies in graph theory. Connectedness Before we can talk about finding the best delivery route, we have to make sure there is a route at all. For example, suppose that you were tasked with visiting every airport on the graph in by plane. Could you accomplish that task, only taking direct flight paths between airports listed on this graph? In other words, are all the airports connected by paths? Or are some of the airports disconnected from the others? Major Central and South Florida Airports In , we can see TPA is adjacent to PBI, FLL, MIA, and EYW. Also, there is a path between TPA and MCO through FLL. This indicates there is a path between each pair of vertices. So, it is possible to travel to each of these airports only taking direct flight paths and visiting no other airports. In other words, the graph is connected because there is a path joining every pair of vertices on the graph. Notice that if one vertex is connected to each of the others in a graph, then all the vertices are connected to each other. So, one way to determine if a graph is connected is to focus on a single vertex and determine if there is a path between that vertex and each of the others. If so, the graph is connected. If not, the graph is disconnected, which means at least one pair of vertices in a graph is not joined by a path. Let’s take a closer look at graph X in . Focus on vertex a . There is a path between vertices a and b , but there is no path between vertex a and vertex c . So, Graph X is disconnected. Connected vs. Disconnected When you are working with a planar graph, you can also determine if a graph is connected by untangling it. If you draw it so that so that none of the edges are overlapping, like we did with Graph X in , it is easier to see that the graph is disconnected. Untangling Graph X Versions 2 and 3 of Graph X in each have the same number of vertices, number of edges, degrees of the vertices, and pairs of adjacent vertices as version 1. In other words, each version retains the same structure as the original graph. Since versions 2 and 3 of Graph X , do not have overlapping edges, it is easier to identify pairs of vertices that do not have paths between them, and it is more obvious that Graph X is disconnected. In fact, there are two completely separate, disconnected subgraphs, one with the vertices in { a, b, e }, and the other with the vertices { c, d, f } These sets of vertices together with all of their edges are called components of Graph X . A component of a graph is a subgraph in which there is a path between each pair of vertices in the subgraph, but no edges between any of the vertices in the subgraph and a vertex that is not in the subgraph. Now let’s focus on Graph Y from . As in Graph X , there is a path between vertices a and b , as well as between vertices a and e , but Graph Y is different from Graph X because of vertex g . Not only is there a path between vertices a and g , but vertex g bridges the gap between a and c with the path a → b → g → c . Similarly, there is a path between vertices a and d and vertices a and f : a → b → g → d , a → b → g → d → f . Since there is a path between vertex a and every other vertex, Graph Y is connected. You can also see this a bit more clearly by untangling Graph Y as in . Even when Y is drawn so that the edges do not overlap, the graph cannot be drawn as two separate, unconnected pieces. In other words, Graph Y has only one component with the vertices { a, b, c, d, e, f }. We can give an alternate definition of connected and disconnected using the idea of components. A graph is connected if it has only one component. A graph is disconnected if it has more than one component. These alternate definitions are equivalent to the previous definitions. This means that you can confirm a graph is connected or disconnected either by checking to see if there is a path between each vertex and each other vertex, or by identifying the number of components. A graph is connected if it has only one component. Untangling Graph Y Determining If a Graph Is Connected or Disconnected Use to answer each question. Graph E Find a path between vertex a and every other vertex on the graph, if possible. Identify all the components of Graph E . Determine whether the graph is connected or disconnected and explain how you know. The paths are a → d → b , a → d → c , and a → d . There is only one component in Graph E . It has the vertices { a, b, c, d }. The graph is connected, because there is a path between vertex a and every other vertex. We can also show that Graph E is connected because it has only one component. Connected and Disconnected Graphs in Graph Theory Applying Connectedness The U.S. Interstate Highway System extends throughout the 48 contiguous states. It also has routes in the states of Hawaii and Alaska, and the commonwealth of Puerto Rico. Consider a graph representing the U.S. Interstate Highway System, in which there is a vertex for each of the 50 states and Puerto Rico, and an edge is drawn between any two vertices representing states that are connected by a highway in that system. Would this graph be connected or disconnected? Explain your reasoning. Hawaii, Alaska, and Puerto Rico are geographically separate from the 48 contiguous states, and each from each other. This means that there are vertices on the graph with no path joining them to the other vertices. So the graph is disconnected. Origin of Euler Circuits The city of Konigsberg, modern day Kaliningrad, Russia, has waterways that divide up the city. In the 1700s, the city had seven bridges over the various waterways. The map of those bridges is shown in . The question as to whether it was possible to find a route that crossed each bridge exactly once and return to the starting point was known as the Konigsberg Bridge Problem. In 1735, one of the most influential mathematicians of all time, Leonard Euler, solved the problem using an area of mathematics that he created himself, graph theory! Map of the Bridges of Konigsberg in 1700s (credit: “Konigsberg Bridge” by Merian Erben/Wikimedia Commons, Public Domain) Euler drew a multigraph in which each vertex represented a land mass, and each edge represented a bridge connecting them, as shown in . Remember from Navigating Graphs that a circuit is a trail, so it never repeats an edge, and it is closed, so it begins and ends at the same vertex. Euler pointed out that the Konigsberg Bridge Problem was the same as asking this graph theory question: Is it possible to find a circuit that crosses every edge? Since then, circuits (or closed trails) that visit every edge in a graph exactly once have come to be known as Euler circuits in honor of Leonard Euler. Recognizing Euler Trails and Euler Circuits Euler was able to prove that, in order to have an Euler circuit, the degrees of all the vertices of a graph have to be even. He also proved that any graph with that characteristic must have an Euler circuit. So, saying that a connected graph is Eulerian is the same as saying it has vertices with all even degrees, known as the Eulerian circuit theorem . Graph of Konigsberg Bridges To understand why the Euler circuit theorem is true, think about a vertex of degree 3 on any graph, as shown in . A Vertex of Degree 3 First imagine the vertex of degree 3 shown in is not the starting vertex. At some point, each edge must be traveled. The first time one of the three edges is traveled, the direction will be toward the vertex, and the second time it will be away from the vertex. Then, at some point, the third edge must be traveled coming in toward the vertex again. This is a problem, because the only way to get back to the starting vertex is to then visit one of the three edges a second time. So, this vertex cannot be part of an Euler circuit. Next imagine the vertex of degree 3 shown in is the starting vertex. The first time one of the edges is traveled, the direction is away from the vertex. At some point, the second edge will be traveled coming in toward the vertex, and the third edge will be the way back out, but the starting vertex is also the ending vertex in a circuit. The only way to return to the vertex is now to travel one of the edges a second time. So, again, this vertex cannot be part of an Euler circuit. For the same reason that a vertex of degree 3 can never be part of an Euler circuit, a vertex of any odd degree cannot either. We can use this fact and the graph in to solve the Konigsberg Bridge Problem. Since the degrees of the vertices of the graph in are not even, the graph is not Eulerian and it cannot have an Euler circuit. This means it is not possible to travel through the city of Konigsberg, crossing every bridge exactly once, and returning to your starting position. Degrees of Vertices in Konigsberg Bridge Multigraph Existence of Euler Circuits in Graph Theory Chinese Postman Problem At Camp Woebegone, campers travel the waterways in canoes. As part of the Camp Woebegone Olympics, you will hold a canoeing race. You have placed a checkpoint on each of the 11 different streams. The competition requires each team to travel each stream, pass through the checkpoints in any order, and return to the starting line, as shown in the . Map of Canoe Event Checkpoints Since the teams want to go as fast as possible, they would like to find the shortest route through the course that visits each checkpoint and returns to the starting line. If possible, they would also like to avoid backtracking. Let’s visualize the course as a multigraph in which the vertices represent turns and the edges represent checkpoints as in . Multigraph of Canoe Event The teams would like to find a closed walk that repeats as few edges as possible while still visiting every edge. If they never repeat an edge, then they have found a closed trail, which is a circuit. That circuit must cover all edges; so, it would be an Euler circuit. The task of finding a shortest circuit that visits every edge of a connected graph is often referred to as the Chinese postman problem . The name Chinese postman problem was coined in honor of the Chinese mathematician named Kwan Mei-Ko in 1960 who first studied the problem. If a graph has an Euler circuit, that will always be the best solution to a Chinese postman problem. Let’s determine if the multigraph of the course has an Euler circuit by looking at the degrees of the vertices in . Since the degrees of the vertices are all even, and the graph is connected, the graph is Eulerian. It is possible for a team to complete the canoe course in such a way that they pass through each checkpoint exactly once and return to the starting line. Degrees of Vertices in Graph of Canoe Event Understanding Eulerian Graphs A postal delivery person must deliver mail to every block on every street in a local subdivision. is a map of the subdivision. Use the map to answer each question. Map of Subdivision Draw a graph or multigraph to represent the subdivision in which the vertices represent the intersections, and the edges represent streets. Is your graph connected? Explain how you know. Determine the degrees of the vertices in the graph. Is your graph an Eulerian graph? Is it possible for the postal delivery person to visit each block on each street exactly once? Justify your answer. The graph is in . Graph of Subdivision The graph is connected. It has only one component and there is a path between each pair of vertices. There are four corner vertices of degree 2, there are eight exterior vertices of degree 3, and there are four interior vertices of degree 4. The graph is not Eulerian because it has vertices of odd degree. No. Since the graph is not Eulerian, there is no Euler circuit, which means that there is no route that would pass through every edge exactly once. Identifying Euler Circuits Solving the Chinese postman problem requires finding a shortest circuit through any graph or multigraph that visits every edge. In the case of Eulerian graphs, this means finding an Euler circuit. The method we will use is to find any circuit in the graph, then find a second circuit starting at a vertex from the first circuit that uses only edges that were not in the first circuit, then find a third circuit starting at a vertex from either of the first two circuits that uses only edges that were not in the first two circuits, and then continue this process until all edges have been used. In the end, you will be able to link all the circuits together into one large Euler circuit. Let’s find an Euler circuit in the map of the Camp Woebegone canoe race. In , we have labeled the edges of the multigraph so that the circuits can be named. In a multigraph it is necessary to name circuits using edges and vertices because there can be more than one edge between adjacent vertices. Multigraph of Canoe Race with Vertices and Edges Labeled We will begin with vertex 1 because it represents the starting line in this application. In general, you can start at any vertex you want. Find any circuit beginning and ending with vertex 1. Remember, a circuit is a trail, so it doesn’t pass through any edge more than once. shows one possible circuit that we could use as the first circuit, 1 → A → 2 → B → 3 → C → 4 → G → 5 → J → 1. First Circuit From the edges that remain, we can form two more circuits that each start at one of the vertices along the first circuit. Starting at vertex 3 we can use 3 → H → 5 → I → 1 → K → 3 and starting at vertex 2 we can use 2 → D → 6 → E → 7 → F → 2, as shown in . Second and Third Circuits Now that each of the edges is included in one and only once circuit, we can create one large circuit by inserting the second and third circuits into the first. We will insert them at their starting vertices 2 and 3 1 → A → 2 → B → 3 → C → 4 → G → 5 → J → 1 becomes 1 → A → 2 → D → 6 → E → 7 → F → 2 → B → 3 → H → 5 → I → 1 → K → 3 → C → 4 → G → 5 → J → 1 Finally, we can name the circuit using vertices, 1 → 2 → 6 → 7 → 2 → 3 → 5 → 1 → 3 → 4 → 5 → 1, or edges, A → D → E → F → B → H → I → K → C → G → J . Let's review the steps we used to find this Eulerian Circuit. Steps to Find an Euler Circuit in an Eulerian Graph Step 1 - Find a circuit beginning and ending at any point on the graph. If the circuit crosses every edges of the graph, the circuit you found is an Euler circuit. If not, move on to step 2. Step 2 - Beginning at a vertex on a circuit you already found, find a circuit that only includes edges that have not previously been crossed. If every edge has been crossed by one of the circuits you have found, move on to Step 3. Otherwise, repeat Step 2. Step 3 - Now that you have found circuits that cover all of the edges in the graph, combine them into an Euler circuit. You can do this by inserting any of the circuits into another circuit with a common vertex repeatedly until there is one long circuit. Finding an Euler Circuit Use to answer each question. Graph F Verify the Graph F is Eulerian. Find an Euler circuit that begins and ends at vertex c . The graph is connected because it has only one component. Also, each of the vertices in graph F has even degree as shown in . So, the graph is Eulerian. Degrees of Vertices in Graph F Step 1: Beginning at vertex c , identify circuit c → e → b → c as shown in . Since this circuit does not cover every edge in the graph, move on to Step 2. First Circuit in Graph F Step 2: Find another circuit beginning at one of the vertices in the first circuit, using only edges that have not been used in the first circuit. It is possible to do this using either vertex c or vertex b . In , we have used vertex b as the starting and ending point for a second circuit, b → d → c → a → b . Since all edges have been crossed by one of the two circuits, move on to Step 3. Second Circuit in Graph F Step 3: Combine the two circuits into one. Replace vertex b in the first circuit with the whole second circuit that begins at vertex b . c → e → b → c becomes c → e → b → d → c → a → b → c . An Euler circuit that begins at vertex c is c → e → b → d → c → a → b → c . Eulerization The Chinese postman problem doesn’t only apply to Eulerian graphs. Recall the postal delivery person who needed to deliver mail to every block of every street in a subdivision. We used the map in to create the graph in . Map of Subdivision Graph of Subdivision Since the graph of the subdivision has vertices of odd degree, there is no Euler circuit. This means that there is no route through the subdivision that visits every block of every street without repeating a block. What should our delivery person do? They need to repeat as few blocks as possible. The technique we will use to find a closed walk that repeats as few edges as possible is called eulerization . This method adds duplicate edges to a graph to create vertices of even degree so that the graph will have an Euler circuit. In , the eight vertices of odd degree in the graph of the subdivision are circled in green. We have added duplicate edges between the pairs of vertices, which changes the degrees of the vertices to even degrees so the resulting multigraph has an Euler circuit. In other words, we have eulerized the graph. An Eulerized Graph The duplicate edges in the eulerized graph correspond to blocks that our delivery person would have to travel twice. By keeping these duplicate edges to a minimum, we ensure the shortest possible route. It can be challenging to determine the fewest duplicate edges needed to eulerize a graph, but you can never do better than half the number of odd vertices. In the graph in , we have found a way to fix the eight vertices of odd degree with only four duplicate edges. Since four is half of eight, we will never do better. IMPORTANT! The duplicate edges that we add indicate places where the route will pass twice. An entirely new edge between two vertices that were not previously adjacent would indicate that our postal delivery person created a new road through someone’s property! So, we can duplicate existing edges, but we cannot create new ones . Eulerizing Graphs Use Graph A and multigraphs B , C , D , and E given in to answer the questions. Graph A and Multigraphs B through E Which of the multigraphs are not eulerizations of Graph A ? Explain your answer. Which eulerization of Graph A uses the fewest duplicate edges? How many does it use? Is it possible to eulerize Graph A using fewer duplicate edges than your answer to part 2? If so, give an example. If not, explain why not. Multigraph B is not an eulerization of A because the edge N between vertices c and d is not a duplicate of an existing edge. Multigraph E is not an eulerization of A because vertices b and e have odd degree. Multigraph C uses 3 duplicate edges while multigraph D only uses 2. So, D uses the fewest. Since there were four vertices of odd degree in Graph A , the fewest number of edges that could possibly eulerize the graph is half of four, which is two. So, it is not possible to eulerize Graph A using fewer edges. IMPORTANT! Since only duplicate edges can be added to eulerize a graph, it is not possible to eulerize a disconnected graph. Map of the Magical Land of Oz (credit: “Map of Oz within the surrounding deserts” by L. Frank Baum/Wikimedia Commons, Public Domain) In The Wonderful Wizard of Oz , written by L. Frank Baum and illustrated by W. W. Denslow, the region of the Emerald City lies at the center of the magical land of Oz, with Gillikin Country to the north, Winkie Country to the east, Munchkin Country to the west, and Quadling Country to the south. Munchkin Country and Winkie Country each shares a border with Gillikin Country and Quadling Country. Let’s apply graph theory to Dorothy’s famous journey through Oz. Draw a graph in which each vertex is one of the regions of Oz. Then answer each question: Is there an Euler trail circuit that Dorothy could follow, instead of the yellow brick road, to lead her from the land of the Munchkins, through all the regions of Oz and back, passing over each border exactly once? If not, how could the graph be Eulerized most efficiently? What is the chromatic number of the graph? Find a graph coloring that demonstrates your answer and use it to draw and color a graph of Oz. Check Your Understanding Key Terms connected component disconnected Euler circuit Eulerian graph Chinese postman problem Eulerization Key Concepts A connected graph has only one component. The Euler circuit theorem states that an Euler circuit exists in every connected graph in which all vertices have even degree, but not in disconnected graphs or any graph with one or more vertices of odd degree. The Chinese postman problem asks how to find the shortest closed trail that visits all edges at least once. If an Euler circuit exists, it is always the best solution to the Chinese postman problem. Eulerization is the process of adding duplicate edges to a graph so that the new multigraph has an Euler circuit. The minimum number of duplicated edges needed to eulerize a graph is half the number of odd vertices or more. Videos Connected and Disconnected Graphs in Graph Theory Recognizing Euler Trails and Euler Circuits Existence of Euler Circuits in Graph Theory", "section": "Euler Circuits", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Euler Trails The Pony Express mail route spanned from California to Missouri. (credit: “Map of Pony Express” by Nathan Hughes Hamiltonh/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and identify Euler trails. Solve applications using Euler trails theorem. Identify bridges in a graph. Apply Fleury’s algorithm. Evaluate Euler trails in real-world applications. We used Euler circuits to help us solve problems in which we needed a route that started and ended at the same place. In many applications, it is not necessary for the route to end where it began. In other words, we may not be looking at circuits, but trails, like the old Pony Express trail that led from Sacramento, California in the west to St. Joseph, Missouri in the east, never backtracking. Euler Trails If we need a trail that visits every edge in a graph, this would be called an Euler trail . Since trails are walks that do not repeat edges, an Euler trail visits every edge exactly once. Recognizing Euler Trails Use to determine if each series of vertices represents a trail, an Euler trail, both, or neither. Explain your reasoning. Graph H a → b → e → g → f → c → d → e a → b → e → g → f → c → d → e → b → a → d → g g → d → a → b → e → d → c → f → g → e It is a trail only. It is a trail because it is a walk that doesn’t cover any edges twice, but it is not an Euler trail because it didn’t cover edges ad or dg . It is neither. It is not a trail because it visits ab and be twice. Since it is not a trail, it cannot be an Euler trail. It is both. It is a trail because it is a walk that doesn’t cover any edges twice, and it is an Euler trail because it visits all the edges. The Five Rooms Puzzle Just as Euler determined that only graphs with vertices of even degree have Euler circuits, he also realized that the only vertices of odd degree in a graph with an Euler trail are the starting and ending vertices. For example, in , Graph H has exactly two vertices of odd degree, vertex g and vertex e . Notice the Euler trail we saw in Excercise 3 of began at vertex g and ended at vertex e . This is consistent with what we learned about vertices off odd degree when we were studying Euler circuits. We saw that a vertex of odd degree couldn't exist in an Euler circuit as depicted in . If it was a starting vertex, at some point we would leave the vertex and not be able to return without repeating an edge. If it was not a starting vertex, at some point we would return and not be able to leave without repeating an edge. Since the starting and ending vertices in an Euler trail are not the same, the start is a vertex we want to leave without returning, and the end is a vertex we want to return to and never leave. Those two vertices must have odd degree, but the others cannot. A Vertex of Degree 3 Let’s use the Euler trail theorem to solve a puzzle so you can amaze your friends! This puzzle is called the “Five Rooms Puzzle.” Suppose that you were in a house with five rooms and the exterior. There is a doorway in every shared wall between any two rooms and between any room and the exterior as shown in . Could you find a route through the house that passes through each doorway exactly once? Five Rooms Puzzle Let’s represent the puzzle with a graph in which vertices are rooms (or the exterior) and an edge indicates a door between two rooms as shown in . Graph of Five Rooms Puzzle To pass through each doorway exactly once means that we cross every edge in the graph exactly once. Since we have not been asked to start and end at the same position, but to visit each edge exactly once, we are looking for an Euler trail. Let’s check the degrees of the vertices. Degrees of Vertices in Five Rooms Puzzle Since there are more than two vertices of odd degree as shown in , the graph of the five rooms puzzle contains no Euler path. Now you can amaze and astonish your friends! Bridges and Local Bridges Now that we know which graphs have Euler trails, let’s work on a method to find them. The method we will use involves identifying bridges in our graphs. A bridge is an edge which, if removed, increases the number of components in a graph. Bridges are often referred to as cut-edges. In , there are several examples of bridges. Notice that an edge that is not part of a cycle is always a bridge, and an edge that is part of a cycle is never a bridge. Graph with Bridges Edges bf , cg , and dg are “bridges” The graph in is connected, which means it has exactly one component. Each time we remove one of the bridges from the graph the number of components increases by one as shown in . If we remove all three, the resulting graph in has four components. Removing a Bridge Increases Number of Components In sociology, bridges are a key part of social network analysis. Sociologists study two kinds of bridges: local bridges and regular bridges. Regular bridges are defined the same in sociology as in graph theory, but they are unusual when studying a large social network because it is very unlikely a group of individuals in a large social network has only one link to the rest of the network. On the other hand, a local bridge occurs much more frequently. A local bridge is a friendship between two individuals who have no other friends in common. If they lose touch, there is no single individual who can pass information between them. In graph theory, a local bridge is an edge between two vertices, which, when removed, increases the length of the shortest path between its vertices to more than two edges. In , a local bridge between vertices b and e has been removed. As a result, the shortest path between b and e is b → i → j → k → e , which is four edges. On the other hand, if edge ab were removed, then there are still paths between a and b that cover only two edges, like a → i → b . Removing a Local Bridge The significance of a local bridge in sociology is that it is the shortest communication route between two groups of people. If the local bridge is removed, the flow of information from one group to another becomes more difficult. Let’s say that vertex b is Brielle and vertex e is Ella. Now, Brielle is less likely to hear about things like job opportunities that Ella may know about. This is likely to impact Brielle as well as the friends of Brielle. Identifying Bridges and Local Bridges Use the graph of a social network in to answer each question. Graph of a Social Network Identify any bridges. If all bridges were removed, how many components would there be in the resulting graph? Identify one local bridge. For the local bridge you identified in part 3, identify the shortest path between the vertices of the local bridge if the local bridge were removed. The edges ku, gh , and hi are bridges. If the bridges were all removed, there would be four components in the resulting graph, { i }, { h }, { u , v , w , x }, and { a , b , c , d , e , f , g , j , k , m , n , o , p , q , r , s , t } as shown in . Graph of Social Network without Bridges Three local bridges are dn, ef , and qt , among others. If dn were removed, the shortest path between d and n would be d → e → f → j → o → m → n . Bridges and Local Bridges in Graph Theory Finding an Euler Trail with Fleury’s Algorithm Now that we are familiar with bridges, we can use a technique called Fleury’s algorithm , which is a series of steps, or algorithm, used to find an Euler trail in any graph that has exactly two vertices of odd degree. Here are the steps involved in applying Fleury’s algorithm. Step 1: Begin at either of the two vertices of odd degree. Step 2: Remove an edge between the vertex and any adjacent vertex that is NOT a bridge, unless there is no other choice, making a note of the edge you removed. Repeat this step until all edges are removed. Step 3: Write out the Euler trail using the sequence of vertices and edges that you found. For example, if you removed ab, bc, cd, de , and ef , in that order, then the Euler trail is a → b → c → d → e → f . shows the steps to find an Euler trail in a graph using Fleury’s algorithm. Using Fleury’s Algorithm To Find Euler Trail The Euler trail that was found in is t → v → w → u → t → w → y → x → v . Finding an Euler Trail with Fleury’s Algorithm Use Fleury’s Algorithm to find an Euler trail for Graph J in . Graph J Step 1: Choose one of the two vertices of odd degree, c or f , as your starting vertex. We will choose c . Step 2: Remove edge ca, cb , or cd . None of these are cut edges so we can select any of the three. We will choose cb as shown in to be the first edge removed. Graph J with cb Removed Repeat Step 2 The next choice is to remove edge ba, bd , or bf as shown in , but bf is not an option since it is a bridge. We will choose ba as shown in to be the second edge removed. Graph J with cb and ba Removed Repeat Step 2 for the third, fourth, fifth, sixth, and seventh edges. As shown in , until we get to the seventh edge there is only one option each time, ac, cd, db , and bf in that order. For the seventh edge, we must choose between fe and fg . Neither of these are bridges. We choose fe . shows that ac, cd, db, bf, and fe have been removed. Graph J with Seven Edges Removed Repeat Step 2 for the eight, ninth, tenth, and eleventh edges. As shown in , there is only one option for each of these edges, eh, hi, ig , and gf , in that order. Step 3: Write out the Euler trail using the vertices in the sequence that the edges were removed. We removed cb, ba, ac, cd, db, bf, fe, eh, hi, ig , and gf , in that order. The Euler trail is c → b → a → c → d → b → f → e → h → i → g → f . TIP! To avoid errors, count the number of edges in your graph and make sure that your Euler trail represents that number of edges. In the previous section, we found Euler circuits using an algorithm that involved joining circuits together into one large circuit. You can also use Fleury’s algorithm to find Euler circuits in any graph with vertices of all even degree. In that case, you can start at any vertex that you would like to use. Step 1: Begin at any vertex. Step 2: Remove an edge between the vertex and any adjacent vertex that is NOT a bridge, unless there is no other choice, making a note of the edge you removed. Repeat this step until all edges are removed. Step 3: Write out the Euler circuit using the sequence of vertices and edges that you found. For example, if you removed ab, bc, cd, de , and ea , in that order, then the Euler circuit is a → b → c → d → e → a . Fluery's Algorithm to Find an Euler Circuit IMPORTANT! Since a circuit is a closed trail, every Euler circuit is also an Euler trail, but when we say Euler trail in this chapter, we are referring to an open Euler trail that begins and ends at different vertices. Finding an Euler Circuit or Euler Trail Using Fleury’s Algorithm Use Fleury’s algorithm to find either an Euler circuit or Euler trail in Graph G in . Graph G Graph G has all vertices of even degree so it has an Euler circuit. Step 1: Choose any vertex. We will choose vertex j . Step 2: Remove one of the four edges that meet at vertex j . Since jn is a bridge, we must remove either jh, ji , or jk. We remove ji as shown in . Graph G with 1 Edge Removed Repeat Step 2: Since id is a bridge, we can remove either ih or ik next. We remove ih , and then the only option is to remove hj as shown in . Graph G with 3 Edges Removed Repeat Step 2: Since jn is a bridge, the next edge removed must be jk , and then the only option is to remove ki followed by id as shown in . Even though id is a bridge, it can be removed because it is the only option at this point. shows Graph G with these additional edges removed. Graph G with 6 Edges Removed Repeat Step 2: Choose any one of the edges db, dc , or de . We remove dc as shown in . Graph G with 7 Edges Removed Repeat Step 2: Since co is a bridge, choose cb next. We remove cb , then bd , and then de as shown in . Graph G with 10 Edges Removed Repeat Step 2: Next, remove ec and co . Then choose any of op, pn , or om . We remove on as shown in . Graph G with 13 Edges Removed Repeat Step 2: Next, remove either nm , np , or nj , but nj is a So, we remove nm as shown in . Graph G with 14 Edges Removed Repeat Step 2: Next, remove mo, op, pn , and nj . And we are done! Step 3: Notice that the algorithm brought us back to the vertex where we started, forming an Euler circuit. Write out the Euler circuit: j → i → h → j → k → i → d → c → b → d → e → c → o → n → m → o → p → n → j We have discussed a lot of subtle concepts in this section. Let’s make sure we are all on the same page. Work with a partner to explain why each of the following facts about bridges are true. Support your explanations with definitions and graphs. When a bridge is removed from a graph, the number of components increases. A bridge is never part of a circuit. An edge that is part of a triangle is never a local bridge. Check Your Understanding Key Terms algorithm Fleury’s algorithm Euler trail bridge local bridge Key Concepts An Euler trail exists whenever a graph has exactly two vertices of odd degree. When a bridge is removed from a graph, the number of components increases. A bridge is never part of a circuit. When a local bridge is removed from a graph, the distance between vertices increases. An edge that is part of a triangle is never a local bridge. Videos Bridges and Local Bridges in Graph Theory Fluery's Algorithm to Find an Euler Circuit", "section": "Euler Trails", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Hamilton Cycles The symmetries of an icosahedron, with 30 edges, 20 faces, and 12 vertices, can be analyzed using graph theory. (credit: \"Really big icosahedron\" by Clayton Shonkwiler/Wikimedia, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and identify Hamilton cycles. Compute the number of Hamilton cycles in a complete graph. Apply and evaluate weighted graphs. In Euler Circuits and Euler Trails , we looked for circuits and paths that visited each edge of a graph exactly once. In this section, we will look for circuits that visit each vertex exactly once. Like many concepts in graph theory, the idea of a circuit that visits each vertex once was inspired by games and puzzles. As early as the 9th century, Indian and Islamic intellectuals wondered whether it was possible for a knight to visit every space on a chess board of a given size, which is equivalent to visiting every vertex of a graph that represents the chess board. In 1857, a mathematician named William Rowan Hamilton invented a puzzle in which players were asked to find a route along the edges of a dodecahedron (see ), which visited every vertex exactly once. Let’s explore how graph theory provides insight into these games as well as practical applications such as the Traveling Salesperson Problem. A Dodecahedron Hamilton’s Puzzle Before we look at the solution to Hamilton's puzzle, let’s review some vocabulary we used in . It will be helpful to remember that directed cycle is a type of circuit that doesn’t repeat any edges or vertices. Closed Walks, Circuits, and Directed Cycles The goal of Hamilton's puzzle was to find a route along the edges of the dodecahedron, which visits each vertex exactly once. A dodecahedron is a three-dimensional space figure with faces that are all pentagons as we saw in . Since it is easier to visualize two dimensions rather than three, we will flatten out the dodecahedron and look at the edges and vertices on a flat surface. Graph A in shows a two-dimensional graph of the edges and vertices, and Graph B shows an untangled version of Graph A in which no edges are crossing. Graph B in is very similar to the design of the game board that Hamilton invented for his puzzle. Untangle graph Graph of Edges and Vertices of Dodecahedron We can see that this is a planar graph because it can be “untangled.” In order to solve Hamilton’s puzzle, we need to find a circuit that visits every vertex once. A solution is shown in . In three Dimensions > A Solution to Hamilton’s Puzzle A circuit that doesn’t repeat any vertices, like the one in , is called a directed cycle. So, we can most accurately say that Hamilton’s puzzle asks us to find a directed cycle that visits every vertex in a graph exactly once. Because Hamilton created and solved this puzzle, these special circuits were named Hamilton cycles , or Hamilton circuits . Icosian Game When Hamilton invented his puzzle in the 19th century, it was originally called the Icosian game. This was a reference to an icosahedron, not a dodecahedron. Why? Hamilton was interested in the symmetries of icosahedrons, specifically regular icosahedrons, which have faces that are all equilateral triangles. It turns out that this space figure has a surprising correspondence with the regular dodecahedron, a space figure with faces that are all regular pentagons. This pair of space figures have the same number of edges, while the number of faces and vertices are swapped. The icosahedron has 30 edges, 20 faces, and 12 vertices, while the dodecahedron has 30 edges, 12 faces, and 20 vertices. By relating the vertices of the dodecahedron to the faces of the icosahedron, Hamilton was able to make the mathematical connections necessary to use graph theory and dodecahedrons to make discoveries about the symmetries of icosahedrons. Hamilton Cycles vs. Euler Circuits Let’s practice naming and identifying Hamilton cycles, as well as distinguishing them from Euler circuits. It is important to remember that Euler circuits visit all edges without repetition, while Hamilton cycles visit all vertices without repetition. Hamilton cycles are named by their vertices just like all circuits. An example is given in . Hamilton Cycle in Graph Z Notice that the Hamilton cycle a → b → c → d for Graph Z in is NOT an Euler circuit, because it does not visit edge ac . Some Hamilton cycles are also Euler circuits while some are not, and some Euler circuits are Hamilton cycles while some are not. TIP! Sometimes students confuse Euler circuits with Hamilton cycles. To help you remember, think of the E in Euler as standing for Edge. Differentiating between Hamilton Cycles or Euler Circuits Use to determine whether the given circuit is a Hamilton cycle, an Euler circuit, both, or neither. Graph Q a → b → c → e → h → g → f → d → a g → e → h → g → f → d → a → b → d → g a → b → c → e → h → g → f → d → b → e → g → d → a This circuit is a Hamilton cycle only. It visits each vertex exactly once, so, it is a Hamilton cycle. It is not an Euler circuit because it doesn’t visit all of the edges. This circuit is neither Hamilton cycle nor an Euler circuit. It doesn’t visit vertex c , so, it is not a Hamilton cycle. It also doesn’t visit edges be and ce , so, it is not an Euler circuit. This circuit is an Euler circuit only. It visits several vertices more than once; so, it is not a Hamilton cycle. It visits every edge exactly once, so, it is an Euler circuit. TIP! Euler circuits never skip a vertex, so, they only fail to be Hamilton cycles when they visit a vertex more than once. Hamilton cycles never visit any edges twice, so, they only fail to be Euler circuits when they skip edges. Notice that the graph is a cycle. A cycle will always be Eulerian because all vertices are degree 2. Moreover, any circuit in the graph will always be both an Euler circuit and a Hamilton cycle. It is not always as easy to determine if a graph has a Hamilton cycle as it is to see that it has an Euler circuit, but there is a large group of graphs that we know will always have Hamilton cycles, the complete graphs. Since all vertices in a complete graph are adjacent, we can always find a directed cycle that visits all the vertices. For example, look at the directed six-cycle, n → o → p → q → r → s , in the complete graph with six vertices in . Directed Cycle in Complete Graph That is not the only directed six-cycle in the graph though. We could find another just be reversing the direction, and we could find even more by using different edges. So, how many Hamilton cycles are in a complete graph with n vertices? Before we tackle this problem, let’s look at a shorthand notation that we use in mathematics which will be helpful to us. Factorials In many areas of mathematics, we must calculate products like 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 or 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 , products that involve multiplying all the counting numbers from a particular number down to 1. Imagine that the product happened to be all the numbers from 100 down to 1. That’s a lot of writing! Instead of writing all of that out, mathematicians came up with a shorthand notation. For example, instead of 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 , we write 7 ! , which is read “7 factorial .” In other words, the product of all the counting numbers from n down to 1 is called n factorial and it is written n ! Evaluating Factorials Evaluate n ! and ( n − 1 ) ! for n = 4 . n ! = 4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 and ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ! = 3 ⋅ 2 ⋅ 1 = 6 A common use for factorials is counting the number of ways to arrange objects. Suppose that there were three students, Aryana, Byron, and Carlos, who wanted to line up in a row. How many arrangements are possible? There are six possibilities: ABC, ACB, BAC, BCA, CAB, or CBA. Notice that there were three students being arranged, and the number of possible arrangements is three. The number of ways to arrange n distinct objects is n ! . Counting Arrangements of Letters Find the number of ways to arrange the letters a, b, c, and d. 4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 Counting Hamilton Cycles in Complete Graphs Now, let’s get back to answering the question of how many Hamilton cycles are in a complete graph. In , we have drawn all the four cycles in a complete graph with four vertices. Remember, cycles can be named starting with any vertex in the cycle, but we will name them starting with vertex a . Complete Graph Cycle Cycle Cycle Cycle Name Clockwise (a, b, c, d) (a, b, d, c) (a, c, b, d) Cycle Name Counterclockwise (a, d, c, b) (a, c, d, b) (a, d, b, c) Four-Cycles in Complete Graph with Four Vertices shows that there are three unique four-cycles in a complete graph with four vertices. Notice that there were two ways to name each cycle, one reading the vertices in a clockwise direction and one reading the vertices in a counterclockwise direction. This is important to us because we are interested in Hamilton cycles, which are directed cycles. Although the cycles (a, b, c, d) and (a, d, c, b) are the same cycle, the directed cycles, a → b → c → d → a and a → d → c → b → a , which travel the same route in reverse order are considered different directed cycles, as shown in . Complete Graph Cycle Cycle Cycle Clockwise Hamilton Cycle a → b → c → d → a a → b → d → c → a a → c → b → d → a Counter-clockwise Hamilton Cycle a → d → c → b → a a → c → d → b → a a → d → b → c → a Hamilton Cycles in a Complete Graph with Four Vertices The six directed four-cycles in are the only distinct Hamilton cycles in a complete graph with four vertices. Six is also the number of ways to arrange the three letters b, c, and d. (Do you see why?) The number of ways to arrange three letters is 3 ! = 3 ⋅ 2 ⋅ 1 = 6 . Similarly, the number of Hamilton cycles in a graph with five vertices is the number of ways to arrange four letters, which is 4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 . In general, to find the number of Hamilton cycles in a graph, we take one less than the number of vertices and find its factorial. The number of distinct Hamilton cycles in a complete graph with n vertices is ( n − 1 ) ! Counting Hamilton Cycles in a Complete Graph How many Hamilton cycles are in the complete graph in ? Complete Graph L There are five vertices in the graph. Using n = 5 , we have ( n − 1 ) ! = ( 5 − 1 ) ! = 4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 Hamilton cycles. Weighted Graphs Suppose that an officer in the U.S. Air Force who is stationed at Vandenberg Air Force base must drive to visit three other California Air Force bases before returning to Vandenberg. The officer needs to visit each base once. The vertices in the graph in represent the four U.S. Air Force bases, Vandenberg, Edwards, Los Angeles, and Beale. The edges are labeled to with the driving distance between each pair of cities. Graph of Four California Air Force Bases The graph in is called a weighted graph , because each edge has been assigned a value or weight. The weights can represent quantities such as time, distance, money, or any quantity associated with the adjacent vertices joined by the edges. The total weight of any walk, trail, or path is the sum of the weights of the edges it visits. Notice that the officer’s trip can be represented as a Hamilton cycle, because each of the four vertices in the graph is visited exactly once. Finding Hamilton Cycles of Lowest Weight Use and the given Hamilton cycles to answer the following questions. V → L → E → B → V V → L → B → E → V V → E → L → B → V V → B → E → L → V Which of the Hamilton cycles (directed cycles) lie on the same cycle (undirected cycle) in the graph? Find the total weight of each cycle. Of the four, which of the Hamilton cycles describes the shortest trip for the officer? Describe the route. V → L → E → B → V and V → B → E → L → V follow the same edges in reverse order. Any Hamilton cycles that lie on the same cycle will have the same edges and the same total weight. V → L → E → B → V and V → B → E → L → V each have total weight 159 + 106 + 410 + 396 = 1071 . V → L → B → E → V has a total weight 159 + 439 + 410 + 207 = 1215 . V → E → L → B → V has a total weight 396 + 439 + 106 + 207 = 1148 . Hamilton cycles V → L → E → B → V and V → B → E → L → V each have the lowest total weight. The officer would take the route from Vandenberg, to Los Angeles, to Edwards, to Beale, and back to Vandenberg, or reverse that route. Check Your Understanding Key Terms Hamilton cycle, or Hamilton circuit n factorial weighted graph total weight Key Concepts A Hamilton cycle is a directed cycle, or circuit, that visits each vertex exactly once. Some Hamilton cycles are also Euler circuits, but some are not. Hamilton cycles that follow the same undirected cycle in the same direction are considered the same cycle even if they begin at a different vertex. The number of unique Hamilton cycles in a complete graph with n vertices is the same as the number of ways to arrange n − 1 distinct objects. Weighted graphs have a value assigned to each edge, which can represent distance, time, money and other quantities. Formulas The number of ways to arrange n distinct objects is n ! . The number of distinct Hamilton cycles in a complete graph with n vertices is ( n − 1 ) ! .", "section": "Hamilton Cycles", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Hamilton Paths A school bus picks up children along a planned route. (credit: “Kids at School Bus Stop” by Ty Hatch/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and identify Hamilton paths. Evaluate Hamilton paths in real-world applications. Distinguish between Hamilton paths and Euler trails. In the United States, school buses carry 25 million children between school and home every day. The total distance they travel is around 6 billion kilometers per year. In the city of Boston, Massachusetts, the 2016 budget for running those buses was $120 million dollars. In 2017, the city held a competition to find ways to cut costs and the Quantum Team from the MIT Operations Research Center came to the rescue, using a computer algorithm to identify the most efficient and least costly routes, which saved the city of Boston $5 million each year and even reduced daily CO 2 emissions by 9,000 kilograms! ( This U.S. city put an algorithm in charge of its school bus routes and saved $5 million , Sean Fleming, World Economic Forum) The problem the Quantum Team tackled involves graph theory. Imagine a graph in which vertices are the bus depot, the school, and the bus stops along a particular route. The bus must start at the depot, visit every stop exactly once, and end at the school. The route is a special kind of path that visits every vertex exactly once. Can you guess what those paths are called? Hamilton Paths Just as circuits that visit each vertex in a graph exactly once are called Hamilton cycles (or Hamilton circuits), paths that visit each vertex on a graph exactly once are called Hamilton paths . As we explore Hamilton paths, you might find it helpful to refresh your memory about the relationships between walks, trails, and paths by looking at . We know that paths are walks that don’t repeat any vertices or edges. So, a Hamilton path visits every vertex without repeating any vertices or edges. shows a path from vertex A to vertex E and a Hamilton path from vertex A to vertex E . Walks, Trails, and Paths Path or Hamilton Path? Identifying Hamilton Paths Which of the following sequences of vertices is a Hamilton path for Graph Q in ? Graph Q a → d → b → c → e → g → f c → b → e → h → g → f → d → a h → e → g → d → b → e → g → f → d → a → b → c Sequence 1 is a path, because it is a walk that doesn’t repeat any vertices or edges, but not a Hamilton path because it skips vertex h . Sequence 2 is a path that visits each vertex exactly once; so, it is a Hamilton path. Sequence 3 is a walk, but it is not a path because it visits vertices g , e , and b each more than once; so, it cannot be a Hamilton path. So, we can see that only sequence 2 is a Hamilton path. TIP! Since a Hamilton path visits each vertex exactly once, it must have the same number of vertices listed as appear in the graph. Finding Hamilton Paths Suppose you were visiting an aquarium with some friends. The map of the aquarium is given in . The letters represent the exhibits. Map of Aquarium Exhibits shows a graph of the aquarium in which each vertex represents an exhibit and each edge is a route between the pair of exhibits that doesn’t bypass another exhibit. Graph of Aquarium Let’s see if we can plan a tour of the exhibits that visits each exhibit exactly once, beginning at exhibit O and ending at exhibit C . Suppose that, after exhibit O , we plan to visit exhibit Q and then exhibit M . After M , should we plan to visit N , L , or R ? Take a look at . If R is not chosen next, that will cause a problem later on. Do you see what it is? Choosing Vertex L , N , or R If L or N is chosen next, the only way to get to R later will be to go from S to R , and then we will not be able to continue without repeating a vertex. So, we will pick R next, and then the only option is S . After S we have another choice to make. As shown in , the next choice is between B and E . Keeping in mind that the goal is to end at C , which would be the better choice? Choosing Vertex B or E If you said vertex B , you are right! Otherwise, we will not be able to visit B later. After B , the only option is E . Then we can choose either D or G . Either will work fine. Let’s choose G as shown in . After G , you must visit H , but should you visit K or L after that? Choose Vertex L or K If you said to go to vertex L next, you are right! Otherwise, it will be impossible to visit N without repeating a vertex. So, next is L , then N , then K , and then at J you have another decision to make we can see in . Should you choose F , I , or P next? Choose Vertex F , I , or P If you said P , you are right! If you choose either of the other two vertices, you will not be able to visit P later without passing through another vertex twice. Once P is chosen, vertex I must be next followed by F . Then you have to choose between A and D as shown in . Choose Vertex A or D In this case, we must go to D then to A so that we can visit C without backtracking. The complete Hamilton path is shown in . Complete Hamilton Path from O to C So, one Hamilton path that begins at O and ends at C is O → Q → M → R → S → B → E → G → H → L → N → K → J → P → I → F → D → A → C . There is no set sequence of steps that can be used to find a Hamilton path if it exists, but it does help to keep in mind where we are headed and avoid choices that will make returning to a particular vertex impossible without repeating vertices. Let’s practice finding Hamilton paths. Finding a Hamilton Path Use to find a Hamilton path between vertices C and D . Graph G If we start at vertex C , A must be next. Then we must choose between B and F . If we choose F , we will have to backtrack to get to include B ; so, we must choose B . Once we choose B , we must choose F next. After F , we choose E , because we want to end at D . So, a Hamilton path between C and D is C → A → B → F → E → D . Existence of a Hamilton Path It turns out that there is no Hamilton path between vertices A and E in Graph G in . To understand why, let’s imagine there is a red apple tree on one side of a bridge and a green apple tree on the other side of the bridge. Now suppose someone asked you to pick up all the fallen apples under each tree without crossing the bridge more than once, and making sure that the first apple you pick up and the last apple you pick up are both red. You would say, that is impossible! To have the first and last apple be red would either require leaving the green apples on the ground or crossing the bridge twice. Let’s see how this relates to finding a Hamilton path between A and E in Graph G . The edge AC is a bridge because, if it were removed, the graph would become disconnected with two components, the component { C } and the component { A , B , D , E , F }. So, we can think of the vertices A , B , D , E , and F as the red apples, vertex C as the green apple, and the edge AC is the bridge between them as in . Bridge between Red and Green Apples The creation of a Hamilton path requires a visit to each vertex, just as picking up all the apples requires a visit to each apple. A and E are both red apples; so, a path from A to E would both start and end at a red apple, just as you were asked to do. And you wouldn’t be able to cross the bridge twice because that would mean visiting A twice, which is not allowed in a Hamilton path. So, it is impossible to find a Hamilton path from A to E just as it was impossible to pick up all the apples without crossing the bridge more than once. By the same reasoning, if a graph has a bridge, there will never be a Hamilton path that begins and ends on the same side of that bridge, meaning beginning and ending at vertices that would be in the same component if the bridge were removed from the graph. Finding a Hamilton Path If One Exists Find a Hamilton path from vertex s to vertex v for each graph in or indicate that there is none. Graphs A , B , C , and D Graph A : Edge uw is a bridge connecting component { s , t , u , v } to the component { w , x , y , z }. There is no Hamilton path from vertex s to vertex v because they would be part of the same component if the bridge uw were removed. Graph B : There are no bridges in Graph B . The only method we have to determine if a Hamilton path from vertex s to vertex v exists is to try every possibility. From vertex s , we can visit either vertex y or vertex t . We will try vertex y first and then come back to see what happens with vertex t . After visiting y , we must visit z and then u , but then we have to decide between vertices r , t , and v next as shown in . Choose Vertex r , t , or v Vertex v is not an option since we want to end at v . Vertex t is not an option since that would force us to go to visit s a second time. So, we must go to vertex r next. After vertex r , we must visit x , then w , then v , but we missed vertex t as shown in . Missed Vertex t Let’s go back to the beginning and choose t instead of y . After t , we must go to u and then we have a choice to make between r , v , and z as shown in . Choose Vertex v , r , or z Vertex v is not an option since we want to end at v . Vertex z is not an option since that would force us to go y and then to visit s a second time. So, we must go to vertex r next. After r , we must go to x then w then v , where we have to stop even though we have missed vertices y and z , as shown in . Missed Vertices y and z So, we have tried every possible route and there are no Hamilton paths between s and v in Graph B . Graph C : In Graph C , there is a Hamilton path, s → t → u → x → w → v . Graph D : In Graph D , there is a bridge, tx , which would form components { r , s , t , u , q } and { v , w , x , y , z } if it were removed. Since s and v would be in different components, it is possible there is a Hamilton path between them. The only way to know is to try all possibilities. If we begin at s , we can go to r then t , or we can go directly to t , either way, we have a problem as you can see in . Vertices Visited Twice or Skipped If we visit all the vertices in the component { r , s , t , u , q }, we will have to visit t a second time in order to cross the bridge. If we visit t only once, we have to skip some of the vertices. So, there is no Hamilton path between s and v . There is not a short way to determine if there is a Hamilton path between two vertices on a graph that works in every situation. However, there are a few common situations that can help us to quickly determine that there is no Hamilton path. Some of these are listed in . Scenario Diagram Scenario 1 If an edge ab is a bridge, then there is no Hamilton path between a pair of vertices that are on the same side of edge ab . We saw this in Graph A of . No Hamilton path between any two vertices in the component { a , c , d , f }. No Hamilton path between any two vertices in { b , e , h , g , i }. Scenario 2 If an edge ab is a bridge with at least three components on each side, then there is no Hamilton path beginning or ending at a or b . We saw this in Graph D of . No Hamilton path beginning or ending at a or b . Scenario 3 If a graph is composed of two cycles joined only at a single vertex p , and v is any vertex that is NOT adjacent to p , then there are no Hamilton paths beginning or ending at either p or v . We saw this in Graph B of . No Hamilton path can be formed starting or ending at vertices, r , v , or u because they are not adjacent to p . Some Impossible Hamilton Paths There are also many other situations in which a Hamilton path is not possible. These are just a few that you will encounter. Hamilton Path or Euler Trail? We learned in Euler Trails that an Euler trail visits each edge exactly once, whereas a Hamilton path visits each vertex exactly once. Let’s practice distinguishing between the two. Distinguishing between Hamilton Path and Euler Trail Use to determine if the given sequence of vertices is a Hamilton path, an Euler trail, both, or neither. Graphs A , F , and K Graph A , e → b → a → e → d → c → b Graph F , f → g → j → h → i Graph K , k → l → m → n → o Since the sequence covers every edge once but visits vertices more than once, it is only an Euler trail. Since the sequence visits every vertex exactly once but skips some edges, it is only a Hamilton path. Since the sequence visits each edge and each vertex exactly once, it is both an Euler trail and a Hamilton path. As we saw in , the Emerald City lies at the center of the Magical Land of Oz, with Gillikin Country to the north, Winkie Country to the east, Munchkin Country to the west, and Quadling Country to the south. Munchkin Country and Winkie Country each shares a border with Gillikin Country and Quadling Country. Let’s apply graph theory to Dorothy’s famous journey through Oz one more time! Draw a graph in which each vertex is one of the regions of Oz. Is there a Hamilton path that Dorothy could follow, instead of the yellow brick road, to lead her from the land of the Munchkins, through all the regions of Oz exactly once, and end in the Emerald City? If so, what might it be? Compare your results with those of a classmate. Check Your Understanding Key Terms Hamilton path Key Concepts A Hamilton path visits every vertex exactly once. Some Hamilton paths are also Euler trails, but some are not.", "section": "Hamilton Paths", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Traveling Salesperson Problem Each door on the route of a traveling salesperson can be represented as a vertex on a graph. (credit: \"Three in a row, Heriot Row\" by Jason Mason/Flickr, CC BY 2.1) Learning Objectives After completing this section, you should be able to: Distinguish between brute force algorithms and greedy algorithms. List all distinct Hamilton cycles of a complete graph. Apply brute force method to solve traveling salesperson applications. Apply nearest neighbor method to solve traveling salesperson applications. We looked at Hamilton cycles and paths in the previous sections Hamilton Cycles and Hamilton Paths . In this section, we will analyze Hamilton cycles in complete weighted graphs to find the shortest route to visit a number of locations and return to the starting point. Besides the many routing applications in which we need the shortest distance, there are also applications in which we search for the route that is least expensive or takes the least time. Here are a few less common applications that you can read about on a website set up by the mathematics department at the University of Waterloo in Ontario, Canada: Design of fiber optic networks Minimizing fuel expenses for repositioning satellites Development of semi-conductors for microchips A technique for mapping mammalian chromosomes in genome sequencing Before we look at approaches to solving applications like these, let's discuss the two types of algorithms we will use. Brute Force and Greedy Algorithms An algorithm is a sequence of steps that can be used to solve a particular problem. We have solved many problems in this chapter, and the procedures that we used were different types of algorithms. In this section, we will use two common types of algorithms, a brute force algorithm and a greedy algorithm . A brute force algorithm begins by listing every possible solution and applying each one until the best solution is found. A greedy algorithm approaches a problem in stages, making the apparent best choice at each stage, then linking the choices together into an overall solution which may or may not be the best solution. To understand the difference between these two algorithms, consider the tree diagram in . Suppose we want to find the path from left to right with the largest total sum. For example, branch A in the tree diagram has a sum of 10 + 2 + 11 + 13 = 36 . Points Along Different Paths To be certain that you pick the branch with greatest sum, you could list each sum from each of the different branches: A : 10 + 2 + 11 + 13 = 36 B : 10 + 2 + 11 + 8 = 31 C : 10 + 2 + 15 + 1 = 28 D : 10 + 2 + 15 + 6 = 33 E : 10 + 7 + 3 + 20 = 40 F : 10 + 7 + 3 + 14 = 34 G : 10 + 7 + 4 + 11 = 32 H : 10 + 7 + 4 + 5 = 26 Then we know with certainty that branch E has the greatest sum. Branch E Now suppose that you wanted to find the branch with the highest value, but you only were shown the tree diagram in phases, one step at a time. Tree Diagram Phase 1 After phase 1, you would have chosen the branch with 10 and 7. So far, you are following the same branch. Let’s look at the next phase. Tree Diagram Phase 2 After phase 2, based on the information you have, you will choose the branch with 10, 7 and 4. Now, you are following a different branch than before, but it is the best choice based on the information you have. Let’s look at the last phase. Tree Diagram Phase 3 After phase 3, you will choose branch G which has a sum of 32. The process of adding the values on each branch and selecting the highest sum is an example of a brute force algorithm because all options were explored in detail. The process of choosing the branch in phases, based on the best choice at each phase is a greedy algorithm. Although a brute force algorithm gives us the ideal solution, it can take a very long time to implement. Imagine a tree diagram with thousands or even millions of branches. It might not be possible to check all the sums. A greedy algorithm, on the other hand, can be completed in a relatively short time, and generally leads to good solutions, but not necessarily the ideal solution. Distinguishing between Brute Force and Greedy Algorithms A cashier rings up a sale for $4.63 cents in U.S. currency. The customer pays with a $5 bill. The cashier would like to give the customer $0.37 in change using the fewest coins possible. The coins that can be used are quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). The cashier starts by selecting the coin of highest value less than or equal to $0.37, which is a quarter. This leaves $ 0.37 − $ 0.25 = $ 0.12 . The cashier selects the coin of highest value less than or equal to $0.12, which is a dime. This leaves $ 0.12 − $ 0.10 = $ 0.02 . The cashier selects the coin of highest value less than or equal to $0.02, which is a penny. This leaves $ 0.02 − $ 0.01 = $ 0.01 . The cashier selects the coin of highest value less than or equal to $0.01, which is a penny. This leaves no remainder. The cashier used one quarter, one dime, and two pennies, which is four coins. Use this information to answer the following questions. Is the cashier’s approach an example of a greedy algorithm or a brute force algorithm? Explain how you know. The cashier’s solution is the best solution. In other words, four is the fewest number of coins possible. Is this consistent with the results of an algorithm of this kind? Explain your reasoning. The approach the cashier used is an example of a greedy algorithm, because the problem was approached in phases and the best choice was made at each phase. Also, it is not a brute force algorithm, because the cashier did not attempt to list out all possible combinations of coins to reach this conclusion. Yes, it is consistent. A greedy algorithm does not always yield the best result, but sometimes it does. The Traveling Salesperson Problem Now let’s focus our attention on the graph theory application known as the traveling salesperson problem (TSP) in which we must find the shortest route to visit a number of locations and return to the starting point. Recall from Hamilton Cycles , the officer in the U.S. Air Force who is stationed at Vandenberg Air Force base and must drive to visit three other California Air Force bases before returning to Vandenberg. The officer needed to visit each base once. We looked at the weighted graph in representing the four U.S. Air Force bases: Vandenberg, Edwards, Los Angeles, and Beal and the distances between them. Graph of Four California Air Force Bases Any route that visits each base and returns to the start would be a Hamilton cycle on the graph. If the officer wants to travel the shortest distance, this will correspond to a Hamilton cycle of lowest weight. We saw in that there are six distinct Hamilton cycles (directed cycles) in a complete graph with four vertices, but some lie on the same cycle (undirected cycle) in the graph. Complete Graph Cycle Cycle Cycle Clockwise Hamilton Cycle a → b → c → d → a a → b → d → c → a a → c → b → d → a Counterclockwise Hamilton Cycle a → d → c → b → a a → c → d → b → a a → d → b → c → a Hamilton Cycles in a Complete Graph with Four Vertices Since the distance between bases is the same in either direction, it does not matter if the officer travels clockwise or counterclockwise. So, there are really only three possible distances as shown in . Three Possible Distances The possible distances are: 396 + 410 + 106 + 159 = 1071 207 + 410 + 439 + 159 = 1215 396 + 439 + 106 + 207 = 1148 So, a Hamilton cycle of least weight is V → B → E → L → V (or the reverse direction). The officer should travel from Vandenberg to Beal to Edwards, to Los Angeles, and back to Vandenberg. Finding Weights of All Hamilton Cycles in Complete Graphs Notice that we listed all of the Hamilton cycles and found their weights when we solved the TSP about the officer from Vandenberg. This is a skill you will need to practice. To make sure you don't miss any, you can calculate the number of possible Hamilton cycles in a complete graph. It is also helpful to know that half of the directed cycles in a complete graph are the same cycle in reverse direction, so, you only have to calculate half the number of possible weights, and the rest are duplicates. In a complete graph with n vertices, The number of distinct Hamilton cycles is ( n − 1 ) ! . There are at most ( n − 1 ) ! 2 different weights of Hamilton cycles. TIP! When listing all the distinct Hamilton cycles in a complete graph, you can start them all at any vertex you choose. Remember, the cycle a → b → c → a is the same cycle as b → c → a → b so there is no need to list both. Calculating Possible Weights of Hamilton Cycles Suppose you have a complete weighted graph with vertices N, M, O , and P . Use the formula ( n − 1 ) ! to calculate the number of distinct Hamilton cycles in the graph. Use the formula ( n − 1 ) ! 2 to calculate the greatest number of different weights possible for the Hamilton cycles. Are all of the distinct Hamilton cycles listed here? How do you know? Cycle 1: N → M → O → P → N Cycle 2: N → M → P → O → N Cycle 3: N → O → M → P → N Cycle 4: N → O → P → M → N Cycle 5: N → P → M → O → N Cycle 6: N → P → O → M → N Which pairs of cycles must have the same weights? How do you know? There are 4 vertices; so, n = 4 . This means there are ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 distinct Hamilton cycles beginning at any given vertex. Since n = 4 , there are ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 possible weights. Yes, they are all distinct cycles and there are 6 of them. Cycles 1 and 6 have the same weight, Cycles 2 and 4 have the same weight, and Cycles 3 and 5 have the same weight, because these pairs follow the same route through the graph but in reverse. TIP! When listing the possible cycles, ignore the vertex where the cycle begins and ends and focus on the ways to arrange the letters that represent the vertices in the middle. Using a systematic approach is best; for example, if you must arrange the letters M, O, and P, first list all those arrangements beginning with M, then beginning with O, and then beginning with P, as we did in Example 12.42. The Brute Force Method The method we have been using to find a Hamilton cycle of least weight in a complete graph is a brute force algorithm, so it is called the brute force method . The steps in the brute force method are: Step 1: Calculate the number of distinct Hamilton cycles and the number of possible weights. Step 2: List all possible Hamilton cycles. Step 3: Find the weight of each cycle. Step 4: Identify the Hamilton cycle of lowest weight. Applying the Brute Force Method On the next assignment, the air force officer must leave from Travis Air Force base, visit Beal, Edwards, and Vandenberg Air Force bases each exactly once and return to Travis Air Force base. There is no need to visit Los Angeles Air Force base. Use to find the shortest route. Distances between Five California Air Force Bases Step 1: Since there are 4 vertices, there will be ( 4 − 1 ) ! = 3 ! = 6 cycles, but half of them will be the reverse of the others; so, there will be ( 4 − 1 ) ! 2 = 6 2 = 3 possible distances. Step 2: List all the Hamilton cycles in the subgraph of the graph in . Subgraph with Cities B, E, T , and V To find the 6 cycles, focus on the three vertices in the middle, B, E, and V . The arrangements of these vertices are BEV, BVE, EBV, EVB, VBE , and VEB . These would correspond to the 6 cycles: 1: T → B → E → V → T 2: T → B → V → E → T 3: T → E → B → V → T 4: T → E → V → B → T 5: T → V → B → E → T 6: T → V → E → B → T Step 3: Find the weight of each path. You can reduce your work by observing the cycles that are reverses of each other. 1: 84 + 410 + 207 + 396 = 1097 2: 84 + 396 + 207 + 370 = 1071 3: 370 + 410 + 396 + 396 = 1572 4: Reverse of cycle 2, 1071 5: Reverse of cycle 3, 1572 6: Reverse of cycle 1, 1097 Step 4: Identify a Hamilton cycle of least weight. The second path, T → B → V → E → T , and its reverse, T → E → V → B → T , have the least weight. The solution is that the officer should travel from Travis Air Force base to Beal Air Force Base, to Vandenberg Air Force base, to Edwards Air Force base, and return to Travis Air Force base, or the same route in reverse. Now suppose that the officer needed a cycle that visited all 5 of the Air Force bases in . There would be ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 different arrangements of vertices and ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 distances to compare using the brute force method. If you consider 10 Air Force bases, there would be ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 different arrangements and ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 distances to consider. There must be another way! The Nearest Neighbor Method When the brute force method is impractical for solving a traveling salesperson problem, an alternative is a greedy algorithm known as the nearest neighbor method , which always visit the closest or least costly place first. This method finds a Hamilton cycle of relatively low weight in a complete graph in which, at each phase, the next vertex is chosen by comparing the edges between the current vertex and the remaining vertices to find the lowest weight. Since the nearest neighbor method is a greedy algorithm, it usually doesn’t give the best solution, but it usually gives a solution that is \"good enough.\" Most importantly, the number of steps will be the number of vertices. That’s right! A problem with 10 vertices requires 10 steps, not 362,880. Let’s look at an example to see how it works. Suppose that a candidate for governor wants to hold rallies around the state. They plan to leave their home in city A , visit cities B, C, D, E , and F each once, and return home. The airfare between cities is indicated in the graph in . Airfares between Cities A, B, C, D, E , and F Let’s help the candidate keep costs of travel down by applying the nearest neighbor method to find a Hamilton cycle that has a reasonably low weight. Begin by marking starting vertex as V 1 for \"visited 1st.\" Then to compare the weights of the edges between A and vertices adjacent to A : $250, $210, $300, $200, and $100 as shown in . The lowest of these is $100, which is the edge between A and F . Finding the Second Vertex Mark F as V 2 for \"visited 2nd\" then compare the weights of the edges between F and the remaining vertices adjacent to F : $170, $330, $150 and $350 as shown in . The lowest of these is $150, which is the edge between F and D . Finding the Third Vertex Mark D as V 3 for \"visited 3rd.\" Next, compare the weights of the edges between D and the remaining vertices adjacent to D : $120, $310, and $270 as shown in . The lowest of these is $120, which is the edge between D and B . Finding the Fourth Vertex So, mark B as V 4 for \"visited 4th.\" Finally, compare the weights of the edges between B and the remaining vertices adjacent to B : $160 and $220 as shown in . The lower amount is $160, which is the edge between B and E . Finding the Fifth Vertex Now you can mark E as V 5 and mark the only remaining vertex, which is C , as V 6 . This is shown in . Make a note of the weight of the edge from E to C , which is $180, and from C back to A , which is $210. Finding the Sixth Vertex The Hamilton cycle we found is A → F → D → B → E → C → A . The weight of the circuit is $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 . This may or may not be the route with the lowest cost, but there is a good chance it is very close since the weights are most of the lowest weights on the graph and we found it in six steps instead of finding 120 different Hamilton cycles and calculating 60 weights. Let’s summarize the procedure that we used. Step 1: Select the starting vertex and label V 1 for \"visited 1st.\" Identify the edge of lowest weight between V 1 and the remaining vertices. Step 2: Label the vertex at the end of the edge of lowest weight that you found in previous step as V n where the subscript n indicates the order the vertex is visited. Identify the edge of lowest weight between V n and the vertices that remain to be visited. Step 3: If vertices remain that have not been visited, repeat Step 2. Otherwise, a Hamilton cycle of low weight is V 1 → V 2 → ⋯ → V n → V 1 . Using the Nearest Neighbor Method Suppose that the candidate for governor wants to hold rallies around the state but time before the election is very limited. They would like to leave their home in city A , visit cities B , C , D , E , and F each once, and return home. The airfare between cities is not as important as the time of travel, which is indicated in . Use the nearest neighbor method to find a route with relatively low travel time. What is the total travel time of the route that you found? Travel Times between Cities A, B, C, D, E and F Step 1: Label vertex A as V 1 . The edge of lowest weight between A and the remaining vertices is 85 min between A and D . Step 2: Label vertex D as V 2 . The edge of lowest weight between D and the vertices that remain to be visited, B, C, E , and F , is 70 min between D and F . Repeat Step 2: Label vertex F as V 3 . The edge of lowest weight between F and the vertices that remain to be visited, B, C, and E , is 75 min between F and C . Repeat Step 2: Label vertex C as V 4 . The edge of lowest weight between C and the vertices that remain to be visited, B and E , is 100 min between C and B . Repeat Step 2: Label vertex B as V 5 . The only vertex that remains to be visited is E . The weight of the edge between B and E is 95 min. Step 3: A Hamilton cycle of low weight is A → D → F → C → B → E → A . So, a route of relatively low travel time is A to D to F to C to B to E and back to A . The total travel time of this route is: 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min Check Your Understanding Key Terms brute force algorithm greedy algorithm traveling salesperson problem (TSP) brute force method nearest neighbor method Key Concepts A brute force algorithm always finds the ideal solution but can be impractical whereas a greedy algorithm is efficient but usually does not lead to the ideal solution. A Hamilton cycle of lowest weight is a solution to the traveling salesperson problem. The brute force method finds a Hamilton cycle of lowest weight in a complete graph. The nearest neighbor method is a greedy algorithm that finds a Hamilton cycle of relatively low weight in a complete graph. Formulas In a complete graph with n vertices, the number of distinct Hamilton cycles is ( n − 1 ) ! . In a complete graph with n vertices, there are at most ( n − 1 ) ! 2 different weights of Hamilton cycles.", "section": "Traveling Salesperson Problem", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Trees In graph theory, graphs known as trees have structures in common with live trees. (credit: “Row of trees in Roslev” by AKA CJ/Flickr, Public Domain) Learning Objectives After completing this section, you should be able to: Describe and identify trees. Determine a spanning tree for a connected graph. Find the minimum spanning tree for a weighted graph. Solve application problems involving trees. We saved the best for last! In this last section, we will discuss arguably the most fun kinds of graphs, trees. Have you every researched your family tree? Family trees are a perfect example of the kind of trees we study in graph theory. One of the characteristics of a family tree graph is that it never loops back around, because no one is their own grandparent! What Is A Tree? Whether we are talking about a family tree or a tree in a forest, none of the branches ever loops back around and rejoins the trunk. This means that a tree has no cyclic subgraphs, or is acyclic . A tree also has only one component. So, a tree is a connected acyclic graph. Here are some graphs that have the same characteristic. Each of the graphs in is a tree. Graphs T , P , and S Let’s practice determining whether a graph is a tree. To do this, check if a graph is connected and has no cycles. Identifying Trees Identify any trees in . If a graph is not a tree, explain how you know. Graphs M , N , and P Graph M is not a tree because it contains the cycle ( b , c , f ). Graph N is not a tree because it is not connected. It has two components, one with vertices h, i, j , and another with vertices k, l, m . Graph P is a tree. It has no cycles and it is connected. Types of Trees Mathematicians have had a lot of fun naming graphs that are trees or that contain trees. For example, the graph in is not a tree, but it contains two components, one containing vertices a through d , and the other containing vertices e through g , each of which would be a tree on its own. This type of structure is called a forest . There are also interesting names for trees with certain characteristics. A path graph or linear graph is a tree graph that has exactly two vertices of degree 1 such that the only other vertices form a single path between them, which means that it can be drawn as a straight line. A star tree is a tree that has exactly one vertex of degree greater than 1 called a root, and all other vertices are adjacent to it. A starlike tree is a tree that has a single root and several paths attached to it. A caterpillar tree is a tree that has a central path that can have vertices of any degree, with each vertex not on the central path being adjacent to a vertex on the central path and having a degree of one. A lobster tree is a tree that has a central path that can have vertices of any degree, with paths consisting of either one or two edges attached to the central path. Examples of each of these types of structures are given in . Forest Graph F Six Types of Trees Identifying Types of Trees Each graph in is one of the special types of trees we have been discussing. Identify the type of tree. Graphs U and V Graph U has a central path a → b → d → f → i → l → o → q . Each vertex that is not on the path has degree 1 and is adjacent to a vertex that is on the path. So, U is a caterpillar tree. Graph V is a path graph because it is a single path connecting exactly two vertices of degree one, r → s → u → v → w . Characteristics of Trees As we study trees, it is helpful to be familiar with some of their characteristics. For example, if you add an edge to a tree graph between any two existing vertices, you will create a cycle, and the resulting graph is no longer a tree. Some examples are shown in . Adding edge bj to Graph T creates cycle ( b , c , i , j ). Adding edge rt to Graph P creates cycle ( r , s , t ). Adding edge tv to Graph S creates cycle ( t , u , v ). Adding Edges to Trees It is also true that removing an edge from a tree graph will increase the number of components and the graph will no longer be connected. In fact, you can see in that removing one or more edges can create a forest. Removing edge qr from Graph P creates a graph with two components, one with vertices o, p and q , and the other with vertices r, s , and t . Removing edge uw from Graph S creates two components, one with just vertex w and the other with the rest of the vertices. When two edges were removed from Graph T , edge bf and edge cd , creates a graph with three components as shown in . Removing Edges from Trees A very useful characteristic of tree graphs is that the number of edges is always one less than the number of vertices. In fact, any connected graph in which the number of edges is one less than the number of vertices is guaranteed to be a tree. Some examples are given in . Number of Vertices and Edges in Trees vs. Other Graphs The number of edges in a tree graph with n vertices is n − 1 . A connected graph with n vertices and n − 1 edges is a tree graph. Exploring Characteristics of Trees Use Graphs I and J in to answer each question. Graphs I and J Which vertices are in each of the components that remain when edge be is removed from Graph I ? Determine the number of edges and the number of vertices in Graph J . Explain how this confirms that Graph J is a tree. What kind of cycle is created if edge im is added to Graph J ? When edge be is removed, there are two components that remain. One component includes vertices a, b , and c . The other component includes vertices d, e , and f . There are seven vertices and six edges in Graph J . This confirms that Graph J is a tree because the number of edges is one less than the number of vertices. The pentagon ( i , h , j , l , m ) is created when edge im is added to Graph J . Graph Theory in the Movies In the 1997 film Good Will Hunting , the main character, Will, played by Matt Damon, solves what is supposed to be an exceptionally difficult graph theory problem, “Draw all the homeomorphically irreducible trees of size n = 10 .” That sounds terrifying! But don’t panic. Watch this great Numberphile video to see why this is actually a problem you can do at home! The Problem in Good Will Hunting by Numberphile Spanning Trees Suppose that you planned to set up your own computer network with four devices. One option is to use a “mesh topology” like the one in , in which each device is connected directly to every other device in the network. Common Network Configurations The mesh topology for four devices could be represented by the complete Graph A 1 in where the vertices represent the devices, and the edges represent network connections. However, the devices could be networked using fewer connections. Graphs A 2 , A 3, and A 4 of show configurations in which three of the six edges have been removed. Each of the Graphs A 2 , A 3 and A 4 in is a tree because it is connected and contains no cycles. Since Graphs A 2 , A 3 and A 4 are also subgraphs of Graph A 1 that include every vertex of the original graph, they are also known as spanning trees . Network Configurations for Four Devices By definition, spanning trees must span the whole graph by visiting all the vertices. Since spanning trees are subgraphs, they may only have edges between vertices that were adjacent in the original graph. Since spanning trees are trees, they are connected and they are acyclic. So, when deciding whether a graph is a spanning tree, check the following characteristics: All vertices are included. No vertices are adjacent that were not adjacent in the original graph. The graph is connected. There are no cycles. Identifying Spanning Trees Use to determine which of graphs M 1 , M 2 , M 3 , and M 4 , are spanning trees of Q . Graphs Q , M 1 , M 2 , M 3, and M 4 Graph M 1 is not a spanning tree of Graph Q because it has a cycle ( c , d , f , e ). Graph M 2 is a spanning tree of Graph Q because it has all the original vertices, no vertices are adjacent in M 2 that weren’t adjacent in Graph Q , Graph M 2 is connected and it contains no cycles. Graph M 3 is not a spanning tree of Graph Q because vertices a and f are adjacent in Graph M 3 but not in Graph Q . Graph M 4 is not a spanning tree of Graph Q because it is not connected. So, only graph M 2 is a spanning tree of Graph Q . Constructing a Spanning Tree Using Paths Suppose that you wanted to find a spanning tree within a graph. One approach is to find paths within the graph. You can start at any vertex, go any direction, and create a path through the graph stopping only when you can’t continue without backtracking as shown in . First Phase to Construct a Spanning Tree Once you have stopped, pick a vertex along the path you drew as a starting point for another path. Make sure to visit only vertices you have not visited before as shown in . Intermediate Phase to Construct a Spanning Tree Repeat this process until all vertices have been visited as shown in . Final Phase to Construct a Spanning Tree The end result is a tree that spans the entire graph as shown in . The Resulting Spanning Tree Notice that this subgraph is a tree because it is connected and acyclic. It also visits every vertex of the original graph, so it is a spanning tree. However, it is not the only spanning tree for this graph. By making different turns, we could create any number of distinct spanning trees. Constructing Spanning Trees Construct two distinct spanning trees for the graph in . Graph L Two possible solutions are given in and . First Spanning Tree for Graph L Second Spanning Tree for Graph L Revealing Spanning Trees Another approach to finding a spanning tree in a connected graph involves removing unwanted edges to reveal a spanning tree. Consider Graph D in . Graph D Graph D has 10 vertices. A spanning tree of Graph D must have 9 edges, because the number of edges is one less than the number of vertices in any tree. Graph D has 13 edges so 4 need to be removed. To determine which 4 edges to remove, remember that trees do not have cycles. There are four triangles in Graph D that we need to break up. We can accomplish this by removing 1 edge from each of the triangles. There are many ways this can be done. Two of these ways are shown in . Removing Four Edges from Graph D Spanning Trees in Graph Theory Removing Edges to Find Spanning Trees Use the graph in to answer each question. Graph V Determine the number of edges that must be removed to reveal a spanning tree. Name all the undirected cycles in Graph V . Find two distinct spanning trees of Graph V . Graph V has nine vertices so a spanning tree for the graph must have 8 edges. Since Graph V has 11 edges, 3 edges must be removed to reveal a spanning tree. ( a , c , d ), ( a , c , f ), ( a , d , c , f ), and ( b , e , h , i , g ) To find the first spanning tree, remove edge ac , which will break up both of the triangles, remove edge cf , which will break up the quadrilateral, and remove be , which will break up the pentagon, to give us the spanning tree shown in . Spanning Tree Formed Removing ac, cf , and be To find another spanning tree, remove ad , which will break up ( a , c , d ) and ( a , d , c , f ), remove af to break up ( a , c , f ), and remove hi to break up ( b , e , h , i , g ). This will give us the spanning tree in . Spanning Tree Formed Removing ad , af , and hi Chains of Affection Here is a strange question to ask in a math class: Have you ever dated your ex’s new partner’s ex? Research suggests that your answer is probably no. When researchers Peter S. Bearman, James Moody, and Katherine Stovel attempted to compare the structure of heterosexual romantic networks at a typical midwestern high school to simulated networks, they found something surprising. The actual social networks were more like spanning trees than other possible models because there were very few short cycles. In particular, there were almost no four-cycles. Chains of Affection “…the prohibition against dating (from a female perspective) one’s old boyfriend’s current girlfriend’s old boyfriend – accounts for the structure of the romantic network at [the highschool].” In their article “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” the researchers went on to explain the implications for the transmission of sexually transmitted diseases. In particular, social structures based on tree graphs are less dense and more likely to fragment. This information can impact social policies on disease prevention. (Peter S. Bearman, James Moody, and Katherine Stovel, “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” American Journal of Sociology Volume 110, Number 1, pp. 44-91, 2004) Kruskal’s Algorithm In many applications of spanning trees, the graphs are weighted and we want to find the spanning tree of least possible weight. For example, the graph might represent a computer network, and the weights might represent the cost involved in connecting two devices. So, finding a spanning tree with the lowest possible total weight, or minimum spanning tree , means saving money! The method that we will use to find a minimum spanning tree of a weighted graph is called Kruskal’s algorithm . The steps for Kruskal’s algorithm are: Step 1: Choose any edge with the minimum weight of all edges. Step 2: Choose another edge of minimum weight from the remaining edges. The second edge does not have to be connected to the first edge. Step 3: Choose another edge of minimum weight from the remaining edges, but do not select any edge that creates a cycle in the subgraph you are creating. Step 4: Repeat step 3 until all the vertices of the original graph are included and you have a spanning tree. Use Kruskal's Algorithm to Find Minimum Spanning Trees in Graph Theory Using Kruskal’s Algorithm A computer network will be set up with six devices. The vertices in the graph in represent the devices, and the edges represent the cost of a connection. Find the network configuration that will cost the least. What is the total cost? Graph of Network Connection Costs A minimum spanning tree will correspond to the network configuration of least cost. We will use Kruskal’s algorithm to find one. Since the graph has six vertices, the spanning tree will have six vertices and five edges. Step 1: Choose an edge of least weight. We have sorted the weights into numerical order. The least is $100. The only edge of this weight is edge AF as shown in . Step 1 Select Edge AF Step 2: Choose the edge of least weight of the remaining edges, which is BD with $120. Notice that the two selected edges do not need to be adjacent to each other as shown in . Step 2 Select Edge BD Step 3: Select the lowest weight edge of the remaining edges, as long as it does not result in a cycle. We select DF with $150 since it does not form a cycle as shown in . Step 3 Select Edge DF Repeat Step 3: Select the lowest weight edge of the remaining edges, which is BE with $160 and it does not form a cycle as shown in . This gives us four edges so we only need to repeat step 3 once more to get the fifth edge. Repeat Step 3 Select Edge DF Repeat Step 3: The lowest weight of the remaining edges is $170. Both BF and CE have a weight of $170, but BF would create cycle ( b , d , f ) and there cannot be a cycle in a spanning tree as shown in . Repeat Step 3 Do Not Select Edge BF So, we will select CE , which will complete the spanning tree as shown in . Repeat Step 3 Select Edge CE The minimum spanning tree is shown in . This is the configuration of the network of least cost. The spanning tree has a total weight of $ 100 + $ 120 + $ 150 + $ 160 + $ 170 = $ 700 , which is the total cost of this network configuration. Final Minimum Spanning Tree Check Your Understanding Key Terms acyclic tree forest path graph or linear graph star tree root starlike tree caterpillar tree lobster tree spanning tree minimum spanning tree Key Concepts A brute force algorithm always finds the ideal solution but can be impractical whereas a greedy algorithm is efficient but usually does not lead to the ideal solution. A Hamilton cycle of lowest weight is a solution to the traveling salesperson problem. The brute force method finds a Hamilton cycle of lowest weight in a complete graph. The nearest neighbor method is a greedy algorithm that finds a Hamilton cycle of relatively low weight in a complete graph. Videos The Problem in Good Will Hunting by Numberphile Spanning Trees in Graph Theory Use Kruskal's Algorithm to Find Minimum Spanning Trees in Graph Theory Formulas The number of edges in a tree graph with n vertices is n − 1 . A connected graph with n vertices and n − 1 edges is a tree graph. Projects Everyone Gets a Turn! – Graph Colorings Let’s put your knowledge of graph colorings to work! Your task is to plan a field day following these steps. Select between seven and ten activities for your field day. You can look online for ideas. Create a survey asking for the participants to select the three to five events in which they would most like to participate. Survey between seven and ten people. Use the results of your survey to create a graph in which each vertex represents one of the events. A pair of vertices will be adjacent if there is at least one participant who would like to participate in both events. Find a minimum coloring for the graph. Explain how you found it and how you know the chromatic number of the graph. Use your solution to part d to determine the minimum number of timeslots you must use to ensure that everyone has the opportunity to participate in their top three events. Find the complement of the graph you created. Explain what the edges in this graph represent. A Beautiful Day in the Neighborhood – Euler Circuits Let’s apply what you have learned to the community in which you live. Using resources such as your county’s property appraiser’s website, create a detailed graph of your neighborhood in which vertices represent turns and intersections. Represent a large enough part of your community to include no fewer than 10 intersections or turns. Then use your graph to answer the following questions. Label the edges of your graph. Determine if your graph is Eulerian. Explain how you know. If it is not, eulerize it. Find an Euler circuit for your graph. Give the sequence of vertices that you found. What does the Euler circuit you found in part c represent for your community? Describe an application for which this Euler circuit might be used. Dream Vacation – Hamilton Cycles and Paths Where in the world would you like to travel most: the Eiffel Tower in Paris, a Broadway musical in New York city, a bike tour of Amsterdam, the Tenerife whale and dolphin cruises in the Canary Islands, the Giza Pyramid in Cairo, or maybe the Jokhang Temple in Tibet? Let's plan your dream vacation! Which four destinations are at the top of your bucket list? Draw a complete weighted graph with five vertices representing the four destinations and your home city, and the weights representing the cost of travel between cities. Use a website (such as Travelocity ) to find the best airfare between each pair of cities. List the airlines and flight numbers along with the prices. Include cost for ground transportation from the nearest airport if there is no airport at the destination you want to visit. Use the nearest neighbor algorithm to find a Hamilton cycle of low weight beginning and ending in your hometown. What is the weight of this circuit and what does it represent? Use the brute force method to find a Hamilton cycle of lowest weight beginning and ending in your hometown. What is the weight of this circuit? Is it the same or different from the weight of the Hamilton cycle you found in Exercise 4? Suppose that instead of returning home, you planned to move to your favorite location on the list, but you wanted to stop at the other three destinations once along the way. Where would you move? List all Hamilton paths between your hometown and your favorite location. Find the weights of all the Hamilton paths you found in Exercise 6. Chapter Review Graph Basics Graph Structures Comparing Graphs Navigating Graphs Euler Circuits Euler Trails Hamilton Cycles Hamilton Paths Traveling Salesperson Problem Trees Chapter Test", "section": "Trees", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Introduction Math can be found in many different areas and subjects. (credit: modification of work \"Chemistry/Physics Library\" by University Libraries/Flickr, CC BY 2.0) Where do we find math around us? Math can be found in areas that are expected and sometimes in areas that are surprising. There are many ways that mathematical concepts, such as those in this text, are infused in the world around us. In this chapter, we will explore a sampling of five distinct areas from everyday life where math’s impact plays a meaningful role. Math's impact on art can be found in numerical relationships that are known to create or enhance beauty. The Fibonacci numbers are one mathematical example that can be found in nature such as in petal count of a rose. On a different note, a mathematical exploration can aid in making a convincing argument on how we can positively impact our environment. Whether looking at the choices of a single individual or the larger impact offered from a collaborative effort, there are measurable responses to positively address climate change. Turning to medicine, which has been a topic of global importance in recent years, we will explore how math is used to determine drug dosage rates and test the validity of a vaccine. Switching back to items of aesthetic nature, we will examine some foundational components of music which, like art, brings beauty and joy to our lives. Finally, we will explore some ways that math is used in sports to predict future performance and analyze tournaments styles.", "section": "Introduction", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Math and Art Sunflower seeds appear in a pattern that involves the Fibonacci sequence. (credit: “Sunflower Surprise” by frankieleon/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify and describe the golden ratio. Identify and describe the Fibonacci sequence and its application to nature. Apply the golden ratio and the Fibonacci sequence relationship. Identify and compute golden rectangles. Art is the expression or application of human creative skill and imagination, typically in a visual form such as painting or sculpture, producing works to be appreciated primarily for their beauty or emotional power. Oxford Dictionary Art, like other disciplines, is an area that combines talent and experience with education. While not everyone considers themself skilled at creating art, there are mathematical relationships commonly found in artistic masterpieces that drive what is considered attractive to the eye. Nature is full of examples of these mathematical relationships. Enroll in a cake decorating class and, when you learn how to create flowers out of icing, you will likely be directed as to the number of petals to use. Depending on the desired size of a rose flower, the recommendation for the number of petals to use is commonly 5, 8, or 13 petals. If learning to draw portraits, you may be surprised to learn that eyes are approximately halfway between the top of a person’s head and their chin. Studying architecture, we find examples of buildings that contain golden rectangles and ratios that add to the beautifying of the design. The Parthenon ( ), which was built around 400 BC, as well as modern-day structures such the Washington Monument are two examples containing these relationships. These seemingly unrelated examples and many more highlight mathematical relationships that we associate with beauty in artistic form. The Parthenon in Greece demonstrates the golden ratio. (credit: “Parthénon” by Julien Maury/Flickr, Public Domain Mark 1.0) Golden Ratio The golden ratio , also known as the golden proportion, is a ratio aspect that can be found in beauty from nature to human anatomy as well as in golden rectangles that are commonly found in building structures. The golden ratio is expressed in nature from plants to creatures such as the starfish, honeybees, seashells, and more. It is commonly noted by the Greek letter ϕ (pronounced “fee”). ϕ = 1 + 5 2 , which has a decimal value approximately equal to 1.618. Consider : Note how the building is balanced in dimension and has a natural shape. The overall structure does not appear as if it is too wide or too tall in comparison to the other dimensions. Vitruvian Man by Leonardo da Vinci (credit: \"Vitruvian Man\" by Leonardo da Vinci/Wikimedia Commons, Public Domain) The golden ratio has been used by artists through the years and can be found in art dating back to 3000 BC. Leonardo da Vinci is considered one of the artists who mastered the mathematics of the golden ratio, which is prevalent in his artwork such as Virtuvian Man ( ). This famous masterpiece highlights the golden ratio in the proportions of an ideal body shape. The golden ratio is approximated in several physical measurements of the human body and parts exhibiting the golden ratio are simply called golden. The ratio of a person’s height to the length from their belly button to the floor is ϕ or approximately 1.618. The bones in our fingers (excluding the thumb), are golden as they form a ratio that approximates ϕ . The human face also includes several ratios and those faces that are considered attractive commonly exhibit golden ratios. Using Golden Ratio and a Person’s Height If a person’s height is 5 ft 6 in, what is the approximate length from their belly button to the floor rounded to the nearest inch, assuming the ratio is golden? Step 1: Convert the height to inches 5 ft 6 in = 66 in Step 2: Calculate the length from the belly button to the floor, L . 66 / L = 1.618 L = 40.8 in The length from the person’s belly button to the floor would be approximately 41 in. Fibonacci Sequence and Application to Nature Rose petals appear in a Fibonacci spiral. (credit: “rilke4” by monchoohcnom/Flickr, Public Domain Mark 1.0) The Fibonacci sequence can be found occurring naturally in a wide array of elements in our environment from the number of petals on a rose flower to the spirals on a pine cone to the spines on a head of lettuce and more. The Fibonacci sequence can be found in artistic renderings of nature to develop aesthetically pleasing and realistic artistic creations such as in sculptures, paintings, landscape, building design, and more. It is the sequence of numbers beginning with 1, 1, and each subsequent term is the sum of the previous two terms in the sequence (1, 1, 2, 3, 5, 8, 13, …). The petal counts on some flowers are represented in the Fibonacci sequence. A daisy is sometimes associated with plucking petals to answer the question “They love me, they love me not.” Interestingly, a daisy found growing wild typically contains 13, 21, or 34 petals and it is noted that these numbers are part of the Fibonacci sequence. The number of petals aligns with the spirals in the flower family. Applying the Fibonacci Sequence to Rose Petals Suppose you were creating a rose out of icing, assuming a Fibonacci sequence in the petals, how many petals would be in the row following a row containing 13 petals? The number of petals on a rose is often modeled with the numbers in the Fibonacci sequence, which is 1, 1, 2, 3, 5, 8, 13,…, where the next number in the sequence is the sum of 8 + 13 = 21 . There would be 21 petals on the next row of the icing rose. Golden Ratio and the Fibonacci Sequence Relationship Mathematicians for years have explored patterns and applications to the world around us and continue to do so today. One such pattern can be found in ratios of two adjacent terms of the Fibonacci sequence. Recall that the Fibonacci sequence = 1, 1, 3, 5, 8, 13,… with 5 and 8 being one example of adjacent terms. When computing the ratio of the larger number to the preceding number such as 8/5 or 13/8, it is fascinating to find the golden ratio emerge. As larger numbers from the Fibonacci sequence are utilized in the ratio, the value more closely approaches ϕ , the golden ratio. Finding Golden Ratio in Adjacent Fibonacci Terms The 24th Fibonacci number is 46,368 and the 25th is 75,025. Show that the ratio of the 25th and 24th Fibonacci numbers is approximately ϕ . Round your answer to the nearest thousandth. 75,025 / 46,368 = 1.618 ; The ratio of the 25th and 24th term is approximately equal to the value of ϕ rounded to the nearest thousandth, 1.618. The pyramids of Giza in Egypt (credit: “Giza Pyramids” by Vincent Brown/Flickr, CC BY 2.0) Golden Rectangles Turning our attention to man-made elements, the golden ratio can be found in architecture and artwork dating back to the ancient pyramids in Egypt ( ) to modern-day buildings such as the UN headquarters. The ancient Greeks used golden rectangles —any rectangles where the ratio of the length to the width is the golden ratio—to create aesthetically pleasing as well as solid structures, with examples of the golden rectangle often being used multiple times in the same building such as the Parthenon, which is shown in . Golden rectangles can be found in twentieth-century buildings as well, such as the Washington Monument. Looking at another man-made element, artists paintings often contain golden rectangles. Well-known paintings such as Leonardo da Vinci’s The Last Supper and the Vitruvian Man contain multiple golden rectangles as do many of da Vinci’s masterpieces. Whether framing a painting or designing a building, the golden rectangle has been widely utilized by artists and are considered to be the most visually pleasing rectangles. Finding Golden Rectangle in Frames A frame has dimensions of 8 in by 6 in. Calculate the ratio of the sides rounded to the nearest thousandth and determine if the size approximates a golden rectangle. 8/6 = 1.333; A golden rectangle’s ratio is approximately 1.618. The frame dimensions are close to a golden rectangle. M.C. Escher M.C. Escher (credit: \"M.C. Escher\" by Hans Peters (ANEFO)/Dutch National Archives, CC0 1.0 Public Domain) Mauritis Cornelis Escher was a Dutch-born world-famous graphic artist and his work can be found in murals, stamps, wallpaper designs, illustrations in books, and even carpets. Over his lifetime, M.C. Escher created hundreds of lithographs and wood engravings as well as more than 2,000 sketches. Escher’s work is characterized with the infusion of geometric designs that obey most of the mathematical rules. If you study his work closely, you can see where he breaks a mathematical relationship to create famous illusions such as soldiers marching around the top of a square turret where the soldiers appear to be always going uphill but are contained on a single set of stairs in a square. Look closely and the golden ratio as well as golden rectangles abound in Escher’s work. Like many famous people, M.C. Escher did not find success in his early school years. Before finding success, Escher failed his final school exam and quit a short stint in architecture. Finding a graphic arts teacher who recognized Escher’s talent, Escher completed art school and enjoyed traveling through Italy, where he found much of his inspiration for his work. Check Your Understanding Key Terms golden ratio ϕ Fibonacci sequence golden rectangle Key Concepts The golden ratio, ϕ , can be found in nature, and the relationship is often associated with beauty and balance. The Fibonacci sequence reflects a pattern of numbers that can be found in various places in nature. The sequence can be used to predict other values that follow the Fibonacci pattern. State some naturally occurring applications of the Fibonacci sequence. State some naturally occurring applications of the golden ratio. Determine if a rectangle is golden. State some artistic applications of the golden rectangle.", "section": "Math and Art", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Math and the Environment Solar panels harness the sun's energy to power homes, businesses, and various methods of transportation. (credit: modification of work “Craters of the Moon solar array” by NPS Climate Change Response/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Compute how conserving water can positively impact climate change. Discuss the history of solar energy. Compute power needs for common devices in a home. Explore advantages of solar power as it applies to home use. Climate change and emissions management have been debated topics in recent years. However, more and more people are recognizing the impacts that have resulted in temperature changes and are seeking timely and effective action. The World Meteorological Organization shared in a June 2021 publication that “2021 is a make-or-break year for climate action, with the window to prevent the worst impacts of climate change—which include ever more frequent more intense droughts, floods and storms—closing rapidly.” The problem no longer belongs to a few countries or regions but rather is a worldwide concern measured with increasing temperatures leading to decreased glacier coverage and resulting rise in sea levels. The good news is, there are small steps that each of us can do that collectively can positively impact climate change. Making a Positive Impact on Climate Change—Water Usage Our use of water is one element that impacts climate change. Having access to clean, potable water is critical for not only our health but also for the health of our ecosystem. About 1 out of 10 people on our planet do not have easy access to clean water to drink. As each of us conserves water, we prolong the life span of fresh water from our lakes and rivers and also reduce the impact on sewer systems and drainage in our communities. Additionally, as we conserve water, we also conserve electricity that is used to bring water to and in our homes. So, what can we do to help conserve water? Brushing Your Teeth (One Person’s Contribution) Brushing your teeth with the water running continually uses about 4 gal of water. Turning the faucet off when you are not rinsing uses less than one-fourth of a gallon of water. Considering the recommendation to brush your teeth twice a day, how much water would be saved in a week if the faucet was off when not rinsing? Leaving the water running continually: Step 1: Calculate gallons used not with water running continually: Brushing twice a day for 7 days using 4 gal of water for each brushing ( 2 times a day ) ( 7 days ) ( 4 gal ) = 56 gal Step 2: Calculate gallons used turning the faucet off when you are not rinsing: Brushing twice a day for 7 days using 0.25 gal of water for each brushing ( 2 times a day ) ( 7 days ) ( 0.25 gal ) = 3.5 gal Step 3: Calculate savings: Savings = 56 gal − 3.5 gal = 52.5 gal During one week, 52.5 gal of water would be saved if one person turned the faucet off except when rinsing when brushing your teeth. Brushing Your Teeth (Multiple People’s Contribution – Town) Using the data in , how much water would be saved in a month if one-fifth of a town’s population of 15,000 turned the faucet off when brushing their teeth except when rinsing? Step 1: From , we found that 1 person saves 52.5 gal per week. Step 2: Calculate the population to save water: One-fifth of 15,000 people = 3,000 people Step 3: One-fifth of a town’s population turning off the faucet when brushing their teeth for a month: ( 3,000 ) ( 52.5 gal per week ) ( 4 weeks ) = 630,000 gal During one month, 630,000 gallons of water would be saved if one-fifth of a town of 15,000 people turned the faucet off except when rinsing when brushing their teeth. Brushing Your Teeth (Multiple People’s Contribution – State) Using the data in , how much water would be saved in a year if one-fourth of the population of the state of Minnesota, which is approximately 5.6 million people, turned the faucet off when brushing their teeth except when rinsing for a year (52 weeks)? Step 1: From , we found that 1 person saves 52.5 gal per week. Step 2: Calculate the population to save water: One-fourth of 5.6 million people = 1.4 million people Step 3: One-fourth of a town’s population turning off the faucet when brushing their teeth for a month: ( 1.4 million ) ( 52.5 gal per week ) ( 52 weeks ) = 3,822 million gal During one year, 3,822 million gal of water would be saved if one-fourth of the state of Minnesota turned the faucet off except when rinsing when brushing their teeth. History of Solar Energy In the mid-1800s, Willoughby Smith discovered photoconductive responsiveness in selenium. Shortly thereafter, William Grylls Adams and Richard Evans Day discovery that selenium can produce electricity if exposed to the sun was a major breakthrough. Less than 10 years later, Charles Fritts invented the first solar cells using selenium. Jumping a mere 100 years later, Bell Labs in the United States produced the first practical photovoltaic cells in the mid-1950s and developed versions used to power satellites in the same decade. Solar panel use has exploded in recent decades and is now used by residences, organizations, businesses, and government buildings such as the White House, space to power satellites, and various methods of transportation. One reason for the expansion is a continuing drop in cost combined with an increase in performance and durability. In the mid-1950s, the cost of a solar panel was around $300 per watt capability. Twenty years later, the cost was a third of the 1950s’ cost. Currently, solar panel cost has dropped to less than $1 per watt while decreasing in size as well as increasing in longevity. The dropping price and improved performance has moved solar to a modest investment that can pay for itself in less than half the time of systems from 15 years ago. Solar Power’s Age The sun has been harnessed by humans for centuries. The earliest recorded use of tapping the sun’s energy for power dates back to the seventh century BC when man focused the sun’s rays through a magnifying glass to create fire. Four thousand years later, we find historical record of using mirrors to focus the sun and light torches, often for ceremonial proceedings. Use of the sun to light torches continued through the centuries and has been recorded by various cultures including the Chinese civilization in 20 AD and beyond. In more recent years, the sun was harnessed to power ovens on ships traversing to oceans in the 1700s. At the same time, the power of the sun was utilized to power steamboats through the 1800s. Mária Telkes, a Hungarian-born American scientist, invented a widely deployed solar seawater distiller used on World War II life rafts. Soon after, she partnered with architect Eleanor Raymond to design the first modern home to be completely heated by solar power. Air warmed on rooftop collectors transferred heat to salts, which stored the heat for later use. Although solar panels as we know them today are relatively new in history, use of the sun to harness power is much older. Compute Power Needs for Common Home Devices A kilowatt (kW) is 1,000 watts (W). A kilowatt-hour (kWh) is a measurement of energy use, which is the amount of energy used by a 1,000-watt device to run for an hour. Using the definition of a kilowatt-hour, to calculate how long it would take to consume 1 kWh of power, we divide 1,000 by the watts use of a device. 1,000 / watts = time needed to use 1 kW For example, a 75 W bulb would take 1,000 ÷ 75 = 13.3 hours to use 1 kW of power. watts / 1 , 000 = kilowatt hours Calculating the Kilowatt-Hours Needed to Run a Television A 48 in plasma television uses about 200 W. How many kilowatt-hours are needed to run the television in a month if the television is one for an average of 2.5 hours a day? Step 1: 1,000 / ( 200 watts ) = 5 hours to use 1 kW Step 2: ( 2.5 hours a day ) ( 30 days ) = 75 hours of use Step 3: 75 hour / 5 hours per kW = 15 kW The television will consume about 15 kW in a month. Calculating the Cost to Run a Refrigerator A medium-sized Energy Star–rated refrigerator uses about 575 W and runs for about 8 hours per day. What is the monthly (30 days) cost of running the refrigerator if the electric rate is 12 cents per kilowatt-hour? Step 1: Calculate the watts per day: ( 575 W ) ( 8 hours ) = 4,600 W per day Step 2: Calculate the kilowatt-hours. ( 4,600 ) / ( 1,000 ) = 4.6 kWh Step 3: Calculate the daily cost. ( 4.6 kWh ) ( 12 cents ) = 55 cents = $ 0.55 Step 4: Calculate the monthly cost. ( $ 0.55 ) ( 30 ) = $ 16.50 It would cost about $16.50 to run the refrigerator for a month. Calculating the Kilowatt-Hours to Run an Oven An electric oven is labeled as 4,000 W. How much would it cost to bake a cake for 30 minutes if the electric rate is 14 cents per kilowatt-hour? Step 1: Determine the time it takes to use 1 kW of power: 1,000 / ( 4,000 watts ) = 0.25 hours to use 1 kW For every 15 minutes, the oven uses 1 kW of power. Step 2: Determine how many kilowatt-hours are needed to bake the cake for 30 minutes: ( 30 minutes ) / ( 15 minutes per kW ) = 2 kW Step 3: Calculate the cost of the oven usage: ( 2 kW ) ( 14 cents per kWh ) = 28 cents It would cost about 28 cents to bake the cake. Solar Advantages There are multiple advantages that solar power can offer us today including reducing greenhouse gas and CO 2 emissions, powering vehicles, reducing water pollution, reducing strain on limited supply of other power options such as fossil fuels. We will look further at reducing greenhouse gas and CO 2 emissions. Any gas that prevents infrared radiation from escaping Earth's atmosphere is a greenhouse gas. There are 24 currently identified greenhouse gases of which carbon dioxide is one. When measuring the impact of any of the greenhouse gases, the measurements are given in units of carbon dioxide emissions. For this reason, greenhouse gas and carbon dioxide have become interchangeable in discussions. Charles Fritts and Mohammad M. Atalla Charles Fritts, a New York inventor, is credited with creating the first solar cell, which he installed on a rooftop in New York City in 1884. While the solar cell was not very efficient, having a rate of conversion between 1 to 2%, this was a major step early in solar power energy. Today’s solar cells have an efficiency on average of 15 to 20%, which yields a notably higher impact. Nonetheless, the work that Fritts successfully completed marked the start of solar energy through the use of photovoltaic solar panels in the United States. Mohamed M. Atalla was an Egyptian-born scientist who moved to the United States to complete his studies, and undertook research and development at Bell Laboratories in New Jersey. Many of the early efficiency gains in solar cells were due to his development of processes for using silicon within electronic devices. Atalla's work led to the invention of silicon transistors and microchips (including his own invention of the MOSFET, the most widely used transistor in the world), and quickly increased the efficiency of solar cells. Calculating the Solar Power for Average Home Use in Kilowatts If a home uses approximately 30 kW of electricity per day, what size solar system would be needed to fuel 80% of a home’s needs for a month (30 days)? ( 30 kW hours ) ( 30 days ) ( 0.80 ) = 720 kW A solar system capable of producing 720 kW a month would be needed. Check Your Understanding Key Terms greenhouse gas CO 2 emissions watt kilowatt (kW) Key Concepts Recognize how water conservation by one person, family, community, or nation can positively impact the world’s freshwater supply. Recall components from the history of solar use by mankind. Calculate electrical demand given watts. Recognize advantages of residential solar power. Formulas 1,000 / watts = time needed to use 1 kWh watts / 1 , 000 = kilowatt hours", "section": "Math and the Environment", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Math and Medicine Shoppers wear masks during the Covid-19 pandemic. (credit: \"True Covid Scene - Mask Buying\" by Joey Zanotti/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute the mathematical factors utilized in concentrations/dosages of drugs. Describe the history of validating effectiveness of a new drug. Describe how mathematical modeling is used to track the spread of a virus. The pandemic that rocked the world starting in 2020 turned attention to finding a cure for the Covid-19 strain into a world race and dominated conversations from major news channels to households around the globe. News reports decreeing the number of new cases and deaths locally as well as around the world were part of the daily news for over a year and progress on vaccines soon followed. How was a vaccine able to be found so quickly? Is the vaccine safe? Is the vaccine effective? These and other questions have been raised through communities near and far and some remain debatable. However, we can educate ourselves on the foundations of these discussions and be more equipped to analyze new information related to these questions as it becomes available. Concentrations and Dosages of Drugs Consider any drug and the recommended dosage varies based on several factors such as age, weight, and degree of illness of a person. Hospitals and medical dispensaries do not stock every possible needed concentration of medicines. Drugs that are delivered in liquid form for intravenous (IV) methods in particular can be easily adjusted to meet the needs of a patient. Whether administering anesthesia prior to an operation or administering a vaccine, calculation of the concentration of a drug is needed to ensure the desired amount of medicine is delivered. The formula to determine the volume needed of a drug in liquid form is a relatively simple formula. The volume needed is calculated based on the required dosage of the drug with respect to the concentration of the drug. For drugs in liquid form, the concentration is noted as the amount of the drug per the volume of the solution that the drug is suspended in which is commonly measured in g/mL or mg/mL. Suppose a doctor writes a prescription for 6 mg of a drug, which a nurse calculates when retrieving the needed prescription from their secure pharmaceutical storage space. On the shelves, the drug is available in liquid form as 2 mg per mL. This means that 1 mg of the drug is found in 0.5 mL of the solution. Multiplying 6 mg by 0.5 mL yields 3 mL, which is the volume of the prescription per single dose. Volume needed = ( medicine dosage required ) ( weight of drug by volume ) . A common calculation for the weight of a liquid drug is measured in grams of a drug per 100 mL of solution and is also called the percentage weight by volume measurement and labeled as % w/v or simply w/v. Note that the units for a desired dose of a drug and the units for a solution containing the drug or pill form of the drug must be the same. If they are not the same, the units must first be converted to be measured in the same units. Suppose you visit your doctor with symptoms of an upset stomach and unrelenting heartburn. One possible recourse is sodium bicarbonate, which aids in reducing stomach acid. Calculating the Quantity in a Mixture How much sodium bicarbonate is there in a 250 mL solution of 1.58% w/v sodium bicarbonate? 1.58 % w / v = 1.58 g sodium bicarbonate in 100 mL. If there is 250 mL of the solution, we have 2.5 times as much sodium bicarbonate as in 100 mL. Thus, we multiply 1.58 by 2.5 to yield 3.95 g sodium bicarbonate in 250 mL solution. Calculating the Quantity of Pills Needed A doctor prescribes 25.5 mg of a drug to take orally per day and pills are available in 8.5 mg. How many pills will be needed each day? The prescription and the pills are in the same units which means no conversions are needed. We can divide the units of the drug prescribed by the units in each pill: 25.5 / 8.5 = 3 . So, 3 pills will be needed each day. Calculating the Drug Dose in Milligrams, Based on Patient Weight A patient is prescribed 2 mg/kg of a drug to be delivered intramuscularly, divided into 3 doses per day. If the patient weighs 45 kg, how many milligrams of the drug should be given per dose? Step 1: Calculate the total daily dose of the drug based on the patient’s weight (measured in kilograms): ( 2 mg / kg ) ( 45 kg ) = 90 mg Step 2: Divide the total daily dose by the number of doses per day: 90 mg / 3 = 30 mg The patient should receive 30 mg of the drug in each dose. Note that the units for a patient’s weight must be compatible with the units used in the medicine measurement. If they are not the same, the units must first be converted to be measured in the same units. Calculating the Drug Dose in Milliliters, Based on Patient Weight A patient is prescribed 2 mg/kg of a drug to be delivered intramuscularly, divided into 3 doses per day. If the drug is available in 20 mg/mL and the patient weighs 60 kg, how many milliliters of the drug should be given per dose? Step 1: Calculate the total daily dose of the drug (measured in milligrams) based on the patient’s weight (measured in kilograms): ( 2 mg / kg ) ( 60 kg ) = 120 mg Step 2: Calculate the volume in each dose: ( 120 mg daily total ) / ( 3 doses a day ) = 40 mg per dose Step 3: Calculate the volume based on the strength of the stock: ( prescribed dose needed ) / ( stock dose ) = volume ( 40 mg per dose ) / ( 20 mg / mL ) = 2 mL The patient should receive 2 mL of the stock drug in each dose. Math Statistics from the CDC The Centers for Disease Control and Prevention (CDC) states that about half the U.S. population in 2019 used at least one prescription drug each month, and about 25% of people used three or more prescription drugs in a month. The resulting overall collective impact of the pharmaceutical industry in the United States exceeded $1.3 trillion a year prior to the 2020 pandemic. Validating Effectiveness of a New Vaccine The process to develop a new vaccine and be able to offer it to the public typically takes 10 to 15 years. In the United States, the system typically involves both public and private participation in a process. During the 1900s, several vaccines were successfully developed, including the following: polio vaccine in the 1950s and chickenpox vaccine in the 1990s. Both of these vaccines took years to be developed, tested, and available to the public. Knowing the typical timeline for a vaccine to move from development to administration, it is not surprising that some people wondered how a vaccine for Covid-19 was released in less than a year’s time. Lesser known is that research on coronavirus vaccines has been in process for approximately 10 years. Back in 2012, concern over the Middle Eastern respiratory syndrome (MERS) broke out and scientists from all over the world began working on researching coronaviruses and how to combat them. It was discovered that the foundation for the virus is a spike protein, which, when delivered as part of a vaccine, causes the human body to generate antibodies and is the platform for coronavirus vaccines. When the Covid-19 pandemic broke out, Operation Warp Speed, fueled by the U.S. federal government and private sector, poured unprecedented human resources into applying the previous 10 years of research and development into targeting a specific vaccine for the Covid-19 strain. Shibo Jiang Dr. Shibo Jiang, MD, PhD, is co-director the Center for Vaccine Development at the Texas Children’s Hospital and head of a virology group at the New York Blood Center. Together with his colleagues, Jiang has been working on vaccines and treatments for a range of viruses and infections including influenzas, HIV, Sars, HPV and more recently Covid-19. His work has been recognized around the world and is marked with receiving grants amounting to over $20 million from U.S. sources as well as the same from foundations in China, producing patents in the United States and China for his antiviral products to combat world concerns. Jiang has been a voice for caution in the search for a vaccine for Covid-19, emphasizing the need for caution to ensure safety in the development and deployment of a vaccine. His work and that of his colleagues for over 10 years on other coronaviruses paved the way for the vaccines that have been shared to combat the Covid-19 pandemic. Mathematical Modeling to Track the Spread of a Vaccine With a large number of people receiving a Covid-19 vaccine, the concern at this time is how to create an affordable vaccine to reach people all over the world. If a world solution is not found, those without access to a vaccine will serve as incubators to variants that might be resistant to the existing vaccines. As we work to vaccinate the world, attention continues with tracking the spread of the Covid-19 and its multiple variants. Mathematical modeling is the process of creating a representation of the behavior of a system using mathematical language. Digital mathematical modeling plays a key role in analyzing the vast amounts of data reported from a variety of sources such as hospitals and apps on cell phones. When attempting to represent an observed quantitative data set, mathematical models can aid in finding patterns and concentrations as well as aid in predicting growth or decline of the system. Mathematical models can also be useful to determine strengths and vulnerabilities of a system, which can be helpful in arresting the spread of a virus. The chapter on Graph Theory explores one such method of mathematical modeling using paths and circuits. Cell phones have been helpful in tracking the spread of the Covid-19 virus using apps regulated by regional government public health authorities to collect data on the network of people exposed to an individual who tests positive for the Covid-19 virus. Gladys West Gladys West (credit: \"Dr. Gladys West Hall\" by The US Air Force/Wikimedia Commons, Public Domain) Dr. Gladys West is a mathematician and hidden figure with a rich résumé of accomplishments spanning Air Force applications and work at NASA. Born in 1930, West rose and excelled both academically and in her professional life at a time when Black women were not embraced in STEM positions. One of her many accomplishments is the Global Positioning System (GPS) used on cell phones for driving directions. West began work as a human computer, someone who computes mathematical computations by hand. Considering the time and complexity of some calculations, she became involved in programming computers to crunch computations. Eventually, West created a mathematical model of Earth with detail and precision that made GPS possible, which is utilized in an array of devices from satellites to cell phones. The next time you tag a photo or obtain driving directions, you are tapping into the mathematical modeling of Earth that West developed. Consider the following graph ( ): Contact Tracing for Math 111 Section 1 At the center of the graph, we find Alyssa, whom we will consider positive for a virus. Utilizing the technology of phone apps voluntarily installed on each phone of the individuals in the graph, tracking of the spread of the virus among the 6 individuals that Alyssa had direct contact with can be implemented, namely Suad, Rocio, Braeden, Soren, and Sandra. Let’s look at José’s exposure risk as it relates to Alyssa. There are multiple paths connecting José with Alyssa. One path includes the following individuals: José to Mikaela to Nate to Sandra to Alyssa. This path contains a length of 4 units, or people, in the contact tracing line. There are 2 more paths connecting José to Alyssa. A second path of the same length consists of José to Lucia to Rocio to Braeden to Alyssa. Path 3 is the shortest and consists of José to Lucia to Rocio to Alyssa. Tracking the spread of positive cases in the line between Alyssa and José aids in monitoring the spread of the infection. Now consider the complexity of tracking a pandemic across the nation. Graphs such as the one above are not practical to be drawn on paper but can be managed by computer programs capable of computing large volumes of data. In fact, a computer-generated mathematical model of contact tracing would look more like a sphere with paths on the exterior as well as on the interior. Mathematical modeling of contact tracing is complex and feasible through the use of technology. Using Mathematical Modeling For the following exercises, use the sample contact tracing graph to identify paths ( ). Contact Tracing for ECON 250 Section 1 How many people have a path of length 2 from Jeffrey? Find 2 paths between Kayla and Rohan. Find the shortest path between Yara and Kalani. State the length and people in the path. 5 (Lura, Naomi, Kalani, Vega, Yara) Answers will vary. Two possible answers are as follows: Kayla, Jeffrey, Rohan Kayla, Lura, Yara, Lev, Vega, Uma, Kalani, Rohan Length is 4. People in path = Yara, Lev, Vega, Uma, Kalani Check Your Understanding Key Terms concentrations/dosages of drugs mathematical modeling Key Concepts Compute volumes of prescription drugs in liquid and pill form. Validate the effectiveness of a new drug. Mathematical modeling can be used to describe and track the spread of a virus. Formula Volume needed = ( medicine dosage required ) ( weight of drug by volume ) .", "section": "Math and Medicine", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Math and Music Friends sing music together around a campfire. (credit: modification of work “Fire is hot! (2)” by Chetan Sarva/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe the basics of frequency related to sound. Describe the basics of pitch as it relates to music. Describe and evaluate musical notes, half-steps, whole steps, and octaves. Describe and find frequencies of octaves. “The world’s most famous and popular language is music.” Psy, South Korean singer, rapper, songwriter, and record producer Imagine a world without music and many of us would struggle to fill the void. Music uplifts, inspires, heals, and generally adds dimension to virtually every aspect of our lives. But what is music? For some it is a song; for others it may be the sounds of birds or the rhythmic sound of drumming or a myriad of other sounds. Whatever you consider music, it is all around us and is an integral part of our lives. “Music can raise someone’s mood, get them excited, or make them calm and relaxed. Music also—and this is important—allows us to feel nearly or possibly all emotions that we experience in our lives. The possibilities are endless” (Galindo, 2009). What music you listen to can impact your mood and emotions. In similar fashion, the music we choose can often tell those around us something about our current moods and emotions. Consider the music you may have been listening to as you today or even as you are reading this text. What cues to your mood do your music selections share? Albert Einstein is quoted as saying, “If I were not a physicist, I would probably be a musician. I often think in music. I live my daydreams in music. I see my life in terms of music.” What clues do your recent music choices say about your mood or how your day is going? Basics of Frequency as It Relates to Sound Every sound is created by an object vibrating and these vibrations travel in waves that are captured by our ears. Some vibrations we may be able to see, such as a plucked guitar string moving, whereas other vibrations we may not be able to see, such as the sound created when we hold our breath when accidentally dropping our cell phone on a hard floor. We don’t see the vibrations of our cell phone hitting the floor; however, any audible sound created in the fall is the result of vibrations in the form of sound waves, which can be pictured similarly to waves moving through the ocean. The waves of sounds each have a frequency , or rate of vibration of sound waves, that measures the number of waves completed in a single second and are measured in hertz (Hz; one Hz is one cycle per second). Louder sounds have stronger vibrations or are created closer to our ear. The further an ear is from the source of the sound, the quieter the sound will appear. Sounds range in frequency from 16 Hz to ultrasonic values, with humans able to hear sounds in a frequency range of about 20 Hz to 20,000 Hz. Adults lose the ability to hear the upper end of the range and typically top out in the ability to hear in a frequency of 15,000–17,000 Hz. Sounds with a frequency above 17,000 Hz are less likely to be heard by adults while still being audible to children. While frequency plays a key role in audible sounds, so too does the sound level, which can be measured in decibels (dB), which are the units of measure for the intensity of a sound or the degree of loudness. As a sound level increases, the decibel level increases. A person with average hearing can hear sounds down to 0 dB. Those with exceptionally good hearing can hear even quieter sounds, down to approximately –5 dB. The following table includes sample sounds with their related decibel values. Sound Decibels (dB) Firecrackers 140 Take-off of military jet from aircraft carrier 130 Clap of thunder 120 Auto horn standing next to the vehicle 110 Outboard motor 100 Motorcycle 90 Noise inside a car in city traffic 80 Typical washing machine 70 Public conversation, such as at a restaurant 60 Private conversation 50 Hum of a computer with fan blowing 40 Quiet whisper 30 Swishing leaves 20 Regular breathing 10 Lowest typical sound audible by teenagers 0 Selecting Decibel Value of Sounds Select the most representative decibel value for each of the following sounds: car wash: 25 dB, 55 dB, 85 dB vacuum cleaner: 15 dB, 70 dB, 90 dB ship’s engine room: 30 dB, 65 dB, 95 dB approaching subway car: 70 dB, 100 dB, 120 dB 85 dB 70 dB 95 dB 100 dB Basics of Pitch When considering the various sound levels the human ear can hear, the ear perceives sound both from the frequency level and the pitch of a sound. The quality of the sound is referred to as pitch , the tonal quality of a sound and how high or low the tone. Sounds with a high frequency have a high pitch, such as 900 Hz, and sounds with a low pitch have a low frequency, such as 50 Hz. Let’s take a look at frequency and pitch using a string instrument such as a guitar or piano. When a string is plucked on a guitar or a key is played on a piano, the related string vibrates at a frequency that is related to the length and thickness of the string. The frequency is measurable and has a singular value. The pitch of the note played is open for interpretation, as the pitch is a function of personal opinion. Graphing Calculators and Music It may be interesting to note that a TI-84 and TI-Nspire graphing calculators can be utilized to tune a musical instrument by measuring the frequency of a note using a small plug-in accessory that captures the sound waves from a note and displays the corresponding frequency. Using the displayed frequency, an instrumentalist can then make the needed adjustment to perfectly tune an instrument. This method of tuning an instrument can be helpful whether a novice player or a seasoned instrumentalist because the instrument can be tuned precisely to the correct frequency. Note Values, Half-Steps, Whole Steps, and Octaves “There are not more than five musical notes, yet the combinations of these five give rise to more melodies than can ever be heard.” Sun Tzu, Chinese strategist Piano Keys and Notes (credit: modification of work \"Contemplate\" by Walt Stoneburner/Flickr, CC BY 2.0) Moving our exploration to note values, the frequency of all notes is well defined by a specific and unique frequency for each note that is measurable. We will explore keys on a keyboard to discuss notes that have the same relationships with any instrument or musical piece. Let’s look at . The white keys are labeled with the letters A–G and the photo begins with middle C, which can be found in the middle of a keyboard. This labeling of the keys repeats across an entire keyboard and keys to the right have a higher pitch and frequency than keys to the left. Each of the keys correlates to a musical note. Movement up or down between any two consecutive keys (black and white) or notes constitutes a half-step . Movement of one half-step sometimes involves a sharp (#) or a flat (♭) symbol. For example, D # is one half-step above D and D ♭ is one half-step below D. Note that this is not always true as one half-step above B is C, and one half-step below F is E. In similar fashion, a whole step is movement up or down between any two half-steps on a keyboard. Identifying Half-Steps Name which keys are one half-step up and one half-step down from the following: D E G# up D # , down D ♭ up F, down E ♭ up A, down G Identifying Whole Steps Name which keys are one whole step up and one whole step down from the following: F # E A ♭ Up G # , down E Up F # , down D Up B ♭ , down G ♭ You may have noticed that there are eight letters of the alphabet used to label notes. Selecting any one note and counting up 12 half-steps you will find that the numbering for notes begins at the same value as you started from. This collection of 12 consecutive half-notes is called an octave and is a basic foundational component in music theory. Listing All Notes in an Octave List the 12 notes forming an octave, beginning with the note C. C, C # , D, D # , E, F, F # , G, G # , A, A # , B Frequencies of Octaves Notes that are one octave apart have the same name and are related in frequency values. Given the frequency of any note, the frequency of same note one octave higher is doubled and this pattern continues as you move up and down the notes on a keyboard or any other musical instrument. Song writers and singers use this knowledge to change the pitch of a note up or down to align with a person’s vocal range. Regardless of which C is played or sung, the pitch is the same and the frequency is related by a power or two. Labeled keys on a keyboard are numbered for ease in identification. For example, middle C is labeled as C 4 on a full keyboard as it is the fourth C from the left in a set of eight notes. The frequency of C 4 is 262 Hz, rounded to the nearest whole number. Calculating the Frequency Values of Octaves Given that the frequency of C 4 is 262 Hz, find the approximate frequency of C 6 . The frequency of each consecutive higher octave doubles. Given that the frequency of C 4 is 262 Hz, the frequency of C 5 is found by doubling the frequency of C 4 , which is 524 Hz. In similar fashion, the frequency of C 6 is found by doubling the frequency of C 5 , which yields 1,048 Hz. David Cope David Cope has a somewhat eclectic list of job titles ranging from author and music professor to scientist and artificial intelligence researcher. Cope combined his interests when he developed software that can analyze a piece of music and create a new and unique musical piece in the same style as the original. Some of his well-known products have been based off of the classical music of Mozart, creating what has been called Mozart’s “42nd Symphony,” as well as other genres including opera and a range of current music styles. Cope has also composed original musical pieces in collaboration with a computer. We have explored some basics components of frequency, pitch, note relationships, and octaves, which are building blocks of music. It may be exciting to learn that the mathematical relationships found in music are vast and grow in complexity beyond the math commonly studied in high school. Spotify Royalty Payments Streaming services have grown exponentially in popularity thanks in large part to customized music listening through cell phone use and devices such as Google as well as Amazon Echo and Alexa devices for home and vehicles, adding to ways that artists are paid royalties. Spotify, which was launched in 2008, typically plays artists $0.06 per time a song is streamed, with some artists receiving up to $0.84 per play amounting to over $9 billion in revenue for Spotify in 2020. Since 2014, Spotify’s revenue has grown over a billion dollars a year, with roughly half of their revenue being paid out in royalties, which was good news for artists during the Covid-19 pandemic when in-person concerts and shopping were hindered. Check Your Understanding Key Terms frequency pitch Hertz decibel half-step whole step flat sharp octave Key Concepts As frequency of a sound increases, the pitch of the sound increases. Hertz is a unit of measurement for frequency. Decibel is a unit of measurement for the intensity of sound. Half-steps and whole-steps describe one type of movement between two notes on a keyboard. Octaves are a collection of any 12 consecutive notes, which is a foundation in music theory.", "section": "Math and Music", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Math and Sports Fans support their team by attending games and wearing team gear. (credit: “Cheering Touchdowns” by Steven Miller/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe why data analytics (statistics) is crucial to advance a team’s success. Describe single round-robin method of tournaments. Describe single-elimination method of tournaments. Explore math in baseball, fantasy football, hockey, and soccer (projects at the end of the section). Sports are big business and entertainment around the world. In the United States alone, the revenue from professional sports is projected to bring in over $77 billion, which includes admission ticket costs, merchandise, media coverage access rights, and advertising. So, whether or not you enjoy watching professional sports, you probably know someone who does. Some celebrities compete to be part of half-time shows and large companies vie for commercial spots that are costly but reach a staggering number of viewers, some who only watch the half-time shows and advertisements. Data Analytics (Statistics) Is Crucial to Advance a Team’s Success Analyzing the vast data that today’s world has amassed to find patterns and to make predictions for future results has created a degree field for data analytics at many colleges, which is in high demand in places that might surprise you. One such place is in sports, where being able to analyze the available data on your team’s players, potential recruits, opposing team strategies, and opposing players can be paramount to your team’s success. Hollywood turned the notion of using data analytics into a major motion picture back in 2011 with the release of Moneyball, starring Brad Pitt, which grossed over $110 million. The critically acclaimed movie, based on a true story as shared in a book by Michael Lewis, follows the story of a general manager for the Oakland Athletics who used data analytics to take a team comprised of relatively unheard of players to ultimately win the American League West title in a year’s time. The win caught the eye of other team managers and owners, which started an avalanche of other teams digging into the data of players and teams. In today’s world of sports, a team has multiple positions utilizing data analytics from road scouts who evaluate a potential recruit’s skills and potential to the ultimate position of general manager who is typically the highest-paid (non-player) employee with the exception of the coaches. Being able to understand and evaluate the available data is big business and is a highly sought after skill set. In college and professional sports, it is no longer sufficient to have a strong playbook and great players. The science to winning is in understanding the math of the data and using it to propel your team to excelling. Single Round-Robin Tournaments Single-Round Robin Tournament A common tournament style is single round-robin tournaments ( ), where each team or opponent plays every other team or opponent, and the champion is determined by the team that wins the most games. Ties are possible and are resolved based on league rules. An advantage of the round-robin tournament style is that no one team has the advantage of seeding , which eliminates some teams from playing against each other based on rank of their prior performance. Rather, each team plays every other team, providing equal opportunity to triumph over each team. In this sense, round-robin tournaments are deemed the fairest tournament style. One hindrance to employing a round-robin-style tournament is the potential for the number of games involved in tournament play to determine a winner. Determining the number of games can be found easily using a formula which, as we will see, can quickly grow in the number of games required for a single round-robin tournament. The number of games in a single round-robin tournament with n teams is n ( n - 1 ) / 2 . Calculating the Number of Games in Single Round-Robin Tournaments Find the number of games in a single round-robin tournament for each of the following numbers of teams: 4 teams 8 teams 20 teams Using the formula with 4 teams yields 4 ( 4 - 1 ) / 2 = 6 tournament games. Using the formula with 8 teams yields 8 ( 8 - 1 ) / 2 = 28 tournament games. Using the formula with 20 teams yields 20 ( 20 - 1 ) / 2 = 190 tournament games. As the examples show, single round-robin tournament play can quickly grow in the number of games required to determine a champion. As such, some tournaments elect to employ variations of single round-robin tournament play as well as other tournament styles such as elimination tournaments. Single-Elimination Tournament (credit: final TK) Single-Elimination Tournaments When desiring a more efficient tournament style to determine a champion, one option is single-elimination tournaments ( ), where teams are paired up and the winner advances to the next round of play. The losing team is defeated from tournament play and does not advance in the tournament, although some leagues offer consolation matches. The number of games in a single-elimination tournament with n teams is ( n - 1 ) . Calculating the Number of Games in Single-Elimination Tournaments Find the number of games in a single-elimination tournament for each of the following numbers of teams: 4 teams 8 teams 20 teams Using the formula with 4 teams yields ( 4 - 1 ) = 3 tournament games. Using the formula with 8 teams yields ( 8 - 1 ) = 7 tournament games. Using the formula with 20 teams yields ( 20 - 1 ) = 19 tournament games. A single-elimination tournament offers an advantage over single round-robin tournament style of play in the number of games needed to complete the tournament. As you can see, in comparing the number of games in a single round-robin tournament in with the number of games in single-elimination tournament as shown in , the number of games required for single round-robin can quickly become unmanageable to schedule. There are modifications to both the round-robin and elimination tournament styles such as double round-robin and double-elimination tournaments. Next time you observe a college or professional sporting event, see if you can determine the tournament style of play. Sports Popularity Shifts Sport has been a popular entertainment venue for hundreds of years and the popularity of various sports shifts over time as well as in different regions of the world today. In today’s world, the most popular sport is, overwhelmingly, soccer, with over 4 billion fans followed by cricket with 2.5 billion fans. It may surprise you to learn that American football doesn’t rank in the top 10 most popular sports in the world today, yet table tennis ranks in sixth place and golf fills the last spot in the top 10 world sports. In the early 1930s, baseball ranked a close second, with basketball virtually tying for third place. By the mid-1940s, hockey slightly led in first place over basketball. Ten years later, hockey remained in the number one sport, but cricket pushed basketball to third place. In the 1960s, soccer dominated in popularity. Over the next 30 years, basketball dropped in popularity and there was much movement in the popularity of sports. By the late 1990s, soccer swept to first place, where it has since continued to grow in popularity. Check Your Understanding Key Terms data analytics seeding Key Concepts Describe how a round-robin tournament is organized. Compute the number of games played in a round-robin tournament. Describe how a single-elimination tournament is organized. Compute the number of games played in a single-elimination tournament. Formulas Number of games in a single round-robin tournament with n teams is n ( n - 1 ) / 2 . Number of games in a single-elimination tournament with n teams is ( n - 1 ) . project Lucas Sequence and Fibonacci Sequence The Lucas numbers bear some similarity to the Fibonacci numbers and exhibit a stronger link to the golden ratio. Edouard Lucas is credited with naming the Fibonacci numbers and the Lucas numbers were so named in his honor. The Lucas numbers play a role in finding prime numbers that are utilized in encrypting data for actions such as using your debit card to obtain money at a cash machine or when making a credit card purchase for point of sale as well as when shopping online. Complete the following questions to explore numbers in the Lucas sequence as well as their relationships to the numbers in the Fibonacci sequence. Conduct an Internet search to find out what a Lucas number is and how the Lucas numbers are related to the Fibonacci numbers. What are the first two numbers in the Lucas sequence? Describe how the next number in the Lucas sequence is determined and compare this to how the next number is determined in the Fibonacci sequence. Complete the following table listing the first 10 terms in the Fibonacci and Lucas sequence: Term 0 1 2 3 4 5 6 7 8 9 Fibonacci Numbers 0 1 1 2 5 13 34 Lucas Numbers 3 7 18 47 Interestingly, many patterns can be found in looking at the relationships in the Fibonacci and Lucas numbers. Look closely at the chart in question 3 to discover one such pattern. Observe the Fibonacci numbers in the third and fifth terms and compare with the Lucas number in the fourth term of the sequence. Describe the pattern found. Does this pattern continue in the table? Research the Fibonacci or the Lucas numbers to find an application in our world distinct from what has been shared in this project and section. Write a paragraph sharing the findings of your research. Solar Array for a Residence One of the first steps in adding solar to a residence is determining the size of a system to achieve the desired output. In this project, we will explore the solar needs of a residence and estimate needs of a solar array to supply electrical output to meet various percentages of electrical need. Step 1: Obtain an electric bill from your apartment/home. Find the average monthly or yearly usage if listed or call the electric company to inquire. If an electric bill is not available, use the Internet to find an average monthly or yearly electric usage for your area. Step 2: Determine a daily and hourly usage. Divide the average monthly usage by 30 or the yearly average by 365. Divide again by 24 to calculate an average hourly electric usage, which will yield the average kilowatt-hours for how much electrical power your is being utilized in an hour. Step 3: Multiply your average hourly use (kilowatts) by 1,000 to convert to watts. Step 4: Use the Internet to determine the average daily peak hours of sunlight where you live. Step 5: Divide your average hourly watts (Step 3) by the average daily peak hours (Step 4) to calculate the average energy needed for a solar array to produce every hour. Step 6: Determine the average energy needed in a solar array per hour to meet each of the following: 100 % coverage of average energy (Step 5) 80 % coverage of average energy 50 % coverage of average energy Step 7: Using the values computed in Step 6, compute the residential savings based on an average cost of 12 cents per watt. Vaccine Validation Validation of vaccines is a topic that exploded in the news when the Covid-19 pandemic spread across the world. As governments and organizations looked for a vaccine to curb the spread and minimize the severity of infection, concern was expressed by some for what appeared to be a quick discovery for a Covid-19 vaccine. Conduct an Internet search to explore the following questions. Pay special attention to the sources you select to ensure that they are credible sources. Research the term efficacy rates . Express what efficacy means in your own words. Using a minimum of two sources, compose a well-developed paragraph describing what validation of a vaccine means. Using a minimum of two sources, compose a well-developed paragraph sharing the steps in validating a vaccine. Using a minimum of two sources, compose a well-developed paragraph describing how a vaccine is determined to be validated. Using your research for Questions 1–3, write a summary paragraph sharing your reflections on the validation of the Covid-19 vaccine or on validating a vaccine in general. What key components did you learn? What would you like to learn more about related to validating a vaccine? Frequency and Ultrasonic Sounds Ultrasonic sounds have been utilized for a variety of reasons, from purportedly repelling rodents and other animals as well as a variety of other applications. Using an Internet search, complete the following questions to explore some of these applications and examine the validity of various claims. Repelling Insects, Rodents, and Small Animals Some radio stations purport to play a high pitch sound dually with their music to aid in deterring insects and other annoying bugs to aid in providing a bug-reduced listening environment. To deter small rodents, some products claim to emit ultrasonic sounds that drive away mice and other similar pests. Research the science behind ultrasonic pest controls, paying attention to the source of the information that you find. Compare the information found on advertisements, reviews, and scientific articles. What frequency ranges do ultrasonic pest deterrent devices utilize and how do these frequencies compare to the audible range that humans can hear? Write a short paragraph comparing the claims in the advertisements with independent reviews and scientific articles. What do you conclude about ultrasonic pest controls and why? Disbursing Teenagers Some business owners and communities have turned to products such as the “mosquito” sonic deterrent device to discourage groups of teenagers from loitering around storefronts and community landmarks, citing a public nuisance issue and public safety concerns. Research the science behind ultrasonic deterrent devices as they apply to dispensing teenagers. Compare the information found on advertisements, reviews, and scientific articles. What frequency ranges do ultrasonic teenager deterrent devices utilize and how do these frequencies compare to the audible range that adults can hear? Write a short paragraph comparing the claims in advertisements with independent reviews and scientific articles. What are some of the ethical debates surrounding the use of ultrasonic teenager deterrent devices? What do you conclude about business owners or communities using ultrasonic teenager deterrent devices and why? Jewelry Cleaner Use of pastes and liquid chemicals to clean jewelry can be harsh on stones as well as metals. So how can we safely obtain the sparkling clean look at home that jewelry stores provide? Some would say the answer is to use an ultrasonic jewelry cleaner, but do these really work? Research the science behind ultrasonic jewelry cleaners, including the phrase “cavitation process” in your search. Compare the information found on advertisements, reviews, and scientific articles. Write a short paragraph detailing how ultrasonic jewelry cleaners work and what role the cavitation process plays in the claims for ultrasonic cleaning. Do ultrasonic jewelry cleaners utilize low or high frequencies? How do these frequencies compare to the audible range that humans can hear? Write a short paragraph comparing the claims in the advertisements with independent reviews and scientific articles. What do you conclude about ultrasonic jewelry cleaners and why? Specialized Ringtones As the use of cell phones has become commonplace and families grow towards each member having their own cell phone, specialized ringtones have become popular and can aid in identifying who is calling just by the ringtone. Ever hear of ringtones that can be heard by teens but often not their teachers? The banning of cell phone use by K–12 students during class time as been implemented across a wide array of schools and some students have purportedly found ways to get around teachers hearing a cell phone ring through the use of ultrasonic ringtones. Research the science behind ultrasonic ringtones. Compare the information found on advertisements, reviews, and scientific articles. Do ultrasonic ringtones utilize low or high frequencies? Write a short paragraph detailing how ultrasonic ringtones work. How do these frequencies compare to the audible range that adults can hear? Include ages that are purported to hear and not hear the ringtones as well as the frequency ranges utilized. Write a short paragraph comparing the claims in the advertisements with independent reviews and scientific articles. What do you conclude about ultrasonic ringtones and why? Streaming Services and Math With the ability to stream music virtually anywhere you are, it is not surprising that Google Play Music, Apple Music, and a slew of other companies such as Spotify, Amazon Music, YouTube Music, Sound Cloud, Pandora, Deezer Music, Tidal, Napster, and Bandcamp have invested heavily to bring streaming service to users worldwide. Streaming services have expanded to offer virtually every genre of music with vast libraries to meet diverse user requests. Considering all of the choices available for streaming music, there is a wide array of options for subscribing. Conduct an Internet research to review your current streaming choices, if any, and evaluate competitors’ products. In this project, you will explore options for music streaming service subscriptions. Select a minimum of five streaming services listed above that you are not currently utilizing and determine the below components. Format your findings in an easy-to-read format such as a table similar to the one shown below. Monthly subscription cost Available features Available Features Service Choice 1: Service Choice 2: Service Choice 3: Service Choice 4: Service Choice 5: Able to stream unlimited music Able to purchase individual songs Able to purchase individual whole albums Ability to create personalized play lists Ability to select specific songs to play Ability to listen Other = Premium cost per month, if available = ______ Feature(s) offered with premium monthly subscription = ______ What is (are) your current music streaming services, if any? What is your current monthly service charge(s)? What features does your current streaming service offer? Write a paragraph summarizing your findings and include if your current streaming service(s) meets your needs or if research for this project has you considering changing streaming services. Support your rationale. Math and Baseball Baseball is known to have one of the largest pools of statistics related to the game and its players. Managers, coaches, and pitchers study the statistics of the players on opposing teams to give their team an edge by knowing what pitches to throw for the best probability to be missed by a batter. In similar fashion, batters study pitchers’ statistics to learn a pitcher’s strength and how to predict what a pitcher will throw and how to best hit against a pitcher. The three primary baseball statistics are batting average, home runs, and runs batted in (RBIs), which are the components of the title of Triple Crown winner that is awarded to players who dominate in these three areas. However, there is a wealth of other statistics to evaluate when studying the performance of a player. Conduct an Internet search to research statistics and how they are calculated in the following categories: Batting Statistics There are about 30 batting statistics. Select a minimum of 10 batting statistics. Compose an organized list including the name of the statistic, abbreviation, explanation of what it represents, as well as how it is calculated. As an example: AB/HR represents at bats per home run and is calculated by the number of times a player is at bat divided by home runs. Pitching Statistics There are about 40 pitching statistics. Select a minimum of 10 pitching statistics. Compose an organized list including the name of the statistic, abbreviation, explanation of what it represents, as well as how it is calculated. As an example: K/9 represents strikeouts per nine innings and is calculated by the number of strikeouts times nine divided by the number of innings pitched. Fielding Statistics There are around 10 fielding statistics. Select a minimum of five fielding statistics. Compose an organized list including the name of the statistic, abbreviation, explanation of what it represents, as well as how it is calculated. As an example: FP represents fielding percentage is calculated by the number of total plays divided by the number of total chances. Overall Select a player from recent years to evaluate. The player can be one that you have followed, one from a favorite team, or any current player. Find the statistics shared in Questions 1–3 to use in evaluating the potential strengths and weaknesses of the player. Write a short paragraph analyzing your selected player, supported by the statistics from the answers to Questions 1–3. Math and Fantasy Football Fantasy football offers spectators an added dimension to football season with a competitive math-based game where the active components are real-life players in the current season. For clarity in this exercise, the actual fantasy football players will be denoted as FFP and actual professional team members will be denoted as players. While some fantasy football leagues have slightly different setups or scoring systems, most share some common elements. Often using a lottery system to determine who picks first, second, and so on, FFPs select 15 current players to comprise their personal fantasy football team. The players selected can be from any professional teams and a FFP can utilize any team recognized by their league. FFPs can elect to keep the same players on their team for the whole league play or trade for any player not selected by another FFP in their league. At the start of each week during football season, each fantasy football player selects their roster of actual players to comprise their roster of starting players. Typically, a starting roster consists of the following players: Number to Select Position Abbreviation Position Title 1 QB Quarterback 1 K Kicker 1 TE Tight End 2 RB Running Back 1 D/ST Defense 2 WR Wide receiver 1 RB or WR Flex As actual professional games are played, points are tallied based on your league’s scoring system. The points the team members on your starting roster make during the week are computed and whichever FFP has the highest score for the week wins that week. The FFPs with the best records of wins versus losses enters fantasy football playoffs to determine the ultimate league champion and collects the pot. The above overview of fantasy football describes the basic game play. The fun comes in understanding and analyzing the math behind the scoring. Talk to people you know who have played fantasy football in the past and interview them. Inquire about how they select players and how league(s) they have played in have computed the weekly scoring. Was a league manager recruited to manage the record keeping of scores? Was an online scoring system utilized or did the fantasy football use a streamlined or specialized scoring system? What strategies were used to select the team of 15 players and who would be on the weekly starting roster? Were any online resources utilized to aid in choosing which players to select? How was math utilized to select team members or who to place as a starter each week? Be as specific as possible, citing examples when available. Conduct an Internet research to find two different resources that a FFP could utilize when selecting players for their team or for their starting roster. Describe the math involved in the resources. Conduct an Internet research to learn more about the scoring utilized by fantasy football leagues. Write a short paragraph sharing how the use of math and knowledge of football statistics is used and how they are calculated to aid a FFP in selecting their team and starting roster. Math and Hockey Hockey is full of math from obvious components such as scoring and statistics to the shape of the rink and the angles involved in puck movement. Collegiate and professional hockey games are 60 minutes long and are divided into three periods of 20 (60/3) minutes each. At any one time, there are five players and one goalie on the ice for each team. If a player is called on a penalty and is placed in the penalty box, that team now has four players, which is 20% less players on the ice competing against five opponents. In some instances, a team may have two players in the penalty box at one time, resulting in three players, or 40% less players on the ice compared to a full team. Being one player down for a 2-minute penalty or potentially 5 minutes for a major penalty leads to an imbalance on the ice and calls for a quick change of offense and defense strategy. Rink Composition—North American An ice rink is comprised of various geometrical shapes, each with precise dimensions. Research the dimensions of a hockey rink and draw an accurate scale model on graph paper. Be sure to include the scale for your model and indicate all units. List the shapes and the numbers of each kind of hockey rink. You should find four different shapes, some with multiple dimensions. Describe the shape and dimensions of a hockey puck using geometric vocabulary. Statistics Using an Internet search, select two top hockey players from the same league to answer the following questions: Write a paragraph sharing a minimum of five statistics for each player you have selected. Describe how each of the statistics are calculated and what each statistic means. Write a paragraph comparing the two players and determine who you believe is the better player. Support your choice. Scoring The basics of scoring in ice hockey is simple, the team with the most goals is the winner. But, how to score the most goals involves much math! Research one of the following components involved in hockey puck movement on the ice and write a paragraph summarizing your findings. Be specific and detailed in your summary. Angles Velocity vectors Angle of incident and angle of return Speed and acceleration (player as well as puck movement) Math and Soccer As the world’s most popular sport, you’ll be excited to confirm that soccer is full of mathematics ranging from scoring and statistics to footwork, angles, and field shape. Soccer requires understanding of mathematical concepts and equations as well as skill, fitness, and game knowledge. One such example is angles, which you all will remember from your geometry class. While players are not carrying protractors and measuring angles during play, mental calculation of angles is a constant in any successful player’s thinking. A goalie is not physically able to cover the entire open net region and a player must calculate an angle to kick the ball consistent with the net opening while predicting the ability of the goalie to stop the ball from entering the net. Research angles as they apply to soccer play. Provide two examples of different plays indicating angle of a player’s body, angle of foot striking the ball, and angle to the net. Include relevant dimensions. Adding a diagram that may aid in clarity is an option. Draw a scale model of a soccer field including dimensions with labels. Using an Internet search, select two top soccer players from the same league to answer the following questions: Write a paragraph sharing a minimum of five statistics for each player you have selected. Describe how each of the statistics are calculated and what each statistic means. Write a paragraph comparing the two players and determine who you believe is the better player. Support your choice. Chapter Review Section 13.1 Math and Art Section 13.2 Math and the Environment Section 13.3 Math and Medicine Section 13.4 Math and Music Section 13.5 Math and Sports Chapter Test", "section": "Math and Sports", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Co-Req Appendix: Integer Powers of 10 Nonnegative Integer Powers of 10 The phrase nonnegative integers refers to the set containing 0, 1, 2, 3, … and so on. In the expression 10 5 , 10 is called the base , and 5 is called the exponent , or power . The exponent 5 is telling us to multiply the base 10 by itself 5 times. So, 10 5 = 10 × 10 × 10 × 10 × 10 = 100,000 . By definition, any number raised to the 0 power is 1. So, 10 0 = 1 . In the following table, there are several nonnegative integer powers of 10 that have been written as a product. Notice that higher exponents result in larger products. What do you notice about the number of zeros in the resulting product? Exponential Form Product Number of Zeros in Product 10 0 1 0 10 1 10 1 10 2 10 × 10 = 100 2 10 3 10 × 10 × 10 = 1,000 3 10 4 10 × 10 × 10 × 10 = 10,000 4 10 5 10 × 10 × 10 × 10 × 10 = 100,000 5 That’s right! The number of zeros is the same as the power each time! Negative Integer Powers of 10 The reciprocal of a number is 1 divided by that number. For example, the reciprocal of 10 is 1 10 . We use negative exponents to indicate a reciprocal. For example, 10 − 1 = 1 10 1 = 1 10 . Similarly, any expression with a negative exponent can be written with a positive exponent by taking the reciprocal. Several negative powers of 10 have been simplified in the table that follows. What do you notice about the number of zeros in the denominator (bottom) of each fraction? Exponential Form Equivalent Simplified Expression Number of Zeros in Denominator 10 − 1 1 10 1 = 1 10 1 10 − 2 1 10 2 = 1 10 × 10 = 1 100 2 10 − 3 1 10 3 = 1 10 × 10 × 10 = 1 1,000 3 10 − 4 1 10 4 = 1 10 × 10 × 10 × 10 = 1 10,000 4 That’s right! The number of zeros is the same as the positive version of the power each time. In the following table, we will write the same powers of 10 as decimals. Count the number of decimal places to the right of the decimal point. What do you notice? Exponential Form Equivalent Simplified Expression Number of Decimal Places to Right of Decimal 10 − 1 1 10 1 = 1 ÷ 10 = 0.1 1 10 − 2 1 10 2 = 1 ÷ 100 = 0.01 2 10 − 3 1 10 3 = 1 ÷ 1,000 = 0.001 3 10 − 4 1 10 4 = 1 ÷ 10,000 = 0.0001 4 That’s right! The number of decimal places to the right of the decimal point is the same as the positive version of the power each time. Multiplying Integers by Positive Powers of 10 Did you know that the distance from the sun to Earth is over 90 million miles? This value can be represented as 90,000,000, or we can write it as a product: 9 × 10,000,000 = 9 × 10 7 , which is actually a more compact way of writing 90 million. Notice that the power of 7 reflects the number of zeros in 90 million. Several products of positive integers and powers of 10 are given in the table that follows. Notice that the number of zeros is the same as the exponent except in one case. Exponential Form Product Number of Zeros in Product 5 × 10 1 5 × 10 = 50 1 13 × 10 2 13 × 100 = 1,300 2 8 × 10 3 8 × 1,000 = 8,000 3 15 × 10 4 15 × 10,000 = 150,000 4 70 × 10 5 70 × 100,000 = 7,000,000 6 The only case in which the number of zeros didn’t equal the exponent was the last case. Why do you think that happened? That’s right! We multiplied by 70 which also had a zero. So, the product had a zero from the 70 and 5 zeros from 10 5 for a total of 6 zeros in 7,000,000. Multiplying by Negative Powers of 10 As we have seen, negative powers of 10 are decimals. Several products of positive integers and powers of 10 are given in the table below. Notice that multiplying an integer by 10 raised to a negative integer power results in a smaller number than you started with. Also, the number of decimal places to the right of the decimal point is the same as the exponent except in one case. Exponential Form Product Number of Decimal Places to Right of Decimal 3 × 10 − 1 3 × 0.1 = 0.3 1 13 × 10 − 2 13 × 0.01 = 0.13 2 9 × 10 − 3 9 × 0.001 = 0.009 3 15 × 10 − 4 15 × 0.0001 = 0.0015 4 70 × 10 − 5 70 × 0.00001 = 0.00070 or 0.0007 5 ( 6 if we leave on the extra 0 ) The only case in which the number of decimal places to the right of the decimal point didn’t equal the positive version of the exponent was the last case. Why do you think that happened? That’s right! We multiplied by 70, which ended in zero. Moving the Decimal Place A helpful shortcut when multiplying a number by a power of 10 is to “move the decimal point.” The following table shows several powers of 10, both positive and negative. Compare the location of the decimal point in the original number to the location of the decimal point in the product. How has it changed? Exponential Form Product How the Position of the Decimal Point Changed 5 × 10 1 5. × 10 = 5 ∧ 0 ⌣ . = 50 1 place to the right 13 × 10 2 13. × 100 = 13 0 ⌣ ∧ 0 ⌣ . = 1,300 2 places to the right 8 × 10 3 8. × 1,000 = 8 ∧ 0 ⌣ 0 ⌣ 0 ⌣ . = 8,000 3 places to the right 15 × 10 4 15. × 10000 = 15 ∧ 0 ⌣ 0 ⌣ 0 ⌣ 0 ⌣ . = 150,000 4 places to the right 70 × 10 5 70. × 100,000 = 70 ∧ 0 ⌣ 0 ⌣ 0 ⌣ 0 ⌣ 0 ⌣ . = 7,000,000 5 places to the right 3 × 10 − 1 3. × 0.1 = . 3 ⌣ ∧ = 0.3 1 place to the left 13 × 10 − 2 13. × 0.01 = . 1 ⌣ 3 ⌣ ∧ = 0.13 2 places to the left 9 × 10 − 3 9. × 0.001 = . 0 ⌣ 0 ⌣ 9 ⌣ ∧ = 0.009 3 places to the left 15 × 10 − 4 15. × 0.0001 = . 0 ⌣ 0 ⌣ 1 ⌣ 5 ⌣ ∧ = 0.0015 4 places to the left 70 × 10 − 5 70 × 0.00001 = . 0 ⌣ 0 ⌣ 0 ⌣ 7 ⌣ 0 ⌣ ∧ = 0.0007 5 places to the left Notice that multiplying by a positive power of 10 moves the decimal point to the right, making the value larger, while multiplying by a negative power of 10 moves the decimal point to the left, making the value smaller. Also, the number of decimal places that the decimal point moves is exactly the positive version of the exponent.", "section": "Co-Req Appendix: Integer Powers of 10", "book": "Contemporary Mathematics", "subject": "Math", "source": "https://openstax.org/details/books/contemporary-mathematics"}
{"text": "Preface Welcome to Introductory Business Statistics 2e , an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost. About OpenStax OpenStax is part of Rice University, which is a 501(c)(3) nonprofit charitable corporation. Our mission is to make an amazing education accessible for all. Through our partnerships with philanthropic organizations and our alliance with other educational resource companies, we’re breaking down the most common barriers to learning. Because we believe that everyone should and can have access to knowledge. About OpenStax resources Customization Introductory Business Statistics 2e is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY-NC-SA) license, which means that you can non-commercially distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors, and distribute all derivatives under the same license. Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book. Instructors also have the option of creating a customized version of their OpenStax book. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit the Instructor Resources section of your book page on OpenStax.org for more information. Art Attribution In Introductory Business Statistics 2e , most art contains attribution to its title, creator or rights holder, host platform, and license within the caption. For any art that is openly licensed, non-commercial users or organizations may reuse the art as long as they provide the same attribution to its original source. (Commercial entities should contact OpenStax to discuss reuse rights and permissions.) To maximize readability and content flow, some art does not include attribution in the text. If you reuse art from this text that does not have attribution provided, use the following attribution: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. Errata All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on OpenStax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on OpenStax.org. Format You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print. About Introductory Business Statistics 2e Introductory Business Statistics 2e is designed to meet the scope and sequence requirements of the one-semester statistics course for business, economics, and related majors. Core statistical concepts and skills have been augmented with practical business examples, scenarios, and exercises. The result is a meaningful understanding of the discipline which will serve students in their business careers and real-world experiences. Coverage and scope Introductory Business Statistics 2e began as a customized version of OpenStax Introductory Statistics by Barbara Illowsky and Susan Dean. Statistics faculty at The University of Oklahoma have used the business statistics adaptation for several years, and the author has continually refined it based on student success and faculty feedback. The book is structured in a similar manner to most traditional statistics textbooks. The most significant topical changes occur in the latter chapters on regression analysis. Discrete probability density functions have been reordered to provide a logical progression from simple counting formulas to more complex continuous distributions. Many additional homework assignments have been added, as well as new, more mathematical examples. Introductory Business Statistics 2e places a significant emphasis on the development and practical application of formulas so that students have a deeper understanding of their interpretation and application of data. To achieve this unique approach, the author included a wealth of additional material and purposely de-emphasized the use of the scientific calculator. Specific changes regarding formula use include: Expanded discussions of the combinatorial formulas, factorials, and sigma notation Adjustments to explanations of the acceptance/rejection rule for hypothesis testing, as well as a focus on terminology regarding confidence intervals Deep reliance on statistical tables for the process of finding probabilities (which would not be required if probabilities relied on scientific calculators) Continual and emphasized links to the Central Limit Theorem throughout the book; Introductory Business Statistics 2e consistently links each test statistic back to this fundamental theorem in inferential statistics Another fundamental focus of the book is the link between statistical inference and the scientific method. Business and economics models are fundamentally grounded in assumed relationships of cause and effect. They are developed to both test hypotheses and to predict from such models. This comes from the belief that statistics is the gatekeeper that allows some theories to remain and others to be cast aside for a new perspective of the world around us. This philosophical view is presented in detail throughout and informs the method of presenting the regression model, in particular. The correlation and regression chapter includes confidence intervals for predictions, alternative mathematical forms to allow for testing categorical variables, and the presentation of the multiple regression model. Pedagogical features Examples are placed strategically throughout the text to show students the step-by-step process of interpreting and solving statistical problems. To keep the text relevant for students, the examples are drawn from a broad spectrum of practical topics; these include examples about college life and learning, health and medicine, retail and business, and sports and entertainment. Practice, Homework, and Bringing It Together give the students problems at various degrees of difficulty while also including real-world scenarios to engage students. Try It practice problems immediately follow many examples and give students the opportunity to practice as they read the text. Changes to the Second Edition The revision of Introductory Business Statistics has been undertaken with the support of many faculty adopters and student users. Most of the edits are focused on currency, clarity, accuracy, and belonging. The author made a number of additions in order to provide more support and practice opportunities for students; these include conceptual explanations, introductory text, new Try It exercises, and more detailed example solution steps. Also, dozens of examples and references have been changed in order to align with contemporary contexts and uses. Accuracy checking and errata resolutions resulted in corrections and clarifications, both in the main text and in the answers/solutions. Finally, the second edition includes several hundred edits related to gender, race, ethnicity, age, academic status, and societal issues, designed to present the most informed and inclusive material. OpenStax only undertakes revisions when pedagogically necessary, and we work to ensure that the text maintains its organization in order to ease transition for instructors. Many of the above changes have been made within problems and examples. However, in nearly all cases, the problems retain their original numbering, so that instructors and online homework providers can move to the new edition with minimal impact on their assignment structures. (This will not be the case for Try It exercises in many chapters, due to the addition of new exercises.) A detailed transition guide is available on the instructor resource page for this textbook. Answers to Questions in the Book Answers to Examples are provided directly under the question. Answers to Try it Questions are provided at the Student Resources page. Answers to an assortment of Practice, Homework, and Bringing it Together questions are available to students in the solutions section at the end of each chapter. All other answers to these questions are provided only to instructors in the Instructor Answer Guide via the Instructor Resources page. Additional resources Student and instructor resources We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructor solution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which you can apply for when you log in or create your account on OpenStax.org. Take advantage of these resources to supplement your OpenStax book. Community Hubs OpenStax partners with the Institute for the Study of Knowledge Management in Education (ISKME) to offer Community Hubs on OER Commons – a platform for instructors to share community-created resources that support OpenStax books, free of charge. Through our Community Hubs, instructors can upload their own materials or download resources to use in their own courses, including additional ancillaries, teaching material, multimedia, and relevant course content. We encourage instructors to join the hubs for the subjects most relevant to your teaching and research as an opportunity both to enrich your courses and to engage with other faculty. To reach the Community Hubs, visit www.oercommons.org/hubs/OpenStax . Technology partners As allies in making high-quality learning materials accessible, our technology partners offer optional low-cost tools that are integrated with OpenStax books. To access the technology options for your text, visit your book page on OpenStax.org. About the authors Senior contributing authors Alexander Holmes, The University of Oklahoma Barbara Illowsky, DeAnza College Susan Dean, DeAnza College Contributing authors Kevin Hadley, Analyst, Federal Reserve Bank of Kansas City Reviewers Birgit Aquilonius, West Valley College Charles Ashbacher, Upper Iowa University - Cedar Rapids Abraham Biggs, Broward Community College Daniel Birmajer, Nazareth College Roberta Bloom, De Anza College Bryan Blount, Kentucky Wesleyan College Ernest Bonat, Portland Community College Sarah Boslaugh, Kennesaw State University David Bosworth, Hutchinson Community College Sheri Boyd, Rollins College George Bratton, University of Central Arkansas Franny Chan, Mt. San Antonio College Jing Chang, College of Saint Mary Laurel Chiappetta, University of Pittsburgh Lenore Desilets, De Anza College Matthew Einsohn, Prescott College Ann Flanigan, Kapiolani Community College Nancy Foreman, West Los Angeles College David French, Tidewater Community College Mo Geraghty, De Anza College Larry Green, Lake Tahoe Community College Michael Greenwich, College of Southern Nevada Inna Grushko, De Anza College Valier Hauber, De Anza College Janice Hector, De Anza College Jim Helmreich, Marist College Robert Henderson, Stephen F. Austin State University Mel Jacobsen, Snow College Mary Jo Kane, De Anza College John Kagochi, University of Houston - Victoria Lynette Kenyon, Collin County Community College Charles Klein, De Anza College Alexander Kolovos Sheldon Lee, Viterbo University Sara Lenhart, Christopher Newport University Wendy Lightheart, Lane Community College Vladimir Logvenenko, De Anza College Jim Lucas, De Anza College Suman Majumdar, University of Connecticut Lisa Markus, De Anza College Miriam Masullo, SUNY Purchase Diane Mathios, De Anza College Robert McDevitt, Germanna Community College John Migliaccio, Fordham University Mark Mills, Central College Cindy Moss, Skyline College Nydia Nelson, St. Petersburg College Benjamin Ngwudike, Jackson State University Jonathan Oaks, Macomb Community College Carol Olmstead, De Anza College Barbara A. Osyk, The University of Akron Adam Pennell, Greensboro College Kathy Plum, De Anza College Lisa Rosenberg, Elon University Sudipta Roy, Kankakee Community College Javier Rueda, De Anza College Yvonne Sandoval, Pima Community College Rupinder Sekhon, De Anza College Travis Short, St. Petersburg College Frank Snow, De Anza College Abdulhamid Sukar, Cameron University Jeffery Taub, Maine Maritime Academy Mary Teegarden, San Diego Mesa College John Thomas, College of Lake County Philip J. Verrecchia, York College of Pennsylvania Dennis Walsh, Middle Tennessee State University Cheryl Wartman, University of Prince Edward Island Carol Weideman, St. Petersburg College Kyle S. Wells, Dixie State University Andrew Wiesner, Pennsylvania State University", "section": "Preface", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction We encounter statistics in our daily lives more often than we probably realize and from many different sources, like the news. (credit: modification of work “New office” by Phil Whitehouse/ Flickr, CC BY 2.0) You are probably asking yourself the question, \"When and where will I use statistics?\" If you read any newspaper, watch television, or use the Internet, you will see statistical information. There are statistics about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or watch a television news program, you are given sample information. With this information, you may make a decision about the correctness of a statement, claim, or \"fact.\" Statistical methods can help you make the \"best educated guess.\" Since you will undoubtedly be given statistical information at some point in your life, you need to know some techniques for analyzing the information thoughtfully. Think about buying a house or managing a budget. Think about your chosen profession. The fields of economics, business, psychology, education, biology, law, computer science, police science, and early childhood development require at least one course in statistics. Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what \"good\" data can be distinguished from \"bad.\"", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Definitions of Statistics, Probability, and Key Terms The science of statistics deals with the collection, analysis, interpretation, and presentation of data . We see and use data in our everyday lives. In this course, you will learn how to organize and summarize data. Organizing and summarizing data is called descriptive statistics . Two ways to summarize data are by graphing and by using numbers (for example, finding an average). After you have studied probability and probability distributions, you will use formal methods for drawing conclusions from \"good\" data. The formal methods are called inferential statistics . Statistical inference uses probability to determine how confident we can be that our conclusions are correct. Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful examination of the data. You will encounter what will seem to be too many mathematical formulas for interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to gain an understanding of your data. The calculations can be done using a calculator or a computer. The understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more confident in the decisions you make in life. Probability Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of an event occurring. For example, if you toss a fair coin four times, the outcomes may not be two heads and two tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails. The expected theoretical probability of heads in any one toss is 1 2 or 0.5. Even though the outcomes of a few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012 heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction 996 2000 is equal to 0.498 which is very close to 0.5, the expected probability. The theory of probability began with the study of games of chance such as poker. Predictions take the form of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of return on a client's investments. You might use probability to decide to buy a lottery ticket or not. In your study of statistics, you will use the power of mathematics through probability calculations to analyze and interpret your data. Key Terms In statistics, we generally want to study a population . You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample . The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population. Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16-ounce can contains 16 ounces of carbonated drink. From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter, in this case the mean. A parameter is a numerical characteristic of the whole population that can be estimated by a statistic. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter. One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample . We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter. A variable , or random variable, usually notated by capital letters such as X and Y , is a characteristic or measurement that can be determined for each member of a population. Variables may be numerical or categorical . Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category. If we let X equal the number of points earned by one math student at the end of a term, then X is a numerical variable. If we let Y be a person's party affiliation, then some examples of Y include Republican, Democrat, and Independent. Y is a categorical variable. We could do some math with values of X (calculate the average number of points earned, for example), but it makes no sense to do math with values of Y (calculating an average party affiliation makes no sense). Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value. Two words that come up often in statistics are mean and proportion . If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is 22 40 and the proportion of women students is 18 40 . Mean and proportion are discussed in more detail in later chapters. NOTE The words \" mean \" and \" average \" are often used interchangeably. The substitution of one word for the other is common practice. The technical term is \"arithmetic mean,\" and \"average\" is technically a center location. However, in practice among non-statisticians, \"average\" is commonly accepted for \"arithmetic mean.\" Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly surveyed 100 first year students at the college. Three of those students spent $150, $200, and $225, respectively. The population is all first year students attending ABC College this term. The sample could be all students enrolled in one section of a beginning statistics course at ABC College (although this sample may not represent the entire population). The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term: the population mean. The statistic is the average (mean) amount of money spent (excluding books) by first year college students in the sample. The variable could be the amount of money spent (excluding books) by one first year student. Let X = the amount of money spent (excluding books) by one first year student attending ABC College. The data are the dollar amounts spent by the first year students. Examples of the data are $150, $200, and $225. Try It Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65, $75, and $95, respectively. Try It Solutions The population is all families with children attending Knoll Academy. The sample is a random selection of 100 families with children attending Knoll Academy. The parameter is the average (mean) amount of money spent on school uniforms by families with children at Knoll Academy. The statistic is the average (mean) amount of money spent on school uniforms by families in the sample. The variable is the amount of money spent by one family. Let X = the amount of money spent on school uniforms by one family with children attending Knoll Academy. The data are the dollar amounts spent by the families. Examples of the data are $65, $75, and $95. Determine what the key terms refer to in the following study. A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below. 1. Population ____ 2. Statistic ____ 3. Parameter ____ 4. Sample ____ 5. Variable ____ 6. Data ____ all students who attended the college last year the cumulative GPA of one student who graduated from the college last year 3.65, 2.80, 1.50, 3.90 a group of students who graduated from the college last year, randomly selected the average cumulative GPA of students who graduated from the college last year all students who graduated from the college last year the average cumulative GPA of students in the study who graduated from the college last year 1. f 2. g 3. e 4. d 5. b 6. c Try It Determine what key terms refer to in the following study. A survey of athletes in a university was conducted to study the heights of athletes, in meters. Fill in the letter of the phrase that best describes each of the items below. Population ____ Statistics ____ Parameter ____ Sample ____ Variable ____ Data _____ the average height of athletes in the university the average height of athletes in the survey all athletes in the university all students in the university the height of one athlete a group of athletes randomly selected 1.82, 1.76, 1.69, 1.93 c b a f e g Determine what the key terms refer to in the following study. As part of a study designed to test the safety of electric automobiles, the National Transportation Safety Board collected and reviewed data about the effects of an automobile crash on test dummies. Here is the criterion they used: Speed at which cars crashed Location of “drivers” (i.e. dummies) 35 miles/hour Front Seat Cars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars. The population is all cars containing dummies in the front seat. The sample is the 75 cars, selected by a simple random sample. The parameter is the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population. The statistic is proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample. The variable X = whether a dummy (if it had been a real person) would have suffered head injuries. The data are either: yes, had head injury, or no, did not. Try It Determine what the key terms refer to in the following study. A survey is conducted to check the time taken by a mobile for charging of battery from 50% to 100%. The criteria used to collect the data are: Wattage of charger used Type of mobile used 30 W Android We want to know the proportion of Android mobiles that are charged to 100% within 30 minutes. We start with a simple random sample of 200 mobiles. The population is all Android mobiles. The sample is the 200 mobiles, selected by a simple random sample. The parameter is the proportion of Android mobiles that are charged to 100% within 30 minutes in the population. The statistic is the proportion of Android mobiles that are charged to 100% within 30 minutes in the survey. The variable X = the number of Android mobiles that are charged to 100% within 30 minutes. The data are either: yes, charged to 100% within 30 minutes; or no, charged to 100% taking more than 30 minutes. Determine what the key terms refer to in the following study. An insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit. The population is all medical doctors listed in the professional directory. The parameter is the proportion of medical doctors who have been involved in one or more malpractice suits in the population. The sample is the 500 doctors selected at random from the professional directory. The statistic is the proportion of medical doctors who have been involved in one or more malpractice suits in the sample. The variable X = whether an individual doctor has been involved in a malpractice suit. The data are either: yes, was involved in one or more malpractice lawsuits, or no, was not. Try It Determine what the key terms refer to in the following study. A study is to be conducted by a news agency to find the proportion of all truck drivers that have no points on their license. The agency selects 1000 truck drivers randomly from the directory of truck drivers and determines the number of truck drivers in the sample who have no points on their license. The population is all truck drivers present in the directory. The parameter is the proportion of truck drivers that have no points on their license in the population. The sample is the 1000 truck drivers selected randomly from the directory. The statistics is the proportion of truck drivers that have no points on their license in the sample. The variable X = the number of truck drivers that have no points on their license. The data are either: yes, they have no points on their license; or no, they have points on their license. Do the following exercise collaboratively with up to four people per group. Find a population, a sample, the parameter, the statistic, a variable, and data for the following study: You want to determine the average (mean) number of glasses of milk college students drink per day. Suppose yesterday, in your English class, you asked five students how many glasses of milk they drank the day before. The answers were 1, 0, 1, 3, and 4 glasses of milk. References The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/CrashTestDummies.html (accessed May 1, 2013). Chapter Review The mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text. HOMEWORK For each of the following eight exercises, identify: a. the population, b. the sample, c. the parameter, d. the statistic, e. the variable, and f. the data. Give examples where appropriate. A fitness center is interested in the mean amount of time a client exercises in the center each week. Ski resorts are interested in the mean age that children take their first ski and snowboard lessons. They need this information to plan their ski classes optimally. all children who take ski or snowboard lessons a group of these children the population mean age of children who take their first snowboard lesson the sample mean age of children who take their first snowboard lesson X = the age of one child who takes their first ski or snowboard lesson values for X , such as 3, 7, and so on A cardiologist is interested in the mean recovery period of their patients who have had heart attacks. Insurance companies are interested in the mean health costs each year of their clients, so that they can determine the costs of health insurance. the clients of the insurance companies a group of the clients the mean health costs of the clients the mean health costs of the sample X = the health costs of one client values for X , such as 34, 9, 82, and so on A politician is interested in the proportion of voters in their district who think the politician is doing a good job. A marriage counselor is interested in the proportion of clients she counsels who stay married. all the clients of this counselor a group of clients of this marriage counselor the proportion of all their clients who stay married the proportion of the sample of the counselor’s clients who stay married X = whether a client stayed married yes, no Political pollsters may be interested in the proportion of people who will vote for a particular cause. A marketing company is interested in the proportion of people who will buy a particular product. all people (maybe in a certain geographic area, such as the United States) a group of the people the proportion of all people who will buy the product the proportion of the sample who will buy the product X = whether a person bought the product buy, not buy Use the following information to answer the next three exercises: A Lake Tahoe Community College instructor is interested in the mean number of days Lake Tahoe Community College math students are absent from class during a quarter. What is the population she is interested in? all Lake Tahoe Community College students all Lake Tahoe Community College English students all Lake Tahoe Community College students in the instructor's classes all Lake Tahoe Community College math students Consider the following: X = number of days a Lake Tahoe Community College math student is absent In this case, X is an example of a: variable. population. statistic. data. a The instructor’s sample produces a mean number of days absent of 3.5 days. This value is an example of a: parameter. data. statistic. variable. Average also called mean or arithmetic mean; a number that describes the central tendency of the data Categorical Variable variables that take on values that are names or labels Data a set of observations (a set of possible outcomes); most data can be put into two groups: qualitative (an attribute whose value is indicated by a label) or quantitative (an attribute whose value is indicated by a number). Quantitative data can be separated into two subgroups: discrete and continuous . Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage) Mathematical Models a description of a phenomenon using mathematical concepts, such as equations, inequalities, distributions, etc. Numerical Variable variables that take on values that are indicated by numbers Parameter a number that is used to represent a population characteristic and that generally cannot be determined easily Population all individuals, objects, or measurements whose properties are being studied Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur Proportion the number of successes divided by the total number in the sample Representative Sample a subset of the population that has the same characteristics as the population Sample a subset of the population studied Statistic a numerical characteristic of the sample; a statistic estimates the corresponding population parameter. Survey a study in which data is collected as reported by individuals. Variable a characteristic of interest for each person or object in a population", "section": "Definitions of Statistics, Probability, and Key Terms", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Data, Sampling, and Variation in Data and Sampling Data may come from a population or from a sample. Lowercase letters like x or y generally are used to represent data values. Most data can be put into the following categories: Qualitative Quantitative Qualitative data are the result of categorizing or describing attributes of a population. Qualitative data are also often called categorical data. Hair color, blood type, ethnic group, the car a person drives, and the street a person lives on are examples of qualitative (categorical) data. Qualitative (categorical) data are generally described by words or letters. For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Researchers often prefer to use quantitative data over qualitative (categorical) data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or blood type. Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes of a population. Amount of money, pulse rate, weight, number of people living in your town, and number of students who take statistics are examples of quantitative data. Quantitative data may be either discrete or continuous . All data that are the result of counting are called quantitative discrete data . These data take on only certain numerical values. If you count the number of phone calls you receive for each day of the week, you might get values such as zero, one, two, or three. Data that are not only made up of counting numbers, but that may include fractions, decimals, or irrational numbers, are called quantitative continuous data . Continuous data are often the results of measurements like lengths, weights, or times. A list of the lengths in minutes for all the phone calls that you make in a week, with numbers like 2.4, 7.5, or 11.0, would be quantitative continuous data. Data Sample of Quantitative Discrete Data The data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data. Try It The data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this? Try It Solutions quantitative discrete data Data Sample of Quantitative Continuous Data The data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data. Try It The data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this? Try It Solutions quantitative continuous data You go to the supermarket and purchase three cans of soup (19 ounces tomato bisque, 14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts and peanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), and two desserts (16 ounces pistachio ice cream and 32 ounces chocolate chip cookies). Name data sets that are quantitative discrete, quantitative continuous, and qualitative(categorical). One Possible Solution: The three cans of soup, two packages of nuts, four kinds of vegetables and two desserts are quantitative discrete data because you count them. The weights of the soups (19 ounces, 14.1 ounces, 19 ounces) are quantitative continuous data because you measure weights as precisely as possible. Types of soups, nuts, vegetables and desserts are qualitative(categorical) data because they are categorical. Try to identify additional data sets in this example. Try It The following list of materials was purchased by a purchasing manager in a company: Two types of nails (2 kg box nails, 3 kg roofing nails) One type of oil (4 L machine oil) Four types of screws (3 kg wood screws, 5 kg machine screws, 1 kg set screws, 2 kg socket screws) Name data sets that are quantitative discrete, quantitative continuous, and qualitative. One possible solution: The two types of nails, one type of oil, and four types of screws are quantitative discrete data because you count them. The weights of materials are quantitative continuous data because you can measure weight as precisely as possible. Types of nails, oil, and screws are qualitative data because they are categorical. Try to identify additional data sets in the example. The data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative(categorical) data. Try It The data are the colors of houses. You sample five houses. The colors of the houses are white, yellow, white, red, and white. What type of data is this? Try It Solutions qualitative(categorical) data NOTE You may collect data as numbers and report it categorically. For example, the quiz scores for each student are recorded throughout the term. At the end of the term, the quiz scores are reported as A, B, C, D, or F. Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words \"the number of.\" the number of pairs of shoes you own the type of car you drive the distance from your home to the nearest grocery store the number of classes you take per school year the type of calculator you use weights of dogs at an animal shelter number of correct answers on a quiz IQ scores (This may cause some discussion.) Items a, d, and g are quantitative discrete; items c, f, and h are quantitative continuous; items b and e are qualitative, or categorical. Try It Determine the correct data type (quantitative or qualitative) for the number of cars in a parking lot. Indicate whether quantitative data are continuous or discrete. Try It Solutions quantitative discrete A statistics professor collects information about the classification of her students as first-year students, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart . What type of data does this graph show? This pie chart shows the students in each year, which is qualitative (or categorical) data . Try It The registrar at State University keeps records of the number of credit hours students complete each semester. The data collected are summarized in the histogram. The class boundaries are 10 to less than 13, 13 to less than 16, 16 to less than 19, 19 to less than 22, and 22 to less than 25. What type of data does this graph show? Try It Solutions A histogram is used to display quantitative data: the numbers of credit hours completed. Because students can complete only a whole number of hours (no fractions of hours allowed), this data is quantitative discrete. Qualitative Data Discussion Below are tables comparing the number of part-time and full-time students at De Anza College and Foothill College enrolled for the most recent spring quarter. The tables display counts (frequencies) and percentages or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is compared to De Anza College. Most Recent Spring Quarter De Anza College Foothill College Number Percent Number Percent Full-time 9,200 40.9% Full-time 4,059 28.6% Part-time 13,296 59.1% Part-time 10,124 71.4% Total 22,496 100% Total 14,183 100% Tables are a good way of organizing and displaying data. But graphs can be even more helpful in understanding the data. There are no strict rules concerning which graphs to use. Two graphs that are used to display qualitative(categorical) data are pie charts and bar graphs. In a pie chart , categories of data are represented by wedges in a circle and are proportional in size to the percent of individuals in each category. In a bar graph , the length of the bar for each category is proportional to the number or percent of individuals in each category. Bars may be vertical or horizontal. A Pareto chart consists of bars that are sorted into order by category size (largest to smallest). Look at and and determine which graph (pie or bar) you think displays the comparisons better. It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the data. We might make different choices of what we think is the “best” graph depending on the data and the context. Our choice also depends on what we are using the data for. Percentages That Add to More (or Less) Than 100% Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages add to more than 100% because students can be in more than one category. A bar graph is appropriate to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the percentages added to less than 100%. De Anza College Most Recent Spring Quarter Characteristic/category Percent Full-time students 40.9% Students who intend to transfer to a 4-year educational institution 48.6% Students under age 25 61.0% TOTAL 150.5% Omitting Categories/Missing Data The table displays Ethnicity of Students but is missing the \"Other/Unknown\" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart. Ethnicity of Students at De Anza College in the Most Recent Fall Term Frequency Percent Asian 8,794 36.1% Black 1,412 5.8% Filipino 1,298 5.3% Hispanic/Latino 4,180 17.1% Native American 146 0.6% Pacific Islander 236 1.0% White 5,978 24.5% TOTAL 22,044 out of 24,382 90.4% out of 100% The following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us. This particular bar graph in can be difficult to understand visually. The graph in is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret. Bar Graph with Other/Unknown Category Pareto Chart With Bars Sorted by Size Pie Charts: No Missing Data The following pie charts have the “Other/Unknown” category included (since the percentages must add to 100%). The chart in (b) is organized by the size of each wedge, which makes it a more visually informative graph than the unsorted, alphabetical graph in (a). Sampling Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling . In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons. The easiest method to describe is called a simple random sample . Any group of n individuals is equally likely to be chosen as any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster sample, and the systematic sample. To choose a stratified sample , divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample. To choose a cluster sample , divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample. To choose a systematic sample , randomly select a starting point and take every n th piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method. A type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others. Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents. True random sampling is done with replacement . That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement . Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low. Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors . A defective counting device can cause a nonsampling error. In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error. In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied. Critical Evaluation We need to evaluate the statistical studies we read about critically and analyze them before accepting the results of the studies. Common problems to be aware of include Problems with samples: A sample must be representative of the population. A sample that is not representative of the population is biased. Biased samples that are not representative of the population give results that are inaccurate and not valid. Self-selected samples: Responses only by people who choose to respond, such as call-in surveys, are often unreliable. Sample size issues: Samples that are too small may be unreliable. Larger samples are better, if possible. In some situations, having small samples is unavoidable and can still be used to draw conclusions. Examples: crash testing cars or medical testing for rare conditions Undue influence: collecting data or asking questions in a way that influences the response Non-response or refusal of subject to participate: The collected responses may no longer be representative of the population. Often, people with strong positive or negative opinions may answer surveys, which can affect the results. Causality: A relationship between two variables does not mean that one causes the other to occur. They may be related (correlated) because of their relationship through a different variable. Self-funded or self-interest studies: A study performed by a person or organization in order to support their claim. Is the study impartial? Read the study carefully to evaluate the work. Do not automatically assume that the study is good, but do not automatically assume the study is bad either. Evaluate it on its merits and the work done. Misleading use of data: improperly displayed graphs, incomplete data, or lack of context Confounding: When the effects of multiple factors on a response cannot be separated. Confounding makes it difficult or impossible to draw valid conclusions about the effect of each factor. A study is done to determine the average tuition that San Jose State undergraduate students pay per semester. Each student in the following samples is asked how much tuition they paid for the Fall semester. What is the type of sampling in each case? A sample of 100 undergraduate San Jose State students is taken by organizing the students’ names by classification (first-year student, sophomore, junior, or senior), and then selecting 25 students from each. A random number generator is used to select a student from the alphabetical listing of all undergraduate students in the Fall semester. Starting with that student, every 50th student is chosen until 75 students are included in the sample. A completely random method is used to select 75 students. Each undergraduate student in the fall semester has the same probability of being chosen at any stage of the sampling process. The first-year, sophomore, junior, and senior years are numbered one, two, three, and four, respectively. A random number generator is used to pick two of those years. All students in those two years are in the sample. An administrative assistant is asked to stand in front of the library one Wednesday and to ask the first 100 undergraduate students he encounters what they paid for tuition the Fall semester. Those 100 students are the sample. a. stratified; b. systematic; c. simple random; d. cluster; e. convenience Try It You are going to use the random number generator to generate different types of samples from the data. This table displays six sets of quiz scores (each quiz counts 10 points) for an elementary statistics class. #1 #2 #3 #4 #5 #6 5 7 10 9 8 3 10 5 9 8 7 6 9 10 8 6 7 9 9 10 10 9 8 9 7 8 9 5 7 4 9 9 9 10 8 7 7 7 10 9 8 8 8 8 9 10 8 8 9 7 8 7 7 8 8 8 10 9 8 7 Instructions: Use the Random Number Generator to pick samples. Create a stratified sample by column. Pick three quiz scores randomly from each column. Number each row one through ten. On your calculator, press Math and arrow over to PRB. For column 1, Press 5:randInt( and enter 1,10). Press ENTER. Record the number. Press ENTER 2 more times (even the repeats). Record these numbers. Record the three quiz scores in column one that correspond to these three numbers. Repeat for columns two through six. These 18 quiz scores are a stratified sample. Create a cluster sample by picking two of the columns. Use the column numbers: one through six. Press MATH and arrow over to PRB. Press 5:randInt( and enter 1,6). Press ENTER. Record the number. Press ENTER and record that number. The two numbers are for two of the columns. The quiz scores (20 of them) in these 2 columns are the cluster sample. Create a simple random sample of 15 quiz scores. Use the numbering one through 60. Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60). Press ENTER 15 times and record the numbers. Record the quiz scores that correspond to these numbers. These 15 quiz scores are the random sample. Create a systematic sample of 12 quiz scores. Use the numbering one through 60. Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60). Press ENTER. Record the number and the first quiz score. From that number, count ten quiz scores and record that quiz score. Keep counting ten quiz scores and recording the quiz score until you have a sample of 12 quiz scores. You may wrap around (go back to the beginning). Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A soccer coach selects six players from a group of boys aged eight to ten, seven players from a group of boys aged 11 to 12, and three players from a group of boys aged 13 to 14 to form a recreational soccer team. A pollster interviews all human resource personnel in five different high tech companies. A high school educational researcher interviews 50 public high school teachers and 50 private high school teachers. A medical researcher interviews every third cancer patient from a list of cancer patients at a local hospital. A high school counselor uses a computer to generate 50 random numbers and then picks students whose names correspond to the numbers. A student interviews classmates in their algebra class to determine how many pairs of jeans a student owns, on the average. a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f.convenience Try It Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A high school principal polls 50 first-year students, 50 sophomores, 50 juniors, and 50 seniors regarding policy changes for after school activities. If we were to examine two samples representing the same population, even if we used random sampling methods for the samples, they would not be exactly the same. Just as there is variation in data, there is variation in samples. As you become accustomed to sampling, the variability will begin to seem natural. NOTE In Confidence Intervals of the text, sample size formulas are provided which will determine sample sizes when sampling from a population. The sample size will be a function of the desired precision and not a function of the population size. It may be somewhat counterintuitive that the sample size does not depend on the population size. However, this implies that a sample size of 1,000 can be adequate to represent a population of 100,000 versus 1,000,000 given that the same level of precision is desired. When working in Confidence Intervals with sample size formulas, the student will notice that population size is not a factor in determining the sample size. Suppose ABC College has 10,000 part-time students (the population). We are interested in the average amount of money a part-time student spends on books in the fall term. Asking all 10,000 students is an almost impossible task. Suppose we take two different samples. First, we use convenience sampling and survey ten students from a first term organic chemistry class. Many of these students are taking first term calculus in addition to the organic chemistry class. The amount of money they spend on books is as follows: $128 $87 $173 $116 $130 $204 $147 $189 $93 $153 The second sample is taken using a list of senior citizens who take P.E. classes and taking every fifth senior citizen on the list, for a total of ten senior citizens. They spend: $50 $40 $36 $15 $50 $100 $40 $53 $22 $22 It is unlikely that any student is in both samples. a. Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population? a. No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are also taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average part-time student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample. b. Since these samples are not representative of the entire population, is it wise to use the results to describe the entire population? b. No. For these samples, each member of the population did not have an equally likely chance of being chosen. Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education, art, and early childhood development. (We assume that these are the only disciplines in which part-time students at ABC College are enrolled and that an equal number of part-time students are enrolled in each of the disciplines.) Each student is chosen using simple random sampling. Using a calculator, random numbers are generated and a student from a particular discipline is selected if they have a corresponding number. The students spend the following amounts: $180 $50 $150 $85 $260 $75 $180 $200 $200 $150 c. Is the sample biased? c. The sample is unbiased, but a larger sample would be recommended to increase the likelihood that the sample will be close to representative of the population. However, for a biased sampling technique, even a large sample runs the risk of not being representative of the population. Students often ask if it is \"good enough\" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes. Try It A local radio station has a fan base of 20,000 listeners. The station wants to know if its audience would prefer more music or more talk shows. Asking all 20,000 listeners is an almost impossible task. The station uses convenience sampling and surveys the first 200 people they meet at one of the station’s music concert events. 24 people said they’d prefer more talk shows, and 176 people said they’d prefer more music. Do you think that this sample is representative of (or is characteristic of) the entire 20,000 listener population? Try It Solutions The sample probably consists more of people who prefer music because it is a concert event. Also, the sample represents only those who showed up to the event earlier than the majority. The sample probably doesn’t represent the entire fan base and is probably biased towards people who would prefer music. Variation in Data Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less than 16 ounces of liquid. In one study, eight 16-ounce cans were measured and produced the following amount (in ounces) of beverage: 15.8 16.1 15.2 14.8 15.8 15.9 16.0 15.5 Measurements of the amount of beverage in a 16-ounce can may vary because different people make the measurements or because the exact amount, 16 ounces of liquid, was not put into the cans. Manufacturers regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range. Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the same purpose. This is completely natural. However, if two or more of you are taking the same data and get very different results, it is time for you and the others to reevaluate your data-taking methods and your accuracy. Variation in Samples It was mentioned previously that two or more samples from the same population , taken randomly, and having close to the same characteristics of the population will likely be different from each other. Suppose Doreen and Jung both decide to study the average amount of time students at their college sleep each night. Doreen and Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample will be different from Jung's sample. Even if Doreen and Jung used the same sampling method, in all likelihood their samples would be different. Neither would be wrong, however. Think about what contributes to making Doreen’s and Jung’s samples different. If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results (the average amount of time a student sleeps) might be closer to the actual population average. But still, their samples would be, in all likelihood, different from each other. This variability in samples cannot be stressed enough. Size of a Sample The size of a sample (often called the number of observations, usually given the symbol n) is important. The examples you have seen in this book so far have been small. Samples of only a few hundred observations, or even smaller, are sufficient for many purposes. In polling, samples that are from 1,200 to 1,500 observations are considered large enough and good enough if the survey is random and is well done. Later we will find that even much smaller sample sizes will give very good results. You will learn why when you study confidence intervals. Be aware that many large samples are biased. For example, call-in surveys are invariably biased, because people choose to respond or not. References Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/default.asp (accessed May 1, 2013). Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/methodology.asp (accessed May 1, 2013). Gallup-Healthways Well-Being Index. http://www.gallup.com/poll/146822/gallup-healthways-index-questions.aspx (accessed May 1, 2013). Data from http://www.bookofodds.com/Relationships-Society/Articles/A0374-How-George-Gallup-Picked-the-President Dominic Lusinchi, “’President’ Landon and the 1936 Literary Digest Poll: Were Automobile and Telephone Owners to Blame?” Social Science History 36, no. 1: 23-54 (2012), http://ssh.dukejournals.org/content/36/1/23.abstract (accessed May 1, 2013). “The Literary Digest Poll,” Virtual Laboratories in Probability and Statistics http://www.math.uah.edu/stat/data/LiteraryDigest.html (accessed May 1, 2013). “Gallup Presidential Election Trial-Heat Trends, 1936–2008,” Gallup Politics http://www.gallup.com/poll/110548/gallup-presidential-election-trialheat-trends-19362004.aspx#4 (accessed May 1, 2013). The Data and Story Library, http://lib.stat.cmu.edu/DASL/Datafiles/USCrime.html (accessed May 1, 2013). LBCC Distance Learning (DL) program data in 2010-2011, http://de.lbcc.edu/reports/2010-11/future/highlights.html#focus (accessed May 1, 2013). Data from San Jose Mercury News Chapter Review Data are individual items of information that come from a population or sample. Data may be classified as qualitative (categorical), quantitative continuous, or quantitative discrete. Because it is not practical to measure the entire population in a study, researchers use samples to represent the population. A random sample is a representative group from the population chosen by using a method that gives each individual in the population an equal chance of being included in the sample. Random sampling methods include simple random sampling, stratified sampling, cluster sampling, and systematic sampling. Convenience sampling is a nonrandom method of choosing a sample that often produces biased data. Samples that contain different individuals result in different data. This is true even when the samples are well-chosen and representative of the population. When properly selected, larger samples model the population more closely than smaller samples. There are many different potential problems that can affect the reliability of a sample. Statistical data needs to be critically analyzed, not simply accepted. HOMEWORK For the following exercises, identify the type of data that would be used to describe a response (quantitative discrete, quantitative continuous, or qualitative), and give an example of the data. number of tickets sold to a concert quantitative discrete, 150 percent of body fat favorite baseball team qualitative, Oakland A’s time in line to buy groceries number of students enrolled at Evergreen Valley College quantitative discrete, 11,234 students most-watched television show brand of toothpaste qualitative, Crest distance to the closest movie theatre age of executives in Fortune 500 companies quantitative continuous, 47.3 years number of competing computer spreadsheet software packages Use the following information to answer the next two exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of resident use of a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every 8th house in the neighborhood around the park was interviewed. “Number of times per week” is what type of data? qualitative (categorical) quantitative discrete quantitative continuous b “Duration (amount of time)” is what type of data? qualitative (categorical) quantitative discrete quantitative continuous Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. Suppose an airline conducts a survey. Over Thanksgiving weekend, it surveys six flights from Boston to Salt Lake City to determine the number of babies on the flights. It determines the amount of safety equipment needed by the result of that study. Using complete sentences, list three things wrong with the way the survey was conducted. Using complete sentences, list three ways that you would improve the survey if it were to be repeated. The survey was conducted using six similar flights. The survey would not be a true representation of the entire population of air travelers. Conducting the survey on a holiday weekend will not produce representative results. Conduct the survey during different times of the year. Conduct the survey using flights to and from various locations. Conduct the survey on different days of the week. Suppose you want to determine the mean number of students per statistics class in your state. Describe a possible sampling method in three to five complete sentences. Make the description detailed. Suppose you want to determine the mean number of cans of soda drunk each month by students in their twenties at your school. Describe a possible sampling method in three to five complete sentences. Make the description detailed. Answers will vary. Sample Answer: You could use a systematic sampling method. Stop the tenth person as they leave one of the buildings on campus at 9:50 in the morning. Then stop the tenth person as they leave a different building on campus at 1:50 in the afternoon. List some practical difficulties involved in getting accurate results from a telephone survey. List some practical difficulties involved in getting accurate results from a mailed survey. Answers will vary. Sample Answer: Many people will not respond to mail surveys. If they do respond to the surveys, you can’t be sure who is responding. In addition, mailing lists can be incomplete. With your classmates, brainstorm some ways you could overcome these problems if you needed to conduct a phone or mail survey. The instructor takes a sample by gathering data on five randomly selected students from each Lake Tahoe Community College math class. The type of sampling used is cluster sampling stratified sampling simple random sampling convenience sampling b A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every eighth house in the neighborhood around the park was interviewed. The sampling method was: simple random systematic stratified cluster Name the sampling method used in each of the following situations: A person in the airport is handing out questionnaires to travelers asking them to evaluate the airport’s service. The person does not ask travelers who are hurrying through the airport with their hands full of luggage, but instead asks all travelers who are sitting near gates and not taking naps while they wait. A teacher wants to know if students are doing homework so they randomly select rows two and five and then call on all students in row two and all students in row five to present the solutions to homework problems to the class. The marketing manager for an electronics chain store wants information about the ages of its customers. Over the next two weeks, at each store location, 100 randomly selected customers are given questionnaires to fill out asking for information about age, as well as about other variables of interest. The librarian at a public library wants to determine what proportion of the library users are children. The librarian has a tally sheet on which they mark whether books are checked out by an adult or a child. The librarian records this data for every fourth patron who checks out books. A political party wants to know the reaction of voters to a debate between the candidates. The day after the debate, the party’s polling staff calls 1,200 randomly selected phone numbers. If a registered voter answers the phone or is available to come to the phone, that registered voter is asked whom they intend to vote for and whether the debate changed their opinion of the candidates. convenience cluster stratified systematic simple random A “random survey” was conducted of 3,274 people of the “microprocessor generation” (people born since 1971, the year the microprocessor was invented). It was reported that 48% of those individuals surveyed stated that if they had $2,000 to spend, they would use it for computer equipment. Also, 66% of those surveyed considered themselves relatively savvy computer users. Do you consider the sample size large enough for a study of this type? Why or why not? Based on your “gut feeling,” do you believe the percents accurately reflect the U.S. population for those individuals born since 1971? If not, do you think the percents of the population are actually higher or lower than the sample statistics? Why? Additional information: The survey, reported by Intel Corporation, was filled out by individuals who visited the Los Angeles Convention Center to see the Smithsonian Institute's road show called “America’s Smithsonian.” With this additional information, do you feel that all demographic and ethnic groups were equally represented at the event? Why or why not? With the additional information, comment on how accurately you think the sample statistics reflect the population parameters. The Well-Being Index is a survey that follows trends of U.S. residents on a regular basis. There are six areas of health and wellness covered in the survey: Life Evaluation, Emotional Health, Physical Health, Healthy Behavior, Work Environment, and Basic Access. Some of the questions used to measure the Index are listed below. Identify the type of data obtained from each question used in this survey: qualitative(categorical), quantitative discrete, or quantitative continuous. Do you have any health problems that prevent you from doing any of the things people your age can normally do? During the past 30 days, for about how many days did poor health keep you from doing your usual activities? In the last seven days, on how many days did you exercise for 30 minutes or more? Do you have health insurance coverage? qualitative(categorical) quantitative discrete quantitative discrete qualitative(categorical) In advance of the 1936 Presidential Election, a magazine titled Literary Digest released the results of an opinion poll predicting that the republican candidate Alf Landon would win by a large margin. The magazine sent post cards to approximately 10,000,000 prospective voters. These prospective voters were selected from the subscription list of the magazine, from automobile registration lists, from phone lists, and from club membership lists. Approximately 2,300,000 people returned the postcards. Think about the state of the United States in 1936. Explain why a sample chosen from magazine subscription lists, automobile registration lists, phone books, and club membership lists was not representative of the population of the United States at that time. What effect does the low response rate have on the reliability of the sample? Are these problems examples of sampling error or nonsampling error? During the same year, George Gallup conducted his own poll of 30,000 prospective voters. These researchers used a method they called \"quota sampling\" to obtain survey answers from specific subsets of the population. Quota sampling is an example of which sampling method described in this module? Crime-related and demographic statistics for 47 US states in 1960 were collected from government agencies, including the FBI's Uniform Crime Report . One analysis of this data found a strong connection between education and crime indicating that higher levels of education in a community correspond to higher crime rates. Which of the potential problems with samples discussed in Data, Sampling, and Variation in Data and Sampling could explain this connection? Causality: The fact that two variables are related does not guarantee that one variable is influencing the other. We cannot assume that crime rate impacts education level or that education level impacts crime rate. Confounding: There are many factors that define a community other than education level and crime rate. Communities with high crime rates and high education levels may have other lurking variables that distinguish them from communities with lower crime rates and lower education levels. Because we cannot isolate these variables of interest, we cannot draw valid conclusions about the connection between education and crime. Possible lurking variables include police expenditures, unemployment levels, region, average age, and size. Imagine you work for a polling company and a member of your team has proposed the following survey question: “Do you feel happy paying your taxes while some politicians are allowed to use loopholes and avoid paying their fair share of taxes?” As part of preliminary data collection, 11 people responded to this question. Each participant answered “NO!” Which of the potential problems with samples discussed in this module could explain this connection? A scholarly article about response rates begins with the following quote: “Declining contact and cooperation rates in random digit dial (RDD) national telephone surveys raise serious concerns about the validity of estimates drawn from such research.” (Scott Keeter et al., “Gauging the Impact of Growing Nonresponse on Estimates from a National RDD Telephone Survey,” Public Opinion Quarterly 70 no. 5 (2006), http://poq.oxfordjournals.org/content/70/5/759.full (accessed May 1, 2013).) The Pew Research Center for People and the Press admits: “The percentage of people we interview – out of all we try to interview – has been declining over the past decade or more.” (Frequently Asked Questions, Pew Research Center for the People & the Press, http://www.people-press.org/methodology/frequently-asked-questions/#dont-you-have-trouble-getting-people-to-answer-your-polls (accessed May 1, 2013).) What are some reasons for the decline in response rate over the past decade? Explain why researchers are concerned with the impact of the declining response rate on public opinion polls. Possible reasons: increased use of caller id, decreased use of landlines, increased use of private numbers, voice mail, privacy managers, hectic nature of personal schedules, decreased willingness to be interviewed When a large number of people refuse to participate, then the sample may not have the same characteristics of the population. Perhaps the majority of people willing to participate are doing so because they feel strongly about the subject of the survey. Cluster Sampling a method for selecting a random sample and dividing the population into groups (clusters); use simple random sampling to select a set of clusters. Every individual in the chosen clusters is included in the sample. Convenience Sampling a nonrandom method of selecting a sample; this method selects individuals that are easily accessible and may result in biased data. Nonsampling Error an issue that affects the reliability of sampling data other than natural variation; it includes a variety of human errors including poor study design, biased sampling methods, inaccurate information provided by study participants, data entry errors, and poor analysis. Qualitative Data a set of observations (a set of possible outcomes); qualitative data has an attribute whose value is indicated by a label. Quantitative Data a set of observations (a set of possible outcomes); quantitative (an attribute whose value is indicated by a number) data can be separated into two subgroups: discrete and continuous . Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage).. Random Sampling a method of selecting a sample that gives every member of the population an equal chance of being selected. Sampling Bias not all members of the population are equally likely to be selected Sampling Error the natural variation that results from selecting a sample to represent a larger population; this variation decreases as the sample size increases, so selecting larger samples reduces sampling error. Sampling with Replacement Once a member of the population is selected for inclusion in a sample, that member is returned to the population for the selection of the next individual. Sampling without Replacement A member of the population may be chosen for inclusion in a sample only once. If chosen, the member is not returned to the population before the next selection. Simple Random Sampling a straightforward method for selecting a random sample; give each member of the population a number. Use a random number generator to select a set of labels. These randomly selected labels identify the members of your sample. Stratified Sampling a method for selecting a random sample used to ensure that subgroups of the population are represented adequately; divide the population into groups (strata). Use simple random sampling to identify a proportionate number of individuals from each stratum. Systematic Sampling a method for selecting a random sample; list the members of the population. Use simple random sampling to select a starting point in the population. Let k = (number of individuals in the population)/(number of individuals needed in the sample). Choose every kth individual in the list starting with the one that was randomly selected. If necessary, return to the beginning of the population list to complete your sample.", "section": "Data, Sampling, and Variation in Data and Sampling", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Levels of Measurement Once you have a set of data, you will need to organize it so that you can analyze how frequently each datum occurs in the set. However, when calculating the frequency, you may need to round your answers so that they are as precise as possible. Levels of Measurement The way a set of data is measured is called its level of measurement . Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level): Nominal scale level Ordinal scale level Interval scale level Ratio scale level Data that is measured using a nominal scale is qualitative (categorical) . Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. For example, trying to classify people according to their favorite food does not make any sense. Putting pizza first and sushi second is not meaningful. Smartphone companies are another example of nominal scale data. The data are the names of the companies that make smartphones, but there is no agreed upon order of these brands, even though people may have personal preferences. Nominal scale data cannot be used in calculations. Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data. Another example of using the ordinal scale is a cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations. Like data measured on an ordinal scale, data that are measured on an interval scale have a definite ordering. While ordinal scale data are categorical, or qualitative, interval scale data are numerical, or quantitative. It is possible to calculate differences between values measured on an interval scale. There is no minimum or \"zero\" value, however. Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not represent a minimum value. In both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0. Data that are measured using the ratio scale give the most information. Ratio scale data is like interval scale data, but there is a minimum value and ratios can be calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points). The exams are machine-graded. The data can be put in order from lowest to highest: 20, 68, 80, 92. The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The minimum score is 0. So 80 is four times 20. The score of 80 is four times better than the score of 20. Frequency Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows: 5 6 3 3 2 4 7 5 2 3 5 6 5 4 4 3 5 2 5 3 . lists the different data values in ascending order and their frequencies. Frequency Table of Student Work Hours Data value Frequency 2 3 3 5 4 3 5 6 6 2 7 1 A frequency is the number of times a value of the data occurs. According to , there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample. A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals. Frequency Table of Student Work Hours with Relative Frequencies Data value Frequency Relative frequency 2 3 3 20 or 0.15 3 5 5 20 or 0.25 4 3 3 20 or 0.15 5 6 6 20 or 0.30 6 2 2 20 or 0.10 7 1 1 20 or 0.05 The sum of the values in the relative frequency column of is 20 20 , or 1. Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in . Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies Data value Frequency Relative frequency Cumulative relative frequency 2 3 3 20 or 0.15 0.15 3 5 5 20 or 0.25 0.15 + 0.25 = 0.40 4 3 3 20 or 0.15 0.40 + 0.15 = 0.55 5 6 6 20 or 0.30 0.55 + 0.30 = 0.85 6 2 2 20 or 0.10 0.85 + 0.10 = 0.95 7 1 1 20 or 0.05 0.95 + 0.05 = 1.00 The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated. NOTE Because of rounding, the relative frequency column may not always sum to one, and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one. represents the heights, in inches, of a sample of 100 semiprofessional soccer players. Frequency Table of Soccer Player Height Heights (inches) Frequency Relative frequency Cumulative relative frequency 59.95–61.95 5 5 100 = 0.05 0.05 61.95–63.95 3 3 100 = 0.03 0.05 + 0.03 = 0.08 63.95–65.95 15 15 100 = 0.15 0.08 + 0.15 = 0.23 65.95–67.95 40 40 100 = 0.40 0.23 + 0.40 = 0.63 67.95–69.95 17 17 100 = 0.17 0.63 + 0.17 = 0.80 69.95–71.95 12 12 100 = 0.12 0.80 + 0.12 = 0.92 71.95–73.95 7 7 100 = 0.07 0.92 + 0.07 = 0.99 73.95–75.95 1 1 100 = 0.01 0.99 + 0.01 = 1.00 Total = 100 Total = 1.00 The data in this table have been grouped into the following intervals: 59.95 to 61.95 inches 61.95 to 63.95 inches 63.95 to 65.95 inches 65.95 to 67.95 inches 67.95 to 69.95 inches 69.95 to 71.95 inches 71.95 to 73.95 inches 73.95 to 75.95 inches In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints. From , find the percentage of heights that are less than 65.95 inches. If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are 5 + 3 + 15 = 23 players whose heights are less than 65.95 inches. The percentage of heights less than 65.95 inches is then 23 100 or 23%. This percentage is the cumulative relative frequency entry in the third row. Try It shows the amount, in inches, of annual rainfall in a sample of towns. Rainfall (inches) Frequency Relative frequency Cumulative relative frequency 2.95–4.97 6 6 50 = 0.12 0.12 4.97–6.99 7 7 50 = 0.14 0.12 + 0.14 = 0.26 6.99–9.01 15 15 50 = 0.30 0.26 + 0.30 = 0.56 9.01–11.03 8 8 50 = 0.16 0.56 + 0.16 = 0.72 11.03–13.05 9 9 50 = 0.18 0.72 + 0.18 = 0.90 13.05–15.07 5 5 50 = 0.10 0.90 + 0.10 = 1.00 Total = 50 Total = 1.00 From , find the percentage of rainfall that is less than 9.01 inches. Try It Solutions 0.56 or 56% From , find the percentage of heights that fall between 61.95 and 65.95 inches. Add the relative frequencies in the second and third rows: 0.03 + 0.15 = 0.18 or 18%. Try It From , find the percentage of rainfall that is between 6.99 and 13.05 inches. Try It Solutions 0.30 + 0.16 + 0.18 = 0.64 or 64% Use the heights of the 100 semiprofessional soccer players in . Fill in the blanks and check your answers. The percentage of heights that are from 67.95 to 71.95 inches is: ____. The percentage of heights that are from 67.95 to 73.95 inches is: ____. The percentage of heights that are more than 65.95 inches is: ____. The number of players in the sample who are between 61.95 and 71.95 inches tall is: ____. What kind of data are the heights? Describe how you could gather this data (the heights) so that the data are characteristic of all semiprofessional soccer players. Remember, you count frequencies . To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row. 29% 36% 77% 87 quantitative continuous get rosters from each team and choose a simple random sample from each Try It From , find the number of towns that have rainfall between 2.95 and 9.01 inches. In your class, have someone conduct a survey of the number of siblings each student has. Create a frequency table. Add to it a relative frequency column and a cumulative relative frequency column. Answer the following questions: What percentage of the students in your class have no siblings? What percentage of the students have from one to three siblings? What percentage of the students have fewer than three siblings? Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are as follows: 2 5 7 3 2 10 18 15 20 7 10 18 5 12 13 12 4 5 10 . was produced: Frequency of Commuting Distances DATA FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 2 2 2 19 2 19 3 1 1 19 3 19 4 1 1 19 4 19 5 3 3 19 7 19 7 2 2 19 9 19 10 3 3 19 12 19 12 2 2 19 14 19 13 1 1 19 15 19 15 1 1 19 16 19 18 1 1 19 17 19 20 1 1 19 19 19 Is the table correct? If it is not correct, what is wrong? True or False: Three percent of the people surveyed commute three miles or less. If the statement is not correct, what should it be? If the table is incorrect, make the corrections. What fraction of the people surveyed commute five or seven miles? What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between five and 13 miles (not including five and 13 miles)? The cumulative relative frequency column should read: 2 19 , 3 19 , 4 19 , 7 19 , 9 19 , 12 19 , 14 19 , 15 19 , 16 19 , 18 19 , 19 19 No. The frequency column sums to 18, not 19. Not all cumulative relative frequencies are correct. False. The frequency for three miles is one; for two miles, two. 5 19 7 19 , 12 19 , 7 19 Try It represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year? Try It Solutions 9 50 contains data for the number of years of service for 70 federal employees. Number of Years of Service Number of Federal Employees 24 2 25 1 26 3 27 0 28 4 29 6 30 11 31 12 32 7 33 8 34 6 35 10 Answer the following questions. What is the cumulative frequency for years of service between 30 and 35 (inclusive)? What is the relative frequency for 30 years of service? What is the relative frequency for 30 years of service or less? What is the relative frequency for 25 years of service or more? Cumulative frequency is 54. 11/70 or 0.157 or 15.7% 27/70 or 0.386 or 38.6% 68/70 or 0.971 or 97.1% Try It contains the total number of fatal motor vehicle traffic crashes in the United States for a period of 18 years. Year Total number of crashes Year Total number of crashes Year 1 36,254 Year 11 38,444 Year 2 37,241 Year 12 39,252 Year 3 37,494 Year 13 38,648 Year 4 37,324 Year 14 37,435 Year 5 37,107 Year 15 34,172 Year 6 37,140 Year 16 30,862 Year 7 37,526 Year 17 30,296 Year 8 37,862 Year 18 29,757 Year 9 38,491 Total 653,782 Year 10 38,477 Answer the following questions. What is the frequency of deaths measured from Year 7 through Year 11? What percentage of deaths occurred after Year 13? What is the relative frequency of deaths that occurred in Year 7 or before? What is the percentage of deaths that occurred in Year 18? What is the cumulative relative frequency for Year 13? Explain what this number tells you about the data. Try It Solutions 190,800 (29.2%) 24.9% 260,086/653,782 or 39.8% 4.6% 75.1% of all fatal traffic crashes for the period from 1994 to 2011 happened from 1994 to 2006. References “State & County QuickFacts,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/download_data.html (accessed May 1, 2013). “State & County QuickFacts: Quick, easy access to facts about people, business, and geography,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/index.html (accessed May 1, 2013). “Table 5: Direct hits by mainland United States Hurricanes (1851-2004),” National Hurricane Center, http://www.nhc.noaa.gov/gifs/table5.gif (accessed May 1, 2013). “Levels of Measurement,” http://infinity.cos.edu/faculty/woodbury/stats/tutorial/Data_Levels.htm (accessed May 1, 2013). Courtney Taylor, “Levels of Measurement,” about.com, http://statistics.about.com/od/HelpandTutorials/a/Levels-Of-Measurement.htm (accessed May 1, 2013). David Lane. “Levels of Measurement,” Connexions, http://cnx.org/content/m10809/latest/ (accessed May 1, 2013). Chapter Review Some calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth. In addition to rounding your answers, you can measure your data using the following four levels of measurement. Nominal scale level: data that cannot be ordered nor can it be used in calculations Ordinal scale level: data that can be ordered; the differences cannot be measured Interval scale level: data with a definite ordering but no starting point; the differences can be measured, but there is no such thing as a ratio. Ratio scale level: data with a starting point that can be ordered; the differences have meaning and ratios can be calculated. When organizing data, it is important to know how many times a value appears. How many statistics students study five hours or more for an exam? What percent of families on our block own two pets? Frequency, relative frequency, and cumulative relative frequency are measures that answer questions like these. HOMEWORK Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below: Part-time Student Course Loads # of courses Frequency Relative frequency Cumulative relative frequency 1 30 0.6 2 15 3 Fill in the blanks in . What percent of students take exactly two courses? What percent of students take one or two courses? Sixty adults with gum disease were asked the number of times per week they used to floss before their diagnosis. The (incomplete) results are shown in . Flossing Frequency for Adults with Gum Disease # flossing per week Frequency Relative frequency Cumulative relative frequency 0 27 0.4500 1 18 3 0.9333 6 3 0.0500 7 1 0.0167 Fill in the blanks in . What percent of adults flossed six times per week? What percent flossed at most three times per week? # flossing per week Frequency Relative frequency Cumulative relative frequency 0 27 0.4500 0.4500 1 18 0.3000 0.7500 3 11 0.1833 0.9333 6 3 0.0500 0.9833 7 1 0.0167 1 5.00% 93.33% Nineteen immigrants to the U.S were asked how many years, to the nearest year, they have lived in the U.S. The data are as follows: 2 5 7 2 2 10 20 15 0 7 0 20 5 12 15 12 4 5 10 . was produced. Frequency of Immigrant Survey Responses Data Frequency Relative frequency Cumulative relative frequency 0 2 2 19 0.1053 2 3 3 19 0.2632 4 1 1 19 0.3158 5 3 3 19 0.4737 7 2 2 19 0.5789 10 2 2 19 0.6842 12 2 2 19 0.7895 15 1 1 19 0.8421 20 1 1 19 1.0000 Fix the errors in . Also, explain how someone might have arrived at the incorrect number(s). Explain what is wrong with this statement: “47 percent of the people surveyed have lived in the U.S. for 5 years.” Fix the statement in b to make it correct. What fraction of the people surveyed have lived in the U.S. five or seven years? What fraction of the people surveyed have lived in the U.S. at most 12 years? What fraction of the people surveyed have lived in the U.S. fewer than 12 years? What fraction of the people surveyed have lived in the U.S. from five to 20 years, inclusive? How much time does it take to travel to work? shows the mean commute time by state for workers at least 16 years old who are not working at home. Find the mean travel time, and round off the answer properly. 24.0 24.3 25.9 18.9 27.5 17.9 21.8 20.9 16.7 27.3 18.2 24.7 20.0 22.6 23.9 18.0 31.4 22.3 24.0 25.5 24.7 24.6 28.1 24.9 22.6 23.6 23.4 25.7 24.8 25.5 21.2 25.7 23.1 23.0 23.9 26.0 16.3 23.1 21.4 21.5 27.0 27.0 18.6 31.7 23.3 30.1 22.9 23.3 21.7 18.6 The sum of the travel times is 1,173.1. Divide the sum by 50 to calculate the mean value: 23.462. Because each state’s travel time was measured to the nearest tenth, round this calculation to the nearest hundredth: 23.46. A leading business magazine publishes data on small businesses (defined as businesses that have been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion). shows the ages of the chief executive officers for the first 60 ranked small businesses. Age Frequency Relative frequency Cumulative relative frequency 40–44 3 45–49 11 50–54 13 55–59 16 60–64 10 65–69 6 70–74 1 What is the frequency for CEO ages between 54 and 65? What percentage of CEOs are 65 years or older? What is the relative frequency of ages under 50? What is the cumulative relative frequency for CEOs younger than 55? Which graph shows the relative frequency and which shows the cumulative relative frequency? Use the following information to answer the next two exercises: contains data on hurricanes that have made direct hits on the U.S. Between 1851 and 2004. A hurricane is given a strength category rating based on the minimum wind speed generated by the storm. Frequency of Hurricane Direct Hits Category Number of direct hits Relative frequency Cumulative frequency 1 109 0.3993 0.3993 2 72 0.2637 0.6630 3 71 0.2601 4 18 0.9890 5 3 0.0110 1.0000 Total = 273 What is the relative frequency of direct hits that were category 4 hurricanes? 0.0768 0.0659 0.2601 Not enough information to calculate b What is the relative frequency of direct hits that were AT MOST a category 3 storm? 0.3480 0.9231 0.2601 0.3370 Cumulative Relative Frequency The term applies to an ordered set of observations from smallest to largest. The cumulative relative frequency is the sum of the relative frequencies for all values that are less than or equal to the given value. Frequency the number of times a value of the data occurs Relative Frequency the ratio of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes", "section": "Levels of Measurement", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Experimental Design and Ethics Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data. The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the independent variable or explanatory variable . The affected variable is called the dependent variable or response variable : stimulus, response. In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments . An experimental unit is a single object or individual to be measured. You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention. Additional variables that can cloud a study are called lurking variables . In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point, the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables. The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted: Results showed that believing one had taken the substance resulted in [ performance ] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment. (McClung, M. Collins, D. “Because I know it will!”: placebo effects of an ergogenic aid on athletic performance. Journal of Sport & Exercise Psychology. 2007 Jun. 29(3):382-94. Web. April 30, 2013.) When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group . This group is given a placebo treatment–an active treatment that cannot directly influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding or masking in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, they don't know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded. The Smell & Taste Treatment and Research Foundation conducted a study to investigate whether smell can affect learning. Subjects completed mazes multiple times while wearing masks. They completed the pencil and paper mazes three times wearing floral-scented masks, and three times with unscented masks. Participants were assigned at random to wear the floral mask during the first three trials or during the last three trials. For each trial, researchers recorded the time it took to complete the maze and the subject’s impression of the mask’s scent: positive, negative, or neutral. Describe the explanatory and response variables in this study. What are the treatments? Identify any lurking variables that could interfere with this study. Is it possible to use blinding in this study? The explanatory variable is scent, and the response variable is the time it takes to complete the maze. There are two treatments: a floral-scented mask and an unscented mask. All subjects experienced both treatments. The order of treatments was randomly assigned so there were no differences between the treatment groups. Random assignment eliminates the problem of lurking variables. Subjects will clearly know whether they can smell flowers or not, so subjects cannot be blinded in this study. Researchers timing the mazes can be blinded, though. The researcher who is observing a subject will not know which mask is being worn. Try It The Placebo Research Group conducted a study to find the extent of placebo effects. A group of people randomly selected were asked to take a test before and after taking a pill that induces a mild headache. The pill in half of the randomly selected people was replaced with a similar pill that has no effect. For each trial, researchers recorded the change in time people took to complete the tests before and after taking the pill. Describe the explanatory and response variable in this study. What are the treatments? Identify any lurking variables that could interfere with this study. Is it possible to use blinding in this study? The explanatory variable is the pill, and the response variable is the change in time taken between the two tests. There are two treatments: a pill that induces a mild headache and a pill that has no effect. All subjects have been given pills. The men for kind of pill intake are randomly selected. Random assignment eliminates the problem of lurking variables. Subjects do not know which pill they are taking, so they are blinded in this study. The researchers know which subject is taking which pill, so they cannot be blinded in this study. References “Vitamin E and Health,” Nutrition Source, Harvard School of Public Health, http://www.hsph.harvard.edu/nutritionsource/vitamin-e/ (accessed May 1, 2013). Stan Reents. “Don’t Underestimate the Power of Suggestion,” athleteinme.com, http://www.athleteinme.com/ArticleView.aspx?id=1053 (accessed May 1, 2013). Ankita Mehta. “Daily Dose of Aspiring Helps Reduce Heart Attacks: Study,” International Business Times, July 21, 2011. Also available online at http://www.ibtimes.com/daily-dose-aspirin-helps-reduce-heart-attacks-study-300443 (accessed May 1, 2013). The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/ScentsandLearning.html (accessed May 1, 2013). M.L. Jackson et al., “Cognitive Components of Simulated Driving Performance: Sleep Loss effect and Predictors,” Accident Analysis and Prevention Journal, Jan no. 50 (2013), http://www.ncbi.nlm.nih.gov/pubmed/22721550 (accessed May 1, 2013). “Fatality Analysis Report Systems (FARS) Encyclopedia,” National Highway Traffic and Safety Administration. http://www-fars.nhtsa.dot.gov/Main/index.aspx (accessed May 1, 2013). Data from www.businessweek.com (accessed May 1, 2013). Data from www.forbes.com (accessed May 1, 2013). U.S. Department of Health and Human Services, Code of Federal Regulations Title 45 Public Welfare Department of Health and Human Services Part 46 Protection of Human Subjects revised January 15, 2009. Section 46.111:Criteria for IRB Approval of Research. “April 2013 Air Travel Consumer Report,” U.S. Department of Transportation, April 11 (2013), http://www.dot.gov/airconsumer/april-2013-air-travel-consumer-report (accessed May 1, 2013). Lori Alden, “Statistics can be Misleading,” econoclass.com, http://www.econoclass.com/misleadingstats.html (accessed May 1, 2013). Maria de los A. Medina, “Ethics in Statistics,” Based on “Building an Ethics Module for Business, Science, and Engineering Students” by Jose A. Cruz-Cruz and William Frey, Connexions, http://cnx.org/content/m15555/latest/ (accessed May 1, 2013). Chapter Review A poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. To counter the influence of the power of suggestion in the study, participants in the control group receive an inactive placebo treatment that looks exactly like the active treatments, but cannot directly influence the response variable. To preserve the integrity of the placebo, both researchers and subjects may be blinded. When a study is designed properly, the only difference between treatment groups is the one imposed by the researcher. Therefore, when groups respond differently to different treatments, the difference must be due to the influence of the explanatory variable. “An ethics problem arises when you are considering an action that benefits you or some cause you support, hurts or reduces benefits to others, and violates some rule.” (Andrew Gelman, “Open Data and Open Methods,” Ethics and Statistics, http://www.stat.columbia.edu/~gelman/research/published/ChanceEthics1.pdf (accessed May 1, 2013).) Ethical violations in statistics are not always easy to spot. Professional associations and federal agencies post guidelines for proper conduct. It is important that you learn basic statistical procedures so that you can recognize proper data analysis. Explanatory Variable the independent variable in an experiment; the value controlled by researchers Treatments different values or components of the explanatory variable applied in an experiment Response Variable the dependent variable in an experiment; the value that is measured for change at the end of an experiment Experimental Unit any individual or object to be measured Lurking Variable a variable that has an effect on a study even though it is neither an explanatory variable nor a response variable Random Assignment the act of organizing experimental units into treatment groups using random methods Control Group a group in a randomized experiment that receives an inactive treatment but is otherwise managed exactly as the other groups Placebo an inactive treatment that cannot directly affect the response variable, used to counter the power of suggestion Blinding not telling participants which treatment a subject is receiving Double-blind experiment an experiment in which both the subjects of an experiment and the researchers who work with the subjects are blinded", "section": "Experimental Design and Ethics", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction When you have large amounts of data, you will need to organize it in a way that makes sense. These ballots from an election are rolled together with similar ballots to keep them organized. (credit: modification of work “HELMAND PROVINCE, Afghanistan (Aug. 21, 2009) Voting ballots organized and arranged for counting by Afghan presidential election workers at a local school in the Nawa District.” by U.S. Marine Corps photo by Staff Sgt. William Greeson/ Wikimedia Commons, Public Domain) Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data. In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called \"Descriptive Statistics.\" You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs. A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied. Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Display Data Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs One simple graph, the stem-and-leaf graph or stemplot , comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit . For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem. For Professor Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest): 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Stem-and-Leaf Graph Stem Leaf 3 3 4 2 9 9 5 3 5 5 6 1 3 7 8 8 9 9 7 2 3 4 8 8 0 3 8 8 8 9 0 2 4 4 4 4 6 10 0 The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% ( 8 31 ) were in the 90s or 100, a fairly high number of As. Try It For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest): 32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61 Construct a stem plot for the data. Stem Leaf 3 2 2 3 4 8 4 0 2 2 3 4 6 7 7 8 8 8 9 5 0 0 1 2 2 2 3 4 6 7 7 6 0 1 The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later. The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data: 1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3 Do the data seem to have any concentration of values? NOTE The leaves are to the right of the decimal. The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers. Stem Leaf 1 1 5 2 3 5 7 3 2 3 3 5 8 4 0 2 5 5 7 8 5 5 6 6 5 7 7 8 9 10 11 12 3 Try It The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers: 0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0 Stem Leaf 0 5 7 1 1 2 2 3 3 5 5 7 7 8 9 2 0 2 5 6 8 8 8 3 5 8 4 4 8 9 5 2 5 7 8 6 7 8 0 The value 8.0 may be an outlier. Values appear to concentrate at one and two miles. A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. and show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data. Presidential Ages at Inauguration President Age President Age President Age Washington 57 Lincoln 52 Hoover 54 J. Adams 61 A. Johnson 56 F. Roosevelt 51 Jefferson 57 Grant 46 Truman 60 Madison 57 Hayes 54 Eisenhower 62 Monroe 58 Garfield 49 Kennedy 43 J. Q. Adams 57 Arthur 51 L. Johnson 55 Jackson 61 Cleveland 47 Nixon 56 Van Buren 54 B. Harrison 55 Ford 61 W. H. Harrison 68 Cleveland 55 Carter 52 Tyler 51 McKinley 54 Reagan 69 Polk 49 T. Roosevelt 42 G.H.W. Bush 64 Taylor 64 Taft 51 Clinton 47 Fillmore 50 Wilson 56 G. W. Bush 54 Pierce 48 Harding 55 Obama 47 Buchanan 65 Coolidge 51 Presidential Age at Death President Age President Age President Age Washington 67 Lincoln 56 Hoover 90 J. Adams 90 A. Johnson 66 F. Roosevelt 63 Jefferson 83 Grant 63 Truman 88 Madison 85 Hayes 70 Eisenhower 78 Monroe 73 Garfield 49 Kennedy 46 J. Q. Adams 80 Arthur 56 L. Johnson 64 Jackson 78 Cleveland 71 Nixon 81 Van Buren 79 B. Harrison 67 Ford 93 W. H. Harrison 68 Cleveland 71 Reagan 93 Tyler 71 McKinley 58 Polk 53 T. Roosevelt 60 Taylor 65 Taft 72 Fillmore 74 Wilson 67 Pierce 64 Harding 57 Buchanan 77 Coolidge 60 Ages at Inauguration Ages at Death 9 9 8 7 7 7 6 3 2 4 6 9 8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 5 3 6 6 7 7 8 9 8 5 4 4 2 1 1 1 0 6 0 0 3 3 4 4 5 6 7 7 7 8 7 0 1 1 1 2 3 4 7 8 8 9 8 0 1 3 5 8 9 0 0 3 3 The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stemand-leaf plot of these wins and losses. Losses Wins Season Losses Wins Season 34 48 1 41 41 22 34 48 2 39 43 23 46 36 3 44 38 24 46 36 4 39 43 25 36 46 5 25 57 26 47 35 6 40 42 27 51 31 7 36 46 28 53 29 8 26 56 29 51 31 9 32 50 30 41 41 10 19 31 31 36 46 11 54 28 32 32 50 12 57 25 33 51 31 13 49 33 34 40 42 14 47 35 35 39 43 15 54 28 36 42 40 16 69 13 37 48 34 17 56 26 38 32 50 18 52 30 39 25 57 19 45 37 40 32 50 20 35 47 41 30 52 21 29 53 42 Another type of graph that is useful for specific data values is a line graph . In the particular line graph shown in , the x -axis (horizontal axis) consists of data values and the y -axis (vertical axis) consists of frequency points . The frequency points are connected using line segments. In a survey, 40 parents were asked how many times per week a teenager must be reminded to do their chores. The results are shown in and in . Number of times teenager is reminded Frequency 0 2 1 5 2 8 3 14 4 7 5 4 Try It In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in . Construct a line graph. Number of times in shop Frequency 0 7 1 10 2 14 3 9 Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in has age groups represented on the x -axis and proportions on the y -axis . The percentage of U.S.-based TikTok users by age is shown in . Construct a bar graph using these data. Age groups Proportion (%) of TikTok users 10–19 32.5% 20–29 29.5% 30–39 16.4% 40–49 13.9% 50+ 7.1% Try It The population in Park City is made up of children, working-age adults, and retirees. shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions. Age groups Number of people Proportion of population Children 67,059 19% Working-age adults 152,198 43% Retirees 131,662 38% The columns in show the projected data for the year 2030 for the number and percentages of high school graduates by geographic region in the United States. Create a bar graph for these data with the geographic region (qualitative data) on the x -axis and the percentage of high school data (quantitative data) on the y -axis. Region Number of Graduates Percentage of Graduates Northeast 517,720 16.1% Midwest 695,170 21.6% South 1,253,540 39.0% West 749,400 23.3% Try It Park City is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district. District Registered voter population Overall city population 1 15.5% 19.4% 2 12.2% 15.6% 3 9.8% 9.0% 4 17.4% 18.5% 5 22.8% 20.7% 6 22.3% 16.8% Below is a two-way table showing the types of pets owned by men and women: Dogs Cats Fish Total Men 4 2 2 8 Women 4 6 2 12 Total 8 8 4 20 Given these data, calculate the conditional distributions for the subpopulation of men who own each pet type. Men who own dogs = 4/8 = 0.5 Men who own cats = 2/8 = 0.25 Men who own fish = 2/8 = 0.25 Note: The sum of all of the conditional distributions must equal one. In this case, 0.5 + 0.25 + 0.25 = 1; therefore, the solution \"checks\". Try It Given the data in , calculate the conditional distributions for the subpopulation of women who own each pet type. Women who own dogs = 4 12 = 0 . 33 Women who own cats = 6 12 = 0 . 5 Women who own fish = 2 12 = 0 . 17 Note: The sum of all of the conditional distributions must equal one. In this case, 0 . 33 + 0 . 5 + 0 . 17 = 1 ; therefore, the solution “checks”. Histograms, Frequency Polygons, and Time Series Graphs For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more. A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data. The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample.(Remember, frequency is defined as the number of times an answer occurs.) If: f = frequency n = total number of data values (or the sum of the individual frequencies), and RF = relative frequency, then: RF = f n For example, if three students in an English class of 40 students received from 90% to 100%, then, f = 3, n = 40, and RF = f n = 3 40 = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures. To construct a histogram , first decide how many bars or intervals , also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data. The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured. 60; 60.5; 61; 61; 61.5 63.5; 63.5; 63.5 64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5 66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5 68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5 70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71 72; 72; 72; 72.5; 72.5; 73; 73.5 74 The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point. 60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95. The largest value is 74, so 74 + 0.05 = 74.05 is the ending value. Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars. 74.05 - 59.95 = 14.1 14.1 ÷ 8 = 1.76 NOTE We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals. The boundaries are: 59.95 59.95 + 2 = 61.95 61.95 + 2 = 63.95 63.95 + 2 = 65.95 65.95 + 2 = 67.95 67.95 + 2 = 69.95 69.95 + 2 = 71.95 71.95 + 2 = 73.95 73.95 + 2 = 75.95 The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95. The following histogram displays the heights on the x -axis and relative frequency on the y -axis. Try It The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars. 9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5 12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14 Smallest value: 9 Largest value: 14 Convenient starting value: 9 – 0.05 = 8.95 Convenient ending value: 14 + 0.05 = 14.05 14.05 − 8.95 6 = 0.85 The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one. Create a histogram for the following data: the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data , since books are counted. 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 2; 2; 2; 2; 2; 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4 5; 5; 5; 5; 5 6; 6 Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books. Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5. Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ . 3.5 to 4.5 4.5 to 5.5 6 5.5 to 6.5 Calculate the number of bars as follows: 6.5 - 0.5 = 6 6 ÷ 1 = 6 where 1 is the width of a bar. Therefore, bars = 6. The following histogram displays the number of books on the x -axis and the frequency on the y -axis. Try It The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted. 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3 20 student athletes play one sport. 22 student athletes play two sports. Eight student athletes play three sports. Fill in the blanks for the following sentence. Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to _____, the 2 in the middle of the interval from _____ to _____, and the 3 in the middle of the interval from _____ to _____. Using this data set, construct a histogram. Number of hours my classmates spent playing video games on weekends 9.95 10 2.25 16.75 0 19.5 22.5 7.5 15 12.75 5.5 11 10 20.75 17.5 23 21.9 24 23.75 18 20 15 22.9 18.8 20.5 Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram. Try It The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram. 22 35 15 26 40 28 18 20 25 34 39 42 24 22 19 27 22 34 40 20 38 and 28 Use 10–19 as the first interval. Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals. Frequency Polygons Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons. To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x -axis and y -axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them. A frequency polygon was constructed from the frequency table below. Frequency distribution for calculus final test scores Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 The first label on the x -axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x -axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x -axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side. Try It Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in . Age at inauguration Frequency 41.5–46.5 4 46.5–51.5 11 51.5–56.5 14 56.5–61.5 9 61.5–66.5 4 66.5–71.5 2 The first label on the x -axis is 39. This represents an interval extending from 36.5 to 41.5. Since there are no ages less than 41.5, this interval is used only to allow the graph to touch the x -axis. The point labeled 44 represents the next interval, or the first “real” interval from the table, and contains four scores. This reasoning is followed for each of the remaining intervals with the point 74 representing the interval from 71.5 to 76.5. Again, this interval contains no data and is only used so that the graph will touch the x -axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side. Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets. We will construct an overlay frequency polygon comparing the scores from with the students’ final numeric grade. Frequency distribution for calculus final test scores Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 Frequency distribution for calculus final grades Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 10 10 59.5 69.5 10 20 69.5 79.5 30 50 79.5 89.5 45 95 89.5 99.5 5 100 Try It We will construct an overlay frequency polygon comparing the scores from with the students’ final test scores in algebra. Frequency Distribution for Algebra Final Test Scores Lower Bound Upper Bound Frequency Cumulative Frequency 49.5 59.5 10 10 59.5 69.5 5 15 69.5 79.5 40 55 79.5 89.5 35 90 89.5 99.5 10 100 Calculus Test Scores v Algebra Test Scores Image Insert Constructing a Time Series Graph Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with these data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected. One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph. To construct a time series graph, we must look at both pieces of our paired data set . We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur. The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only. Year Jan Feb Mar Apr May Jun Jul 1 181.7 183.1 184.2 183.8 183.5 183.7 183.9 2 185.2 186.2 187.4 188.0 189.1 189.7 189.4 3 190.7 191.8 193.3 194.6 194.4 194.5 195.4 4 198.3 198.7 199.8 201.5 202.5 202.9 203.5 5 202.416 203.499 205.352 206.686 207.949 208.352 208.299 6 211.080 211.693 213.528 214.823 216.632 218.815 219.964 7 211.143 212.193 212.709 213.240 213.856 215.693 215.351 8 216.687 216.741 217.631 218.009 218.178 217.965 218.011 9 220.223 221.309 223.467 224.906 225.964 225.722 225.922 10 226.665 227.663 229.392 230.085 229.815 229.478 229.104 Year Aug Sep Oct Nov Dec Annual 1 184.6 185.2 185.0 184.5 184.3 184.0 2 189.5 189.9 190.9 191.0 190.3 188.9 3 196.4 198.8 199.2 197.6 196.8 195.3 4 203.9 202.9 201.8 201.5 201.8 201.6 5 207.917 208.490 208.936 210.177 210.036 207.342 6 219.086 218.783 216.573 212.425 210.228 215.303 7 215.834 215.969 216.177 216.330 215.949 214.537 8 218.312 218.439 218.711 218.803 219.179 218.056 9 226.545 226.889 226.421 226.230 225.672 224.939 10 230.379 231.407 231.317 230.221 229.601 229.594 Try It The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO 2 emissions for the United States. CO 2 emissions Year Ukraine United Kingdom United States 1 352,259 540,640 5,681,664 2 343,121 540,409 5,790,761 3 339,029 541,990 5,826,394 4 327,797 542,045 5,737,615 5 328,357 528,631 5,828,697 6 323,657 522,247 5,656,839 7 272,176 474,579 5,299,563 Uses of a Time Series Graph Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot. How NOT to Lie with Statistics It is important to remember that the very reason we develop a variety of methods to present data is to develop insights into the subject of what the observations represent. We want to get a \"sense\" of the data. Are the observations all very much alike or are they spread across a wide range of values, are they bunched at one end of the spectrum or are they distributed evenly and so on. We are trying to get a visual picture of the numerical data. Shortly we will develop formal mathematical measures of the data, but our visual graphical presentation can say much. It can, unfortunately, also say much that is distracting, confusing and simply wrong in terms of the impression the visual leaves. Many years ago Darrell Huff wrote the book How to Lie with Statistics . It has been through 25 plus printings and sold more than one and one-half million copies. His perspective was a harsh one and used many actual examples that were designed to mislead. He wanted to make people aware of such deception, but perhaps more importantly to educate so that others do not make the same errors inadvertently. Again, the goal is to enlighten with visuals that tell the story of the data. Pie charts have a number of common problems when used to convey the message of the data. Too many pieces of the pie overwhelm the reader. More than perhaps five or six categories ought to give an idea of the relative importance of each piece. This is after all the goal of a pie chart, what subset matters most relative to the others. If there are more components than this then perhaps an alternative approach would be better or perhaps some can be consolidated into an \"other\" category. Pie charts cannot show changes over time, although we see this attempted all too often. In federal, state, and city finance documents pie charts are often presented to show the components of revenue available to the governing body for appropriation: income tax, sales tax motor vehicle taxes and so on. In and of itself this is interesting information and can be nicely done with a pie chart. The error occurs when two years are set side-by-side. Because the total revenues change year to year, but the size of the pie is fixed, no real information is provided and the relative size of each piece of the pie cannot be meaningfully compared. Histograms can be very helpful in understanding the data. Properly presented, they can be a quick visual way to present probabilities of different categories by the simple visual of comparing relative areas in each category. Here the error, purposeful or not, is to vary the width of the categories. This of course makes comparison to the other categories impossible. It does embellish the importance of the category with the expanded width because it has a greater area, inappropriately, and thus visually \"says\" that that category has a higher probability of occurrence. Time series graphs perhaps are the most abused. A plot of some variable across time should never be presented on axes that change part way across the page either in the vertical or horizontal dimension. Perhaps the time frame is changed from years to months. Perhaps this is to save space or because monthly data was not available for early years. In either case this confounds the presentation and destroys any value of the graph. If this is not done to purposefully confuse the reader, then it certainly is either lazy or sloppy work. Changing the units of measurement of the axis can smooth out a drop or accentuate one. If you want to show large changes, then measure the variable in small units, penny rather than thousands of dollars. And of course to continue the fraud, be sure that the axis does not begin at zero, zero. If it begins at zero, zero, then it becomes apparent that the axis has been manipulated. Perhaps you have a client that is concerned with the volatility of the portfolio you manage. An easy way to present the data is to use long time periods on the time series graph. Use months or better, quarters rather than daily or weekly data. If that doesn't get the volatility down then spread the time axis relative to the rate of return or portfolio valuation axis. If you want to show \"quick\" dramatic growth, then shrink the time axis. Any positive growth will show visually \"high\" growth rates. Do note that if the growth is negative then this trick will show the portfolio is collapsing at a dramatic rate. Again, the goal of descriptive statistics is to convey meaningful visuals that tell the story of the data. Purposeful manipulation is fraud and unethical at the worst, but even at its best, making these type of errors will lead to confusion on the part of the analysis. References Doyle, Brandon. \"TikTok Statistics – Updated Mar 2023.\" Wallaroo. March 21, 2023. https://wallaroomedia.com/blog/ social-media/tiktok-statistics/. Table 219.20. Public High School Graduates, by Region, State, and Jurisdiction: Selected Years, 1980-81 through 2030-31.” National Center for Education Statistics. Accessed May 24, 2023. https://nces.ed.gov/programs/digest/d21/tables/dt21_219.20.asp. “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker “Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013). “Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013). “Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013). “Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013). “CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013). “Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013). “Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013). Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach . CRC Press: 1980. “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). Chapter Review A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used. A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y -axis with the frequency being graphed on the x -axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time. For the next three exercises, use the data to construct a line graph. In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in . Number of times in store Frequency 1 4 2 10 3 16 4 6 5 4 In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in . Years since last purchase Frequency 0 2 1 8 2 13 3 22 4 16 5 9 Several children were asked how many TV shows they watch each day. The results of the survey are shown in . Number of TV shows Frequency 0 12 1 18 2 36 3 7 4 2 The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students. Seasons Number of students Proportion of population Spring 8 24% Summer 9 26% Autumn 11 32% Winter 6 18% Using the data from Mrs. Ramirez’s math class supplied in , construct a bar graph showing the percentages. David County has six high schools. Each school sent students to participate in a county-wide science competition. shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school. High school Science competition population Overall student population Alabaster 28.9% 8.6% Concordia 7.6% 23.2% Genoa 12.1% 15.0% Mocksville 18.5% 14.3% Tynneson 24.2% 10.1% West End 8.7% 28.8% Use the data from the David County science competition supplied in . Construct a bar graph that shows the county-wide population percentage of students at each school. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table. Data value (# cars) Frequency Relative frequency Cumulative relative frequency What does the frequency column in sum to? Why? 65 What does the relative frequency column in sum to? Why? What is the difference between relative frequency and frequency for each data value in ? The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value. What is the difference between cumulative relative frequency and relative frequency for each data value? To construct the histogram for the data in , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling. Answers will vary. One possible histogram is shown: Construct a frequency polygon for the following: Pulse rates for Females Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 Actual speed in a 30 MPH zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 Tar (mg) in nonfiltered cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2 Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger. Depth of hunger Frequency 230–259 21 260–289 13 290–319 5 320–349 7 350–379 1 380–409 1 410–439 1 Find the midpoint for each class. These will be graphed on the x -axis. The frequency values will be graphed on the y -axis values. Use the two frequency tables to compare the life expectancy of males and females from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of females compared to males? Life expectancy at birth – Females Frequency 49–55 3 56–62 3 63–69 1 70–76 3 77–83 8 84–90 2 Life expectancy at birth – Males Frequency 49–55 3 56–62 3 63–69 1 70–76 1 77–83 7 84–90 5 Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births. Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009 Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354 Sex/Year 1870 1871 1872 1873 1874 1875 Female 56,431 56,099 57,472 58,233 60,109 60,146 Male 58,959 60,029 61,293 61,467 63,602 63,432 Total 115,390 116,128 118,765 119,700 123,711 123,578 The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for a city during the period from 1961 to 1973. Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36 Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33 Construct a double time series graph using a common x -axis for both sets of data. Which variable increased the fastest? Explain. Did the city's increase in police officers have an impact on the murder rate? Explain. Homework contains the percentages of adults who were never married in the 50 U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Use a random number generator to randomly pick eight states. Construct a bar graph of the rates of those eight states. Construct a bar graph for all the states beginning with the letter \"A.\" Construct a bar graph for all the states beginning with the letter \"M.\" Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows. Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically) Press MATH Arrow over to PRB Press 5:randInt( Enter 51,1,8) Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}. Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}. Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows: Publisher A # of books Freq. Rel. freq. 0 10 1 12 2 16 3 12 4 8 5 6 6 2 8 2 Publisher B # of books Freq. Rel. freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Publisher C # of books Freq. Rel. freq. 0–1 20 2–3 35 4–5 12 6–7 2 8–9 1 Find the relative frequencies for each survey. Write them in the charts. Use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? Make new histograms for Publisher A and Publisher B. This time, make bar widths of two. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group. Singles Amount($) Frequency Rel. frequency 51–100 5 101–150 10 151–200 15 201–250 15 251–300 10 301–350 5 Couples Amount($) Frequency Rel. frequency 100–150 5 201–250 5 251–300 5 301–350 5 351–400 10 401–450 10 451–500 10 501–550 10 551–600 5 601–650 5 Fill in the relative frequency for each group. Construct a histogram for the singles group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis. Construct a histogram for the couples group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis. Compare the two graphs: List two similarities between the graphs. List two differences between the graphs. Overall, are the graphs more similar or different? Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x -axis by $50, scale it by $100. Use relative frequency on the y -axis. Compare the graph for the singles with the new graph for the couples: List two similarities between the graphs. Overall, are the graphs more similar or different? How did scaling the couples graph differently change the way you compared it to the singles graph? Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences. Singles Amount($) Frequency Relative frequency 51–100 5 0.08 101–150 10 0.17 151–200 15 0.25 201–250 15 0.25 251–300 10 0.17 301–350 5 0.08 Couples Amount($) Frequency Relative frequency 100–150 5 0.07 201–250 5 0.07 251–300 5 0.07 301–350 5 0.07 351–400 10 0.14 401–450 10 0.14 451–500 10 0.14 501–550 10 0.14 551–600 5 0.07 601–650 5 0.07 See and . In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included). In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included). Compare the two graphs: Answers may vary. Possible answers include: Both graphs have a single peak. Both graphs use class intervals with width equal to $50. Answers may vary. Possible answers include: The couples graph has a class interval with no values. It takes almost twice as many class intervals to display the data for couples. Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same. Answers may vary. Compare the graph for the Singles with the new graph for the Couples: Both graphs have a single peak. Both graphs display 6 class intervals. Both graphs show the same general pattern. Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different. Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison. Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows. # of movies Frequency Relative frequency Cumulative relative frequency 0 5 1 9 2 6 3 4 4 1 Construct a histogram of the data. Complete the columns of the chart. Use the following information to answer the next two exercises: Suppose 111 people who shopped in a special T-shirt store were asked the number of T-shirts they own costing more than $19 each. The percentage of people who own at most three T-shirts costing more than $19 each is approximately: 21 59 41 Cannot be determined c If the data were collected by asking the first 111 people who entered the store, then the type of sampling is: cluster simple random stratified convenience Following are the rates of unmarried adults for the 50 U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Construct a bar graph of the unmarried adult rates of your state and the four states closest to your state. Hint: Label the x -axis with the states. Answers will vary. Frequency the number of times a value of the data occurs Histogram a graphical representation in x - y form of the distribution of data in a data set; x represents the data and y represents the frequency, or relative frequency. The graph consists of contiguous rectangles. Relative Frequency the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes", "section": "Display Data", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Measures of the Location of the Data The common measures of location are quartiles and percentiles . Quartiles divide an ordered data set into four equal parts. The three quartiles of a data set are labeled as Q1, Q2, and Q3. About one-fourth of the data falls on or below the first quartile Q1. About one-half of the data falls on or below the second quartile Q2. About three-fourths of the data falls on or below the third quartile Q3. In the same way, percentiles divide a data set into 100 equal parts. To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90 th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score. Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75 th percentile. That translates into a score of at least 1220. Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant. The median is a number that measures the \"center\" of the data. You can think of the median as the \"middle value,\" but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data. 1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1 Ordered from smallest to largest: 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5 Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two. 6.8 + 7.2 = 14 14 ÷ 2 = 7 The median is seven. Half of the values are smaller than seven and half of the values are larger than seven. Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q 1 , is the middle value of the lower half of the data, and the third quartile, Q 3 , is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set: 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5 The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two. 1; 1; 2; 2; 4; 6; 6.8 The number two, which is part of the data, is the first quartile . One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two. The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine. The third quartile , Q 3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example. The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile ( Q 3 ) and the first quartile ( Q 1 ). IQR = Q 3 – Q 1 The IQR can help to determine potential outliers . A value is suspected to be a potential outlier if it is less than (1.5)( IQR ) below the first quartile or more than (1.5)( IQR ) above the third quartile . Potential outliers always require further investigation. NOTE A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data. For the following 13 real estate prices, calculate the IQR and determine if any prices are potential outliers. Prices are in dollars. 389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000 Order the data from smallest to largest. 114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000 M = 488,800 Q 1 = 230,500 + 387,000 2 = 308,750 Q 3 = 639,000 + 659,000 2 = 649,000 IQR = 649,000 – 308,750 = 340,250 (1.5)( IQR ) = (1.5)(340,250) = 510,375 Q 1 – (1.5)( IQR ) = 308,750 – 510,375 = –201,625 Q 3 + (1.5)( IQR ) = 649,000 + 510,375 = 1,159,375 No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier . Try It For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars. $33,000 $64,500 $28,000 $54,000 $72,000 $68,500 $69,000 $42,000 $54,000 $120,000 $40,500 Test scores for a college statistics class held during the day are: 99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90 Test scores for a college statistics class held during the evening are: 98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5 For the two data sets, find the following: The interquartile range. Compare the two interquartile ranges. Any outliers in either set. The five number summary for the day and night classes is Minimum Q 1 Median Q 3 Maximum Day 32 56 74.5 82.5 99 Night 25.5 78 81 89 98 The IQR for the day group is Q 3 – Q 1 = 82.5 – 56 = 26.5 The IQR for the night group is Q 3 – Q 1 = 89 – 78 = 11 The interquartile range (the spread or variability) for the day class is larger than the night class IQR . This suggests more variation will be found in the day class’s class test scores. Day class outliers are found using the IQR times 1.5 rule. So, Q 1 - IQR (1.5) = 56 – 26.5(1.5) = 16.25 Q 3 + IQR (1.5) = 82.5 + 26.5(1.5) = 122.25 Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers. Night class outliers are calculated as: Q 1 – IQR (1.5) = 78 – 11(1.5) = 61.5 Q 3 + IQR(1.5) = 89 + 11(1.5) = 105.5 For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier. Try It Find the interquartile range for the following two data sets and compare them. Test Scores for Class A 69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94 Test Scores for Class B 90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100 Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were: Amount of sleep per school night (hours) Frequency Relative frequency Cumulative relative frequency 4 2 0.04 0.04 5 5 0.10 0.14 6 7 0.14 0.28 7 12 0.24 0.52 8 14 0.28 0.80 9 7 0.14 0.94 10 3 0.06 1.00 Find the 28 th percentile . Notice the 0.28 in the \"cumulative relative frequency\" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28 th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28 th percentile is between the last six and the first seven. The 28 th percentile is 6.5. Find the median . Look again at the \"cumulative relative frequency\" column and find 0.52. The median is the 50 th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50 th percentile is between the 25 th , or seven, and 26 th , or seven, values. The median is seven. Find the third quartile . The third quartile is the same as the 75 th percentile. You can \"eyeball\" this answer. If you look at the \"cumulative relative frequency\" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75 th percentile, then, must be an eight . Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q 3 , is the 38 th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.) Try it Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65 th percentile. Amount of time spent on route (hours) Frequency Relative frequency Cumulative relative frequency 2 12 0.30 0.30 3 14 0.35 0.65 4 10 0.25 0.90 5 4 0.10 1.00 The 65 th percentile is between the last three and the first four. The 65 th percentile is 3.5. A Formula for Finding the k th Percentile If you were to do a little research, you would find several formulas for calculating the k th percentile. Here is one of them. k = the k th percentile. It may or may not be part of the data. i = the index (ranking or position of a data value) n = the total number of data points, or observations Order the data from smallest to largest. Calculate i = k 100 ( n + 1 ) If i is an integer, then the k th percentile is the data value in the i th position in the ordered set of data. If i is not an integer, then round i up and round i down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example. Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the 70 th percentile. Find the 83 rd percentile. k = 70 i = the index n = 29 i = k 100 ( n + 1) = ( 70 100 )(29 + 1) = 21. Twenty-one is an integer, and the data value in the 21 st position in the ordered data set is 64. The 70 th percentile is 64 years. k = 83 rd percentile i = the index n = 29 i = k 100 ( n + 1) = ( 83 100 )(29 + 1) = 24.9, which is NOT an integer. Round it down to 24 and up to 25. The age in the 24 th position is 71 and the age in the 25 th position is 72. Average 71 and 72. The 83 rd percentile is 71.5 years. Try It Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Calculate the 20 th percentile and the 55 th percentile. k = 20. Index = i = k 100 ( n + 1 ) = 20 100 (29 + 1) = 6. The age in the sixth position is 27. The 20 th percentile is 27 years. k = 55. Index = i = k 100 ( n + 1 ) = 55 100 (29 + 1) = 16.5. Round down to 16 and up to 17. The age in the 16 th position is 52 and the age in the 17 th position is 55. The average of 52 and 55 is 53.5. The 55 th percentile is 53.5 years. NOTE You can calculate percentiles using calculators and computers. There are a variety of online calculators. Using : Find the 80 th percentile. Find the 90 th percentile. Find the first quartile. What is another name for the first quartile? Using the data from the frequency table, we have: Notice there are 50 data values in the table, so n = 50. Calculate the index i as follows: i = 80 100 ( 50 + 1 ) = 40 . 8 Since i = 40.8, calculate the mean of the 40 th and 41 st data values. The 40 th data value is 8, the 41 st data value is 9, and the mean of these two data values is 8.5. Thus, the 80 th percentile is 8.5. Calculate the index i as follows: i = 90 100 ( 50 + 1 ) = 45 . 9 Since i = 45.9, calculate the mean of the 45 th and 46 th data values. The 45 th data value is 9, the 46 th data value is 9, and the mean of these two data values is 9. Thus, the 90 th percentile is 9. Another name for the first quartile is the 25 th percentile. Proceed to calculate the 25 th percentile: Calculate the index i as follows: i = 25 100 ( 50 + 1 ) = 12 . 75 Since i = 12.75, calculate the mean of the 12 th and 13 th data values. The 12 th data value is 6, the 13 th data value is 6, and the mean of these two data values is 6. Thus, the first quartile is 6. Try It Refer to the . Find the third quartile. What is another name for the third quartile? Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions: How many students were surveyed? What kind of sampling did you do? Construct two different histograms. For each, starting value = _____ ending value = ____. Find the median, first quartile, and third quartile. Construct a table of the data to find the following: the 10 th percentile the 70 th percentile the percent of students who own less than four sweaters A Formula for Finding the Percentile of a Value in a Data Set Order the data from smallest to largest. x = the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile. y = the number of data values equal to the data value for which you want to find the percentile. n = the total number of data. Calculate x + 0.5 y n (100). Then round to the nearest integer. Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentile for 58. Find the percentile for 25. Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58. x = 18 and y = 1. x + 0.5 y n (100) = 18 + 0.5 ( 1 ) 29 (100) = 63.80. 58 is the 64 th percentile. Counting from the bottom of the list, there are three data values less than 25. There is one value of 25. x = 3 and y = 1. x + 0.5 y n (100) = 3 + 0.5 ( 1 ) 29 (100) = 12.07. Twenty-five is the 12 th percentile. Try It Listed are 30 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentiles for 47 and 31. Interpreting Percentiles, Quartiles, and Median A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15 th percentile. Low percentiles always correspond to lower data values. High percentiles always correspond to higher data values. A percentile may or may not correspond to a value judgment about whether it is \"good\" or \"bad.\" The interpretation of whether a certain percentile is \"good\" or \"bad\" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered \"good;\" in other contexts a high percentile might be considered \"good\". In many situations, there is no value judgment that applies. Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text. NOTE When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information. information about the context of the situation being considered the data value (value of the variable) that represents the percentile the percent of individuals or items with data values below the percentile the percent of individuals or items with data values above the percentile. On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation. Twenty-five percent of students finished the exam in 35 minutes or less. Seventy-five percent of students finished the exam in 35 minutes or more. A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.) Try It For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation. On a 20 question math test, the 70 th percentile for number of correct answers was 16. Interpret the 70 th percentile in the context of this situation. Seventy percent of students answered 16 or fewer questions correctly. Thirty percent of students answered 16 or more questions correctly. A higher percentile could be considered good, as answering more questions correctly is desirable. Try It On a 60 point written assignment, the 80 th percentile for the number of points earned was 49. Interpret the 80 th percentile in the context of this situation. Eighty percent of students earned 49 points or fewer. Twenty percent of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable. At a community college, it was found that the 30 th percentile of credit units that students are enrolled for is seven units. Interpret the 30 th percentile in the context of this situation. Thirty percent of students are enrolled in seven or fewer credit units. Seventy percent of students are enrolled in seven or more credit units. In this example, there is no \"good\" or \"bad\" value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs. Try It During a season, the 40 th percentile for points scored per player in a game is eight. Interpret the 40 th percentile in the context of this situation. Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown. 0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes 10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes; 30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes Determine the following five values. Min = 0 Q 1 = 20 Med = 40 Q 3 = 60 Max = 300 If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment. However, the principal needs to be careful. The value 300 appears to be a potential outlier. Q 3 + 1.5( IQR ) = 60 + (1.5)(40) = 120. The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values: Min = 0 Q 1 = 20 Q 3 = 60 Max = 120 We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results. Try It A college statistics instructor is investigating the amount of time students spend working on a final project in the course. The instructor would like students to spend approximately 3 to 4 hours as the typical amount of time to be spent on the project. The instructor collects data from a random sample of 10 students for the number of hours spent working on the final project. The results obtained are as follows: 2 hours; 3 hours; 5 hours; 3 hours; 4 hours; 4 hours; 3 hours; 11 hours; 3 hours; 2 hours. Determine the following five values: Min , Q 1 , Med , Q 3 , Max . Should the instructor modify the final project or leave as is? The five values are obtained as follows: Min = 2 Q 1 = 3 Med = 3 Q 3 = 4 Max = 11 Given that the 3 rd quartile is 4 hours, this indicates that 75% of the students spend 4 hours or less on the project. Also for the interquartile range: IQR = 4 – 3 = 1, we know that half the students spent between 3 and 4 hours on the project. This seems like a reasonable amount of time for students to spend on the final project. However, the value 11 hours appears to be a potential outlier. Q 3 + 1 . 5 ( I Q R ) = 4 + 1 . 5 ( 1 ) = 5 . 5 The value 11 is greater than 5.5, so it is an outlier. If we delete it and recalculate the five values, we get the following values: Min = 2 Q 1 = 2.5 Q 3 = 4 Max = 5 We still have 75% of students spending 4 hours or less on the project. Half of the students spend between 2.5 hours and 4 hours on the project. However, 10 students may be a relatively small sample size. The instructor might consider a larger sample size. References Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at http://usatoday30.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1 (accessed April 3, 2013). Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013). “1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013). Data from San Jose Mercury News . Data from Time Magazine ; survey by Yankelovich Partners, Inc. Chapter Review The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50 th percentile would be greater than 50 percent of the other observations in the set. Quartiles divide data into quarters. The first quartile ( Q 1 ) is the 25 th percentile,the second quartile ( Q 2 or median) is 50 th percentile, and the third quartile ( Q 3 ) is the 75 th percentile. The interquartile range, or IQR , is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q 1 from Q 3 , and can help determine outliers by using the following two expressions. Q 3 + IQR (1.5) Q 1 – IQR (1.5) Formula Review i = ( k 100 ) ( n + 1 ) where i = the ranking or position of a data value, k = the kth percentile, n = total number of data. Expression for finding the percentile of a data value: ( x + 0.5 y n ) (100) where x = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile, y = the number of data values equal to the data value for which you want to find the percentile, n = total number of data Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the 40 th percentile. Find the 78 th percentile. The 40 th percentile is 37 years. The 78 th percentile is 70 years. Listed are 32 ages for Academy Award winning best actors in order from smallest to largest. 18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentile of 37. Find the percentile of 72. Jesse was ranked 37 th in his graduating class of 180 students. At what percentile is Jesse’s ranking? Jesse graduated 37 th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37. x = 143 and y = 1. x + 0.5 y n (100) = 143 + 0.5 ( 1 ) 180 (100) = 79.72. Jesse’s rank of 37 puts him at the 80 th percentile. For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race? The 20 th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20 th percentile in the context of the situation. A bicyclist in the 90 th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90 th percentile in the context of the situation. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race? The 40 th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40 th percentile in the context of the situation. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed which is faster. 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster). On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain. Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85 th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85 th percentile in the context of this situation. When waiting in line at the DMV, the 85 th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer. In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78 th percentile. Should Li be pleased or upset by this result? Explain. In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1,700 in damage and was in the 90 th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90 th percentile in the context of this problem. The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of $1700 or more. The University of Wisconsin has two criteria used to set admission standards for students to be admitted to a college in the UW system: Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an \"admissions index\" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent? Students whose GPAs are at or above the 96 th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percentage of students from each high school are \"eligible in the local context\"? Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34 th percentile. The 34 th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses? You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more. Use the following information to answer the next six exercises. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. First quartile = _______ Second quartile = median = 50 th percentile = _______ 4 Third quartile = _______ Interquartile range ( IQR ) = _____ – _____ = _____ 6 – 4 = 2 10 th percentile = _______ 70 th percentile = _______ 6 Homework The median age for Black people in the United States currently is 30.9 years; for White people in the United States it is 42.3 years. Based upon this information, give two reasons why the median age for Black people could be lower than the median age for White people. Does the lower median age for Black people necessarily mean that Black people die younger than White people? Why or why not? How might it be possible for Black people and White people to die at approximately the same age, but for the median age for White people to be higher? Six hundred adult Americans were asked by telephone poll, \"What do you think constitutes a middle-class income?\" The results are in . Also, include left endpoint, but not the right endpoint. Salary ($) Relative frequency < 20,000 0.02 20,000–25,000 0.09 25,000–30,000 0.19 30,000–40,000 0.26 40,000–50,000 0.18 50,000–75,000 0.17 75,000–99,999 0.02 100,000+ 0.01 What percentage of the survey answered \"not sure\"? What percentage think that middle-class is from $25,000 to $50,000? Construct a histogram of the data. Should all bars have the same width, based on the data? Why or why not? How should the <20,000 and the 100,000+ intervals be handled? Why? Find the 40 th and 80 th percentiles Construct a bar graph of the data 1 – (0.02+0.09+0.19+0.26+0.18+0.17+0.02+0.01) = 0.06 0.19+0.26+0.18 = 0.63 Answers may vary. 40 th percentile will fall between 30,000 and 40,000 80 th percentile will fall between 50,000 and 75,000 Answers may vary. Interquartile Range or IQR , is the range of the middle 50 percent of the data values; the IQR is found by subtracting the first quartile from the third quartile. Outlier an observation that does not fit the rest of the data Percentile a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50 th percentile. The first and third quartiles are the 25 th and the 75 th percentiles, respectively. Quartiles the numbers that separate the data into quarters; quartiles may or may not be part of the data. The second quartile is the median of the data.", "section": "Measures of the Location of the Data", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Measures of the Center of the Data The \"center\" of a data set is also a way of describing location. The two most widely used measures of the \"center\" of the data are the mean (average) and the median . To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. Technically this is the arithmetic mean. We will discuss the geometric mean later. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts meaning an equal number of observations on each side. The weight of 25 people are below this weight and 25 people are heavier than this weight. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center. NOTE The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. Formally, the arithmetic mean is called the first moment of the distribution by mathematicians. However, in practice among non-statisticians, “average\" is commonly accepted for “arithmetic mean.” When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an x with a bar over it (pronounced “ x bar”): x – . The Greek letter μ (pronounced \"mew\") represents the population mean . One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random. To see that both ways of calculating the mean are the same, consider the sample: 1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4 x – = 1 + 1 + 1 + 2 + 2 + 3 + 4 + 4 + 4 + 4 + 4 11 = 2.7 x ¯ = 3 ( 1 ) + 2 ( 2 ) + 1 ( 3 ) + 5 ( 4 ) 11 = 2.7 In the second calculation, the frequencies are 3, 2, 1, and 5. You can quickly find the location of the median by using the expression n + 1 2 . The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is 97, then n + 1 2 = 97 + 1 2 = 49. The median is the 49 th value in the ordered data. If the total number of data values is 100, then n + 1 2 = 100 + 1 2 = 50.5. The median occurs midway between the 50 th and 51 st values. The location of the median and the value of the median are not the same. The upper case letter M is often used to represent the median. The next example illustrates the location of the median and the value of the median. A hospital administrator keeps track of the ages (in years) of patients visiting the emergency room over a one-week period (data are sorted from smallest to largest): 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47; Calculate the mean and the median. The calculation for the mean is: x – = [ 3 + 4 + ( 8 ) ( 2 ) + 10 + 11 + 12 + 13 + 14 + ( 15 ) ( 2 ) + ( 16 ) ( 2 ) + ... + 35 + 37 + 40 + ( 44 ) ( 2 ) + 47 ] 40 = 23.6 To find the median, M , first use the formula for the location. The location is: n + 1 2 = 40 + 1 2 = 20.5 Starting at the smallest value, the median is located between the 20 th and 21 st values (the two 24s): 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47; M = 24 + 24 2 = 24 Try It The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median. 3 4 5 7 7 7 7 8 8 9 9 10 10 10 10 10 11 12 12 13 14 14 15 15 17 17 18 19 19 19 21 21 22 22 23 24 24 24 24 Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the \"center\": the mean or the median? x ¯ = 5 , 000 , 000 + 49 ( 30 , 000 ) 50 = 129,400 M = 30,000 (There are 49 people who earn $30,000 and one person who earns $5,000,000.) The median is a better measure of the \"center\" than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data. Try It In a sample of 60 households, one house is worth $2,500,000. Twenty-nine houses are worth $280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median? Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal. Statistics exam scores for 20 students are as follows: 50 53 59 59 63 63 72 72 72 72 72 76 78 81 83 84 84 84 90 93 Find the mode. The most frequent score is 72, which occurs five times. Mode = 72. Try It The number of books checked out from the library from 25 students are as follows: 0 0 0 1 2 3 3 4 4 5 5 7 7 7 7 8 8 8 9 10 10 11 11 12 12 Find the mode. Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice. When is the mode the best measure of the \"center\"? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing. NOTE The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red. Try It Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is $50,000 and the mean is $47,500. What would be the best measure of the “center”? Calculating the Arithmetic Mean of Grouped Frequency Tables When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean: mean = data sum number of data values We simply need to modify the definition to fit within the restrictions of a frequency table. Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is lower boundary + upper boundary 2 . We can now modify the mean definition to be Mean of Frequency Table = ∑ f m ∑ f where f = the frequency of the interval and m = the midpoint of the interval. A frequency table displaying professor Blount’s last statistic test is shown. Find the best estimate of the class mean. Grade interval Number of students 50–56.5 1 56.5–62.5 0 62.5–68.5 4 68.5–74.5 4 74.5–80.5 2 80.5–86.5 3 86.5–92.5 4 92.5–98.5 1 Find the midpoints for all intervals Grade interval Midpoint 50–56.5 53.25 56.5–62.5 59.5 62.5–68.5 65.5 68.5–74.5 71.5 74.5–80.5 77.5 80.5–86.5 83.5 86.5–92.5 89.5 92.5–98.5 95.5 Calculate the sum of the product of each interval frequency and midpoint. ∑ ​ f m 53.25 ( 1 ) + 59.5 ( 0 ) + 65.5 ( 4 ) + 71.5 ( 4 ) + 77.5 ( 2 ) + 83.5 ( 3 ) + 89.5 ( 4 ) + 95.5 ( 1 ) = 1460.25 μ = ∑ f m ∑ f = 1460.25 19 = 76.86 Try It A researcher conducted a study on the effect that playing video games has on memory recall. As part of the study, they compiled the following data: Hours teenagers spend on video games Number of teenagers 0–3.5 3 3.5–7.5 7 7.5–11.5 12 11.5–15.5 7 15.5–19.5 9 What is the best estimate for the mean number of hours spent playing video games? Find the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagers The midpoints are 1.75, 5.5, 9.5, 13.5,17.5. Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75/38 = 10.78 References Data from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013). “Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013). Chapter Review The mean and the median can be calculated to help you find the \"center\" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set. Formula Review μ = ∑ f m ∑ f Where f = interval frequencies and m = interval midpoints. The arithmetic mean for a sample (denoted by x ¯ ) is x ¯ = Sum of all values in the sample Number of values in the sample The arithmetic mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population Find the mean for the following frequency tables. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Daily low temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 Points per game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2 Use the following information to answer the next three exercises: The following data show the lengths of boats moored in a marina. The data are ordered from smallest to largest: 16 17 19 20 20 21 23 24 25 25 25 26 26 27 27 27 28 29 30 32 33 33 34 35 37 39 40 Calculate the mean. Mean: 16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 + 34 + 35 + 37 + 39 + 40 = 738; 738 27 = 27.33 Identify the median. Identify the mode. The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27 Use the following information to answer the next three exercises: Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Calculate the following: sample mean = x ¯ = _______ median = _______ 4 mode = _______ Homework The countries with the highest rates of obesity in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table. Percent of population Number of countries 11.4–20.45 29 20.45–29.45 13 29.45–38.45 4 38.45–47.45 0 47.45–56.45 2 56.45–65.45 1 65.45–74.45 0 74.45–83.45 1 What is the best estimate of the average obesity percentage for these countries? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How does the United States compare to other countries? gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children? Percent of underweight children Number of countries 16–21.45 23 21.45–26.9 4 26.9–32.35 9 32.35–37.8 7 37.8–43.25 6 43.25–48.7 1 The mean percentage, x – = 1328.65 50 = 26.57 Bringing It Together Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information. Javier Ercilia x ¯ 6.0 miles 6.0 miles s 4.0 miles 7.0 miles How can you determine which survey was correct ? Explain what the difference in the results of the surveys implies about the data. If the two histograms depict the distribution of values for each supervisor, which one depicts Ercilia's sample? How do you know? Use the following information to answer the next three exercises : We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section. Number of years Frequency Number of years Frequency 7 1 22 1 14 3 23 1 15 1 26 1 18 1 40 2 19 4 42 2 20 3 Total = 20 What is the IQR ? 8 11 15 35 a What is the mode? 19 19.5 14 and 20 22.65 Is this a sample or the entire population? sample entire population neither b Frequency Table a data representation in which grouped data is displayed along with the corresponding frequencies Mean (arithmetic) a number that measures the central tendency of the data; a common name for mean is 'average.' The term 'mean' is a shortened form of 'arithmetic mean.' By definition, the mean for a sample (denoted by x ¯ ) is x ¯ = Sum of all values in the sample Number of values in the sample , and the mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population . Mean (geometric) a measure of central tendency that provides a measure of average geometric growth over multiple time periods. Median a number that separates ordered data into halves; half the values are the same number or smaller than the median and half the values are the same number or larger than the median. The median may or may not be part of the data. Midpoint the mean of an interval in a frequency table Mode the value that appears most frequently in a set of data", "section": "Measures of the Center of the Data", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Sigma Notation and Calculating the Arithmetic Mean Formula for Population Mean μ = 1 N ∑ i = 1 N x i Formula for Sample Mean x – = 1 n ∑ i = 1 n x i This unit is here to remind you of material that you once studied and said at the time “I am sure that I will never need this!” Here are the formulas for a population mean and the sample mean. The Greek letter μ is the symbol for the population mean and x ¯ is the symbol for the sample mean. Both formulas have a mathematical symbol that tells us how to make the calculations. It is called Sigma notation because the symbol is the Greek capital letter sigma: Σ. Like all mathematical symbols it tells us what to do: just as the plus sign tells us to add and the x tells us to multiply. These are called mathematical operators. The Σ symbol tells us to add a specific list of numbers. Let’s say we have a sample of animals from the local animal shelter and we are interested in their average age. If we list each value, or observation, in a column, you can give each one an index number. The first number will be number 1 and the second number 2 and so on. Animal Age 1 9 2 1 3 8.5 4 10.5 5 10 6 8.5 7 12 8 8 9 1 10 9.5 Each observation represents a particular animal in the sample. Purr is animal number one and is a 9 year old cat, Toto is animal number 2 and is a 1 year old puppy and so on. To calculate the mean we are told by the formula to add up all these numbers, ages in this case, and then divide the sum by 10, the total number of animals in the sample. Animal number one, the cat Purr, is designated as X 1 , animal number 2, Toto, is designated as X 2 and so on through Dundee who is animal number 10 and is designated as X 10 . The i in the formula tells us which of the observations to add together. In this case it is X 1 through X 10 which is all of them. We know which ones to add by the indexing notation, the i = 1 and the n or capital N for the population. For this example the indexing notation would be i = 1 and because it is a sample we use a small n on the top of the Σ which would be 10. The standard deviation requires the same mathematical operator and so it would be helpful to recall this knowledge from your past. The sum of the ages is found to be 78 and dividing by 10 gives us the sample mean age as 7.8 years. A group of 10 children are on a scavenger hunt to find different color rocks. The results are shown in the below. The column on the right shows the number of colors of rocks each child has. What is the mean number of rocks? Child Rock colors 1 5 2 5 3 6 4 2 5 4 6 3 7 7 8 2 9 1 10 10 A group of children are measured to determine the average height of the group. The results are in below. What is the mean height of the group to the nearest hundredth of an inch? Child Height in inches Adam 45.21 Betina 39.45 Chen 43.78 Donna 48.76 Edhas 37.39 Fran 39.90 George 45.56 Heather 46.24 39.48 in. A person compares prices for five automobiles. The results are in . What is the mean price of the cars the person has considered? Price $20,987 $22,008 $19,998 $23,433 $21,444 $21,574 A customer protection service has obtained 8 bags of candy that are supposed to contain 16 ounces of candy each. The candy is weighed to determine if the average weight is at least the claimed 16 ounces. The results are in given in . What is the mean weight of a bag of candy in the sample? Weight in ounces 15.65 16.09 16.01 15.99 16.02 16.00 15.98 16.08 15.98 ounces A teacher records grades for a class of 70, 72, 79, 81, 82, 82, 83, 90, and 95. What is the mean of these grades? 81.56 A family is polled to see the mean of the number of hours per day the television set is on. The results, starting with Sunday, are 6, 3, 2, 3, 1, 3, and 7 hours. What is the average number of hours the family had the television set on to the nearest whole number? 4 hours A city received the following rainfall for a recent year. What is the mean number of inches of rainfall the city received monthly, to the nearest hundredth of an inch? Use . Month Rainfall in inches January 2.21 February 3.12 March 4.11 April 2.09 May 0.99 June 1.08 July 2.99 August 0.08 September 0.52 October 1.89 November 2.00 December 3.06 2.01 inches A football team scored the following points in its first 8 games of the new season. Starting at game 1 and in order the scores are 14, 14, 24, 21, 7, 0, 38, and 28. What is the mean number of points the team scored in these eight games? 18.25 Homework A sample of 10 prices is chosen from a population of 100 similar items. The values obtained from the sample, and the values for the population, are given in and respectively. Is the mean of the sample within $1 of the population mean? What is the difference in the sample and population means? Prices of the sample $21 $23 $21 $24 $22 $22 $25 $21 $20 $24 Prices of the population Frequency $20 20 $21 35 $22 15 $23 10 $24 18 $25 2 Yes The sample is 0.5 higher. A standardized test is given to ten people at the beginning of the school year with the results given in below. At the end of the year the same people were again tested. What is the average improvement? Does it matter if the means are subtracted, or if the individual values are subtracted? Student Beginning score Ending score 1 1100 1120 2 980 1030 3 1200 1208 4 998 1000 5 893 948 6 1015 1030 7 1217 1224 8 1232 1245 9 967 988 10 988 997 20 No A small class of 7 students has a mean grade of 82 on a test. If six of the grades are 80, 82,86, 90, 90, and 95, what is the other grade? 51 A class of 20 students has a mean grade of 80 on a test. Nineteen of the students has a mean grade between 79 and 82, inclusive. What is the lowest possible grade of the other student? What is the highest possible grade of the other student? 42 99 If the mean of 20 prices is $10.39, and 5 of the items with a mean of $10.99 are sampled, what is the mean of the other 15 prices? $10.19", "section": "Sigma Notation and Calculating the Arithmetic Mean", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Geometric Mean The mean (Arithmetic), median and mode are all measures of the “center” of the data, the “average”. They are all in their own way trying to measure the “common” point within the data, that which is “normal”. In the case of the arithmetic mean this is solved by finding the value from which all points are equal linear distances. We can imagine that all the data values are combined through addition and then distributed back to each data point in equal amounts. The sum of all the values is what is redistributed in equal amounts such that the total sum remains the same. The geometric mean redistributes not the sum of the values but the product of multiplying all the individual values and then redistributing them in equal portions such that the total product remains the same. This can be seen from the formula for the geometric mean, x ~ : (Pronounced x-tilde) x ~ = ( ∏ i = 1 n x i ) 1 n = x 1 · x 2 ··· x n n = ( x 1 · x 2 ··· x n ) 1 n where π is another mathematical operator, that tells us to multiply all the x i numbers in the same way capital Greek sigma tells us to add all the x i numbers. Remember that a fractional exponent is calling for the nth root of the number thus an exponent of 1/3 is the cube root of the number. The geometric mean answers the question, \"if all the quantities had the same value, what would that value have to be in order to achieve the same product?” The geometric mean gets its name from the fact that when redistributed in this way the sides form a geometric shape for which all sides have the same length. To see this, take the example of the numbers 10, 51.2 and 8. The geometric mean is the product of multiplying these three numbers together (4,096) and taking the cube root because there are three numbers among which this product is to be distributed. Thus the geometric mean of these three numbers is 16. This describes a cube 16x16x16 and has a volume of 4,096 units. The geometric mean is relevant in Economics and Finance for dealing with growth: growth of markets, in investment, population and other variables the growth in which there is an interest. Imagine that our box of 4,096 units (perhaps dollars) is the value of an investment after three years and that the investment returns in percents were the three numbers in our example. The geometric mean will provide us with the answer to the question, what is the average rate of return: 16 percent. The arithmetic mean of these three numbers is 23.6 percent. The reason for this difference, 16 versus 23.6, is that the arithmetic mean is additive and thus does not account for the interest on the interest, compound interest, embedded in the investment growth process. The same issue arises when asking for the average rate of growth of a population or sales or market penetration, etc., knowing the annual rates of growth. The formula for the geometric mean rate of return, or any other growth rate, is: r s = ( x 1 · x 2 ··· x n ) 1 n - 1 Manipulating the formula for the geometric mean can also provide a calculation of the average rate of growth between two periods knowing only the initial value a 0 and the ending value a n and the number of periods, n . The following formula provides this information: ( a n a 0 ) 1 n = x ~ Finally, we note that the formula for the geometric mean requires that all numbers be positive, greater than zero. The reason of course is that the root of a negative number is undefined for use outside of mathematical theory. There are ways to avoid this problem however. In the case of rates of return and other simple growth problems we can convert the negative values to meaningful positive equivalent values. Imagine that the annual returns for the past three years are +12%, -8%, and +2%. Using the decimal multiplier equivalents of 1.12, 0.92, and 1.02, allows us to compute a geometric mean of 1.0167. Subtracting 1 from this value gives the geometric mean of +1.67% as a net rate of population growth (or financial return). From this example we can see that the geometric mean provides us with this formula for calculating the geometric (mean) rate of return for a series of annual rates of return: r s = x ~ - 1 where r s is average rate of return and x ~ is the geometric mean of the returns during some number of time periods. Note that the length of each time period must be the same. As a general rule one should convert the percent values to its decimal equivalent multiplier. It is important to recognize that when dealing with percents, the geometric mean of percent values does not equal the geometric mean of the decimal multiplier equivalents and it is the decimal multiplier equivalent geometric mean that is relevant. Formula Review The Geometric Mean: x ~ = ( ∏ i = 1 n x i ) 1 n = x 1 · x 2 ··· x n n = ( x 1 · x 2 ··· x n ) 1 n What is the geometric mean of the data set given? 5, 10, 20 10 What is the geometric mean of the data set given? 9.000, 15.00, 21.00 14.15 What is the geometric mean of the data set given? 7.0, 10.0, 39.2 14 What is the geometric mean of the data set given? 17.00, 10.00, 19.00 14.78 What is the average rate of return for the values that follow? 1.0, 2.0, 1.5 44% What is the average rate of return for the values that follow? 0.80, 2.0, 5.0 100% What is the average rate of return for the values that follow? 0.90, 1.1, 1.2 6% What is the average rate of return for the values that follow? 4.2, 4.3, 4.5 33% Homework An investment grows from $10,000 to $22,000 in five years. What is the average rate of return? 17% An initial investment of $20,000 grows at a rate of 9% for five years. What is its final value? $30,772.48 A culture contains 1,300 bacteria. The bacteria grow to 2,000 in 10 hours. What is the rate at which the bacteria grow per hour to the nearest tenth of a percent? 4.4% An investment of $3,000 grows at a rate of 5% for one year, then at a rate of 8% for three years. What is the average rate of return to the nearest hundredth of a percent? 7.24% An investment of $10,000 goes down to $9,500 in four years. What is the average return per year to the nearest hundredth of a percent? -1.27%", "section": "Geometric Mean", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Skewness and the Mean, Median, and Mode Consider the following data set. 4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10 This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval. The histogram in displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. The histogram in for the data: 4; 5; 6; 6; 6; 7; 7; 7; 7; 8 is not symmetrical. The right-hand side seems \"chopped off\" compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left. We can formally measure the skewness of a distribution just as we can mathematically measure the center weight of the data or its general \"speadness\". The mathematical formula for skewness is: a 3 = ∑ ( x i − x – ) 3 n s 3 . The greater the deviation from zero indicates a greater degree of skewness. If the skewness is negative then the distribution is skewed left as in . A positive measure of skewness indicates right skewness such as . The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so. The histogram for the data: 6 7 7 7 7 8 8 8 9 10 shown in , is also not symmetrical. It is skewed to the right . The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest . Again, the mean reflects the skewing the most. The mean is affected by outliers that do not influence the median. Therefore, when the distribution of data is skewed to the left, the mean is often less than the median. When the distribution is skewed to the right, the mean is often greater than the median. In symmetric distributions, we expect the mean and median to be approximately equal in value. This is an important connection between the shape of the distribution and the relationship of the mean and median. It is not, however, true for every data set. The most common exceptions occur in sets of discrete data. As with the mean, median and mode, and as we will see shortly, the variance, there are mathematical formulas that give us precise measures of these characteristics of the distribution of the data. Again looking at the formula for skewness we see that this is a relationship between the mean of the data and the individual observations cubed. a 3 = ∑ ( x i − x – ) 3 n s 3 where s is the sample standard deviation of the data, X i , and x ¯ is the arithmetic mean and n is the sample size. Formally the arithmetic mean is known as the first moment of the distribution. The second moment we will see is the variance, and skewness is the third moment. The variance measures the squared differences of the data from the mean and skewness measures the cubed differences of the data from the mean. While a variance can never be a negative number, the measure of skewness can and this is how we determine if the data are skewed right of left. The skewness for a normal distribution is zero, and any symmetric data should have skewness near zero. Negative values for the skewness indicate data that are skewed left and positive values for the skewness indicate data that are skewed right. By skewed left, we mean that the left tail is long relative to the right tail. Similarly, skewed right means that the right tail is long relative to the left tail. The skewness characterizes the degree of asymmetry of a distribution around its mean. While the mean and standard deviation are dimensional quantities (this is why we will take the square root of the variance ) that is, have the same units as the measured quantities X i , the skewness is conventionally defined in such a way as to make it nondimensional . It is a pure number that characterizes only the shape of the distribution. A positive value of skewness signifies a distribution with an asymmetric tail extending out towards more positive X and a negative value signifies a distribution whose tail extends out towards more negative X. A zero measure of skewness will indicate a symmetrical distribution. Skewness and symmetry become important when we discuss probability distributions in later chapters. Chapter Review Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A left (or negative) skewed distribution has a shape like . A right (or positive) skewed distribution has a shape like . A symmetrical distribution looks like . Formula Review Formula for skewness: a 3 = ∑ ( x i − x ¯ ) 3 n s 3 Formula for Coefficient of Variation: C V = s x ¯ · 100 conditioned upon x ¯ ≠ 0 Use the following information to answer the next three exercises: State whether the data are symmetrical, skewed to the left, or skewed to the right. 1 1 1 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 5 5 The data are symmetrical. The median is 3 and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical. 16 17 19 22 22 22 22 22 23 87 87 87 87 87 88 89 89 90 91 The data are skewed right. The median is 87.5 and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right. When the data are skewed left, what is the typical relationship between the mean and median? When the data are symmetrical, what is the typical relationship between the mean and median? When the data are symmetrical, the mean and median are close or the same. What word describes a distribution that has two modes? Describe the shape of this distribution. The distribution is skewed right because it looks pulled out to the right. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean is 4.1 and is slightly greater than the median, which is four. Describe the shape of this distribution. Describe the relationship between the mode and the median of this distribution. The mode and the median are the same. In this case, they are both five. Are the mean and the median the exact same in this distribution? Why or why not? Describe the shape of this distribution. The distribution is skewed left because it looks pulled out to the left. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean and the median are both six. The mean and median for the data are the same. 3 4 5 5 6 6 6 6 7 7 7 7 7 7 7 Is the data perfectly symmetrical? Why or why not? Which is the greatest, the mean, the mode, or the median of the data set? 11 11 12 12 12 12 13 15 17 22 22 22 The mode is 12, the median is 12.5, and the mean is 15.1. The mean is the largest. Which is the least, the mean, the mode, and the median of the data set? 56 56 56 58 59 60 62 64 64 65 67 Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why? The mean tends to reflect skewing the most because it is affected the most by outliers. In a perfectly symmetrical distribution, when would the mode be different from the mean and median? Homework The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years. What does it mean for the median age to rise? Give two reasons why the median age could rise. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?", "section": "Skewness and the Mean, Median, and Mode", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Measures of the Spread of the Data An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean. The standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean. The standard deviation provides a measure of the overall variation in a data set The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation. Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B . The average wait time at both supermarkets is five minutes. At supermarket A , the standard deviation for the wait time is two minutes; at supermarket B . The standard deviation for the wait time is four minutes. Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B . Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average. Calculating the Standard Deviation If x is a number, then the difference \" x minus the mean\" is called its deviation . In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is x – μ . For sample data, in symbols a deviation is x – x – . The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ . To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – x – values for a sample, or the x – μ values for a population). The symbol σ 2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s 2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations. Formally, the variance is the second moment of the distribution or the first moment around the mean. Remember that the mean is the first moment of the distribution. If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N , the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1 , one less than the number of items in the sample. Formulas for the Sample Standard Deviation s = Σ ( x − x – ) 2 n − 1 or s = Σ f ( x − x – ) 2 n − 1 or s = ( ∑ i = 1 n x 2 ) - n x ¯ 2 n - 1 For the sample standard deviation, the denominator is n - 1 , that is the sample size minus 1. Formulas for the Population Standard Deviation σ = Σ ( x − μ ) 2 N or σ = Σ f ( x – μ ) 2 N or σ = ∑ i = 1 N x i 2 N - μ 2 For the population standard deviation, the denominator is N , the number of items in the population. In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three. Two important observations concerning the variance and standard deviation: the deviations are measured from the mean and the deviations are squared. In principle, the deviations could be measured from any point, however, our interest is measurement from the center weight of the data, what is the \"normal\" or most usual value of the observation. Later we will be trying to measure the \"unusualness\" of an observation or a sample mean and thus we need a measure from the mean. The second observation is that the deviations are squared. This does two things, first it makes the deviations all positive and second it changes the units of measurement from that of the mean and the original observations. If the data are weights then the mean is measured in pounds, but the variance is measured in pounds-squared. One reason to use the standard deviation is to return to the original units of measurement by taking the square root of the variance. Further, when the deviations are squared it explodes their value. For example, a deviation of 10 from the mean when squared is 100, but a deviation of 100 from the mean is 10,000. What this does is place great weight on outliers when calculating the variance. Types of Variability in Samples When trying to study a population, a sample is often used, either for convenience or because it is not possible to access the entire population. Variability is the term used to describe the differences that may occur in these outcomes. Common types of variability include the following: Observational or measurement variability Natural variability Induced variability Sample variability Here are some examples to describe each type of variability. Example 1: Measurement variability Measurement variability occurs when there are differences in the instruments used to measure or in the people using those instruments. If we are gathering data on how long it takes for a ball to drop from a height by having students measure the time of the drop with a stopwatch, we may experience measurement variability if the two stopwatches used were made by different manufacturers: For example, one stopwatch measures to the nearest second, whereas the other one measures to the nearest tenth of a second. We also may experience measurement variability because two different people are gathering the data. Their reaction times in pressing the button on the stopwatch may differ; thus, the outcomes will vary accordingly. The differences in outcomes may be affected by measurement variability. Example 2: Natural variability Natural variability arises from the differences that naturally occur because members of a population differ from each other. For example, if we have two identical corn plants and we expose both plants to the same amount of water and sunlight, they may still grow at different rates simply because they are two different corn plants. The difference in outcomes may be explained by natural variability. Example 3: Induced variability Induced variability is the counterpart to natural variability; this occurs because we have artificially induced an element of variation (that, by definition, was not present naturally): For example, we assign people to two different groups to study memory, and we induce a variable in one group by limiting the amount of sleep they get. The difference in outcomes may be affected by induced variability. Example 4: Sample variability Sample variability occurs when multiple random samples are taken from the same population. For example, if I conduct four surveys of 50 people randomly selected from a given population, the differences in outcomes may be affected by sample variability. In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of the students. The following data are the ages for a SAMPLE of n = 20 fifth grade students. The ages are rounded to the nearest half year: 9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; x ¯ = 9 + 9 .5(2) + 10(4) + 10 .5(4) + 11(6) + 11 .5(3) 20 = 10.525 The average age is 10.53 years, rounded to two places. The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s . Data Freq. Deviations Deviations 2 (Freq.)(Deviations 2 ) x f ( x – x – ) ( x – x – ) 2 ( f )( x – x – ) 2 9 1 9 – 10.525 = –1.525 (–1.525) 2 = 2.325625 1 × 2.325625 = 2.325625 9.5 2 9.5 – 10.525 = –1.025 (–1.025) 2 = 1.050625 2 × 1.050625 = 2.101250 10 4 10 – 10.525 = –0.525 (–0.525) 2 = 0.275625 4 × 0.275625 = 1.1025 10.5 4 10.5 – 10.525 = –0.025 (–0.025) 2 = 0.000625 4 × 0.000625 = 0.0025 11 6 11 – 10.525 = 0.475 (0.475) 2 = 0.225625 6 × 0.225625 = 1.35375 11.5 3 11.5 – 10.525 = 0.975 (0.975) 2 = 0.950625 3 × 0.950625 = 2.851875 The total is 9.7375 The sample variance, s 2 , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s 2 = 9.7375 20 − 1 = 0.5125 The sample standard deviation s is equal to the square root of the sample variance: s = 0.5125 = 0.715891 , which is rounded to two decimal places, s = 0.72. Try It On a baseball team, the ages of each of the players are as follows: 21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40 Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean. Explanation of the standard deviation calculation shown in the table The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero . (For , there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation. By squaring the deviations we are placing an extreme penalty on observations that are far from the mean; these observations get greater weight in the calculations of the variance. We will see later on that the variance (standard deviation) plays the critical role in determining our conclusions in inferential statistics. We can begin now by using the standard deviation as a measure of \"unusualness.\" \"How did you do on the test?\" \"Terrific! Two standard deviations above the mean.\" This, we will see, is an unusually good exam grade. The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data. Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one ( n – 1). Why not divide by n ? The answer has to do with the population variance. The sample variance is an estimate of the population variance. This estimate requires us to use an estimate of the population mean rather than the actual population mean. Based on the theoretical mathematics that lies behind these calculations, dividing by ( n – 1) gives a better estimate of the population variance. The standard deviation, s or σ , is either zero or larger than zero. Describing the data with reference to the spread is called \"variability\". The variability in data depends upon the method by which the outcomes are obtained; for example, by measuring or by random sampling. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large. Use the following data (first exam scores) from Professor Dean's spring pre-calculus class: 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. Calculate the following to one decimal place: The sample mean The sample standard deviation The median The first quartile The third quartile IQR See The sample mean = 73.5 The sample standard deviation = 17.9 The median = 73 The first quartile = 61 The third quartile = 90 IQR = 90 – 61 = 29 Data Frequency Relative frequency Cumulative relative frequency 33 1 0.032 0.032 42 1 0.032 0.064 49 2 0.065 0.129 53 1 0.032 0.161 55 2 0.065 0.226 61 1 0.032 0.258 63 1 0.032 0.29 67 1 0.032 0.322 68 2 0.065 0.387 69 2 0.065 0.452 72 1 0.032 0.484 73 1 0.032 0.516 74 1 0.032 0.548 78 1 0.032 0.580 80 1 0.032 0.612 83 1 0.032 0.644 88 3 0.097 0.741 90 1 0.032 0.773 92 1 0.032 0.805 94 4 0.129 0.934 96 1 0.032 0.966 100 1 0.032 0.998 (Why isn't this value 1? ANSWER: Rounding) Try It The following data show the different types of pet food stores in the area carry. 6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12; Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator. Standard deviation of Grouped Frequency Tables Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: M e a n o f F r e q u e n c y T a b l e = ∑ f m ∑ f where f = interval frequencies and m = interval midpoints. Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean. Find the standard deviation for the data in . Class Frequency, f Midpoint, m f · m f ( m - x ¯ ) 2 0–2 1 1 1 · 1 = 1 1 ( 1 - 6.88 ) 2 = 34.57 3–5 6 4 6 · 4 = 24 6 ( 4 - 6.88 ) 2 = 49.77 6-8 10 7 10 · 7 = 70 10 ( 7 - 6.88 ) 2 = 0.14 9-11 7 10 7 · 10 = 70 7 ( 10 - 6.88 ) 2 = 68.14 12-14 0 13 0 · 13 = 0 0 ( 13 - 6.88 ) 2 = 0 n = 24 x ¯ = 165 24 = 6.88 s 2 = 152.62 24 - 1 = 6.64 For this data set, we have the mean, x – = 6.88 and the standard deviation, s x = 2.58. This means that a randomly selected data value would be expected to be 2.58 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost three standard deviations from the mean. While the formula for calculating the standard deviation is not complicated, s x = Σ ( m − x – ) 2 f n − 1 where s x = sample standard deviation, x – = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations. Try It Find the standard deviation for the data from the previous example Class Frequency, f 0–2 1 3–5 6 6–8 10 9–11 7 12–14 0 15–17 2 Comparing Values from Different Data Sets The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading. For each data value x, calculate how many standard deviations away from its mean the value is. Use the formula: x = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs. # o f S T D E V s = x – mean standard deviation Compare the results of this calculation. #ofSTDEVs is often called a \" z -score\"; we can use the symbol z . In symbols, the formulas become: Sample x = x ¯ + zs z = x − x – s Population x = μ + zσ z = x − μ σ Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school? Student GPA School mean GPA School standard deviation John 2.85 3.0 0.7 Ali 77 80 10 For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer. z = # of STDEVs = value – mean standard deviation = x - μ σ For John, z = # o f S T D E V s = 2.85 - 3.0 0.7 = - 0.21 For Ali, z = # o f S T D E V s = 77 - 80 10 = - 0.3 John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean. John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. Try It Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team? Swimmer Time (seconds) Team mean time Team standard deviation Angie 26.2 27.2 0.8 Beth 27.3 30.1 1.4 For Angie: z = 26 .2 – 27 .2 0 .8 = –1.25 For Beth: z = 27 .3 – 30 .1 1. 4 = –2 The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data. For ANY data set, no matter what the distribution of the data is: At least 75% of the data is within two standard deviations of the mean. At least 89% of the data is within three standard deviations of the mean. At least 95% of the data is within 4.5 standard deviations of the mean. This is known as Chebyshev's Rule. For data having a Normal Distribution, which we will examine in great detail later: Approximately 68% of the data is within one standard deviation of the mean. Approximately 95% of the data is within two standard deviations of the mean. More than 99% of the data is within three standard deviations of the mean. This is known as the Empirical Rule. It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the \"Normal\" or \"Gaussian\" probability distribution in later chapters. Coefficient of Variation Another useful way to compare distributions besides simple comparisons of means or standard deviations is to adjust for differences in the scale of the data being measured. Quite simply, a large variation in data with a large mean is different than the same variation in data with a small mean. To adjust for the scale of the underlying data the Coefficient of Variation (CV) has been developed. Mathematically: C V = s x – * 100 conditioned upon x – ≠ 0, where s is the standard deviation of the data and x – is the mean. We can see that this measures the variability of the underlying data as a percentage of the mean value; the center weight of the data set. This measure is useful in comparing risk where an adjustment is warranted because of differences in scale of two data sets. In effect, the scale is changed to common scale, percentage differences, and allows direct comparison of the two or more magnitudes of variation of different data sets. References Data from Microsoft Bookshelf. King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013). Chapter Review The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population. The Standard Deviation allows us to compare individual data or classes to the data set mean numerically. s = ∑ ​ ( x − x – ) 2 n − 1 or s = ∑ ​ f ( x − x – ) 2 n − 1 is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, μ , and the formula σ = ∑ ​ ( x − μ ) 2 N or σ = ∑ ​ f ( x − μ ) 2 N . Formula Review s x = ∑ f m 2 n − x – 2 where s x = sample standard deviation x – = sample mean Formulas for Sample Standard Deviation s = Σ ( x − x – ) 2 n − 1 or s = Σ f ( x − x – ) 2 n − 1 or s = ( ∑ i = 1 n x 2 ) - n x ¯ 2 n - 1 For the sample standard deviation, the denominator is n - 1 , that is the sample size - 1. Formulas for Population Standard Deviation σ = Σ ( x − μ ) 2 N or σ = Σ f ( x – μ ) 2 N or σ = ∑ i = 1 N x i 2 N - μ 2 For the population standard deviation, the denominator is N , the number of items in the population. Use the following information to answer the next two exercises : The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles. 29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150 Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth. s = 34.5 Find the value that is one standard deviation below the mean. Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team? Baseball player Batting average Team batting average Team standard deviation Fredo 0.158 0.166 0.012 Karl 0.177 0.189 0.015 For Fredo: z = 0.158 – 0.166 0.012 = –0.67 For Karl: z = 0.177 – 0.189 0.015 = –0.8 Fredo’s z -score of –0.67 is higher than Karl’s z -score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team. Use to find the value that is three standard deviations: a above the mean b below the mean Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84 . Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Daily low temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 Points per game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2 s x = ∑ f m 2 n − x – 2 = 193157.45 30 − 79.5 2 = 10.88 s x = ∑ f m 2 n − x – 2 = 380945.3 101 − 60.94 2 = 7.62 s x = ∑ f m 2 n − x – 2 = 440051.5 86 − 70.66 2 = 11.14 Homework Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College over a 29-year period. μ = 1000 FTES median = 1,014 FTES σ = 474 FTES first quartile = 528.5 FTES third quartile = 1,447.5 FTES n = 29 years A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer. The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or above the median. 75% of all years have an FTES: at or below: _____ at or above: _____ The population standard deviation = _____ 474 FTES What percent of the FTES were from 528.5 to 1447.5? How do you know? What is the IQR ? What does the IQR represent? 919 How many standard deviations away from the mean is the median? Additional Information: The population for a specific six-year period was given in an updated report. The data are reported here. Year 1 2 3 4 5 6 Total FTES 1,585 1,690 1,735 1,935 2,021 1,890 Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR . Round to one decimal place. mean = 1,809.3 median = 1,812.5 standard deviation = 151.2 first quartile = 1,690 third quartile = 1,935 IQR = 245 Compare the IQR for the FTES for the previous 29-year period with the IQR for the six-year period shown in . Why do you suppose the IQR s are so different? Hint: Think about the number of years covered by each time period and what happened to higher education during those periods. Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer. Student GPA School Average GPA School Standard Deviation Thuy 2.7 3.2 0.8 Vichet 87 75 20 Kamala 8.6 8 0.4 A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer. For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type. An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he? Who is the fastest runner with respect to their class? Explain why. A selection of countries has poverty rates that range from 11.4% to 74.6%. This data is summarized in Table 14 . Percent of population Number of countries 11.4–20.45 29 20.45–29.45 13 29.45–38.45 4 38.45–47.45 0 47.45–56.45 2 56.45–65.45 1 65.45–74.45 0 74.45–83.45 1 What is the best estimate of the average poverty percentage for these countries? What is the standard deviation for the listed poverty rates? The United States has an average poverty rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ poverty rate compared to the average rate? Explain. x – = 23.32 Using the TI 83/84, we obtain a standard deviation of: s x = 12.95. The poverty rate of the United States is 10.58% higher than the average poverty rate. Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the poverty percentage that is one standard deviation from the mean. The United States poverty rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States does not have an unusually high percentage of people experiencing poverty. gives the percent of children under five considered to be underweight. Percent of underweight children Number of countries 16–21.45 23 21.45–26.9 4 26.9–32.35 9 32.35–37.8 7 37.8–43.25 6 43.25–48.7 1 What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain. Bringing It Together Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: # of movies Frequency 0 5 1 9 2 6 3 4 4 1 Find the sample mean x ¯ . Find the approximate sample standard deviation, s . 1.48 1.12 Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows: X Frequency 1 2 2 5 3 8 4 12 5 12 6 0 7 1 Find the sample mean x ¯ Find the sample standard deviation, s Construct a histogram of the data. Complete the columns of the chart. Find the first quartile. Find the median. Find the third quartile. What percent of the students owned at least five pairs? Find the 40 th percentile. Find the 90 th percentile. Construct a line graph of the data Construct a stemplot of the data Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year. 177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265 Organize the data from smallest to largest value. Find the median. Find the first quartile. Find the third quartile. The middle 50% of the weights are from _______ to _______. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? Assume the population was the San Francisco 49ers. Find: the population mean, μ . the population standard deviation, σ . the weight that is two standard deviations below the mean. When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he? That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer? 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302 241 205.5 272.5 205.5, 272.5 sample 236.34 37.50 161.34 0.84 std. dev. below the mean Young One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows: 3 8 –1 2 0 5 –3 1 –1 6 5 –2 What is the mean change score? What is the standard deviation for this population? What is the median change score? Find the change score that is 2.2 standard deviations below the mean. Refer to determine which of the following are true and which are false. Explain your solution to each part in complete sentences. The medians for both graphs are the same. We cannot determine if any of the means for both graphs is different. The standard deviation for graph b is larger than the standard deviation for graph a. We cannot determine if any of the third quartiles for both graphs is different. True True True False In a recent issue of the IEEE Spectrum , 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference. Organize the data in a chart. Find the median, the first quartile, and the third quartile. Find the 65 th percentile. Find the 10 th percentile. The middle 50% of the conferences last from _______ days to _______ days. Calculate the sample mean of days of engineering conferences. Calculate the sample standard deviation of days of engineering conferences. Find the mode. If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences. A survey of enrollment at 35 community colleges across the United States yielded the following figures: 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622 Organize the data into a chart with six intervals of equal width from 0 to 30,000. Label the two columns \"Enrollment\" and \"Frequency.\" Construct a histogram of the data. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean? Calculate the sample mean. Calculate the sample standard deviation. A school with an enrollment of 8000 would be how many standard deviations away from the mean? Enrollment Frequency 1000-5000 10 5000-10000 16 10000-15000 3 15000-20000 3 20000-25000 1 25000-30000 2 Answers may vary. mode 8628.74 6943.88 –0.09 Use the following information to answer the next two exercises. X = the number of days per week that 100 clients use a particular exercise facility. x Frequency 0 3 1 12 2 33 3 28 4 11 5 9 6 4 The 80 th percentile is _____ 5 80 3 4 The number that is 1.5 standard deviations BELOW the mean is approximately _____ 0.7 4.8 –2.8 Cannot be determined a Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the . # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion. If a data value is identified as an outlier, what should be done about it? Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.) Do parts a and c of this problem give the same answer? Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data? Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode? Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Variance mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as x – x – where x is a value of the data and x – is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.", "section": "Measures of the Spread of the Data", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction Meteor showers are rare, but the probability of them occurring can be calculated. (credit: modification of work “A meteor during the peak of the 2009 Leonid Meteor Shower” by Navicore/ Wikimedia Commons, CC BY 3.0) It is often necessary to \"guess\" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs. You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Terminology Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment. A result of an experiment is called an outcome . The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = { H , T } where H = heads and T = tails are the outcomes. An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P ( A ). The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P ( A ) = 0 means the event A can never happen. P ( A ) = 1 means the event A always happens. P ( A ) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads). Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair , six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head ( H ) and a Tail ( T ) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer. To calculate the probability of an event A when all outcomes in the sample space are equally likely , count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is { HH , TH , HT , TT } where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition { HT , TH }, so P ( A ) = 2 4 = 0.5. Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P ( E ) = 2 6 . If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 2 6 of the rolls would result in an outcome of \"at least five\". You would not expect exactly 2 6 . The long-term relative frequency of obtaining this result would approach the theoretical probability of 2 6 as the number of repetitions grows larger and larger. This important characteristic of probability experiments is known as the law of large numbers (discussed in detail later), which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.) It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair , or biased . Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely. \"OR\" Event: An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B . The event A OR B can also be written as A Union B, with notation as follows: A ∪ B For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. Then A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice. \"AND\" Event: An outcome is in the event A AND B if the outcome is in both A and B at the same time. The event A AND B can also be written as A Intersection B , with notation as follows: A ∩ B For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A ∩ B = {4, 5}. The complement of event A is denoted A′ (read \" A prime\"). A′ consists of all outcomes that are NOT in A . Notice that P ( A ) + P ( A′ ) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A′ = {5, 6}. P ( A ) = 4 6 , P ( A′ ) = 2 6 , and P ( A ) + P ( A′ ) = 4 6 + 2 6 = 1 The conditional probability of A given B is written P ( A | B ). P ( A | B ) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space . We calculate the probability of A from the reduced sample space B . The formula to calculate P ( A | B ) is P ( A | B ) = P ( A ∩ B ) P ( B ) where P ( B ) is greater than zero. For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P ( A | B ), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S ). We get the same result by using the formula. Remember that S has six outcomes. P ( A | B ) = P ( A ∩ B ) P ( B ) = ( the number of outcomes that are 2 or 3 and even in S ) 6 ( the number of outcomes that are even in S ) 6 = 1 6 3 6 = 1 3 Odds The odds of an event presents the probability as a ratio of success to failure. This is common in various gambling formats. Mathematically, the odds of an event can be defined as: P ( A ) 1 − P ( A ) where P ( A ) is the probability of success and of course 1 − P ( A ) is the probability of failure. Odds are always quoted as \"numerator to denominator,\" e.g. 2 to 1. Here the probability of winning is twice that of losing; thus, the probability of winning is 0.66. A probability of winning of 0.60 would generate odds in favor of winning of 3 to 2. While the calculation of odds can be useful in gambling venues in determining payoff amounts, it is not helpful for understanding probability or statistical theory. Understanding Terminology and Symbols It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any. The sample space S is the whole numbers starting at one and less than 20. S = _____________________________ Let event A = the even numbers and event B = numbers greater than 13. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A ∩ B = ____________________, A OR B = ________________ P ( A ∩ B ) = _________, P ( A ∪ B ) = _____________ A′ = _____________, P ( A′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19} P ( A ) = 9 19 , P ( B ) = 6 19 A ∩ B = {14,16,18}, A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19} P ( A ∩ B ) = 3 19 , P ( A ∪ B ) = 12 19 A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P ( A′ ) = 10 19 P ( A ) + P ( A′ ) = 1 ( 9 19 + 10 19 = 1) P ( A | B ) = P ( A ∩ B ) P ( B ) = 3 6 , P ( B | A ) = P ( A ∩ B ) P ( A ) = 3 9 , No Try It The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)). S = _____________________________ Let event A = the sum is even and event B = the first number is prime. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A ∩ B = ____________________, A ∪ B = ________________ P ( A ∩ B ) = _________, P ( A ∪ B ) = _____________ B′ = _____________, P ( B′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)} B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} P ( A ) = 1 2 , P ( B ) = 2 3 A ∩ B = {(2,2), (2,4), (3,1), (3,3)} A ∪ B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} P ( A ∩ B ) = 1 3 , P ( A ∪ B ) = 5 6 B′ = {(1,1), (1,2), (1,3), (1,4)}, P ( B′ ) = 1 3 P ( B ) + P ( B′ ) = 1 P ( A | B ) = P ( A ∩ B ) P ( B ) = 1 2 , P ( B | A ) = P ( A ∩ B ) P ( B ) = 2 3 , No. A fair, six-sided die is rolled. Describe the sample space S , identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up). Event T = the outcome is two. Event A = the outcome is an even number. Event B = the outcome is less than four. The complement of A . A | B B | A A ∩ B A ∪ B A ∪ B′ Event N = the outcome is a prime number. Event I = the outcome is seven. T = {2}, P ( T ) = 1 6 A = {2, 4, 6}, P ( A ) = 1 2 B = {1, 2, 3}, P ( B ) = 1 2 A′ = {1, 3, 5}, P ( A′ ) = 1 2 A | B = {2}, P ( A | B ) = 1 3 B | A = {2}, P ( B | A ) = 1 3 A ∩ B = {2}, P ( A ∩ B ) = 1 6 A ∪ B = {1, 2, 3, 4, 6}, P ( A ∪ B ) = 5 6 A ∪ B′ = {2, 4, 5, 6}, P ( A ∪ B′ ) = 2 3 N = {2, 3, 5}, P ( N ) = 1 2 A six-sided die does not have seven dots. P (7) = 0. Try It A number is randomly selected between 1 to 10. Describe the sample space S , identify each of the following events with a subset of S, and compute its probability (an outcome is the number selected). Event F = the outcome is 5. Event A = the number is more than 6. Event B = the number is odd. The complement of B . A GIVEN B . B GIVEN A . A OR B . A’ OR B . Event P = the outcome is a composite number. Event M = the number is a multiple of 3. F = 5 , P ( F ) = 1 10 A = 7 , 8 , 9 , P ( A ) = 3 10 B = 1 , 3 , 5 , 7 , 9 , P ( B ) = 5 10 B ' = 2 , 4 , 6 , 8 , 10 , P ( B ' ) = 5 10 A | B = 7 , 9 , P A | B = 2 5 B | A = 7 , 9 , P ( B | A ) = 1 2 A O R B = 1 , 3 , 5 , 7 , 8 , 9 , 10 , P ( A O R B ) = 7 10 A ' O R B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 9 , P ( A ' O R B ) = 4 5 P = 4 , 6 , 8 , 9 , 10 , P ( P ) = 1 2 M = 3 , 6 , 9 , P ( M ) = 3 10 describes the distribution of a random sample S of 100 individuals, organized by sex assigned at birth and whether they are right- or left-handed. Right-handed Left-handed Males 43 9 Females 44 4 Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities: P ( M ) P ( F ) P ( R ) P ( L ) P ( M ∩ R ) P ( F ∩ L ) P ( M ∪ F ) P ( M ∪ R ) P ( F ∪ L ) P ( M' ) P ( R | M ) P ( F | L ) P ( L | F ) P ( M ) = 0.52 P ( F ) = 0.48 P ( R ) = 0.87 P ( L ) = 0.13 P ( M ∩ R ) = 0.43 P ( F ∩ L ) = 0.04 P ( M ∪ F ) = 1 P ( M ∪ R ) = 0.96 P ( F ∪ L ) = 0.57 P ( M' ) = 0.48 P ( R | M ) = 0.8269 (rounded to four decimal places) P ( F | L ) = 0.3077 (rounded to four decimal places) P ( L | F ) = 0.0833 Try It The table describes the distribution of a random sample S of 100 individuals, organized by gender and whether they like tea or coffee. Tea Coffee Men 22 26 Women 16 36 Let's denote the events M = the subject is a man, W = the subject is a woman, T = the subject likes tea, and C = the subject likes coffee. Compute the following probabilities: P ( M ) P ( W ) P ( T ) P ( C ) P ( M AND T ) P ( W and C ) P ( M OR W ) P ( M OR T ) P ( W OR C ) P ( M ') P ( T | M ) P ( W | C ) P ( C | W ) P ( M ) = 0 . 48 P ( W ) = 0 . 52 P ( T ) = 0 . 38 P ( C ) = 0 . 62 P ( M A N D T ) = 0 . 22 P ( W a n d C ) = 0 . 36 P ( M O R W ) = 1 P ( M O R T ) = 0 . 64 P ( W O R C ) = 0 . 78 P ( M ' ) = 0 . 52 P ( T | M ) = 0 . 4583 (rounded to four decimal places) P ( W | C ) = 0 . 5806 (rounded to four decimal places) P ( C | W ) = 0 . 6923 (rounded to four decimal places) References “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013). Chapter Review In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive. Formula Review A and B are events P ( S ) = 1 where S is the sample space 0 ≤ P ( A ) ≤ 1 P ( A | B ) = P ( A ∩ B ) P ( B ) In a particular college class, there are both men and women students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) Let W be the event that a student is a woman. Let M be the event that a student is a man. Let S be the event that a student has short hair. Let L be the event that a student has long hair. The probability that a student does not have long hair. The probability that a student is a man or has short hair. The probability that a student is a woman and has long hair. The probability that a student is a man, given that the student has long hair. The probability that a student has long hair, given that the student is a man. Of all the women students, the probability that a student has short hair. Of all students with long hair, the probability that a student is a woman. The probability that a student is a woman or has long hair. The probability that a randomly selected student is a man with short hair. The probability that a student is a woman. P ( L′ ) = P ( S ) P ( M ∪ S ) P ( W ∩ L ) P ( M | L ) P ( L | M ) P ( S | W ) P ( W | L ) P ( W ∪ L ) P ( M ∩ S ) P ( W ) Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. One party favor is chosen from the box at random. Let H = the event of getting a hat. Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap. Let C = the event of getting a bag of confetti. Find P ( H ). Find P ( N ). P ( N ) = 15 42 = 5 14 = 0.36 Find P ( F ). Find P ( C ). P ( C ) = 5 42 = 0.12 Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. One jelly bean is chosen from the box at random. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. Find P ( B ). Find P ( G ). P ( G ) = 20 150 = 2 15 = 0.13 Find P ( P ). Find P ( R ). P ( R ) = 22 150 = 11 75 = 0.15 Find P ( Y ). Find P ( O ). P ( O ) = 150 - 22 - 38 - 20 - 28 - 26 150 = 16 150 = 8 75 = 0.11 Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. Find P ( A ). Find P ( E ). P ( E ) = 47 194 = 0.24 Find P ( F ). Find P ( N ). P ( N ) = 23 194 = 0.12 Find P ( O ). Find P ( S ). P ( S ) = 12 194 = 6 97 = 0.06 What is the probability of drawing a red card in a standard deck of 52 cards? What is the probability of drawing a club in a standard deck of 52 cards? 13 52 = 1 4 = 0.25 What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six? What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six? 3 6 = 1 2 = 0.5 Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area. Let B = the event of landing on blue. Let R = the event of landing on red. Let G = the event of landing on green. Let Y = the event of landing on yellow. If you land on Y , you get the biggest prize. Find P ( Y ). If you land on red, you don’t get a prize. What is P ( R )? P ( R ) = 4 8 = 0.5 Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. Write the symbols for the probability that a player is not an outfielder. Write the symbols for the probability that a player is an outfielder or is a great hitter. P ( O ∪ H ) Write the symbols for the probability that a player is an infielder and is not a great hitter. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder. P ( H | I ) Write the symbols for the probability that a player is an infielder, given that the player is a great hitter. Write the symbols for the probability that of all the outfielders, a player is not a great hitter. P ( N | O ) Write the symbols for the probability that of all the great hitters, a player is an outfielder. Write the symbols for the probability that a player is an infielder or is not a great hitter. P ( I ∪ N ) Write the symbols for the probability that a player is an outfielder and is a great hitter. Write the symbols for the probability that a player is an infielder. P ( I ) What is the word for the set of all possible outcomes? What is conditional probability? The likelihood that an event will occur given that another event has already occurred. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space? What is the sum of the probabilities of an event and its complement? 1 Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three. What does P ( E | M ) mean in words? What does P ( E ∪ M ) mean in words? the probability of landing on an even number or a multiple of three Homework The graph in displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045. Define three events in the graph. Describe in words what the entry 40 means. Describe in words the complement of the entry in question 2. Describe in words what the entry 30 means. Out of the men and women, what percent are men? Out of the women, what percent disapprove of Mayor Ford? Out of all the age groups, what percent approve of Mayor Ford? Find P (Approve | Men). Out of the age groups, what percent are more than 44 years old? Find P (Approve | Age < 35). Explain what is wrong with the following statements. Use complete sentences. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100% A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs. Conditional Probability the likelihood that an event will occur given that another event has already occurred Equally Likely Each outcome of an experiment has the same probability. Event a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S . An event is an arbitrary subset in S . It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A , B , C , and so on. Experiment a planned activity carried out under controlled conditions Outcome a particular result of an experiment Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S . Then: 0 ≤ P ( A ) ≤ 1 If A and B are any two mutually exclusive events, then P ( A ∪ B ) = P ( A ) + P (B). P ( S ) = 1 Sample Space the set of all possible outcomes of an experiment Intersection: the ∩ Event An outcome is in the event A ∩ B if the outcome is in both A ∩ B at the same time. Conditional Probability of A | B P ( A | B ) is the probability that event A will occur given that the event B has already occurred. Union: the ∪ Event An outcome is in the event A ∪ B if the outcome is in A or is in B or is in both A and B .", "section": "Terminology", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Independent and Mutually Exclusive Events Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if one of the following are true: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ B ) = P ( A ) P ( B ) Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent . Sampling may be done with replacement or without replacement . With replacement : If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. Without replacement : When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent. If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise . You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. a. Sampling with replacement: Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are { Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement: Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are { K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. The probability of picking the three of diamonds is called a conditional probability because it is conditioned on what was picked first. This is true also of the probability of picking the J of spades. The probability of picking the J of spades is actually conditioned on both the previous picks. Try It You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? With replacement No You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD . Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH . Which of a. or b. did you sample with replacement and which did you sample without replacement? a. Without replacement; b. With replacement Try It You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement. QS , 1 D , 1 C , QD KH , 7 D , 6 D , KH QS , 7 D , 6 D , KS without replacement: 1. Possible; 2. Impossible, 3. Possible with replacement: 1. Possible; 2. Possible, 3. Possible Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. Said another way, If A occurred then B cannot occur and vise-a-versa. This means that A and B do not share any outcomes and P ( A ∩ B ) = 0 . For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A ∩ B = {4, 5}. P ( A ∩ B ) = 2 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P ( A ∩ C ) = 0 . Therefore, A and C are mutually exclusive. If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise . The following examples illustrate these definitions and terms. Flip two fair coins. (This is an experiment.) The sample space is { HH , HT , TH , TT } where T = tails and H = heads. The outcomes are HH , HT , TH , and TT . The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads. Let A = the event of getting at most one tail . (At most one tail means zero or one tail.) Then A can be written as { HH , HT , TH }. The outcome HH shows zero tails. HT and TH each show one tail. Let B = the event of getting all tails. B can be written as { TT }. B is the complement of A , so B = A′ . Also, P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1. The probabilities for A and for B are P ( A ) = 3 4 and P ( B ) = 1 4 . Let C = the event of getting all heads. C = { HH }. Since B = { TT }, P ( B ∩ C ) = 0 . B and C are mutually exclusive. ( B and C have no members in common because you cannot have all tails and all heads at the same time.) Let D = event of getting more than one tail. D = { TT }. P ( D ) = 1 4 Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = { HT , HH }. P ( E ) = 2 4 Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = { HT , TH , TT }. P ( F ) = 3 4 Try It Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Try It Solutions The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is { BB , BR , RB , RR }. Event A = Getting at least one black card = { BB , BR , RB } P ( A ) = 3 4 = 0.75 Flip two fair coins. Find the probabilities of the events. Let F = the event of getting at most one tail (zero or one tail). Let G = the event of getting two faces that are the same. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. Are F and G mutually exclusive? Let J = the event of getting all tails. Are J and H mutually exclusive? Look at the sample space in . Zero (0) or one (1) tails occur when the outcomes HH , TH , HT show up. P ( F ) = 3 4 Two faces are the same if HH or TT show up. P ( G ) = 2 4 A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P ( H ) = 2 4 F and G share HH so P ( F ∩ G ) is not equal to zero (0). F and G are not mutually exclusive. Getting all tails occurs when tails shows up on both coins ( TT ). H ’s outcomes are HH and HT . J and H have nothing in common so P ( J ∩ H ) = 0. J and H are mutually exclusive. Try It A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events: Let F = the event of getting the white ball twice. Let G = the event of getting two balls of different colors. Let H = the event of getting white on the first pick. Are F and G mutually exclusive? Are G and H mutually exclusive? P ( F ) = 1 4 P ( G ) = 1 2 P ( H ) = 1 2 Yes No Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}. Find the complement of A , A′ . The complement of A , A′ , is B because A and B together make up the sample space. P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1. Also, P ( A ) = 3 6 and P ( B ) = 3 6 . Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P ( C ∩ D ) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. Let event E = all faces less than five. E = {1, 2, 3, 4}. Are C and E mutually exclusive events? (Answer yes or no.) Why or why not? No. C = {3, 5} and E = {1, 2, 3, 4}. P ( C ∩ E ) = 1 6 . To be mutually exclusive, P ( C ∩ E ) must be zero. Find P ( C | A ) . This is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P ( C | A ) , find the probability of C using the sample space A . You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P ( C | A ) = 2 3 . Try It Let event A = learning Spanish. Let event B = learning German. Then A ∩ B = learning Spanish and German. Suppose P ( A ) = 0.4 and P ( B ) = 0.2 . P ( A ∩ B ) = 0.08 . Are events A and B independent? Hint: You must show ONE of the following: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ B ) = P ( A ) P ( B ) P ( A | B ) = P ( A ∩ B ) P ( B ) = 0. 08 0.2 = 0 .4 = P ( A ) The events are independent because P ( A | B ) = P ( A ). Let event G = taking a math class. Let event H = taking a science class. Then, G ∩ H = taking a math class and a science class. Suppose P ( G ) = 0.6 , P ( H ) = 0.5 , and P ( G ∩ H ) = 0.3 . Are G and H independent? If G and H are independent, then you must show ONE of the following: P ( G | H ) = P ( G ) P ( H | G ) = P ( H ) P ( G ∩ H ) = P ( G ) P ( H ) NOTE The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information. a. Show that P ( G | H ) = P ( G ) . P ( G | H ) = P ( G ∩ H ) P ( H ) = 0 .3 0 .5 = 0.6 = P ( G ) b. Show P ( G ∩ H ) = P ( G ) P ( H ) . P ( G ) P ( H ) = ( 0.6 ) ( 0.5 ) = 0.3 = P ( G ∩ H ) Since G and H are independent, knowing that a person is taking a science class does not change the chance that they are taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance they are taking math. For practice, show that P ( H | G ) = P ( H ) to show that G and H are independent events. Try It In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4. R = a red marble G = a green marble O = an odd-numbered marble The sample space is S = { R 1, R 2, R 3, R 4, R 5, R 6, G 1, G 2, G 3, G 4}. S has ten outcomes. What is P ( G ∩ O ) ? Event G and O = { G 1, G 3} P ( G ∩ O ) = 2 10 = 0.2 Let event C = taking an English class. Let event D = taking a speech class. Suppose P ( C ) = 0.75 , P ( D ) = 0.3 , P ( C | D ) = 0.75 and P ( C ∩ D ) = 0.225 . Justify your answers to the following questions numerically. Are C and D independent? Are C and D mutually exclusive? What is P ( D | C ) ? Yes, because P ( C | D ) = P ( C ) . No, because P ( C ∩ D ) is not equal to zero. P ( D | C ) = P ( C ∩ D ) P ( C ) = 0 .225 0.75 = 0.3 Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40 , P ( D ) = 0.30 and P ( B ∩ D ) = 0.20 . Find P ( B | D ) . Find P ( D | B ) . Are B and D independent? Are B and D mutually exclusive? P ( B | D ) = 0.6667 P ( D | B ) = 0.5 No No In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R 1, R 2, R 3, B 1, B 2, B 3, B 4, B 5. S has eight outcomes. P ( R ) = 3 8 . P ( B ) = 5 8 . P ( R ∩ B ) = 0 . (You cannot draw one card that is both red and blue.) P ( E ) = 3 8 . (There are three even-numbered cards, R 2, B 2, and B 4.) P ( E | B ) = 2 5 . (There are five blue cards: B 1, B 2, B 3, B 4, and B 5. Out of the blue cards, there are two even cards; B 2 and B 4.) P ( B | E ) = 2 3 . (There are three even-numbered cards: R 2, B 2, and B 4. Out of the even-numbered cards, to are blue; B 2 and B 4.) The events R and B are mutually exclusive because P ( R ∩ B ) = 0 . Let G = card with a number greater than 3. G = { B 4, B 5}. P ( G ) = 2 8 . Let H = blue card numbered between one and four, inclusive. H = { B 1, B 2, B 3, B 4}. P ( G | H ) = 1 4 . (The only card in H that has a number greater than three is B 4.) Since 2 8 = 1 4 , P ( G ) = P ( G | H ) , which means that G and H are independent. Try It In a basketball arena, 70% of the fans are rooting for the home team. 25% of the fans are wearing blue. 20% of the fans are wearing blue and are rooting for the away team. Of the fans rooting for the away team, 67% are wearing blue. Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? P ( B | A ) = 0.67 P ( B ) = 0.25 So P ( B ) does not equal P ( B | A ) which means that B and A are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because P ( B ∩ A ) = 0.20 , not 0. In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent? The following probabilities are given in this example: P ( F ) = 0.60 ; P ( L ) = 0.50 P ( F ∩ L ) = 0.45 P ( L | F ) = 0.75 NOTE The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P ( F | L ) yet, so you cannot use the second condition. Check whether P ( F ∩ L ) = P ( F ) P ( L ) . We are given that P ( F ∩ L ) = 0.45 , but P ( F ) P ( L ) = ( 0.60 ) ( 0.50 ) = 0.30 . The events of being female and having long hair are not independent because P ( F ∩ L ) does not equal P ( F ) P ( L ) . Check whether P ( L | F ) equals P ( L ) . We are given that P ( L | F ) = 0.75 , but P ( L ) = 0.50 ; they are not equal. The events of being female and having long hair are not independent. Interpretation of Results: The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair. Try It Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street. P ( I ) = 0.44 and P ( F ) = 0.56 P ( I ∩ F ) = 0 because Mark will take only one route to work. What is the probability of P ( I ∪ F ) ? Because P ( I ∩ F ) = 0 , P ( I ∪ F ) = P ( I ) + P ( F ) − P ( I ∩ F ) = 0.44 + 0.56 − 0 = 1 Toss one fair coin (the coin has two sides, H and T ). The outcomes are ________. Count the outcomes. There are ____ outcomes. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes. Multiply the two numbers of outcomes. The answer is _______. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer to c is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H 1 and T 6.) Event A = heads ( H ) on the coin followed by an even number (2, 4, 6) on the die. A = {_________________}. Find P ( A ). Event B = heads on the coin followed by a three on the die. B = {________}. Find P ( B ). Are A and B mutually exclusive? (Hint: What is P ( A ∩ B ) ? If P ( A ∩ B ) = 0 , then A and B are mutually exclusive.) Are A and B independent? (Hint: Is P ( A ∩ B ) = P ( A ) P ( B ) ? If P ( A ∩ B ) = P ( A ) P ( B ) , then A and B are independent. If not, then they are dependent). H and T ; 2 1, 2, 3, 4, 5, 6; 6 2(6) = 12 T 1, T 2, T 3, T 4, T 5, T 6, H 1, H 2, H 3, H 4, H 5, H 6 A = { H 2, H 4, H 6}; P ( A ) = 3 12 B = { H 3}; P ( B ) = 1 12 Yes, because P ( A ∩ B ) = 0 P ( A ∩ B ) = 0 . P ( A ) P ( B ) = 3 12 · 1 12 = 1 48 . P ( A ∩ B ) does not equal P ( A ) P ( B ) , so A and B are dependent. Try It A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing. Compute P ( T ) . Compute P ( T | F ) . Are T and F independent?. Are F and S mutually exclusive? Are F and S independent? P ( T ) = 1 4 P ( T | F ) = 1 2 No No Yes References Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013). Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013). Chapter Review Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. Formula Review If A and B are independent, P ( A ∩ B ) = P ( A ) P ( B ) , P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ) . If A and B are mutually exclusive, P ( A ∪ B ) = P ( A ) + P ( B ) and P ( A ∩ B ) = 0 . E and F are mutually exclusive events. P ( E ) = 0.4 ; P ( F ) = 0.5 . Find P ( E ∣ F ) . J and K are independent events. P ( J | K ) = 0.3 . Find P ( J ) . P ( J ) = 0.3 U and V are mutually exclusive events. P ( U ) = 0.26; P ( V ) = 0.37. Find: P ( U ∩ V ) = P ( U | V ) = P ( U ∪ V ) = Q and R are independent events. P ( Q ) = 0.4 and P ( Q ∩ R ) = 0.1 . Find P ( R ) . P ( Q ∩ R ) = P ( Q ) P ( R ) 0.1 = (0.4) P ( R ) P ( R ) = 0.25 Homework Use the following information to answer the next 12 exercises. In a certain year, Gallup interviewed more than 170,000 employed Americans 18 years of age or older. The graph shown is compiled from data collected by Gallup. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score. Find the probability that the selected occupation has an Emotional Health Index Score of 82.7. Find the probability that the selected occupation has an Emotional Health Index Score of 81.0. 0 Find the probability that the selected occupation has an Emotional Health Index Score more than 81. Find the probability that the selected occupation has an Emotional Health Index Score between 80.5 and 82. 0.3571 If we know an occupation has an Emotional Health Index Score of 81.5 or more, what is the probability that it is 82.7? What is the probability that an occupation has an Emotional Health Index Score of 80.7 or 82.7? 0.2142 What is the probability that an occupation has an Emotional Health Index Score less than 80.2 given that it is already less than 81? What occupation has the highest emotional index score? Physician (83.7) What occupation has the lowest emotional index score? What is the range of the data? 83.7 − 79.6 = 4.1 Compute the average EHIS. If all occupations are equally likely for a certain individual, what is the probability that they will have an occupation with lower than average EHIS? P (Occupation < 81.3) = 0.5 Bringing It Together A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data are compiled into . Shirt # ≤ 210 211–250 251–290 290≤ 1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P (Shirt# 1–33|≤ 210 pounds)? The probability that someone develops cancer during their lifetime is 0.4567. The probability that a person has at least one false positive test result (meaning the test comes back for cancer when the person does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a person develops cancer and P = person has at least one false positive. P ( C ) = ______ P ( P | C ) = ______ P ( P | C' ) = ______ If a test comes up positive, based upon numerical values, can you assume that person has cancer? Justify numerically and explain why or why not. P ( C ) = 0.4567 not enough information not enough information No, because over half (0.51) of men have at least one false positive text Given events G and H : P ( G ) = 0.43; P ( H ) = 0.26; P ( H ∩ G ) = 0.14 Find P ( H ∪ G ) . Find the probability of the complement of event ( H ∩ G ) . Find the probability of the complement of event ( H ∪ G ) . Given events J and K : P ( J ) = 0.18 ; P ( K ) = 0.37 ; P ( J ∪ K ) = 0.45 Find P ( J ∩ K ) . Find the probability of the complement of event ( J ∩ K ) . Find the probability of the complement of event ( J ∩ K ) . ( J ∪ K ) = P ( J ) + P ( K ) − P ( J ∩ K ) ; 0.45 = 0.18 + 0.37 − P ( J ∩ K ) ; solve to find P ( J ∩ K ) = 0.10 P ( NOT ( J ∩ K ) ) = 1 − P ( J ∩ K ) = 1 − 0.10 = 0.90 P ( NOT ( J ∪ K ) ) = 1 − P ( J ∪ K ) = 1 − 0.45 = 0.55", "section": "Independent and Mutually Exclusive Events", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. The Multiplication Rule If A and B are two events defined on a sample space , then: P ( A ∩ B ) = P ( B ) P ( A | B ) . We can think of the intersection symbol as substituting for the word \"and\". This rule may also be written as: P ( A | B ) = P ( A ∩ B ) P ( B ) This equation is read as the probability of A given B equals the probability of A and B divided by the probability of B . If A and B are independent , then P ( A | B ) = P ( A ) . Then P ( A ∩ B ) = P ( A | B ) P ( B ) becomes P ( A ∩ B ) = P ( A ) ( B ) because the P ( A | B ) = P ( A ) if A and B are independent. One easy way to remember the multiplication rule is that the word \"and\" means that the event has to satisfy two conditions. For example the name drawn from the class roster is to be both a woman and a sophomore. It is harder to satisfy two conditions than only one and of course when we multiply fractions the result is always smaller. This reflects the increasing difficulty of satisfying two conditions. The Addition Rule If A and B are defined on a sample space, then: P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) . We can think of the union symbol substituting for the word \"or\". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B . If A and B are mutually exclusive , then P ( A ∩ B ) = 0 . Then P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) becomes P ( A ∪ B ) = P ( A ) + P ( B ) . Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska Klaus can only afford one vacation. The probability that he chooses A is P ( A ) = 0.6 and the probability that he chooses B is P ( B ) = 0.35. P ( A ∩ B ) = 0 because Klaus can only afford to take one vacation Therefore, the probability that he chooses either New Zealand or Alaska is P ( A ∪ B ) = P ( A ) + P ( B ) = 0.6 + 0.35 = 0.95 . Note that the probability that he does not choose to go anywhere on vacation must be 0.05. Try It Anna has to buy a new car. She has two choices, car A and car B. Anna can afford only one car. The probability that Anna will buy car A is P ( A ) = 0 . 25 , and the probability that Anna will buy car B is P ( B ) = 0 . 65 . Find: P ( A A N D B ) P ( A O R B ) P ( A A N D B ) = 0 because it is given that Anna can afford only one car. Since P ( A A N D B ) = 0 , then P ( A O R B ) = P ( A ) + P ( B ) = 0 . 25 + 0 . 65 = 0 . 9 . Note that the probability that Anna does not a car must be 0.1. Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P ( A ) = 0.65. B = the event Carlos is successful on his second attempt. P ( B ) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal | that he made the first goal is 0.90. What is the probability that he makes both goals? What is the probability that Carlos makes either the first goal or the second goal? Are A and B independent? Are A and B mutually exclusive? a. The problem is asking you to find P ( A ∩ B ) = P ( B ∩ A ) . Since P ( B | A ) = 0.90: P ( B ∩ A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find P ( A ∪ B ). P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. No, they are not, because P ( B ∩ A ) = 0.585. P ( B ) P ( A ) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P ( B ∩ A ) So, P ( B ∩ A ) is not equal to P ( B ) P ( A ). d. No, they are not because P ( A ∩ B ) = 0.585. To be mutually exclusive, P ( A ∩ B ) must equal zero. Try It Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = 0.75. D = the event Helen makes the second shot. P ( D ) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? P ( D | C ) = 0.85 P ( C ∩ D ) = P ( D ∩ C ) P ( D ∩ C ) = P ( D | C ) P ( C ) = (0.85)(0.75) = 0.6375 Helen makes the first and second free throws with probability 0.6375. A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. What is the probability that the member is a novice swimmer? What is the probability that the member practices four times a week? What is the probability that the member is an advanced swimmer and practices four times a week? What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Are being a novice swimmer and practicing four times a week independent events? Why or why not? 28 150 80 150 40 150 P (advanced ∩ intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. No, these are not independent events. P (novice ∩ practices four times per week) = 0.0667 P (novice) P (practices four times per week) = 0.0996 0.0667 ≠ 0.0996 Try It A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? P = 200 − 140 − 40 200 = 20 200 = 0.1 Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class | that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M | S = math given speech What is the probability that Felicity enrolls in math and speech? Find P ( M ∩ S ) = P ( M | S ) P ( S ). What is the probability that Felicity enrolls in math or speech classes? Find P ( M ∪ S ) = P ( M ) + P ( S ) - P ( M ∩ S ). Are M and S independent? Is P ( M | S ) = P ( M )? Are M and S mutually exclusive? Is P ( M ∩ S ) = 0? a. 0.1625, b. 0.6875, c. No, d. No Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5. Find P ( B ∩ D ). Find P ( B ∪ D ). P ( B ∩ D ) = P ( D | B ) P ( B ) = (0.5)(0.4) = 0.20. P ( B ∪ D ) = P ( B ) + P ( D ) − P ( B ∩ D ) = 0.40 + 0.30 − 0.20 = 0.50 Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? Given that the woman has breast cancer, what is the probability that she tests negative? What is the probability that the woman has breast cancer AND tests negative? What is the probability that the woman has breast cancer or tests negative? Are having breast cancer and testing negative independent events? Are having breast cancer and testing negative mutually exclusive? P ( B ) = 0.143; P ( N ) = 0.85 P ( N | B ) = 0.02 P ( B ∩ N ) = P ( B ) P ( N | B ) = (0.143)(0.02) = 0.0029 P ( B ∪ N ) = P ( B ) + P ( N ) - P ( B ∩ N ) = 0.143 + 0.85 - 0.0029 = 0.9901 No. P ( N ) = 0.85; P ( N | B ) = 0.02. So, P ( N | B ) does not equal P ( N ). No. P ( B ∩ N ) = 0.0029. For B and N to be mutually exclusive, P ( B ∩ N ) must be zero. Try It A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Let A = student is a senior going to college. Let B = student plays sports. P ( B ) = 140 200 P ( B | A ) = 50 140 P ( A ∩ B ) = P ( B | A ) P ( A ) P ( A ∩ B ) = ( 140 200 ) ( 50 140 ) = 1 4 Refer to the information in . P = tests positive. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P ( P | B ) = 1 - P ( N | B ). What is the probability that a woman develops breast cancer and tests positive. Find P ( B ∩ P ) = P ( P | B ) P ( B ). What is the probability that a woman does not develop breast cancer. Find P ( B′ ) = 1 - P ( B ). What is the probability that a woman tests positive for breast cancer. Find P ( P ) = 1 - P ( N ). a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5. Find P ( B′ ). Find P ( D ∩ B ). Find P ( B | D ). Find P ( D ∩ B′ ). Find P ( D | B′ ). P ( B′ ) = 0.60 P ( D ∩ B ) = P ( D | B ) P ( B ) = 0.20 P ( B | D ) = P ( B ∩ D ) P ( D ) = ( 0.20 ) (0 .30) = 0.66 P ( D ∩ B′ ) = P ( D ) - P ( D ∩ B ) = 0.30 - 0.20 = 0.10 P ( D | B′ ) = P ( D ∩ B′ ) P ( B′ ) = ( P ( D ) - P ( D ∩ B ))(0.60) = (0.10)(0.60) = 0.06 References DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013). “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). “Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Data from U.S. Census Bureau. www.census.gov/library/stories/2022/12/languages-we-speak-in-united-states.html (accessed June 19, 2023). Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013). Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). Chapter Review The multiplication rule and the addition rule are used for computing the probability of A and B , as well as the probability of A or B for two given events A , B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. Formula Review The multiplication rule: P ( A ∩ B ) = P ( A | B ) P ( B ) The addition rule: P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) Use the following information to answer the next ten exercises. A local restaurant knows that the probability that a customer will order a pizza is 87%. The restaurant also knows that the probability that a customer will order a salad is 32%. Of the customers who order pizzas, 55% of them also order a salad In this problem, let: Z = event that a customer orders a pizza S = event that a customer orders a salad Suppose that one customer is randomly selected. Find P ( Z ). Find P ( Z ) : P ( Z ) = 0.87 Find P ( S ). Find P ( S ) : P ( S ) = 0.32 Find P ( S | Z ). Find P ( S | Z ) : P ( S | Z ) = 0.55 In words, what is S | Z ? In words, what is S | Z ? : S | Z means, given that the customer has ordered pizza, the person also orders a salad. Find P ( Z ∩ S ) . Find P ( Z A N D S ) : P ( Z A N D S ) = 0.4785 In words, what is Z ∩ S ? In words, what is Z A N D S ? : Z A N D S represents the event that a customer orders a pizza and salad. Are P and S independent events? Show why or why not. Are P and S independent events? Show why or why not. No, because P ( S ) does not equal P ( S | Z ). Find P ( Z ∪ S ) . Find P ( Z O R S ) : P ( Z O R S ) = 0.7115 In words, what is Z ∪ S ? In words, what is Z O R S ? Z O R S represents the event that a customer orders a pizza or salad. Are Z and S mutually exclusive events? Show why or why not. Are Z and S mutually exclusive events? Show why or why not. No, because P ( S and Z ) does not equal 0. Homework Prior to the 2015 Supreme Court decision legalizing same-sex marriage nationwide, a survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them. In this problem, let: C = California registered voters who support same-sex marriage. B = California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them A = California registered voters who are 18 to 39 years old. Find P ( C ). Find P ( B ). Find P ( C | A ). Find P ( B | C ). In words, what is C | A ? In words, what is B | C ? Find P ( C ∩ B ). In words, what is C ∩ B ? Find P ( C ∪ B ). Are C and B mutually exclusive events? Show why or why not. A survey was conducted in a large city to measure the popularity of that city’s mayor. The survey was repeated every year for three years. The survey polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results the poll produced: In Year 1, 60% of the population approved of the mayor’s actions in office. In Year 2, 57% of the population approved of his actions. In Year 3, the percentage of popular approval was measured at 42%. What is the sample size for this study? What proportion in the poll disapproved of the mayor, according to the results from Year 3? How many people polled responded that they approved the mayor based on results from Year 3? What is the probability that a person supported the mayor, based on the data collected in Year 2? What is the probability that a person supported the mayor, based on the data collected in Year 1? The Forum Research surveyed 1,046 Torontonians. 58% 42% of 1,046 = 439 (rounding to the nearest integer) 0.57 0.60. Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range. (credit: film8ker/wikibooks) List the sample space of the 38 possible outcomes in roulette. You bet on red. Find P (red). You bet on -1st 12- (1st Dozen). Find P (-1st 12-). You bet on an even number. Find P (even number). Is getting an odd number the complement of getting an even number? Why? Find two mutually exclusive events. Are the events Even and 1st Dozen independent? Compute the probability of winning the following types of bets: Betting on two lines that touch each other on the table as in 1-2-3-4-5-6 Betting on three numbers in a line, as in 1-2-3 Betting on one number Betting on four numbers that touch each other to form a square, as in 10-11-13-14 Betting on two numbers that touch each other on the table, as in 10-11 or 10-13 Betting on 0-00-1-2-3 Betting on 0-1-2; or 0-00-2; or 00-2-3 P (Betting on two line that touch each other on the table) = 6 38 P (Betting on three numbers in a line) = 3 38 P (Betting on one number) = 1 38 P (Betting on four number that touch each other to form a square) = 4 38 P (Betting on two number that touch each other on the table ) = 2 38 P (Betting on 0-00-1-2-3) = 5 38 P (Betting on 0-1-2; or 0-00-2; or 00-2-3) = 3 38 Compute the probability of winning the following types of bets: Betting on a color Betting on one of the dozen groups Betting on the range of numbers from 1 to 18 Betting on the range of numbers 19–36 Betting on one of the columns Betting on an even or odd number (excluding zero) Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card. G = card drawn is green E = card drawn is even-numbered List the sample space. P ( G ) = _____ P ( G | E ) = _____ P ( G ∩ E ) = _____ P ( G ∪ E ) = _____ Are G and E mutually exclusive? Justify your answer numerically. { G 1, G 2, G 3, G 4, G 5, Y 1, Y 2, Y 3} 5 8 2 3 2 8 6 8 No, because P ( G ∩ E ) does not equal 0. Roll two fair dice separately. Each die has six faces. List the sample space. Let A be the event that either a three or four is rolled first, followed by an even number. Find P ( A ). Let B be the event that the sum of the two rolls is at most seven. Find P ( B ). In words, explain what “ P ( A | B )” represents. Find P ( A | B ). Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification. A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin. List the sample space. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P ( A ). Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. NOTE The coin toss is independent of the card picked first. {( G , H ) ( G , T ) ( B , H ) ( B , T ) ( R , H ) ( R , T )} P ( A ) = P (blue) P (head) = ( 3 10 ) ( 1 2 ) = 3 20 Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P ( A ∩ B ) = 0 No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A ; if the card chosen is blue it is also (red or blue). P ( A ∩ C ) = P ( A ) = 3 20 An experiment consists of first rolling a die and then tossing a coin. List the sample space. Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P ( A ). Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on. List the sample space. Let A be the event that there are at least two tails. Find P ( A ). Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including justification. S = {( HHH ), ( HHT ), ( HTH ), ( HTT ), ( THH ), ( THT ), ( TTH ), ( TTT )} 4 8 Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P ( A ∩ B ) = 0. Consider the following scenario: Let P ( C ) = 0.4. Let P ( D ) = 0.5. Let P ( C | D ) = 0.6. Find P ( C ∩ D ). Are C and D mutually exclusive? Why or why not? Are C and D independent events? Why or why not? Find P ( C ∪ D ). Find P ( D | C ). Y and Z are independent events. Rewrite the basic Addition Rule P ( Y ∪ Z ) = P ( Y ) + P ( Z ) - P ( Y ∩ Z ) using the information that Y and Z are independent events. Use the rewritten rule to find P ( Z ) if P ( Y ∪ Z ) = 0.71 and P ( Y ) = 0.42. If Y and Z are independent, then P ( Y ∩ Z ) = P ( Y ) P ( Z ), so P ( Y ∪ Z ) = P ( Y ) + P ( Z ) - P ( Y ) P ( Z ). 0.5 G and H are mutually exclusive events. P ( G ) = 0.5 P ( H ) = 0.3 Explain why the following statement MUST be false: P ( H | G ) = 0.4. Find P ( H ∪ G ). Are G and H independent or dependent events? Explain in a complete sentence. According to the 2019 U.S. Census, approximately 331,449,281 people live in the United States. Of these people, 67,800,000 speak a language other than English at home. Of those who speak another language at home, 61.6% speak Spanish. Let: E = speaks English at home; E' = speaks another language at home; S = speaks Spanish. Finish each probability statement by matching the correct answer. Probability Statements Answers a. P ( E ′ ) = i. 0.7954 b. P ( E ) = ii. 0.616 c. P ( S ∩ E ′ ) = iii. 0.2046 d. P ( S | E ′ ) = iv. 0.1260 iii i iv ii 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won green card. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let F = was a finalist. Are G and F independent or dependent events? Justify your answer numerically and also explain why. Are G and F mutually exclusive events? Justify your answer numerically and explain why. Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned. Let: R = money returned; E = economics classes; O = other classes Write a probability statement for the overall percent of money returned. Write a probability statement for the percent of money returned out of the economics classes. Write a probability statement for the percent of money returned out of the other classes. Is money being returned independent of the class? Justify your answer numerically and explain it. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer. P ( R ) = 0.44 P ( R | E ) = 0.56 P ( R | O ) = 0.31 No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P ( R | E ) ≠ P ( R ). No, this study definitely does not support that notion; in fact , it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P ( R | E ) > P ( R ). The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home run Total hits Babe Ruth 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,053 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 Total 7,918 2,127 583 1,723 12,351 Are \"the hit being made by Hank Aaron\" and \"the hit being a double\" independent events? Yes, because P (hit by Hank Aaron | hit is a double) = P (hit by Hank Aaron) No, because P (hit by Hank Aaron | hit is a double) ≠ P (hit is a double) No, because P (hit is by Hank Aaron | hit is a double) ≠ P (hit by Hank Aaron) Yes, because P (hit is by Hank Aaron | hit is a double) = P (hit is a double) According to the American Red Cross, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any bloodtype. Their data show that 45% of people have type O blood and 7% of people have Rh- factor; 38% of people have type O or Rh+ factor. Find the probability that a person has both type O blood and the Rh- factor. Find the probability that a person does NOT have both type O blood and the Rh- factor. P (type O OR Rh-) = P (type O) + P (Rh-) – P (type O AND Rh-) 0.38 = 0.45 + 0.07 – P (type O AND Rh-); Solve to find P (type O AND Rh-) = 0.14. 14% of people have type O, Rh- blood. P (NOT(type O AND Rh-)) = 1 – P (type O AND Rh-) = 1 – 0.14 = 0.86 86% of people do not have type O, Rh- blood. At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper. Find the probability that a course has a final exam or a research project. Find the probability that a course has NEITHER of these two requirements. In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. In the box, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts. Find the probability that a cookie contains chocolate or nuts (he can't eat it). Find the probability that a cookie does not contain chocolate or nuts (he can eat it). Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts. P ( C ∪ N ) = P ( C ) + P ( N ) - P ( C ∩ N ) = 0.36 + 0.12 - 0.08 = 0.40 P (NEITHER chocolate NOR nuts) = 1 - P ( C ∪ N ) = 1 - 0.40 = 0.60 A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student Find P ( D ∩ E ). Find P ( E | D ). Find P ( D ∪ E ). Using an appropriate test, show whether D and E are independent. Using an appropriate test, show whether D and E are mutually exclusive. Independent Events The occurrence of one event has no effect on the probability of the occurrence of another event. Events A and B are independent if one of the following is true: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ B ) = P ( A ) P ( B ) Mutually Exclusive Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events A and B are mutually exclusive, then P ( A ∩ B ) = 0.", "section": "Two Basic Rules of Probability", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Contingency Tables and Probability Trees Contingency Tables A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner. Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: Speeding violation in the last year No speeding violation in the last year Total Uses cell phone while driving 25 280 305 Does not use cell phone while driving 45 405 450 Total 70 685 755 The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. Calculate the following probabilities using the table. Find P (Driver is a cell phone user). Find P (Driver had no violation in the last year). Find P (Driver had no violation in the last year ∩ was a cell phone user). Find P (Driver is a cell phone user ∪ driver had no violation in the last year). Find P (Driver is a cell phone user | driver had a violation in the last year). Find P (Driver had no violation last year | driver was not a cell phone user) number of cell phone users total number in study = 305 755 number that had no violation total number in study = 685 755 280 755 ( 305 755 + 685 755 ) − 280 755 = 710 755 25 70 (The sample space is reduced to the number of drivers who had a violation.) 405 450 (The sample space is reduced to the number of drivers who were not cell phone users.) Try it shows the number of athletes who stretch before exercising and how many had injuries within the past year. Injury in last year No injury in last year Total Stretches 55 295 350 Does not stretch 231 219 450 Total 286 514 800 What is P (athlete stretches before exercising)? What is P (athlete stretches before exercising | no injury in the last year)? P (athlete stretches before exercising) = 350 800 = 0.4375 P (athlete stretches before exercising | no injury in the last year) = 295 514 = 0.5739 shows a random sample of 100 hikers and the areas of hiking they prefer. Hiking Area Preference Sex The coastline Near lakes and streams On mountain peaks Total Women 18 16 ___ 45 Men ___ ___ 14 55 Total ___ 41 ___ ___ a. Complete the table. a. Hiking Area Preference Sex The coastline Near lakes and streams On mountain peaks Total Women 18 16 11 45 Men 16 25 14 55 Total 34 41 25 100 b. Are the events \"being a woman\" and \"preferring the coastline\" independent events? Let F = being a woman and let C = preferring the coastline. Find P ( F ∩ C ) . Find P ( F ) P ( C ) Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent. b. P ( F ∩ C ) = 18 100 = 0.18 P ( F ) P ( C ) = ( 45 100 ) ( 34 100 ) = (0.45)(0.34) = 0.153 P ( F ∩ C ) ≠ P ( F ) P ( C ), so the events F and C are not independent. c. Find the probability that a person is a man given that the person prefers hiking near lakes and streams. Let M = being a man, and let L = prefers hiking near lakes and streams. What word tells you this is a conditional? Fill in the blanks and calculate the probability: P (___ | ___) = ___. Is the sample space for this problem all 100 hikers? If not, what is it? c. The word 'given' tells you that this is a conditional. P ( M | L ) = 25 41 No, the sample space for this problem is the 41 hikers who prefer lakes and streams. d. Find the probability that a person is a woman or prefers hiking on mountain peaks. Let F = being a woman, and let P = prefers mountain peaks. Find P ( F ). Find P ( P ). Find P ( F ∩ P ) . Find P ( F ∪ P ) . d. P ( F ) = 45 100 P ( P ) = 25 100 P ( F ∩ P ) = 11 100 P ( F ∪ P ) = 45 100 + 25 100 - 11 100 = 59 100 Try It shows a random sample of 200 cyclists and the routes they prefer. Let O = older and H = hilly path. Age Group Lake Path Hilly Path Wooded Path Total Younger 45 38 27 110 Older 26 52 12 90 Total 71 90 39 200 Out of the older group, what is the probability that the cyclist prefers a hilly path? Are the events “being older” and “preferring the hilly path” independent events? P ( H | M ) = 52 90 = 0.5778 For M and H to be independent, show P ( H | M ) = P ( H ) P ( H | M ) = 0.5778, P ( H ) = 90 200 = 0.45 P ( H | M ) does not equal P ( H ) so M and H are NOT independent. Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 and the probability he is not caught is 4 5 . If he goes out the second door, the probability he gets caught by Alissa is 1 4 and the probability he is not caught is 3 4 . The probability that Alissa catches Muddy coming out of the third door is 1 2 and the probability she does not catch Muddy is 1 2 . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1 3 . Door Choice Caught or not Door one Door two Door three Total Caught 1 15 1 12 1 6 ____ Not caught 4 15 3 12 1 6 ____ Total ____ ____ ____ 1 The first entry 1 15 = ( 1 5 ) ( 1 3 ) is P ( Door One ∩ Caught ) The entry 4 15 = ( 4 5 ) ( 1 3 ) is P ( Door One ∩ Not Caught ) Verify the remaining entries. a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. a. Door Choice Caught or not Door one Door two Door three Total Caught 1 15 1 12 1 6 19 60 Not caught 4 15 3 12 1 6 41 60 Total 5 15 4 12 2 6 1 b. What is the probability that Alissa does not catch Muddy? b. 41 60 c. What is the probability that Muddy chooses Door One ∪ Door Two given that Muddy is caught by Alissa? c. 9 19 Try It Fred's preference to drink coffee or iced tea depends on the season. When it is summer, Fred's preference to drink coffee is 1 4 , and his preference to drink iced tea is 3 4 . When it is rainy season, Fred's preference to drink coffee is 1 2 , and his preference to drink iced tea is 1 2 . When it is winter, Fred's preference to drink coffee is 1 5 , and his preference to drink iced tea is 1 5 . A day randomly selected in a year has a probability of 1 3 for each season. The event here is selection of a day randomly in a year. Season Summer Rainy Winter Total Coffee 1 12 1 6 4 15 __ Iced tea 1 4 1 6 1 15 __ Total __ __ __ 1 The first entry is 1 4 1 3 = 1 12 . The second entry is 3 4 1 3 = 1 4 . Verify the remaining entries. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. What is the probability that Fred will drink iced tea? What is the probability that the day is in summer or rainy season given that Fred drinks iced tea? Season Summer Rainy Winter Total Coffee 1 12 1 6 4 15 31 60 Iced tea 1 4 1 6 1 15 29 60 Total 4 12 2 6 5 15 1 29 60 25 29 contains the number of crimes per 100,000 inhabitants in the United States over the span of several years. United States Crime Index Rates Per 100,000 Inhabitants Year Robbery Burglary Vandalism Vehicle Total 1 145.7 732.1 29.7 314.7 2 133.1 717.7 29.1 259.2 3 119.3 701 27.7 239.1 4 113.7 702.2 26.8 229.6 Total TOTAL each column and each row. Total data = 4,520.7 Find P (Year 2 AND Robbery). Find P (Year 3 AND Burglary). Find P (Year 3 OR Burglary). Find P (Year 4 | Vandalism). Find P (Vehicle | Year 1). a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575 Try It relates the weights and heights of a group of individuals participating in an observational study. Weight/height Tall Medium Short Totals Overweight 18 28 14 Typical Weight Range 20 51 28 Underweight 12 25 9 Totals Find the total for each row and column Find the probability that a randomly chosen individual from this group is Tall. Find the probability that a randomly chosen individual from this group is Overweight and Tall. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Overweight. Find the probability that a randomly chosen individual from this group is Overweight given that the individual is Tall. Find the probability a randomly chosen individual from this group is Tall and Underweight. Are the events Overweight and Tall independent? Weight/height Tall Medium Short Totals Obese 18 28 14 60 Normal 20 51 28 99 Underweight 12 25 9 46 Totals 50 104 51 205 Row Totals: 60, 99, 46. Column totals: 50, 104, 51. P (Tall) = 50 205 = 0.244 P ( Obese ∩ Tall ) = 18 205 = 0.088 P (Tall | Obese) = 18 60 = 0.3 P (Obese | Tall) = 18 50 = 0.36 P ( Tall ∩ Underweight ) = 12 205 = 0.0585 No. P (Tall) does not equal P (Tall | Obese). Tree Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities. Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of \"branches\" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram. In an urn, there are 11 balls. Three balls are red ( R ) and eight balls are blue ( B ). Draw two balls, one at a time, with replacement . \"With replacement\" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. Total = 64 + 24 + 24 + 9 = 121 The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R 1, R 2, and R 3 and each blue ball as B 1, B 2, B 3, B 4, B 5, B 6, B 7, and B 8. Then the nine RR outcomes can be written as: R 1 R 1 R 1 R 2 R 1 R 3 R 2 R 1 R 2 R 2 R 2 R 3 R 3 R 1 R 3 R 2 R 3 R 3 The other outcomes are similar. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space . a. List the 24 BR outcomes: B 1 R 1, B 1 R 2, B 1 R 3, ... a. B 1 R 1 B 1 R 2 B 1 R 3 B 2 R 1 B 2 R 2 B 2 R 3 B 3 R 1 B 3 R 2 B 3 R 3 B 4 R 1 B 4 R 2 B 4 R 3 B 5 R 1 B 5 R 2 B 5 R 3 B 6 R 1 B 6 R 2 B 6 R 3 B 7 R 1 B 7 R 2 B 7 R 3 B 8 R 1 B 8 R 2 B 8 R 3 b. Using the tree diagram, calculate P ( RR ). b. P ( RR ) = ( 3 11 ) ( 3 11 ) = 9 121 c. Using the tree diagram, calculate P ( RB ∪ BR ) . c. P ( RB ∪ BR ) = ( 3 11 ) ( 8 11 ) + ( 8 11 ) ( 3 11 ) = 48 121 d. Using the tree diagram, calculate P ( R on 1st draw ∩ B on 2nd draw ) . d. P ( R on 1st draw ∩ B on 2nd draw ) = ( 3 11 ) ( 8 11 ) = 24 121 e. Using the tree diagram, calculate P ( R on 2nd draw | B on 1st draw). e. P ( R on 2nd draw | B on 1st draw) = P ( R on 2nd | B on 1st) = 24 88 = 3 11 This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB ). Twenty-four of the 88 possible outcomes are BR . 24 88 = 3 11 . f. Using the tree diagram, calculate P ( BB ). f. P ( BB ) = 64 121 g. Using the tree diagram, calculate P ( B on the 2nd draw | R on the first draw). g. P ( B on 2nd draw | R on 1st draw) = 8 11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB ). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 24 33 . Try It In a standard deck, there are 52 cards. 12 cards are face cards (event F ) and 40 cards are not face cards (event N ). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P ( FF ). Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704. P ( FF ) = 144 144 + 480 + 480 + 1,600 = 144 2 , 704 = 9 169 An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. \"Without replacement\" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, ( 3 11 ) ( 2 10 ) = 6 110 . Total = 56 + 24 + 24 + 6 110 = 110 110 = 1 NOTE If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement , so that on the second draw there are ten marbles left in the urn. Calculate the following probabilities using the tree diagram. a. P ( RR ) = ________ a. P ( RR ) = ( 3 11 ) ( 2 10 ) = 6 110 b. Fill in the blanks: P ( RB ∪ BR ) = ( 3 11 ) ( 8 10 ) + (___)(___) = 48 110 b. P ( RB ∪ BR ) = ( 3 11 ) ( 8 10 ) + ( 8 11 ) ( 3 10 ) = 48 110 c. P ( R on 2nd | B on 1st) = c. P ( R on 2nd | B on 1st) = 3 10 d. Fill in the blanks. P ( R on 1st ∩ B on 2nd ) = (___)(___) = 24 110 d. P ( R on 1st ∩ B on 2nd ) = ( 3 11 ) ( 8 10 ) = 24 110 e. Find P ( BB ). e. P ( BB ) = ( 8 11 ) ( 7 10 ) f. Find P ( B on 2nd | R on 1st). f. Using the tree diagram, P ( B on 2nd | R on 1st) = P ( R | B ) = 8 10 . If we are using probabilities, we can label the tree in the following general way. P ( R | R ) here means P ( R on 2nd | R on 1st) P ( B | R ) here means P ( B on 2nd | R on 1st) P ( R | B ) here means P ( R on 2nd | B on 1st) P ( B | B ) here means P ( B on 2nd | B on 1st) Try It In a standard deck, there are 52 cards. Twelve cards are face cards ( F ) and 40 cards are not face cards ( N ). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities. Find P ( FN ∪ NF ) . Find P ( N | F ). Find P (at most one face card). Hint: \"At most one face card\" means zero or one face card. Find P (at least one face card). Hint: \"At least one face card\" means one or two face cards. P ( FN ∪ NF ) = 480 2,652 + 480 2,652 = 960 2,652 = 80 221 P ( N | F ) = 40 51 P (at most one face card) = (480 + 480 + 1,560) 2,652 = 2 , 520 2 , 652 P (at least one face card) = (132 + 480 + 480) 2,652 = 1,092 2,652 A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption. What is the probability that both kittens are tabby? a. ( 1 2 ) ( 1 2 ) b. ( 4 9 ) ( 4 9 ) c. ( 4 9 ) ( 3 8 ) d. ( 4 9 ) ( 5 9 ) What is the probability that one kitten of each coloring is selected? a. ( 4 9 ) ( 5 9 ) b. ( 4 9 ) ( 5 8 ) c. ( 4 9 ) ( 5 9 ) + ( 5 9 ) ( 4 9 ) d. ( 4 9 ) ( 5 8 ) + ( 5 9 ) ( 4 8 ) What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? What is the probability of choosing two kittens of the same color? a. c, b. d, c. 4 8 , d. 32 72 Try It Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? ( 4 7 ) ( 3 6 ) + ( 3 7 ) ( 4 6 ) References “Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (accessed May 3, 2013). Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services. Data from United States Senate. Available online at www.senate.gov (accessed May 2, 2013). Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcolm C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013). Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013). “United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013). Data from Clara County Public H.D. Data from the American Cancer Society. Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013). Data from the Federal Highway Administration, part of the United States Department of Transportation. Data from the United States Census Bureau, part of the United States Department of Commerce. Data from USA Today. “Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013). “Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at https://ropercenter.cornell.edu/?s=Search+for+Datasets (accessed February 6, 2019). Chapter Review There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilities that have multiple dependent variables. A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. Tree Diagram the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies) Contingency Table the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities.", "section": "Contingency Tables and Probability Trees", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Venn Diagrams A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Venn diagrams also help us to convert common English words into mathematical terms that help add precision. Venn diagrams are named for their inventor, John Venn, a mathematics professor at Cambridge and an Anglican minister. His main work was conducted during the late 1870's and gave rise to a whole branch of mathematics and a new way to approach issues of logic. We will develop the probability rules just covered using this powerful way to demonstrate the probability postulates including the Addition Rule, Multiplication Rule, Complement Rule, Independence, and Conditional Probability. Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A intersect B = A ∩ B = {6} and A union B = A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. . The Venn diagram is as follows: shows the most basic relationship among these numbers. First, the numbers are in groups called sets; set A and set B. Some number are in both sets; we say in set A ∩ in set B. The English word \"and\" means inclusive, meaning having the characteristics of both A and B, or in this case, being a part of both A and B. This condition is called the INTERSECTION of the two sets. All members that are part of both sets constitute the intersection of the two sets. The intersection is written as A ∩ B where ∩ is the mathematical symbol for intersection. The statement A ∩ B is read as \"A intersect B.\" You can remember this by thinking of the intersection of two streets. There are also those numbers that form a group that, for membership, the number must be in either one or the other group. The number does not have to be in BOTH groups, but instead only in either one of the two. These numbers are called the UNION of the two sets and in this case they are the numbers 1-5 (from A exclusively), 7-9 (from set B exclusively) and also 6, which is in both sets A and B. The symbol for the UNION is ∪ , thus A ∪ B = numbers 1-9, but excludes number 10, 11, and 12. The values 10, 11, and 12 are part of the universe, but are not in either of the two sets. Translating the English word \"AND\" into the mathematical logic symbol ∩ , intersection, and the word \"OR\" into the mathematical symbol ∪ , union, provides a very precise way to discuss the issues of probability and logic. The general terminology for the three areas of the Venn diagram in is shown in . Try It Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C ∩ P = {blue} and C ∪ P = {green, blue, purple, red, yellow} . Draw a Venn diagram representing this situation. Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = { TT , TH } and B = { TT , HT }. Therefore, A ∩ B = {TT} . A ∪ B = {TH, TT, HT} . The sample space when you flip two fair coins is X = { HH , HT , TH , TT }. The outcome HH is in NEITHER A NOR B . The Venn diagram is as follows: Try It Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A ∩ B = {3, 5} . A ∪ B = {1, 2, 3, 5} . The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood. The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor. We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor. P ( O ) = ___________ P ( R ) = ___________ P ( O ∩ R ) = ___________ P ( O ∪ R ) = ____________ In the Venn Diagram, describe the overlapping area using a complete sentence. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence. a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor. Try It Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time. If a student is selected at random, find the probability that the student belongs to a club. P ( C ) = 0.40 the probability that the student works part time. P ( PT ) = 0.50 the probability that the student belongs to a club AND works part time. P ( C ∩ PT ) = 0.05 the probability that the student belongs to a club given that the student works part time. P ( C | P T ) = P ( C ∩ P T ) P ( P T ) = 0.05 0.50 = 0.1 the probability that the student belongs to a club OR works part time. P ( C ∪ PT ) = P ( C ) + P ( PT ) - P ( C ∩ PT ) = 0.40 + 0.50 - 0.05 = 0.85 In order to solve we had to draw upon the concept of conditional probability from the previous section. There we used tree diagrams to track the changes in the probabilities, because the sample space changed as we drew without replacement. In short, conditional probability is the chance that something will happen given that some other event has already happened. Put another way, the probability that something will happen conditioned upon the situation that something else is also true. In the probability P ( C | PT ) is the conditional probability that the randomly drawn student is a member of the club, conditioned upon the fact that the student also is working part time. This allows us to see the relationship between Venn diagrams and the probability postulates. Try It Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works. In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2. Draw a Venn diagram representing the situation. Find the probability that the customer buys either a novel or a non-fiction book. In the Venn diagram, describe the overlapping area using a complete sentence. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks. b. P (novel or non-fiction) = P (Blue ∪ Red) = P (Blue) + P (Red) - P (Blue ∩ Red) = 0.6 + 0.4 - 0.2 = 0.8. c. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book. A set of 20 German Shepherd dogs is observed. 12 are male, 8 are female, 10 have some brown coloring, and 5 have some white sections of fur. Answer the following using Venn Diagrams. Draw a Venn diagram simply showing the sets of male and female dogs. The Venn diagram below demonstrates the situation of mutually exclusive events where the outcomes are independent events. If a dog cannot be both male and female, then there is no intersection. Being male precludes being female and being female precludes being male: in this case, the characteristic gender is therefore mutually exclusive. A Venn diagram shows this as two sets with no intersection. The intersection is said to be the null set using the mathematical symbol ∅. Draw a second Venn diagram illustrating that 10 of the male dogs have brown coloring. The Venn diagram below shows the overlap between male and brown where the number 10 is placed in it. This represents Male ∩ Brown : both male and brown. This is the intersection of these two characteristics. To get the union of Male and Brown, then it is simply the two circled areas minus the overlap. In proper terms, Male ∪ Brown = Male + Brown − Male ∩ Brown will give us the number of dogs in the union of these two sets. If we did not subtract the intersection, we would have double counted some of the dogs. Now draw a situation depicting a scenario in which the non-shaded region represents \"No white fur and female,\" or White fur′ ∩ Female. the prime above \"fur\" indicates \"not white fur.\" The prime above a set means not in that set, e.g. A ′ means not A . Sometimes, the notation used is a line above the letter. For example, A ¯ = A ′ . The Addition Rule of Probability We met the addition rule earlier but without the help of Venn diagrams. Venn diagrams help visualize the counting process that is inherent in the calculation of probability. To restate the Addition Rule of Probability: P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) Remember that probability is simply the proportion of the objects we are interested in relative to the total number of objects. This is why we can see the usefulness of the Venn diagrams. shows how we can use Venn diagrams to count the number of dogs in the union of brown and male by reminding us to subtract the intersection of brown and male. We can see the effect of this directly on probabilities in the addition rule. Try It A set of 40 people are selected randomly. 24 are male, 16 are female, 21 like coffee, and 19 like tea. Answer the following using Venn diagrams. Draw a Venn diagram simply showing the sets of male and female. Draw a second Venn diagram illustrating 14 males liking coffee. Draw a situation with a region depicting female and not liking tea. IBS2e_UnFigure_03_Ex31_001 IBS2e_UnFigure_03_Ex31_002 IBS2e_UnFigure_03_Ex31_003 Let's sample 50 students who are in a statistics class. 20 are first-year students and 30 are sophomores. 15 students get a \"B\" in the course, and 5 students both get a \"B\" and are first-year students. Find the probability of selecting a student who either earns a \"B\" OR is a first-year student. We are translating the word OR to the mathematical symbol for the addition rule, which is the union of the two sets. We know that there are 50 students in our sample, so we know the denominator of our fraction to give us probability. We need only to find the number of students that meet the characteristics we are interested in, i.e. any first-year students and any student who earned a grade of \"B.\" With the Addition Rule of probability, we can skip directly to probabilities. Let \"A\" = the number of first-year students, and let \"B\" = the grade of \"B.\" Below we can see the process for using Venn diagrams to solve this. The P ( A ) = 20 50 = 0.40 , P ( B ) = 15 50 = 0.30 , and P ( A ∩ B ) = 5 50 = 0.10 . Therefore, P ( A ∪ B ) = 0.40 + 0.30 − 0.10 = 0.60 . If two events are mutually exclusive, then, like the example where we diagram the male and female dogs, the addition rule is simplified to just P ( A ∪ B ) = P ( A ) + P ( B ) − 0 . This is true because, as we saw earlier, the union of mutually exclusive events is the null set, ∅. The diagrams below demonstrate this. Try It A set of 40 random people is taken. 15 are male, and 25 are female. 18 people like coffee. 10 people are male and like coffee. Find the probability of selecting a person that likes coffee or is male. Let “A” = the number of males and let “B” = people who like coffee. The probabilities obtained are: P ( A ) = 15 40 = 0 . 375 , P ( B ) = 18 40 = 0 . 45 , P ( A ∩ B ) = 10 40 = 0 . 25 So, P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) = 0 . 375 + 0 . 45 - 0 . 25 = 0 . 575 . So the probability that a person likes coffee or is male is 0.575. The Multiplication Rule of Probability Restating the Multiplication Rule of Probability using the notation of Venn diagrams, we have: P ( A ∩ B ) = P ( A | B ) ⋅ P ( B ) The multiplication rule can be modified with a bit of algebra into the following conditional rule. Then Venn diagrams can then be used to demonstrate the process. The conditional rule: P ( A | B ) = P ( A ∩ B ) P ( B ) Using the same facts from above, find the probability that someone will earn a \"B\" if they are a \"first-year students.\" P ( A | B ) = 0.10 0.30 = 1 3 The multiplication rule must also be altered if the two events are independent. Independent events are defined as a situation where the conditional probability is simply the probability of the event of interest. Formally, independence of events is defined as P ( A | B ) = P ( A ) or P ( B | A ) = P ( B ) . When flipping coins, the outcome of the second flip is independent of the outcome of the first flip; coins do not have memory. The Multiplication Rule of Probability for independent events thus becomes: P ( A ∩ B ) = P ( A ) ⋅ P ( B ) One easy way to remember this is to consider what we mean by the word \"and.\" We see that the Multiplication Rule has translated the word \"and\" to the Venn notation for intersection. Therefore, the outcome must meet the two conditions of first-year and grade of \"B\" in the above example. It is harder, less probable, to meet two conditions than just one or some other one. We can attempt to see the logic of the Multiplication Rule of probability due to the fact that fractions multiplied times each other become smaller. The development of the Rules of Probability with the use of Venn diagrams can be shown to help as we wish to calculate probabilities from data arranged in a contingency table. is from a sample of 200 people who were asked how much education they completed. The columns represent the highest education they completed, and the rows separate the individuals by men and women. Less than high school grad High school grad Some college College grad Total Men 5 15 40 60 120 Women 8 12 30 30 80 Total 13 27 70 90 200 Now, we can use this table to answer probability questions. The following examples are designed to help understand the format above while connecting the knowledge to both Venn diagrams and the probability rules. What is the probability that a selected person both finished college and is a woman? This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes. P ( College Grad ∩ Female ) = 30 200 = 0.15 What is the probability of selecting either a woman or someone who finished college? This task involves the use of the addition rule to solve for this probability. P ( College Grad ∪ Female ) = P ( F ) + P ( CG )− P ( F ∩ CG ) P ( College Grad ∪ Female ) = 80 200 + 90 200 − 30 200 = 140 200 = 0.70 What is the probability of selecting a high school graduate if we only select from the group of men? Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability. P ( HS Grad | Man = P ( HS Grad ∩ Man ) P ( Man ) = ( 15 200 ) ( 120 200 ) = 15 120 = 0.125 Can we conclude that the level of education attained by these 200 people is independent of the gender of the person? There are two ways to approach this test. The first method seeks to test if the intersection of two events equals the product of the events separately remembering that if two events are independent than P ( A )* P ( B ) = P ( A ∩ B ). For simplicity's sake, we can use calculated values from above. Does P(College Grad ∩ Woman) = P(CG) ⋅ P(F)? 30 200 ≠ 90 200 ⋅ 80 200 because 0.15 ≠ 0.18. Therefore, gender and education here are not independent. The second method is to test if the conditional probability of A given B is equal to the probability of A . Again for simplicity, we can use an already calculated value from above. Does P(HS Grad | Man) = P(HS Grad)? 15 120 ≠ 27 200 because 0.125 ≠ 0.135. Therefore, again gender and education here are not independent. Try It The table shows a spread of what 100 random people like to eat. The columns represent the fruit that they like to eat, and the rows separate the individuals by men and women. Grape Watermelon Apple Pomegranate Total Men 12 18 20 10 60 Women 13 7 15 5 40 Total 25 25 35 15 100 Answer the following questions: What is the probability that a selected person likes pomegranate and is a man? What is the probability of selecting either a man or someone who likes apple? What is the probability of selecting a person who likes watermelon if we select from the group of women? P ( P o m e g a n a t e ∩ m a l e ) = 10 100 = 0 . 1 P ( M a l e ∪ A p p l e ) = P ( M a l e ) + P ( A p p l e ) - P ( M a l e ∩ A p p l e ) = 60 100 + 35 100 - 20 100 = 0 . 75 P ( W a t e r m e l o n | F e m a l e ) = P ( W a t e r m e l o n ∩ F e m a l e ) P ( F e m a l e ) = 7 100 40 100 = 7 40 = 0 . 175 Chapter Review A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S or universe of the objects of interest together with circles or ovals. The circles or ovals represent groups of events called sets. A Venn diagram is especially helpful for visualizing the ∪ event, the ∩ event, and the complement of an event and for understanding conditional probabilities. A Venn diagram is especially helpful for visualizing an Intersection of two events, a Union of two events, or a Complement of one event. A system of Venn diagrams can also help to understand Conditional probabilities. Venn diagrams connect the brain and eyes by matching the literal arithmetic to a picture. It is important to note that more than one Venn diagram is needed to solve the probability rule formulas introduced in Section 3.3 . Use the following information to answer the next four exercises. shows a random sample of musicians and how they learned to play their instruments. Gender Self-taught Studied in school Private instruction Total Women 12 38 22 72 Men 19 24 15 58 Total 31 62 37 130 Find P (musician is a woman). Find P (musician is a man ∩ had private instruction). P (musician is a male ∩ had private instruction) = 15 130 = 3 26 = 0.12 Find P (musician is a woman ∪ is self taught). Are the events “being a woman musician” and “learning music in school” mutually exclusive events? The events are not mutually exclusive. It is possible to be a female musician who learned music in school. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation. Bringing It Together Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine , reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Latino people, 8,378 Japanese American people, and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people. Complete the table using the data provided. Smoking Levels by Ethnicity Smoking level Black people Native Hawaiian Latino Japanese American White people TOTALS 1–10 11–20 21–30 31+ TOTALS Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day. 35,065 100,450 Find the probability that the person was Latino. In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability. To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is 4,715 100,450 . In words, explain what it means to pick one person from the study who is “Japanese American ∪ smokes 21 to 30 cigarettes per day.” Also, find the probability. In words, explain what it means to pick one person from the study who is “Japanese American | that person smokes 21 to 30 cigarettes per day.” Also, find the probability. To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is 4715 15,273 . Prove that smoking level/day and ethnicity are dependent events. Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. Suppose that you randomly draw two cards, one at a time, with replacement . Let G 1 = first card is green Let G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 ∩ G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 independent events? Explain why or why not. P ( GG ) = ( 5 8 ) ( 5 8 ) = 25 64 P (at least one green) = P ( GG ) + P ( GY ) + P ( YG ) = 25 64 + 15 64 + 15 64 = 55 64 P ( G | G ) = 5 8 Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two. Suppose that you randomly draw two cards, one at a time, without replacement . G 1 = first card is green G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 ∩ G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 independent events? Explain why or why not. Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are women is 48.60. Of the women, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. men drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. Complete the following. Construct a table or a tree diagram of the situation. Find P (driver is a woman). Find P (driver is age 65 or over | driver is a woman). Find P (driver is age 65 or over ∩ a woman). In words, explain the difference between the probabilities in part c and part d. Find P (driver is age 65 or over). Are being age 65 or over and being a woman mutually exclusive events? How do you know? <20 20–64 >64 Totals Female 0.0244 0.3954 0.0661 0.486 Male 0.0259 0.4186 0.0695 0.514 Totals 0.0503 0.8140 0.1356 1 P ( F ) = 0.486 P (>64 | F ) = 0.1361 P (>64 and F ) = P ( F ) P (>64| F ) = (0.486)(0.1361) = 0.0661 P (>64 | F ) is the percentage of female drivers who are 65 or older and P (>64 ∩ F ) is the percentage of drivers who are female and 65 or older. P (> 64 ) = P (>64 ∩ F ) + P (>64 ∩ M ) = 0.1356 No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 ∩ F ) = 0.0661. Suppose that 10,000 U.S. licensed drivers are randomly selected. How many would you expect to be men? Using the table or tree diagram, construct a contingency table of gender versus age group. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is a woman. Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work. Assuming that the walkers walk alone, what percent of all commuters travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool? Car, truck or van Walk Public transportation Other Totals Alone 0.7318 Not alone 0.1332 Totals 0.8650 0.0390 0.0530 0.0430 1 If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P (Alone) = 0.7318 + 0.0390 = 0.7708. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771 (0.1332)(1,000) = 133 When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H ) while 110 showed a tail (event T ). On that basis, they claimed that it is not a fair coin. Based on the given data, find P ( H ) and P ( T ). Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin. Use the tree to find the probability of obtaining at least one head. Homework Use the information in the to answer the next eight exercises. The table shows the political party affiliation for various members of the U.S. Senate during two separate years when they are up for reelection. Up for reelection: Democratic party Republican party Other Total Year A 20 13 0 Year B 10 24 0 Total What is the probability that a randomly selected senator has an “Other” affiliation? 0 What is the probability that a randomly selected senator is up for reelection in Year B? What is the probability that a randomly selected senator is a Democrat and up for reelection in Year B? 10 67 What is the probability that a randomly selected senator is a Republican or is up for reelection in Year A? Suppose that a member of the U.S. Senate is randomly selected. Given that the randomly selected senator is up for reelection in Year B, what is the probability that this senator is a Democrat? 10 34 Suppose that a member of the U.S. Senate is randomly selected. What is the probability that the senator is up for reelection in Year A, knowing that this senator is a Republican? The events “Republican” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. d The events “Other” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. gives the number of participants in the recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted by age, race (Black or White), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population. Race and sex 15–24 25–40 41–65 Over 65 TOTALS White, male 1,165 2,036 3,703 8,395 White, female 1,076 2,242 4,060 9,129 Black, male 142 194 384 824 Black, female 131 290 486 1,061 All others TOTALS 2,792 5,279 9,354 21,081 Do not include \"all others\" for parts f and g. Fill in the column for cancer treatment for individuals over age 65. Fill in the row for all other races. Find the probability that a randomly selected individual was a White male. Find the probability that a randomly selected individual was a Black female. Find the probability that a randomly selected individual was Black Find the probability that a randomly selected individual was male. Out of the individuals over age 65, find the probability that a randomly selected individual was a Black or White male. Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 156 TOTALS 2,792 5,279 9,354 3,656 21,081 Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 278 517 721 156 1672 TOTALS 2,792 5,279 9,354 3,656 21,081 8,395 21,081 ≈ 0.3982 1,061 21,081 ≈ 0.0503 1,885 21,081 ≈ 0.0894 9,219 19,409 ≈ 0.475 1,595 3,500 ≈ 0.456 Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home run TOTAL HITS Babe Ruth 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,053 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 TOTAL 7,918 2,127 583 1,723 12,351 Find P (hit was made by Babe Ruth). 1518 2873 2873 12351 583 12351 4189 12351 Find P (hit was made by Ty Cobb | The hit was a Home Run). 4189 12351 117 1723 1723 4189 117 12351 b identifies a group of children by one of four hair colors, and by type of hair. Hair type Brown Blond Black Red Totals Wavy 20 15 3 43 Straight 80 15 12 Totals 20 215 Complete the table. What is the probability that a randomly selected child will have wavy hair? What is the probability that a randomly selected child will have either brown or blond hair? What is the probability that a randomly selected child will have wavy brown hair? What is the probability that a randomly selected child will have red hair, given that they have straight hair? If B is the event of a child having brown hair, find the probability of the complement of B . In words, what does the complement of B represent? In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data were compiled into the following table. Shirt # ≤ 210 211–250 251–290 > 290 1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. Find the probability that his shirt number is from 1 to 33. Find the probability that he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds. 26 106 33 106 21 106 ( 26 106 ) + ( 33 106 ) - ( 21 106 ) = ( 38 106 ) 21 33 Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red ( R ), four yellow ( Y ) and five blue ( B ) beads. For the coin, P ( H ) = 2 3 and P ( T ) = 1 3 where H is heads and T is tails. Find P (tossing a Head on the coin AND a Red bead) 2 3 5 15 6 36 5 36 Find P (Blue bead). 15 36 10 36 10 12 6 36 a A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?) Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain. For each complete path through the tree, write the event it represents and find the probabilities. Let S be the event that both cookies selected were the same flavor. Find P ( S ). Let T be the event that the cookies selected were different flavors. Find P ( T ) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods. Let U be the event that the second cookie selected is a butter cookie. Find P ( U ). Venn Diagram the visual representation of a sample space and events in the form of circles or ovals showing their intersections", "section": "Venn Diagrams", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction You can use probability and discrete random variables to calculate the likelihood of lightning striking the ground five times during a half-hour thunderstorm. (credit: modification of work “CG lightning strike” by Axel Rouvin/ Flickr, CC BY 2.0) A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, they could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%? The manager of an auto dealership might be interested in the color preferences for new car buyers. Suppose on average the dealership sells 20 cars per month. What is the probability that a customer prefers red cars? These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count, that is, the random variable can only take on whole number values. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment, often called a trial. Random Variable Notation The upper case letter X denotes a random variable. Lower case letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is given as a number. For example, let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT ; THH ; HTH ; HHT ; HTT ; THT ; TTH ; HHH . Then, x = 0, 1, 2, 3. X is in words and x is a number. Notice that for this example, the x values are countable outcomes. Because you can count the possible values as whole numbers that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is a discrete random variable. Probability Density Functions (PDF) for a Random Variable A probability density function or probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one. A probability density function is a mathematical formula that calculates probabilities for specific types of events, what we have been calling experiments. There is a sort of magic to a probability density function (Pdf) partially because the same formula often describes very different types of events. For example, the binomial Pdf will calculate probabilities for flipping coins, yes/no questions on an exam, opinions of voters in an up or down opinion poll, indeed any binary event. Other probability density functions will provide probabilities for the time until a part will fail, when a customer will arrive at the turnpike booth, the number of telephone calls arriving at a central switchboard, the growth rate of a bacterium, and on and on. There are whole families of probability density functions that are used in a wide variety of applications, including medicine, business and finance, physics and engineering, among others. For our needs here we will concentrate on only a few probability density functions as we develop the tools of inferential statistics. Counting Formulas and the Combinational Formula To repeat, the probability of event A , P(A), is simply the number of ways the experiment will result in A, relative to the total number of possible outcomes of the experiment. As an equation this is: P ( A ) = number of ways to get A Total number of possible outcomes When we looked at the sample space for flipping 3 coins we could easily write the full sample space and thus could easily count the number of events that met our desired result, e.g. x = 1 , where X is the random variable defined as the number of heads. As we have larger numbers of items in the sample space, such as a full deck of 52 cards, the ability to write out the sample space becomes impossible. We see that probabilities are nothing more than counting the events in each group we are interested in and dividing by the number of elements in the universe, or sample space. This is easy enough if we are counting sophomores in a Stat class, but in more complicated cases listing all the possible outcomes may take a life time. There are, for example, 36 possible outcomes from throwing just two six-sided dice where the random variable is the sum of the number of spots on the up-facing sides. If there were four dice then the total number of possible outcomes would become 1,296. There are more than 2.5 MILLION possible 5 card poker hands in a standard deck of 52 cards. Obviously keeping track of all these possibilities and counting them to get at a single probability would be tedious at best. An alternative to listing the complete sample space and counting the number of elements we are interested in, is to skip the step of listing the sample space, and simply figuring out the number of elements in it and doing the appropriate division. If we are after a probability we really do not need to see each and every element in the sample space, we only need to know how many elements are there. Counting formulas were invented to do just this. They tell us the number of unordered subsets of a certain size that can be created from a set of unique elements. By unordered it is meant that, for example, when dealing cards, it does not matter if you got {ace, ace, ace, ace, king} or {king, ace, ace, ace, ace} or {ace, king, ace, ace, ace} and so on. Each of these subsets are the same because they each have 4 aces and one king. Combinational Formula n x = n C x = n ! x ! ( n - x ) ! This is the formula that tells the number of unique unordered subsets of size x that can be created from n unique elements. The formula is read “n combinatorial x”. Sometimes it is read as “n choose x.\" The exclamation point \"!\" is called a factorial and tells us to take all the numbers from 1 through the number before the ! and multiply them together thus 4! is 1·2·3·4=24. By definition 0! = 1. The formula is called the Combinatorial Formula. It is also called the Binomial Coefficient, for reasons that will be clear shortly. While this mathematical concept was understood long before 1653, Blaise Pascal is given major credit for his proof that he published in that year. Further, he developed a generalized method of calculating the values for combinatorials known to us as the Pascal Triangle. Pascal was one of the geniuses of an era of extraordinary intellectual advancement which included the work of Galileo, Rene Descartes, Isaac Newton, William Shakespeare and the refinement of the scientific method, the very rationale for the topic of this text. Let’s find the hard way the total number of combinations of the four aces in a deck of cards if we were going to take them two at a time. The sample space would be: S={Spade,Heart),(Spade, Diamond),(Spade,Club), (Diamond,Club),(Heart,Diamond),(Heart,Club)} There are 6 combinations; formally, six unique unordered subsets of size 2 that can be created from 4 unique elements. To use the combinatorial formula we would solve the formula as follows: 4 2 = 4 ! ( 4 - 2 ) ! 2 ! = 4 · 3 · 2 · 1 2 · 1 · 2 · 1 = 6 If we wanted to know the number of unique 5 card poker hands that could be created from a 52 card deck we simply compute: 52 5 where 52 is the total number of unique elements from which we are drawing and 5 is the size group we are putting them into. With the combinatorial formula we can count the number of elements in a sample space without having to write each one of them down, truly a lifetime's work for just the number of 5 card hands from a deck of 52 cards. We can now apply this tool to a very important probability density function, the hypergeometric distribution. Remember, a probability density function computes probabilities for us. We simply put the appropriate numbers in the formula and we get the probability of specific events. However, for these formulas to work they must be applied only to cases for which they were designed. Chapter Review The characteristics of a probability distribution or density function (PDF) are as follows: Each probability is between zero and one, inclusive ( inclusive means to include zero and one). The sum of the probabilities is one. Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution. Let X = the number of years a new hire will stay with the company. Let P ( x ) = the probability that a new hire will stay with the company x years. Complete using the data provided. x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 5 0.10 6 0.05 x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 0.10 5 0.10 6 0.05 P ( x = 4) = _______ P ( x ≥ 5) = _______ 0.10 + 0.05 = 0.15 On average, how long would you expect a new hire to stay with the company? What does the column “ P ( x )” sum to? 1 Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution. x P ( x ) 1 0.15 2 0.35 3 0.40 4 0.10 Define the random variable X . What is the probability the baker will sell more than one batch? P ( x > 1) = _______ 0.35 + 0.40 + 0.10 = 0.85 What is the probability the baker will sell exactly one batch? P ( x = 1) = _______ On average, how many batches should the baker make? 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45 Use the following information to answer the next four exercises: Emery has music practice three days a week. They practice for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. Define the random variable X . Construct a probability distribution table for the data. x P ( x ) 0 0.03 1 0.04 2 0.08 3 0.85 We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic? Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time. Define the random variable X . Let X = the number of events Javier volunteers for each month. What values does x take on? Construct a PDF table. x P ( x ) 0 0.05 1 0.05 2 0.10 3 0.20 4 0.25 5 0.35 Find the probability that Javier volunteers for less than three events each month. P ( x < 3) = _______ Find the probability that Javier volunteers for at least one event each month. P ( x > 0) = _______ 1 – 0.05 = 0.95 Random Variable (RV) a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters X , Y , Z ,...; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters x, y, and z . For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in the two following ways. The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange}. We can tell what specific value x the random variable X takes only after performing the experiment. Probability Distribution Function (PDF) a mathematical description of a discrete random variable ( RV ), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Hypergeometric Distribution The simplest probability density function is the hypergeometric. This is the most basic one because it is created by combining our knowledge of probabilities from Venn diagrams, the addition and multiplication rules, and the combinatorial counting formula. To find the number of ways to get 2 aces from the four in the deck we computed: 4 2 = 4 ! 2 ! ( 4 - 2 ) ! = 6 And if we did not care what else we had in our hand for the other three cards we would compute: 48 3 = 48 ! 3 ! 45 ! = 17,296 Putting this together, we can compute the probability of getting exactly two aces in a 5 card poker hand as: 4 2 48 3 52 5 = .0399 This solution is really just the probability distribution known as the Hypergeometric. The generalized formula is: h ( x ) = A x N - A n - x N n where x = the number we are interested in coming from the group with A objects. h(x) is the probability of x successes, in n attempts, when A successes (aces in this case) are in a population that contains N elements. The hypergeometric distribution is an example of a discrete probability distribution because there is no possibility of partial success, that is, there can be no poker hands with 2 1/2 aces. Said another way, a discrete random variable has to be a whole, or counting, number only. This probability distribution works in cases where the probability of a success changes with each draw. Another way of saying this is that the events are NOT independent. In using a deck of cards, we are sampling WITHOUT replacement. If we put each card back after it was drawn then the hypergeometric distribution be an inappropriate Pdf. For the hypergeometric to work, the population must be dividable into two and only two independent subsets (aces and non-aces in our example). The random variable X = the number of items from the group of interest. the experiment must have changing probabilities of success with each experiment (the fact that cards are not replaced after the draw in our example makes this true in this case). Another way to say this is that you sample without replacement and therefore each pick is not independent. the random variable must be discrete, rather than continuous. A candy dish contains 30 jelly beans and 20 gumdrops. Ten candies are picked at random. What is the probability that 5 of the 10 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group A in the formula) is gumdrops. The size of the group of interest (first group) is 30. The size of the second group is 20. The size of the sample is 10 (jelly beans or gumdrops). Let X = the number of gumdrops in the sample of 10. X takes on the values x = 0, 1, 2, ..., 10. a. What is the probability statement written mathematically? b. What is the hypergeometric probability density function written out to solve this problem? c. What is the answer to the question \"What is the probability of drawing 5 gumdrops in 10 picks from the dish?\" a. P ( x = 5 ) b. P ( x = 5 ) = ( 5 30 ) ( 5 20 ) ( 10 50 ) c. P ( x = 5 ) = 0.215 Try It A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample? The group of interest is the vowel letter tiles. The size of the group of interest is 44. The size of the sample is seven. TSA officers claim that they randomly select boarding passengers for additional search of carry-on luggage. During the past hour, of the 100 passengers passing through the security point, 5 were silver-haired grandmothers, of which 3 were chosen for additional screening among the 20 total chosen for “enhanced vetting.” Calculate the relevant probability and comment. Step 1. Determine what the random variable is. Or asked another way, what are we interested in? In this case it is the number of silver-haired grandmothers who were stopped for additional search of carry-on luggage. We also determine that this is a discrete random variable because it is a counting number. Further, we determine that X can take on only the values X = 0, 1, 2, 3, 4, 5. Step 2. What is the experiment? Here passengers are selected we are told randomly for additional baggage search. This would be selection without replacement. Step 3. Can the objects, the passengers here, be divide into at most two categories? Yes, here because we are interested in only the category silver-haired grandmothers and everyone else. These data can be analyzed with the hypergeometric probability density function because it meets all the requirements of that distribution. Step 4. What is the question asked? Unlike Example 4.1 about drawing candies from a bowl, there is no obvious question being asked. The candy bowl problem is mechanical with an obvious question directly stated concerning the probability of drawing 5 gumdrops. In short, an obvious question with no implications. Do we really care about the probability of drawing 5 gumdrops? Here there is an interesting question because the security office made a claim that we may indirectly test using simple probability. The security office claims they randomly select passengers for enhanced vetting. From the data we see immediately that of the 5 silver-haired grandmother passengers, 3 were selected for additional search. This is 60 percent of all silver-haired grandmothers. We can form our own question to determine the probability of selecting 3 of 5 in this category of passengers if we are selecting randomly. Remember that random selection means that each object has an equal probability of being selecting. Our probability question in words is thus “What is the probability of selecting 3 silver-haired grandmothers from a group of 100 passengers if the group included only 5 silver-haired grandmothers?” Formally, what is P ( X = 3) using the hypergeometric probability distribution? From the formula: X = 3, A = 5 N = 100 (the total number of passengers from which those will be drawn for enhanced vetting) n= 20 (the number of passengers that will be selected for enhanced vetting) P ( x ) = ( 5 3 ) ( 95 17 ) ( 100 20 ) = 0.048 Our conclusion is that the probability of the outcome of 3 of 5 silver-haired grandmothers being randomly selected for additional luggage search is only 0.048. This is only one one-thousandth better than drawing 5 cards from a 52-card deck and getting a pair of aces. In short, an unrealistically small probability. A comment here would be that we can with only just under 5% confidence believe that the passengers are selected on a purely random basis. This example demonstrates the ability to answer questions of some interest and policy implication with the use of the tools of probability analysis. Try It There are a total of 200 balls in a bag. This includes 25 red balls. 40 balls are selected randomly out of the bag. 8 of them are found to be red. Find the relevant probability. From the given values, it is obtained that X = 8 , A = 25 , N = 200 and n = 40 . P ( x ) = 25 8 175 32 200 40 = 0 . 058 Chapter Review The combinatorial formula can provide the number of unique subsets of size x that can be created from n unique objects to help us calculate probabilities. The combinatorial formula is n x = n C x = n ! x ! ( n - x ) ! A hypergeometric experiment is a statistical experiment with the following properties: You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. Each pick is not independent, since sampling is without replacement. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. h ( x ) = A x N - A n - x N n . Formula Review h ( x ) = A x N - A n - x N n Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. In words, define the random variable X . X = the number of business majors in the sample. What values does X take on? 2, 3, 4, 5, 6, 7, 8, 9 HOMEWORK A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration. In words, define the random variable X . List the values that X may take on. How many Shotokan Karate students do we expect to be in that first demonstration? In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once. In words, define the random variable X . List the values that X may take on. How many pages do you expect to advertise footwear on them? Calculate the standard deviation. X = the number of pages that advertise footwear 0, 1, 2, 3, ..., 20 3.03 1.5197 Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient. In words, define the random variable X . List the values that X may take on. How many instructors do you expect on the committee who are not technically proficient? Find the probability that at least five on the committee are not technically proficient. Find the probability that at most three on the committee are not technically proficient. Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked. In words, define the random variable X . List the values that X may take on. Are you choosing the nine athletes with or without replacement? X = the number of Patriots picked 0, 1, 2, 3, 4 Without replacement A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart? What is the group of interest? How many are in the group of interest? How many are in the other group? Let X = _________. What values does X take on? The probability question is P (_______). Find the probability in question. Find the (i) mean and (ii) standard deviation of X . Hypergeometric Experiment a statistical experiment with the following properties: You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. Each pick is not independent, since sampling is without replacement. Hypergeometric Probability a discrete random variable (RV) that is characterized by: A fixed number of trials. The probability of success is not the same from trial to trial. We sample from two groups of items when we are interested in only one group. X is defined as the number of successes out of the total number of items chosen.", "section": "Hypergeometric Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Binomial Distribution A more valuable probability density function with many applications is the binomial distribution. This distribution will compute probabilities for any binomial process. A binomial process, often called a Bernoulli process after the first person to fully develop its properties, is any case where there are only two possible outcomes in any one trial, called successes and failures. It gets its name from the binary number system where all numbers are reduced to either 1's or 0's, which is the basis for computer technology and CD music recordings. Binomial Formula b ( x ) = n x p x q n - x where b(x) is the probability of X successes in n trials when the probability of a success in ANY ONE TRIAL is p. And of course q=(1-p) and is the probability of a failure in any one trial. We can see now why the combinatorial formula is also called the binomial coefficient because it reappears here again in the binomial probability function. For the binomial formula to work, the probability of a success in any one trial must be the same from trial to trial, or in other words, the outcomes of each trial must be independent. Flipping a coin is a binomial process because the probability of getting a head in one flip does not depend upon what has happened in PREVIOUS flips. (At this time it should be noted that using p for the parameter of the binomial distribution is a violation of the rule that population parameters are designated with Greek letters. In many textbooks θ (pronounced theta) is used instead of p and this is how it should be.) Just like a set of data, a probability density function has a mean and a standard deviation that describes the data set. For the binomial distribution these are given by the formulas: μ = np σ = n p q Notice that p is the only parameter in these equations. The binomial distribution is thus seen as coming from the one-parameter family of probability distributions. In short, we know all there is to know about the binomial distribution once we know p, the probability of a success in any one trial. In probability theory, under certain circumstances, one probability distribution can be used to approximate another. We say that one is the limiting distribution of the other. If a small number is to be drawn from a large population, even if there is no replacement, we can still use the binomial even thought this is not a binomial process. If there is no replacement it violates the independence rule of the binomial. Nevertheless, we can use the binomial to approximate a probability that is really a hypergeometric distribution if we are drawing fewer than 10 percent of the population, i.e. n is less than 10 percent of N in the formula for the hypergeometric function. The rationale for this argument is that when drawing a small percentage of the population we do not alter the probability of a success from draw to draw in any meaningful way. Imagine drawing from not one deck of 52 cards but from 6 decks of cards. The probability of say drawing an ace does not change the conditional probability of what happens on a second draw in the same way it would if there were only 4 aces rather than the 24 aces now to draw from. This ability to use one probability distribution to estimate others will become very valuable to us later. There are three characteristics of a binomial experiment. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials. The random variable, x , number of successes, is discrete. There are only two possible outcomes, called \"success\" and \"failure,\" for each trial. The letter p denotes the probability of a success on any one trial, and q denotes the probability of a failure on any one trial. p + q = 1. The n trials are independent and are repeated using identical conditions. Think of this as drawing WITH replacement. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p , of a success and probability, q , of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with a probability p = 0.6. Then, q = 0.4. This means that for every true-false statistics question Joe answers, his probability of success ( p = 0.6) and his probability of failure ( q = 0.4) remain the same. The outcomes of a binomial experiment fit a binomial probability distribution . The random variable X = the number of successes obtained in the n independent trials. The mean, μ , and variance, σ 2 , for the binomial probability distribution are μ = np and σ 2 = npq . The standard deviation, σ , is then σ = n p q . Any experiment that has characteristics three and four and where n = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials. Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can be stated mathematically as P ( x = 15). P ( x = 15 ) stated more carefully in English would be stated as “the probability of exactly 15 wins in 20 trials.” P ( x < 15 ) would need again a more careful statement because if we desire P ( x < 15 ) we need to calculate all the values of x from P : ( x = 0 , 1 , 2 , … . 14 ) . Similarly we may wish to know the probability “greater than” some number, which requires multiple calculations of the values of x . Try It A trainer is teaching a rescued dolphin to catch live fish before returning it to the wild. The probability that the dolphin successfully catches a fish is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. Find the P(X=12) using the binomial Pdf. P ( x = 12) A coin has been altered to weight the outcome from 0.5 to 0.25 and is flipped 5 times. Each flip is independent. What is the probability of getting more than 3 heads? Let X = the number of heads in 5 flips of the fair coin. X takes on the values 0, 1, 2, 3, 4, 5. Since the coin is altered to result in p = 0.25, q is 0.75. The number of trials is n = 5. State the probability question mathematically. P ( x > 10) First develop fully the probability density function and graph the probability density function. With the fully developed probability density function we can simply read the solution to the question P ( x > 3 ) heads. P ( x > 3 ) = P ( x = 4 ) + P ( x = 5 ) = 0 . 0146 + 0 . 0007 = 0 . 0153 . We have added the two individual probabilities because of the addition rule from Probability Topics . also allows us to see the link between the probability density function and probability and area. We also see in the skew of the binomial distribution when p is not equal to 0.5. In the distribution is skewed right as a result of μ = n p = 1 . 25 because p = 0 . 25 . P x = x 0 = n x p x 1 - p n - x = 5 x 0 · 25 x 0 · 75 5 - x 0 etc . μ = np = 1 . 25 Try It A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically. Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly. a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial. b. If we are interested in the number of students who do their homework on time, then how do we define X ? c. What values does x take on? d. What is a \"failure,\" in words? e. If p + q = 1, then what is q ? f. The words \"at least\" translate as what kind of inequality for the probability question P ( x ____ 40). a. failure b. X = the number of statistics students who do their homework on time c. 0, 1, 2, …, 50 d. Failure is defined as a student who does not complete their homework on time. The probability of a success is p = 0.70. The number of trials is n = 50. e. q = 0.30 f. greater than or equal to (≥) The probability question is P ( x ≥ 40). Try It Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem. During a certain NBA season, a player for the Los Angeles Clippers had the highest field goal completion rate in the league. This player scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by this player during the season. Let X = the number of shots that scored points. What is the probability distribution for X ? Using the formulas, calculate the (i) mean and (ii) standard deviation of X . Find the probability that this player scored with 60 of these shots. Find the probability that this player scored with more than 50 of these shots. This is a binomial problem because there is only a success or a failure, and there are a definite number of trials. The probability of a success stays the same for each trial. X ~ B (80, 0.613) Mean = np = 80(0.613) = 49.04 Standard Deviation = n p q = 80 ( 0.613 ) ( 0.387 ) ≈ 4.3564 P ( x = 60)= 0.0036 P ( x > 50) = 1 – P ( x ≤ 50) = 1 – 0.6282 = 0.3718 References “Access to electricity (% of population),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc (accessed May 15, 2015). “Distance Education.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Distance_education (accessed May 15, 2013). “NBA Statistics – 2013,” ESPN NBA, 2013. Available online at http://espn.go.com/nba/statistics/_/seasontype/2 (accessed May 15, 2013). Newport, Frank. “Americans Still Enjoy Saving Rather than Spending: Few demographic differences seen in these views other than by income,” GALLUP® Economy, 2013. Available online at http://www.gallup.com/poll/162368/americans-enjoy-saving-rather-spending.aspx (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011 . Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013). “What are the key statistics about pancreatic cancer?” American Cancer Society, 2013. Available online at http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics (accessed May 15, 2013). Chapter Review A statistical experiment can be classified as a binomial experiment if the following conditions are met: There are a fixed number of trials, n . There are only two possible outcomes, called \"success\" and, \"failure\" for each trial. The letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial. The n trials are independent and are repeated using identical conditions. The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np , and the standard deviation is given by the formula σ = n p q . The formula for the Binomial probability density function is P ( x ) = n ! x ! ( n - x ) ! · p x q ( n - x ) Formula Review X ~ B ( n , p ) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p . X = the number of successes in n independent trials n = the number of independent trials X takes on the values x = 0, 1, 2, 3, ..., n p = the probability of a success for any trial q = the probability of a failure for any trial p + q = 1 q = 1 – p The mean of X is μ = np . The standard deviation of X is σ = n p q . P ( x ) = n ! x ! ( n - x ) ! · p x q ( n - x ) where P(X) is the probability of X successes in n trials when the probability of a success in ANY ONE TRIAL is p. Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time first-year students from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time first-year students from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status. In words, define the random variable X . X = the number that reply “yes” X ~ _____(_____,_____) What values does the random variable X take on? 0, 1, 2, 3, 4, 5, 6, 7, 8 Construct the probability distribution function (PDF). x P ( x ) On average ( μ ), how many would you expect to answer yes? 5.7 What is the standard deviation ( σ )? What is the probability that at most five of the first-year students reply “yes”? 0.4151 What is the probability that at least two of the first-year students reply “yes”? HOMEWORK According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that they truly have the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Define the random variable and list its possible values. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, ...25 State the distribution of X . Find the probability that at least four of the 25 patients actually have the flu. 0.0165 On average, for every 25 patients calling in, how many do you expect to have the flu? People visiting video game rental stores often rent more than one game at a time. The probability distribution for game rentals per customer at Game Stop is given . There is five-game limit per customer at this store, so nobody ever rents more than five video games. x P ( x ) 0 0.03 1 0.50 2 0.24 3 4 0.07 5 0.04 Describe the random variable X in words. Find the probability that a customer purchases three games. Find the probability that a customer purchases at least four games. Find the probability that a customer purchases at most two games. X = the number of video games a Game Stop customer rents 0.12 0.11 0.77 A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many of the 12 students do we expect to attend the festivities? Find the probability that at most four students will attend. Find the probability that more than two students will attend. Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. The expected number of wins for that upcoming month is: 1.67 12 382 1043 4.43 d. 4.43 Let X = the number of games won in that upcoming month. What is the probability that the San Jose Sharks win six games in that upcoming month? 0.1476 0.2336 0.7664 0.8903 What is the probability that the San Jose Sharks win at least five games in that upcoming month 0.3694 0.5266 0.4734 0.2305 c A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct. A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly. X = number of questions answered correctly X ~ B ( 32, 1 3 ) We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P ( x > 24). The event \"more than 24\" is the complement of \"less than or equal to 24.\" P ( x > 24) = 0 The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero. Six different colored dice are rolled. Of interest is the number of dice that show a one. In words, define the random variable X . List the values that X may take on. On average, how many dice would you expect to show a one? Find the probability that all six dice show a one. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically. More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many schools would you expect to offer such courses? Find the probability that at most ten offer such courses. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence. X = the number of college and universities that offer online offerings. 0, 1, 2, …, 13 X ~ B (13, 0.96) 12.48 0.0135 P ( x = 12) = 0.3186 P ( x = 13) = 0.5882 More likely to get 13. Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to attend their graduation? Find the probability that 17 or 18 attend. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically. At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to not to use the foil as their main weapon? Find the probability that six do not use the foil as their main weapon. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically. X = the number of fencers who do not use the foil as their main weapon 0, 1, 2, 3,... 25 X ~ B (25,0.40) 10 0.0442 The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising. Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many seniors are expected to have participated in after-school sports all four years of high school? Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many audits are expected in a 20-year period? Find the probability that a person is not audited at all. Find the probability that a person is audited more than twice. X = the number of audits in a 20-year period 0, 1, 2, …, 20 X ~ B (20, 0.02) 0.4 0.6676 0.0071 It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) What is the probability that at least eight have adequate earthquake supplies? Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? How many residents do you expect will have adequate earthquake supplies? There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The “house” rolls three dice. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back their $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back their $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back their $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game. In words, define the random variable X . List the values that X may take on. List the values that Y may take on. Then, construct one PDF table that includes both X and Y and their probabilities. Calculate the average expected matches over the long run of playing this game for the player. Calculate the average expected earnings over the long run of playing this game for the player. Determine who has the advantage, the player or the house. X = the number of matches 0, 1, 2, 3 In dollars: −1, 1, 2, 3 1 2 The answer is −0.0787. You lose about eight cents, on average, per game. The house has the advantage. Only 9% of the population of Niger had access to the internet. Suppose we randomly sample 150 people in Niger. Let X = the number of people who have access the internet. What is the probability distribution for X ? Using the formulas, calculate the mean and standard deviation of X . Find the probability that 15 people in the sample have access to the internet. Find the probability that at most ten people in the sample have access to the internet. Find the probability that more than 25 people in the sample have access to the internet. The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in a certain country is 28.1%. Suppose you choose 15 people in a certain country at random. Let X = the number of people who are literate. Sketch a graph of the probability distribution of X . Using the formulas, calculate the (i) mean and (ii) standard deviation of X . Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate. X ~ B (15, 0.281) Mean = μ = np = 15(0.281) = 4.215 Standard Deviation = σ = n p q = 15 ( 0.281 ) ( 0.719 ) = 1.7409 P ( x > 5)=1 – 0.7754 = 0.2246 P ( x = 3) = 0.1927 P ( x = 4) = 0.2259 It is more likely that four people are literate that three people are. Binomial Experiment a statistical experiment that satisfies the following three conditions: There are a fixed number of trials, n . There are only two possible outcomes, called \"success\" and, \"failure,\" for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. The n trials are independent and are repeated using identical conditions. Bernoulli Trials an experiment with the following characteristics: There are only two possible outcomes called “success” and “failure” for each trial. The probability p of a success is the same for any trial (so the probability q = 1 − p of a failure is the same for any trial). Binomial Probability Distribution a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The mean is μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = ( n x ) p x q n − x .", "section": "Binomial Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Geometric Distribution There are four main characteristics of a geometric experiment. A trial is repeated until a success occurs. Think of this as one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a \"success,\" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP. In theory, the number of trials could go on forever. The repeated trials are independent of each other. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 - p . For example, the probability of rolling a three when you throw one fair die is 1 / 6 . This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = 5 / 6 , the probability of a failure. The probability of getting a first three on the fifth roll is ( 5 / 6 ) ( 5 / 6 ) ( 5 / 6 ) ( 5 / 6 ) ( 1 / 6 ) = 0 . 0804 The random variable X represents the number of the trial in which the first success occurs. That is, X = the number of independent trials until the first success. The following are additional attributes of the geometric distribution: The random variable is discrete. The random variable may be defined in two ways depending upon the analyst’s interest. In the example above for throwing a die, the question was “What is the probability that the first success will be on the fifth throw?” Alternatively, the question could be asked as “What is the probability that it takes four failures before a success?” We will see that each way of asking the question will alter the form of the geometric probability density function slightly and will change the mean and standard deviation of the geometric pdf. Implicit in the random variable is that the probability of a success is constant and therefore so is the probability of a failure. For flipping a coin this is obvious, but in experiments that require skill, such as hitting a baseball or throwing a dart, one might consider that learning during the experiment would alter the probability of a success. The geometric distribution cannot capture “learning” thus the historical probability of success is assumed to be constant. The geometric distribution is “memoryless.” There are very few probability density functions that are what is known as “memoryless,” and the Geometric distribution is the only one with a discrete random variable that is memoryless. As an example, historically Major League Baseball player Jones has a record of hitting the ball for at least an advance to first base with a probability of 0.20. Jones has not had a hit in his last 10 times at bat. What is the probability that Jones will get a hit in his third time at bat? The answer ignores his 10 previous failures. All events prior to the events in current time are irrelevant and thus are considered “memoryless.” Formally: P x = n + k | x ≥ k + 1 = P ( x = n ) , where k = number of previous failures Jones’s probability of a hit begins anew each time he comes to bat. This feature of the geometric distribution results in a curious result: Drawing parts from a manufacturing process to test for parts that are defective, the geometric distribution begins with a clean slate each time the tests begin with no consideration of previous test results. More on this when we get to the exponential probability density function. You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P ( x = 5). Try It You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? 1, 2, 3, 4, … n . It can go on indefinitely. A safety engineer feels that 35% of all industrial accidents in the plant are caused by failure of employees to follow instructions. They decide to look at the accident reports (selected randomly and replaced in the pile after reading) until they find one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until they find a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions? Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P ( x ≥ 3). (\"At least\" translates to a \"greater than or equal to\" symbol). On average, how many reports would the safety engineer expect to review until they find a report showing an accident caused by something other than failure to follow instructions? This question asks you to find the expected value or the mean. This is the average number of failures from something other than following instructions. The formula for the mean of this geometric distribution is: μ = E x = 1 p ≈ 2 . 8 Note that the mean does not need to be a whole number although the random variable is discrete and must be a counting number What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions? This question is answered by first defining the random variable. Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X can take on the values 1, 2, 3, 4, … ∞. Unlike the binomial distribution with a fixed number of trials, the geometric distribution may have an infinite number of failed trials before a success occurs. This second question asks you to find P ( x ≥ 3 ) . (\"At least\" translates to a \"greater than or equal to\" symbol.) In a binomial distribution, we saw the solution to questions of “more than” or “less than” would be to calculate each probability individually and add from zero to three. If the probability of interest is “greater than or equal to” (≥) the individual probabilities are added from zero to three and then subtracted from one. P x ≥ 3 = 1 - P ( x < 3 ) Because we cannot add the full range of the geometric distribution random variable because it goes through infinity, an alternative solution is developed. An alternative geometric distribution pair of formulas provides solutions for questions asking for probabilities “more than” and “less than.” If the question is: What is the probability it takes MORE THAN n events to get first success? P x > n = ( 1 - p ) n What is the probability it takes LESS THAN n event for success? P x < n = 1 - ( 1 - p ) n Try It An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically? P ( x ≥ 10) Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says they live within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays approximately the same each time you ask a student if they live within five miles of you. There is no definite number of trials (number of times you ask a student). a. Let X = the number of ____________ you must ask ____________ one says yes. b. What values does X take on? c. What are p and q ? d. The probability question is P (_______). a. Let X = the number of students you must ask until one says yes. b. X can be any positive integer: 1, 2, 3, 4, …, c. p = 0.55; q = 0.45 d. P ( x = 4) Try It You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q ? Notation for the Geometric: G = Geometric Probability Distribution Function Read this as \" X is a random variable with a geometric distribution .\" The parameter is p ; p = the probability of a success for each trial. CASE I: Random Variable X Is Event of First Success In this case we ask, “What is the probability that we will have some number x of events of interest to us of failures before a success?” The geometric pdf tells us the probability that the first occurrence of success requires x number of failure independent trials, each with probability (1-p). If the probability of success on each trial is p, then the probability that the x th trial (out of x trials) is the first success is: P ( X = x ) = ( 1 - p ) x - 1 p for x = 1 , 2 , 3 , . . . . Like the binomial distribution, the geometric distribution has the parameters of the mean and standard deviation. The expected value of X, the mean for Case I, is μ = 1 p . This tells us how many failed trials to expect until we get the first success. This count includes in the count of trials the trial that results in success. The above form of the geometric distribution is used for modeling the number of trials until the first success. The number of trials includes the one that is a success: x = all trials including the one that is a success. This can be seen in the form of the formula. If X = number of trials including the success, then we must multiply the probability of failure, ( 1 - p ) , times the number of failures, that is x - 1 . The standard deviation of Case I of the geometric distribution is: σ = 1 p 1 p - 1 CASE II: Random Variable X Is Number of Failures BEFORE a Success By contrast to Case I, the following form of the geometric distribution used for modeling number of failures until the first success is: P X = x = 1 - p x p for x = 0 , 1 , 2 , 3 , . . . . In this case the trial that is the success is not counted as a trial in the formula: x = number of failures. The expected value, the mean, of this distribution is μ = 1 - p p . This tells us how many failures to expect before we have a success. In either case, the sequence of probabilities is a geometric sequence. In Case II the standard deviation parameter is: σ = 1 - p p . GEOMETRIC: P = . 02 , Common ratio = r = . 98 The y -axis in contains the probability of x , and the x -axis is the random number components tested. For example, at x = 1 the probability it will be found to be defective is 0.0196. With two components tested, the probability the second component is defective is graphed at a probability of 0.0196 at x = 2 on the x axis. For the probability that the third component is defective we find P X = 3 = 0 . 019208 . (The first two are the same because of rounding in the computations.) Notice in that the probabilities decline by the same step down with each change in the value of x. This increment is called the common ratio. This exists for the geometric probability distribution uniquely. The common ratio, called r, can be calculated by dividing any value by the previous value, e.g., P x = 5 P x = 4 = 0 . 018447 0 . 018823 = 0 . 98 For this set of data, therefore, the common ratio is 0.98. The common ratio then multiplied by any other probability value will provide the next probability value in the sequence. For example, the probability that the sixth component tested is a failure is 0.018078. Check this using the formula from Case I. Now we have the P x = 6 . Knowing this and the common ratio we can calculate the P x = 7 by simple multiplication: P x = 7 = 0 . 018078 × 0 . 98 = 0 . 017716 , the same value we found earlier by using the geometric probability distribution. This common ratio increment is the same ratio between every number and is called a geometric progression and thus the name for this probability density function. Once the common ratio is calculated, any P x = x a one desires to know can be easily found. The number of components that you would expect to test until you find the first defective component is the mean, μ = 50 for this case of defective components. The formula for the mean of the geometric distribution for the random variable defined as number of failures before first success is μ = 1 p = 1 0 . 02 = 50 . See for an example where the geometric random variable is defined as number of trials until first success. The expected value of this formula for the geometric distribution will be different from this version of the distribution. Case II also has a variance but is changed from the Case I formula. This formula for the variance is: P X = x = 1 - p x - 1 p for x = 1 , 2 , 3 , . . . . The formula for the variance is σ 2 = 1 p 1 p - 1 = 1 0 . 02 1 0 . 02 - 1 = 2 , 450 The standard deviation is σ = 1 p 1 p - 1 = 1 0 . 02 1 0 . 02 - 1 = 49 . 5 This tells us how many failures to expect before we have a success. In either case, the sequence of probabilities is a geometric sequence. Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3, ... where p = 0.02. Find P ( x = 7). Answer: P ( x = 7) = (1 - 0.02) 7-1 × 0.02 = 0.0177. The probability that the seventh component is the first defect is 0.0177. The graph of X ~ G(0.02) is: The y -axis contains the probability of x , where X = the number of computer components tested. Notice that the probabilities decline by a common increment. This increment is the same ratio between each number and is called a geometric progression and thus the name for this probability density function. See for an example where the geometric random variable is defined as number of trials until first success. The expected value of this formula for the geometric will be different from this version of the distribution. The formula for the variance is σ 2 = ( 1 p ) ( 1 p − 1 ) = ( 1 0.02 ) ( 1 0.02 − 1 ) = 2,450 The standard deviation is σ = ( 1 p ) ( 1 p − 1 ) = ( 1 0. 02 ) ( 1 0. 02 − 1 ) = 49.5 Try It The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. The lifetime risk of developing cancer is about one in 67 (1.5%). Let X = the number of people you ask before one says they have cancer. The random variable X in this case includes only the number of trials that were failures and does not count the trial that was a success in finding a person who had the disease. The appropriate formula for this random variable is the second one presented above. Then X is a discrete random variable with a geometric distribution: What is the probability that you ask 9 people before one says they have cancer? This is asking, what is the probability that you ask 9 people who say no and the tenth person says yes? What is the probability that you must ask 20 people? Find the (i) mean and (ii) standard deviation of X . P x = 9 = 1 - 0 . 015 9 × 0 . 015 = 0 . 0131 P x = 20 = 1 - 0 . 015 19 × 0 . 015 = 0 . 0113 Mean = μ = 1 - p p = 1 - 0 . 015 0 . 015 = 65 . 67 Standard Deviation = σ = 1 - p p 2 = 1 - 0 . 015 0 . 015 2 = 66 . 16 Try It The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in The United Colonies of Independence is 12%. Let X = the number of women you ask until one says that she is literate. What is the probability distribution of X ? What is the probability that you ask five women before one says she is literate? What is the probability that you must ask ten women? X ~ G (0.12) P ( x = 5) = 0.0720 P ( x = 10) = 0.0380 A baseball player has a batting average of 0.320. This is the general probability that he gets a hit each time he is at bat. What is the probability that he gets his first hit in the third trip to bat? P ( x =3) = (1-0.32) 3-1 × .32 = 0.1480 In this case the sequence is failure, failure success. How many trips to bat do you expect the hitter to need before getting a hit? μ = 1 p = 1 0.320 = 3.125 ≈ 3 This is simply the expected value of successes and therefore the mean of the distribution. Try It A runner has a probability of 0.426 to come in first in a race. What is the probability that they will come in first in the fourth race? How many races do you expect the runner to need before they come in first? P x = 4 = 1 - 0 . 426 4 - 1 = 0 . 189 μ = 1 p = 1 0 . 426 = 2 . 347 ≈ 2 There is an 80% chance that a Dalmatian dog has 13 black spots. You go to a dog show and count the spots on Dalmatians. What is the probability that you will review the spots on 3 dogs before you find one that has 13 black spots? P ( x =3) = (1 - 0.80) 3 × 0.80 = 0.0064 Try It There is a 75% chance that a parrot is green. You go to a pet shop and observe the parrots. What is the probability that you will observe 4 parrots before you find one that is a green parrot? P x = 4 = 1 - 0 . 75 4 × 0 . 75 = 0 . 0117 References “Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at http://www.pewsocialtrends.org/files/2010/10/millennials-confident-connected-open-to-change.pdf (accessed May 15, 2013). “Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-to-change/ (accessed May 15, 2013). “Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013). “Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at http://ec.europa.eu/europeaid/where/asia/documents/afgh_brochure_summary_en.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013). “UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/video/afghan-female-literacy-centers.html (accessed May 15, 2013). Chapter Review There are three characteristics of a geometric experiment: There are one or more Bernoulli trials with all failures except the last one, which is a success. In theory, the number of trials could go on forever. There must be at least one trial. The probability, p , of a success and the probability, q , of a failure are the same for each trial. In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G ( p ) where p is the probability of success in a single trial. The mean of the geometric distribution X ~ G ( p ) is μ = 1 / p where x = number of trials until first success for the formula P ( X = x ) = ( 1 - p ) x - 1 p where the number of trials is up and including the first success. An alternative formulation of the geometric distribution asks the question: what is the probability of x failures until the first success? In this formulation the trial that resulted in the first success is not counted. The formula for this presentation of the geometric is: P ( X = x ) = p ( 1 − p ) x The expected value in this form of the geometric distribution is μ = 1 − p p The easiest way to keep these two forms of the geometric distribution straight is to remember that p is the probability of success and (1−p) is the probability of failure. In the formula the exponents simply count the number of successes and number of failures of the desired outcome of the experiment. Of course the sum of these two numbers must add to the number of trials in the experiment. Formula Review P ( X = x ) = p ( 1 − p ) x − 1 X ~ G( p ) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p . X = the number of independent trials until the first success X takes on the values x = 1, 2, 3, ... p = the probability of a success for any trial q = the probability of a failure for any trial p + q = 1 q = 1 – p The mean is μ = 1 p . The standard deviation is σ = 1 – p p 2 = 1 p ( 1 p − 1 ) . Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time first-year students from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select first-year students from the study until you find one who replies “yes.” You are interested in the number of first-year students you must ask. In words, define the random variable X . X = the number of first-year students selected from the study until one replied \"yes\" that same-sex couples should have the right to legal marital status. X ~ _____(_____,_____) What values does the random variable X take on? 1,2,… Construct the probability distribution function (PDF). Stop at x = 6. x P ( x ) 1 2 3 4 5 6 On average ( μ ), how many first-year students would you expect to have to ask until you found one who replies \"yes?\" 1.4 What is the probability that you will need to ask fewer than three first-year students? HOMEWORK A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many dealerships would we expect her to have to call until she finds one that has the car? Find the probability that she must call at most four dealerships. Find the probability that she must call three or four dealerships. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many adults in America do you expect to survey until you find one who will watch the Super Bowl? Find the probability that you must ask seven people. Find the probability that you must ask three or four people. X = the number of adults in America who are surveyed until one says they will watch the Super Bowl. X ~ G (0.40) 2.5 0.0187 0.2304 It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies? What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies? How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies? How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies? In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many pages do you expect to advertise footwear on them? Is it probable that all twenty will advertise footwear on them? Why or why not? What is the probability that fewer than ten will advertise footwear on them? Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it? How many pages do you expect to need to survey in order to find one that advertises footwear? X = the number of pages that advertise footwear X takes on the values 0, 1, 2, ..., 20 X ~ B (20, 29 192 ) 3.02 No 0.9997 X = the number of pages we must survey until we find one that advertises footwear. X ~ G ( 29 192 ) 0.3881 6.6207 pages Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. p = probability of success (event F occurs) q = probability of failure (event F does not occur) Write the description of the random variable X . What are the values that X can take on? Find the values of p and q . Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on? 0, 1, 2, and 3 Medical researchers in a certain country have documented that the percentage of people ages 15 to 49 who are infected with a certain respiratory virus is 17.3%. Let X = the number of people you test until you find a person infected with the virus. Sketch a graph of the distribution of the discrete random variable X . What is the probability that you must test 30 people to find one with the virus? What is the probability that you must ask ten people? Find the (i) mean and (ii) standard deviation of the distribution of X . According to a recent Pew Research poll, 75% of millennials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millennials you ask until you find a person without a profile on a social networking site. Describe the distribution of X . Find the (i) mean and (ii) standard deviation of X . What is the probability that you must ask ten people to find one person without a social networking site? What is the probability that you must ask 20 people to find one person without a social networking site? What is the probability that you must ask at most five people? X ~ G (0.25) Mean = μ = 1 p = 1 0.25 = 4 Standard Deviation = σ = 1 − p p 2 = 1 − 0 .25 0.25 2 ≈ 3.4641 P ( x = 10) = 0.0188 P ( x = 20) = 0.0011 P ( x ≤ 5) = 0.7627 Geometric Distribution a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable X is defined as the number of trials until the first success. The mean is μ = 1 p and the standard deviation is σ = 1 p ( 1 p − 1 ) . The probability of exactly x failures before the first success is given by the formula: P ( X = x ) = p (1 – p ) x – 1 where one wants to know the probability for the number of trials until the first success: the xth trial is the first success. An alternative formulation of the geometric distribution asks the question: what is the probability of x failures until the first success? In this formulation the trial that resulted in the first success is not counted. The formula for this presentation of the geometric is: P ( X = x ) = p ( 1 − p ) x The expected value in this form of the geometric distribution is μ = 1 − p p The easiest way to keep these two forms of the geometric distribution straight is to remember that p is the probability of success and (1−p) is the probability of failure. In the formula the exponents simply count the number of successes and number of failures of the desired outcome of the experiment. Of course the sum of these two numbers must add to the number of trials in the experiment. Geometric Experiment a statistical experiment with the following properties: There are one or more Bernoulli trials with all failures except the last one, which is a success. In theory, the number of trials could go on forever. There must be at least one trial. The probability, p , of a success and the probability, q , of a failure do not change from trial to trial.", "section": "Geometric Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Poisson Distribution Another useful probability distribution is the Poisson distribution, or waiting time distribution. This distribution is used to determine how many checkout clerks are needed to keep the waiting time in line to specified levels, how may telephone lines are needed to keep the system from overloading, and many other practical applications. A modification of the Poisson, the Pascal, invented nearly two centuries ago, is used today by telecommunications companies worldwide for load factors, satellite hookup levels and Internet capacity problems. The distribution gets its name from Simeon Poisson who presented it in 1837 as an extension of the binomial distribution which we will see can be estimated with the Poisson. There are two main characteristics of a Poisson experiment. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate. The events are independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages and it is assumed that there is no relationship between when misspellings occur. The random variable X = the number of occurrences in the interval of interest. A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question. P ( x < 5) Try It An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically. You notice that a news reporter says \"uh,\" on average, two times per broadcast. What is the probability that the news reporter says \"uh\" more than two times per broadcast. This is a Poisson problem because you are interested in knowing the number of times the news reporter says \"uh\" during a broadcast. a. What is the interval of interest? a. one broadcast measured in minutes b. What is the average number of times the news reporter says \"uh\" during one broadcast? b. 2 c. Let X = ____________. What values does X take on? c. Let X = the number of times the news reporter says \"uh\" during one broadcast. x = 0, 1, 2, 3, ... d. The probability question is P (______). d. P ( x > 2) Try It An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution. Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P ( μ ) Read this as \" X is a random variable with a Poisson distribution.\" The parameter is μ (or λ ); μ (or λ ) = the mean for the interval of our interest. The mean is the number of occurrences that occur on average during the interval period. The formula for computing probabilities that are from a Poisson process is: P ( x ) = μ x e - μ x ! where P(X) is the probability of X successes, μ is the expected number of successes based upon historical data, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time. Remember your algebra class: x - n = 1 x n . The Poisson distribution has both a mean, µ , the average number of occurrences per unit of time, and also a standard deviation, σ = μ . In order to use the Poisson distribution, certain assumptions must hold. These are: the probability of a success, μ, is unchanged within the interval, there cannot be simultaneous successes within the interval, and finally, that the probability of a success among intervals is independent, the same assumption of the binomial distribution. In a way, the Poisson distribution can be thought of as a clever way to convert a continuous random variable, usually time, into a discrete random variable by breaking up time into discrete independent intervals. This way of thinking about the Poisson helps us understand why it can be used to estimate the probability for the discrete random variable from the binomial distribution. The Poisson is asking for the probability of a number of successes during a period of time while the binomial is asking for the probability of a certain number of successes for a given number of trials. Leah receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes? Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or 1 4 hour.) x = 0, 1, 2, 3, ... If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives ( 1 8 ) (6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem. Find P ( x > 1). The Poisson distribution is discrete, and thus > 1 includes all whole numbers through infinity. The solution is to subtract the probability less than 1 thus 1 - P x = 0 + P x = 1 P x = μ x e - μ x ! P x > 1 = 1 - P x ≤ 1 = 1 - P x = 0 + P x = 1 = 1 - 0 . 75 0 e - 0 . 75 0 ! + 0 . 75 1 e - 0 . 75 1 ! = 1 - 1 0 . 4724 1 + 0 . 75 0 . 4724 1 = 1 - 0 . 4724 + 0 . 3543 = 0 . 1733 The y -axis in contains the probability of x where X = the number of calls in 15 minutes. Try It A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (147). The mean is 147 emails. What is the probability that an email user receives exactly 160 emails per day? What is the probability that an email user receives at most 160 emails per day? What is the standard deviation? P ( x = 160) = poissonpdf(147, 160) ≈ 0.0180 P ( x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666 Standard Deviation = σ = μ = 147 ≈ 12.1244 Try It According to a recent poll by the Pew Internet Project, people between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (187). The mean is 187 text messages. What is the probability that a person sends exactly 175 texts per day? What is the probability that a person sends at most 150 texts per day? What is the standard deviation? Text message users receive or send an average of 41.5 text messages per day. How many text messages does a text message user receive or send per hour? What is the probability that a text message user receives or sends two messages per hour? What is the probability that a text message user receives or sends more than two messages per hour? Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is 41.5 24 ≈ 1.7292. P ( x = 2 ) = μ x e -μ x! = 1.729 2 e -1.729 2! = 0.265 P ( x > 2 ) = 1 - P ( x ≤ 2 ) = 1 - [ 7 0 e -7 0! + 7 1 e -7 1! + 7 2 e -7 2! ] = 0.250 Try It Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,700 arrivals and departures each day. How many airplanes arrive and depart the airport per hour? What is the probability that there are exactly 100 arrivals and departures in one hour? What is the probability that there are at most 100 arrivals and departures in one hour? On a specific day in May starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? Let X = the number of days with low seismic activity. Using the binomial distribution: P ( x = 10 ) = 200! 10! ( 200 - 10 ) ! × .0102 10 × .9898 190 = 0.000039 Using the Poisson distribution: Calculate μ = np = 200(0.0102) ≈ 2.04 P ( x = 10 ) = μ x e -μ x! = 2.04 10 e -2.04 10! = 0.000045 We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0. Try It On a specific day in May starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? The Poisson distribution is often referred to as a “waiting time” distribution. In a sense this is tied to the link between the discrete distribution measuring the number of occurrences and the treatment of the continuous random variable time. The Poisson takes the continuous random variable time and breaks it into a discrete random variable by measuring the discrete number of occurrences. In Example 4.16 we were measuring the number of messages sent, an occurrence, a discrete random variable, per day, a continuous random variable. The Poisson distribution can be used for many other applications. Any continuous random variable that can be broken into discrete measures can use the Poisson distribution to calculate probabilities of the number of a given discrete value of x, the number of occurrences in which we have an interest. We are interested in the number of potholes in a highway to evaluate the quality of the paving job. In this case the pothole is the occurrence and can be counted per square miles of highway. Mile is a continuous random variable, and potholes is the discrete random variable. We may want to know the probability that a particular paving technique results in more than 20 potholes in 10 miles of paving. If that is the case, we may discard that paving technique. Alternatively, our chocolate cookies claim that each cookie has more than 5 chocolate pieces in each cookie. Sampling the cookie dough, we can calculate the probability that the resulting cookies will meet the claim of 5 chocolate pieces per cookie. Perhaps the production of glass for automobile windshields suffers from a flaw known as the “dot.” In the production process, air infrequently becomes lodged in the hot glass and upon cooling leaves a dot that distracts the driver. This would also be a case where use of the Poisson probability distribution would be the appropriate tool. To calculate the probability that the results have fewer than 3 “dots,” the number acceptable to the manufacturing process, the Poisson distribution would be useful. By way of warning, the Poisson is a discrete random variable. We are counting occurrences. It occurred or it did not during the time, the miles of road, or the batch of cookie dough. In this sense the Poisson is like the binomial in that the occurrence is binary, happened or did not. Because of this link to the binomial, we can use the Poisson to estimate a binomial distribution, and this is discussed below. Try It In a small city, a survey 500 vehicles reveals that there is a 2% chance of an accident occurring in a week. Calculate the probability that at most 2 accidents occur in any given week. This is a binomial probability distribution problem. Binomial solution: n * p = 500 * 0 . 02 = 10 = μ ( 0 ) = 500 ! 0 ! ( 500 - 0 ) ! ( 0 . 02 ) 0 ( 1 - 0 . 02 ) 500 - 0 = 0 . 00004 P ( 1 ) = 500 ! 1 ! ( 500 - 1 ) ! ( 0 . 02 ) 1 ( 1 - 0 . 02 ) 500 - 1 = 0 . 00041 P ( 2 ) = 500 ! 2 ! ( 500 - 2 ) ! ( 0 . 02 ) 2 ( 1 - 0 . 02 ) 500 - 2 = 0 . 00213 The probability that at most 2 accidents will occur in any week is obtained as follows: P ( 0 ) + P ( 1 ) + P ( 2 ) = 0 . 00004 + 0 . 00041 + 0 . 00213 = 0 . 00258 Estimating the Binomial Distribution with the Poisson Distribution We found before that the binomial distribution provided an approximation for the hypergeometric distribution. Now we find that the Poisson distribution can provide an approximation for the binomial. We say that the binomial distribution approaches the Poisson. The binomial distribution approaches the Poisson distribution is as n gets larger and p is small such that np becomes a constant value. There are several rules of thumb for when one can say they will use a Poisson to estimate a binomial. One suggests that np , the mean of the binomial, should be less than 25. Another author suggests that it should be less than 7. And another, noting that the mean and variance of the Poisson are both the same, suggests that np and npq , the mean and variance of the binomial, should be greater than 5. There is no one broadly accepted rule of thumb for when one can use the Poisson to estimate the binomial. As we move through these probability distributions we are getting to more sophisticated distributions that, in a sense, contain the less sophisticated distributions within them. This proposition has been proven by mathematicians. This gets us to the highest level of sophistication in the next probability distribution which can be used as an approximation to all of those that we have discussed so far. This is the normal distribution. A survey of 500 seniors in the Price Business School yields the following information. 75% go straight to work after graduation. 15% go on to work on their MBA. 9% stay to get a minor in another program. 1% go on to get a Master's in Finance. What is the probability that more than 2 seniors go to graduate school for their Master's in finance? This is clearly a binomial probability distribution problem. The choices are binary when we define the results as \"Graduate School in Finance\" versus \"all other options.\" The random variable is discrete, and the events are, we could assume, independent. Solving as a binomial problem, we have: Binomial Solution n · p = 500 · 0.01 = 5 = µ P ( 0 ) = 500 ! 0 ! ( 500 − 0 ) ! 0.01 0 ( 1 − 0.01 ) 500 − 0 = 0.00657 P ( 1 ) = 500 ! 1 ! ( 500 − 1 ) ! 0.01 1 ( 1 − 0.01 ) 500 − 1 = 0.03318 P ( 2 ) = 500 ! 2 ! ( 500 − 2 ) ! 0.01 2 ( 1 − 0.01 ) 500 − 2 = 0.08363 Adding all 3 together = 0.12339 1 − 0.12339 = 0.87661 Poisson approximation n · p = 500 · 0.01 = 5 = μ n · p · ( 1 − p ) = 500 · 0.01 · ( 0.99 ) ≈ 5 = σ 2 = μ P ( X ) = e −np ( n p ) x x ! = { P ( 0 ) = e −5 · 5 0 0 ! } + { P ( 1 ) = e −5 · 5 1 1 ! } + { P ( 2 ) = e −5 · 5 2 2 ! } 0.0067 + 0.0337 + 0.0842 = 0.1247 1 − 0.1247 = 0.8753 An approximation that is off by 1 one thousandth is certainly an acceptable approximation. Try It In , what is the probability that less than 4 seniors will go to graduate school for their master’s in finance? The problem is to be solved as a binomial problem: n * p = 500 * 0 . 09 = 45 = μ P 0 = 500 ! 0 ! 500 - 0 ! 0 . 01 0 1 - 0 . 01 500 - 0 = 0 . 00657 P 1 = 500 ! 1 ! 500 - 1 ! 0 . 01 1 1 - 0 . 01 500 - 1 = 0 . 03318 P 2 = 500 ! 2 ! 500 - 2 ! 0 . 01 2 1 - 0 . 01 500 - 2 = 0 . 08363 P 3 = 500 ! 3 ! 500 - 3 ! 0 . 01 3 1 - 0 . 01 500 - 3 = 0 . 14022 Adding all 3 together = 0.12339. So, the required probability that less than 4 seniors will go to graduate school for their master’s in finance is 0.26361. References “ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atl.com/about-atl/atl-factsheet/ (accessed February 6, 2019). Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013). “Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html (accessed May 15, 2013). “Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www.state.sc.us/dmh/anorexia/statistics.htm (accessed May 15, 2013). “Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where there is an average of 60 births a day,” theguardian, 2013. Available online at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013). “How Americans Use Text Messaging,” Pew Internet, 2013. Available online at http://pewinternet.org/Reports/2011/Cell-Phone-Texting-2011/Main-Report.aspx (accessed May 15, 2013). Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volume is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013). “One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline. Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-day-mothers-bed.html (accessed May 15, 2013). Vanderkam, Laura. “Stop Checking Your Email, Now.” CNNMoney, 2013. Available online at http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15, 2013). “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013). Chapter Review A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is \"small\" (less than or equal to 0.01) and the number of trials is \"large\" (greater than or equal to 25). Other rules of thumb are also suggested by different authors, but all recognize that the Poisson distribution is the limiting distribution of the binomial as n increases and p approaches zero. The formula for computing probabilities that are from a Poisson process is: P ( x ) = μ x e - μ x ! where P(X) is the probability of successes, μ (pronounced mu) is the expected number of successes, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time. Formula Review X ~ P ( μ ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest. X takes on the values x = 0, 1, 2, 3, ... The mean μ or λ is typically given. The variance is σ 2 = μ , and the standard deviation is σ = μ . The probability of having exactly x successes in r trials is P X = x = e - μ μ x x ! . When P ( μ ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial. P ( x ) = μ x e - μ x ! Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day. Assume the event occurs independently in any given day. Define the random variable X . What values does X take on? 0, 1, 2, 3, 4, … What is the probability of getting 150 customers in one day? What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day. 0.0485 What is the probability that the store will have more than 12 customers in the first hour? What is the probability that the store will have fewer than 12 customers in the first two hours? 0.0214 Which type of distribution can the Poisson model be used to approximate? When would you do this? Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age. Assume the event occurs independently in any given day. In words, define the random variable X . X = the number of U.S. teens who die from motor vehicle injuries per day. X ~ _____(_____,_____) What values does X take on? 0, 1, 2, 3, 4, ... For the given values of the random variable X , fill in the corresponding probabilities. Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically. No Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically. HOMEWORK The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon. Find the mean and standard deviation of X . What is the probability that the office receives at most six calls at noon on Monday? Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon? What is the probability that the office receives more than eight calls at noon? X ~ P (5.5); μ = 5.5; σ = 5.5 ≈ 2.3452 P ( x ≤ 6) ≈ 0.6860 There is a 15.7% probability that the law staff will receive more calls than they can handle. P ( x > 8) = 1 – P ( x ≤ 8) ≈ 1 – 0.8944 = 0.1056 The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour. Find the mean and standard deviation of X . Sketch a graph of the probability distribution of X . What is the probability that the maternity ward will deliver three babies in one hour? What is the probability that the maternity ward will deliver at most three babies in one hour? What is the probability that the maternity ward will deliver more than five babies in one hour? A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions. Let X = the number of defective bulbs in a string. Using the Poisson distribution: μ = np = 100(0.03) = 3 X ~ P (3) P ( x ≤ 4) ≈ 0.8153 Using the binomial distribution: X ~ B (100, 0.03) P ( x ≤ 4) = 0.8179 The Poisson approximation is very good—the difference between the probabilities is only 0.0026. The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen. In words, define the random variable X . List the values that X may take on. Find the probability that she has no children. Find the probability that she has fewer children than the Japanese average. Find the probability that she has more children than the Japanese average. The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen. In words, define the Random Variable X . List the values that X may take on. Find the probability that she has no children. Find the probability that she has fewer children than the Spanish average. Find the probability that she has more children than the Spanish average . X = the number of children for a Spanish woman 0, 1, 2, 3,... 0.2299 0.5679 0.4321 Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces: In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _______ Find the probability that she has no litters in one year. Find the probability that she has at least two litters in one year. Find the probability that she has exactly three litters in one year. The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem. In words, define the random variable X . List the values that X may take on. How many cookies do we expect to have an extra fortune? Find the probability that none of the cookies have an extra fortune. Find the probability that more than three have an extra fortune. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences. X = the number of fortune cookies that have an extra fortune 0, 1, 2, 3,... 144 4.32 0.0124 or 0.0133 0.6300 or 0.6264 As n gets larger, the probabilities get closer together. For every 200 U.S. kids and teens, the average number who have obsessive compulsive disorder is one. Out of a randomly chosen group of 600 U.S. kids and teens determine the following. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to have obsessive compulsive disorder? Find the probability that no one has obsessive compulsive disorder. Find the probability that more than four have obsessive compulsive disorder. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to answer the following questions. In words, define the random variable X . List the values that X may take on. How many are expected to be audited? Find the probability that no one was audited. Find the probability that at least three were audited. X = the number of people audited in one year 0, 1, 2, ..., 100 2 0.1353 0.3233 Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school. In words, define the random variable X . List the values that X may take on. How many seniors are expected to have participated in after-school sports all four years of high school? Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes. In words, define the random variable X . List the values that X may take on. On average, how many pieces of egg shell do you expect to be in the cake? What is the probability that there will not be any pieces of egg shell in the cake? Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes? Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why? X = the number of shell pieces in one cake 0, 1, 2, 3,... 1.5 0.2231 0.0001 Yes Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week. In words, the random variable X = _________________ the number of times Mrs. Plum’s cats wake her up each week. the number of times Mrs. Plum’s cats wake her up each hour. the number of times Mrs. Plum’s cats wake her up each night. the number of times Mrs. Plum’s cats wake her up. Find the probability that her cats will wake her up no more than five times next week. 0.5000 0.9329 0.0378 0.0671 d Poisson Probability Distribution a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable: The probability that the event occurs in a given interval is the same for all intervals. The events occur with a known mean and independently of the time since the last event. The distribution is defined by the mean μ of the event in the interval. The mean is μ = np . The standard deviation is σ = μ . The probability of having exactly x successes in r trials is P ( x ) = μ x e - μ x ! . The Poisson distribution is often used to approximate the binomial distribution, when n is “large” and p is “small” (a general rule is that np should be greater than or equal to 25 and p should be less than or equal to 0.01).", "section": "Poisson Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction The heights of these radish plants are continuous random variables. (credit: modification of work “Radish (Raphanus sativus): White rust caused by Albugo candida” by Scot Nelson/ Flickr, Public Domain) Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, rates of return from an investment, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables, as do all areas of risk analysis. NOTE The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Properties of Continuous Probability Density Functions The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. We have already met this concept when we developed relative frequencies with histograms in Descriptive Statistics . The relative area for a range of values was the probability of drawing at random an observation in that group. Again with the Poisson distribution in Random Discrete Variables , the graph in Example 4.15 used boxes to represent the probability of specific values of the random variable. In this case, we were being a bit casual because the random variables of a Poisson distribution are discrete, whole numbers, and a box has width. Notice that the horizontal axis, the random variable x, purposefully did not mark the points along the axis. The probability of a specific value of a continuous random variable will be zero because the area under a point is zero. Probability is area. The curve is called the probability density function (abbreviated as pdf ). We use the symbol f ( x ) to represent the curve. f ( x ) is the function that corresponds to the graph; we use the density function f ( x ) to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf ). The cumulative distribution function is used to evaluate probability as area. Mathematically, the cumulative probability density function is the integral of the pdf, and the probability between two values of a continuous random variable will be the integral of the pdf between these two values: the area under the curve between these values. Remember that the area under the pdf for all possible values of the random variable is one, certainty. Probability thus can be seen as the relative percent of certainty between the two values of interest. The outcomes are measured, not counted. The entire area under the curve and above the x-axis is equal to one. Probability is found for intervals of x values rather than for individual x values. P(c < x < d) is the probability that the random variable X is in the interval between the values c and d . P(c < x < d) is the area under the curve, above the x -axis, to the right of c and the left of d . P(x = c) = 0 The probability that x takes on any single individual value is zero. The area below the curve, above the x -axis, and between x = c and x = c has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero. P(c < x < d) is the same as P(c ≤ x ≤ d) because probability is equal to area. We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, integral calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus. There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way. In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions. The graph shows a Uniform Distribution with the area between x = 3 and x = 6 shaded to represent the probability that the value of the random variable X is in the interval between three and six. The graph shows an Exponential Distribution with the area between x = 2 and x = 4 shaded to represent the probability that the value of the random variable X is in the interval between two and four. The graph shows the Standard Normal Distribution with the area between x = 1 and x = 2 shaded to represent the probability that the value of the random variable X is in the interval between one and two. For continuous probability distributions, PROBABILITY = AREA. Consider the function f ( x ) = 1 20 for 0 ≤ x ≤ 20. x = a real number. The graph of f ( x ) = 1 20 is a horizontal line. However, since 0 ≤ x ≤ 20, f ( x ) is restricted to the portion between x = 0 and x = 20, inclusive. f ( x ) = 1 20 for 0 ≤ x ≤ 20. The graph of f ( x ) = 1 20 is a horizontal line segment when 0 ≤ x ≤ 20. The area between f ( x ) = 1 20 where 0 ≤ x ≤ 20 and the x -axis is the area of a rectangle with base = 20 and height = 1 20 . AREA = 20 ( 1 20 ) = 1 Suppose we want to find the area between f( x ) = 1 20 and the x -axis where 0 < x < 2. AREA = ( 2 – 0 ) ( 1 20 ) = 0.1 ( 2 – 0 ) = 2 = base of a rectangle REMINDER area of a rectangle = (base)(height). The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as P (0 < x < 2) = P ( x < 2) = 0.1. Suppose we want to find the area between f ( x ) = 1 20 and the x -axis where 4 < x < 15. AREA = ( 15 – 4 ) ( 1 20 ) = 0.55 ( 15 – 4 ) = 11 = the base of a rectangle The area corresponds to the probability P (4 < x < 15) = 0.55. Suppose we want to find P ( x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P ( x = 15) = (base)(height) = (0) ( 1 20 ) = 0 P ( X ≤ x ), which can also be written as P ( X < x ) for continuous distributions, is called the cumulative distribution function or CDF. Notice the \"less than or equal to\" symbol. We can also use the CDF to calculate P ( X > x ). The CDF gives \"area to the left\" and P ( X > x ) gives \"area to the right.\" We calculate P ( X > x ) for continuous distributions as follows: P ( X > x ) = 1 – P ( X < x ). Label the graph with f ( x ) and x . Scale the x and y axes with the maximum x and y values. f ( x ) = 1 20 , 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. P ( 2.3 < x < 12.7 ) = ( base ) ( height ) = ( 12.7 − 2.3 ) ( 1 20 ) = 0.52 Try It Consider the function f ( x ) = 1 8 for 0 ≤ x ≤ 8. Draw the graph of f ( x ) and find P (2.5 < x < 7.5). P (2.5 < x < 7.5) = 0.625 Chapter Review The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P ( a < x < b ). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f ( x ), is used to draw the graph of the probability distribution. The total area under the graph of f ( x ) is one. The area under the graph of f ( x ) and between values a and b gives the probability P ( a < x < b ). The cumulative distribution function (cdf) of X is defined by P ( X ≤ x ). It is a function of x that gives the probability that the random variable is less than or equal to x . Formula Review Probability density function (pdf) f ( x ): f ( x ) ≥ 0 The total area under the curve f ( x ) is one. Cumulative distribution function (cdf): P ( X ≤ x ) Which type of distribution does the graph illustrate? Uniform Distribution Which type of distribution does the graph illustrate? Which type of distribution does the graph illustrate? Normal Distribution What does the shaded area represent? P (___< x < ___) What does the shaded area represent? P (___< x < ___) P (6 < x < 7) For a continuous probability distribution, 0 ≤ x ≤ 15. What is P ( x > 15)? What is the area under f ( x ) if the function is a continuous probability density function? one For a continuous probability distribution, 0 ≤ x ≤ 10. What is P ( x = 7)? A continuous probability function is restricted to the portion between x = 0 and 7. What is P ( x = 10)? zero f ( x ) for a continuous probability function is 1 5 , and the function is restricted to 0 ≤ x ≤ 5. What is P ( x < 0)? f ( x ), a continuous probability function, is equal to 1 12 , and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)? one Find the probability that x falls in the shaded area. Find the probability that x falls in the shaded area. 0.625 Find the probability that x falls in the shaded area. f ( x ), a continuous probability function, is equal to 1 3 and the function is restricted to 1 ≤ x ≤ 4. Describe P ( x > 3 2 ) . The probability is equal to the area from x = 3 2 to x = 4 above the x-axis and up to f ( x ) = 1 3 . Homework For each probability and percentile problem, draw the picture. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor. What part of the experiment will yield discrete data? What part of the experiment will yield continuous data? When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why? Age is a measurement, regardless of the accuracy used. Uniform Distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b ; it is often referred as the rectangular distribution because the graph of the pdf has the form of a rectangle. The mean is μ = a + b 2 and the standard deviation is σ = ( b − a ) 2 12 . The probability density function is f ( x ) = 1 b − a for a < x < b or a ≤ x ≤ b . The cumulative distribution is P ( X ≤ x ) = x − a b − a . Exponential Distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital. The mean is μ = 1 m and the standard deviation is σ = 1 m . The probability density function is f ( x ) = m e - m x or f ( x ) = 1 μ e - 1 μ x , x ≥ 0 and the cumulative distribution function is P ( X ≤ x ) = 1 - e − m x or P ( X ≤ x ) = 1 - e − 1 μ x .", "section": "Properties of Continuous Probability Density Functions", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints. The mathematical statement of the uniform distribution is f ( x ) = 1 b − a for a ≤ x ≤ b where a = the lowest value of x and b = the highest value of x . Formulas for the theoretical mean and standard deviation are μ = a + b 2 and σ = ( b − a ) 2 12 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. a. What is the probability that a person waits fewer than 12.5 minutes? a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U (0, 15). Write the probability density function. f ( x ) = 1 15 − 0 = 1 15 for 0 ≤ x ≤ 15. Find P ( x < 12.5). Draw a graph. P ( x < k ) = ( base ) ( height ) = ( 12.5 - 0 ) ( 1 15 ) = 0.8333 The probability a person waits less than 12.5 minutes is 0.8333. b. On the average, how long must a person wait? Find the mean, μ , and the standard deviation, σ . b. μ = a + b 2 = 15 + 0 2 = 7.5. On the average, a person must wait 7.5 minutes. σ = ( b - a ) 2 12 = ( 15 - 0 ) 2 12 = 4.3. The Standard deviation is 4.3 minutes. c. Ninety percent of the time, the time a person must wait falls below what value? NOTE This asks for the 90 th percentile. c. Find the 90 th percentile. Draw a graph. Let k = the 90 th percentile. P ( x < k ) = ( base ) ( height ) = ( k − 0 ) ( 1 15 ) 0.90 = ( k ) ( 1 15 ) k = ( 0.90 ) ( 15 ) = 13.5 The 90 th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. Try It The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive. Find a and b and describe what they represent. Write the distribution. Find the mean and the standard deviation. What is the probability that the duration of games for a team in a single season is between 480 and 500 hours? a is 447, and b is 521. a is the minimum duration of games for a team for the 2011 season, and b is the maximum duration of games for a team for the 2011 season. X ~ U (447, 521). μ = 484, and σ = 21.36 P (480 < x < 500) = 0.2703 Chapter Review If X has a uniform distribution where a < x < b or a ≤ x ≤ b , then X takes on values between a and b (may include a and b ). All values x are equally likely. We write X ∼ U ( a , b ). The mean of X is μ = a + b 2 . The standard deviation of X is σ = ( b − a ) 2 12 . The probability density function of X is f ( x ) = 1 b − a for a ≤ x ≤ b . The cumulative distribution function of X is P ( X ≤ x ) = x − a b − a . X is continuous. The probability P ( c < X < d ) may be found by computing the area under f ( x ), between c and d . Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. Formula Review X = a real number between a and b (in some instances, X can take on the values a and b ). a = smallest X ; b = largest X X ~ U (a, b) The mean is μ = a + b 2 The standard deviation is σ = ( b – a ) 2 12 Probability density function: f ( x ) = 1 b − a for a ≤ X ≤ b Area to the Left of x : P ( X < x ) = ( x – a ) ( 1 b − a ) Area to the Right of x : P ( X > x ) = ( b – x ) ( 1 b − a ) Area Between c and d : P ( c < x < d ) = (base)(height) = ( d – c ) ( 1 b − a ) pdf: f ( x ) = 1 b − a for a ≤ x ≤ b cdf: P ( X ≤ x ) = x − a b − a mean µ = a + b 2 standard deviation σ = ( b − a ) 2 12 P ( c < X < d ) = ( d – c ) ( 1 b – a ) References McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 1.5 2.4 3.6 2.6 1.6 2.4 2.0 3.5 2.5 1.8 2.4 2.5 3.5 4.0 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as X ~ U (1.5, 4.5). What type of distribution is this? In this distribution, outcomes are equally likely. What does this mean? It means that the value of x is just as likely to be any number between 1.5 and 4.5. What is the height of f ( x ) for the continuous probability distribution? What are the constraints for the values of x ? 1.5 ≤ x ≤ 4.5 Graph P (2 < x < 3). What is P (2 < x < 3)? 0.3333 What is P (x < 3.5 | x < 4)? What is P ( x = 1.5)? zero Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. 0.6 Use the following information to answer the next eight exercises. A distribution is given as X ~ U (0, 12). What is a ? What does it represent? What is b ? What does it represent? b is 12, and it represents the highest value of x . What is the probability density function? What is the theoretical mean? six What is the theoretical standard deviation? Draw the graph of the distribution for P ( x > 9). Find P ( x > 9). Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. What is being measured here? In words, define the random variable X . X = The age (in years) of cars in the staff parking lot Are the data discrete or continuous? The interval of values for x is ______. 0.5 to 9.5 The distribution for X is ______. Write the probability density function. f ( x ) = 1 9 where x is between 0.5 and 9.5, inclusive. Graph the probability distribution. Sketch the graph of the probability distribution. Identify the following values: Lowest value for x – : _______ Highest value for x – : _______ Height of the rectangle: _______ Label for x -axis (words): _______ Label for y -axis (words): _______ Find the average age of the cars in the lot. μ = 5 Find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, and shade the area of interest. Find the probability. P ( x < 4) = _______ Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, shade the area of interest. Find the probability. P ( x < 4 | x < 7.5) = _______ Answers may vary. 3.5 7 What has changed in the previous two problems that made the solutions different? Find the third quartile of ages of cars in the lot. This means you will have to find the value such that 3 4 , or 75%, of the cars are at most (less than or equal to) that age. Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.75. The third quartile is _______ Answers may vary. k = 7.25 7.25 Homework For each probability and percentile problem, draw the picture. Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks). Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that a person is born at the exact moment week 19 starts. That is, find P ( x = 19) = _________ P (2 < x < 31) = _________ Find the probability that a person is born after week 40. P (12 < x | x < 28) = _________ A random number generator picks a number from one to nine in a uniform manner. Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ P (3.5 < x < 7.25) = _________ P ( x > 5.67) P ( x > 5 | x > 3) = _________ Answers may vary. f ( x ) = 1 8 where 1 ≤ x ≤ 9 five 2.3 15 32 333 800 2 3 According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. Define the random variable. X = _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that the individual lost more than ten pounds in a month. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month. P (7 < x < 13 | x > 9) = __________. State this in a probability question, similarly to parts g and h, draw the picture, and find the probability. A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. Define the random variable. X = _______ Graph the probability distribution. f ( x ) = _______ μ = _______ σ = _______ Find the probability that the commuter waits less than one minute. Find the probability that the commuter waits between three and four minutes. X represents the length of time a commuter must wait for a train to arrive on the Red Line. Graph the probability distribution. f ( x ) = 1 8 where 0 ≤ x ≤ 8 four 2.31 1 8 1 8 The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class. Define the random variable. X = _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that she is over 6.5 years old. Find the probability that she is between four and six years old. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. What is the average waiting time (in minutes)? zero two three four d The probability of waiting more than seven minutes given a person has waited more than four minutes is? 0.125 0.25 0.5 0.75 b The time (in minutes) until the next bus departs a major bus depot follows a distribution with f ( x ) = 1 20 where x goes from 25 to 45 minutes. Define the random variable. X = ________ Graph the probability distribution. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous). μ = ________ σ = ________ Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. P (25 < x < 55) = _________. State this in a probability statement, similarly to parts g and h, draw the picture, and find the probability. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution. Find the probability that the value of the stock is more than $19. Find the probability that the value of the stock is between $19 and $22. Given that the stock is greater than $18, find the probability that the stock is more than $21. The probability density function of X is 1 25 − 16 = 1 9 . P ( X > 19) = (25 – 19) ( 1 9 ) = 6 9 = 2 3 . P (19 < X < 22) = (22 – 19) ( 1 9 ) = 3 9 = 1 3 . This is a conditional probability question. P(x > 21 | x > 18). You can do this two ways: Draw the graph where a is now 18 and b is still 25. The height is 1 ( 25 − 18 ) = 1 7 So, P ( x > 21 | x > 18) = (25 – 21) ( 1 7 ) = 4/7. Use the formula: P ( x > 21 | x > 18) = P ( x > 21 ∩ x > 18 ) P ( x > 18 ) = P ( x > 21 ) P ( x > 18 ) = ( 25 − 21 ) ( 25 − 18 ) = 4 7 . A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution. Find the average time between fireworks. Find probability that the time between fireworks is greater than four seconds. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. Find the probability that the truck driver goes more than 650 miles in a day. Find the probability that the truck drivers goes between 400 and 650 miles in a day. P ( X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 = 0.125 P (400 < X < 650) = 650 − 400 700 − 300 = 250 400 = 0.625 Conditional Probability the likelihood that an event will occur given that another event has already occurred.", "section": "The Uniform Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Exponential Distribution The exponential probability density function is built upon the general exponential function where the variable is an exponent: f ( x ) = a ( b ) x . This equation can be converted to a natural system of logarithms with a base e that has an approximate value of 2.71828. The exponential probability density function is valuable with a number of practical applications. The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length of time, in minutes, of long-distance business telephone calls and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, marketing studies have shown that the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money. Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts. A firm will use the exponential distribution analysis to set the number of months they will provide a warranty based upon the probability of a specific time until the failure of the product. The random variable for the exponential distribution is continuous and often measures a passage of time, although it can be used in other continuous random variable applications. Typical questions may be “What is the probability that some event will occur within the next x hours or days?” or “What is the probability that some event will occur between x 1 hours and x 2 hours?” or “What is the probability that the event will take more than x 1 hours to perform?” In short, the random variable X equals ( a ) the time between events or ( b ) the passage of time to complete an action (e.g., wait on a customer). Be sure to differentiate the exponential distribution random variable and that of the Poisson distribution. The Poisson has a discrete random variable that gives us the probability that x number of occurrences will occur in the next time period. The continuous exponential distribution provides us the probability that an X -long time period will occur between now and the next occurrence. We will explore more this link between the exponential distribution and the Poisson distribution. The exponent probability density function is given by: f ( x ) = 1 μ e - x μ where μ is the historical average waiting time. The exponential distribution has a mean and standard deviation both equal to μ . To find a probability with a continuous random variable, the cumulative distribution is integrated between the relevant values of x . The definite integral of a function provides the area under the function between two values of x . For the exponential distribution, there are three cases that depend upon the question asked. Case I would answer the question “What is the probability that the time to completion is les s than X ?” The probability question asked would be written as P ( X ≤ x ) . For example, to calculate the probability that a customer will be served within the next hour given that the historical waiting time was 30 minutes, we integrate the exponential probability distribution function 1 μ e - x μ from x = 0 to x = 60 . F ( x ) = P ( X ≤ x ) = ∫ 0 x 1 μ e - x μ d x = 1 - e - x μ Case I graphs the exponential probability density function for this example. The area under the function from zero to 60 minutes is the probability for the question asked. Note that the exponential probability distribution at zero is at the value of 1 / µ . Solving for the probability density function equation above where µ = 30 is the historical waiting time, and x = 60 , which is the time we are interested in. P ( x ≤ 60 ) = 1 - e - x μ = 1 - e - 60 30 = 0 . 865 With a historical waiting time of 30 minutes, certainly we would expect the high probability of 0.865 of being waited on in an hour’s time. (A probability of 0.865 is almost a certainty.) Case II would answer questions as “What is the probability more than X ?” What is the probability that a person will have to wait MORE THAN x minutes? The relevant probability density function would be P ( X ≥ x α ) = 1 - 1 - e - x μ = e - x μ and is presented in . Case II Imagine an electric component is guaranteed to last two years. If it fails within the two-year guarantee, the purchaser will receive their money back in full. The manufacture knows by careful monitoring of their products the historical life of this component is 750 days, or two-and-one-half years. Are they engaged in a risky business policy? Here we can calculate the probability that a component will last longer than 730 days (two years) and thus no payment is required by the firm. The historical mean is 750 days, and the relevant time period is x = 730 days. Using the exponential probability density function, it is found that the probability this electric component will last more than two years is only 0.378, as shown on . P ( X ≥ x ) = e - x μ = e - 730 750 = 0 . 378 Because the working life of each of the electric components is an independent random variable, we can say that approximately 38 percent of these components survive two years and thus 62 percent of purchasers of these components will be asking for a full refund. In short, the low production quality and thus high refund requirements will result in significant financial costs. We could expect that the two-year guarantee will be eliminated soon. Case III allows for knowing the probability that an occurrence will occur within a window of time. For example, when asking for the arrival time of an airplane, one might be given a range of time such as “plus or minus from the historical expected arrival time measured in hours from the time of take-off.” In this case, the mean is the historical expected arrival time given the historical travel time. Assume that the historical travel time between these two points is four hours. Case III requires calculating the probability between two points. : Case III shows the two points that are one hour longer and one hour sooner than the historical arrival time, which is midway between these points. If the historical travel time is four hours, then x o is three hours of travel time and x 1 is five hours of travel time. The mathematical expression is: P ( x 0 ≤ x ≤ x 1 ) = P ( X ≥ x 0 ) - P ( X ≥ x 1 ) = e - x 0 μ - e - x 1 μ P ( 3 ≤ x ≤ 5 ) = e - 3 4 - e - 5 4 = 0 . 186 The probability the flight will arrive plus or minus the historical arrival time of four hours from takeoff is 0.186. Intuitively, this seems a very small probability to hit such a broad window. At the core, the standard deviation of average flight time must be very large. Exponential Distribution and Decay Factor The core of the exponential distribution is e, the natural logarithm in an exponential function with the variable - x μ in the exponent. The exponent is negative, and thus it describes an exponentially declining function as we have seen in , , and above. This gives rise to an alternative formula for the exponential probability distribution: f ( x ) = m e - m F ( x ) = ∫ 0 x m e - m t d t = 1 - e - m x Where m = 1 μ is given the name decay factor and measures the speed of decay. The decay factor simply measures how rapidly the probability of an event declines as the random variable X increases. shows the exponential distribution for three different decay factors. As before, the intercept on the vertical axis, the probability density, is the decay factor. A decay factor of 1.5, as in the first exponential function, is three time the speed of the third exponential distribution with a decay factor of 0.5. Looking at the probability for each of the three functions at the same value of x 1 , it is clear that the first function has a significantly greater probability of occurrence at x 1 than the other two (i.e., P ( X α ≤ x 1 ) > P ( X b ≤ x 1 ) > P ( X c ≤ x 1 ) comparing the relevant probabilities for each of the functions a, b, c in order). Finally, the exponential distribution shows with drama the true meaning of “exponential.” In each unit, a change in x in an exponential distribution has a significant change in probability. For example, if the simple exponential distribution function Y = 2 x changes from x = 4 to x = 5 , the y -value increases from 16 to 32, but a one-unit change from x = 10 to x = 11 causes the y -value to increase from 1,024 to 2,048. Likewise, in an exponential function with a negative exponent such as the exponential probability distribution, the changes are startling for each unit change at higher levels of X . At X = – 4 to x = - 5 , the y -value changes from 0.063 to 0.031, while a one-unit change from X = – 10 to X = – 11 results in a change in the y -value from 0.00097 to 0.00048. The conclusion is that probabilities in an exponential probability distribution at large x -values are scarce almost to the point of nonexistence. Let X = amount of time (in minutes) a postal clerk spends with a customer. The time is known from historical data to have an average amount of time equal to four minutes. It is given that μ = 4 minutes, that is, the average time the clerk spends with a customer is 4 minutes. Remember that we are still doing probability and thus we have to be told the population parameters such as the mean. To do any calculations, we need to know the mean of the distribution: the historical time to provide a service, for example. Knowing the historical mean allows the calculation of the decay parameter, m . m = 1 μ . Therefore, m = 1 4 = 0.25 When the notation used the decay parameter, m , the probability density function is presented as f ( x ) = m e − m x , which is simply the original formula with m substituted for 1 μ , or f ( x ) = 1 μ e − 1 μ x . To calculate probabilities for an exponential probability density function, we need to use the cumulative density function. As shown below, the curve for the cumulative density function is: f ( x ) = 0.25 e –0.25 x where x is at least zero and m = 0.25. For example, f (5) = 0.25 e (-0.25)(5) = 0.072. In other words, the function has a value of .072 when x = 5. The graph is as follows: Notice the graph is a declining curve. When x = 0, f ( x ) = 0.25 e (−0.25)(0) = (0.25)(1) = 0.25 = m . The maximum value on the y -axis is always m , one divided by the mean. Try It The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution. X ~ Exp (0.125); f ( x ) = 0.125e –0.125 x ; a. Using the information in , find the probability that a clerk spends four to five minutes with a randomly selected customer. a. Find P (4 < x < 5). The cumulative distribution function (CDF) gives the area to the left. P ( x < x ) = 1 – e –mx P ( x < 5) = 1 – e (–0.25)(5) = 0.7135 and P ( x < 4) = 1 – e (–0.25)(4) = 0.6321 P (4 < x < 5)= 0.7135 – 0.6321 = 0.0814 Try It The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait? P ( x < 10) = 0.4866 50 th percentile = 10.40 On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than 7 years? a. Let x = the amount of time (in years) a computer part lasts. μ = 10 so m = 1 μ = 1 10 = 0.1 Find P ( x > 7). Draw the graph. P ( x > 7) = 1 – P ( x < 7). Since P ( X < x ) = 1 – e –mx then P ( X > x ) = 1 – ( 1 – e –mx ) = e –mx P ( x > 7) = e (–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966. b. On the average, how long would five computer parts last if they are used one after another? b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years. d. What is the probability that a computer part lasts between nine and 11 years? d. Find P (9 < x < 11). Draw the graph. P (9 < x < 11) = P ( x < 11) – P ( x < 9) = (1 – e (–0.1)(11) ) – (1 – e (–0.1)(9) ) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737. Try It On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day? P ( x > 15) = 0.4346 Six pairs of running shoes would last 108 months on average. 80 th percentile = 28.97 months Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12 . The decay p[parameter is another way to view 1/λ. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes. What is m , μ , and σ ? The probability that you must wait more than five minutes is _______ . m = 1 12 μ = 12 σ = 12 P ( x > 5) = 0.6592 Try It Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1 20 . Let X = the distance people are willing to commute in miles. What is m , μ , and σ ? What is the probability that a person is willing to commute more than 25 miles? The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed. On average, how many minutes elapse between two successive arrivals? When the store first opens, how long on average does it take for three customers to arrive? After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive. Is an exponential distribution reasonable for this situation? Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive. Let X = the time between arrivals, in minutes. By part a, μ = 2, so m = 1 2 = 0.5. The cumulative distribution function is P ( X < x ) = 1 – e (-0.5)(x) Therefore P ( X < 1) = 1 – e (–0.5)(1) = 0.3935. P ( X > 5) = 1 – P ( X < 5) = 1 – (1 – e (-0.5)(5) ) = e –2.5 ≈ 0.0821. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others. Try It Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution. On average, how many seconds elapse between two successive cars? After a car passes by, how long on average will it take for another seven cars to pass by? Find the probability that after a car passes by, the next car will pass within the next 20 seconds. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds. Memorylessness of the Exponential Distribution Recall that the amount of time between customers for the postal clerk discussed earlier is exponentially distributed with a mean of two minutes. Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property . The exponential and geometric probability density functions are the only probability functions that have the memoryless property. Specifically, the memoryless property says that P ( X > r + t | X > r ) = P ( X > t ) for all r ≥ 0 and t ≥ 0 For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation. P ( X > 5 + 1 | X > 5) = P ( X > 1) = e ( – 0.5 ) ( 1 ) = 0.6065. This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival. The exponential distribution is often used to model the longevity of an electrical or mechanical device. In , the lifetime of a certain computer part has the exponential distribution with a mean of ten years. The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P ( X > 17| X > 10) = P ( X > 7) = 0.4966, where the vertical line is read as \"given\". Refer back to the postal clerk again where the time a postal clerk spends with a customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that the customer will spend at least an additional three minutes with the postal clerk? The decay parameter of X is m = 1 4 = 0.25, so X ∼ Exp (0.25). The cumulative distribution function is P ( X < x ) = 1 – e –0.25 x . We want to find P ( X > 7| X > 4). The memoryless property says that P ( X > 7| X > 4) = P ( X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk. This is P ( X > 3) = 1 – P ( X < 3) = 1 – (1 – e –0.25⋅3 ) = e –0.75 ≈ 0.4724. Try It Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years. Relationship between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean μ . Recall that if X has the Poisson distribution with mean μ , then P ( X = x ) = μ x e − μ x ! . The formula for the exponential distribution: P ( X = x ) = m e - m x = 1 μ e - 1 μ x Where m = the rate parameter, or μ = average time between occurrences. We see that the exponential is the cousin of the Poisson distribution and they are linked through this formula. There are important differences that make each distribution relevant for different types of probability problems. First, the Poisson has a discrete random variable, x, where time; a continuous variable is artificially broken into discrete pieces. We saw that the number of occurrences of an event in a given time interval, x, follows the Poisson distribution. For example, the number of times the telephone rings per hour. By contrast, the time between occurrences follows the exponential distribution. For example. The telephone just rang, how long will it be until it rings again? We are measuring length of time of the interval, a continuous random variable, exponential, not events during an interval, Poisson. The Exponential Distribution v. the Poisson Distribution A visual way to show both the similarities and differences between these two distributions is with a time line. The random variable for the Poisson distribution is discrete and thus counts events during a given time period, t 1 to t 2 on , and calculates the probability of that number occurring. The number of events, four in the graph, is measured in counting numbers; therefore, the random variable of the Poisson is a discrete random variable. The exponential probability distribution calculates probabilities of the passage of time, a continuous random variable. In this is shown as the bracket from t 1 to the next occurrence of the event marked with a triangle. Classic Poisson distribution questions are \"how many people will arrive at my checkout window in the next hour?\". Classic exponential distribution questions are \"how long it will be until the next person arrives,\" or a variant, \"how long will the person remain here once they have arrived?\". Again, the formula for the exponential distribution is: f ( x ) = m e - m x or f ( x ) = 1 μ e - 1 μ x We see immediately the similarity between the exponential formula and the Poisson formula. P ( x ) = μ x e − μ x ! Both probability density functions are based upon the relationship between time and exponential growth or decay. The “e” in the formula is a constant with the approximate value of 2.71828 and is the base of the natural logarithmic exponential growth formula. When people say that something has grown exponentially this is what they are talking about. An example of the exponential and the Poisson will make clear the differences been the two. It will also show the interesting applications they have. Poisson Distribution Suppose that historically 10 customers arrive at the checkout lines each hour. Remember that this is still probability so we have to be told these historical values. We see this is a Poisson probability problem. We can put this information into the Poisson probability density function and get a general formula that will calculate the probability of any specific number of customers arriving in the next hour. The formula is for any value of the random variable we chose, and so the x is put into the formula. In this example, μ = 10 because we expect 10 customers in line each hour. This is the formula: f ( x ) = 10 x e -10 x ! As an example, the probability of 15 people arriving at the checkout counter in the next hour would be P ( x = 15 ) = 10 15 e -10 15 ! = 0.0347 Here we have inserted x = 15 and calculated the probability that in the next hour 15 people will arrive is 0.035. Exponential Distribution If we keep the same historical facts that 10 customers arrive each hour, but we now are interested in the service time a person spends at the counter, then we would use the exponential distribution. The exponential probability function for any value of x, the random variable, for this particular checkout counter historical data is: f ( x ) = 1 .1 e -x .1 = 10 e -10 x To calculate µ , the historical average service time, we simply divide the number of people that arrive per hour, 10 , into the time period, one hour, and have µ = 0.1. Historically, people spend 0.1 of an hour at the checkout counter, or 6 minutes. This explains the .1 in the formula. There is a natural confusion with µ in both the Poisson and exponential formulas. They have different meanings, although they have the same symbol. The mean of the exponential is one divided by the mean of the Poisson. If you are given the historical number of arrivals you have the mean of the Poisson. If you are given an historical length of time between events you have the mean of an exponential. Continuing with our example at the checkout clerk; if we wanted to know the probability that a person would spend 9 minutes or less checking out, then we use this formula. First, we convert to the same time units which are parts of one hour. Nine minutes is 0.15 of one hour. Next we note that we are asking for a range of values. This is always the case for a continuous random variable. We write the probability question as: p ( x ≤ 9 ) = 1 - 10 e -10 x We can now put the numbers into the formula and we have our result. p ( x = .15 ) = 1 - 10 e -10 ( .15 ) = 0.7769 The probability that a customer will spend 9 minutes or less checking out is 0.7769. We see that we have a high probability of getting out in less than nine minutes and a tiny probability of having 15 customers arriving in the next hour. At a police station in a large city, calls come in at an average rate of four calls per minute. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has a Poisson distribution. Find the average time between two successive calls. Find the probability that after a call is received, the next call occurs in less than ten seconds. Find the probability that exactly five calls occur within a minute. Find the probability that fewer than five calls occur within a minute. Find the probability that more than 40 calls occur in an eight-minute period. On average, four calls occur per minute, so one call occurs every 15 seconds, or = 0.25 minutes occur between successive calls on average. Let T = time elapsed between calls. From part a, μ = 0 . 25 , so m = 4 . Thus, T ∼ Exp (4). The cumulative distribution function is P ( T < t ) = 1 - e - 4 t ) . The probability that the next call occurs in less than ten seconds ( ten seconds = 1 6 minute ) is P T < 1 6 = 1 - e - 4 ( 1 6 ) = 0 . 4866 . Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. Therefore, X ~ P o i s s o n ( 4 ) and so P ( X = 5 ) = 4 5 e - 4 5 ! ≈ 0 . 1563 . 5 ! = ( 5 ) ( 4 ) ( 3 ) ( 2 ) ( 1 ) Keep in mind that X must be a whole number, so P ( X < 5 ) = P ( X ≤ 4 ) . To compute this, we could take P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) . Using technology, we see that P ( X ≤ 4 ) = 0 . 6288 . Let Y = the number of calls that occur during an eight-minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight-minute period. Hence, Y ~ P o i s s o n ( 32 ) . Therefore, P ( Y > 40 ) = 1 - P ( Y ≤ 40 ) = 1 - 0 . 9294 = 0 . 0707 . Try It In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. Calculate the probability that there are at most 2 accidents occur in any given week. What is the probability that there is at least two weeks between any 2 accidents? Chapter Review If X has an exponential distribution with mean μ , then the decay parameter is m = 1 μ . The probability density function of X is f ( x ) = me -mx (or equivalently f ( x ) = 1 μ e − x / μ . The cumulative distribution function of X is P ( X ≤ x ) = 1 – e – mx . Formula Review pdf: f ( x ) = me (– mx ) where x ≥ 0 and m > 0 cdf: P ( X ≤ x ) = 1 – e (– mx ) mean µ = 1 m standard deviation σ = µ Additionally P ( X > x ) = e (– mx ) P ( a < X < b ) = e (– ma ) – e (– mb ) Poisson probability: P ( X = x ) = μ x e − μ x ! with mean and variance of μ References Data from the United States Census Bureau. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013). “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013). Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf‎ (accessed June 11, 2013). Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp (0.2) What type of distribution is this? Are outcomes equally likely in this distribution? Why or why not? No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. What is m ? What does it represent? What is the mean? five What is the standard deviation? State the probability density function. f ( x ) = 0.2e -0.2 x Graph the distribution. Find P (2 < x < 10). 0.5350 Find P ( x > 6). Find the 70 th percentile. 6.02 Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp (0.75). What is m ? What is the probability density function? f ( x ) = 0.75 e -0.75 x What is the cumulative distribution function? Draw the distribution. Find P ( x < 4). Find the 30 th percentile. 0.4756 Find the median. Which is larger, the mean or the median? The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33 , which is greater than 0.9242. Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. What is being measured here? Are the data discrete or continuous? continuous In words, define the random variable X . What is the decay rate ( m )? m = 0.000121 The distribution for X is ______. Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P ( x < 5,730). Sketch the graph, and shade the area of interest. Find the probability. P ( x < 5,730) = __________ Answers may vary. P ( x < 5,730) = 0.5001 Find the percentage of carbon-14 lasting longer than 10,000 years. Sketch the graph, and shade the area of interest. Find the probability. P ( x > 10,000) = ________ Thirty percent (30%) of carbon-14 will decay within how many years? Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.30. Answers may vary. k = 2947.73 Homework Suppose that the length of phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. Define the random variable. X = ________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a phone call lasts less than nine minutes. Find the probability that a phone call lasts more than nine minutes. Find the probability that a phone call lasts between seven and nine minutes. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. Define the random variable. X = _________________________________. Is X continuous or discrete? On average, how long would you expect one car battery to last? On average, how long would you expect nine car batteries to last, if they are used one after another? Find the probability that a car battery lasts more than 36 months. Seventy percent of the batteries last at least how long? X = the useful life of a particular car battery, measured in months. X is continuous. 40 months 360 months 0.4066 14.27 At one point in time, the percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. Define the random variable. X = _________________________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the percent is less than 12. Find the probability that the percent is between eight and 14. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. Why is this number different from 9.848%? What would make this number higher than 9.848%? The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. Define the random variable. X = _________________________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the person retired after age 70. Do more people retire before age 65 or after age 65? In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet? X = the time (in years) after reaching age 60 that it takes an individual to retire X is continuous. five five Answers may vary. 0.1353 before 18.3 The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150. Define the random variable. X = _________________________________. μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a car required over $300 for maintenance during its first year. Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution. The decay rate is: 0.3333 0.5000 2 3 a What is the probability that a smartphones will fail within two years of the date of purchase? 0.8647 0.4866 0.2212 0.9997 What is the median lifetime of these smartphones (in years)? 0.1941 1.3863 2.0794 5.5452 c At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution. On average, how much time occurs between five consecutive calls? Find the probability that after a call is received, it takes more than three minutes for the next call to occur. Ninety-percent of all calls occur within how many minutes of the previous call? Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute. Find the probability that less than 20 calls occur within an hour. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. What is the probability that an entire season elapses with a single no-hitter? If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? What is the probability that there are more than 3 no-hitters in a single season? Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, ( X = 0) = 3 0 e – 3 0 ! = e –3 ≈ 0.0498 NOTE You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 1 3 season. For the exponential, µ = 1 3 . Therefore, m = 1 μ = 3 and T ∼ Exp (3). The desired probability is P ( T > 1) = 1 – P ( T < 1) = 1 – (1 – e –3 ) = e –3 ≈ 0.0498. Let T = duration of time between no-hitters. We find P ( T > 2| T > 1), and by the memoryless property this is simply P ( T > 1), which we found to be 0.0498 in part a. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P ( X > 3) = 1 – P ( X ≤ 3) = 0.3528. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential. What is the probability that the next earthquake occurs within the next three months? Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes? What is the probability of zero earthquakes occurring in 2014? What is the probability that at least two earthquakes will occur in 2014? According to the American Red Cross, about one out of nine people in the U.S. have Type B positive blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B positive blood types that arrive roughly follows the Poisson distribution. If 100 people arrive, how many on average would be expected to have Type B positive blood? What is the probability that over 10 people out of these 100 have type B positive blood? What is the probability that more than 20 people arrive before a person with type B positive blood is found? 100 9 = 11.11 P ( X > 10) = 1 – P ( X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532. The number of people with Type B positive blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 . The cumulative distribution function of X is P ( X < x ) = 1 − e − x 9 . Thus, P ( X > 20) = 1 - P ( X ≤ 20) = 1 − ( 1 − e − 20 9 ) ≈ 0.1084. NOTE We could also deduce that each person arriving has a 8/9 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive blood is ( 8 9 ) 20 ≈ 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between Type B positive people is discrete instead of continuous.) A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution. Find the probability that the duration between two successive visits to the web site is more than ten minutes. The top 25% of durations between visits are at least how long? Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes? Find the probability that less than 7 visits occur within a one-hour period. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed. Find the probability that the time between two successive visits to the urgent care facility is less than 2 minutes. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes? Find the probability that more than eight patients arrive during a half-hour period. Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 . The cdf is P ( T < t ) = 1 − e t 7 P ( T < 2) = 1 - 1 − e − 2 7 ≈ 0.2485. P ( T > 15) = 1 − P ( T < 15 ) = 1 − ( 1 − e − 15 7 ) ≈ e − 15 7 ≈ 0.1173 . P ( T > 15| T > 10) = P ( T > 5) = 1 − ( 1 − e − 5 7 ) = e − 5 7 ≈ 0.4895 . Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 , X ∼ Poisson ( 30 7 ) . Find P ( X > 8) = 1 – P ( X ≤ 8) ≈ 0.0311. decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x . It is the value m in the probability density function f ( x ) = me (- mx ) of an exponential random variable. It is also equal to m = 1 μ , where μ is the mean of the random variable. memoryless property For an exponential random variable X , the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that X exceeds x + t , given that it has exceeded x , is the same as the probability that X would exceed t if we had no knowledge about it. In symbols we say that P ( X > x + t | X > x ) = P ( X > t ). Poisson distribution If there is a known average of μ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of x events occurring in one unit time is equal to P ( X = x ) = μ x e − μ x ! .", "section": "The Exponential Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction If you ask enough people about their shoe size, you will find that your graphed data is shaped like a bell curve and can be described as normally distributed. (credit: modification of work “Shoe shop” by Lo/ Flickr, Public Domain) The normal probability density function, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. Remember here that we are still talking about the distribution of population data. This is a discussion of probability and thus it is the population data that may be normally distributed, and if it is, then this is how we can find probabilities of specific events just as we did for population data that may be binomially distributed or Poisson distributed. This caution is here because in the next chapter we will see that the normal distribution describes something very different from raw data and forms the foundation of inferential statistics. The normal distribution has two parameters (two numerical descriptive measures): the mean ( μ ) and the standard deviation ( σ ). If X is a quantity to be measured that has a normal distribution with mean ( μ ) and standard deviation ( σ ), we designate this by writing the following formula of the normal probability density function: The probability density function is a rather complicated mathematical function. Do not memorize it . It is not necessary. f ( x ) = 1 σ ⋅ 2 ⋅ π ⋅ e − 1 2 ⋅ ( x − μ σ ) 2 The curve is symmetric about a vertical line drawn through the mean, μ . The mean is the same as the median, which is the same as the mode, because the graph is symmetric about μ . As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Note that this is unlike several probability density functions we have already studied, such as the Poisson, where the mean is equal to μ and the standard deviation simply the square root of the mean, or the binomial, where p is used to determine both the mean and standard deviation. Since the area under the curve must equal one, a change in the standard deviation, σ , causes a change in the shape of the normal curve; the curve becomes fatter and wider or skinnier and taller depending on σ . A change in μ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution . Formula Review X ∼ N ( μ , σ ) μ = the mean σ = the standard deviation Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 σ 2 2 , where μ is the mean of the distribution and σ is the standard deviation; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV, Z, is called the standard normal distribution .", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Standard Normal Distribution The standard normal distribution is a normal distribution of standardized values called z -scores . A z -score is measured in units of the standard deviation. The mean for the standard normal distribution is zero, and the standard deviation is one. What this does is dramatically simplify the mathematical calculation of probabilities. Take a moment and substitute zero and one in the appropriate places in the above formula and you can see that the equation collapses into one that can be much more easily solved using integral calculus. The transformation z = x − μ σ produces the distribution Z ~ N (0, 1). The value x in the given equation comes from a known normal distribution with known mean μ and known standard deviation σ . The z -score tells how many standard deviations a particular x is away from the mean. Z -Scores If X is a normally distributed random variable and X ~ N(μ, σ) , then the z -score for a particular x is: z = x – μ σ The z -score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ . Values of x that are larger than the mean have positive z -scores, and values of x that are smaller than the mean have negative z -scores. If x equals the mean, then x has a z -score of zero. Suppose X ~ N(5, 6) . This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then: z = x – μ σ = 17 – 5 6 = 2 This means that x = 17 is two standard deviations (2 σ ) above or to the right of the mean μ = 5. Now suppose x = 1. Then: z = x – μ σ = 1 – 5 6 = –0.67 (rounded to two decimal places) This means that x = 1 is 0.67 standard deviations (–0.67 σ ) below or to the left of the mean μ = 5. Try It Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. X ~ N (16,4). Suppose Jerome scores ten points in a game. The z –score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?). The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ , then the Empirical Rule states the following: About 68% of the x values lie between –1 σ and +1 σ of the mean µ (within one standard deviation of the mean). About 95% of the x values lie between –2 σ and +2 σ of the mean µ (within two standard deviations of the mean). About 99.7% of the x values lie between –3 σ and +3 σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. The z -scores for +1 σ and –1 σ are +1 and –1, respectively. The z -scores for +2 σ and –2 σ are +2 and –2, respectively. The z -scores for +3 σ and –3 σ are +3 and –3 respectively. Suppose x has a normal distribution with mean 50 and standard deviation 6. About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1 σ = (–1)(6) = –6 and 1 σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z -scores are –1 and +1 for 44 and 56, respectively. About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2 σ = (–2)(6) = –12 and 2 σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z -scores are –2 and +2 for 38 and 62, respectively. About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 99.7% of the x values lie between –3 σ = (–3)(6) = –18 and 3 σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations from the mean 50. The z -scores are –3 and +3 for 32 and 68, respectively. Try It Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie? References “Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013). “The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013). “2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013). “Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013). Data from the San Jose Mercury News . Data from The World Almanac and Book of Facts . “List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013). Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013). Chapter Review A z -score is a standardized value. Its distribution is the standard normal, Z ~ N (0, 1). The mean of the z -scores is zero and the standard deviation is one. If z is the z -score for a value x from the normal distribution N ( µ , σ ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ . Formula Review Z ~ N (0, 1) z = a standardized value ( z -score) mean = 0; standard deviation = 1 To find the observed value, x , when the z -scores is known: x = μ + ( z ) σ z -score: z = x – μ σ or z = | x – μ | σ Z = the random variable for z -scores Z ~ N (0, 1) A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________. ounces of water in a bottle A normal distribution has a mean of 61 and a standard deviation of 15. What is the median? X ~ N (1, 2) σ = _______ 2 A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________. X ~ N (–4, 1) What is the median? –4 X ~ N (3, 5) σ = _______ X ~ N (–2, 1) μ = _______ –2 What does a z -score measure? What does standardizing a normal distribution do to the mean? The mean becomes zero. Is X ~ N (0, 1) a standardized normal distribution? Why or why not? What is the z -score of x = 12, if it is two standard deviations to the right of the mean? z = 2 What is the z -score of x = 9, if it is 1.5 standard deviations to the left of the mean? What is the z -score of x = –2, if it is 2.78 standard deviations to the right of the mean? z = 2.78 What is the z -score of x = 7, if it is 0.133 standard deviations to the left of the mean? Suppose X ~ N (2, 6). What value of x has a z -score of three? x = 20 Suppose X ~ N (8, 1). What value of x has a z -score of –2.25? Suppose X ~ N (9, 5). What value of x has a z -score of –0.5? x = 6.5 Suppose X ~ N (2, 3). What value of x has a z -score of –0.67? Suppose X ~ N (4, 2). What value of x is 1.5 standard deviations to the left of the mean? x = 1 Suppose X ~ N (4, 2). What value of x is two standard deviations to the right of the mean? Suppose X ~ N (8, 9). What value of x is 0.67 standard deviations to the left of the mean? x = 1.97 Suppose X ~ N (–1, 2). What is the z -score of x = 2? Suppose X ~ N (12, 6). What is the z -score of x = 2? z = –1.67 Suppose X ~ N (9, 3). What is the z -score of x = 9? Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z -score of x = 5.5? z ≈ –0.33 In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean. 0.67, right In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean. 3.14, left In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean. About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution? about 68% About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution? About what percent of x values lie between the second and third standard deviations (both sides)? about 4% Suppose X ~ N (15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15). Suppose X ~ N (–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3). between –5 and –1 Suppose X ~ N (–3, 1). Between what x values does 34.14% of the data lie? About what percent of x values lie between the mean and three standard deviations? about 50% About what percent of x values lie between the mean and one standard deviation? About what percent of x values lie between the first and second standard deviations from the mean (both sides)? about 27% About what percent of x values lie between the first and third standard deviations(both sides)? Use the following information to answer the next two exercises: The life of wearable fitness devicess is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Define the random variable X in words. X = _______________. The lifetime of a wearable fitness device measured in years. X ~ _____(_____,_____) Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the median recovery time? 2.7 5.3 7.4 2.1 What is the z -score for a patient who takes ten days to recover? 1.5 0.2 2.2 7.3 c The length of time to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? The data cannot follow the uniform distribution. The data cannot follow the exponential distribution.. The data cannot follow the normal distribution. I only II only III only I, II, and III The heights of the 430 National Basketball Association players were listed on team rosters at the start of a recent season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z -score and interpret it using complete sentences. 77 inches 85 inches If an NBA player reported his height had a z -score of 3.5, would you believe him? Explain your answer. Use the z -score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. Use the z -score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely. The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution. Calculate the z -scores for the male systolic blood pressures 100 and 150 millimeters. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him? Kyle’s doctor told him that the z -score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14). Which answer(s) is/are correct? Kyle’s systolic blood pressure is 175. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. Calculate Kyle’s blood pressure. iv Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N (10.2, 0.8). Calculate the z -scores that correspond to the following weights and interpret them. 11 kg 7.9 kg 12.2 kg During a certain year, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115. Calculate the z -score for an SAT score of 720. Interpret it using a complete sentence. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? During a different year, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took? Let X = an SAT math score and Y = an ACT math score. X = 720 : z = 720 - 520 115 = 1 . 74 . The exam score of 720 is 1.74 standard deviations above the mean of 520. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. X – μ σ = 700 – 514 117 ≈ 1.59, the z -score for the SAT. Y – μ σ = 30 – 21 5.3 ≈ 1.70, the z -scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z -score). Standard Normal Distribution a continuous random variable (RV) X ~ N (0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N (0, 1). z-score the linear transformation of the form z = x - μ σ or written as z = | x – μ | σ ; if this transformation is applied to any normal distribution X ~ N ( μ , σ ) the result is the standard normal distribution Z ~ N (0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ , the result is called the z -score of x . The z -score allows us to compare data that are normally distributed but scaled differently. A z -score is the number of standard deviations a particular x is away from its mean value.", "section": "The Standard Normal Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Using the Normal Distribution The shaded area in the following graph indicates the area to the right of x . This area is represented by the probability P ( X > x ). Normal tables provide the probability between the mean, zero for the standard normal distribution, and a specific value such as x 1 . This is the unshaded part of the graph from the mean to x 1 . Because the normal distribution is symmetrical , if x 1 were the same distance to the left of the mean the area, probability, in the left tail, would be the same as the shaded area in the right tail. Also, bear in mind that because of the symmetry of this distribution, one-half of the probability is to the right of the mean and one-half is to the left of the mean. Calculations of Probabilities To find the probability for probability density functions with a continuous random variable we need to calculate the area under the function across the values of X we are interested in. For the normal distribution this seems a difficult task given the complexity of the formula. There is, however, a simply way to get what we want. Here again is the formula for the normal distribution: f ( x ) = 1 σ ⋅ 2 ⋅ π ⋅ e − 1 2 ⋅ ( x − μ σ ) 2 Looking at the formula for the normal distribution it is not clear just how we are going to solve for the probability doing it the same way we did it with the previous probability functions. There we put the data into the formula and did the math. To solve this puzzle we start knowing that the area under a probability density function is the probability. This shows that the area between X 1 and X 2 is the probability as stated in the formula: P (X 1 ≤ x ≤ X 2 ) The mathematical tool needed to find the area under a curve is integral calculus. The integral of the normal probability density function between the two points x 1 and x 2 is the area under the curve between these two points and is the probability between these two points. Doing these integrals is no fun and can be very time consuming. But now, remembering that there are an infinite number of normal distributions out there, we can consider the one with a mean of zero and a standard deviation of 1. This particular normal distribution is given the name Standard Normal Distribution. Putting these values into the formula it reduces to a very simple equation. We can now quite easily calculate all probabilities for any value of x, for this particular normal distribution, that has a mean of zero and a standard deviation of 1. These have been produced and are available here in the appendix to the text or everywhere on the web. They are presented in various ways. The table in this text is the most common presentation and is set up with probabilities for one-half the distribution beginning with zero, the mean, and moving outward. The shaded area in the graph at the top of the table in Statistical Tables represents the probability from zero to the specific Z value noted on the horizontal axis, Z. The only problem is that even with this table, it would be a ridiculous coincidence that our data had a mean of zero and a standard deviation of one. The solution is to convert the distribution we have with its mean and standard deviation to this new Standard Normal Distribution. The Standard Normal has a random variable called Z. Using the standard normal table, typically called the normal table, to find the probability of one standard deviation, go to the Z column, reading down to 1.0 and then read at column 0. That number, 0.3413 is the probability from zero to 1 standard deviation. At the top of the table is the shaded area in the distribution which is the probability for one standard deviation. The table has solved our integral calculus problem. But only if our data has a mean of zero and a standard deviation of 1. However, the essential point here is, the probability for one standard deviation on one normal distribution is the same on every normal distribution. If the population data set has a mean of 10 and a standard deviation of 5 then the probability from 10 to 15, one standard deviation, is the same as from zero to 1, one standard deviation on the standard normal distribution. To compute probabilities, areas, for any normal distribution, we need only to convert the particular normal distribution to the standard normal distribution and look up the answer in the tables. As review, here again is the standardizing formula : Z = x - μ σ where Z is the value on the standard normal distribution, X is the value from a normal distribution one wishes to convert to the standard normal, μ and σ are, respectively, the mean and standard deviation of that population. Note that the equation uses μ and σ which denotes population parameters. This is still dealing with probability so we always are dealing with the population, with known parameter values and a known distribution. It is also important to note that because the normal distribution is symmetrical it does not matter if the z-score is positive or negative when calculating a probability. One standard deviation to the left (negative Z-score) covers the same area as one standard deviation to the right (positive Z-score). This fact is why the Standard Normal tables do not provide areas for the left side of the distribution. Because of this symmetry, the Z-score formula is sometimes written as: Z = | x - μ | σ Where the vertical lines in the equation means the absolute value of the number. What the standardizing formula is really doing is computing the number of standard deviations X is from the mean of its own distribution. The standardizing formula and the concept of counting standard deviations from the mean is the secret of all that we will do in this statistics class. The reason this is true is that all of statistics boils down to variation, and the counting of standard deviations is a measure of variation. This formula, in many disguises, will reappear over and over throughout this course. The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. a. Find the probability that a randomly selected student scored more than 65 on the exam. b. Find the probability that a randomly selected student scored less than 85. a. Let X = a score on the final exam. X ~ N (63, 5), where μ = 63 and σ = 5. Draw a graph. Then, find P ( x > 65). P ( x > 65) = 0.3446 Z 1 = x 1 − μ σ = 65 − 63 5 = 0.4 P ( x ≥ x 1 ) = P ( Z ≥ Z 1 ) = 0.3446 The probability that any student selected at random scores more than 65 is 0.3446. Here is how we found this answer. The normal table provides probabilities from zero to the value Z 1 . For this problem the question can be written as: P(X ≥ 65) = P(Z ≥ Z 1 ), which is the area in the tail. To find this area the formula would be 0.5 – P(X ≤ 65). One half of the probability is above the mean value because this is a symmetrical distribution. The graph shows how to find the area in the tail by subtracting that portion from the mean, zero, to the Z 1 value. The final answer is: P(X ≥ 63) = P(Z ≥ 0.4) = 0.3446 z = 65 – 63 5 = 0.4 Area to the left of Z 1 to the mean of zero is 0.1554 P ( x > 65) = P ( z > 0.4) = 0.5 – 0.1554 = 0.3446 b. Z = x - μ σ = 85 - 63 5 = 4.4 which is larger than the maximum value on the Standard Normal Table. Therefore, the probability that one student scores less than 85 is approximately one or 100%. A score of 85 is 4.4 standard deviations from the mean of 63 which is beyond the range of the standard normal table. Therefore, the probability that one student scores less than 85 is approximately one (or 100%). Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65. normalcdf(0,65,68,3) = 0.1587 A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day. a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N (2, 0.5) where μ = 2 and σ = 0.5. Find P (1.8 < x < 2.75). The probability for which you are looking is the area between x = 1.8 and x = 2.75. P (1.8 < x < 2.75) = 0.5886 P (1.8 ≤ x ≤ 2.75) = P ( Z i ≤ Z ≤ Z 2 ) The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment. b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25 th percentile, k , where P ( x < k ) = 0.25. f ( Z ) = 0.5 - 0.25 = 0.25 , therefore Z ≈ -0.675 ( or just 0.67 using the table ) Z = x - μ σ = x - 2 0.5 = -0.675 , therefore x = -0.675 * 0.5 + 2 = 1.66 hours. The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours. Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70. normalcdf(66,70,68,3) = 0.4950 In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old. a. 0.8186 b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. b. 0.8413 Try It Use the information in to answer the following questions. Find the 30 th percentile, and interpret it in a complete sentence. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old. A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on their farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph. Z 1 = 6 − 5.85 .24 = .625 P ( x ≥ 6) = P ( z ≥ 0.625) = 0.2670 b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______. f ( Z ) = 0.20 2 = 0.10 , therefore Z ≈ ± 0.25 Z = x - μ σ = x - 5.85 0.24 = ± 0.25 → ± 0.25 · 0.24 + 5.85 = ( 5.79 , 5.91 ) Try It Using the information from , answer the following: The middle 40% of mandarin oranges from this farm are between ______ and ______. Find the 16 th percentile and interpret it in a complete sentence. References “Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013). “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013). “Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013). “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013). “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013).", "section": "Using the Normal Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Estimating the Binomial with the Normal Distribution We found earlier that various probability density functions are the limiting distributions of others; thus, we can estimate one with another under certain circumstances. We will find here that the normal distribution can be used to estimate a binomial process. The Poisson was used to estimate the binomial previously, and the binomial was used to estimate the hypergeometric distribution. In the case of the relationship between the hypergeometric distribution and the binomial, we had to recognize that a binomial process assumes that the probability of a success remains constant from trial to trial: a head on the last flip cannot have an effect on the probability of a head on the next flip. In the hypergeometric distribution this is the essence of the question because the experiment assumes that any \"draw\" is without replacement. If one draws without replacement, then all subsequent \"draws\" are conditional probabilities. We found that if the hypergeometric experiment draws only a small percentage of the total objects, then we can ignore the impact on the probability from draw to draw. Imagine that there are 312 cards in a deck comprised of 6 normal decks. If the experiment called for drawing only 10 cards, less than 5% of the total, than we will accept the binomial estimate of the probability, even though this is actually a hypergeometric distribution because the cards are presumably drawn without replacement. The Poisson likewise was considered an appropriate estimate of the binomial under certain circumstances. In Random Discrete Variables we found that if the number of trials of interest is large and the probability of success is small, such that μ = n p < 7 , the Poisson can be used to estimate the binomial with good results. Again, these rules of thumb do not in any way claim that the actual probability is what the estimate determines, only that the difference is in the third or fourth decimal and is thus de minimus . Here, again, we find that the normal distribution makes particularly accurate estimates of a binomial process under certain circumstances. is a frequency distribution of a binomial process for the experiment of flipping three coins where the random variable is the number of heads. The sample space is listed below the distribution. The experiment assumed that the probability of a success is 0.5; the probability of a failure, a tail, is thus also 0.5. In observing we are struck by the fact that the distribution is symmetrical. The root of this result is that the probabilities of success and failure are the same, 0.5. If the probability of success were smaller than 0.5, the distribution becomes skewed right. Indeed, as the probability of success diminishes, the degree of skewness increases. If the probability of success increases from 0.5, then the skewness increases in the lower tail, resulting in a left-skewed distribution. The reason the skewness of the binomial distribution is important is because if it is to be estimated with a normal distribution, then we need to recognize that the normal distribution is symmetrical. The closer the underlying binomial distribution is to being symmetrical, the better the estimate that is produced by the normal distribution. shows a symmetrical normal distribution transposed on a graph of a binomial distribution where p = 0.2 and n = 5. The discrepancy between the estimated probability using a normal distribution and the probability of the original binomial distribution is apparent. The criteria for using a normal distribution to estimate a binomial thus addresses this problem by requiring BOTH np AND n (1 − p ) are greater than five. Again, this is a rule of thumb, but is effective and results in acceptable estimates of the binomial probability. Imagine that it is known that only 10% of Australian Shepherd puppies are born with what is called \"perfect symmetry\" in their three colors, black, white, and copper. Perfect symmetry is defined as equal coverage on all parts of the dog when looked at in the face and measuring left and right down the centerline. A kennel would have a good reputation for breeding Australian Shepherds if they had a high percentage of dogs that met this criterion. During the past 5 years and out of the 100 dogs born to Dundee Kennels, 16 were born with this coloring characteristic. What is the probability that, in 100 births, more than 16 would have this characteristic? If we assume that one dog's coloring is independent of other dogs' coloring, a bit of a brave assumption, this becomes a classic binomial probability problem. The statement of the probability requested is 1 − [ p ( X = 0) + p ( X = 1) + p ( X = 2)+ … + p ( X = 16)]. This requires us to calculate 17 binomial formulas and add them together and then subtract from one to get the right hand part of the distribution. Alternatively, we can use the normal distribution to get an acceptable answer and in much less time. First, we need to check if the binomial distribution is symmetrical enough to use the normal distribution. We know that the binomial for this problem is skewed because the probability of success, 0.1, is not the same as the probability of failure, 0.9. Nevertheless, both np = 10 and n ( 1 − p ) = 90 are larger than 5, the cutoff for using the normal distribution to estimate the binomial. below shows the binomial distribution and marks the area we wish to know. The mean of the binomial, 10, is also marked, and the standard deviation is written on the side of the graph: σ = n p q = 3. The area under the distribution from zero to 16 is the probability requested, and has been shaded in. Below the binomial distribution is a normal distribution to be used to estimate this probability. That probability has also been shaded. Standardizing from the binomial to the normal distribution as done in the past shows where we are asking for the probability from 16 to positive infinity, or 100 in this case. We need to calculate the number of standard deviations 16 is away from the mean: 10. Z = x − μ σ = 16 − 10 3 = 2 We are asking for the probability beyond two standard deviations, a very unlikely event. We look up two standard deviations in the standard normal table and find the area from zero to two standard deviations is 0.4772. We are interested in the tail, however, so we subtract 0.4772 from 0.5 and thus find the area in the tail. Our conclusion is the probability of a kennel having 16 dogs with \"perfect symmetry\" is 0.0228. Dundee Kennels has an extraordinary record in this regard. Mathematically, we write this as: 1 − [ p ( X = 0 ) + p ( X = 1 ) + p ( X = 2 ) + … + p ( X = 16 ) ] = p ( X > 16 ) = p ( Z > 2 ) = 0.0228 Try It Let’s assume that only 10% of the wooden furniture made by a carpenter is made to the exact dimensions in the blueprint. During the last 3 years, the carpenter has made 100 pieces of wooden furniture, and 16 were found to be made to the exact dimensions in the blueprint. What is the probability that, in 100 wooden furniture items, more than 16 would be made to the exact dimensions of the blueprint? The required probability is 1 - [ p ( X = 0 ) + p ( X = 1 ) + p ( X = 2 ) + . . . + p ( X = 16 ) ] . The mean of the binomial is 10, since np = 10. The standard deviation is obtained as σ = n p q = 3 . The number of standard deviations 16 is away from 10 is obtained as follows: Z = x - μ σ = 16 - 10 3 = 2 From the standard normal table, the probability for Z = 2 is obtained as 0.4772. Since the required probability is only above 16, the required probability is obtained as 0.5 – 0.4772 = 0.0228. Chapter Review The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one. Formula Review Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation. Standard Normal Distribution: Z ~ N (0, 1). How would you represent the area to the left of one in a probability statement? P ( x < 1) What is the area to the right of one? Is P ( x < 1) equal to P ( x ≤ 1)? Why? Yes, because they are the same in a continuous distribution: P ( x = 1) = 0 How would you represent the area to the left of three in a probability statement? What is the area to the right of three? 1 – P ( x < 3) or P ( x > 3) If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x ? If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x ? 1 – 0.543 = 0.457 Use the following information to answer the next four exercises: X ~ N (54, 8) Find the probability that x > 56. Find the probability that x < 30. 0.0013 X ~ N (6, 2) Find the probability that x is between three and nine. X ~ N (–3, 4) Find the probability that x is between one and four. 0.1186 X ~ N (4, 5) Find the maximum of x in the bottom quartile. Use the following information to answer the next three exercise: The life of wearable fitness devices is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Find the probability that a wearable fitness device will break down during the guarantee period. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (0 < x < ____________) = ___________ (Use zero for the minimum value of x .) Answers may vary. 3, 0.1979 Find the probability that a wearable fitness device will last between 2.8 and six years. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (__________ < x < __________) = __________ An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 45 successes. 0.154 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 22 successes. 0.874 An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 35 to 45 successes. 0.693 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 26 to 30 successes. 0.346 An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes. 0.110 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes. 0.946 A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 30, but no more than 32, questions will be guessed correctly. 0.071 A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 24, but no more than 28, questions will be guessed correctly. 0.347 Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the probability of spending more than two days in recovery? 0.0580 0.8447 0.0553 0.9420 Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? Yes No Unable to determine Find the probability that it takes at least eight minutes to find a parking space. 0.0001 0.9270 0.1862 0.0668 d Seventy percent of the time, it takes more than how many minutes to find a parking space? 1.24 2.41 3.95 6.05 According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual. X ~ _____(_____,_____) Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically. The middle 40% of heights fall between what two values? Sketch the graph, and write the probability statement. X ~ N (66, 2.5) 0.5404 No, the probability that an Asian male is over 72 inches tall is 0.0082 IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. X ~ _____(_____,_____) Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories. X ~ _____(_____,_____) Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement. X ~ N (36, 10) The probability that a person consumes more than 40% of their calories as fat is 0.3446. Approximately 25% of people consume less than 29.26% of their calories as fat. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X . Shade the region corresponding to the probability. Find the probability. In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. In words, define the random variable X . X ~ _____(_____,_____) Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. What percent of the children spend over ten hours per day unsupervised? Seventy percent of the children spend at least how long per day unsupervised? X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day. X ~ N (3, 1.5) The probability that the child spends less than one hour a day unsupervised is 0.0918. The probability that a child spends over ten hours a day unsupervised is less than 0.0001. 2.21 hours In a certain presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for Candidate A. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for Candidate A was bell-shaped. Let X = number of votes for Candidate A for an election district. State the approximate distribution of X . Is 1,956.8 a population mean or a sample mean? How do you know? Find the probability that a randomly selected district had fewer than 1,600 votes for Candidate A. Sketch the graph and write the probability statement. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for Candidate A. Find the third quartile for votes for Candidate A. Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days. In words, define the random variable X . X ~ _____(_____,_____) If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement. Sixty percent of all trials of this type are completed within how many days? X = the distribution of the number of days a particular type of criminal trial will take X ~ N (21, 7) The probability that a randomly selected trial will last more than 24 days is 0.3336. 22.77 A motorcycle racer averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of their race times is normally distributed. We are interested in one of their randomly selected laps. In words, define the random variable X . X ~ _____(_____,_____) Find the percent of the racer's laps that are completed in less than 130 seconds. The fastest 3% of the racer's laps are under _____. The middle 80% of the racer's laps are from _______ seconds to _______ seconds. Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. displays the ordered real data (in minutes): 0.50 4.25 5 6 7.25 1.75 4.25 5.25 6 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 6.5 8 2.5 4.75 5.5 6.5 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 6 6.75 9.75 3.75 5 6 6.75 10.75 Calculate the sample mean and the sample standard deviation. Construct a histogram. Draw a smooth curve through the midpoints of the tops of the bars. In words, describe the shape of your histogram and smooth curve. Let the sample mean approximate μ and the sample standard deviation approximate σ . The distribution of X can then be approximated by X ~ _____(_____,_____) Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes. Determine the cumulative relative frequency for waiting less than 6.1 minutes. Why aren’t the answers to part f and part g exactly the same? Why are the answers to part f and part g as close as they are? If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion. mean = 5.51, s = 2.15 Answers may vary. Answers may vary. Answers may vary. X ~ N (5.51, 2.15) 0.6029 The cumulative frequency for less than 6.1 minutes is 0.64. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well. Suppose that Ric and Anita attend different colleges. Ric’s GPA is the same as the average GPA at their school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false. Ric’s actual GPA is lower than Anita’s actual GPA. Ric is not passing because their z -score is zero. Anita is in the 70 th percentile of students at her college. An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. The person who is being sued for lack of child support was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z -scores first, and then use those to calculate the probability. An automotive factory can build an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample. n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = n p q = (100)(0 .1)(0 .9) = 3 z = ± 1 : x 1 = µ + zσ = 10 + 1(3) = 13 and x 2 = µ – zσ = 10 – 1(3) = 7.68% of the defective cars will fall between seven and 13. z = ± 2 : x 1 = µ + zσ = 10 + 2(3) = 16 and x 2 = µ – zσ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16 z = ± 3 : x 1 = µ + zσ = 10 + 3(3) = 19 and x 2 = µ – zσ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19. We flip a coin 100 times ( n = 100) and note that it only comes up heads 20% ( p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following: There is about a 68% chance that the number of heads will be somewhere between ___ and ___. There is about a ____chance that the number of heads will be somewhere between 12 and 28. There is about a ____ chance that the number of heads will be somewhere between eight and 32. A $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are somewhere between 34 and 54 prizes. somewhere between 54 and 64 prizes. more than 64 prizes. n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = n p q = (190)(0 .2)(0 .8) = 5.5136 For this problem: P (34 < x < 54) = 0.7641 For this problem: P (54 < x < 64) = 0.0018 For this problem: P ( x > 64) = 0.0000012 (approximately 0) On average, 28% of 18 to 34 year olds check social media before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of 5%. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is at least 30%. Find the 95th percentile, and express it in a sentence. A hospital has 49 births in a year. It is considered equally likely that a birth be a boy as it is the birth be a girl. What is the mean? What is the standard deviation? Can this binomial distribution be approximated with a normal distribution? If so, use the normal distribution to find the probability that at least 23 of the 49 births were boys. 24.5 3.5 Yes 0.67 Historically, a final exam in a course is passed with a probability of 0.9. The exam is given to a group of 70 students. What is the mean of the binomial distribution? What is the standard deviation? Can this binomial distribution be approximate with a normal distribution? If so, use the normal distribution to find the probability that at least 60 of the students pass the exam? 63 2.5 Yes 0.88 A tree in an orchard has 200 oranges. Of the oranges, 40 are not ripe. Use the normal distribution to approximate the binomial distribution, and determine the probability a box containing 35 oranges has at most two oranges that are not ripe. 0.02 In a large city one in ten fire hydrants are in need of repair. If a crew examines 100 fire hydrants in a week, what is the probability they will find nine of fewer fire hydrants that need repair? Use the normal distribution to approximate the binomial distribution. 0.37 On an assembly line it is determined 85% of the assembled products have no defects. If one day 50 items are assembled, what is the probability at least 4 and no more than 8 are defective. Use the normal distribution to approximate the binomial distribution. 0.50", "section": "Estimating the Binomial with the Normal Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction If you want to figure out the distribution of the change people carry in their pockets, using the Central Limit Theorem and assuming your sample is large enough, you will find that the distribution is the normal probability density function. (credit: modification of work “american coins” by Paula R. Lively/ Flickr, CC BY 2.0) Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem . The Central Limit Theorem is one of the most powerful and useful ideas in all of statistics. The Central Limit Theorem is a theorem which means that it is NOT a theory or just somebody's idea of the way things work. As a theorem it ranks with the Pythagorean Theorem, or the theorem that tells us that the sum of the angles of a triangle must add to 180. These are facts of the ways of the world rigorously demonstrated with mathematical precision and logic. As we will see this powerful theorem will determine just what we can, and cannot say, in inferential statistics. The Central Limit Theorem is concerned with drawing finite samples of size n from a population with a known mean, μ , and a known standard deviation, σ . The conclusion is that if we collect samples of size n with a \"large enough n ,\" calculate each sample's mean, and create a histogram (distribution) of those means, then the resulting distribution will tend to have an approximate normal distribution. The astounding result is that it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means tend to follow the normal distribution. The size of the sample, n , that is required in order to be \"large enough\" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means. Sampling is done randomly and with replacement in the theoretical model. Sampling Distribution Given simple random samples of size n from a given population with a measured characteristic such as mean, proportion, or standard deviation for each sample, the probability distribution of all the measured characteristics is called a sampling distribution.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Central Limit Theorem for Sample Means The sampling distribution is a theoretical distribution. It is created by taking many samples of size n from a population. Each sample mean is then treated like a single observation of this new distribution, the sampling distribution. The genius of thinking this way is that it recognizes that when we sample we are creating an observation and that observation must come from some particular distribution. The Central Limit Theorem answers the question: from what distribution did a sample mean come? If this is discovered, then we can treat a sample mean just like any other observation and calculate probabilities about what values it might take on. We have effectively moved from the world of statistics where we know only what we have from the sample, to the world of probability where we know the distribution from which the sample mean came and the parameters of that distribution. The reasons that one samples a population are obvious. The time and expense of checking every invoice to determine its validity or every shipment to see if it contains all the items may well exceed the cost of errors in billing or shipping. For some products, sampling would require destroying them, called destructive sampling. One such example is measuring the ability of a metal to withstand saltwater corrosion for parts on ocean going vessels. Sampling thus raises an important question; just which sample was drawn. Even if the sample were randomly drawn, there are theoretically an almost infinite number of samples. With just 100 items, there are more than 75 million unique samples of size five that can be drawn. If six are in the sample, the number of possible samples increases to just more than one billion. Of the 75 million possible samples, then, which one did you get? If there is variation in the items to be sampled, there will be variation in the samples. One could draw an \"unlucky\" sample and make very wrong conclusions concerning the population. This recognition that any sample we draw is really only one from a distribution of samples provides us with what is probably the single most important theorem is statistics: the Central Limit Theorem . Without the Central Limit Theorem it would be impossible to proceed to inferential statistics from simple probability theory. In its most basic form, the Central Limit Theorem states that regardless of the underlying probability density function of the population data, the theoretical distribution of the means of samples from the population will be normally distributed. In essence, this says that the mean of a sample should be treated like an observation drawn from a normal distribution. The Central Limit Theorem only holds if the sample size is \"large enough\" which has been shown to be only 30 observations or more. graphically displays this very important proposition. Notice that the horizontal axis in the top panel is labeled X. These are the individual observations of the population. This is the unknown distribution of the population values. The graph is purposefully drawn all squiggly to show that it does not matter just how odd ball it really is. Remember, we will never know what this distribution looks like, or its mean or standard deviation for that matter. The horizontal axis in the bottom panel is labeled X – 's. This is the theoretical distribution called the sampling distribution of the means. Each observation on this distribution is a sample mean. All these sample means were calculated from individual samples with the same sample size. The theoretical sampling distribution contains all of the sample mean values from all the possible samples that could have been taken from the population. Of course, no one would ever actually take all of these samples, but if they did this is how they would look. And the Central Limit Theorem says that they will be normally distributed. The Central Limit Theorem goes even further and tells us the mean and standard deviation of this theoretical distribution. Parameter Population distribution Sample Sampling distribution of X – 's Mean μ X ¯ μ x ¯ and E ( μ x ¯ ) = μ Standard deviation σ s σ x ¯ = σ n The practical significance of The Central Limit Theorem is that now we can compute probabilities for drawing a sample mean, X – , in just the same way as we did for drawing specific observations, X's, when we knew the population mean and standard deviation and that the population data were normally distributed.. The standardizing formula has to be amended to recognize that the mean and standard deviation of the sampling distribution, sometimes, called the standard error of the mean, are different from those of the population distribution, but otherwise nothing has changed. The new standardizing formula is Z = X – − μ X – σ X – = X – - μ σ n Notice that µ X – in the first formula has been changed to simply µ in the second version. The reason is that mathematically it can be shown that the expected value of µ X – is equal to µ. This was stated in above. Mathematically, the E(x) symbol read the “expected value of x”. This formula will be used in the next unit to provide estimates of the unknown population parameter μ. References Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013). Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013). Data from the United States Department of Agriculture. Chapter Review In a population whose distribution may be known or unknown, if the size ( n ) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ( n ). Formula Review The Central Limit Theorem for Sample Means: X – ~ N ( μ x – , σ n ) Z = X – − μ X – σ X – = X – - μ σ / n The Mean X – : μ x ¯ Central Limit Theorem for Sample Means z-score z = x – − μ x ¯ ( σ n ) Standard Error of the Mean (Standard Deviation ( X – )): σ n Finite Population Correction Factor for the sampling distribution of means: Z = x ¯ − μ σ n · N − n N − 1 Finite Population Correction Factor for the sampling distribution of proportions: σ p' = p ( 1 − p ) n × N − n N − 1 Homework Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. In words, Χ = ____________ Χ ~ _____(_____,_____) In words, X – = ____________ X – ~ ______ (______, ______) Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Explain why there is a difference in part e and part f. Χ = amount of change a student carries E x p ( 1 0 . 88 ) o r Χ ~ E x p ( 1 . 1364 ) X – = average amount of change carried by a sample of 25 students. X – ~ N (0.88, 0.176) 0.0819 0.4276 The distributions are different. Part e asks about the probability for an individual and part f asks about the probability of the mean value for a sample of 25. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. If X – = average distance in feet for 49 fly balls, then X – ~ _______(_______,_______) What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X – . Shade the region corresponding to the probability. Find the probability. Find the 80 th percentile of the distribution of the average of 49 fly balls. According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. In words, Χ = _____________ In words, X – = _____________ X – ~ _____(_____,_____) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. Would you be surprised if one taxpayer finished their Form 1040 in more than 12 hours? In a complete sentence, explain why. length of time for an individual to complete IRS form 1040, in hours. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours. N ( 10 .53, 1 3 ) Yes. I would be surprised, because the probability is almost 0. No. I would not be totally surprised because the probability is 0.2312 Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let X – the average of the 49 races. X – ~ _____(_____,_____) Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. Find the 80 th percentile for the average of these 49 marathons. Find the median of the average running times. The length of songs in a collector’s Apple Music album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. In words, Χ = _________ Χ ~ _____________ In words, X – = _____________ X – ~ _____(_____,_____) Find the first quartile for the average song length. The IQR(interquartile range) for the average song length is from _______–_______. the length of a song, in minutes, in the collection U (2, 3.5) the average length, in minutes, of the songs from a sample of five albums from the collection N (2.75, 0.066) 2.74 minutes 0.03 minutes In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. In words, Χ = _____________ In words, X – = _____________ X – ~ _____(_____,_____) The IQR for X – is from _______ acres to _______ acres. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. When the sample size is large, the mean of X – is approximately equal to the mean of Χ . When the sample size is large, X – is approximately normally distributed. When the sample size is large, the standard deviation of X – is approximately the same as the standard deviation of Χ . True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. False. The standard deviation of the sample distribution of the means will decrease as the sample size increases; however, the standard deviation of the sample distribution of the means will not equal the standard deviation of X. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X – = average percent of fat calories. X – ~ ______(______, ______) For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. Find the first quartile for the average percent of fat calories. The distribution of income in some developing countries is considered wedge shaped (many low income people, very few middle income people, and even fewer high income people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country. In words, Χ = _____________ In words, X – = _____________ X – ~ _____(_____,_____) How is it possible for the standard deviation to be greater than the average? Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200? X = the yearly income of someone in a developing country the average salary from samples of 1,000 residents of a developing country X – ∼ N ( 2000, 8000 1000 ) Very wide differences in data values can have averages smaller than standard deviations. The distribution of the sample mean will have higher probabilities closer to the population mean. P (2000 < X – < 2100) = 0.1537 P (2100 < X – < 2200) = 0.1317 Which of the following is NOT TRUE about the distribution for averages? The mean, median, and mode are equal. The area under the curve is one. The curve never touches the x -axis. The curve is skewed to the right. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is: X – ~ N (4.59, 0.10) X – ~ N ( 4 .59, 0.10 16 ) X – ~ N ( 4 .59, 16 0.10 ) X – ~ N ( 4 .59, 16 0.10 ) b Average a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean. Central Limit Theorem Given a random variable with known mean μ and known standard deviation, σ , we are sampling with size n , and we are interested in two new RVs: the sample mean, X – . If the size ( n ) of the sample is sufficiently large, then X – ~ N ( μ , σ n ). If the size ( n ) of the sample is sufficiently large, then the distribution of the sample means will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean. Standard Error of the Mean the standard deviation of the distribution of the sample means, or σ n .", "section": "The Central Limit Theorem for Sample Means", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Using the Central Limit Theorem The Law of Large Numbers , along with the Central Limit Theorem, provides another critical piece of information to allow us to engage in inferential statistics. In short, the Law of Large Numbers proves that the expected value of the sampling distribution of the sample mean is the population mean: E ( μ x ¯ ) = μ The proof is through the concept of large numbers. Suppose you were to take a sample and calculate a sample mean. Then you take another sample, combine it with the previous sample, and calculate the sample mean of the combined sample. Then you repeat this process over and over, creating bigger and bigger samples and calculating a sample mean each time along the way. The sample means from larger and larger samples will get closer and closer to the population mean, μ. shows the running average as more sample means are added and then averaged. The proof of the Law of Large Numbers mathematically was perfected during a period of 20 years and was presented by Jacob Bernoulli in 1713. Stated mathematically lim n → ∞ P ( | X ¯ n - μ | ) > ε = 0 or alternatively presented as P lim n → ∞ | X ¯ n = μ | < ε = 1 where X ¯ n is the running average as additional sample means are added to the previous sample means. In summary: There are three critical mathematical conclusions that flow from the Central Limit Theorem and the application of the Law of Large Numbers. By the Central Limit Theorem, for large enough sample sizes, the sampling distribution of sample means tends to be normally distributed regardless of the underlying distribution of the population data. As the sample size, n, gets larger and larger, the sampling distribution standard deviation gets smaller. Remember that the standard deviation for the sampling distribution of X ¯ n is σ n . The sample mean, x ¯ , is more likely to be closer to μ as n increases. By the Law of Large Numbers, the expected value of the sampling distribution of the sample mean is the population mean. Law of Large Numbers The law of large numbers says that if you take samples of larger and larger size from any population, then the mean of the sampling distribution, μ x – tends to get closer and closer to the true population mean, μ . From the Central Limit Theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation of the sampling distribution gets. (Remember that the standard deviation for the sampling distribution of X – is σ n .) This means that the sample mean x – must be closer to the population mean μ as n increases. We can say that μ is the value that the sample means approach as n gets larger. The Central Limit Theorem illustrates the law of large numbers. Examples of the Central Limit Theorem This concept is so important and plays such a critical role in what follows it deserves to be developed further. Indeed, there are two critical issues that flow from the Central Limit Theorem and the application of the Law of Large numbers to it. These are The probability density function of the sampling distribution of means is normally distributed regardless of the underlying distribution of the population observations and standard deviation of the sampling distribution decreases as the size of the samples that were used to calculate the means for the sampling distribution increases. Taking these in order. It would seem counterintuitive that the population may have any distribution and the distribution of means coming from it would be normally distributed. With the use of computers, experiments can be simulated that show the process by which the sampling distribution changes as the sample size is increased. These simulations show visually the results of the mathematical proof of the Central Limit Theorem. Here are three examples of very different population distributions and the evolution of the sampling distribution to a normal distribution as the sample size increases. The top panel in these cases represents the histogram for the original data. The three panels show the histograms for 1,000 randomly drawn samples for different sample sizes: n=10, n= 25 and n=50. As the sample size increases, and the number of samples taken remains constant, the distribution of the 1,000 sample means becomes closer to the smooth line that represents the normal distribution. is for a normal distribution of individual observations and we would expect the sampling distribution to converge on the normal quickly. The results show this and show that even at a very small sample size the distribution is close to the normal distribution. is a uniform distribution which, a bit amazingly, quickly approached the normal distribution even with only a sample of 10. is a skewed distribution. This last one could be an exponential, geometric, or binomial with a small probability of success creating the skew in the distribution. For skewed distributions our intuition would say that this will take larger sample sizes to move to a normal distribution and indeed that is what we observe from the simulation. Nevertheless, at a sample size of 50, not considered a very large sample, the distribution of sample means has very decidedly gained the shape of the normal distribution. The Central Limit Theorem provides more than the proof that the sampling distribution of means is normally distributed. It also provides us with the mean and standard deviation of this distribution. Further, as discussed above, the expected value of the mean, μ x – , is equal to the mean of the population of the original data which is what we are interested in estimating from the sample we took. We have already inserted this conclusion of the Central Limit Theorem into the formula we use for standardizing from the sampling distribution to the standard normal distribution. And finally, the Central Limit Theorem has also provided the standard deviation of the sampling distribution, σ x – = σ n , and this is critical to have to calculate probabilities of values of the new random variable, x ¯ . shows a sampling distribution. The mean has been marked on the horizontal axis of the x ¯ 's and the standard deviation has been written to the right above the distribution. Notice that the standard deviation of the sampling distribution is the original standard deviation of the population, divided by the sample size. We have already seen that as the sample size increases the sampling distribution becomes closer and closer to the normal distribution. As this happens, the standard deviation of the sampling distribution changes in another way; the standard deviation decreases as n increases. At very large n, the standard deviation of the sampling distribution becomes very small and at infinity it collapses on top of the population mean. This is what it means that the expected value of µ x – is the population mean, µ. At non-extreme values of n,this relationship between the standard deviation of the sampling distribution and the sample size plays a very important part in our ability to estimate the parameters we are interested in. shows three sampling distributions. The only change that was made is the sample size that was used to get the sample means for each distribution. As the sample size increases, n goes from 10 to 30 to 50, the standard deviations of the respective sampling distributions decrease because the sample size is in the denominator of the standard deviations of the sampling distributions. The implications for this are very important. shows the effect of the sample size on the confidence we will have in our estimates. These are two sampling distributions from the same population. One sampling distribution was created with samples of size 10 and the other with samples of size 50. All other things constant, the sampling distribution with sample size 50 has a smaller standard deviation that causes the graph to be higher and narrower. The important effect of this is that for the same probability of one standard deviation from the mean, this distribution covers much less of a range of possible values than the other distribution. One standard deviation is marked on the X ¯ axis for each distribution. This is shown by the two arrows that are plus or minus one standard deviation for each distribution. If the probability that the true mean is one standard deviation away from the mean, then for the sampling distribution with the smaller sample size, the possible range of values is much greater. A simple question is, would you rather have a sample mean from the narrow, tight distribution, or the flat, wide distribution as the estimate of the population mean? Your answer tells us why people intuitively will always choose data from a large sample rather than a small sample. The sample mean they are getting is coming from a more compact distribution. This concept will be the foundation for what will be called level of confidence in the next unit. Chapter Review The Central Limit Theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean x – gets to μ . Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation? What is the distribution for the mean weight of 100 25-pound lifting weights? Find the probability that the mean actual weight for the 100 weights is less than 24.9. U (24, 26), 25, 0.5774 N (25, 0.0577) 0.0416 Draw the graph from Find the probability that the mean actual weight for the 100 weights is greater than 25.2. 0.0003 Draw the graph from Find the 90 th percentile for the mean weight for the 100 weights. 25.07 Draw the graph from What is the distribution for the sum of the weights of 100 25-pound lifting weights? Find P ( Σx < 2,450). N (2,500, 5.7735) 0 Draw the graph from Find the 90 th percentile for the total weight of the 100 weights. 2,507.40 Draw the graph from Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken. What is the standard deviation? What is the parameter m ? 10 1 10 What is the distribution for the length of time one battery lasts? What is the distribution for the mean length of time 64 batteries last? N ( 10, 10 8 ) What is the distribution for the total length of time 64 batteries last? Find the probability that the sample mean is between seven and 11. 0.7799 Find the 80 th percentile for the total length of time 64 batteries last. Find the IQR for the mean amount of time 64 batteries last. 1.69 Find the middle 80% for the total amount of time 64 batteries last. Use the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken. Find P ( Σx > 420). 0.0072 Find the 90 th percentile for the sums. Find the 15 th percentile for the sums. 391.54 Find the first quartile for the sums. Find the third quartile for the sums. 405.51 Find the 80 th percentile for the sums. A population has a mean of 25 and a standard deviation of 2. If it is sampled repeatedly with samples of size 49, what is the mean and standard deviation of the sample means? Mean = 25, standard deviation = 2/7 A population has a mean of 48 and a standard deviation of 5. If it is sampled repeatedly with samples of size 36, what is the mean and standard deviation of the sample means? Mean = 48, standard deviation = 5/6 A population has a mean of 90 and a standard deviation of 6. If it is sampled repeatedly with samples of size 64, what is the mean and standard deviation of the sample means? Mean = 90, standard deviation = 3/4 A population has a mean of 120 and a standard deviation of 2.4. If it is sampled repeatedly with samples of size 40, what is the mean and standard deviation of the sample means? Mean = 120, standard deviation = 0.38 A population has a mean of 17 and a standard deviation of 1.2. If it is sampled repeatedly with samples of size 50, what is the mean and standard deviation of the sample means? Mean = 17, standard deviation = 0.17 A population has a mean of 17 and a standard deviation of 0.2. If it is sampled repeatedly with samples of size 16, what is the expected value and standard deviation of the sample means? Expected value = 17, standard deviation = 0.05 A population has a mean of 38 and a standard deviation of 3. If it is sampled repeatedly with samples of size 48, what is the expected value and standard deviation of the sample means? Expected value = 38, standard deviation = 0.43 A population has a mean of 14 and a standard deviation of 5. If it is sampled repeatedly with samples of size 60, what is the expected value and standard deviation of the sample means? Expected value = 14, standard deviation = 0.65 Homework A large population of 5,000 students take a practice test to prepare for a standardized test. The population mean is 140 questions correct, and the standard deviation is 80. What size samples should a researcher take to get a distribution of means of the samples with a standard deviation of 10? 64 A large population has skewed data with a mean of 70 and a standard deviation of 6. Samples of size 100 are taken, and the distribution of the means of these samples is analyzed. Will the distribution of the means be closer to a normal distribution than the distribution of the population? Will the mean of the means of the samples remain close to 70? Will the distribution of the means have a smaller standard deviation? What is that standard deviation? Yes Yes Yes 0.6 A researcher is looking at data from a large population with a standard deviation that is much too large. In order to concentrate the information, the researcher decides to repeatedly sample the data and use the distribution of the means of the samples? The first effort used sample sized of 100. But the standard deviation was about double the value the researcher wanted. What is the smallest size samples the researcher can use to remedy the problem? 400 A researcher looks at a large set of data, and concludes the population has a standard deviation of 40. Using sample sizes of 64, the researcher is able to focus the mean of the means of the sample to a narrower distribution where the standard deviation is 5. Then, the researcher realizes there was an error in the original calculations, and the initial standard deviation is really 20. Since the standard deviation of the means of the samples was obtained using the original standard deviation, this value is also impacted by the discovery of the error. What is the correct value of the standard deviation of the means of the samples? 2.5 A population has a standard deviation of 50. It is sampled with samples of size 100. What is the variance of the means of the samples? 25 Mean a number that measures the central tendency; a common name for mean is \"average.\" The term \"mean\" is a shortened form of \"arithmetic mean.\" By definition, the mean for a sample (denoted by x – ) is x ¯ = Sum of all values in the sample Number of values in the sample , and the mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population . Finite Population Correction Factor adjusts the variance of the sampling distribution if the population is known and more than 5% of the population is being sampled. Normal Distribution a continuous random variable with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation.; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the random variable, Z, is called the standard normal distribution . Standard Error of the Proportion the standard deviation of the sampling distribution of proportions", "section": "Using the Central Limit Theorem", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Central Limit Theorem for Proportions The Central Limit Theorem tells us that the point estimate for the sample mean, x ¯ , comes from a normal distribution of x ¯ 's. This theoretical distribution is called the sampling distribution of x ¯ 's. We now investigate the sampling distribution for another important parameter we wish to estimate; p from the binomial probability density function. If the random variable is discrete, such as for categorical data, then the parameter we wish to estimate is the population proportion. This is, of course, the probability of drawing a success in any one random draw. Unlike the case just discussed for a continuous random variable where we did not know the population distribution of X's, here we actually know the underlying probability density function for these data; it is the binomial distribution. The random variable is X = the number of successes and the parameter we wish to know is p, the probability of drawing a success which is of course the proportion of successes in the population. The question at issue is: from what distribution was the sample proportion, p' = x n drawn? The sample size is n and X is the number of successes found in that sample. This is a parallel question that was just answered by the Central Limit Theorem: from what distribution was the sample mean, x ¯ , drawn? We saw that once we knew that the distribution was the Normal distribution then we were able to create confidence intervals for the population parameter, µ. We will also use this same information to test hypotheses about the population mean later. We wish now to be able to develop confidence intervals for the population parameter \"p\" from the binomial probability density function. In order to find the distribution from which sample proportions come we need to develop the sampling distribution of sample proportions just as we did for sample means. So again imagine that we randomly sample say 50 people and ask them if they support the new school bond issue. From this we find a sample proportion, p', and graph it on the axis of p's. We do this again and again etc., etc. until we have the theoretical distribution of p's. Some sample proportions will show high favorability toward the bond issue and others will show low favorability because random sampling will reflect the variation of views within the population. What we have done can be seen in . The top panel is the population distributions of probabilities for each possible value of the random variable X. While we do not know what the specific distribution looks like because we do not know p, the population parameter, we do know that it must look something like this. In reality, we do not know either the mean or the standard deviation of this population distribution, the same difficulty we faced when analyzing the X's previously. places the mean on the distribution of population probabilities as µ = n p but of course we do not actually know the population mean because we do not know the population probability of success, p . Below the distribution of the population values is the sampling distribution of p 's. Again the Central Limit Theorem tells us that this distribution is normally distributed just like the case of the sampling distribution for x ¯ 's. This sampling distribution also has a mean, the mean of the p 's, and a standard deviation, σ p ' . Importantly, in the case of the analysis of the distribution of sample means, the Central Limit Theorem told us the expected value of the mean of the sample means in the sampling distribution, and the standard deviation of the sampling distribution. Again the Central Limit Theorem provides this information for the sampling distribution for proportions. The answers are: The expected value of the mean of sampling distribution of sample proportions, µ p' , is the population proportion, p. The standard deviation of the sampling distribution of sample proportions, σ p' , is the population standard deviation divided by the square root of the sample size, n. Both these conclusions are the same as we found for the sampling distribution for sample means. However in this case, because the mean and standard deviation of the binomial distribution both rely upon p , the formula for the standard deviation of the sampling distribution requires algebraic manipulation to be useful. We will take that up in the next chapter. The proof of these important conclusions from the Central Limit Theorem is provided below. E ( p ' ) = E ( x n ) = ( 1 n ) E ( x ) = ( 1 n ) n p = p (The expected value of X, E(x), is simply the mean of the binomial distribution which we know to be np.) σ p' 2 = Var ( p ' ) = Var ( x n ) = 1 n 2 ( Var ( x ) ) = 1 n 2 ( n p ( 1 − p ) ) = p ( 1 − p ) n The standard deviation of the sampling distribution for proportions is thus: σ p' = p ( 1 − P ) n Parameter Population distribution Sample Sampling distribution of p's Mean µ = np p ' = x n p' and E(p') = p Standard Deviation σ = n p q σ p' = p ( 1 − p ) n summarizes these results and shows the relationship between the population, sample and sampling distribution. Notice the parallel between this Table and Table 7.1 for the case where the random variable is continuous and we were developing the sampling distribution for means. Reviewing the formula for the standard deviation of the sampling distribution for proportions we see that as n increases the standard deviation decreases. This is the same observation we made for the standard deviation for the sampling distribution for means. Again, as the sample size increases, the point estimate for either µ or p is found to come from a distribution with a narrower and narrower distribution. We concluded that with a given level of probability, the range from which the point estimate comes is smaller as the sample size, n, increases. Figure 7.8 shows this result for the case of sample means. Simply substitute p ' for x ¯ and we can see the impact of the sample size on the estimate of the sample proportion. Chapter Review The Central Limit Theorem can also be used to illustrate that the sampling distribution of sample proportions is normally distributed with the expected value of p and a standard deviation of σ p' = p ( 1 − p ) n A question is asked of a class of 200 first-year students, and 23% of the students know the correct answer. If a sample of 50 students is taken repeatedly, what is the expected value of the mean of the sampling distribution of sample proportions? 0.23 A question is asked of a class of 200 first-year students, and 23% of the students know the correct answer. If a sample of 50 students is taken repeatedly, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.060 A game is played repeatedly. A player wins one-fifth of the time. If samples of 40 times the game is played are taken repeatedly, what is the expected value of the mean of the sampling distribution of sample proportions? 1/5 A game is played repeatedly. A player wins one-fifth of the time. If samples of 40 times the game is played are taken repeatedly, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.063 A virus attacks one in three of the people exposed to it. An entire large city is exposed. If samples of 70 people are taken, what is the expected value of the mean of the sampling distribution of sample proportions? 1/3 A virus attacks one in three of the people exposed to it. An entire large city is exposed. If samples of 70 people are taken, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.056 A company inspects products coming through its production process, and rejects detected products. One-tenth of the items are rejected. If samples of 50 items are taken, what is the expected value of the mean of the sampling distribution of sample proportions? 1/10 A company inspects products coming through its production process, and rejects detected products. One-tenth of the items are rejected. If samples of 50 items are taken, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.042 Homework A farmer picks pumpkins from a large field. The farmer makes samples of 260 pumpkins and inspects them. If one in fifty pumpkins are not fit to market and will be saved for seeds, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.0087 A store surveys customers to see if they are satisfied with the service they received. Samples of 25 surveys are taken. One in five people are unsatisfied. What is the variance of the mean of the sampling distribution of sample proportions for the number of unsatisfied customers? What is the variance for satisfied customers? 0.0064, 0.0064 A company gives an anonymous survey to its employees to see what percent of its employees are happy. The company is too large to check each response, so samples of 50 are taken, and the tendency is that three-fourths of the employees are happy. For the mean of the sampling distribution of sample proportions, answer the following questions, if the sample size is doubled. How does this affect the mean? How does this affect the standard deviation? How does this affect the variance? It has no effect. It is divided by 2 . It is divided by 2. A pollster asks a single question with only yes and no as answer possibilities. The poll is conducted nationwide, so samples of 100 responses are taken. There are four yes answers for each no answer overall. For the mean of the sampling distribution of sample proportions, find the following for yes answers. The expected value. The standard deviation. The variance. 4/5 0.04 0.0016 The mean of the sampling distribution of sample proportions has a value of p of 0.3, and sample size of 40. Is there a difference in the expected value if p and q reverse roles? Is there a difference in the calculation of the standard deviation with the same reversal? Yes No", "section": "The Central Limit Theorem for Proportions", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Finite Population Correction Factor We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is: Z = x ¯ − µ σ n · N − n N − 1 and for the variance of proportions is: σ p' = p ( 1 − p ) n × N − n N − 1 The following examples show how to apply the factor. Sampling variances get adjusted using the above formula. It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 225 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds. N = 4000 , n = 225 , σ = 10.37 , µ = 75.45 , ( x ¯ − µ ) = ± 2 Z = x ¯ − µ σ n · N − n N − 1 = ± 2 10.37 225 · 4000 − 225 4000 − 1 ≈ 2.978 f ( Z ) ≈ 2.978 · 2 = 0.997 Note that \"differs by less\" references the area on both sides of the mean within 2 pounds right or left. Try It For the 5,000 students in University A, the average height of the students is 170 cm. The population standard deviation is 5 cm. If the sample size is 100 students, find the probability that the sample mean is within 1 cm of the true mean. N = 5000 , n = 100 , σ = 5 , μ = 170 , ( x ¯ - μ ) = ± 1 Z = x ¯ - μ σ n * N - n N - 1 = ± 1 5 100 * 5000 - 100 5000 - 1 = ± 2 . 02 P ( - 1 ≤ x ¯ - μ ≤ 1 ) = P ( - 2 . 02 ≤ Z ≤ 2 . 02 ) = 2 · P ( 0 ≤ Z ≤ 2 . 02 ) = 2 · 0 . 4783 = 0 . 9566 So, the probability that the sample mean is within 1 cm of the true mean is 0.9566. When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded their credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit? N = 3000 , n = 360 , p = 0.06 σ p' = p ( 1 − p ) n × N − n N − 1 = 0.06 ( 1 − 0.06 ) 360 × 3000 − 360 3000 − 1 = 0.0117 p 1 = 10 360 = 0.0278 , p 2 = 20 360 = 0.0556 Z = p ' − p p ( 1 − p ) n · N − n N − 1 = 0.0278 − 0.06 0.011744 = −2.74 Z = p ' − p p ( 1 − p ) n · N − n N − 1 = 0.0556 − 0.06 0.011744 = −0.38 p ( 0.0278 − 0.06 0.011744 < z < 0.0556 − 0.06 0.011744 ) = p ( −2.74 < z < −0.38 ) = 0.4969 − 0.1480 = 0.3489 Try It There is a 0.01 probability that a student gets food poisoning eating food in a cafeteria. Suppose there are 2,000 orders in a day in the cafeteria. If we select 200 orders randomly, what is the probability that between 5 and 10 students will get food poisoning? N = 2000 , n = 200 , p = 0 . 01 σ p ' = p ( 1 - p ) n × N - n N - 1 = 0 . 01 ( 1 - 0 . 01 ) 200 × 2000 - 200 2000 - 1 = 0 . 00668 p 1 = 5 200 = 0 . 025 , p 2 = 10 200 = 0 . 05 Z = p ' - p σ p ' = 0 . 025 - 0 . 01 0 . 00668 = 2 . 2455 Z = p ' - p σ p ' = 0 . 05 - 0 . 01 0 . 00668 = 5 . 9880 P ( 2 . 2455 ≤ z ≤ 5 . 9880 ) = 0 . 0124 So, the probability that between 5 and 10 students will get food poisoning is 0.0124. A fishing boat has 1,000 fish on board, with an average weight of 120 pounds and a standard deviation of 6.0 pounds. If sample sizes of 50 fish are checked, what is the probability the fish in a sample will have mean weight within 2.8 pounds the true mean of the population? 0.999 An experimental garden has 500 sunflowers plants. The plants are being treated so they grow to unusual heights. The average height is 9.3 feet with a standard deviation of 0.5 foot. If sample sizes of 60 plants are taken, what is the probability the plants in a given sample will have an average height within 0.1 foot of the true mean of the population? 0.901 A company has 800 employees. The average number of workdays between absence for illness is 123 with a standard deviation of 14 days. Samples of 50 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 124 days? 0.301 Cars pass an automatic speed check device that monitors 2,000 cars on a given day. This population of cars has an average speed of 67 miles per hour with a standard deviation of 2 miles per hour. If samples of 30 cars are taken, what is the probability a given sample will have an average speed within 0.50 mile per hour of the population mean? 0.832 A town keeps weather records. From these records it has been determined that it rains on an average of 37% of the days each year. If 30 days are selected at random from one year, what is the probability that at least 5 and at most 11 days had rain? 0.483 A maker of yardsticks has an ink problem that causes the markings to smear on 4% of the yardsticks. The daily production run is 2,000 yardsticks. What is the probability if a sample of 100 yardsticks is checked, there will be ink smeared on at most 4 yardsticks? 0.500 A school has 300 students. Usually, there are an average of 21 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at most 2 students in the sample will be absent? 0.502 A college gives a placement test to 5,000 incoming students each year. On the average 1,213 place in one or more developmental courses. If a sample of 50 is taken from the 5,000, what is the probability at most 12 of those sampled will have to take at least one developmental course? 0.519 Homework A company has 1,000 employees. The average number of workdays between absence for illness is 80 with a standard deviation of 11 days. Samples of 80 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 78 days and at most 84 days? 0.955 Trucks pass an automatic scale that monitors 2,000 trucks. This population of trucks has an average weight of 20 tons with a standard deviation of 2 tons. If a sample of 50 trucks is taken, what is the probability the sample will have an average weight within one-half ton of the population mean? 0.927 A town keeps weather records. From these records it has been determined that it rains on an average of 12% of the days each year. If 30 days are selected at random from one year, what is the probability that at most 3 days had rain? 0.648 A maker of greeting cards has an ink problem that causes the ink to smear on 7% of the cards. The daily production run is 500 cards. What is the probability that if a sample of 35 cards is checked, there will be ink smeared on at most 5 cards? 0.101 A school has 500 students. Usually, there are an average of 20 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at least 2 students in the sample will be absent? 0.273", "section": "Finite Population Correction Factor", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction Have you ever wondered what the average number of M&Ms in a bag at the grocery store is? You can use confidence intervals to answer this question. (credit: modification of work “sweet, orange, food, green, red, color, brown, blue, colorful, yellow, chocolate, snack, dessert, toy, plain, candy, sweetness, treat, confectionery, coated, m ms, hard shell, snack food, jelly bean”/ Pxhere, Public Domain) Suppose you were trying to determine the rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion the parameter p in the binomial probability density function. We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics . The sample data help us to make an estimate of a population parameter . We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals. What statistics provides us beyond a simple average, or point estimate, is an estimate to which we can attach a probability of accuracy, what we will call a confidence level. We make inferences with a known level of probability. In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed. If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from Apple Music. If so, you could conduct a survey and calculate the sample mean, x – , and the sample standard deviation, s . You would use x – to estimate the population mean and s to estimate the population standard deviation. The sample mean, x – , is the point estimate for the population mean, μ . The sample standard deviation, s , is the point estimate for the population standard deviation, σ . x – and s are each called a statistic. A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include the unknown population parameter. Suppose, for the Apple Music example, we do not know the population mean μ , but we do know the sample mean and that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation of the sampling distribution of the sample means is σ n = 1 100 = 0.1 The Empirical Rule , which applies to the normal distribution, says that in approximately 95% of the samples, the sample mean, x – , will be within two standard deviations of the population mean μ . For our Apple Music example, two standard deviations is (2)(0.1) = 0.2. The sample mean x – is likely to be within 0.2 units of μ . Because x – is within 0.2 units of μ , which is unknown, then μ is likely to be within 0.2 units of x – with 95% probability. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between x ¯ − 0 .2 and x ¯ + 0 .2 in 95% of all the samples. For the Apple Music example, suppose that a sample produced a sample mean x ¯ = 2 . Then with 95% probability the unknown population mean μ is between x ¯ − 0.2 = 2 − 0.2 = 1.8 and x ¯ + 0.2 = 2 + 0.2 = 2.2 We say that we are 95% confident that the unknown population mean number of songs downloaded from Apple Music per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2). Please note that we talked in terms of 95% confidence using the empirical rule. The empirical rule for two standard deviations is only approximately 95% of the probability under the normal distribution. To be precise, two standard deviations under a normal distribution is actually 95.44% of the probability. To calculate the exact 95% confidence level we would use 1.96 standard deviations. The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ, or our sample produced an x – that is not within 0.2 units of the true mean μ . The first possibility happens for 95% of well-chosen samples. It is important to remember that the second possibility happens for 5% of samples, even though correct procedures are followed. Remember that a confidence interval is created for an unknown population parameter like the population mean, μ . For the confidence interval for a mean the formula would be: μ = X ¯ ± Z α σ n Or written another way as: X ¯ − Z α σ n ≤ μ ≤ X ¯ + Z α σ n Where X ¯ is the sample mean. Z α is determined by the level of confidence desired by the analyst, and σ n is the standard deviation of the sampling distribution for means given to us by the Central Limit Theorem. Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: the desired confidence level, information that is known about the distribution (for example, known standard deviation), the sample and its size. Inferential Statistics also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if four out of the 100 calculators sampled are defective we might infer that four percent of the production is defective. Parameter a numerical characteristic of a population Point Estimate a single number computed from a sample and used to estimate a population parameter", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "A Confidence Interval When the Population Standard Deviation Is Known or Large Sample Size A confidence interval for a population mean, when the population standard deviation is known based on the conclusion of the Central Limit Theorem that the sampling distribution of the sample means follow an approximately normal distribution. Calculating the Confidence Interval Consider the standardizing formula for the sampling distribution developed in the discussion of the Central Limit Theorem: Z 1 = X – − μ X – σ X – = X – − μ σ n Notice that µ is substituted for µ x − because we know that the expected value of µ x − is µ from the Central Limit theorem and σ x − is replaced with σ n , also from the Central Limit Theorem. In this formula we know X – , σ x − and n, the sample size. (In actuality we do not know the population standard deviation, but we do have a point estimate for it, s, from the sample we took. More on this later.) What we do not know is μ or Z 1 . We can solve for either one of these in terms of the other. Solving for μ in terms of Z 1 gives: μ = X – ± Z 1 σ n Remembering that the Central Limit Theorem tells us that the distribution of the X ¯ 's, the sampling distribution for means, is normal, and that the normal distribution is symmetrical, we can rearrange terms thus: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) This is the formula for a confidence interval for the mean of a population. Notice that Z α has been substituted for Z 1 in this equation. This is where a choice must be made by the statistician. The analyst must decide the level of confidence they wish to impose on the confidence interval. α is the probability that the interval will not contain the true population mean. The confidence level is defined as (1-α). Z α is the number of standard deviations X ¯ lies from the mean with a certain probability. If we chose Z α = 1.96 we are asking for the 95% confidence interval because we are setting the probability that the true mean lies within the range at 0.95. If we set Z α at 1.64 we are asking for the 90% confidence interval because we have set the probability at 0.90. These numbers can be verified by consulting the Standard Normal table. Divide either 0.95 or 0.90 in half and find that probability inside the body of the table. Then read on the top and left margins the number of standard deviations it takes to get this level of probability. In reality, we can set whatever level of confidence we desire simply by changing the Z α value in the formula. It is the analyst's choice. Common convention in Economics and most social sciences sets confidence intervals at either 90, 95, or 99 percent levels. Levels less than 90% are considered of little value. The level of confidence of a particular interval estimate is called by (1-α). A good way to see the development of a confidence interval is to graphically depict the solution to a problem requesting a confidence interval. This is presented in for the example in the introduction concerning the number of downloads from Apple Music. That case was for a 95% confidence interval, but other levels of confidence could have just as easily been chosen depending on the need of the analyst. However, the level of confidence MUST be pre-set and not subject to revision as a result of the calculations. μ = X ¯ ± Z α σ n = 2 ± 1 . 96 ( 0 . 1 ) = 2 ± 0 . 196 1 . 804 ≤ μ ≤ 2 . 196 For this example, let's say we know that the actual population mean number of Apple Music downloads is 2.1. The true population mean falls within the range of the 95% confidence interval. There is absolutely nothing to guarantee that this will happen. Further, if the true mean falls outside of the interval we will never know it. We must always remember that we will never ever know the true mean. Statistics simply allows us, with a given level of probability (confidence), to say that the true mean is within the range calculated. This is what was called in the introduction, the \"level of ignorance admitted\". Changing the Confidence Level or Sample Size Here again is the formula for a confidence interval for an unknown population mean assuming we know the population standard deviation: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) It is clear that the confidence interval is driven by two things, the chosen level of confidence, Z α , and the standard deviation of the sampling distribution. The Standard deviation of the sampling distribution is further affected by two things, the standard deviation of the population and the sample size we chose for our data. Here we wish to examine the effects of each of the choices we have made on the calculated confidence interval, the confidence level and the sample size. For a moment we should ask just what we desire in a confidence interval. Our goal was to estimate the population mean from a sample. We have forsaken the hope that we will ever find the true population mean, and population standard deviation for that matter, for any case except where we have an extremely small population and the cost of gathering the data of interest is very small. In all other cases we must rely on samples. With the Central Limit Theorem we have the tools to provide a meaningful confidence interval with a given level of confidence, meaning a known probability of being wrong. By meaningful confidence interval we mean one that is useful. Imagine that you are asked for a confidence interval for the ages of your classmates. You have taken a sample and find a mean of 19.8 years. You wish to be very confident so you report an interval between 9.8 years and 29.8 years. This interval would certainly contain the true population mean and have a very high confidence level. However, it hardly qualifies as meaningful. The very best confidence interval is narrow while having high confidence. There is a natural tension between these two goals. The higher the level of confidence the wider the confidence interval as the case of the students' ages above. We can see this tension in the equation for the confidence interval. μ = x _ ± Z α ( σ n ) The confidence interval will increase in width as Z α increases, Z α increases as the level of confidence increases. There is a tradeoff between the level of confidence and the width of the interval. Now let's look at the formula again and we see that the sample size also plays an important role in the width of the confidence interval. The sample size, n , shows up in the denominator of the standard deviation of the sampling distribution. As the sample size increases, the standard deviation of the sampling distribution decreases and thus the width of the confidence interval, while holding constant the level of confidence. Again we see the importance of having large samples for our analysis although we then face a second constraint, the cost of gathering data. Suppose we are interested in the mean scores on an exam. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68 ( X ¯ = 68). In this example we have the unusual knowledge that the population standard deviation is 3 points. Do not count on knowing the population parameters outside of textbook examples. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90% confidence interval for the true (population) mean of statistics exam scores. The solution is shown step by step: The formula for a confidence interval for an unknown population mean assuming we know the population standard deviation is: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) For a 90% confidence interval, visualize an area of 0.90 centered under the normal curve (See ). The remaining area for the two tails of the normal distribution is then 0.10, which indicates that the area in the left tail is one-half of 0.10, which is 0.05. The corresponding z-score that cuts off an area of 0.05 in the left tail is 1.645. In this example we are given that the population standard deviation σ = 3 . We are also given that the sample size n = 36 and the sample mean X ¯ = 68 . Substituting these values in the confidence interval formula results in the following: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) 68 - 1 . 645 3 36 ≤ μ ≤ 68 + 1 . 645 3 36 68 - 0 . 8225 ≤ μ ≤ 68 + 0 . 8225 67 . 1775 ≤ μ ≤ 68 . 8225 We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Try It Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time. (34.1347, 37.8653) Suppose we change the original problem in by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score. μ = x _ ± Z α ( σ n ) μ = 68 ± 1.96 ( 3 36 ) 67.02 ≤ μ ≤ 68.98 σ = 3; n = 36; The confidence level is 95% ( CL = 0.95). CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05 Z α 2 = Z 0.025 = 1.96 Notice that the plus/minus term in the equation is larger for a 95% confidence level in the original problem. Comparing the results: The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. This demonstrates a very important principle of confidence intervals. There is a trade off between the level of confidence and the width of the interval. Our desire is to have a narrow confidence interval, huge wide intervals provide little information that is useful. But we would also like to have a high level of confidence in our interval. This demonstrates that we cannot have both. Summary: Effect of Changing the Confidence Level Increasing the confidence level makes the confidence interval wider. Decreasing the confidence level makes the confidence interval narrower. And again here is the formula for a confidence interval for an unknown mean assuming we have the population standard deviation: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) The standard deviation of the sampling distribution was provided by the Central Limit Theorem as σ n . While we infrequently get to choose the sample size it plays an important role in the confidence interval. Because the sample size is in the denominator of the equation, as n increases it causes the standard deviation of the sampling distribution to decrease and thus the width of the confidence interval to decrease. We have met this before as we reviewed the effects of sample size on the Central Limit Theorem. There we saw that as n increases the sampling distribution narrows until in the limit it collapses on the true population mean. Try It Refer back to the pizza-delivery Try It 8.1 exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time. Changing the Sample Size Suppose we change the original problem in to see what happens to the confidence interval if the sample size is changed. Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36? μ = x _ ± Z α ( σ n ) μ = 68 ± 1.645 ( 3 100 ) 67.5065 ≤ μ ≤ 68.4935 If we increase the sample size n to 100, we decrease the width of the confidence interval relative to the original sample size of 36 observations. μ = x _ ± Z α ( σ n ) μ = 68 ± 1.645 ( 3 25 ) 67.013 ≤ μ ≤ 68.987 If we decrease the sample size n to 25, we increase the width of the confidence interval by comparison to the original sample size of 36 observations. Summary: Effect of Changing the Sample Size Increasing the sample size makes the confidence interval narrower. Decreasing the sample size makes the confidence interval wider. Try It Refer back to the pizza-delivery Try It 8.1 exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time. We have already seen this effect when we reviewed the effects of changing the size of the sample, n , on the Central Limit Theorem. See to see this effect. Before we saw that as the sample size increased the standard deviation of the sampling distribution decreases. This was why we choose the sample mean from a large sample as compared to a small sample, all other things held constant. Thus far we assumed that we knew the population standard deviation. This will virtually never be the case. We will have the sample standard deviation, s , however. This is a point estimate for the population standard deviation and can be substituted into the formula for confidence intervals for a mean under certain circumstances. We just saw the effect the sample size has on the width of confidence interval and the impact on the sampling distribution for our discussion of the Central Limit Theorem. We can invoke this to substitute the point estimate for the standard deviation if the sample size is large \"enough\". Simulation studies indicate that 30 observations or more will be sufficient to eliminate any meaningful bias in the estimated confidence interval. Spring break can be a very expensive holiday. A sample of 80 students is surveyed, and the average amount spent by students on travel and beverages is $593.84. The sample standard deviation is approximately $369.34. Construct a 92% confidence interval for the population mean amount of money spent by spring breakers. We begin with the confidence interval for a mean. We use the formula for a mean because the random variable is dollars spent and this is a continuous random variable. The point estimate for the population standard deviation, s , has been substituted for the true population standard deviation because with 80 observations there is no concern for bias in the estimate of the confidence interval. μ = x ¯ ± [ Z ( a / 2 ) s n ] Substituting the values into the formula, we have: μ = 593.84 ± [ 1.75 369.34 80 ] Z ( a / 2 ) is found on the standard normal table by looking up 0.46 in the body of the table and finding the number of standard deviations on the side and top of the table; 1.75. The solution for the interval is thus: μ = 593.84 ± 72.2636 = ( 521.57 , 666.10 ) $ 521.58 ≤ μ ≤ $ 666.10 Try It The price of a chair is a large range of cost. The average cost of 25 chairs in a store is $100. The sample standard deviation is $50. Construct a 92% confidence interval for the population mean of the cost of chairs. Since there is no bias in the estimate of the confidence interval, the confidence interval is obtained as follows: μ = x ¯ ± Z ( α / 2 ) s n μ = 100 ± 1 . 4 50 25 μ = 100 ± 14 So, the confidence interval is obtained as ($86, $114). References “American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t (accessed July 2, 2013). “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013). “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm (accessed September 30,2013). Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/2000growthchart-us.pdf (accessed July 2, 2013). La, Lynn, Kent German. \"Cell Phone Radiation Levels.\" c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013). “Mean Income in the Past 12 Months (in 2011 Inflation-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013). “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013). “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013). Formula Review The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) This formula is used when the population standard deviation is known. CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter z α 2 = the z -score with the property that the area to the right of the z-score is ∝ 2 this is the z -score used in the calculation where α = 1 – CL . Confidence Level (CL) the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter. Error Bound for a Population Mean ( EBM ) the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation.", "section": "A Confidence Interval When the Population Standard Deviation Is Known or Large Sample Size", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case In practice, we rarely know the population standard deviation . In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. This is what we did in above. The point estimate for the standard deviation, s, was substituted in the formula for the confidence interval for the population standard deviation. In this case the 80 observations are well above the suggested 30 observations to eliminate any bias from a small sample. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval. William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to \"discover\" what is called the Student's t-distribution . The name comes from the fact that Gosset wrote under the pen name \"A Student.\" Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30 observations. If you draw a simple random sample of size n from a population with mean μ and unknown population standard deviation σ and calculate the t -score t = x – – μ ( s n ) , then the t -scores follow a Student's t-distribution with n – 1 degrees of freedom . The t -score has the same interpretation as the z -score . It measures how far in standard deviation units x – is from its mean μ . For each sample size n , there is a different Student's t-distribution. The degrees of freedom , n – 1 , come from the calculation of the sample standard deviation s . Remember when we first calculated a sample standard deviation we divided the sum of the squared deviations by n − 1, but we used n deviations ( x – x ¯ values ) to calculate s . Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df) in recognition that one is lost in the calculations. The effect of losing a degree of freedom is that the t-value increases and the confidence interval increases in width. Properties of the Student's t-Distribution The graph for the Student's t-distribution is similar to the standard normal curve and at infinite degrees of freedom it is the normal distribution. You can confirm this by reading the bottom line at infinite degrees of freedom for a familiar level of confidence, e.g. at column 0.05, 95% level of confidence, we find the t-value of 1.96 at infinite degrees of freedom. The mean for the Student's t-distribution is zero and the distribution is symmetric about zero, again like the standard normal distribution. The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution. The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution. The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . This assumption comes from the Central Limit theorem because the individual observations in this case are the x ¯ s of the sampling distribution. The size of the underlying population is generally not relevant unless it is very small. If it is normal then the assumption is met and doesn't need discussion. A probability table for the Student's t-distribution is used to calculate t-values at various commonly-used levels of confidence. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). When using a t -table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. Notice that at the bottom the table will show the t-value for infinite degrees of freedom. Mathematically, as the degrees of freedom increase, the t -distribution approaches the standard normal distribution. You can find familiar Z-values by looking in the relevant alpha column and reading value in the last row. A Student's t table (See ) gives t -scores given the degrees of freedom and the right-tailed probability. The Student's t -distribution has one of the most desirable properties of the normal distribution: it is symmetrical. What the Student's t -distribution does is spread out the horizontal axis so it takes a larger number of standard deviations to capture the same amount of probability. In reality there are an infinite number of Student's t -distributions, one for each adjustment to the sample size. As the sample size increases, the Student's t -distribution become more and more like the normal distribution. When the sample size reaches 30 the normal distribution is usually substituted for the Student's t because they are so much alike. This relationship between the Student's t -distribution and the normal distribution is shown in . This is another example of one distribution limiting another one, in this case the normal distribution is the limiting distribution of the Student's t when the degrees of freedom in the Student's t approaches infinity. This conclusion comes directly from the derivation of the Student's t -distribution by Mr. Gosset. He recognized the problem as having few observations and no estimate of the population standard deviation. He was substituting the sample standard deviation and getting volatile results. He therefore created the Student's t -distribution as a ratio of the normal distribution and Chi squared distribution. The Chi squared distribution is itself a ratio of two variances, in this case the sample variance and the unknown population variance. The Student's t -distribution thus is tied to the normal distribution, but has degrees of freedom that come from those of the Chi squared distribution. The algebraic solution demonstrates this result. Development of Student's t-distribution: t = z χ 2 v where z is the standard normal variable and χ 2 is the chi-squared distribution with v degrees of freedom. Substitute values and simplify: t = ( x ¯ − μ ) σ s 2 ( n − 1 ) σ 2 ( n − 1 ) = ( x ¯ − μ ) σ s 2 σ 2 = ( x ¯ − μ ) σ s σ = x ¯ − μ s n t = s n Restating the formula for a confidence interval for the mean for cases when the sample size is smaller than 30 and we do not know the population standard deviation, σ: x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) Here the point estimate of the population standard deviation, s has been substituted for the population standard deviation, σ, and t ν ,α has been substituted for Z α . The Greek letter ν (pronounced nu) is placed in the general formula in recognition that there are many Student t v distributions, one for each sample size. ν is the symbol for the degrees of freedom of the distribution and depends on the size of the sample. Often df is used to abbreviate degrees of freedom. For this type of problem, the degrees of freedom is ν = n-1, where n is the sample size. To look up a probability in the Student's t table we have to know the degrees of freedom in the problem. The average earnings per share (EPS) for 10 industrial stocks randomly selected from those listed on the Dow-Jones Industrial Average was found to be X ¯ = 1.85 with a standard deviation of s=0.395. Calculate a 99% confidence interval for the average EPS of all the industrials listed on the DJIA. x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) To help visualize the process of calculating a confident interval we draw the appropriate distribution for the problem. In this case this is the Student’s t because we do not know the population standard deviation and the sample is small, less than 30. To find the appropriate t-value requires two pieces of information, the level of confidence desired and the degrees of freedom. The question asked for a 99% confidence level. On the graph this is shown where (1-α) , the level of confidence , is in the unshaded area. The tails, thus, have .005 probability each, α/2. The degrees of freedom for this type of problem is n-1= 9. From the Student’s t table, at the row marked 9 and column marked .005, is the number of standard deviations to capture 99% of the probability, 3.2498. These are then placed on the graph remembering that the Student’s t is symmetrical and so the t-value is both plus or minus on each side of the mean. Inserting these values into the formula gives the result. These values can be placed on the graph to see the relationship between the distribution of the sample means, X ¯ 's and the Student’s t -distribution. μ = X ¯ ± t α/2,df=n-1 s n = 1.851 ± 3.2498 0.395 10 = 1.8551 ± 0.406 1.445 ≤ μ ≤ 2.257 We state the formal conclusion as : With 99% confidence level, the average EPS of all the industries listed at DJIA is from $1.44 to $2.26. Try It You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5 (8.1634, 9.8032) References “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013). Data from Microsoft Bookshelf . Data from http://www.businessweek.com/. Data from http://www.forbes.com/. “Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013). “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013). “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013). Chapter Review In many cases, the researcher does not know the population standard deviation, σ , of the measure being studied. In these cases, it is common to use the sample standard deviation, s , as an estimate of σ . The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: t = x – - μ s n The t -score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with x ¯ ± ( t α 2 ) s n where t α 2 is the t -score with area to the right equal to α 2 , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 for a given α . Formula Review s = the standard deviation of sample values. t = x – − μ s n is the formula for the t -score which measures how far away a measure is from the population mean in the Student’s t-distribution df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample T ~ t df the random variable, T , has a Student’s t-distribution with df degrees of freedom The general form for a confidence interval for a single mean, population standard deviation unknown, and sample size less than 30 Student's t is given by: x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. Identify the following: x – =_______ s x =_______ n =_______ n – 1 =_______ Define the random variables X and X – in words. X is the number of hours a patient waits in the emergency room before being called back to be examined. X – is the mean wait time of 70 patients in the emergency room. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound. CI: (1.3808, 1.6192) EBM = 0.12 Explain in complete sentences what the confidence interval means. Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Identify the following: x – =_______ s x =_______ n =_______ n – 1 =_______ x – = 151 s x = 32 n = 108 n – 1 = 107 Define the random variable X in words. Define the random variable X – in words. X – is the mean number of hours spent watching television per month from a sample of 108 Americans. Which distribution should you use for this problem? Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound. CI: (142.92, 159.08) EBM = 8.08 Why would the error bound change if the confidence level were lowered to 95%? Use the following information to answer the next 13 exercises: The data in are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 1 2 7 3 18 4 7 5 6 Calculate the following: x – =______ s x =______ n =______ 3.26 1.02 39 Define the random variable X – in words. What is x – estimating? μ Is σ x known? As a result of your answer to , state the exact distribution to use when calculating the confidence interval. t 38 Construct a 95% confidence interval for the true mean number of colors on national flags. How much area is in both tails (combined)? How much area is in each tail? 0.025 Calculate the following: lower limit upper limit error bound The 95% confidence interval is_____. (2.93, 3.59) Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean. In one complete sentence, explain what the interval means. We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. Using the same x – , s x , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know? The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean. Using the same x - , s x , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why? Homework In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice Calculate p ′. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag. State the confidence interval. Sketch the graph. Calculate the error bound. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal. x – = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. State the confidence interval. Sketch the graph. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why? 8629 6944 35 34 t 34 CI: (6244, 11,014) It will become smaller Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. x – = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean time wasted. State the confidence interval. Sketch the graph. Explain in a complete sentence what the confidence interval means. A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X - in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean length of time. State the confidence interval. Sketch the graph. What does it mean to be “95% confident” in this problem? x – = 2.51 s x = 0.318 n = 9 n - 1 = 8 the effective length of time for a tranquilizer the mean effective length of time of tranquilizers from a sample of nine patients We need to use a Student’s-t distribution, because we do not know the population standard deviation. CI: (2.27, 2.76) Answers may vary. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal. x - = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 99% confidence interval for the population mean length of time using training wheels. State the confidence interval. Sketch the graph. Why would the error bound change if the confidence level were lowered to 90%? The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns. The FEC has reported financial information for 556 Leadership PACs that were operating during a certain election cycle. The following table shows the total receipts during this cycle for a random selection of 30 Leadership PACs. $46,500.00 $0 $40,966.50 $105,887.20 $5,175.00 $29,050.00 $19,500.00 $181,557.20 $31,500.00 $149,970.80 $2,555,363.20 $12,025.00 $409,000.00 $60,521.70 $18,000.00 $61,810.20 $76,530.80 $119,459.20 $0 $63,520.00 $6,500.00 $502,578.00 $705,061.10 $708,258.90 $135,810.00 $2,000.00 $2,000.00 $0 $1,287,933.80 $219,148.30 x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Use this sample data to construct a 95% confidence interval for the mean amount of money raised by all Leadership PACs during that specific election cycle. Use the Student's t-distribution. x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29 CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04 α 2 = 0.02 t α 2 = t 0.02 = 2.150 E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 ) ~ $ 204 , 561.66 x - - EBM = $251,854.23 - $204,561.66 = $47,292.57 x - + EBM = $251,854.23+ $204,561.66 = $456,415.89 We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during that specific election cycle lies between $47,292.57 and $456,415.89. Forbes magazine published data on the best small firms in a certain year. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 49 Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. x ¯ = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. State the confidence interval. Sketch the graph. x ¯ = 11.6 s x = 4.1 n = 225 n - 1 = 224 X is the number of unoccupied seats on a single flight. X – is the mean number of unoccupied seats from a sample of 225 flights. We will use a Student’s-t distribution, because we do not know the population standard deviation. CI: (11.12 , 12.08) Answers may vary. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Explain your choice. Define the random variable X - in words. Construct a 95% confidence interval for the population mean cost of a used car. State the confidence interval. Sketch the graph. Explain what a “95% confidence interval” means for this study. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. State the confidence interval. Sketch the graph. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies. Calculate the mean. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not? CI: (7.64 , 9.36) The sample should have been increased. Answers will vary. Answers will vary. Answers will vary. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. x - = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X - in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean worth of coupons. State the confidence interval. Sketch the graph. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. Find the 95% Confidence Interval for the true population mean for the amount of soda served. (12.42, 14.18) (12.32, 14.29) (12.50, 14.10) Impossible to determine b Degrees of Freedom ( df ) the number of objects in a sample that are free to vary Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 / 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation, notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t -Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student; the major characteristics of this random variable (RV) are: It is continuous and assumes any real values. The pdf is symmetrical about its mean of zero. It approaches the standard normal distribution as n get larger. There is a \"family\" of t–distributions: each representative of the family is completely defined by the number of degrees of freedom, which depends upon the application for which the t is being used.", "section": "A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "A Confidence Interval for A Population Proportion During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43. Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval for a population proportion is similar to that for the population mean, but the formulas are a bit different although conceptually identical. While the formulas are different, they are based upon the same mathematical foundation given to us by the Central Limit Theorem. Because of this we will see the same basic format using the same three pieces of information: the sample value of the parameter in question, the standard deviation of the relevant sampling distribution, and the number of standard deviations we need to have the confidence in our estimate that we desire. How do you know you are dealing with a proportion problem? First, the underlying distribution has a binary random variable and therefore is a binomial distribution . (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B ( n , p ) where n is the number of trials and p is the probability of a success. To form a sample proportion, take X , the random variable for the number of successes and divide it by n , the number of trials (or the sample size). The random variable p ′ (read \"P prime\") is the sample proportion, P ′ = X n (Sometimes the random variable is denoted as P ^ , read \"P hat\".) p ′ = the estimated proportion of successes or sample proportion of successes ( p ′ is a point estimate for p , the true population proportion, and thus q is the probability of a failure in any one trial.) x = the number of successes in the sample n = the size of the sample The formula for the confidence interval for a population proportion follows the same format as that for an estimate of a population mean. Remembering the sampling distribution for the proportion from The Central Limit Theorem , the standard deviation was found to be: σ p' = p ( 1 − p ) n The confidence interval for a population proportion, therefore, becomes: p = p ′ ± [ Z ( a 2 ) p ′ ( 1 − p ′ ) n ] Z ( a 2 ) is set according to our desired degree of confidence and p ′ ( 1 − p ′ ) n is the standard deviation of the sampling distribution. The sample proportions p ′ and q ′ are estimates of the unknown population proportions p and q . The estimated proportions p ′ and q ′ are used because p and q are not known. Remember that as p moves further from 0.5 the binomial distribution becomes less symmetrical. Because we are estimating the binomial with the symmetrical normal distribution the further away from symmetrical the binomial becomes the less confidence we have in the estimate. This conclusion can be demonstrated through the following analysis. Proportions are based upon the binomial probability distribution. The possible outcomes are binary, either “success” or “failure”. This gives rise to a proportion, meaning the percentage of the outcomes that are “successes”. It was shown that the binomial distribution could be fully understood if we knew only the probability of a success in any one trial, called p. The mean and the standard deviation of the binomial were found to be: μ = np σ = np q It was also shown that the binomial could be estimated by the normal distribution if BOTH np AND nq were greater than 5. From the discussion above, it was found that the standardizing formula for the binomial distribution is: Z = p' - p ( p q n ) which is nothing more than a restatement of the general standardizing formula with appropriate substitutions for μ and σ from the binomial. We can use the standard normal distribution, the reason Z is in the equation, because the normal distribution is the limiting distribution of the binomial. This is another example of the Central Limit Theorem. We have already seen that the sampling distribution of means is normally distributed. Recall the extended discussion in The Central Limit Theorem concerning the sampling distribution of proportions and the conclusions of the Central Limit Theorem. We can now manipulate this formula in just the same way we did for finding the confidence intervals for a mean, but to find the confidence interval for the binomial population parameter, p. p ′ - Z α 2 p ′ q ′ n ≤ p ≤ p ′ + Z α 2 p ′ q ′ n Where p ′ = x/n, the point estimate of p taken from the sample. Notice that p ′ has replaced p in the formula. This is because we do not know p , indeed, this is just what we are trying to estimate. Unfortunately, there is no correction factor for cases where the sample size is small so np ′ and nq ′ must always be greater than 5 to develop an interval estimate for p . Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have smartphones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have smartphones. Of the 500 people sampled, 421 responded yes - they own smartphones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have smartphones. The solution step-by-step. Let X = the number of people in the sample who have smartphones. X is binomial: the random variable is binary, people either have a smartphone or they do not. To calculate the confidence interval, we must find p ′ , q ′ . n = 500 x = the number of successes in the sample = 421 p ′ = x n = 421 500 = 0.842 p ′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. q ′ = 1 – p ′ = 1 – 0.842 = 0.158 Since the requested confidence level is CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ( α 2 ) = 0.025. Then z α 2 = z 0.025 = 1.96 This can be found using the Standard Normal probability table in . This can also be found in the students t table at the 0.025 column and infinity degrees of freedom because at infinite degrees of freedom the students t -distribution becomes the standard normal distribution, Z. The confidence interval for the true binomial population proportion is p ′ - Z α p ′ q ′ n ≤ p ≤ p ′ + Z α p ′ q ′ n Substituting in the values from above we find the confidence interval is : 0.810 ≤ p ≤ 0.874 Interpretation: We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have smartphones. Explanation of 95% Confidence Level: Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have smartphones. Try It Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. (0.3315, 0.4525) The Dundee Dog Training School has a larger than average proportion of clients who compete in competitive professional events. A confidence interval for the population proportion of dogs that compete in professional events from 150 different training schools is constructed. The lower limit is determined to be 0.08 and the upper limit is determined to be 0.16. Determine the level of confidence used to construct the interval of the population proportion of dogs that compete in professional events. We begin with the formula for a confidence interval for a proportion because the random variable is binary; either the client competes in professional competitive dog events or they don't. p = p ′ ± [ Z ( a 2 ) p ′ ( 1 − p ′ ) n ] Next we find the sample proportion: p ′ = 0.08 + 0.16 2 = 0.12 The ± that makes up the confidence interval is thus 0.04; 0.12 + 0.04 = 0.16 and 0.12 − 0.04 = 0.08, the boundaries of the confidence interval. Finally, we solve for Z . [ Z ⋅ 0.12 ( 1 − 0.12 ) 150 ] = 0.04 , therefore Z = 1.51 And then look up the probability for 1.51 standard deviations on the standard normal table. p ( Z = 1.51 ) = 0.4345 , p ( Z ) ⋅ 2 = 0.8690 or 86.90 % . Try It A student polls their school to see if students in the school district are for or against the new legislation regarding school uniforms. They survey 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. A financial officer for a company wants to estimate the percent of accounts receivable that are more than 30 days overdue. They survey 500 accounts and find that 300 are more than 30 days overdue. Compute a 90% confidence interval for the true percent of accounts receivable that are more than 30 days overdue, and interpret the confidence interval. The solution is step-by-step: x = 300 and n = 500 p ′ = x n = 300 500 = 0.600 q ′ = 1 - p ′ = 1 - 0.600 = 0.400 Since confidence level = 0.90, then α = 1 – confidence level = (1 – 0.90) = 0.10 ( α 2 ) = 0.05 Z α 2 = Z 0.05 = 1.645 This Z-value can be found using a standard normal probability table. The student's t-table can also be used by entering the table at the 0.05 column and reading at the line for infinite degrees of freedom. The t-distribution is the normal distribution at infinite degrees of freedom. This is a handy trick to remember in finding Z-values for commonly used levels of confidence. We use this formula for a confidence interval for a proportion: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n Substituting in the values from above we find the confidence interval for the true binomial population proportion is 0.564 ≤ p ≤ 0.636 Interpretation: We estimate with 90% confidence that the true percent of all accounts receivable overdue 30 days is between 56.4% and 63.6%. Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL accounts are overdue 30 days. Explanation of 90% Confidence Level: Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of accounts receivable that are overdue 30 days. Try It A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. (0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. Solution Sixty-eight percent (68%) of students own an iPod and a smart phone. p ′ = 0.68 q ′ = 1 – p ′ = 1 – 0.68 = 0.32 Since CL = 0.97, we know α = 1 – 0.97 = 0.03 and α 2 = 0.015. The area to the left of z 0.015 is 0.015, and the area to the right of z 0.015 is 1 – 0.015 = 0.985. Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1), z 0.015 = 2.17 E P B = ( z α 2 ) p ′ q ′ n = 2.17 0.68 ( 0.32 ) 300 ≈ 0.0584 p ′ – EPB = 0.68 – 0.0584 = 0.0584 p ′ + EPB = 0.68 + 0.0584 = 0.0584 We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6216 and 0.7384. References Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling. Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013). Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, and Meredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx (accessed July 2, 2013). Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.” Pew Research Center: Internet and American Life Project. Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013). Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at http://www.gallup.com/poll/162758/three-four-workers-plan-work-past-retirement-age.aspx (accessed July 2, 2013). The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2, 2013). Zogby. “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. Available online at http://www.zogbyanalytics.com/news/299-americans-neither-worried-nor-prepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). “52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process (accessed July 2, 2013). Chapter Review Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning. Let p ′ represent the sample proportion, x/n , where x represents the number of successes and n represents the sample size. Let q ′ = 1 – p ′ . Then the confidence interval for a population proportion is given by the following formula: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n Formula Review p′= x n where x represents the number of successes in a sample and n represents the sample size. The variable p ′ is the sample proportion and serves as the point estimate for the true population proportion. q ′ = 1 – p ′ The variable p ′ has a binomial distribution that can be approximated with the normal distribution shown here. The confidence interval for the true population proportion is given by the formula: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n n = Z α 2 2 p ′ q ′ e 2 provides the number of observations needed to sample to estimate the population proportion, p , with confidence 1 - α and margin of error e . Where e = the acceptable difference between the actual population proportion and the sample proportion. Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05? If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why? The sample size needed would increase. As the confidence level increases, α decreases and z ( a 2 ) increases. To maintain the same error bound, the size of the sample needs to increase. Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. Identify the following: x = ______ n = ______ p ′ = ______ Define the random variables X and p ′ in words. X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P ′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. Which distribution should you use for this problem? Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.5321, 0.6679) EBM : 0.0679 List two difficulties the company might have in obtaining random results, if this survey were done by email. Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P ′ in words. X is the number of “successes” where an executive prefers a truck. P ′ is the percentage of executives sampled who prefer a truck. Which distribution should you use for this problem? Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.19432, 0.33068) Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given in the survey is ±2%. Explain what the ±2% means. The sampling error means that the true mean can be 2% above or below the sample mean. Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important. Define the random variable X in words. Define the random variable P ′ in words. P ′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. Which distribution should you use for this problem? Construct a 90% confidence interval, and state the confidence interval and the error bound. CI: (0.62735, 0.67265) EBM : 0.02265 What would happen to the confidence interval if the level of confidence were 95%? Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. What is being counted? The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class. In words, define the random variable X . Calculate the following: x = _______ n = _______ p ′ = _______ x = 64 n = 80 p ′ = 0.8 State the estimated distribution of X . X ~________ Define a new random variable P ′. What is p ′ estimating? p In words, define the random variable P ′. State the estimated distribution of P ′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet. P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 ) . (0.72171, 0.87829). How much area is in both tails (combined)? How much area is in each tail? 0.04 Calculate the following: lower limit upper limit error bound The 92% confidence interval is _______. (0.72; 0.88) Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. In one complete sentence, explain what the interval means. With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%. Using the same p ′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know? Using the same p ′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why? The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? Homework Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03? If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 1,068 The sample size would need to be increased since the critical value increases as the confidence level increases. Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. x = __________ n = __________ p ′ = __________ Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion who claim they always buckle up. State the confidence interval. Sketch the graph. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job. State the confidence interval. Sketch the graph. X = the number of people who feel that the president is doing an acceptable job; P ′ = the proportion of people in a sample who feel that the president is doing an acceptable job. N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 ) CI: (0.59, 0.63) Answers may vary. An article regarding interracial dating and marriage recently appeared in the Washington Post . Of the 1,709 randomly selected adults, 315 identified themselves as Hispanic/Latino, 323 identified themselves as Black, 254 identified themselves as Asian, and 779 identified themselves as White. In this survey, 86% of Black people said that they would welcome a White person into their families. Among Asians, 77% would welcome a White person into their families, 71% would welcome a Hispanic/Latino, and 66% would welcome a Black person. We are interested in finding the 95% confidence interval for the percent of all Black adults who would welcome a White person into their families. Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval. State the confidence interval. Sketch the graph. Refer to the information in . Construct three 95% confidence intervals. percent of all Asians who would welcome a White person into their families. percent of all Asians who would welcome a Hispanic/Latino into their families. percent of all Asians who would welcome a Black person into their families. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? (0.72, 0.82) (0.65, 0.76) (0.60, 0.72) Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Hispanic/Latino person into their families. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families. Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period. State the confidence interval. Sketch the graph. Explain what a “97% confidence interval” means for this study. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine . One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem. State the confidence interval. Sketch the graph. Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. X = the number of adult Americans who feel that crime is the main problem; p ′ = the proportion of adult Americans who feel that crime is the main problem Since we are estimating a proportion, given p ′ = 0.2 and n = 1000, the distribution we should use is N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) . CI: (0.18, 0.22) Answers may vary. One way to lower the sampling error is to increase the sample size. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%. Refer to . Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools. State the confidence interval. Sketch the graph. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. Use the following information to answer the next three exercises: According to a Field Poll, 79% of Georgia adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing Georgia. We wish to construct a 90% confidence interval for the true proportion of Georgia adults who feel that education and the schools is one of the top issues facing Georgia. A point estimate for the true population proportion is: 0.90 1.27 0.79 400 c A 90% confidence interval for the population proportion is _______. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not. Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. 0.6614 0.3386 173 338 a In a specific year, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error. Determine the estimated proportion from the sample. Determine the sample size. Identify CL and α . Calculate the error bound based on the information provided. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values. Create a confidence interval for the results of this study. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? A national survey of 1,000 adults was conducted by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. Find the point estimate and the error bound for this confidence interval. Can we (with 95% confidence) conclude that more than half of all American adults believe this? Use the point estimate from part a and n = 1,000 to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education. Can we (with 75% confidence) conclude that at least half of all American adults believe this? p ′ = (0 .55 + 0 .49) 2 = 0.52; EBP = 0.55 - 0.52 = 0.03 No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125 z α 2 = 1.150 . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) E B P = ( 1.150 ) 0.52 ( 0.48 ) 1 , 000 ≈ 0.018 ( p ′ - EBP , p ′ + EBP ) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence. Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. This survey was conducted through automated telephone interviews. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2020 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview? Binomial Distribution a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B ( n , p ). The mean is μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = n x p x q n − x . Error Bound for a Population Proportion (EBP) the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes.", "section": "A Confidence Interval for A Population Proportion", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Calculating the Sample Size n: Continuous and Binary Random Variables Continuous Random Variables Usually we have no control over the sample size of a data set. However, if we are able to set the sample size, as in cases where we are taking a survey, it is very helpful to know just how large it should be to provide the most information. Sampling can be very costly in both time and product. Simple telephone surveys will cost approximately $30.00 each, for example, and some sampling requires the destruction of the product. If we go back to our standardizing formula for the sampling distribution for means, we can see that it is possible to solve it for n. If we do this we have ( X ¯ - μ ) in the denominator. n = Z α 2 2 σ 2 ( X ¯ - μ ) 2 = Z α 2 2 σ 2 e 2 Because we have not taken a sample yet we do not know any of the variables in the formula except that we can set Z α to the level of confidence we desire just as we did when determining confidence intervals. If we set a predetermined acceptable error, or tolerance, for the difference between X ¯ and μ, called e in the formula, we are much further in solving for the sample size n. We still do not know the population standard deviation, σ. In practice, a pre-survey is usually done which allows for fine tuning the questionnaire and will give a sample standard deviation that can be used. In other cases, previous information from other surveys may be used for σ in the formula. While crude, this method of determining the sample size may help in reducing cost significantly. It will be the actual data gathered that determines the inferences about the population, so caution in the sample size is appropriate calling for high levels of confidence and small sampling errors. Binary Random Variables What was done in cases when looking for the mean of a distribution can also be done when sampling to determine the population parameter p for proportions. Manipulation of the standardizing formula for proportions gives: n = Z α 2 2 p q e 2 where e = (p′-p), and is the acceptable sampling error, or tolerance, for this application. This will be measured in percentage points. In this case the very object of our search is in the formula, p, and of course q because q =1-p. This result occurs because the binomial distribution is a one parameter distribution. If we know p then we know the mean and the standard deviation. Therefore, p shows up in the standard deviation of the sampling distribution which is where we got this formula. If, in an abundance of caution, we substitute 0.5 for p we will draw the largest required sample size that will provide the level of confidence specified by Zα and the tolerance we have selected. This is true because of all combinations of two fractions that add to one, the largest multiple is when each is 0.5. Without any other information concerning the population parameter p, this is the common practice. This may result in oversampling, but certainly not under sampling, thus, this is a cautious approach. There is an interesting trade-off between the level of confidence and the sample size that shows up here when considering the cost of sampling. shows the appropriate sample size at different levels of confidence and different level of the acceptable error, or tolerance. Required sample size (90%) Required sample size (95%) Tolerance level 1691 2401 2% 752 1067 3% 271 384 5% 68 96 10% This table is designed to show the maximum sample size required at different levels of confidence given an assumed p= 0.5 and q=0.5 as discussed above. The acceptable error, called tolerance in the table, is measured in plus or minus values from the actual proportion. For example, an acceptable error of 5% means that if the sample proportion was found to be 26 percent, the conclusion would be that the actual population proportion is between 21 and 31 percent with a 90 percent level of confidence if a sample of 271 had been taken. Likewise, if the acceptable error was set at 2%, then the population proportion would be between 24 and 28 percent with a 90 percent level of confidence, but would require that the sample size be increased from 271 to 1,691. If we wished a higher level of confidence, we would require a larger sample size. Moving from a 90 percent level of confidence to a 95 percent level at a plus or minus 5% tolerance requires changing the sample size from 271 to 384. A very common sample size often seen reported in political surveys is 384. With the survey results it is frequently stated that the results are good to a plus or minus 5% level of “accuracy”. Suppose a mobile phone company wants to determine the percentage of customers who would upgrade to the latest-version smartphone from their current smartphone. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers who would upgrade to the latest-version smartphone? From the problem, we know that the acceptable error, e , is 0.03 (3%=0.03) and z α 2 z 0.05 = 1.645 because the confidence level is 90%. The acceptable error, e , is the difference between the actual population proportion p , and the sample proportion we expect to get from the sample. However, in order to find n , we need to know the estimated (sample) proportion p ′. Remember that q ′ = 1 – p ′. But, we do not know p ′ yet. Since we multiply p ′ and q ′ together, we make them both equal to 0.5 because p ′ q ′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n . This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size n , use the formula and make the substitutions. n = z 2 p ′ q ′ e 2 gives n = 1.645 2 ( 0.5 ) ( 0.5 ) 0.03 2 = 751.7 Round the answer to the next higher value. The sample size should be 752 smartphone customers in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers who would upgrade their smartphone. Try It Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones? 271 customers should be surveyed.Check the Real Estate section in your local Chapter Review Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the relevant confidence interval formula for n to discover the size of the sample that is needed to achieve this goal: n = Z α 2 σ 2 ( x ¯ - μ ) 2 If the random variable is binary then the formula for the appropriate sample size to maintain a particular level of confidence with a specific tolerance level is given by n = Z α 2 pq e 2 Formula Review n = Z 2 σ 2 ( x ¯ - μ ) 2 = the formula used to determine the sample size ( n ) needed to achieve a desired margin of error at a given level of confidence for a continuous random variable n = Z α 2 pq e 2 = the formula used to determine the sample size if the random variable is binary Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. Identify the following: x - = _____ σ = _____ n = _____ 244 15 50 In words, define the random variables X and X - . Which distribution should you use for this problem? N ( 244 , 15 50 ) Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why? As the sample size increases, there will be less variability in the mean, so the interval size decreases. Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Identify the following: x - = _____ σ = _____ n = _____ In words, define the random variables X and X - . X is the time in minutes it takes to complete the U.S. Census short form. X - is the mean time it took a sample of 200 people to complete the U.S. Census short form. Which distribution should you use for this problem? Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound. CI: (7.9441, 8.4559) If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why? The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases. Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not? Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. Identify the following: x - = ______ σ = ______ n = ______ x - = 2.2 σ = 0.2 n = 20 In words, define the random variable X . In words, define the random variable X - . X - is the mean weight of a sample of 20 heads of lettuce. Which distribution should you use for this problem? Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. EBM = 0.07 CI: (2.1264, 2.2736) Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. In complete sentences, explain why the confidence interval in is larger than in . The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals. In complete sentences, give an interpretation of what the interval in means. What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same? The confidence level would increase. What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same? Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student. x - = _____ 30.4 n = _____ ________ = 15 σ In words, define the random variable X - . What is x - estimating? μ Is σ x known? As a result of your answer to , state the exact distribution to use when calculating the confidence interval. normal Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises . How much area is in both tails (combined)? α =________ How much area is in each tail? α 2 =________ 0.025 Identify the following specifications: lower limit upper limit error bound The 95% confidence interval is:__________________. (24.52,36.28) Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. In one complete sentence, explain what the interval means. We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28. Using the same mean, standard deviation, and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know? Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why? The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean. Find the value of the sample size needed to be 90% confident that the sample proportion and the population proportion are within 4% of each other. The sample proportion is 0.60. Note: Round all fractions up for n . Find the value of the sample size needed to be 95% confident that the sample proportion and the population proportion are within 2% of each other. The sample proportion is 0.650. Note: Round all fractions up for n . 2,185 Find the value of the sample size needed to be 96% confident that the sample proportion and the population proportion are within 5% of each other. The sample proportion is 0.70. Note: Round all fractions up for n . Find the value of the sample size needed to be 90% confident that the sample proportion and the population proportion are within 1% of each other. The sample proportion is 0.50. Note: Round all fractions up for n . 6,765 Find the value of the sample size needed to be 94% confident that the sample proportion and the population proportion are within 2% of each other. The sample proportion is 0.65. Note: Round all fractions up for n . Find the value of the sample size needed to be 95% confident that the sample proportion and the population proportion are within 4% of each other. The sample proportion is 0.45. Note: Round all fractions up for n . 595 Find the value of the sample size needed to be 90% confident that the sample proportion and the population proportion are within 2% of each other. The sample proportion is 0.3. Note: Round all fractions up for n . Homework Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches. x - =________ σ =________ n =________ In words, define the random variables X and X - . Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean height of male Swedes. State the confidence interval. Sketch the graph. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why? 71 2.8 48 X is the height of a male Swede, and x _ is the mean height from a sample of 48 male Swedes. Normal. We know the standard deviation for the population, and the sample size is greater than 30. CI: (70.151, 71.85) The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean. Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal. In words, define the random variables X and X - . Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean length of engineering conferences. State the confidence interval. Sketch the graph. Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. x - =________ σ =________ n =________ In words, define the random variables X and X – . Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean time to complete the tax forms. State the confidence interval. Sketch the graph. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why? Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why? x ¯ = 23.6 σ = 7 n = 100 X is the time needed to complete an individual tax form. X ¯ is the mean time to complete tax forms from a sample of 100 customers. N ( 23.6 , 7 100 ) because we know sigma. (22.228, 24.972) It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce. x – =________ σ =________ s x =________ In words, define the random variable X . In words, define the random variable X - . Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean weight of the candies. State the confidence interval. Sketch the graph. Construct a 98% confidence interval for the population mean weight of the candies. State the confidence interval. Sketch the graph. Calculate the error bound. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e. In complete sentences, give an interpretation of what the interval in part f means. A camp director is interested in the mean number of letters each child sends during their camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8. x - =________ σ =________ n =________ Define the random variables X and X - in words. Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean number of letters campers send home. State the confidence interval. Sketch the graph. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why? 7.9 2.5 20 X is the number of letters a single camper will send home. X ¯ is the mean number of letters sent home from a sample of 20 campers. N 7.9 ( 2.5 20 ) CI: (6.98, 8.82) The error bound and confidence interval will decrease. What is meant by the term “90% confident” when constructing a confidence interval for a mean? If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During a certain campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200. $3,600 $1,243,900 $10,900 $385,200 $581,500 $7,400 $2,900 $400 $3,714,500 $632,500 $391,000 $467,400 $56,800 $5,800 $405,200 $733,200 $8,000 $468,700 $75,200 $41,000 $13,300 $9,500 $953,800 $1,113,500 $1,109,300 $353,900 $986,100 $88,600 $378,200 $13,200 $3,800 $745,100 $5,800 $3,072,100 $1,626,700 $512,900 $2,309,200 $6,600 $202,400 $15,800 Find the point estimate for the population mean. Using 95% confidence, calculate the error bound. Create a 95% confidence interval for the mean total individual contributions. Interpret the confidence interval in the context of the problem. x - = $568,873 CL = 0.95 α = 1 – 0.95 = 0.05 z α 2 = 1.96 EBM = z 0.025 σ n = 1.96 909200 40 = $281,764 x - − EBM = 568,873 − 281,764 = 287,109 x - + EBM = 568,873 + 281,764 = 850,637 We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The most recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income. The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure? If the confidence interval is change to a higher probability, would this cause a lower, or a higher, minimum sample size? Higher If the tolerance is reduced by half, how would this affect the minimum sample size? It would increase to four times the prior value. If the value of p is reduced, would this necessarily reduce the sample size needed? No, It could have no affect if it were to change to 1 – p , for example. If it gets closer to 0.5 the minimum sample size would increase. Is it acceptable to use a higher sample size than the one calculated by z 2 p q e 2 ? Yes A company has been running an assembly line with 97.42%% of the products made being acceptable. Then, a critical piece broke down. After the repairs the decision was made to see if the number of defective products made was still close enough to the long standing production quality. Samples of 500 pieces were selected at random, and the defective rate was found to be 0.025%. Is this sample size adequate to claim the company is checking within the 90% confidence interval? The 95% confidence interval? No No", "section": "Calculating the Sample Size n: Continuous and Binary Random Variables", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (credit: modification of work “black and white, view, dog, animal, fur, mammal, garden, head, vertebrate, beautiful, elegant, rush, attention, dalmatian, dog breed, dog head, noble, stains, animal portrait, head drawing, spotty, hundeportrait, dalmatians”/ Pxhere, Public Domain) Now we are down to the bread and butter work of the statistician: developing and testing hypotheses. It is important to put this material in a broader context so that the method by which a hypothesis is formed is understood completely. Using textbook examples often clouds the real source of statistical hypotheses. Statistical testing is part of a much larger process known as the scientific method. This method was developed more than two centuries ago as the accepted way that new knowledge could be created. Until then, and unfortunately even today, among some, \"knowledge\" could be created simply by some authority saying something was so, ipso dicta . Superstition and conspiracy theories were (are?) accepted uncritically. The scientific method, briefly, states that only by following a careful and specific process can some assertion be included in the accepted body of knowledge. This process begins with a set of assumptions upon which a theory, sometimes called a model, is built. This theory, if it has any validity, will lead to predictions; what we call hypotheses. As an example, in Microeconomics the theory of consumer choice begins with certain assumption concerning human behavior. From these assumptions a theory of how consumers make choices using indifference curves and the budget line. This theory gave rise to a very important prediction, namely, that there was an inverse relationship between price and quantity demanded. This relationship was known as the demand curve. The negative slope of the demand curve is really just a prediction, or a hypothesis, that can be tested with statistical tools. Unless hundreds and hundreds of statistical tests of this hypothesis had not confirmed this relationship, the so-called Law of Demand would have been discarded years ago. This is the role of statistics, to test the hypotheses of various theories to determine if they should be admitted into the accepted body of knowledge; how we understand our world. Once admitted, however, they may be later discarded if new theories come along that make better predictions. Not long ago two scientists claimed that they could get more energy out of a process than was put in. This caused a tremendous stir for obvious reasons. They were on the cover of Time and were offered extravagant sums to bring their research work to private industry and any number of universities. It was not long until their work was subjected to the rigorous tests of the scientific method and found to be a failure. No other lab could replicate their findings. Consequently they have sunk into obscurity and their theory discarded. It may surface again when someone can pass the tests of the hypotheses required by the scientific method, but until then it is just a curiosity. Many pure frauds have been attempted over time, but most have been found out by applying the process of the scientific method. This discussion is meant to show just where in this process statistics falls. Statistics and statisticians are not necessarily in the business of developing theories, but in the business of testing others' theories. Hypotheses come from these theories based upon an explicit set of assumptions and sound logic. The hypothesis comes first, before any data are gathered. Data do not create hypotheses; they are used to test them. If we bear this in mind as we study this section the process of forming and testing hypotheses will make more sense. One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about the value of a specific parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year. A statistician will make a decision about these claims. This process is called \" hypothesis testing .\" A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests. Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: The desired confidence level. Information that is known about the distribution (for example, known standard deviation). The sample and its size. Hypothesis Testing Based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Null and Alternative Hypotheses The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints. H 0 : The null hypothesis: It is a statement of no difference between the variables–they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action. H a : The alternative hypothesis: It is a claim about the population that is contradictory to H 0 and what we conclude when we cannot accept H 0 . This is usually what the researcher is trying to prove. The alternative hypothesis is the contender and must win with significant evidence to overthrow the status quo. This concept is sometimes referred to the tyranny of the status quo because as we will see later, to overthrow the null hypothesis takes usually 90 or greater confidence that this is the proper decision. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are \"cannot accept H 0 \" if the sample information favors the alternative hypothesis or \"do not reject H 0 \" or \"decline to reject H 0 \" if the sample information is insufficient to reject the null hypothesis. These conclusions are all based upon a level of probability, a significance level, that is set by the analyst. Table 9.1 presents the various hypotheses in the relevant pairs. For example, if the null hypothesis is equal to some value, the alternative has to be not equal to that value. H 0 H a equal (=) not equal (≠) greater than or equal to (≥) less than (<) less than or equal to (≤) more than (>) NOTE As a mathematical convention H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ .30 H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30 Try It A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses. We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are: H 0 : μ = 2.0 H a : μ ≠ 2.0 Try It We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66 We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are: H 0 : μ ≥ 5 H a : μ < 5 Try It We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45 Chapter Review In a hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using not equal, less than or greater than symbols, i.e., (≠, <, or > ). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words. The random variable is the mean Internet speed in Megabits per second. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses. The American family has an average of two children. What is the random variable? Describe in words. The random variable is the mean number of children an American family has. The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words. The random variable is the proportion of people picked at random in Times Square visiting the city. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses. In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses. H 0 : p = 0.42 H a : p < 0.42 Suppose that a previous study stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A revised study was then done to see if the mean time has increased. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal. H 0 : ________ H a : ________ A random survey of 75 people with student loans revealed that the mean length of time in repayment is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time in repayment could likely be 15 years, what would the null and alternative hypotheses be? H 0 : __________ H a : __________ H 0 : μ = 15 H a : μ ≠ 15 The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5% of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population, what would the null and alternative hypotheses be? H 0 : ________ H a : ________ Homework Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, H 0 , and the alternative hypothesis. H a , in terms of the appropriate parameter ( μ or p ). The mean number of years Americans work before retiring is 34. At most 60% of Americans vote in presidential elections. The mean starting salary for San Jose State University graduates is at least $100,000 per year. Twenty-nine percent of high school seniors get drunk each month. Fewer than 5% of adults ride the bus to work in Los Angeles. The mean number of cars a person owns in her lifetime is not more than ten. About half of Americans prefer to live away from cities, given the choice. Europeans have a mean paid vacation each year of six weeks. The chance of developing breast cancer is under 11% for females. Private universities' mean tuition cost is more than $20,000 per year. H 0 : μ = 34; H a : μ ≠ 34 H 0 : p ≤ 0.60; H a : p > 0.60 H 0 : μ ≥ 100,000; H a : μ < 100,000 H 0 : p = 0.29; H a : p ≠ 0.29 H 0 : p = 0.05; H a : p < 0.05 H 0 : μ ≤ 10; H a : μ > 10 H 0 : p = 0.50; H a : p ≠ 0.50 H 0 : μ = 6; H a : μ ≠ 6 H 0 : p ≥ 0.11; H a : p < 0.11 H 0 : μ ≤ 20,000; H a : μ > 20,000 Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin? The alternative hypothesis is: p < 0.30 p ≤ 0.30 p ≥ 0.30 p > 0.30 A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is: p = 0.20 p > 0.20 p < 0.20 p ≤ 0.20 c Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are: H o : x ¯ = 4.5, H a : x ¯ > 4.5 H o : μ ≥ 4.5, H a : μ < 4.5 H o : μ = 4.75, H a : μ > 4.75 H o : μ = 4.5, H a : μ > 4.5 References Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm. Hypothesis a statement about the value of a population parameter, in case of two hypotheses, the statement assumed to be true is called the null hypothesis (notation H 0 ) and the contradictory statement is called the alternative hypothesis (notation H a ).", "section": "Null and Alternative Hypotheses", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Outcomes and the Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis H 0 and the decision to reject or not. The outcomes are summarized in the following table: Statistical Decision H 0 is actually... True False Cannot reject H 0 Correct outcome Type II error Cannot accept H 0 Type I error Correct outcome The four possible outcomes in the table are: The decision is cannot reject H 0 when H 0 is true (correct decision). The decision is cannot accept H 0 when H 0 is true (incorrect decision known as a Type I error ). This case is described as \"rejecting a good null\". As we will see later, it is this type of error that we will guard against by setting the probability of making such an error. The goal is to NOT take an action that is an error. The decision is cannot reject H 0 when, in fact, H 0 is false (incorrect decision known as a Type II error ). This is called \"accepting a false null\". In this situation you have allowed the status quo to remain in force when it should be overturned. As we will see, the null hypothesis has the advantage in competition with the alternative. The decision is cannot accept H 0 when H 0 is false ( correct decision ). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. α = probability of a Type I error = P (Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true: rejecting a good null. β = probability of a Type II error = P (Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. (1 − β ) is called the Power of the Test . α and β should be as small as possible because they are probabilities of errors. Statistics allows us to set the probability that we are making a Type I error. The probability of making a Type I error is α. Recall that the confidence intervals in the last unit were set by choosing a value called Z α (or t α ) and the alpha value determined the confidence level of the estimate because it was the probability of the interval failing to capture the true mean (or proportion parameter p). This alpha and that one are the same. The easiest way to see the relationship between the alpha error and the level of confidence is with the following figure. In the center of is a normally distributed sampling distribution marked H 0 . This is a sampling distribution of X ¯ and by the Central Limit Theorem it is normally distributed. The distribution in the center is marked H 0 and represents the distribution for the null hypotheses H 0 : µ = 100. This is the value that is being tested. The formal statements of the null and alternative hypotheses are listed below the figure. The distributions on either side of the H 0 distribution represent distributions that would be true if H 0 is false, under the alternative hypothesis listed as H a . We do not know which is true, and will never know. There are, in fact, an infinite number of distributions from which the data could have been drawn if H a is true, but only two of them are on representing all of the others. To test a hypothesis we take a sample from the population and determine if it could have come from the hypothesized distribution with an acceptable level of significance. This level of significance is the alpha error and is marked on as the shaded areas in each tail of the H 0 distribution. (Each area is actually α/2 because the distribution is symmetrical and the alternative hypothesis allows for the possibility for the value to be either greater than or less than the hypothesized value--called a two-tailed test). If the sample mean marked as X ¯ 1 is in the tail of the distribution of H 0 , we conclude that the probability that it could have come from the H 0 distribution is less than alpha. We consequently state, \"the null hypothesis cannot be accepted with (α) level of significance\". The truth may be that this X ¯ 1 did come from the H 0 distribution, but from out in the tail. If this is so then we have falsely rejected a true null hypothesis and have made a Type I error. What statistics has done is provide an estimate about what we know, and what we control, and that is the probability of us being wrong, α. We can also see in that the sample mean could be really from an H a distribution, but within the boundary set by the alpha level. Such a case is marked as X ¯ 2 . There is a probability that X ¯ 2 actually came from H a but shows up in the range of H 0 between the two tails. This probability is the beta error, the probability of accepting a false null. Our problem is that we can only set the alpha error because there are an infinite number of alternative distributions from which the mean could have come that are not equal to H 0 . As a result, the statistician places the burden of proof on the alternative hypothesis. That is, we will not reject a null hypothesis unless there is a greater than 90, or 95, or even 99 percent probability that the null is false: the burden of proof lies with the alternative hypothesis. This is why we called this the tyranny of the status quo earlier. By way of example, the American judicial system begins with the concept that a defendant is \"presumed innocent\". This is the status quo and is the null hypothesis. The judge will tell the jury that they cannot find the defendant guilty unless the evidence indicates guilt beyond a \"reasonable doubt\" which is usually defined in criminal cases as 95% certainty of guilt. If the jury cannot accept the null, innocent, then action will be taken, jail time. The burden of proof always lies with the alternative hypothesis. (In civil cases, the jury needs only to be more than 50% certain of wrongdoing to find culpability, called \"a preponderance of the evidence\"). The example above was for a test of a mean, but the same logic applies to tests of hypotheses for all statistical parameters one may wish to test. The following are examples of Type I and Type II errors. Suppose the null hypothesis, H 0 , is: Navah's rock climbing equipment is safe. Type I error : Navah thinks that his rock climbing equipment may not be safe when, in fact, it really is safe. Type II error : Navah thinks that her rock climbing equipment may be safe when, in fact, it is not safe. α = probability that Navah thinks her rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Navah thinks her rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Navah thinks her rock climbing equipment is safe, she will go ahead and use it.) This is a situation described as \"accepting a false null\". Try It Suppose the null hypothesis, H 0 , is: the blood cultures contain no traces of pathogen X . State the Type I and Type II errors. Suppose the null hypothesis, H 0 , is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital. This is the status quo and requires no action if it is true. If the null hypothesis cannot be accepted then action is required and the hospital will begin appropriate procedures. Type I error : The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error : The emergency crew does not know if the victim is alive when, in fact, the victim is dead. α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P (Type I error). β = probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P (Type II error). The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.) Try It Suppose the null hypothesis, H 0 , is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II? The error with the greater consequence is the Type II error: the patient will be thought well when, in fact, he is sick, so he will not get treatment. A company called Genetic Labs claims to be able to increase the likelihood that a pregnancy will result in a male being born. Statisticians want to test the claim. Suppose that the null hypothesis, H 0 , is: Genetic Labs has no effect on sex outcome. The status quo is that the claim is false. The burden of proof always falls to the person making the claim, in this case the Genetics Lab. Type I error : This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that Genetic Labs influences the sex outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α . Type II error : This results when we fail to reject a false null hypothesis. In context, we would state that Genetic Labs does not influence the sex outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β . The error of greater consequence would be the Type I error since people would use the Genetic Labs product in hopes of increasing the chances of having a male. Try It “Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence. In this scenario, an appropriate null hypothesis would be H 0 : the mean level of toxins is at most 800 μ g, H 0 : μ 0 ≤ 800 μ g. Type I error : The DMF believes that toxin levels are still too high when, in fact, toxin levels are at most 800 μ g. The DMF continues the harvesting ban. Type II error : The DMF believes that toxin levels are within acceptable levels (are at least 800 μ g) when, in fact, toxin levels are still too high (more than 800 μ g). The DMF lifts the harvesting ban. This error could be the most serious. If the ban is lifted and clams are still toxic, consumers could possibly eat tainted food. In summary, the more dangerous error would be to commit a Type II error, because this error involves the availability of tainted clams for consumption. A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious? Type I : A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%. Type II : A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%. In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. Try It Determine both Type I and Type II errors for the following scenario: Assume a null hypothesis, H 0 , that states the percentage of adults with jobs is at least 88%. Identify the Type I and Type II errors from these four statements. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88% Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%. Chapter Review In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters α and β , for a Type I and a Type II error respectively. The power of the test, 1 – β , quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable. The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000. A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences. For Exercise 9.12 , what are α and β in words? α = the probability that you think the bag cannot withstand -15 degrees F, when in fact it can β = the probability that you think the bag can withstand -15 degrees F, when in fact it cannot In words, describe 1 – β For Exercise 9.12 . A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H 0 , is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences. Type I: The procedure will go well, but the doctors think it will not. Type II: The procedure will not go well, but the doctors think it will. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H 0 , is: the surgical procedure will go well. Which is the error with the greater consequence? The power of a test is 0.981. What is the probability of a Type II error? 0.019 A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H 0 , is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H 0 , is: the sample does not contain E-coli. The probability that the sample does not contain E-coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E-coli, but the microbiologist thinks it does not is 0.002. What is the power of this test? 0.998 A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H 0 , is: the sample contains E-coli. Which is the error with the greater consequence? Homework State the Type I and Type II errors in complete sentences given the following statements. The mean number of years Americans work before retiring is 34. At most 60% of Americans vote in presidential elections. The mean starting salary for San Jose State University graduates is at least $100,000 per year. Twenty-nine percent of high school seniors get drunk each month. Fewer than 5% of adults ride the bus to work in Los Angeles. The mean number of cars a person owns in their lifetime is not more than ten. About half of Americans prefer to live away from cities, given the choice. Europeans have a mean paid vacation each year of six weeks. The chance of developing breast cancer is under 11% for females. Private universities mean tuition cost is more than $20,000 per year. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years. Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do. Type I error: We conclude that the mean number of cars a person owns in their lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in their lifetime is not more than 10 when, in fact, it is more than 10. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of females who develop breast cancer is at least 11%, when in fact it is less than 11%. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000. For statements a-j in Exercise 9.49 , answer the following in complete sentences. State a consequence of committing a Type I error. State a consequence of committing a Type II error. When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error? To conclude the drug is safe when in, fact, it is unsafe. Not to conclude the drug is safe when, in fact, it is safe. To conclude the drug is safe when, in fact, it is safe. Not to conclude the drug is unsafe when, in fact, it is unsafe. b A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Marvel Universe movie. The instructor surveys 84 of students in the class and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________. at least 20%, when in fact, it is less than 20%. 20%, when in fact, it is 20%. less than 20%, when in fact, it is at least 20%. less than 20%, when in fact, it is less than 20%. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours is more than seven hours. is at most seven hours. is at least seven hours. is less than seven hours. d Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is: to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same to conclude that the mean hours per week currently is 4.5, when in fact, it is higher to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher Type I Error The decision is to reject the null hypothesis when, in fact, the null hypothesis is true. Type II Error The decision is not to reject the null hypothesis when, in fact, the null hypothesis is false.", "section": "Outcomes and the Type I and Type II Errors", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Probability Distribution Needed for Hypothesis Testing Earlier in the course, we discussed sampling distributions. Particular distributions are associated with various types of hypothesis testing. The following table summarizes various hypothesis tests and corresponding probability distributions that will be used to conduct the test (based on the assumptions shown below): Type of Hypothesis Test Population Parameter Estimated value (point estimate) Probability Distribution Used Hypothesis test for the mean, when the population standard deviation is known Population mean μ Sample mean x ¯ Normal distribution, X ¯ ~ N ( μ X , σ X n ) Hypothesis test for the mean, when the population standard deviation is unknown and the distribution of the sample mean is approximately normal Population mean μ Sample mean x ¯ Student’s t-distribution, t d f Hypothesis test for proportions Population proportion p Sample proportion p ' Normal distribution, P ' ~ N ( p , p · q n ) Assumptions When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed , or your sample size is sufficiently large. You know the value of the population standard deviation , which, in reality, is rarely known. When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t -test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t -test will work even if the population is not approximately normally distributed). When you perform a hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution : there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( n p > 5 and n q > 5 ). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and σ = p q n . Remember that q = 1 - p . Hypothesis Test for the Mean Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means. Z c = x ¯ - μ 0 σ / n The standardizing formula cannot be solved as it is because we do not have μ, the population mean. However, if we substitute in the hypothesized value of the mean, μ 0 in the formula as above, we can compute a Z value. This is the test statistic for a test of hypothesis for a mean and is presented in . We interpret this Z value as the associated probability that a sample with a sample mean of X ¯ could have come from a distribution with a population mean of H 0 and we call this Z value Z c for “calculated”. and show this process. In two of the three possible outcomes are presented. X ¯ 1 and X ¯ 3 are in the tails of the hypothesized distribution of H 0 . Notice that the horizontal axis in the top panel is labeled X ¯ 's. This is the same theoretical distribution of X ¯ 's, the sampling distribution, that the Central Limit Theorem tells us is normally distributed. This is why we can draw it with this shape. The horizontal axis of the bottom panel is labeled Z and is the standard normal distribution. Z α 2 and -Z α 2 , called the critical values , are marked on the bottom panel as the Z values associated with the probability the analyst has set as the level of significance in the test, (α). The probabilities in the tails of both panels are, therefore, the same. Notice that for each X ¯ there is an associated Z c , called the calculated Z, that comes from solving the equation above. This calculated Z is nothing more than the number of standard deviations that the hypothesized mean is from the sample mean. If the sample mean falls \"too many\" standard deviations from the hypothesized mean we conclude that the sample mean could not have come from the distribution with the hypothesized mean, given our pre-set required level of significance. It could have come from H 0 , but it is deemed just too unlikely. In both X ¯ 1 and X ¯ 3 are in the tails of the distribution. They are deemed \"too far\" from the hypothesized value of the mean given the chosen level of alpha. If in fact this sample mean it did come from H 0 , but from in the tail, we have made a Type I error: we have rejected a good null. Our only real comfort is that we know the probability of making such an error, α, and we can control the size of α. shows the third possibility for the location of the sample mean, x _ . Here the sample mean is within the two critical values. That is, within the probability of (1-α) and we cannot reject the null hypothesis. This gives us the decision rule for testing a hypothesis for a two-tailed test: Decision rule: two-tail test If | Z c | < Z α 2 : then do not REJECT H 0 If | Z c | > Z α 2 : then REJECT H 0 This rule will always be the same no matter what hypothesis we are testing or what formulas we are using to make the test. The only change will be to change the Z c to the appropriate symbol for the test statistic for the parameter being tested. Stating the decision rule another way: if the sample mean is unlikely to have come from the distribution with the hypothesized mean we cannot accept the null hypothesis. Here we define \"unlikely\" as having a probability less than alpha of occurring. P-Value Approach An alternative decision rule can be developed by calculating the probability that a sample mean could be found that would give a test statistic larger than the test statistic found from the current sample data assuming that the null hypothesis is true. Here the notion of \"likely\" and \"unlikely\" is defined by the probability of drawing a sample with a mean from a population with the hypothesized mean that is either larger or smaller than that found in the sample data. Simply stated, the p-value approach compares the desired significance level, α, to the p-value which is the probability of drawing a sample mean further from the hypothesized value than the actual sample mean. A large p -value calculated from the data indicates that we should not reject the null hypothesis . The smaller the p -value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. The relationship between the decision rule of comparing the calculated test statistics, Z c , and the Critical Value, Z α , and using the p -value can be seen in . The calculated value of the test statistic is Z c in this example and is marked on the bottom graph of the standard normal distribution because it is a Z value. In this case the calculated value is in the tail and thus we cannot accept the null hypothesis, the associated X ¯ is just too unusually large to believe that it came from the distribution with a mean of µ 0 with a significance level of α. If we use the p -value decision rule we need one more step. We need to find in the standard normal table the probability associated with the calculated test statistic, Z c . We then compare that to the α associated with our selected level of confidence. In we see that the p -value is less than α and therefore we cannot accept the null. We know that the p -value is less than α because the area under the p-value is smaller than α/2. It is important to note that two researchers drawing randomly from the same population may find two different P-values from their samples. This occurs because the P-value is calculated as the probability in the tail beyond the sample mean assuming that the null hypothesis is correct. Because the sample means will in all likelihood be different this will create two different P-values. Nevertheless, the conclusions as to the null hypothesis should be different with only the level of probability of α. Here is a systematic way to make a decision of whether you cannot accept or cannot reject a null hypothesis if using the p -value and a preset or preconceived α (the \" significance level \"). A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. In any case, the value of α is the decision of the analyst. When you make a decision to reject or not reject H 0 , do as follows: If α > p -value, cannot accept H 0 . The results of the sample data are significant. There is sufficient evidence to conclude that H 0 is an incorrect belief and that the alternative hypothesis , H a , may be correct. If α ≤ p -value, cannot reject H 0 . The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis, H a , may be correct. In this case the status quo stands. When you \"cannot reject H 0 \", it does not mean that you should believe that H 0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H 0 . Remember that the null is the status quo and it takes high probability to overthrow the status quo. This bias in favor of the null hypothesis is what gives rise to the statement \"tyranny of the status quo\" when discussing hypothesis testing and the scientific method. Both decision rules will result in the same decision and it is a matter of preference which one is used. One and Two-tailed Tests The discussion of - was based on the null and alternative hypothesis presented in . This was called a two-tailed test because the alternative hypothesis allowed that the mean could have come from a population which was either larger or smaller than the hypothesized mean in the null hypothesis. This could be seen by the statement of the alternative hypothesis as μ ≠ 100, in this example. It may be that the analyst has no concern about the value being \"too\" high or \"too\" low from the hypothesized value. If this is the case, it becomes a one-tailed test and all of the alpha probability is placed in just one tail and not split into α/2 as in the above case of a two-tailed test. Any test of a claim will be a one-tailed test. For example, a car manufacturer claims that their Model 17B provides gas mileage of greater than 25 miles per gallon. The null and alternative hypothesis would be: H 0 : µ ≤ 25 H a : µ > 25 The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent confidence that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo. This is a one-tailed test and all of the alpha probability is placed in just one tail and not split into α/2 as in the above case of a two-tailed test. shows the two possible cases and the form of the null and alternative hypothesis that give rise to them. where μ 0 is the hypothesized value of the population mean. Test Statistics for Test of Means, Varying Sample Size, Population Standard Deviation Known or Unknown Sample size Test statistic < 30 (σ unknown) t c = X ¯ - μ 0 s / n < 30 (σ known) Z c = X ¯ - μ 0 σ / n > 30 (σ unknown) Z c = X ¯ - μ 0 s / n > 30 (σ known) Z c = X ¯ - μ 0 σ / n Effects of Sample Size on Test Statistic In developing the confidence intervals for the mean from a sample, we found that most often we would not have the population standard deviation, σ. If the sample size were less than 30, we could simply substitute the point estimate for σ, the sample standard deviation, s, and use the student's t -distribution to correct for this lack of information. When testing hypotheses we are faced with this same problem and the solution is exactly the same. Namely: If the population standard deviation is unknown, and the sample size is less than 30, substitute s, the point estimate for the population standard deviation, σ, in the formula for the test statistic and use the student's t -distribution. All the formulas and figures above are unchanged except for this substitution and changing the Z distribution to the student's t -distribution on the graph. Remember that the student's t -distribution can only be computed knowing the proper degrees of freedom for the problem. In this case, the degrees of freedom is computed as before with confidence intervals: df = (n-1). The calculated t-value is compared to the t-value associated with the pre-set level of confidence required in the test, t α , df found in the student's t tables. If we do not know σ, but the sample size is 30 or more, we simply substitute s for σ and use the normal distribution. summarizes these rules. A Systematic Approach for Testing a Hypothesis A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test. Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test? Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for businesses are 80, 90, 95, 98, and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error. Next, on the basis of the hypotheses and sample size, select the appropriate test statistic and find the relevant critical value: Z α , t α , etc. Drawing the relevant probability distribution and marking the critical value is always big help. Be sure to match the graph with the hypothesis, especially if it is a one-tailed test. Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated. Compare the calculated test statistic and the critical value. Marking these on the graph will give a good visual picture of the situation. There are now only two situations: The test statistic is in the tail: Cannot Accept the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data. The test statistic is not in the tail: Cannot Reject the null, the sample data are compatible with the hypothesized population parameter. Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 5 % level of significance we cannot accept the null hypotheses that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”. All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question. Chapter Review In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied. When testing for a single population mean: A Student's t -test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large. When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > 5 where n is the sample size, p is the probability of a success, and q is the probability of a failure. Formula Review Test Statistics for Test of Means, Varying Sample Size, Population Known or Unknown Sample size Test statistic < 30 (σ unknown) t c = X ¯ - μ 0 s / n < 30 (σ known) Z c = X ¯ - μ 0 σ / n > 30 (σ unknown) Z c = X ¯ - μ 0 s / n > 30 (σ known) Z c = X ¯ - μ 0 σ / n Which two distributions can you use for hypothesis testing for this chapter? A normal distribution or a Student’s t -distribution Which distribution do you use when you are testing a population mean and the population standard deviation is known? Assume sample size is large. Assume a normal distribution with n ≥ 30. Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume a normal distribution, with n ≥ 30. Use a Student’s t -distribution A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal. A population has a mean is 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test? a normal distribution for a single population mean It is thought that 42% of respondents in a taste test would prefer Brand A . In a particular test of 100 people, 39% preferred Brand A . What distribution should you use to perform a hypothesis test? You are performing a hypothesis test of a single population mean using a Student’s t -distribution. What must you assume about the distribution of the data? It must be approximately normally distributed. You are performing a hypothesis test of a single population mean using a Student’s t -distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq ? They must both be greater than five. You are performing a hypothesis test of a single population proportion. You find out that np is less than five. What must you do to be able to perform a valid hypothesis test? You are performing a hypothesis test of a single population proportion. The data come from which distribution? binomial distribution Homework It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X – ~ ________________ N ( 7.24 , 1.93 22 ) N ( 7.24 , 1.93 ) t 22 t 21 d Binomial Distribution a discrete random variable (RV) that arises from Bernoulli trials. There are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in n trials. The notation is: X ~ B(n, p) μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = ( n x ) p x q n − x . Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 , where μ is the mean of the distribution, and σ is the standard deviation, notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Student's t -Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. The major characteristics of the random variable (RV) are: It is continuous and assumes any real values. The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. It approaches the standard normal distribution as n gets larger. There is a \"family\" of t -distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items. Test Statistic The formula that counts the number of standard deviations on the relevant distribution that estimated parameter is away from the hypothesized value. Critical Value The t or Z value set by the researcher that measures the probability of a Type I error, α.", "section": "Probability Distribution Needed for Hypothesis Testing", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Full Hypothesis Test Examples Tests on Means Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean . H 0 : μ = 16.43 H a : μ < 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The \"<\" tells you this is left-tailed. Determine the distribution needed: Random variable: X ¯ = the mean time to swim the 25-yard freestyle. Distribution for the test: X ¯ is normal (population standard deviation is known: σ = 0.8) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15. Calculate the p -value using the normal distribution for a mean: p -value = P ( x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16. Graph: μ = 16.43 comes from H 0 . Our assumption is μ = 16.43. Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p -value: α = 0.05 p -value = 0.0187 α > p -value Make a decision: Since α > p -value, reject H 0 . This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds. Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles. The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) Try It The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal. First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion. Since the problem is about a mean, this is a test of a single population mean. H 0 : μ = 40 H a : μ > 40 p = 0.0062 Because p < α , we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance. Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the confidence level of 95%? H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not \"too high\" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis. Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 Critical value: t a = 1.753 with n-1 degrees of freedom= 15 The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards. Try It It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors. H 0 : μ = 5 H a : μ < 5 p = 0.0082 Because p < α , we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week. Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true). Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false). A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of confidence in the analysis should be 99%. Again we will follow the steps in our analysis of this problem. STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus: H 0 : μ = 8 H a : μ ≠ 8 STEP 2 : Decide the level of confidence and draw the graph showing the critical value. This problem has already set the level of confidence at 99%. The decision seems an appropriate one and shows the thought process when setting the confidence level at 1%. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table. We draw the graph and mark these points. STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph. Z c = x ¯ - μ 0 s n = 7.91 - 8 .173 35 = -3.07 STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis. STEP 5 : Reach a Conclusion Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of confidence we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of confidence we conclude that the machine is under filling the bottles and is in need of repair”. Try It A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2). Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 . Set up the Hypothesis Test: H 0 : μ = 475 H 0 : μ > 475 This is a right-tailed test. The random variable X is the mean time of employees working in a day. Distribution for the test: σ is known. X ~ N 475 , 45 20 x ¯ = 475 . 5 minutes (from the data). σ = 45 minutes. We assume μ = 475 minutes unless our data shows otherwise. The p -value is calculated: p -value = P ( x ¯ > 475 . 5 ) = 0 . 5044 Interpretation of p -value: If H 0 is true, then there is a 0.5044, or 50.44%, chance that the mean time of employees working is 475.5 minutes or more. Since a 50.44% chance is large enough, the mean time of 475.5 is not a rare event. Compare α and p values: α = 0 . 025 and p = 0 . 5044 Make a decision: Since α < p value, do not reject H 0 . Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean time of employees is more than 475 minutes. Hypothesis Test for Proportions Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p . The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size. When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np and σ = np . Remember that q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p . Again, we begin with the standardizing formula modified because this is the distribution of a binomial. Z = p' - p pq n Substituting p 0 , the hypothesized value of p , we have: Z c = p' - p 0 p 0 q 0 n This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms: Two-tailed test One-tailed test One-tailed test H 0 : p = p 0 H 0 : p ≤ p 0 H 0 : p ≥ p 0 H a : p ≠ p 0 H a : p > p 0 H a : p < p 0 The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is \"too many\" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is \"too many\" is pre-determined by the analyst depending on the level of significance required in the test. The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance. STEP 1 : Set the null and alternative hypothesis. H 0 : p = 0.50 H a : p ≠ 0.50 The words \"is the same or different from\" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.) STEP 2 : Decide the level of significance and draw the graph showing the critical value The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations. STEP 3 : Calculate the sample parameters and critical value of the test statistic. The test statistic is a normal distribution, Z, for testing proportions and is: Z = p' - p 0 p 0 q 0 n = .53 - .50 .5 ( .5 ) 100 = 0.60 For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations. STEP 4 : Compare the test statistic and the critical value. The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations. STEP 5 : Reach a conclusion The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data. Try It A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance. Since the problem is about percentages, this is a test of single population proportions. H 0 : p = 0.85 H a : p ≠ 0.85 p = 0.7554 Because p > α , we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%. Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones. Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions. H 0 : p = 0.3 H a : p ≠ 0.3 n = 150 p' = x n = 43 150 = 0.287 Z c = p' - p 0 p 0 q 0 n = 0.287 - 0.3 .3 ( .7 ) 150 = 0.347 Try It Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors. The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass. 1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Let’s follow a four-step process to answer this statistical question. State the Question : We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be H 0 : μ ≤ 1 H a : μ > 1 Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal. Do the calculations and draw the graph . State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one. Try It The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows: 205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205 Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal. The hypotheses, at a significance level of 0.1, will be as follows: H 0 : μ ≤ 200 H 0 : μ > 200 Since the standard deviation is not known, Student’s t -distribution is to be used. The data is input into the TI-83 as follows: Since the p -value is less than alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average boiling point is greater than 200. In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be H 0 : p ≤ 0.00034 H a : p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population. Try It In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. The test hypotheses are as follows: H 0 : p ≤ 0 . 00041 H 0 : p > 0 . 00041 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes disease-causing environments, we want to minimize the chances of incorrectly identifying causes of diseases. The testing will be done with x = 138 and n = 390,000. The sample is sufficiently large because we have np = 390,000(0.00041) = 159.9, nq = 390,000(0.99959) = 389,840.1, two independent outcomes, and a fixed probability of success p = 0.00041. Thus, we will be able to generalize our results to the population. The associated TI results are: Since the p -value = 0.0832 is greater than alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher skin diseases rates for moisturizer users. Chapter Review The hypothesis test itself has an established process. This can be summarized as follows: Determine H 0 and H a . Remember, they are contradictory. Determine the random variable. Determine the distribution for the test. Draw a graph and calculate the test statistic. Compare the calculated test statistic with the Z critical value determined by the level of significance required by the test and make a decision (cannot reject H 0 or cannot accept H 0 ), and write a clear conclusion using English sentences. Formula Review Test statistic for a hypothesis test of proportions: Z c = p ' – p 0 p 0 ( 1 – p 0 ) n Assume H 0 : μ = 9 and H a : μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume H 0 : μ ≤ 6 and H a : μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? Assume H 0 : p = 0.25 and H a : p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? This is a two-tailed test. Draw the general graph of a left-tailed test. Draw the graph of a two-tailed test. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use? Your friend claims that their mean golf score is 63. You want to show that it is higher than that. What type of test would you use? a right-tailed test A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use? You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use? a left-tailed test If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test? This is a two-tailed test. Homework A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim? H 0 : μ ≥ 50,000 H a : μ < 50,000 Let X ¯ = the average lifespan of a brand of tires. normal distribution z = -2.315 p -value = 0.0103 Answers may vary. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. (43,537, 49,463) From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level? H 0 : μ = $1.00 H a : μ ≠ $1.00 Let X ¯ = the average cost of a daily newspaper. normal distribution z = –0.866 p -value = 0.3865 Answers may vary. Alpha: 0.01 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.01. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. ($0.84, $1.06) An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten? H 0 : μ = 10 H a : μ ≠ 10 Let X ¯ the mean number of sick days an employee takes per year. Student’s t -distribution t = –1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten. (4.9443, 11.806) In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level? Your statistics instructor claims that 60% of the students who take their Elementary Statistics class go through life feeling more enriched. For some reason that the instructor can't quite figure out, most people don't believe them. You decide to check this out on your own. You randomly survey 64 of their past Elementary Statistics students and find that 34 feel more enriched as a result of taking the class. Now, what do you think? H 0 : p ≥ 0.6 H a : p < 0.6 Let P′ = the proportion of students who feel more enriched as a result of taking Elementary Statistics. normal for a single proportion 1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: There is insufficient evidence to conclude that less than 60 percent of the students feel more enriched. Confidence Interval: (0.409, 0.654) The “plus-4s” confidence interval is (0.411, 0.648) A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief. Refer to Exercise 9.63 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. H 0 : μ = 4 H a : μ ≠ 4 Let X ¯ the average I.Q. of a set of brown trout. two-tailed Student's t-test t = 1.95 p -value = 0.076 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is greater than 0.05 Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. (3.8865,5.9468) According to an article in Newsweek , the birth ratio of females to males is 100:105. In China, the birth ratio is 100: 114 (46.7% females). Suppose you want to explore the reported figures of the percent of females born in China. You conduct a study. In this study, you count the number of females and males born in 150 randomly chosen recent births. There are 60 females and 90 males born of the 150. Based on your study, do you believe that the percent of females born in China is 46.7? A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. H 0 : p ≥ 0.13 H a : p < 0.13 Let P′ = the proportion of Americans who have seen or sensed angels normal for a single proportion –2.688 p -value = 0.0036 Answers may vary. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%. (0, 0.0623). The“plus-4s” confidence interval is (0.0022, 0.0978) The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. They ask ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should they count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55. Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data. According to New York City's Community Health Survey, the most recent adult smoking rate is 12%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 12% or if it has decreased. H 0 : p = 0.12 H a : p < 0.12 Let P ' = the proportion of NYC residents that smoke. Normal for a single proportion 0.2207 p -value = 0.5873 Answers may vary alpha 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: The p -value is greater than 0.05. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14. Confidence interval: Confidence Interval: (0.0502, 0.2070): The “plus-4s” confidence interval (see Confidence Intervals ) is (0.0676, 0.2297). The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. They randomly survey 56 online students and find that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test. H 0 : μ = 69,110 H a : μ > 69,110 Let X ¯ = the mean salary in dollars for California registered nurses. Student's t -distribution t = 1.719 p -value: 0.0466 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. ($68,757, $73,485) La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old. Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin? After conducting the test, your decision and conclusion are Reject H 0 : There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Reject H 0 : There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. c A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is: There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is less than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is more than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is less than 20%. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is at least 20%. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion? There is enough evidence to conclude that the mean number of hours is more than 4.75 There is enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.75 c Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question. State the null and alternate hypothesis. State the p -value. State alpha. What is your decision? Write a conclusion. Answer any other questions asked in the problem. According to the Center for Disease Control website, in 2022 at least 2% of high school students reported they had smoked a cigarette. An Introduction to Statistics class conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. They surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? H 0 : p = 0.488 H a : p ≠ 0.488 p -value = 0.0114 alpha = 0.05 Reject the null hypothesis. There is enough evidence to conclude that the percent of families owning stocks is not 48.8%. The survey does not appear to be accurate. Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate? The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level in Kentucky? Are the results applicable across the country? Why? H 0 : p = 0.517 H a : p ≠ 0.517 p -value = 0.9203. alpha = 0.05. Do not reject the null hypothesis. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is different from the proportion of homes heated by natural gas across the country. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation. For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so they asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library? The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. H 0 : µ ≥ 11.52 H a : µ < 11.52 p -value = 0.000002 which is almost 0. alpha = 0.05. Reject the null hypothesis. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average. We would make the same conclusion if alpha was 1% because the p -value is almost 0. A survey finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities? A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals 3 2 1 3 7 2 9 4 6 6 8 0 5 6 4 2 1 3 4 1 At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year? H 0 : µ ≤ 5.8 H a : µ > 5.8 p -value = 0.9987 alpha = 0.05 Do not reject the null hypothesis. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year. According to the U.S. Census Bureau , the mean family size in the U.S. is 3.13. A sample of a college math class resulted in the following family sizes: 5 4 5 4 4 3 6 4 3 3 5 5 6 3 3 2 7 4 5 2 2 2 3 2 At α = 0.05 level, is the class’ mean family size greater than the national average? Does the census result remain valid? Why? The student academic group on a college campus claims that first-year students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 first-year students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? H 0 : µ ≥ 150 H a : µ < 150 p -value = 0.0622 alpha = 0.01 Do not reject the null hypothesis. At the 1% significance level, there is not enough evidence to conclude that first-year students study less than 2.5 hours per day, on average. The student academic group’s claim appears to be correct. References Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC. Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013). Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013). Data from Growing by Degrees by Allen and Seaman. Data from La Leche League International. Available online at http://www.lalecheleague.org/Law/BAFeb01.html. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013). Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013) Data from the U.S. Census Bureau, available online at http://quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013). Data from the United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/. Data from Toastmasters International. Available online at http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1. Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013). Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013). “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf. Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013). Central Limit Theorem Given a random variable (RV) with known mean μ and known standard deviation σ. We are sampling with size n and we are interested in two new RVs - the sample mean, X ¯ . If the size n of the sample is sufficiently large, then X ¯ ~ N ( μ , σ n ) . If the size n of the sample is sufficiently large, then the distribution of the sample means will approximate a normal distribution regardless of the shape of the population. The expected value of the mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean.", "section": "Full Hypothesis Test Examples", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River) you can use a slightly different technique when conducting a hypothesis test. (credit: modification of work “scrambled egg, toast and home-made berry jam (no added sugar) for breakfast” by Chloe Lim/ Flickr, CC BY 2.0) Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores. Many business applications require comparing two groups. It may be the investment returns of two different investment strategies, or the differences in production efficiency of different management styles. To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs . Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions of each group. Independent Groups two samples that are selected from two populations, and the values from one population are not related in any way to the values from the other population. Matched Pairs two samples that are dependent. Differences between a before and after scenario are tested by testing one population mean of differences.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Comparing Two Independent Population Means The comparison of two independent population means is very common and provides a way to test the hypothesis that the two groups differ from each other. Is the night shift less productive than the day shift, are the rates of return from fixed asset investments different from those from common stock investments, and so on? An observed difference between two sample means depends on both the means and the sample standard deviations. Very different means can occur by chance if there is great variation among the individual samples. The test statistic will have to account for this fact. The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula we will see later was developed by Aspin-Welch. When we developed the hypothesis test for the mean and proportions we began with the Central Limit Theorem. We recognized that a sample mean came from a distribution of sample means, and sample proportions came from the sampling distribution of sample proportions. This made our sample parameters, the sample means and sample proportions, into random variables. It was important for us to know the distribution that these random variables came from. The Central Limit Theorem gave us the answer: the normal distribution. Our Z and t statistics came from this theorem. This provided us with the solution to our question of how to measure the probability that a sample mean came from a distribution with a particular hypothesized value of the mean or proportion. In both cases that was the question: what is the probability that the mean (or proportion) from our sample data came from a population distribution with the hypothesized value we are interested in? Now we are interested in whether or not two samples have the same mean. Our question has not changed: Do these two samples come from the same population distribution? To approach this problem we create a new random variable. We recognize that we have two sample means, one from each set of data, and thus we have two random variables coming from two unknown distributions. To solve the problem we create a new random variable, the difference between the sample means. This new random variable also has a distribution and, again, the Central Limit Theorem tells us that this new distribution is normally distributed, regardless of the underlying distributions of the original data. A graph may help to understand this concept. Pictured are two distributions of data, X 1 and X 2 , with unknown means and standard deviations. The second panel shows the sampling distribution of the newly created random variable ( X ¯ 1 - X ¯ 2 ). This distribution is the theoretical distribution of many sample means from population 1 minus sample means from population 2. The Central Limit Theorem tells us that this theoretical sampling distribution of differences in sample means is normally distributed, regardless of the distribution of the actual population data shown in the top panel. Because the sampling distribution is normally distributed, we can develop a standardizing formula and calculate probabilities from the standard normal distribution in the bottom panel, the Z distribution. We have seen this same analysis before in The Central Limit Theorem Figure 7.2 . The Central Limit Theorem, as before, provides us with the standard deviation of the sampling distribution, and further, that the expected value of the mean of the distribution of differences in sample means is equal to the differences in the population means. Mathematically this can be stated: E ( µ x ¯ 1 - µ x ¯ 2 ) = µ 1 - µ 2 Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means , X ¯ 1 – X ¯ 2 . The standard error is: ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 We remember that substituting the sample variance for the population variance when we did not have the population variance was the technique we used when building the confidence interval and the test statistic for the test of hypothesis for a single mean back in Confidence Intervals and Hypothesis Testing with One Sample . The test statistic ( t -score) is calculated as follows: t c = ( x – 1 – x – 2 ) – δ 0 ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 where: s 1 and s 2 , the sample standard deviations, are estimates of σ 1 and σ 2 , respectively and σ 1 and σ 2 are the unknown population standard deviations. x – 1 and x – 2 are the sample means. μ 1 and μ 2 are the unknown population means. δ = the hypothesized differences between the two populations means. Substituting the sample variances of the two sample groups when the populations variances are unknown and the sample size is large, the t -distribution will provide good estimates using degrees of freedom: v = n 1 + n 2 - 2 . Large in this case would be greater than 100 observations from the combined two groups ( n 1 + n 2 ) although other statisticians suggest that the definition of large is greater than 30 observations for each of the two groups. The number of degrees of freedom ( df ) requires a somewhat complicated calculation. The df are not always a whole number. The test statistic above is approximated by the Student's t -distribution with df as follows: Degrees of Freedom The standard error is: d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 – 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 – 1 ) ( ( s 2 ) 2 n 2 ) 2 When both sample sizes n 1 and n 2 are 30 or larger, the Student's t approximation is very good. If each sample has more than 30 observations then the degrees of freedom can be calculated as n 1 + n 2 − 2 . The format of the sampling distribution, differences in sample means, specifies that the format of the null and alternative hypothesis is: The test hypothesis for differences between two means H 0 : µ 1 - µ 2 = δ 0 H a : µ 1 - µ 2 ≠ δ 0 where δ 0 is the hypothesized difference between the two means. If the question is simply “is there any difference between the means?” then δ 0 = 0 and the null and alternative hypotheses becomes: H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 An example of when δ 0 might not be zero is when the comparison of the two groups requires a specific difference for the decision to be meaningful. Imagine that you are making a capital investment. You are considering changing from your current model machine to another. You measure the productivity of your machines by the speed they produce the product. It may be that a contender to replace the old model is faster in terms of product throughput, but is also more expensive. The second machine may also have more maintenance costs, setup costs, etc. The null hypothesis would be set up so that the new machine would have to be better than the old one by enough to cover these extra costs in terms of speed and cost of production. This form of the null and alternative hypothesis shows how valuable this particular hypothesis test can be. For most of our work we will be testing simple hypotheses asking if there is any difference between the two distribution means. Independent groups The Kona Iki Corporation produces coconut milk. They take coconuts and extract the milk inside by drilling a hole and pouring the milk into a vat for processing. They have both a day shift (called the B shift) and a night shift (called the G shift) to do this part of the process. They would like to know if the day shift and the night shift are equally efficient in processing the coconuts. A study is done sampling 9 shifts of the G shift and 16 shifts of the B shift. The results of the number of hours required to process 100 pounds of coconuts is presented in . A study is done and data are collected, resulting in the data in . Sample Size Average Number of Hours to Process 100 Pounds of Coconuts Sample Standard Deviation G Shift 9 2 0.866 B Shift 16 3.2 1.00 Is there a difference in the mean amount of time for each shift to process 100 pounds of coconuts? Test at the 5% level of significance. The population standard deviations are not known and cannot be assumed to equal each other. Let g be the subscript for the G Shift and b be the subscript for the B Shift. Then, μ g is the population mean for G Shift and μ b is the population mean for B Shift. This is a test of two independent groups , two population means . Random variable : X ¯ g − X ¯ b = difference in the sample mean amount of time between the G Shift and the B Shift takes to process the coconuts. H 0 : μ g = μ b H 0 : μ g – μ b = 0 H a : μ g ≠ μ b H a : μ g – μ b ≠ 0 The words \"the same\" tell you H 0 has an \"=\". Since there are no other words to indicate H a , is either faster or slower. This is a two tailed test. Distribution for the test: Use t df where df is calculated using the df formula for independent groups, two population means above. Using a calculator, df is approximately 18.8462. Graph: t c = ( X ¯ 1 − X ¯ 2 ) − δ 0 S 1 2 n 1 + S 2 2 n 2 = -3.01 We next find the critical value on the t-table using the degrees of freedom from above. The critical value, 2.093, is found in the .025 column, this is α/2, at 19 degrees of freedom. (The convention is to round up the degrees of freedom to make the conclusion more conservative.) Next we calculate the test statistic and mark this on the t-distribution graph. Make a decision: Since the calculated t-value is in the tail we cannot accept the null hypothesis that there is no difference between the two groups. The means are different. The graph has included the sampling distribution of the differences in the sample means to show how the t-distribution aligns with the sampling distribution data. We see in the top panel that the calculated difference in the two means is -1.2 and the bottom panel shows that this is 3.01 standard deviations from the mean. Typically we do not need to show the sampling distribution graph and can rely on the graph of the test statistic, the t-distribution in this case, to reach our conclusion. Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that the G Shift takes to process 100 pounds of coconuts is different from the B Shift (mean number of hours for the B Shift is greater than the mean number of hours for the G Shift). Try It Two samples are shown in . Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 5 1 Population B 16 4.7 1.2 NOTE When the sum of the sample sizes is larger than 30 ( n 1 + n 2 > 30) you can use the normal distribution to approximate the Student's t . Use the Student's t distribution, however, whenever possible. A study is done to determine if Company A retains its workers longer than Company B. It is believed that Company A has a higher retention than Company B. The study finds that in a sample of 11 workers at Company A their average time with the company is four years with a standard deviation of 1.5 years. A sample of 9 workers at Company B finds that the average time with the company was 3.5 years with a standard deviation of 1 year. Test this proposition at the 1% level of significance. a. Is this a test of two means or two proportions? a. two means because time is a continuous random variable. b. Are the populations standard deviations known or unknown? b. unknown c. Which distribution do you use to perform the test? c. Student's t d. What is the random variable? d. X ¯ A - X ¯ B e. What are the null and alternate hypotheses? e. H Ø : μ A ≤ μ B H a : μ A > μ B f. Is this test right-, left-, or two-tailed? f. right one-tailed test g. What is the value of the test statistic? t c = ( X ¯ 1 − X ¯ 2 ) − δ 0 S 1 2 n 1 + S 2 2 n 2 = 0.89 h. Can you accept/reject the null hypothesis? h. Cannot reject the null hypothesis that there is no difference between the two groups. Test statistic is not in the tail. The critical value of the t -distribution is 2.764 with 10 degrees of freedom. This example shows how difficult it is to reject a null hypothesis with a very small sample. The critical values require very large test statistics to reach the tail. i. Conclusion: i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the retention of workers at Company A is longer than Company B, on average. Try It A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed. Are the population standard deviations known? Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion? An interesting research question is the effect, if any, that different types of teaching formats have on the grade outcomes of students. To investigate this issue one sample of students' grades was taken from a hybrid class and another sample taken from a standard lecture format class. Both classes were for the same subject. The mean course grade in percent for the 35 hybrid students is 74 with a standard deviation of 16. The mean grades of the 40 students form the standard lecture class was 76 percent with a standard deviation of 9. Test at 5% to see if there is any significant difference in the population mean grades between standard lecture course and hybrid class. We begin by noting that we have two groups, students from a hybrid class and students from a standard lecture format class. We also note that the random variable, what we are interested in, is students' grades, a continuous random variable. We could have asked the research question in a different way and had a binary random variable. For example, we could have studied the percentage of students with a failing grade, or with an A grade. Both of these would be binary and thus a test of proportions and not a test of means as is the case here. Finally, there is no presumption as to which format might lead to higher grades so the hypothesis is stated as a two-tailed test. H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 As would virtually always be the case, we do not know the population variances of the two distributions and thus our test statistic is: t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 s 2 n 1 + s 2 n 2 = ( 74 − 76 ) − 0 16 2 35 + 9 2 40 = −0.65 To determine the critical value of the Student's t we need the degrees of freedom. For this case we use: df = n1 + n2 - 2 = 35 + 40 -2 = 73. This is large enough to consider it the normal distribution thus t a/2 = 1.96. Again as always we determine if the calculated value is in the tail determined by the critical value. In this case we do not even need to look up the critical value: the calculated value of the difference in these two average grades is not even one standard deviation apart. Certainly not in the tail. Conclusion: Cannot reject the null at α=5%. Therefore, evidence does not exist to prove that the grades in hybrid and standard classes differ. Try It Two professors, A and B, teach classes on the same subjects. Scores of 10 students from each class are selected randomly. The final exam scores of the students are as follows: Professor A: 97 62 73 58 84 74 66 93 73 85 Professor B: 85 64 74 55 76 67 72 84 71 98 Professor A says that the mean score of their class is more than the mean of professor B’s class. Is professor A correct? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right-, left-, or two-tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ___ (is/is not) sufficient evidence to conclude that ___. two means unknown Student’s t X 1 - X 2 H 0 : μ 1 = μ 2 Null hypothesis: the means of final exam scores are equal for both professors. H 0 : μ 2 < μ 1 Alternative hypothesis: the mean of final exam scores for class of professor B is less than mean of final exam scores for class of professor A. left-tailed p -value = 0.03 Reject the null hypothesis. Professor A is correct. The evidence shows that the mean of final exam scores of the class of professor A is more than that of the class of professor B. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of final exam scores of the class of professor A is more than that of the class of professor B. References Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com Data from Microsoft Bookshelf . Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013). “List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age (accessed June 17, 2013). “Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013). “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013). Chapter Review Two population means from independent samples where the population standard deviations are not known Random Variable: X – 1 − X – 2 = the difference of the sampling means Distribution: Student's t -distribution with degrees of freedom (variances not pooled) Formula Review Standard error: SE = ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Test statistic ( t -score): t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Degrees of freedom: d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 − 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 − 1 ) ( ( s 2 ) 2 n 2 ) 2 where: s 1 and s 2 are the sample standard deviations, and n 1 and n 2 are the sample sizes. x ¯ 1 and x ¯ 2 are the sample means. Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for independent group means, population standard deviations, and/or variances known independent group means, population standard deviations, and/or variances unknown matched or paired samples single mean two proportions single proportion It is believed that 70% of men pass their drivers test in the first attempt, while 65% of women pass the test in the first attempt. Of interest is whether the proportions are in fact equal. two proportions A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. matched or paired samples The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their mid-level professionals are paid differently, on average. The average worker in Germany gets eight weeks of paid vacation. single mean According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market. It is believed that the average grade on an English essay in a particular school system for women is higher than for men. A random sample of 31 women had a mean score of 82 with a standard deviation of three, and a random sample of 25 men had a mean score of 76 with a standard deviation of four. independent group means, population standard deviations and/or variances unknown The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? two proportions A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6. independent group means, population standard deviations and/or variances unknown Varsity athletes practice five times a week, on average. A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? independent group means, population standard deviations and/or variances unknown A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? two proportions Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. Are standard deviations known or unknown? What is the random variable? The random variable is the difference between the mean amounts of sugar in the two soft drinks. Is this a one-tailed or two-tailed test? Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for White people born in 1900 and 33.0 years for non-White people. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 White people, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 non-White people, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for White people and non-White people. Is this a test of means or proportions? means State the null and alternative hypotheses. H 0 : __________ H a : __________ Is this a right-tailed, left-tailed, or two-tailed test? two-tailed In symbols, what is the random variable of interest for this test? In words, define the random variable of interest for this test. the difference between the mean life spans of White and non-White people Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for . This is a comparison of two population means with unknown population standard deviations. Calculate the test statistic. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p -value. Answers may vary. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Cannot accept the null hypothesis p -value < 0.05 At the 5% level of significance, the evidence supports the claim that life expectancy in the 1900s was different between White people and non-White people. Does it appear that the means are the same? Why or why not? Homework The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from 29 men and 16 women. The men took an average of three English courses with a standard deviation of 0.8. The women took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. Subscripts: 1: two-year colleges; 2: four-year colleges H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 X ¯ 1 – X ¯ 2 is the difference between the mean enrollments of the two-year colleges and the four-year colleges. Student’s- t test statistic: -0.2480 p -value: 0.4019 Answers may vary. Alpha: 0.05 Decision: Cannot reject Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. At Rachel’s 11 th birthday party, eight guests were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The children thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 21 47 40 30 28 22 21 23 25 45 43 37 35 29 32 Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. Subscripts: 1: mechanical engineering; 2: electrical engineering H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 X ¯ 1 − X ¯ 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers. t 108 test statistic: t = –0.82 p -value: 0.2061 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers. Marketing companies have collected data implying that teenage girls use more social media apps on their smartphones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with smartphones, the mean number of social media apps for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from Data Sets in Appendix A: Statistical Tables to answer the next four exercises. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 X ¯ 1 − X ¯ 2 is the difference between the mean times for completing a lap in races and in practices. t 20.32 test statistic: –4.70 p -value: 0.0001 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. Repeat the test in , but use Lap 5 data this time. Repeat the test in , but this time combine the data from Laps 1 and 5. H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 is the difference between the mean times for completing a lap in races and in practices. t 40.94 test statistic: –5.08 p -value: zero Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?” Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. Western Eastern Los Angeles 9 D.C. United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Conduct a hypothesis test to answer the next two exercises. The exact distribution for the hypothesis test is: the normal distribution the Student's t -distribution the uniform distribution the exponential distribution If the level of significance is 0.05, the conclusion is: There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score. Unable to determine c Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. They take random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is: There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. Elijah wants to know whether textbook costs are different for different courses of study. He selects a random sample of 33 sociology textbooks offered on a popular online site. The mean price of his sample is $74.64 with a standard deviation of $49.36. He then selects a random sample of 33 math and science textbooks from the same site. The mean price of this sample is $111.56 with a standard deviation of $66.90. Is the mean price of a sociology textbook lower than the mean price of a math or science textbook? Test at a 1% significance level. Test: two independent sample means, population standard deviations unknown. μ 1 = the mean price of a sociology text on the selected site μ 2 = the mean price of a math/science text on the selected site Random variable: X 1 - X 2 = the difference in the sample mean textbook price between sociology texts and math/science texts. Hypotheses: H 0 : μ 1 - μ 2 = 0 , H a : μ 1 - μ 2 < 0 which can be expressed as H 0 s : μ 1 = μ 2 , H a : μ 1 < μ 2 Distribution for the test: Use tdf ; because each sample has more than 30 observations, d f = n 1 + n 2 - 2 = 33 + 33 - 2 = 64 . Estimate the critical value on the t -table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column. Calculate the test statistic: t c = ( X 1 - X 2 ) - 0 s 1 2 n 1 + s 2 2 n 2 = ( 74 . 64 - 111 . 56 ) - 0 49 . 36 2 33 + 66 . 90 2 33 = - 2 . 55 Using a calculator with t c = - 2 . 55 and d f = 64 , the left-tailed p -value: 0.0066. Decision: Reject H 0 . Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. They take random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: μ day > μ night μ day < μ night μ day = μ night μ day ≠ μ night d Cohen’s d a measure of effect size based on the differences between two means. If d is between 0 and 0.2 then the effect is small. If d approaches is 0.5, then the effect is medium, and if d approaches 0.8, then it is a large effect. Pooled Variance a weighted average of two variances that can then be used when calculating standard error.", "section": "Comparing Two Independent Population Means", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Cohen's Standards for Small, Medium, and Large Effect Sizes Cohen's d is a measure of \"effect size\" based on the differences between two means. Cohen’s d , named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes. Cohen's Standard Effect Sizes Size of effect d Small 0.2 Medium 0.5 Large 0.8 Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: d = x ¯ 1 – x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 – 1 ) s 1 2 + ( n 2 – 1 ) s 2 2 n 1 + n 2 – 2 It is important to note that Cohen's d does not provide a level of confidence as to the magnitude of the size of the effect comparable to the other tests of hypothesis we have studied. The sizes of the effects are simply indicative. Calculate Cohen’s d for Example 10.2 . Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. x̅ 1 = 4 s 1 = 1.5 n 1 = 11 x̅ 2 = 3.5 s 2 = 1 n 2 = 9 d = 0.384 The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two companies is small indicating that there is not a significant difference between them. Try It Calculate Cohen’s d for Try It 10.4. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. μ 1 = 5 s 1 = 1 . 2 n 1 = 15 μ 2 = 4 . 5 s 2 = 0 . 8 n 2 = 20 d = 0.49 The effect is medium because 0.49 is very close to Cohen’s value of 0.5 for medium effect size. The size of the differences of the means for the two companies is small, indicating that there is not a significant difference between them. Chapter Review Cohen's d is a measure of “effect size” based on the differences between two means. It is important to note that Cohen's d does not provide a level of confidence as to the magnitude of the size of the effect comparable to the other tests of hypothesis we have studied. The sizes of the effects are simply indicative. Formula Review Cohen’s d is the measure of effect size: d = x ¯ 1 − x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2", "section": "Cohen's Standards for Small, Medium, and Large Effect Sizes", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Test for Differences in Means: Assuming Equal Population Variances Typically we can never expect to know any of the population parameters, mean, proportion, or standard deviation. When testing hypotheses concerning differences in means we are faced with the difficulty of two unknown variances that play a critical role in the test statistic. We have been substituting the sample variances just as we did when testing hypotheses for a single mean. And as we did before, we used a Student's t to compensate for this lack of information on the population variance. There may be situations, however, when we do not know the population variances, but we can assume that the two populations have the same variance. If this is true then the pooled sample variance will be smaller than the individual sample variances. This will give more precise estimates and reduce the probability of discarding a good null. The null and alternative hypotheses remain the same, but the test statistic changes to: t c = ( x – 1 − x – 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) where S p 2 is the pooled variance given by the formula: s p 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2 A drug trial is attempted using a real drug and a pill made of just sugar. 18 people are given the real drug in hopes of increasing the production of endorphins. The increase in endorphins is found to be on average 8 micrograms per person, and the sample standard deviation is 5.4 micrograms. 11 people are given the sugar pill, and their average endorphin increase is 4 micrograms with a standard deviation of 2.4. From previous research on endorphins it is determined that it can be assumed that the variances within the two samples can be assumed to be the same. Test at 5% to see if the population mean for the real drug had a significantly greater impact on the endorphins than the population mean with the sugar pill. First we begin by designating one of the two groups Group 1 and the other Group 2. This will be needed to keep track of the null and alternative hypotheses. Let's set Group 1 as those who received the actual new medicine being tested and therefore Group 2 is those who received the sugar pill. We can now set up the null and alternative hypothesis as: H 0 : µ 1 ≤ µ 2 H 1 : µ 1 > µ 2 This is set up as a one-tailed test with the claim in the alternative hypothesis that the medicine will produce more endorphins than the sugar pill. We now calculate the test statistic which requires us to calculate the pooled variance, S p 2 using the formula above. t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) = ( 8 − 4 ) − 0 20.4933 ( 1 18 + 1 11 ) = 2.31 t α , allows us to compare the test statistic and the critical value. t α = 1.703 at d f = n 1 + n 2 − 2 = 18 + 11 − 2 = 27 The test statistic is clearly in the tail, 2.31 is larger than the critical value of 1.703, and therefore we cannot maintain the null hypothesis. Thus, we conclude that there is significant evidence at the 95% level of confidence that the new medicine produces the effect desired. Try It Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in and , respectively. Northeast 94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6 77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5 76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0 West 126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6 116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6 78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5 Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right, left, or two tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. Calculate Cohen’s d and interpret it. Chapter Review In situations when we do not know the population variances but assume the variances are the same, the pooled sample variance will be smaller than the individual sample variances. This will give more precise estimates and reduce the probability of discarding a good null. Formula Review t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) where S p 2 is the pooled variance given by the formula: S p 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2", "section": "Test for Differences in Means: Assuming Equal Population Variances", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Comparing Two Independent Population Proportions Proportions have their base in the binomial probability distribution. As a probability distribution, we know the mean and standard deviation of the distribution. As a binary or categorical sample data set, we lose this knowledge from when we knew the population parameters. We do not know the population mean, µ = n p , or variance, σ ∠ = n p q . We can gather data that we know comes from a binary distribution but not know the specific parameter. It is then that we have moved from probability to inferential statistics. When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: The two independent samples are random samples that are independent. The number of successes is at least five, and the number of failures is at least five, for each of the samples. Growing literature states that the population must be at least ten or even perhaps 20 times the size of the sample. This keeps each population from being over-sampled and causing biased results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance in the sampling. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the two population proportions. Like the case of differences in sample means, we construct a sampling distribution for differences in sample proportions: ( p A ' - p B ' ) where p ' A = X A n A and p ' B = X ' B n B are the sample proportions for the two sets of data in question. X A and X B are the number of successes in each sample group respectively, and n A and n B are the respective sample sizes from the two groups. Again we go the Central Limit theorem to find the distribution of this sampling distribution for the differences in sample proportions. And again we find that this sampling distribution, like the ones past, are normally distributed as proved by the Central Limit Theorem, as seen in . Generally, the null hypothesis allows for the test of a difference of a particular value, 𝛿 0 , just as we did for the case of differences in means. H 0 : p 1 − p 2 = 𝛿 0 H 1 : p 1 − p 2 ≠ 𝛿 0 Most common, however, is the test that the two proportions are the same. That is, H 0 : p A = p B H a : p A ≠ p B To conduct the test, we use a pooled proportion, p c . The pooled proportion is calculated as follows: p c = x A + x B n A + n B The test statistic ( z -score) is: Z c = ( p A ′ − p B ′ ) − δ 0 p c ( 1 − p c ) ( 1 n A + 1 n B ) where δ 0 is the hypothesized differences between the two proportions and p c is the pooled variance from the formula above. A bank has recently acquired a new branch and thus has customers in this new territory. They are interested in the default rate in their new territory. They wish to test the hypothesis that the default rate is different from their current customer base. They sample 200 files in area A, their current customers, and find that 20 have defaulted. In area B, the new customers, another sample of 200 files shows 12 have defaulted on their loans. At a 10% level of significance can we say that the default rates are the same or different? This is a test of proportions. We know this because the underlying random variable is binary, default or not default. Further, we know it is a test of differences in proportions because we have two sample groups, the current customer base and the newly acquired customer base. Let A and B be the subscripts for the two customer groups. Then p A and p B are the two population proportions we wish to test. Random Variable: P′ A – P′ B = difference in the proportions of customers who defaulted in the two groups. H 0 : p A = p B H a : p A ≠ p B The words \"is a difference\" tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: p c = x A + x B n A + n B = 20 + 12 200 + 200 = 0.08 1 – p c = 0.92 ( p′ A – p′ B ) = 0.04 follows an approximate normal distribution. Estimated proportion for group A: p ′ A = x A n A = 20 200 = 0.1 Estimated proportion for group B: p ′ B = x B n B = 12 200 = 0.06 The estimated difference between the two groups is : p′ A – p′ B = 0.1 – 0.06 = 0.04. Z c = ( P′ A − P′ B ) − δ 0 P c ( 1 − P c ) ( 1 n A + 1 n B ) = 1.47 The calculated test statistic is 1.47 and is not in the tail of the distribution. Make a decision: Since the calculate test statistic is not in the tail of the distribution we cannot reject H 0 . Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference between the proportions of customers who defaulted in the two groups. Try It Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. The p -value is 0.0379, so we can reject the null hypothesis. At the 5% significance level, the data support that there is a difference in the pressure tolerances between the two valves. References Data from Educational Resources , December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994. “State of the States.” Gallup, 2013. Available online at http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive (accessed June 17, 2013). “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/westnile/index.htm (accessed June 17, 2013). Chapter Review Test of two population proportions from independent samples. Random variable: p' A – p' B = difference between the two estimated proportions Distribution: normal distribution Formula Review Pooled Proportion: p c = x A + x B n A + n B Test Statistic ( z -score): Z c = ( p ′ A − p ′ B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) where p A ' and p B ' are the sample proportions, p A and p B are the population proportions, P c is the pooled proportion, and n A and n B are the sample sizes. Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS 1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS 2 had system failures within the first eight hours of operation. OS 2 is believed to be more stable (have fewer crashes) than OS 1 . Is this a test of means or proportions? What is the random variable? P ′ OS1 – P ′ OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS 1 and OS 2 . State the null and alternative hypotheses. What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. Is this a test of means or proportions? proportions State the null and alternative hypotheses. H 0 : _________ H a : _________ Is this a right-tailed, left-tailed, or two-tailed test? How do you know? right-tailed What is the random variable of interest for this test? In words, define the random variable for this test. The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for the Exercise 10.39 . Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. Calculate the test statistic. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Cannot accept the null hypothesis. p -value < alpha At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? Homework A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. We are interested in whether the proportions of conferred physical science degrees for people aged 21 to 24 are the same for White people and Black people in the United States. The number of conferred degrees for White people in a given year is 4,930. Five hundred eighty were aged 21 to 24. The estimate for Black people is 330. Forty were aged 21 to 24. We will let people with conferred physical science degrees be our population. H 0 : P W = P B H a : P W ≠ P B The random variable is the difference in the proportions of White and Black people with conferred degrees in a given year, aged 21 to 24. normal for two proportions test statistic: –0.1944 p -value: 0.8458 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of White and Black people with conferred physical science degrees, aged 21 to 24, are different. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic/Latino students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic/Latino students out of a total of 2,441 students. In general, do you think that the percent of Hispanic/Latino students at the two colleges is basically the same or different? Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The random variable is the difference between the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College. normal for two proportions test statistic: 4.29 p -value: 0.00002 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College are different. Use the following information to answer the next two exercises. Some individuals who were exposed to the COVID-19 virus experienced a loss of taste. In a sample of COVID patients during 2022, there were 629 reported cases of loss of taste out of a total of 1,021 reported cases. In 2021, there were 486 reported cases of loss of taste out of a sample of 712 cases. Is the 2022 proportion of loss of taste more than the 2021 proportion? Using a 1% level of significance, conduct an appropriate hypothesis test. “2022” subscript: 2022 group. “2021” subscript: 2021 group This is: a test of two proportions a test of two independent means a test of a single mean a test of matched pairs An appropriate null hypothesis is: p 2022 ≤ p 2021 p 2022 ≥ p 2021 μ 2022 ≤ μ 2021 p 2022 > p 2021 a Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. Test: two independent sample proportions. Random variable: p ′ 1 - p ′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed Adults aged 18 years old and older from a certain city were randomly selected for a survey on working from home. The researchers wanted to determine if the proportion of women who work from home is less than the proportion of men who work from home. The results are shown in . Test at the 1% level of significance. Number who work at home Sample size Men 42,769 155,525 Women 67,169 248,775 Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a tablet 69 231 Sample size 628 2,309 Test: two independent sample proportions Random variable: p′ 1 − p′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 > p 2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: right-tailed Do not reject the H 0 . Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. A group of friends debated whether more people in their twenties use wearable fitness devices than people in their thirties. They consulted a research study of wearable fitness device use among adults. The results of the survey indicate that of the 973 randomly sampled people in their twenties, 379 use such devices. For people in their thirties, 404 of the 1,304 who were randomly sampled use wearable fitness devices. Test at the 5% level of significance. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Subscripts: 1: men; 2: women H 0 : p 1 ≤ p 2 H a : p 1 > p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. normal for two proportions test statistic: 0.22 p -value: 0.4133 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. Joan Nguyen recently claimed that the proportion of college-age men with at least one pierced ear is as high as the proportion of college-age women. She conducted a survey in her classes. Out of 107 men, 20 had at least one pierced ear. Out of 92 women, 47 had at least one pierced ear. Do you believe that the proportion of men has reached the proportion of women? H 0 : p 1 = p 2 H a : p 1 ≠ p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women that have at least one pierced ear. normal for two proportions test statistic: –4.82 p -value: zero Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different. \"To Breakfast or Not to Breakfast?\" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what they have achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in , solve our problem. Work hours with breakfast Work hours without breakfast 8 6 7 5 9 5 5 4 9 7 8 7 10 7 7 5 6 6 9 5 H 0 : µ d = 0 H a : µ d > 0 The random variable X d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t 9 test statistic: 4.8963 p -value: 0.0004 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.", "section": "Comparing Two Independent Population Proportions", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is X 1 – – X 2 – . The normal distribution has the following format: The standard deviation is: ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 The test statistic ( z -score) is: Z c = ( x – 1 – x – 2 ) – δ 0 ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax . Both populations have a normal distributions. The data are recorded in . Wax Sample mean number of months floor wax lasts Population standard deviation 1 3 0.33 2 2.9 0.36 Does the data indicate that wax 1 is more effective than wax 2 ? Test at a 5% level of significance. This is a test of two independent groups, two population means, population standard deviations known. Random Variable : X – 1 – X – 2 = difference in the mean number of months the competing floor waxes last. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The words \"is more effective\" says that wax 1 lasts longer than wax 2 , on average. \"Longer\" is a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula for the test statistic we find the calculated value for the problem. Z c = ( μ 1 - μ 2 ) - δ 0 σ 1 2 n 1 + σ 2 2 n 2 = 0.1 The estimated difference between he two means is : X – 1 – X – 2 = 3 – 2.9 = 0.1 Compare calculated value and critical value and Z α : We mark the calculated value on the graph and find the calculated value is not in the tail therefore we cannot reject the null hypothesis. Make a decision: the calculated value of the test statistic is not in the tail, therefore you cannot reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts. Try It The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines of each type are randomly assigned to be tested. Both populations have normal distributions. shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine Sample mean number of RPM Population standard deviation 1 1,500 50 2 1,600 60 The p -value is almost zero, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1. An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance. This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators Random variable: X – 1 – X – 2 = difference in the mean age of Democratic and Republican U.S. senators. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H a : μ 1 > μ 2 H a : μ 1 - μ 2 > 0 The words \"older than\" translates as a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Make a decision: The p-value is larger than 5%, therefore we cannot reject the null hypothesis. By calculating the test statistic we would find that the test statistic does not fall in the tail, therefore we cannot reject the null hypothesis. We reach the same conclusion using either method of a making this statistical decision. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. Try It The average age of 10 professors selected randomly in university A is 46.672 with a standard deviation of 8.53. The average age of 10 professors selected randomly in university B is 47.531 with a standard deviation of 7.83. Does the data indicate that university A has older professors than university B, on average? Test at a 5% level of significance. The total sample size is 10 + 10 = 20., which is greater than 10, so we can use the Student’s t -distribution. The random variable is X 1 - X 2 = difference between the mean age of universities A and B. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H 0 : μ 1 > μ 2 H 0 : μ 1 - μ 2 > 0 The phrase “older than” translates as a “>” symbol and goes into H 0 . Therefore, this is a right-tailed test. The distribution is: X 1 - X 2 ~ N 0 , ( 8 . 53 ) 2 10 + ( 7 . 83 ) 2 10 The mean for the normal distribution is 0 since μ 1 ≤ μ 2 and μ 1 - μ 2 ≤ 0 . The p -value is obtained as 0.4258. Since alpha is less than the p -value, do not reject the H 0 . At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of professors in university A is more than the mean age of professors in university B. References Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbullying Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). “State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013). Chapter Review A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics: Random variable: X – 1 − X – 2 = the difference of the means Distribution: normal distribution Formula Review Test Statistic ( z -score): Z c = ( x – 1 − x – 2 ) − δ 0 ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 where: σ 1 and σ 2 are the known population standard deviations. n 1 and n 2 are the sample sizes. x – 1 and x – 2 are the sample means. μ 1 and μ 2 are the population means. Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample mean speed of pitches (mph) Population standard deviation Wesley 86 3 Rodriguez 91 7 What is the random variable? The difference in mean speeds of the fastball pitches of the two pitchers State the null and alternative hypotheses. What is the test statistic? –2.46 At the 1% significance level, what is your conclusion? At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant group Sample mean height of plants (inches) Population standard deviation Food 16 2.5 No food 14 1.5 Is the population standard deviation known or unknown? State the null and alternative hypotheses. Subscripts: 1 = Food, 2 = No Food H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample mean melting temperatures (°F) Population standard deviation Alloy Gamma 800 95 Alloy Zeta 900 105 State the null and alternative hypotheses. Subscripts: 1 = Gamma, 2 = Zeta H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 Is this a right-, left-, or two-tailed test? At the 1% significance level, what is your conclusion? There is sufficient evidence so we cannot accept the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. Homework NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. Subscripts: 1 = boys, 2 = girls H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The random variable is the difference in the mean auto insurance costs for boys and girls. normal test statistic: z = 2.50 p -value: 0.0062 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. normal test statistic: 6.36 p -value: 0 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s score 2 2 3 3 4 2 1 1 2 4 Husband’s score 2 2 1 3 2 1 1 1 2 4 H 0 : µ d = 0 H a : µ d < 0 The random variable X d is the average difference between husband’s and wife’s satisfaction level. t 9 test statistic: t = –1.86 p -value: 0.0479 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis, but run another test. Reason for Decision: p -value < alpha Conclusion: This is a weak test because alpha and the p -value are close. However, there is insufficient evidence to conclude that the mean difference is negative.", "section": "Two Population Means with Known Standard Deviations", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Matched or Paired Samples In most cases of economic or business data we have little or no control over the process of how the data are gathered. In this sense the data are not the result of a planned controlled experiment. In some cases, however, we can develop data that are part of a controlled experiment. This situation occurs frequently in quality control situations. Imagine that the production rates of two machines built to the same design, but at different manufacturing plants, are being tested for differences in some production metric such as speed of output or meeting some production specification such as strength of the product. The test is the same in format to what we have been testing, but here we can have matched pairs for which we can test if differences exist. Each observation has its matched pair against which differences are calculated. First, the differences in the metric to be tested between the two lists of observations must be calculated, and this is typically labeled with the letter \"d.\" Then, the average of these matched differences, X – d = ∑ ( x 1 - x 2 ) n is calculated as is its standard deviation, S d . We expect that the standard deviation of the differences of the matched pairs will be smaller than unmatched pairs because presumably fewer differences should exist because of the correlation between the two groups. When using a hypothesis test for matched or paired samples, the following characteristics may be present: Simple random sampling is used. Sample sizes are often small. Two measurements (samples) are drawn from the same pair of individuals or objects. Differences are calculated from the matched or paired samples. The differences from the sample that is used for the hypothesis test. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal, as has been the case using the Central Limit Theorem. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences, that is, the number of pairs not the number of observations. The null and alternative hypotheses for this test are: H 0 : µ d = 0 H a : µ d ≠ 0 The test statistic is: t c = x – d − μ d ( s d n ) A company has developed a training program for its entering employees because management has become concerned with the results of the six-month employee review. They hope that the training program can result in better six-month reviews. Each trainee constitutes a “pair”, the entering score the employee received when first entering the firm and the score given at the six-month review. The difference in the two scores were calculated for each employee and the means for before and after the training program was calculated. The sample mean before the training program was 20.4 and the sample mean after the training program was 23.9. The standard deviation of the differences in the two scores across the 20 employees was 3.8 points. Test at the 10% significance level the null hypothesis that the two population means are equal against the alternative that the training program helps improve the employees’ scores. The first step is to identify this as a two sample case: before the training and after the training. This differentiates this problem from simple one sample issues. Second, we determine that the two samples are \"paired.\" Each observation in the first sample has a paired observation in the second sample. This information tells us that the null and alternative hypotheses should be: H 0 : µ d ≤ 0 H a : µ d > 0 This form reflects the implied claim that the training course improves scores; the test is one-tailed and the claim is in the alternative hypothesis. Because the experiment was conducted as a matched paired sample rather than simply taking scores from people who took the training course those who didn't, we use the matched pair test statistic: Test Statistic: t c = X ¯ d − µ d S d n = ( 23.9 − 20.4 ) − 0 ( 3.8 20 ) = 4.12 In order to solve this equation, the individual scores, pre-training course and post-training course need to be used to calculate the individual differences. These scores are then averaged and the average difference is calculated: X ¯ d = ∑ ( x 1 - x 2 ) n From these differences we can calculate the standard deviation across the individual differences: S d = ∑ d i - X ¯ d 2 n - 1 w h e r e d i = x 1 i - x 2 i We can now compare the calculated value of the test statistic, 4.12, with the critical value. The critical value is a Student's t with degrees of freedom equal to the number of pairs, not observations, minus 1. In this case 20 pairs and at 90% confidence level t a/2 = ±1.729 at df = 20 - 1 = 19. The calculated test statistic is most certainly in the tail of the distribution and thus we cannot accept the null hypothesis that there is no difference from the training program. Evidence seems indicate that the training aids employees in gaining higher scores. Try-It Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 111 135 140 125 Off-hand 105 109 98 111 99 A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in . A lower score indicates less pain. The \"before\" value is matched to an \"after\" value and the differences are calculated. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Corresponding \"before\" and \"after\" values form matched pairs. (Calculate \"after\" – \"before.\") After data Before data Difference 6.8 6.6 0.2 2.4 6.5 -4.1 7.4 9 -1.6 8.5 10.3 -1.8 8.1 11.3 -3.2 6.1 8.1 -2 3.4 6.3 -2.9 2 11.6 -9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: x – d = –3.13 and s d = 2.91 Verify these values. Let μ d be the population mean for the differences. We use the subscript d to denote \"differences.\" Random variable: X – d = the mean difference of the sensory measurements H 0 : μ d ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ d is the population mean of the differences.) H a : μ d < 0 The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t 7 . (Notice that the test is for a single population mean.) Calculate the test statistic and look up the critical value using the Student's-t distribution: The calculated value of the test statistic is 3.06 and the critical value of the t -distribution with 7 degrees of freedom at the 5% level of confidence is 1.895 with a one-tailed test. X – d is the random variable for the differences. The sample mean and sample standard deviation of the differences are: x – d = –3.13 s – d = 2.91 Compare the critical value for alpha against the calculated test statistic. The conclusion from using the comparison of the calculated test statistic and the critical value will gives us the result. In this question the calculated test statistic is 3.06 and the critical value is 1.895. The test statistic is clearly in the tail and thus we cannot accept the null hypotheses that there is no difference between the two situations, hypnotized and not hypnotized. Make a decision: Cannot accept the null hypothesis, H 0 . This means that μ d < 0 and there is a statistically significant improvement. Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain. Try It A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level. Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 A college softball coach was interested in whether the college's strength development class increased their players' maximum lift (in pounds). Four players were asked to participate in the study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 241 338 368 Amount of weight lifted after the class 295 252 330 360 The coach wants to know if the strength development class makes the players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. x – d = 21.3, s d = 46.7 Using the difference data, this becomes a test of a single mean. Define the random variable: X – d mean difference in the maximum lift per player. The distribution for the hypothesis test is a student's t with 3 degrees of freedom. H 0 : μ d ≤ 0, H a : μ d > 0 Calculate the test statistic look up the critical value: The calculated value of the test statistic is 0.91. The critical value of the student's t at 5% level of significance and 3 degrees of freedom is 2.353. Decision: If the level of significance is 5%, we cannot reject the null hypothesis, because the calculated value of the test statistic is not in the tail. What is the conclusion? At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. Be sure to note in your conclusion that a sample size of only 4 leaves a degree of freedom of only 3 and sets the critical value very large. In short, sample size so small cannot result in always meaningful conclusions. Try It A new prep class was designed to improve SAT test scores. Four students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in . Are the scores, on average, higher after the class? Test at a 5% level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 1960 1920 2150 Score after class 1920 2160 2200 2100 Chapter Review A hypothesis test for matched or paired samples (t-test) has these characteristics: Test the differences by subtracting one measurement from the other measurement Random Variable: x – d = mean of the differences Distribution: Student’s-t distribution with n – 1 degrees of freedom If the number of differences is small (less than 30), the differences must follow a normal distribution. Two samples are drawn from the same set of objects. Samples are dependent. Formula Review Test Statistic ( t -score): t c = x – d − μ d ( s d n ) where: x – d is the mean of the sample differences. μ d is the mean of the population differences. s d is the sample standard deviation of the differences. n is the sample size. Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in . The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. Installation A B C D E F G H Before 3 6 4 2 5 8 2 6 After 1 5 2 0 1 0 2 2 What is the random variable? the mean difference of the system failures State the null and alternative hypotheses. What conclusion can you draw about the software patch? With a p -value 0.0067, we can cannot accept the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. Subject A B C D E F Before 3 4 3 2 4 5 After 4 5 6 4 5 7 State the null and alternative hypotheses. What is the sample mean difference? What conclusion can you draw about the juggling class? Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 State the null and alternative hypotheses. H 0 : μ d ≥ 0 H a : μ d < 0 What is the test statistic? What is the sample mean difference? What is the conclusion? We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. Homework Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in . Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 140 220 230 110 120 240 220 200 190 180 150 190 200 360 300 280 300 260 240 p -value = 0.1494 At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. A new heart disease prevention drug was tried on a group of 224 patients. Forty-five patients developed heart disease after four years. In a control group of 224 patients, 68 developed heart disease after four years. We want to test whether the method of treatment reduces the proportion of patients that develop heart disease after four years or if the proportions of the treated group and the untreated group stay the same. Let the subscript t = treated patient and ut = untreated patient. The appropriate hypotheses are: H 0 : p t < p ut and H a : p t ≥ p ut H 0 : p t ≤ p ut and H a : p t > p ut H 0 : p t = p ut and H a : p t ≠ p ut H 0 : p t = p ut and H a : p t < p ut An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: x – d = −10.2 s d = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training. The distribution for the test is: t 5 t 6 N (−10.2, 8.4) N(−10.2, 8.4 6 ) A golf instructor is interested in determining if their new technique for improving players’ golf scores is effective. The instructor takes four new students and records their 18-hole scores before learning the technique and then after having taken the class. The instructor conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 The correct decision is: Reject H 0 . Do not reject the H 0 . A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. The group compared the estimates of new female breast cancer cases by southern state in Year 1 and in Year 2. The results are in . Southern states Year 1 Year 2 Alabama 3,450 3,720 Arkansas 2,150 2,280 Florida 15,540 15,710 Georgia 6,970 7,310 Kentucky 3,160 3,300 Louisiana 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas 15,050 14,980 Virginia 6,190 6,280 Test: two matched pairs or paired samples ( t -test) Random variable: X – d Distribution: t 12 H 0 : μ d = 0 H a : μ d > 0 The mean of the differences of new female breast cancer cases in the south between Year 2 and Year 1 is greater than zero. The estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. Graph: right-tailed p -value: 0.0004 Decision: Cannot accept H 0 Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in Year 2 than in Year 1. A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in . Test at the 1% level of significance. Cities Hyatt Regency prices in dollars Hilton prices in dollars Atlanta 107 169 Boston 358 289 Chicago 209 299 Dallas 209 198 Denver 167 169 Indianapolis 179 214 Los Angeles 179 169 New York City 625 459 Philadelphia 179 159 Washington, DC 245 239 A politician asked their staff to determine whether the underemployment rate in the northeast decreased year over year. The results are in . Northeastern states Year 1 Year 2 Connecticut 17.3 16.4 Delaware 17.4 13.7 Maine 19.3 16.1 Maryland 16.0 15.5 Massachusetts 17.6 18.2 New Hampshire 15.4 13.5 New Jersey 19.2 18.7 New York 18.5 18.7 Ohio 18.2 18.8 Pennsylvania 16.5 16.9 Rhode Island 20.7 22.4 Vermont 14.7 12.3 West Virginia 15.5 17.3 Test: matched or paired samples ( t -test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8} Random Variable: X – d Distribution: H 0 : μ d = 0 H a : μ d < 0 The mean of the differences of the rate of underemployment in the northeastern states between Year 2 and Year 1 is less than zero. The underemployment rate went down from Year 1 to Year 2. Graph: left-tailed. Decision: Cannot reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from Year 1 to Year 2. Bringing It Together Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. independent group means, population standard deviations and/or variances known independent group means, population standard deviations and/or variances unknown matched or paired samples single mean two proportions single proportion A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. e The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from nine men and 16 women. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. d A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively. According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this. f According to a recent study, U.S. companies have a mean maternity-leave of six weeks. A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. e A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 300 960 920 1010 1100 840 880 1100 1070 1250 1320 860 860 1330 1370 790 770 990 1040 1110 1200 740 850 University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked. f Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown . Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size 41 41 Sample mean 97.5 98.1 Sample standard deviation 17.5 19.2 Two independent means, normal distribution Two independent means, Student’s-t distribution Matched or paired samples, Student’s-t distribution Two population proportions, normal distribution A golf instructor is interested in determining if a new technique for improving players’ golf scores is effective. The instructor takes four (4) new students and records their 18-hole scores before learning the technique and then after having taken the class. The instructor conducts a hypothesis test. The data are as . Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 This is: a test of two independent means. a test of two proportions. a test of a single mean. a test of a single proportion. a", "section": "Matched or Paired Samples", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: modification of work “The first few times I used my Asda 'bag for life' I left the receipts in the bottom of the bag” by Pete/ Flickr, Public domain) Have you ever wondered if lottery winning numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example the test of independence, which determines if events are independent, such as in the movie example the test of a single variance, which tests variability, such as in the coffee example", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Facts About the Chi-Square Distribution The notation for the chi-square distribution is: X ∼ χ d f 2 where df = degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chi-square probabilities then use df = n - 1. The degrees of freedom for the three major uses are each calculated differently.) For the χ 2 distribution, the population mean is μ = df and the population standard deviation is σ = 2 ( d f ) . The random variable is shown as χ 2 . The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables. χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + ... + ( Z k ) 2 The curve is nonsymmetrical and skewed to the right. There is a different chi-square curve for each df . The test statistic for any test is always greater than or equal to zero. When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ 1,000 2 the mean, μ = df = 1,000 and the standard deviation, σ = 2 ( 1,000 ) = 44.7. Therefore, X ~ N (1,000, 44.7), approximately. The mean, μ , is located just to the right of the peak. References Data from Parade Magazine . “HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011. Chapter Review The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df . For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. Formula Review χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + … ( Z df ) 2 chi-square distribution random variable μ χ 2 = df chi-square distribution population mean σ χ 2 = 2 ( d f ) Chi-Square distribution population standard deviation If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? mean = 25 and standard deviation = 7.0711 If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. When does the chi-square curve approximate a normal distribution? when the number of degrees of freedom is greater than 90 Where is μ located on a chi-square curve? Is it more likely the df is 90, 20, or two in the graph? df = 2 Homework Decide whether the following statements are true or false. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. true The standard deviation of the chi-square distribution is twice the mean. The mean and the median of the chi-square distribution are the same if df = 24. false", "section": "Facts About the Chi-Square Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Test of a Single Variance Thus far our interest has been exclusively on the population parameter μ or it's counterpart in the binomial, p. Surely the mean of a population is the most critical piece of information to have, but in some cases we are interested in the variability of the outcomes of some distribution. In almost all production processes quality is measured not only by how closely the machine matches the target, but also the variability of the process. If one were filling bags with potato chips not only would there be interest in the average weight of the bag, but also how much variation there was in the weights. No one wants to be assured that the average weight is accurate when their bag has no chips. Electricity voltage may meet some average level, but great variability, spikes, can cause serious damage to electrical machines, especially computers. I would not only like to have a high mean grade in my classes, but also low variation about this mean. In short, statistical tests concerning the variance of a distribution have great value and many applications. A test of a single variance assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population variance . The test statistic is: χ c 2 = ( n - 1 ) s 2 σ 0 2 where: n = the total number of observations in the sample data s 2 = sample variance σ 0 2 = hypothesized value of the population variance H 0 : σ 2 = σ 0 2 H a : σ 2 ≠ σ 0 2 You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Even though we are given the population standard deviation, we can set up the test using the population variance as follows. H 0 : σ 2 ≤ 5 2 H a : σ 2 > 5 2 Try It A SCUBA instructor wants to record the collective depths each of his students' dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? H 0 : σ 2 = 3 2 H a : σ 2 < 3 2 With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon. With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers . Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ 2 . Random Variable: The sample standard deviation, s , is the random variable. Let s = standard deviation for the waiting times. H 0 : σ 2 ≥ 7.2 2 H a : σ 2 < 7.2 2 The word \"less\" tells you this is a left-tailed test. Distribution for the test: χ 24 2 , where: n = the number of customers sampled df = n – 1 = 25 – 1 = 24 Calculate the test statistic: χ c 2 = ( n − 1 ) s 2 σ 2 = ( 25 − 1 ) ( 3.5 ) 2 7.2 2 = 5.67 where n = 25, s = 3.5, and σ = 7.2. The graph of the Chi-square shows the distribution and marks the critical value with 24 degrees of freedom at 95% level of confidence, α = 0.05, 13.85. The critical value of 13.85 came from the Chi squared table which is read very much like the students t table. The difference is that the students t -distribution is symmetrical and the Chi squared distribution is not. At the top of the Chi squared table we see not only the familiar 0.05, 0.10, etc. but also 0.95, 0.975, etc. These are the columns used to find the left hand critical value. The graph also marks the calculated χ 2 test statistic of 5.67. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Make a decision: Because the calculated test statistic is in the tail we cannot accept H 0 . This means that you reject σ 2 ≥ 7.2 2 . In other words, you do not think the variation in waiting times is 7.2 minutes or more; you think the variation in waiting times is less. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. Try It The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p -value, and draw a conclusion. Test at the 1% significance level. Professor Holmes has a weakness for cream filled donuts, but they believe that some bakeries are not properly filling the donuts. A sample of 24 donuts reveals a mean amount of filling equal to 0.04 cups, and the sample standard deviation is 0.11 cups. Professor Holmes has an interest in the average quantity of filling, of course, but he is particularly distressed if one donut is radically different from another. Professor Holmes does not like surprises. Test at 95% the null hypothesis that the population variance of donut filling is significantly different from the average amount of filling 0.04 cups. This is clearly a problem dealing with variances. In this case we are testing a single sample rather than comparing two samples from different populations. The null and alternative hypotheses are thus: H 0 : σ 2 = 0.04 H a : σ 2 ≠ 0.04 The test is set up as a two-tailed test because Professor Holmes has shown concern with too much variation in filling as well as too little: their dislike of a surprise is a significant difference greater than or less than the expected average of 0.04 cups. Conduct the hypothesis test with n = 24 , s 2 = 0 . 11 , and σ 2 = 0 . 04 . The test statistic is calculated to be: χ c 2 = ( n − 1 ) s 2 σ o 2 = ( 24 − 1 ) 0.11 0.04 = 63.25 The calculated χ 2 test statistic, 63.25, is in the tail. Therefore, at a 0.05 level of significance, we cannot accept the null hypothesis that the variance in the donut filling is equal to 0.04 cups. It seems that Professor Holmes is destined to meet disappointment with each bite. Try It The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the distribution and mark the area associated with the level of confidence, and draw a conclusion. Test at the 1% significance level. H 0 : σ 2 = 12.2 2 H a : σ 2 > 12.2 2 df = 14 chi 2 test statistic = 16.39 The p -value is 0.2902, so we decline to reject the null hypothesis. There is not enough evidence to suggest that the variance is greater than 12.2 2 . In 2nd DISTR , use7: χ2cdf . The syntax is (lower, upper, df) for the parameter list. χ2cdf(16.39,10^99,14) . The p -value = 0.2902. References “AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013). Data from the World Bank, June 5, 2012. Chapter Review To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation). Formula Review χ 2 = ( n − 1 ) s 2 σ 0 2 Test of a single variance statistic where: n : sample size s : sample standard deviation σ 0 : hypothesized value of the population standard deviation df = n – 1 Degrees of freedom Test of a Single Variance Use the test to determine variation. The degrees of freedom is the number of samples – 1. The test statistic is ( n – 1 ) s 2 σ 0 2 , where n = sample size, s 2 = sample variance, and σ 2 = population variance. The test may be left-, right-, or two-tailed. Use the following information to answer the next three exercises: An archer’s standard deviation for hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less. What type of test should be used? a test of a single variance State the null and alternative hypotheses. Is this a right-tailed, left-tailed, or two-tailed test? a left-tailed test Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. What type of test should be used? State the null and alternative hypotheses. H 0 : σ 2 = 0.81 2 ; H a : σ 2 > 0.81 2 df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. What type of test should be used? a test of a single variance What is the test statistic? What can you conclude at the 5% significance level? Homework Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. Is the traveler disputing the claim about the average or about the variance? A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 225 Is this a right-tailed, left-tailed, or two-tailed test? H 0 : __________ H 0 : σ 2 ≤ 150 df = ________ chi-square test statistic = ________ 36 Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the area associated with the level of confidence. Answers may vary. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________ How did you know to test the variance instead of the mean? The claim is that the variance is no more than 150 minutes. If an additional test were done on the claim of the average delay, which distribution would you use? If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? a Student's t - or normal distribution A plant manager is concerned their equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. H 0 : σ = 15 H a : σ > 15 df = 42 chi-square with df = 42 test statistic = 26.88 Answers may vary. Alpha = 0.05 Decision: Cannot reject null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. H 0 : σ ≤ 3 H a : σ > 3 df = 17 chi-square distribution with df = 17 test statistic = 28.73 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in . Does the students’ survey indicate that the standard deviation is greater than 0.75? # of births Frequency 0 5 1 30 2 10 3 5 According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. They count the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11 H 0 : σ = 2 H a : σ ≠ 2 df = 14 chi-square distribution with df = 14 chi-square test statistic = 5.2094 Answers may vary. Alpha = 0.05 Decision: Cannot accept the null hypothesis Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2. The manager of \"Frenchies\" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. The manager randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis? The sample standard deviation is $34.29. H 0 : σ 2 = 25 2 H a : σ 2 > 25 2 df = n – 1 = 7. test statistic: x 2 = x 7 2 = ( n – 1 ) s 2 25 2 = ( 8 – 1 ) ( 34.29 ) 2 25 2 = 13.169 ; Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test. (a) at the 5% significance level (b) at the 1% significance level Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172;", "section": "Test of a Single Variance", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Goodness-of-Fit Test In this type of hypothesis test, you determine whether the data \"fit\" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: Σ k ( O − E ) 2 E where: O = observed values (data) E = expected values (from theory) k = the number of different data cells or categories The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form ( O − E ) 2 E . The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The number of expected values inside each cell needs to be at least five in order to use this test. Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to . Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9–11 6 12+ 2 A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in displays the results of that survey. Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9–11 1 12+ 4 Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test. H 0 : Student absenteeism fits faculty perception. The alternative hypothesis is the opposite of the null hypothesis. H a : Student absenteeism does not fit faculty perception. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test? a. No. Notice that the expected number of absences for the \"12+\" entry is less than five (it is two). Combine that group with the \"9–11\" group to create new tables where the number of students for each entry are at least five. The new results are in and . Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9+ 8 Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9+ 5 b. What is the number of degrees of freedom ( df )? b. There are four \"cells\" or categories in each of the new tables. df = number of cells – 1 = 4 – 1 = 3 Try It A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in . Number produced Number defective 0–100 5 101–200 6 201–300 7 301–400 8 401–500 10 A random sample was taken to determine the actual number of defects. shows the results of the survey. Number produced Number defective 0–100 5 101–200 7 201–300 8 301–400 9 401–500 11 State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom. H 0 :The number of defaults fits expectations. H a :The number of defaults does not fit expectations. df = 4 Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in . For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level. Day of the Week Employees were Most Absent Monday Tuesday Wednesday Thursday Friday Number of absences 15 12 9 9 15 The null and alternative hypotheses are: H 0 : The absent days occur with equal frequencies, that is, they fit a uniform distribution. H a : The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected ( E ) values. The values in the table are the observed ( O ) values or data. This time, calculate the χ 2 test statistic by hand. Make a chart with the following headings and fill in the columns: Expected ( E ) values (12, 12, 12, 12, 12) Observed ( O ) values (15, 12, 9, 9, 15) ( O – E ) ( O – E ) 2 ( O – E ) 2 E Now add (sum) the last column. The sum is three. This is the χ 2 test statistic. The calculated test statistics is 3 and the critical value of the χ 2 distribution at 4 degrees of freedom the 0.05 level of confidence is 9.48. This value is found in the χ 2 table at the 0.05 column on the degrees of freedom row 4. The degrees of freedom are the number of cells – 1 = 5 – 1 = 4 Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.) χ c 2 = Σ k ( O − E ) 2 E = 3 The decision is not to reject the null hypothesis because the calculated value of the test statistic is not in the tail of the distribution. Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies. Try It Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in . Sunday Monday Tuesday Wednesday Thursday Friday Saturday Number of students 11 8 10 7 10 5 5 From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use? df = 6 p -value = 0.6093 We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week. One study indicates that the number of streaming services that American families have is distributed (this is the given distribution for the American population) as in . Number of Streaming Services Percent 0 10 1 16 2 55 3 11 4+ 8 The table contains expected ( E ) percents. A random sample of 600 families in the far western United States resulted in the data in . Number of Streaming Services Frequency 0 66 1 119 2 340 3 60 4+ 15 Total = 600 The table contains observed ( O ) frequency values. At the 1% significance level, does it appear that the distribution \"number of streaming services\" of far western United States families is different from the distribution for the American population as a whole? This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected ( E ) frequencies, multiply the percentage by 600. The expected frequencies are shown in . Number of Streaming Services Percent Expected frequency 0 10 (0.10)(600) = 60 1 16 (0.16)(600) = 96 2 55 (0.55)(600) = 330 3 11 (0.11)(600) = 66 over 3 8 (0.08)(600) = 48 Therefore, the expected frequencies are 60, 96, 330, 66, and 48. H 0 : The \"number of streaming services\" distribution of far western United States families is the same as the \"number of streaming services\" distribution of the American population. H a : The \"number of streaming services\" distribution of far western United States families is different from the \"number of streaming services\" distribution of the American population. Distribution for the test: χ 4 2 where df = (the number of cells) – 1 = 5 – 1 = 4. Calculate the test statistic: χ 2 = 29.65 Graph: The graph of the Chi-square shows the distribution and marks the critical value with four degrees of freedom at 99% level of confidence, α = .01, 13.277. The graph also marks the calculated chi squared test statistic of 29.65. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Make a decision: Because the test statistic is in the tail of the distribution we cannot accept the null hypothesis. This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole. Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the \"number of streaming services\" distribution for the far western United States is different from the \"number of streaming services\" distribution for the American population as a whole. Try It The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in . Number of pets Percent 0 18 1 25 2 30 3 18 4+ 9 A random sample of 1,000 students from the Eastern United States resulted in the data in . Number of pets Frequency 0 210 1 240 2 320 3 140 4+ 90 At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? p -value = 0.0036 We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole. Suppose you flip two coins 100 times. The results are 20 HH , 27 HT , 30 TH , and 23 TT . Are the coins fair? Test at a 5% significance level. This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is { HH , HT , TH , TT }. Out of 100 flips, you would expect 25 HH , 25 HT , 25 TH , and 25 TT . This is the expected distribution from the binomial probability distribution. The question, \"Are the coins fair?\" is the same as saying, \"Does the distribution of the coins (20 HH , 27 HT , 30 TH , 23 TT ) fit the expected distribution?\" Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three . Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed. H 0 : The coins are fair. H a : The coins are not fair. Distribution for the test: χ 2 2 where df = 3 – 1 = 2. Calculate the test statistic: χ 2 = 2.14 Graph: The graph of the Chi-square shows the distribution and marks the critical value with two degrees of freedom at 95% level of confidence, α = 0.05, 5.991. The graph also marks the calculated χ 2 test statistic of 2.14. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Conclusion: There is insufficient evidence to conclude that the coins are not fair: we cannot reject the null hypothesis that the coins are fair. Try It Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom? MDG Region Adult Literacy Rate (%) Developed Regions 99.0 Commonwealth of Independent States 99.5 Northern Africa 67.3 Sub-Saharan Africa 62.5 Latin America and the Caribbean 91.0 Eastern Asia 93.8 Southern Asia 61.9 South-Eastern Asia 91.9 Western Asia 84.5 Oceania 66.4 References Data from the U.S. Census Bureau Data from the College Board. Available online at http://www.collegeboard.com. Data from the U.S. Census Bureau, Current Population Reports. Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92. Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013). Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013). Chapter Review To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five. Formula Review ∑ k ( O − E ) 2 E goodness-of-fit test statistic where: O : observed values E : expected values k : number of different data cells or categories df = k − 1 degrees of freedom Determine the appropriate test to be used in the next three exercises. An archaeologist is calculating the distribution of the frequency of the number of artifacts they find in a dig site. Based on previous digs, the archaeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, the archeologist compares the actual number of artifacts found in each grid section to see if their expectation was accurate. An economist is deriving a model to predict outcomes on the stock market. They create a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, the economist records the actual points on the index. They want to see how well the model matched what actually happened. a goodness-of-fit test A personal trainer is putting together a weight-lifting program for clients. For a 90-day program, the trainer expects each client to lift a specific maximum weight each week. As the program goes along, the trainer records the actual maximum weights her clients lifted. They want to know how well their expectations met with what was observed. Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in . Grade Proportion A 0.25 B 0.30 C 0.35 D 0.10 The actual distribution for a class of 20 is in . Grade Frequency A 7 B 7 C 5 D 1 d f = ______ 3 State the null and alternative hypotheses. χ 2 test statistic = ______ 2.04 At the 5% significance level, what can you conclude? We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores. Use the following information to answer the next eight exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in . Ethnicity Number of cases White 2,229 Hispanic/Latino 1,157 Black/African-American 457 Asian, Pacific Islander 232 Total = 4,075 The percentage of each ethnic group in Santa Clara County is as in . Ethnicity Percentage of total county population Number expected (round to two decimal places) White 42.9% 1748.18 Hispanic/Latino 26.7% Black/African-American 2.6% Asian, Pacific Islander 27.8% Total = 100% If the ethnicities of COVID-19 related cases followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group. Perform a goodness-of-fit test to determine whether the occurrence of COVID-19 cases follows the ethnicities of the general population of Santa Clara County. H 0 : _______ H 0 : the distribution of COVID-19 cases follows the ethnicities of the general population of Santa Clara County. H a : _______ Is this a right-tailed, left-tailed, or two-tailed test? right-tailed degrees of freedom = _______ χ 2 test statistic = _______ 2016.136 Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the confidence level. Let α = 0.05 Decision: ________________ Reason for the Decision: ________________ Conclusion (write out in complete sentences): ________________ Graph: Answers may vary. Decision: Cannot accept the null hypothesis. Reason for the Decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion (write out in complete sentences): The make-up of COVID-19 cases does not fit the ethnicities of the general population of Santa Clara County. Does it appear that the pattern of COVID-19 cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? Homework A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in are the result of the 120 rolls. Face value Frequency Expected frequency 1 15 2 29 3 16 4 15 5 30 6 15 The marital status distribution of the U.S. male population, ages 15 and older, is as shown in . Marital status Percent Expected frequency Never married 31.3 Married 56.1 Widowed 2.5 Divorced/Separated 10.1 Suppose that a random sample of 400 U.S. young adult men, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in , rounding to two decimal places. Marital status Frequency Never married 140 Married 238 Widowed 2 Divorced/Separated 20 Marital status Percent Expected frequency Never married 31.3 125.2 Married 56.1 224.4 Widowed 2.5 10 Divorced/Separated 10.1 40.4 The data fits the distribution. The data does not fit the distribution. 3 chi-square distribution with df = 3 19.27 0.0002 Answers may vary. Alpha = 0.05 Decision: Cannot accept null hypothesis at the 5% level of significance Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: Data does not fit the distribution. Use the following information to answer the next two exercises: The columns in contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for students using mass transit to get to school, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who used mass transit to get to school. Race/Ethnicity Percentage Who Use Mass Transit to Get to School Overall student population Survey frequency Asian, Asian American, or Pacific Islander 10.2% 5.4% 113 Black or African-American 8.2% 14.5% 94 Hispanic or Latino 15.5% 15.9% 136 American Indian or Alaska Native 0.6% 1.2% 10 White 59.4% 61.6% 604 Not reported/other 6.1% 1.4% 43 Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity. Perform a goodness-of-fit test to determine whether the local results follow the distribution of those who use mass transit, based on ethnicity. H 0 : The local results follow the distribution of the percentage of students who use mass transit to get to school H a : The local results do not follow the distribution of the percentage of students who use mass transit to get to school df = 5 chi-square distribution with df = 5 chi-square test statistic = 13.4 Answers may vary. Alpha = 0.05 Decision: Cannot accept null when a = 0.05 Reason for Decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: Local data do not fit the mass transit Distribution. Decision: Do not reject null when a = 0.01 Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of students who use mass transit distribution. The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in . Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area. Race Lake Tahoe frequency Manhattan frequency Asian Indian 131 174 Chinese 118 557 Filipino 1,045 518 Japanese 80 54 Korean 12 29 Vietnamese 9 21 Other 24 66 Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 first-year students from 385 colleges in a certain year. Suppose a survey of 5,000 graduating women and 5,000 graduating men was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for and . The second column in each table does not add to 100% because of rounding. Conduct a goodness-of-fit test to determine if the actual college majors of graduating women fit the distribution of their expected majors. Major Women - expected major Women - actual major Arts & Humanities 14.0% 670 Biological Sciences 8.4% 410 Business 13.1% 685 Education 13.0% 650 Engineering 2.6% 145 Physical Sciences 2.6% 125 Professional 18.9% 975 Social Sciences 13.0% 605 Technical 0.4% 15 Other 5.8% 300 Undecided 8.0% 420 H 0 : The actual college majors of graduating women fit the distribution of their expected majors H a : The actual college majors of graduating women do not fit the distribution of their expected majors df = 10 chi-square distribution with df = 10 test statistic = 11.48 Answers may vary. Alpha = 0.05 Decision: Cannot reject null when a = 0.05 and a = 0.01 Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating women fits the distribution of their expected majors. Conduct a goodness-of-fit test to determine if the actual college majors of graduating men fit the distribution of their expected majors. Major Men - expected major Men - actual major Arts & Humanities 11.0% 600 Biological Sciences 6.7% 330 Business 22.7% 1130 Education 5.8% 305 Engineering 15.6% 800 Physical Sciences 3.6% 175 Professional 9.3% 460 Social Sciences 7.6% 370 Technical 1.8% 90 Other 8.2% 400 Undecided 6.6% 340 Read the statement and decide whether it is true or false. In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not. true The test to use to determine if a six-sided die is fair is a goodness-of-fit test. In a goodness-of fit test, if the p -value is 0.0113, in general, do not reject the null hypothesis. false A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values. Business type Number in class Observed number that recycle one commodity Expected number that recycle one commodity Office 35 19 17.5 Retail/Wholesale 48 27 24 Food/Restaurants 53 35 26.5 Manufacturing/Medical 52 21 26 Hotel/Mixed 24 9 12 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population. Age class (years) Obese (percentage) Expected USA average (percentage) 20–30 15.0 32.6 31–40 26.5 32.6 41–50 13.6 36.6 51–60 21.9 36.6 61–70 21.0 39.7 The hypotheses for the goodness-of-fit test are: H 0 : Surveyed obese fit the distribution of expected obese H a : Surveyed obese do not fit the distribution of expected obese Use a chi-square distribution with df = 4 to evaluate the data. The test statistic is X 2 = 9.85 The P -value = 0.0431 At 5% significance level, α = 0.05. For this data, P < α. Reject the null hypothesis. At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese. Goodness-of-Fit a hypothesis test that compares expected and observed values in order to look for significant differences within one non-parametric variable. The degrees of freedom used equals the (number of categories – 1).", "section": "Goodness-of-Fit Test", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Test of Independence Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test: Σ ( i ⋅ j ) ( O – E ) 2 E where: O = observed values E = expected values i = the number of rows in the table j = the number of columns in the table There are i ⋅ j terms of the form ( O – E ) 2 E . A test of independence determines whether two factors are independent or not. You first encountered the term independence in Independent and Mutually Exclusive Events earlier. As a review, consider the following example. NOTE The expected value inside each cell needs to be at least five in order for you to use this test. Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P ( A ∩ B ) = P ( A ) P ( B ). A ∩ B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not. Let y = expected number of drivers who used a cell phone while driving and received speeding violations. If A and B are independent, then P ( A ∩ B ) = P ( A ) P ( B ). By substitution, y 755 = ( 70 755 ) ( 305 755 ) Solve for y : y = ( 70 ) ( 305 ) 755 = 28.3 About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations. In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors , the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent) . If we do a test of independence using the example, then the null hypothesis is: H 0 : Being a cell phone user while driving and receiving a speeding violation are independent events; in other words, they have no effect on each other. If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit. The number of degrees of freedom for the test of independence is: df = (number of columns - 1)(number of rows - 1) The following formula calculates the expected number ( E ): E = (row total)(column total) total number surveyed Try It A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven of the 300 surveyed were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll? About 16 students are expected to be music students and on the honor roll. A volunteer group, provides from one to nine hours each week with disabled senior citizens. The program recruits among community college students, four-year college students, and nonstudents. In is a sample of the adult volunteers and the number of hours they volunteer per week. Number of Hours Worked Per Week by Volunteer Type (Observed) Type of volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row total Community college students 111 96 48 255 Four-year college students 96 133 61 290 Nonstudents 91 150 53 294 Column total 298 379 162 839 The table contains observed (O) values (data). Is the number of hours volunteered independent of the type of volunteer? The observed table and the question at the end of the problem, \"Is the number of hours volunteered independent of the type of volunteer?\" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer . This test is always right-tailed. H 0 : The number of hours volunteered is independent of the type of volunteer. H a : The number of hours volunteered is dependent on the type of volunteer. The expected results are in . Number of Hours Worked Per Week by Volunteer Type (Expected) Type of volunteer 1-3 Hours 4-6 Hours 7-9 Hours Community college students 90.57 115.19 49.24 Four-year college students 103.00 131.00 56.00 Nonstudents 104.42 132.81 56.77 The table contains expected ( E ) values (data). For example, the calculation for the expected frequency for the top left cell is E = ( row total ) ( column total ) total number surveyed = ( 255 ) ( 298 ) 839 = 90.57 Calculate the test statistic: χ 2 = 12.99 (calculator or computer) Distribution for the test: χ 4 2 df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4 Graph: The graph of the Chi-square shows the distribution and marks the critical value with four degrees of freedom at 95% level of confidence, α = 0.05, 9.488. The graph also marks the calculated χ c 2 test statistic of 12.99. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Make a decision: Because the calculated test statistic is in the tail we cannot accept H 0 . This means that the factors are not independent. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another. For the example in , if there had been another type of volunteer, teenagers, what would the degrees of freedom be? Try It The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. shows the results: Industry sector 2000 2010 2020 Total Nonagriculture wage and salary 13,243 13,044 15,018 41,305 Goods-producing, excluding agriculture 2,457 1,771 1,950 6,178 Services-providing 10,786 11,273 13,068 35,127 Agriculture, forestry, fishing, and hunting 240 214 201 655 Nonagriculture self-employed and unpaid family worker 931 894 972 2,797 Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36 Secondary jobs as a self-employed or unpaid family worker 196 144 152 492 Total 27,867 27,351 31,372 86,590 We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom. H 0 : The number of jobs is independent of the year. H a : The number of jobs is dependent on the year. df = 12 Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 3 ENTER 3 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 227.73 and the p −value = 5.90E - 42 = 0. Do the procedure a second time but arrow down to Draw instead of calculate . De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. Need to Succeed in School vs. Anxiety Level Need to Succeed in School High Anxiety Med-high Anxiety Medium Anxiety Med-low Anxiety Low Anxiety Row Total High Need 35 42 53 15 10 155 Medium Need 18 48 63 33 31 193 Low Need 4 5 11 15 17 52 Column Total 57 95 127 63 58 400 a. How many high anxiety level students are expected to have a high need to succeed in school? a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 155 ⋅ 57 400 = 22.09 The expected number of students who have a high anxiety level and a high need to succeed in school is about 22. b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety? b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 8.19 The expected number of students who have a low need to succeed in school and a med-low level of anxiety is 8. Try It Refer back to the information in . How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020? References DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/Rls2436.pdf (accessed May 24, 2013). Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-ice-cream (accessed May 24, 2013) “Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013). Chapter Review To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5. Formula Review Test of Independence The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1). The test statistic is ∑ i ⋅ j ( O − E ) 2 E where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table. If the null hypothesis is true, the expected number E = (row total)(column total) total surveyed . Determine the appropriate test to be used in the next three exercises. A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups. a test of independence The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. They take a random sample of 100 players from different organizations. A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. They take a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing. a test of independence Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel. Traveling distance Third class Second class First class Total 1–100 miles 21 14 6 41 101–200 miles 18 16 8 42 201–300 miles 16 17 15 48 301–400 miles 12 14 21 47 401–500 miles 6 6 10 22 Total 73 67 60 200 State the hypotheses. H 0 : _______ H a : _______ df = _______ 8 How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets? How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets? 6.6 What is the test statistic? What can you conclude at the 5% level of significance? Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Hispanic/Latino people, 2,305 Japanese American people, and 3,970 White people. Complete the table. Smoking Levels by Ethnicity (Observed) Smoking level per day Black Native Hawaiian Hispanic/Latinos Japanese American White Totals 1-10 11-20 21-30 31+ Totals Smoking level per day Black Native Hawaiian Hispanic/Latino Japanese Americans White Totals 1-10 9,886 2,745 12,831 8,378 7,650 41,490 11-20 6,514 3,062 4,932 10,680 9,877 35,065 21-30 1,671 1,419 1,406 4,715 6,062 15,273 31+ 759 788 800 2,305 3,970 8,622 Totals 18,830 8,014 19,969 26,078 27,559 10,0450 State the hypotheses. H 0 : _______ H a : _______ Enter expected values in . Round to two decimal places. Calculate the following values: Smoking level per day Black Native Hawaiian Hispanic/Latino Japanese Americans White 1-10 7777.57 3310.11 8248.02 10771.29 11383.01 11-20 6573.16 2797.52 6970.76 9103.29 9620.27 21-30 2863.02 1218.49 3036.20 3965.05 4190.23 31+ 1616.25 687.87 1714.01 2238.37 2365.49 df = _______ χ 2 test statistic = ______ 10,301.8 Is this a right-tailed, left-tailed, or two-tailed test? Explain why. right Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the confidence level. State the decision and conclusion (in a complete sentence) for the following preconceived levels of α . α = 0.05 Decision: ___________________ Reason for the decision: ___________________ Conclusion (write out in a complete sentence): ___________________ Cannot accept the null hypothesis. Calculated value of test statistics is either in or out of the tail of the distribution. There is sufficient evidence to conclude that smoking level is dependent on ethnic group. α = 0.01 Decision: ___________________ Reason for the decision: ___________________ Conclusion (write out in a complete sentence): ___________________ Homework A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier. U.S. ski area Beginner Intermediate Advanced Tahoe 20 30 40 Utah 10 30 60 Colorado 10 40 50 Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in . Conduct a test of independence. Family Size Sub & Compact Mid-size Full-size Van & Truck 1 20 35 40 35 2 20 50 70 80 3–4 20 50 100 90 5+ 20 30 70 70 H 0 : Car size is independent of family size. H a : Car size is dependent on family size. df = 9 chi-square distribution with df = 9 test statistic = 15.8284 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent. College students may be interested in whether or not their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. shows the data. Conduct a test of independence. Major < $50,000 $50,000 – $68,999 $69,000 + English 5 20 5 Engineering 10 30 60 Nursing 10 15 15 Business 10 20 30 Psychology 20 30 20 Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in . Conduct a test of independence. Location 20–29 30–39 40–49 50 and over Niagara Falls 15 25 25 20 Poconos 15 25 25 10 Europe 10 25 15 5 Virgin Islands 20 25 15 5 H 0 : Honeymoon locations are independent of bride’s age. H a : Honeymoon locations are dependent on bride’s age. df = 9 chi-square distribution with df = 9 test statistic = 15.7027 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent. A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and their choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence. Sport 18 - 25 26 - 30 31 - 40 41 and over Racquetball 42 58 30 46 Tennis 58 76 38 65 Swimming 72 60 65 33 A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in . Conduct a test of independence. Type of Fries Northeast South Central West Skinny fries 70 50 20 25 Curly fries 100 60 15 30 Steak fries 20 40 10 10 H 0 : The types of fries sold are independent of the location. H a : The types of fries sold are dependent on the location. df = 6 chi-square distribution with df = 6 test statistic =18.8369 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent. According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by men in the following age groups. He is interested in whether the age of the man and the amount of life insurance purchased are independent events. Conduct a test for independence. Age of Men None < $200,000 $200,000–$400,000 $401,001–$1,000,000 $1,000,001+ 20–29 40 15 40 0 5 30–39 35 5 20 20 10 40–49 20 0 30 0 30 50+ 40 30 15 15 10 Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence. Annual salary Not a high school graduate High school graduate College graduate Masters or doctorate < $30,000 15 25 10 5 $30,000–$40,000 20 40 70 30 $40,000–$50,000 10 20 40 55 $50,000–$60,000 5 10 20 60 $60,000+ 0 5 10 150 H 0 : Salary is independent of level of education. H a : Salary is dependent on level of education. df = 12 chi-square distribution with df = 12 test statistic = 255.7704 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent. Read the statement and decide whether it is true or false. The number of degrees of freedom for a test of independence is equal to the sample size minus one. The test for independence uses tables of observed and expected data values. true The test to use when determining if the college or university a student chooses to attend is related to their socioeconomic status is a test for independence. In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed. true An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on , do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5% significance level. U.S. region/Flavor Strawberry Chocolate Vanilla Rocky road Mint chocolate chip Pistachio Row total West 12 21 22 19 15 8 97 Midwest 10 32 22 11 15 6 96 East 8 31 27 8 15 7 96 South 15 28 30 8 15 6 102 Column total 45 112 101 46 60 27 391 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level. Age group\\ Net worth value (in millions of US dollars) 1–5 6–24 ≥25 Row total 17–25 8 7 5 20 26–30 6 5 9 20 Column total 14 12 14 40 H 0 : Age is independent of the youngest online entrepreneurs’ net worth. H a : Age is dependent on the net worth of the youngest online entrepreneurs. df = 2 chi-square distribution with df = 2 test statistic = 1.76 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent. A poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in , and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5% significance level. Opinion/Ethnicity Asian-American White/Non-Hispanic Black Hispanic/Latino Row total Against tax 48 433 41 160 682 In favor of tax 54 234 24 147 459 No opinion 16 43 16 19 94 Column total 118 710 81 326 1235 Contingency Table a table that displays sample values for two different factors that may be dependent or contingent on one another; it facilitates determining conditional probabilities. Test of Independence a hypothesis test that compares expected and observed values for contingency tables in order to test for independence between two variables. The degrees of freedom used equals the (number of columns – 1) multiplied by the (number of rows – 1).", "section": "Test of Independence", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Test for Homogeneity The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity , can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. NOTE The expected value inside each cell needs to be at least five in order for you to use this test. Hypotheses H 0 : The distributions of the two populations are the same. H a : The distributions of the two populations are not the same. Test Statistic Use a χ 2 test statistic. It is computed in the same way as the test for independence. Degrees of Freedom ( df ) df = number of columns - 1 Requirements All values in the table must be greater than or equal to five. Common Uses Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values. Do men and women college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected men college students and 300 randomly selected women college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in . Do men and women college students have the same distribution of living arrangements? Distribution of living arrangements for college men and college women Dormitory Apartment With Parents Other Men 72 84 49 45 Women 91 86 88 35 H 0 : The distribution of living arrangements for men college students is the same as the distribution of living arrangements for women college students. H a : The distribution of living arrangements for men college students is not the same as the distribution of living arrangements for women college students. Degrees of Freedom ( df ): df = number of columns – 1 = 4 – 1 = 3 Distribution for the test: χ 3 2 Calculate the test statistic: χ c 2 = 10.129 The graph of the Chi-square shows the distribution and marks the critical value with three degrees of freedom at 95% level of confidence, α = 0.05, 7.815. The graph also marks the calculated χ 2 test statistic of 10.129. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Make a decision: Because the calculated test statistic is in the tail we cannot accept H 0 . This means that the distributions are not the same. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for men and women college students are not the same. Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. Try It Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in . Do families and singles have the same distribution of cars? Test at a level of significance of 0.05. Sport Sedan Hatchback Truck Van/SUV Family 5 15 35 17 28 Single 45 65 37 46 7 Ivy League schools receive many applications, but only some can be accepted. At the schools listed in , two types of applications are accepted: regular and early decision. Application type accepted Brown Columbia Cornell Dartmouth Penn Yale Regular 2,115 1,792 5,306 1,734 2,685 1,245 Early decision 577 627 1,228 444 1,195 761 We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the χ 2 distribution and show the critical value and the calculated value of the test statistic, and draw a conclusion about the test of homogeneity. With a p -value of almost zero, we reject the null hypothesis. The data show that the distribution of cars is not the same for families and singles. H 0 : The distribution of regular applications accepted is the same as the distribution of early applications accepted. H a : The distribution of regular applications accepted is not the same as the distribution of early applications accepted. df = 5 χ 2 test statistic = 430.06 Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 3 ENTER 3 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 430.06 and the p -value = 9.80E-91. Do the procedure a second time but arrow down to Draw instead of calculate . References Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013). “Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/pubsinfo.asp?pubid=2009030 (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/2009030_sup.pdf (accessed May 24, 2013). Chapter Review To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five. Formula Review ∑ i ⋅ j ( O − E ) 2 E Homogeneity test statistic where: O = observed values E = expected values i = number of rows in data contingency table j = number of columns in data contingency table df = ( i −1)( j −1) Degrees of freedom A math teacher wants to see if two of their classes have the same distribution of test scores. What test should they use? test for homogeneity What are the null and alternative hypotheses for ? A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should they use? test for homogeneity A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should they use? What condition must be met to use the test for homogeneity? All values in the table must be greater than or equal to five. Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in . 20–30 30–40 40–50 50–60 Private practice 16 40 38 6 Hospital 8 44 59 39 State the null and alternative hypotheses. df = _______ 3 What is the test statistic? What can you conclude at the 5% significance level? Homework A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in . Conduct a test of homogeneity. Test at a 5% level of significance. Open Conscientious Extrovert Agreeable Neurotic Business 41 52 46 61 58 Social Science 72 75 63 80 65 H 0 : The distribution for personality types is the same for both majors H a : The distribution for personality types is not the same for both majors df = 4 chi-square with df = 4 test statistic = 3.01 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in . Conduct a test for homogeneity at a 5% level of significance. French toast Pancakes Waffles Omelettes Men 47 35 28 53 Women 65 59 55 60 A fish and wildlife technician is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance. H 0 : The distribution for fish caught is the same in Green Valley Lake and in Echo Lake. H a : The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake. 3 chi-square with df = 3 11.75 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake In a recent year, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5% significance level. Did you expect the result you obtained? Reasons for Homeschooling Applicable reason (in thousands of respondents) Most important reason (in thousands of respondents) Row total Concern about the environment of other schools 1,321 309 1,630 Dissatisfaction with academic instruction at other schools 1,096 258 1,354 To provide religious or moral instruction 1,257 540 1,797 Child has special needs, other than physical or mental 315 55 370 Nontraditional approach to child’s education 984 99 1,083 Other reasons (e.g., finances, travel, family time, etc.) 485 216 701 Column total 5,458 1,477 6,935 When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for a specific six-year period. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level. Year European Union United States Row total 6 3,413 7,164 10,557 5 3,302 7,057 10,359 4 3,505 7,488 10,993 3 3,537 7,758 11,295 2 3,595 7,697 11,292 1 3,613 7,847 11,460 Column total 20,965 45,011 65,976 H 0 : The distribution of average energy use in the USA is the same as in Europe between Year 1 and Year 6. H a : The distribution of average energy use in the USA is not the same as in Europe between Year 1 and Year 6. df = 4 chi-square with df = 4 test statistic = 2.7434 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from Year 1 to Year 6. The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. presents the number of Top Safety Picks in six car categories for specific two years. Analyze the table data to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between Year 1 and Year 2. Derive your results at the 5% significance level. Year \\ Car type Small Mid-size Large Small SUV Mid-size SUV Large SUV Row total Year 1 12 22 10 10 27 6 87 Year 2 31 30 19 11 29 4 124 Column total 43 52 29 21 56 10 211 Test for Homogeneity a test used to draw a conclusion about whether two populations have the same distribution. The degrees of freedom used equals the (number of columns – 1).", "section": "Test for Homogeneity", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Comparison of the Chi-Square Tests Above the χ 2 test statistic was used in three different circumstances. The following bulleted list is a summary of which χ 2 test is the appropriate one to use in different circumstances. Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an unknown distribution \"fits\" a known distribution. In this case there will be a single qualitative survey question or a single outcome of an experiment from a single population. Goodness-of-Fit is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution. The null and alternative hypotheses are: H 0 : The population fits the given distribution. H a : The population does not fit the given distribution. Independence: Use the test for independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a contingency table will be constructed. The goal is to see if the two variables are unrelated (independent) or related (dependent). The null and alternative hypotheses are: H 0 : The two variables (factors) are independent. H a : The two variables (factors) are dependent. Homogeneity: Use the test for homogeneity to decide if two populations with unknown distributions have the same distribution as each other. In this case there will be a single qualitative survey question or experiment given to two different populations. The null and alternative hypotheses are: H 0 : The two populations follow the same distribution. H a : The two populations have different distributions. Chapter Review The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown. Which test do you use to decide whether an observed distribution is the same as an expected distribution? a goodness-of-fit test What is the null hypothesis for the type of test from ? Which test would you use to decide whether two factors have a relationship? a test for independence Which test would you use to decide if two populations have the same distribution? How are tests of independence similar to tests for homogeneity? Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ ( i j ) ( O - E ) 2 E . In addition, all values must be greater than or equal to five. How are tests of independence different from tests for homogeneity? Homework Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance. H 0 : The distribution for technology use is the same for community college students and university students. H a : The distribution for technology use is not the same for community college students and university students. 2 chi-square with df = 2 7.05 p -value = 0.0294 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities. Read the statement and decide whether it is true or false. If df = 2, the chi-square distribution has a shape that reminds us of the exponential. Bringing It Together Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests. If you did a left-tailed test, what would you be testing? The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected. Testing to see if the data fits the distribution “too well” or is too perfect.", "section": "Comparison of the Chi-Square Tests", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction One-way ANOVA is used to measure information from several groups. (credit: modification of work “Magazine Stack” by thebittenword.com/ Flickr, CC BY 2.0) Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to their upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests comparing averages among more than two groups, statisticians have developed a method called \" Analysis of Variance \" (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or one-way ANOVA . You will also study the F distribution, used for one-way ANOVA, and the test for differences between two variances. This is just a very brief overview of one-way ANOVA. One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Test of Two Variances This chapter introduces a new probability density function, the F distribution. This distribution is used for many applications including ANOVA and for testing equality across multiple means. We begin with the F distribution and the test of hypothesis of differences in variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation in the lid and the container should be approximately the same. A supermarket might be interested in the variability of check-out times for two checkers. In finance, the variance is a measure of risk and thus an interesting question would be to test the hypothesis that two different investment portfolios have the same variance, the volatility. In order to perform a F test of two variances, it is important that the following are true: The populations from which the two samples are drawn are approximately normally distributed. The two populations are independent of each other. Unlike most other hypothesis tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, or close, the test can give a biased result for the test statistic. Suppose we sample randomly from two independent normal populations. Let σ 1 2 and σ 2 2 be the unknown population variances and s 1 2 and s 2 2 be the sample variances. Let the sample sizes be n 1 and n 2 . Since we are interested in comparing the two sample variances, we use the F ratio: F = [ s 1 2 σ 1 2 ] [ s 2 2 σ 2 2 ] F has the distribution F ~ F ( n 1 – 1, n 2 – 1) where n 1 – 1 are the degrees of freedom for the numerator and n 2 – 1 are the degrees of freedom for the denominator. If the null hypothesis is σ 1 2 = σ 2 2 , then the F Ratio, test statistic, becomes F c = [ s 1 2 σ 1 2 ] [ s 2 2 σ 2 2 ] = s 1 2 s 2 2 H 0 : σ 1 2 σ 2 2 = δ 0 H a : σ 1 2 σ 2 2 ≠ δ 0 if δ 0 =1 then H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 ≠ σ 2 Test statistic is : F c = S 1 2 S 2 2 The various forms of the hypotheses tested are: Two-Tailed Test One-Tailed Test One-Tailed Test H 0 : σ 1 2 = σ 2 2 H 0 : σ 1 2 ≤ σ 2 2 H 0 : σ 1 2 ≥ σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 H 1 : σ 1 2 > σ 2 2 H 1 : σ 1 2 < σ 2 2 A more general form of the null and alternative hypothesis for a two tailed test would be : H 0 : σ 1 2 σ 2 2 = δ 0 H a : σ 1 2 σ 2 2 ≠ δ 0 Where if δ 0 = 1 it is a simple test of the hypothesis that the two variances are equal. This form of the hypothesis does have the benefit of allowing for tests that are more than for simple differences and can accommodate tests for specific differences as we did for differences in means and differences in proportions. This form of the hypothesis also shows the relationship between the F distribution and the χ 2 : the F is a ratio of two chi squared distributions a distribution we saw in the The Chi-Square Distribution . This is helpful in determining the degrees of freedom of the resultant F distribution. If the two populations have equal variances, then s 1 2 and s 2 2 are close in value and the test statistic, F c = s 1 2 s 2 2 is close to one. But if the two population variances are very different, s 1 2 and s 2 2 tend to be very different, too. Choosing s 1 2 as the larger sample variance causes the ratio s 1 2 s 2 2 to be greater than one. If s 1 2 and s 2 2 are far apart, then F c = s 1 2 s 2 2 is a large number. Therefore, if F is close to one, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than one, then the evidence is against the null hypothesis. In essence, we are asking if the calculated F statistic, test statistic, is significantly different from one. To determine the critical points we have to find F α , df1 , df2 . See Appendix A for the F table. This F table has values for various levels of significance from 0.1 to 0.001 designated as \"p\" in the first column. To find the critical value choose the desired significance level and follow down and across to find the critical value at the intersection of the two different degrees of freedom. The F distribution has two different degrees of freedom, one associated with the numerator, df1 , and one associated with the denominator, df2 and to complicate matters the F distribution is not symmetrical and changes the degree of skewness as the degrees of freedom change. The degrees of freedom in the numerator is n 1 -1, where n 1 is the sample size for group 1, and the degrees of freedom in the denominator is n 2 -1, where n 2 is the sample size for group 2. F α , df1 , df2 will give the critical value on the upper end of the F distribution. To find the critical value for the lower end of the distribution, reverse the degrees of freedom and divide the F-value from the table into one. Upper tail critical value : F α , df1 , df2 Lower tail critical value : 1/F α , df2 , df1 When the calculated value of F is between the critical values, not in the tail, we cannot reject the null hypothesis that the two variances came from a population with the same variance. If the calculated F-value is in either tail we cannot accept the null hypothesis just as we have been doing for all of the previous tests of hypothesis. An alternative way of finding the critical values of the F distribution makes the use of the F-table easier. We note in the F-table that all the values of F are greater than one therefore the critical F value for the left hand tail will always be less than one because to find the critical value on the left tail we divide an F value into the number one as shown above. We also note that if the sample variance in the numerator of the test statistic is larger than the sample variance in the denominator, the resulting F value will be greater than one. The shorthand method for this test is thus to be sure that the larger of the two sample variances is placed in the numerator to calculate the test statistic. This will mean that only the right hand tail critical value will have to be found in the F-table. Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 10 exams. The first instructor's grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 1%. Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively. n 1 = n 2 = 10. H 0 : σ 1 2 ≥ σ 2 2 and H a : σ 1 2 < σ 2 2 Calculate the test statistic: By the null hypothesis ( σ 1 2 ≥ σ 2 2 ) , the F statistic is: F c = s 2 2 s 1 2 = 89.9 52.3 = 1.719 Critical value for the test: F 9,9 = 5.35 where n 1 – 1 = 9 and n 2 – 1 = 9. Make a decision: Since the calculated F value is not in the tail we cannot reject H 0 . Conclusion: With a 1% level of significance, from the data, there is insufficient evidence to conclude that the variance in grades for the first instructor is smaller. Try It The New York Choral Society divides male singers up into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different? Tenor1 Bass 2 Tenor 1 Bass 2 Tenor 1 Bass 2 69 72 67 72 68 67 72 75 70 74 67 70 71 67 65 70 64 70 66 75 72 66 69 76 74 70 68 72 74 72 68 75 71 71 72 64 68 74 66 74 73 70 75 68 72 66 72 The histograms are not as normal as one might like. Plot them to verify. However, we proceed with the test in any case. Subscripts: T1= tenor1 and B2 = bass 2 The standard deviations of the samples are s T 1 = 3.3302 and s B 2 = 2.7208. The hypotheses are H 0 : σ T 1 2 = σ B 2 2 and H 0 : σ T 1 2 ≠ σ B 2 2 (two tailed test) The F statistic is 1.4894 with 20 and 25 degrees of freedom. The p -value is 0.3430. If we assume alpha is 0.05, then we cannot reject the null hypothesis. We have no good evidence from the data that the heights of Tenor1 and Bass2 singers have different variances (despite there being a significant difference in mean heights of about 2.5 inches.) References “MLB Vs. Division Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/order/true. Chapter Review The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom. Assumptions: The populations from which the two samples are drawn are normally distributed. The two populations are independent of each other. Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an F test of two variances. Name one assumption that must be true. The populations from which the two samples are drawn are normally distributed. What is the other assumption that must be true? Use the following information to answer the next five exercises. Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times. Test the claim at the 10% level. Assume that commute times are normally distributed. State the null and alternative hypotheses. H 0 : σ 1 = σ 2 H a : σ 1 < σ 2 or H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 < σ 2 2 What is s 1 in this problem? What is s 2 in this problem? 4.11 What is n ? What is the F statistic? 0.7159 What is the critical value? Is the claim accurate? No, at the 10% level of significance, we cannot reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker. Use the following information to answer the next four exercises. Two students are interested in whether or not there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are more consistent. State the null and alternative hypotheses. What is the F Statistic? 2.8674 What is the critical value? At the 5% significance level, do we reject the null hypothesis? Cannot accept the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student. Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records their speeds going up 35 hills. The first cyclist has a variance of 23.8 and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. Assume that commute times are normally distributed. State the null and alternative hypotheses. What is the F Statistic? 0.7414 At the 5% significance level, what can we say about the cyclists’ variances? Homework Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Linda's rats Tuan's rats Javier's rats 43.5 47.0 51.2 39.4 40.5 40.9 41.3 38.9 37.9 46.0 46.3 45.0 38.2 44.2 48.6 Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%. H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 ≠ σ 1 2 df ( num ) = 4; df ( denom ) = 4 F 4, 4 3.00 Answers may vary. Decision: Cannot reject the null hypothesis; Conclusion: There is insufficient evidence to conclude that the variances are different. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows. Working-class Professional (middle incomes) Professional (wealthy) 17.8 16.5 8.5 26.7 17.4 6.3 49.4 22.0 4.6 9.4 7.4 12.6 65.4 9.4 11.0 47.1 2.1 28.6 19.5 6.4 15.4 51.2 13.9 9.3 Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups. Use a 5% significance level. Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines. Home decorating News Health Computer 172 87 82 104 286 94 153 136 163 123 87 98 205 106 103 207 197 101 96 146 Which two magazine types do you think have the same variance in length? Which two magazine types do you think have different variances in length? The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances. Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that the shows the results of a study. Saturday Sunday Saturday Sunday 75 44 62 137 18 58 0 82 150 61 124 39 94 19 50 127 62 99 31 141 73 60 118 73 89 Are the variances for incomes on the East Coast and the West Coast the same? Suppose that shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05. East West 38 71 47 126 30 42 82 51 75 44 52 90 115 88 67 H 0 : = σ 1 2 = σ 2 2 H a : σ 1 2 ≠ σ 1 2 df ( n ) = 7, df ( d ) = 6 F 7,6 0.8117 0.7825 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: calculated test statistics is not in the tail of the distribution Conclusion: There is not sufficient evidence to conclude that the variances are different. Thirty college students were taught a method of finger tapping. They were randomly assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100 mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours after ingesting the caffeine, the students had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how? Here are the data: 0 mg 100 mg 200 mg 0 mg 100 mg 200 mg 242 248 246 245 246 248 244 245 250 248 247 252 247 248 248 248 250 250 242 247 246 244 246 248 246 243 245 242 244 250 King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul) during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but declined significantly afterwards. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from a fourth. These spanned most of his reign. We have data on the silver content of the coins: First coinage Second coinage Third coinage Fourth coinage 5.9 6.9 4.9 5.3 6.8 9.0 5.5 5.6 6.4 6.6 4.6 5.5 7.0 8.1 4.5 5.1 6.6 9.3 6.2 7.7 9.2 5.8 7.2 8.6 5.8 6.9 6.2 Did the silver content of the coins change over the course of Manuel’s reign? Here are the means and variances of each coinage. The data are unbalanced. First Second Third Fourth Mean 6.7444 8.2429 4.875 5.6143 Variance 0.2953 1.2095 0.2025 0.1314 Here is a strip chart of the silver content of the coins: While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table: Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) 37.748 4 – 1 = 3 12.5825 26.272 Error (Within) 11.015 27 – 4 = 23 0.4789 Total 48.763 27 – 1 = 26 P ( F > 26.272) = 0; Cannot accept the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth. The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in a certain year Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in this specific season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same or not. Note that the data are not balanced, as two divisions had five teams, while one had only four. Division Team Wins East NY Yankees 95 East Baltimore 93 East Tampa Bay 90 East Toronto 73 East Boston 69 Division Team Wins Central Detroit 88 Central Chicago Sox 85 Central Kansas City 72 Central Cleveland 68 Central Minnesota 66 Division Team Wins West Oakland 94 West Texas 93 West LA Angels 89 West Seattle 75 Here is a stripchart of the number of wins for the 14 teams in the AL for a specific season. While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust. Here is the ANOVA table for the data: Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) 344.16 3 – 1 = 2 172.08 Error (Within) 1,219.55 14 – 3 = 11 110.87 1.5521 Total 1,563.71 14 – 1 = 13 P ( F > 1.5521) = 0.2548 Since the p -value is so large, there is not good evidence against the null hypothesis of equal means. We cannot reject the null hypothesis. Thus, for this specific season, there is not any have any good evidence of a significant difference in mean number of wins between the divisions of the American League.", "section": "Test of Two Variances", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a one-way ANOVA test, there are five basic assumptions to be fulfilled: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances) . The factor is a categorical variable. The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups: H 0 : μ 1 = μ 2 = μ 3 = . . . μ k H a : At least two of the group means μ 1 , μ 2 , μ 3 , . . . , μ k are not equal. That is, μ i ≠ μ j for some i ≠ j . The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (blue box plots), H 0 : μ 1 = μ 2 = μ 3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots). (a) H 0 is true. All means are the same; the differences are due to random variation. (b) H 0 is not true. The means are not all the same; the differences are too large to be due to random variation. Chapter Review Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances). Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they? Write one assumption. Each population from which a sample is taken is assumed to be normal. Write another assumption. Write a third assumption. The populations are assumed to have equal standard deviations (or variances). Write a fourth assumption. Homework Three different traffic routes are tested for mean driving time. The entries in the are the driving times in minutes on the three different routes. Route 1 Route 2 Route 3 30 27 16 32 29 41 27 28 22 35 36 31 State SS between , SS within , and the F statistic. SS between = 26 SS within = 441 F = 0.2653 Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x – = ________ ________ ________ ________ ________ s 2 = ________ ________ ________ ________ ________ State the hypotheses. H 0 : ____________ H a : ____________ Analysis of Variance also referred to as ANOVA, is a method of testing whether or not the means of three or more populations are equal. The method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. there is one independent variable and one dependent variable. The test statistic for analysis of variance is the F -ratio. One-Way ANOVA a method of testing whether or not the means of three or more populations are equal; the method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F -ratio. Variance mean of the squared deviations from the mean; the square of the standard deviation. For a set of data, a deviation can be represented as x – x – where x is a value of the data and x – is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.", "section": "One-Way ANOVA", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The F Distribution and the F-Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution , invented by George Snedecor but named in honor of Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F 4,10 . To calculate the F ratio , two estimates of the variance are made. Variance between samples : An estimate of σ 2 that is the variance of the sample means multiplied by n (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. Variance within samples : An estimate of σ 2 that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. SS between = the sum of squares that represents the variation among the different samples SS within = the sum of squares that represents the variation within samples that is due to chance. To find a \"sum of squares\" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics . MS means \" mean square .\" MS between is the variance between groups, and MS within is the variance within groups. Calculation of Sum of Squares and Mean Square k = the number of different groups n j = the size of the j th group s j = the sum of the values in the j th group n = total number of all the values combined (total sample size: ∑ n j ) x = one value: ∑ x = ∑ s j Sum of squares of all values from every group combined: ∑ x 2 Total sum of squares: S S total = ∑ x 2 ( ∑ x ) 2 n Explained variation: sum of squares representing variation among the different samples: SS between = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n Unexplained variation: sum of squares representing variation within samples due to chance: S S within = S S total – S S between df 's for different groups ( df 's for the numerator): d f between = k – 1 Equation for errors within samples ( df 's for the denominator): df within = n – k Mean square (variance estimate) explained by the different groups: MS between = S S between d f between Mean square (variance estimate) that is due to chance (unexplained): MS within = S S within d f within MS between and MS within can be written as follows: M S between = S S between d f between = S S between k − 1 M S w i t h i n = S S w i t h i n d f w i t h i n = S S w i t h i n n − k The one-way ANOVA test depends on the fact that MS between can be influenced by population differences among means of the several groups. Since MS within compares values of each group to its own group mean, the fact that group means might be different does not affect MS within . The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MS between and MS within should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances. F -Ratio or F Statistic F = M S between M S within If MS between and MS within estimate the same value (following the belief that H 0 is true), then the F -ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MS between consists of the population variance plus a variance produced from the differences between the samples. MS within is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MS between will generally be larger than MS within .Then the F -ratio will be larger than one. However, if the population effect is small, it is not unlikely that MS within will be larger in a given sample. The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F -ratio can be written as: F -Ratio Formula when the groups are the same size F = n ⋅ s x – 2 s 2 pooled where ... n = the sample size df numerator = k – 1 df denominator = n – k s 2 pooled = the mean of the sample variances (pooled variance) s x – 2 = the variance of the sample means Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software. Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) SS (Factor) k – 1 MS (Factor) = SS (Factor)/( k – 1) F = MS (Factor)/ MS (Error) Error (Within) SS (Error) n – k MS (Error) = SS (Error)/( n – k ) Total SS (Total) n – 1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in . Plan 1: n 1 = 4 Plan 2: n 2 = 3 Plan 3: n 3 = 3 5 3.5 8 4.5 7 4 4 3.5 3 4.5 s 1 = 16.5, s 2 =15, s 3 = 15.5 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. S S ( b e t w e e n ) = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n = s 1 2 4 + s 2 2 3 + s 3 2 3 − ( s 1 + s 2 + s 3 ) 2 10 where n 1 = 4, n 2 = 3, n 3 = 3 and n = n 1 + n 2 + n 3 = 10 = ( 16.5 ) 2 4 + ( 15 ) 2 3 + ( 15.5 ) 2 3 − ( 16.5 + 15 + 15.5 ) 2 10 S S ( b e t w e e n ) = 2.2458 S ( t o t a l ) = ∑ x 2 − ( ∑ x ) 2 n = ( 5 2 + 4.5 2 + 4 2 + 3 2 + 3.5 2 + 7 2 + 4.5 2 + 8 2 + 4 2 + 3.5 2 ) − ( 5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5 ) 2 10 = 244 − 47 2 10 = 244 − 220.9 S S ( t o t a l ) = 23.1 S S ( w i t h i n ) = S S ( t o t a l ) − S S ( b e t w e e n ) = 23.1 − 2.2458 S S ( w i t h i n ) = 20.8542 Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) SS (Factor) = SS (Between) = 2.2458 k – 1 = 3 groups – 1 = 2 MS (Factor) = SS (Factor)/( k – 1) = 2.2458/2 = 1.1229 F = MS (Factor)/ MS (Error) = 1.1229/2.9792 = 0.3769 Error (Within) SS (Error) = SS (Within) = 20.8542 n – k = 10 total data – 3 groups = 7 MS (Error) = SS (Error)/( n – k ) = 20.8542/7 = 2.9792 Total SS (Total) = 2.2458 + 20.8542 = 23.1 n – 1 = 10 total data – 1 = 9 Try It As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments bare soil a commercial ground cover black plastic straw compost All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants: Bare: n 1 = 3 Ground Cover: n 2 = 3 Plastic: n 3 = 3 Straw: n 4 = 3 Compost: n 5 = 3 2,625 5,348 6,583 7,285 6,277 2,997 5,682 8,560 6,897 7,818 4,915 5,482 3,830 9,230 8,677 Create the one-way ANOVA table. Enter the data into lists L1, L2, L3, L4 and L5. Press STAT and arrow over to TESTS. Arrow down to ANOVA. Press ENTER and enter L1, L2, L3, L4, L5). Press ENTER. The table was filled in with the results from the calculator. One-Way ANOVA table: Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) 36,648,561 5 – 1 = 4 36 , 648 , 561 4 = 9 , 162 , 140 9 , 162 , 140 2 , 044 , 672.6 = 4.4810 Error (Within) 20,446,726 15 – 5 = 10 20 , 446 , 726 10 = 2 , 044 , 672.6 Total 57,095,287 15 – 1 = 14 The one-way ANOVA hypothesis test is always right-tailed because larger F -values are way out in the right tail of the F -distribution curve and tend to make us reject H 0 . Let’s return to the slicing tomato exercise in . The means of the tomato yields under the five mulching conditions are represented by μ 1 , μ 2 , μ 3 , μ 4 , μ 5 . We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. The null and alternative hypotheses are: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j some i ≠ j The one-way ANOVA results are shown in Source of variation Sum of squares ( SS ) Degrees of freedom ( df ) Mean square ( MS ) F Factor (Between) 36,648,561 5 – 1 = 4 36,648,561 4 = 9,162,140 9,162,140 2,044,672 .6 = 4 .4810 Error (Within) 20,446,726 15 – 5 = 10 20,446,726 10 = 2,044,672 .6 Total 57,095,287 15 – 1 = 14 Distribution for the test: F 4,10 df ( num ) = 5 – 1 = 4 df ( denom ) = 15 – 5 = 10 Test statistic: F = 4.4810 Probability Statement: p -value = P ( F > 4.481) = 0.0248. Compare α and the p -value: α = 0.05, p -value = 0.0248 Make a decision: Since α > p -value, we cannot accept H 0 . Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields. Try It There are multiple variants of the virus that causes COVID-19. The length of hospital stays for patients afflicted with various strains of COVID-19 is shown in . Delta Strain Omicron Strain Alpha Strain Gamma Strain Beta Strain 13.9 11.7 18.2 16.9 9.3 14.9 15.1 14.6 12.8 15.8 16.8 9.9 10.1 11.2 16.4 Test whether the mean length of hospital stay is the same or different for the various strains of COVID-19. Construct the ANOVA table, find the p -value, and state your conclusion. Use a 5% significance level. We test for equality of mean length of hospital stay: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j , some i ≠ j The one-way ANOVA table results are shown: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F P-value Factor (Between) 14.056 5 – 1 = 4 3.514 0.353095 0.8362 Error (Within) 99.52 15 – 5 = 10 9.952 Total 113.576 15 – 1 = 14 Distribution for the test: F 4 , 10 Probability Statement: p - v a l u e = P ( F > 0 . 3531 ) = 0 . 8362 Compare  and the p -value: α = 0 . 05 , p - v a l u e = 08362 , α > p - v a l u e Make a decision: Since  > p -value, we do not reject H 0 . Conclusion: At the 5% significance level, there is insufficient evidence from these data that different strains of COVID-19 virus will cause a significant difference in the length of a hospital stay. Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in . Mean grades for four sororities Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 2.63 2.63 3.79 1.85 1.77 3.78 3.45 2.83 3.25 4.00 3.08 1.69 1.86 2.55 2.26 3.33 2.21 2.45 3.18 Using a significance level of 1%, is there a difference in mean grades among the sororities? Let μ 1 , μ 2 , μ 3 , μ 4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design , because each factor (i.e., sorority) has the same number of observations. H 0 : μ 1 = μ 2 = μ 3 = μ 4 H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 are equal. Distribution for the test: F 3,16 where k = 4 groups and n = 20 samples in total df ( num )= k – 1 = 4 – 1 = 3 df ( denom ) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph: Probability statement: p -value = P ( F > 2.23) = 0.1241 Compare α and the p -value: α = 0.01 p -value = 0.1241 α < p -value Make a decision: Since α < p -value, you cannot reject H 0 . Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Try It Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in . GPAs for four sports teams Basketball Baseball Hockey Lacrosse 3.6 2.1 4.0 2.0 2.9 2.6 2.0 3.6 2.5 3.9 2.6 3.9 3.3 3.1 3.2 2.7 3.8 3.4 3.2 2.5 Use a significance level of 5%, and determine if there is a difference in GPA among the teams. With a p -value of 0.9271, we decline to reject the null hypothesis. There is not sufficient evidence to conclude that there is a difference among the GPAs for the sports teams. A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in . Tommy's plants Tara's plants Nick's plants 24 25 23 21 31 27 23 23 22 30 20 30 23 28 20 Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance. This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = n ⋅ s x – 2 s 2 pooled . First, calculate the sample mean and sample variance of each group. Tommy's plants Tara's plants Nick's plants Sample mean 24.2 25.4 24.4 Sample variance 11.7 18.3 16.3 Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = s x – 2 Then MS between = n s x – 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s 2 pooled Then MS within = s 2 pooled = 15.433. The F statistic (or F ratio) is F = M S between M S within = n s x – 2 s 2 p o o l e d = ( 5 ) ( 0.413 ) 15.433 = 0.134 The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12 The distribution for the test is F 2,12 and the F statistic is F = 0.134 The p -value is P ( F > 0.134) = 0.8759. Decision: Since α = 0.03 and the p -value = 0.8759, then you cannot reject H 0 . (Why?) Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. Try It Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in . Notation The notation for the F distribution is F ~ F df ( num ), df ( denom ) where df ( num ) = df between and df ( denom ) = df within The mean for the F distribution is μ = d f ( n u m ) d f ( d e n o m ) – 2 References Tomato Data, Marist College School of Science (unpublished student research) Chapter Review Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. Formula Review S S between = ∑ ​ [ ( s j ) 2 n j ] − ( ∑ ​ s j ) 2 n S S total = ∑ ​ x 2 − ( ∑ ​ x ) 2 n S S within = S S total − S S between df between = df ( num ) = k – 1 df within = df(denom) = n – k MS between = S S between d f between MS within = S S within d f within F = M S between M S within k = the number of groups n j = the size of the j th group s j = the sum of the values in the j th group n = the total number of all values (observations) combined x = one value (one observation) from the data s x – 2 = the variance of the sample means s 2 p o o l e d = the mean of the sample variances (pooled variance) Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in are the weights for the different groups. Group 1 Group 2 Group 3 216 202 170 198 213 165 240 284 182 187 228 197 176 210 201 What is the Sum of Squares Factor? 4,939.2 What is the Sum of Squares Error? What is the df for the numerator? 2 What is the df for the denominator? What is the Mean Square Factor? 2,469.6 What is the Mean Square Error? What is the F statistic? 3.7416 Use the following information to answer the next eight exercises. Players from four different soccer teams are to be tested for mean goals scored per game. The entries in are the goals per game for the different teams. Team 1 Team 2 Team 3 Team 4 1 2 0 3 2 3 1 4 0 2 1 4 3 4 0 3 2 4 0 2 What is SS between ? What is the df for the numerator? 3 What is MS between ? What is SS within ? 13.2 What is the df for the denominator? What is MS within ? 0.825 What is the F statistic? Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p -value, so it is likely that we cannot accept the null hypothesis. Homework Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x – = ________ ________ ________ ________ ________ s 2 = ________ ________ ________ ________ ________ H 0 : µ 1 = µ 2 = µ 3 = µ 4 = µ 5 Hα : At least any two of the group means µ 1 , µ 2 , …, µ 5 are not equal. degrees of freedom – numerator: df ( num ) = _________ degrees of freedom – denominator: df ( denom ) = ________ df ( denom ) = 15 F statistic = ________", "section": "The F Distribution and the F-Ratio", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Facts About the F Distribution Here are some facts about the F distribution. The curve is not symmetrical but skewed to the right. There is a different curve for each set of degrees of freedom. The F statistic is greater than or equal to zero. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal as can be seen in the two figures below. Figure (b) with more degrees of freedom is more closely approaching the normal distribution, but remember that the F cannot ever be less than zero so the distribution does not have a tail that goes to infinity on the left as the normal distribution does. Other uses for the F distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter. References Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall, 1994. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 50. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118. “MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012. Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), \"A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich,\" Journal of the American Medical Association , 268, 1578-1580. Chapter Review The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p -value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p -value) to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from need to be used for hand calculations. In the case of balanced data (the groups are the same size) however, simplified calculations based on group means and variances may be used. In practice, of course, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look at your data! An F statistic can have what values? What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? The curves approximate the normal distribution. Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in . Team 1 Team 2 Team 3 Team 4 Team 5 36 32 48 38 41 42 35 50 44 39 51 38 39 46 40 What is the df(num) ? What is the df(denom) ? ten What are the Sum of Squares and Mean Squares Factors? What are the Sum of Squares and Mean Squares Errors? SS = 237.33; MS = 23.73 What is the F statistic? What is the p -value? 0.1614 At the 5% significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Group A Group B Group C 101 151 101 108 149 109 98 160 198 107 112 186 111 126 160 What is the df(num) ? two What is the df(denom) ? What are the SS between and MS between ? SS = 5,700.4; MS = 2,850.2 What are the SS within and MS within ? What is the F Statistic? 3.6101 What is the p -value? At the 10% significance level, are the scores among the different groups different? Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level. Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = ________ ________ ________ ________ ________ s 2 = ________ ________ ________ ________ ________ Enter the data into your calculator or computer. p -value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α . α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ Homework Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. Weights of Student Lab Rats Linda's rats Tuan's rats Javier's rats 43.5 47.0 51.2 39.4 40.5 40.9 41.3 38.9 37.9 46.0 46.3 45.0 38.2 44.2 48.6 H 0 : µ L = µ T = µ J H a : at least any two of the means are different df ( num ) = 2; df ( denom ) = 12 F distribution 0.67 0.5305 Answers may vary. Decision: Cannot reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in . Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. Working-class Professional (middle incomes) Professional (wealthy) 17.8 16.5 8.5 26.7 17.4 6.3 49.4 22.0 4.6 9.4 7.4 12.6 65.4 9.4 11.0 47.1 2.1 28.6 19.5 6.4 15.4 51.2 13.9 9.3 Use the following information to answer the next two exercises. lists the number of pages in four different types of magazines. Home decorating News Health Computer 172 87 82 104 286 94 153 136 163 123 87 98 205 106 103 207 197 101 96 146 Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? H a : µ c = µ n = µ h At least any two of the magazines have different mean lengths. df ( num ) = 2, df ( denom ) = 12 F distribution F = 15.28 p -value = 0.001 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that shows the results of a study. CNN FOX Local 45 15 72 12 43 37 18 68 56 38 50 60 23 31 51 35 22 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the means for the final exams the same for all statistics class delivery types? shows the scores on final exams from several randomly selected classes that used the different delivery types. Online Hybrid Face-to-Face 72 83 80 84 73 78 77 84 84 80 81 81 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ o = μ h = μ f At least two of the means are different. df ( n ) = 2, df ( d ) = 13 F 2,13 0.64 0.5437 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: The mean scores of different class delivery are not different. Are the mean number of times a month a person eats out the same for White people, Black people, Hispanic/Latinos and Asian people? Suppose that shows the results of a study. White Black Hispanic/Latino Asian 6 4 7 8 8 1 3 3 2 5 5 5 4 2 4 1 6 6 7 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that shows the results of a study. Powder Machine Made Hard Packed 1,210 2,107 2,846 1,080 1,149 1,638 1,537 862 2,019 941 1,870 1,178 1,528 2,233 1,382 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ p = μ m = μ h At least any two of the means are different. df ( n ) = 2, df ( d ) = 12 F 2,12 3.13 0.0807 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew. Paper type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy 5.1 meters 3.1 meters 4.7 meters 5.3 meters Medium 4 meters 3.5 meters 4.5 meters 6.1 meters Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? Yes or No. Why is this a balanced design? Calculate the sample mean and sample standard deviation for each group. Does the weight of the paper have an effect on how far the plane will travel? Use a 1% level of significance. Complete the test using the method shown in the bean plant example in . variance of the group means __________ MS between = ___________ mean of the three sample variances ___________ MS within = _____________ F statistic = ____________ df(num) = __________, df(denom) = ___________ number of groups _______ number of observations _______ p -value = __________ ( P ( F > _______) = __________) Graph the p -value. decision: _______________________ conclusion: _______________________________________________________________ DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 16.4 26.7 28.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 14.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three groups: The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H 0 : µ 1 = µ 2 = µ 3 ; H a : µ i ≠ ; some i ≠ j Define μ 1 , μ 2 , μ 3 , as the population mean number of eggs laid by the three groups of fruit flies. F statistic = 8.6657; p -value = 0.0004 Decision: Since the p -value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t -test to compare the RS and NS groups they are significantly different ( p = 0.0013). Similarly, SS and NS are significantly different ( p = 0.0006). However, the two selected groups, RS and SS are not significantly different ( p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 98 97.3 97.4 98.7 98.7 98.2 98.8 97.6 98 97.4 97.5 98.8 98.8 98.2 98.8 97.7 98 97.4 97.6 98.8 98.8 98.3 98.9 97.8 98 97.4 97.7 98.8 98.8 98.4 99 97.8 98.1 97.5 97.8 98.8 98.9 98.4 99 97.9 98.3 97.6 97.9 99.2 99 98.5 99 97.9 98.3 97.6 98 99.3 99 98.5 99.2 98 98.3 97.8 98 99.1 98.6 99.5 98.2 98.4 97.8 98 99.1 98.6 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98 98.4 99.9 99.3 98.2 98.5 98 98.6 100 99.4 98.2 98.6 98 98.6 100.8", "section": "Facts About the F Distribution", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Introduction Linear regression and correlation can help you determine if an auto mechanic’s salary is related to his work experience. (credit: modification of work “USPS commissions local repair-shop for some needed work on its older trucks” by Joshua Rothhaas/ Flickr, CC BY 2.0) Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability, or your gender or color. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee. These examples may or may not be tied to a model, meaning that some theory suggested that a relationship exists. This link between a cause and an effect, often referred to as a model, is the foundation of the scientific method and is the core of how we determine what we believe about how the world works. Beginning with a theory and developing a model of the theoretical relationship should result in a prediction, what we have called a hypothesis earlier. Now the hypothesis concerns a full set of relationships. As an example, in Economics the model of consumer choice is based upon assumptions concerning human behavior: a desire to maximize something called utility, knowledge about the benefits of one product over another, likes and dislikes, referred to generally as preferences, and so on. These combined to give us the demand curve. From that we have the prediction that as prices rise the quantity demanded will fall. Economics has models concerning the relationship between what prices are charged for goods and the market structure in which the firm operates, monopoly verse competition, for example. Models for who would be most likely to be chosen for an on-the-job training position, the impacts of Federal Reserve policy changes and the growth of the economy and on and on. Models are not unique to Economics, even within the social sciences. In political science, for example, there are models that predict behavior of bureaucrats to various changes in circumstances based upon assumptions of the goals of the bureaucrats. There are models of political behavior dealing with strategic decision making both for international relations and domestic politics. The so-called hard sciences are, of course, the source of the scientific method as they tried through the centuries to explain the confusing world around us. Some early models today make us laugh; spontaneous generation of life for example. These early models are seen today as not much more than the foundational myths we developed to help us bring some sense of order to what seemed chaos. The foundation of all model building is the perhaps the arrogant statement that we know what caused the result we see. This is embodied in the simple mathematical statement of the functional form that y = f(x). The response, Y, is caused by the stimulus, X. Every model will eventually come to this final place and it will be here that the theory will live or die. Will the data support this hypothesis? If so then fine, we shall believe this version of the world until a better theory comes to replace it. This is the process by which we moved from flat earth to round earth, from earth-center solar system to sun-center solar system, and on and on. The scientific method does not confirm a theory for all time: it does not prove “truth”. All theories are subject to review and may be overturned. These are lessons we learned as we first developed the concept of the hypothesis test earlier in this book. Here, as we begin this section, these concepts deserve review because the tool we will develop here is the cornerstone of the scientific method and the stakes are higher. Full theories will rise or fall because of this statistical tool; regression and the more advanced versions call econometrics. In this chapter we will begin with correlation, the investigation of relationships among variables that may or may not be founded on a cause and effect model. The variables simply move in the same, or opposite, direction. That is to say, they do not move randomly. Correlation provides a measure of the degree to which this is true. From there we develop a tool to measure cause and effect relationships; regression analysis. We will be able to formulate models and tests to determine if they are statistically sound. If they are found to be so, then we can use them to make predictions: if as a matter of policy we changed the value of this variable what would happen to this other variable? If we imposed a gasoline tax of 50 cents per gallon how would that effect the carbon emissions, sales of Hummers/Hybrids, use of mass transit, etc.? The ability to provide answers to these types of questions is the value of regression as both a tool to help us understand our world and to make thoughtful policy decisions.", "section": "Introduction", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Correlation Coefficient r As we begin this section we note that the type of data we will be working with has changed. Perhaps unnoticed, all the data we have been using is for a single variable. It may be from two samples, but it is still a univariate variable. The type of data described in the examples above and for any model of cause and effect is bivariate data — \"bi\" for two variables. In reality, statisticians use multivariate data, meaning many variables. For our work we can classify data into three broad categories, time series data, cross-section data, and panel data. We met the first two very early on. Time series data measures a single unit of observation; say a person, or a company or a country, as time passes. What are measured will be at least two characteristics, say the person’s income, the quantity of a particular good they buy and the price they paid. This would be three pieces of information in one time period, say 1985. If we followed that person across time we would have those same pieces of information for 1985,1986, 1987, etc. This would constitute a times series data set. If we did this for 10 years we would have 30 pieces of information concerning this person’s consumption habits of this good for the past decade and we would know their income and the price they paid. A second type of data set is for cross-section data. Here the variation is not across time for a single unit of observation, but across units of observation during one point in time. For a particular period of time we would gather the price paid, amount purchased, and income of many individual people. A third type of data set is panel data. Here a panel of units of observation is followed across time. If we take our example from above we might follow 500 people, the unit of observation, through time, ten years, and observe their income, price paid and quantity of the good purchased. If we had 500 people and data for ten years for price, income and quantity purchased we would have 15,000 pieces of information. These types of data sets are very expensive to construct and maintain. They do, however, provide a tremendous amount of information that can be used to answer very important questions. As an example, what is the effect on the labor force participation rate of women as their family of origin, mother and father, age? Or are there differential effects on health outcomes depending upon the age at which a person started smoking? Only panel data can give answers to these and related questions because we must follow multiple people across time. The work we do here however will not be fully appropriate for data sets such as these. Beginning with a set of data with two independent variables we ask the question: are these related? One way to visually answer this question is to create a scatter plot of the data. We could not do that before when we were doing descriptive statistics because those data were univariate. Now we have bivariate data so we can plot in two dimensions. Three dimensions are possible on a flat piece of paper, but become very hard to fully conceptualize. Of course, more than three dimensions cannot be graphed although the relationships can be measured mathematically. To provide mathematical precision to the measurement of what we see we use the correlation coefficient. The correlation tells us something about the co-movement of two variables, but nothing about why this movement occurred. Formally, correlation analysis assumes that both variables being analyzed are independent variables. This means that neither one causes the movement in the other. Further, it means that neither variable is dependent on the other, or for that matter, on any other variable. Even with these limitations, correlation analysis can yield some interesting results. The correlation coefficient , ρ (pronounced rho), is the mathematical statistic for a population that provides us with a measurement of the strength of a linear relationship between the two variables. For a sample of data, the statistic, r, developed by Karl Pearson in the early 1900s, is an estimate of the population correlation and is defined mathematically as: r = 1 n − 1 Σ ( X 1 i − X ¯ 1 ) ( X 2 i − X ¯ 2 ) s x 1 s x 2 OR r = ∑ ( X 1 i - X ¯ 1 ) ( X 2 i - X ¯ 2 ) ∑ ( X 1 i - X ¯ 1 ) 2 ∑ ( X 2 i - X ¯ 2 ) 2 where s x1 and s x2 are the standard deviations of the two independent variables X 1 and X 2 , X ¯ 1 and X ¯ 2 are the sample means of the two variables, and X 1i and X 2i are the individual observations of X 1 and X 2 . The correlation coefficient r ranges in value from -1 to 1. The second equivalent formula is often used because it may be computationally easier. As scary as these formulas look they are really just the ratio of the covariance between the two variables and the product of their two standard deviations. That is to say, it is a measure of relative variances. In practice all correlation and regression analysis will be provided through computer software designed for these purposes. Anything more than perhaps one-half a dozen observations creates immense computational problems. It was because of this fact that correlation, and even more so, regression, were not widely used research tools until after the advent of “computing machines”. Now the computing power required to analyze data using regression packages is deemed almost trivial by comparison to just a decade ago. To visualize any linear relationship that may exist review the plot of a scatter diagrams of the standardized data. presents several scatter diagrams and the calculated value of r. In panels (a) and (b) notice that the data generally trend together, (a) upward and (b) downward. Panel (a) is an example of a positive correlation and panel (b) is an example of a negative correlation, or relationship. The sign of the correlation coefficient tells us if the relationship is a positive or negative (inverse) one. If all the values of X 1 and X 2 are on a straight line the correlation coefficient will be either 1 or -1 depending on whether the line has a positive or negative slope and the closer to one or negative one the stronger the relationship between the two variables. BUT ALWAYS REMEMBER THAT THE CORRELATION COEFFICIENT DOES NOT TELL US THE SLOPE. Remember, all the correlation coefficient tells us is whether or not the data are linearly related. In panel (d) the variables obviously have some type of very specific relationship to each other, but the correlation coefficient is zero, indicating no linear relationship exists. If you suspect a linear relationship between X 1 and X 2 then r can measure how strong the linear relationship is. What the VALUE of r tells us: The value of r is always between –1 and +1: –1 ≤ r ≤ 1. The size of the correlation r indicates the strength of the linear relationship between X 1 and X 2 . Values of r close to –1 or to +1 indicate a stronger linear relationship between X 1 and X 2 . If r = 0 there is absolutely no linear relationship between X 1 and X 2 (no linear correlation) . If r = 1, there is perfect positive correlation. If r = –1, there is perfect negative correlation. In both these cases, all of the original data points lie on a straight line: ANY straight line no matter what the slope. Of course, in the real world, this will not generally happen. What the SIGN of r tells us A positive value of r means that when X 1 increases, X 2 tends to increase and when X 1 decreases, X 2 tends to decrease (positive correlation) . A negative value of r means that when X 1 increases, X 2 tends to decrease and when X 1 decreases, X 2 tends to increase (negative correlation) . NOTE Strong correlation does not suggest that X 1 causes X 2 or X 2 causes X 1 . We say \"correlation does not imply causation.\" In order to have a correlation coefficient between traits A and B, it is necessary to have: one group of subjects, some of whom possess characteristics of trait A, the remainder possessing those of trait B measures of trait A on one group of subjects and of trait B on another group two groups of subjects, one which could be classified as A or not A, the other as B or not B two groups of subjects, one which could be classified as A or not A, the other as B or not B d Define the Correlation Coefficient and give a unique example of its use. A measure of the degree to which variation of one variable is related to variation in one or more other variables. The most commonly used correlation coefficient indicates the degree to which variation in one variable is described by a straight line relation with another variable. Suppose that sample information is available on family income and Years of schooling of the head of the household. A correlation coefficient = 0 would indicate no linear association at all between these two variables. A correlation of 1 would indicate perfect linear association (where all variation in family income could be associated with schooling and vice versa). If the correlation between age of an auto and money spent for repairs is +.90 81% of the variation in the money spent for repairs is explained by the age of the auto 81% of money spent for repairs is unexplained by the age of the auto 90% of the money spent for repairs is explained by the age of the auto none of the above a. 81% of the variation in the money spent for repairs is explained by the age of the auto Suppose that college grade-point average and verbal portion of an IQ test had a correlation of .40. What percentage of the variance do these two have in common? 20 16 40 80 b. 16 True or false? If false, explain why: The coefficient of determination can have values between -1 and +1. The coefficient of determination is r··2 with 0 ≤ r··2 ≤ 1, since -1 ≤ r ≤ 1. True or False: Whenever r is calculated on the basis of a sample, the value which we obtain for r is only an estimate of the true correlation coefficient which we would obtain if we calculated it for the entire population. True Under a \"scatter diagram\" there is a notation that the coefficient of correlation is .10. What does this mean? plus and minus 10% from the means includes about 68% of the cases one-tenth of the variance of one variable is shared with the other variable one-tenth of one variable is caused by the other variable on a scale from -1 to +1, the degree of linear relationship between the two variables is +.10 d. on a scale from -1 to +1, the degree of linear relationship between the two variables is +.10 The correlation coefficient for X and Y is known to be zero. We then can conclude that: X and Y have standard distributions the variances of X and Y are equal there exists no relationship between X and Y there exists no linear relationship between X and Y none of these d. there exists no linear relationship between X and Y What would you guess the value of the correlation coefficient to be for the pair of variables: \"number of man-hours worked\" and \"number of units of work completed\"? Approximately 0.9 Approximately 0.4 Approximately 0.0 Approximately -0.4 Approximately -0.9 Approximately 0.9 In a given group, the correlation between height measured in feet and weight measured in pounds is +.68. Which of the following would alter the value of r? height is expressed centimeters. weight is expressed in Kilograms. both of the above will affect r. neither of the above changes will affect r. d. neither of the above changes will affect r. Bivariate two variables are present in the model where one is the “cause” or independent variable and the other is the “effect” of dependent variable. Multivariate a system or model where more than one independent variable is being used to predict an outcome. There can only ever be one dependent variable, but there is no limit to the number of independent variables. R – Correlation Coefficient A number between −1 and 1 that represents the strength and direction of the relationship between “X” and “Y.” The value for “ r ” will equal 1 or −1 only if all the plotted points form a perfectly straight line. Linear a model that takes data and regresses it into a straight line equation.", "section": "The Correlation Coefficient r", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Testing the Significance of the Correlation Coefficient The correlation coefficient, r , tells us about the strength and direction of the linear relationship between X 1 and X 2 . The sample data are used to compute r , the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r , is our estimate of the unknown population correlation coefficient. ρ = population correlation coefficient (unknown) r = sample correlation coefficient (known; calculated from sample data) The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is \"close to zero\" or \"significantly different from zero\". We decide this based on the sample correlation coefficient r and the sample size n . If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is \"significant.\" Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between X 1 and X 2 because the correlation coefficient is significantly different from zero. What the conclusion means: There is a significant linear relationship X 1 and X 2 . If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is \"not significant\". Performing the Hypothesis Test Null Hypothesis: H 0 : ρ = 0 Alternate Hypothesis: H a : ρ ≠ 0 What the Hypotheses Mean in Words Null Hypothesis H 0 : The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship (correlation) between X 1 and X 2 in the population. Alternate Hypothesis H a : The population correlation coefficient is significantly different from zero. There is a significant linear relationship (correlation) between X 1 and X 2 in the population. Drawing a Conclusion There are two methods of making the decision concerning the hypothesis. The test statistic to test this hypothesis is: t c = r ( 1 − r 2 ) ( n − 2 ) OR t c = r n − 2 1 − r 2 Where the second formula is an equivalent form of the test statistic, n is the sample size and the degrees of freedom are n-2. This is a t-statistic and operates in the same way as other t tests. Calculate the t-value and compare that with the critical value from the t-table at the appropriate degrees of freedom and the level of confidence you wish to maintain. If the calculated value is in the tail then cannot accept the null hypothesis that there is no linear relationship between these two independent random variables. If the calculated t-value is NOT in the tailed then cannot reject the null hypothesis that there is no linear relationship between the two variables. A quick shorthand way to test correlations is the relationship between the sample size and the correlation. If: | r | ≥ 2 n then this implies that the correlation between the two variables demonstrates that a linear relationship exists and is statistically significant at approximately the 0.05 level of significance. As the formula indicates, there is an inverse relationship between the sample size and the required correlation for significance of a linear relationship. With only 10 observations, the required correlation for significance is 0.6325, for 30 observations the required correlation for significance decreases to 0.3651 and at 100 observations the required level is only 0.2000. Correlations may be helpful in visualizing the data, but are not appropriately used to \"explain\" a relationship between two variables. Perhaps no single statistic is more misused than the correlation coefficient. Citing correlations between health conditions and everything from place of residence to eye color have the effect of implying a cause and effect relationship. This simply cannot be accomplished with a correlation coefficient. The correlation coefficient is, of course, innocent of this misinterpretation. It is the duty of the analyst to use a statistic that is designed to test for cause and effect relationships and report only those results if they are intending to make such a claim. The problem is that passing this more rigorous test is difficult so lazy and/or unscrupulous \"researchers\" fall back on correlations when they cannot make their case legitimately. Define a t Test of a Regression Coefficient, and give a unique example of its use. Definition: A t test is obtained by dividing a regression coefficient by its standard error and then comparing the result to critical values for Students' t with Error df . It provides a test of the claim that β i = 0 when all other variables have been included in the relevant regression model. Example: Suppose that 4 variables are suspected of influencing some response. Suppose that the results of fitting Y i = β 0 + β 1 X 1 i + β 2 X 2 i + β 3 X 3 i + β 4 X 4 i + e i include: Variable Regression coefficient Standard error of regular coefficient .5 1 -3 .4 2 +2 .02 3 +1 .6 4 -.5 t calculated for variables 1, 2, and 3 would be 5 or larger in absolute value while that for variable 4 would be less than 1. For most significance levels, the hypothesis β 1 = 0 would be rejected. But, notice that this is for the case when X 2 , X 3 , and X 4 have been included in the regression. For most significance levels, the hypothesis β 4 = 0 would be continued (retained) for the case where X 1 , X 2 , and X 3 are in the regression. Often this pattern of results will result in computing another regression involving only X 1 , X 2 , X 3 , and examination of the t ratios produced for that case. The correlation between scores on a neuroticism test and scores on an anxiety test is high and positive; therefore anxiety causes neuroticism those who score low on one test tend to score high on the other. those who score low on one test tend to score low on the other. no prediction from one test to the other can be meaningfully made. c. those who score low on one test tend to score low on the other.", "section": "Testing the Significance of the Correlation Coefficient", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Linear Equations Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form: y = a + bx where a and b are constant numbers. The variable x is the independent variable, and y is the dependent variable. Another way to think about this equation is a statement of cause and effect. The X variable is the cause and the Y variable is the hypothesized effect. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable. The following examples are linear equations. y = 3 + 2x y = –0.01 + 1.2x Try It Is the following an example of a linear equation? y = –0.125 – 3.5 x The graph of a linear equation of the form y = a + bx is a straight line . Any line that is not vertical can be described by this equation. Graph the equation y = –1 + 2 x . Try It Is the following an example of a linear equation? Why or why not? No, the graph is not a straight line; therefore, it is not a linear equation. A local small business completes federal tax returns for customers. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job. Find the equation that expresses the total cost in terms of the number of hours required to complete the job. Let x = the number of hours it takes to get the job done. Let y = the total cost to the customer. The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)( x ) is the cost of the tax return processing only. The total cost is: y = 31.50 + 32 x Try It Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class. Slope and Y -Intercept of a Linear Equation For the linear equation y = a + bx , b = slope and a = y -intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the y -intercept is the y coordinate of the point (0, a ) where the line crosses the y -axis. From calculus the slope is the first derivative of the function. For a linear function the slope is dy / dx = b where we can read the mathematical expression as \"the change in y ( dy ) that results from a change in x ( dx ) = b * dx \". Three possible graphs of y = a + bx . (a) If b > 0, the line slopes upward to the right. (b) If b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right. Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15 x . What are the independent and dependent variables? What is the y -intercept and what is the slope? Interpret them using complete sentences. The independent variable ( x ) is the number of hours Svetlana tutors each session. The dependent variable ( y ) is the amount, in dollars, Svetlana earns for each session. The y -intercept is 25 ( a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 ( b = 15). For each session, Svetlana earns $15 for each hour she tutors. Try It Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20 x . What are the independent and dependent variables? What is the y -intercept and what is the slope? Interpret them using complete sentences. True or False? If False, correct it: Suppose a 95% confidence interval for the slope β of the straight line regression of Y on X is given by -3.5 < β < -0.5. Then a two-sided test of the hypothesis H 0 : β = −1 would result in rejection of H 0 at the 1% level of significance. False. Since H 0 : β = −1 would not be rejected at α = 0.05 , it would not be rejected at α = 0.01 . True or False: It is safer to interpret correlation coefficients as measures of association rather than causation because of the possibility of spurious correlation. True We are interested in finding the linear relation between the number of widgets purchased at one time and the cost per widget. The following data has been obtained: X: Number of widgets purchased – 1, 3, 6, 10, 15 Y: Cost per widget(in dollars) – 55, 52, 46, 32, 25 Suppose the regression line is y ^ = −2.5 x + 60 . We compute the average price per widget if 30 are purchased and observe which of the following? y ^ = 15 dollars ; obviously, we are mistaken; the prediction y ^ is actually +15 dollars. y ^ = 15 dollars , which seems reasonable judging by the data. y ^ = −15 dollars , which is obvious nonsense. The regression line must be incorrect. y ^ = −15 dollars , which is obvious nonsense. This reminds us that predicting Y outside the range of X values in our data is a very poor practice. d Discuss briefly the distinction between correlation and causality. Some variables seem to be related, so that knowing one variable's status allows us to predict the status of the other. This relationship can be measured and is called correlation. However, a high correlation between two variables in no way proves that a cause-and-effect relation exists between them. It is entirely possible that a third factor causes both variables to vary together. True or False: If r is close to + or -1, we shall say there is a strong correlation, with the tacit understanding that we are referring to a linear relationship and nothing else. True Chapter Review The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form y = mx + b , where m and b are constants, x is the independent variable, y is the dependent variable. In a statistical context, a linear equation is written in the form y = a + bx , where a and b are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation y = a + bx , the constant b , called the coefficient, represents the slope . The constant a is called the y-intercept . The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable ( y ) changes for every one unit increase in the independent ( x ) variable, on average. The y -intercept is used to describe the dependent variable when the independent variable equals zero. Y – the dependent variable Also, using the letter “y” represents actual values while y ^ represents predicted or estimated values. Predicted values will come from plugging in observed “x” values into a linear model. X – the independent variable This will sometimes be referred to as the “predictor” variable, because these values were measured in order to determine what possible outcomes could be predicted. a is the symbol for the Y-Intercept Sometimes written as b 0 , because when writing the theoretical linear model β 0 is used to represent a coefficient for a population. b is the symbol for Slope The word coefficient will be used regularly for the slope, because it is a number that will always be next to the letter “x.” It will be written as b 1 when a sample is used, and β 1 will be used with a population or when writing the theoretical linear model.", "section": "Linear Equations", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "The Regression Equation Regression analysis is a statistical technique that can test the hypothesis that a variable is dependent upon one or more other variables. Further, regression analysis can provide an estimate of the magnitude of the impact of a change in one variable on another. This last feature, of course, is all important in predicting future values. Regression analysis is based upon a functional relationship among variables and further, assumes that the relationship is linear. This linearity assumption is required because, for the most part, the theoretical statistical properties of non-linear estimation are not well worked out yet by the mathematicians and econometricians. This presents us with some difficulties in economic analysis because many of our theoretical models are nonlinear. The marginal cost curve, for example, is decidedly nonlinear as is the total cost function, if we are to believe in the effect of specialization of labor and the Law of Diminishing Marginal Product. There are techniques for overcoming some of these difficulties, exponential and logarithmic transformation of the data for example, but at the outset we must recognize that standard ordinary least squares (OLS) regression analysis will always use a linear function to estimate what might be a nonlinear relationship. The general linear regression model can be stated by the equation: y i = β 0 + β 1 X 1 i + β 2 X 2 i + ⋯ + β k X k i + ε i where β 0 is the intercept, β i 's are the slope between Y and the appropriate X i , and ε (pronounced epsilon), is the error term that captures errors in measurement of Y and the effect on Y of any variables missing from the equation that would contribute to explaining variations in Y. This equation is the theoretical population equation and therefore uses Greek letters. The equation we will estimate will have the Roman equivalent symbols. This is parallel to how we kept track of the population parameters and sample parameters before. The symbol for the population mean was µ and for the sample mean X ¯ and for the population standard deviation was σ and for the sample standard deviation was s. The equation that will be estimated with a sample of data for two independent variables will thus be: y i = b 0 + b 1 x 1 i + b 2 x 2 i + e i As with our earlier work with probability distributions, this model works only if certain assumptions hold. These are that the Y is normally distributed, the errors are also normally distributed with a mean of zero and a constant standard deviation, and that the error terms are independent of the size of X and independent of each other. Assumptions of the Ordinary Least Squares Regression Model Each of these assumptions needs a bit more explanation. If one of these assumptions fails to be true, then it will have an effect on the quality of the estimates. Some of the failures of these assumptions can be fixed while others result in estimates that quite simply provide no insight into the questions the model is trying to answer or worse, give biased estimates. The independent variables, X , are all measured without error, and are all measured without error, and are uncorrelated with the error term. This assumption is saying in effect that that the error term, ϵ , captures only what the X variables cannot explain about Y . The independent variables should not be mixed up with those unexplained leftovers. (The X variables can be related to Y ; that's what we want to measure.) The error term is a random variable with a mean of zero and a constant variance. This means the spread (variance) of the errors around the predicted line should be about the same, no matter the value of X . Consider the relationship between personal income and the quantity of a good purchased as an example of a case where the variance might depend upon the value of the independent variable, income. It is plausible that as income increases, the variation in the amount purchased (the errors) will also increase simply because of the flexibility provided with higher levels of income. The assumption is for constant variance of the errors with respect to the magnitude of X called homoscedasticity. If the assumption fails, then it is called heteroscedasticity. shows the case of homoscedasticity where all three distributions have the same variance around the predicted value of Y regardless of the magnitude of X . Error terms should be normally distributed. This can be seen in by the shape of the distributions placed on the predicted line at the expected value of the relevant value of Y . The independent variables are not perfectly correlated with each other (no perfect multicollinearity). The model is designed to estimate the effects of independent variables on some dependent variable in accordance with a proposed theory. The case where some or more of the independent variables are correlated is not unusual, and it's perfectly okay for regression analysis. There may be no cause and effect relationship among the independent variables, but nevertheless they move together. Take the case of a simple supply curve where quantity supplied is theoretically related to the price of the product and the prices of inputs. There may be multiple inputs that may over time move together from general inflationary pressure. High, but not perfect, correlations like this are called multicollinearity, which makes it harder to precisely separate the individual effects of each X (inflated standard errors and unstable estimates), but will be taken up in detail later. The error terms are uncorrelated with each other. This situation arises when one error term is related to another, such as from omitted factors that affect multiple observations. While not exclusively a time series problem, it is here that we most often see this case. An omitted factor in time period one might affect the Y variable, and then carry over to affect the next time period. This effect gives rise to a relationship among the error terms. This case is called autocorrelation, “self-correlated.” The error terms are now not independent of each other, but rather influence subsequent error terms. shows the case where the assumptions of the regression model are being satisfied. The estimated line is y ^ = a + b x. Three values of X are shown. A normal distribution is placed at each point where X equals the estimated line and the associated error at each value of Y. Notice that the three distributions are normally distributed around the point on the line, and further, the variation, variance, around the predicted value is constant indicating homoscedasticity from assumption 2. does not show all the assumptions of the regression model, but it helps visualize these important ones. This is the general form that is most often called the multiple regression model. So-called \"simple\" regression analysis has only one independent (right-hand) variable rather than many independent variables. Simple regression is just a special case of multiple regression. There is some value in beginning with simple regression: it is easy to graph in two dimensions, difficult to graph in three dimensions, and impossible to graph in more than three dimensions. Consequently, our graphs will be for the simple regression case. presents the regression problem in the form of a scatter plot graph of the data set where it is hypothesized that Y is dependent upon the single independent variable X. A basic relationship from Macroeconomic Principles is the consumption function. This theoretical relationship states that as a person's income rises, their consumption rises, but by a smaller amount than the rise in income. If Y is consumption and X is income in the equation below , the regression problem is, first, to establish that this relationship exists, and second, to determine the impact of a change in income on a person's consumption. The parameter β 1 was called the Marginal Propensity to Consume in Macroeconomics Principles. Each \"dot\" in represents the consumption and income of different individuals at some point in time. This was called cross-section data earlier; observations on variables at one point in time across different people or other units of measurement. This analysis is often done with time series data, which would be the consumption and income of one individual or country at different points in time. For macroeconomic problems it is common to use times series aggregated data for a whole country. For this particular theoretical concept these data are readily available in the annual report of the President’s Council of Economic Advisors. The regression problem comes down to determining which straight line would best represent the data in . Regression analysis is sometimes called \"least squares\" analysis because the method of determining which line best \"fits\" the data is to minimize the sum of the squared residuals of a line put through the data. Population Equation: C = β 0 + β 1 Income + ε Estimated Equation: C = b 0 + b 1 Income + e This figure shows the assumed relationship between consumption and income from macroeconomic theory. Here the data are plotted as a scatter plot and an estimated straight line has been drawn. From this graph we can see an error term, e 1 . Each data point also has an error term. Again, the error term is put into the equation to capture effects on consumption that are not caused by income changes. Such other effects might be a person’s savings or wealth, or periods of unemployment. We will see how by minimizing the sum of these errors we can get an estimate for the slope and intercept of this line. Consider the graph below. The notation has returned to that for the more general model rather than the specific case of the Macroeconomic consumption function in our example. The ŷ is read \" y hat\" and is the estimated value of y . (In C ^ represents the estimated value of consumption because it is on the estimated line.) It is the value of y obtained using the regression line. ŷ is not generally equal to y from the data. The term y 0 - ŷ 0 = e 0 is called the residual or \"error\" . It is not an error in the sense of a mistake. The error term was put into the estimating equation to capture missing variables and errors in measurement that may have occurred in the dependent variables. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y . In other words, it measures the vertical distance between the actual data point and the predicted point on the line as can be seen on the graph at point X 0 . If the observed data point lies above the line, the residual is positive, and the line underestimates the actual data value for y . If the observed data point lies below the line, the residual is negative, and the line overestimates that actual data value for y . In the graph, y 0 - ŷ 0 = e 0 is the residual for the point shown. Here the point lies above the line and the residual is positive. For each data point the residuals, or errors, are calculated y i – ŷ i = e i for i = 1, 2, 3, ..., n where n is the sample size. Each |e| is a vertical distance. The sum of the errors squared is the term obviously called Sum of Squared Errors (SSE) . Using calculus, you can determine the straight line that has the parameter values of b 0 and b 1 that minimizes the SSE . When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation: ŷ = b 0 + b 1 x where b 0 = y – − b 1 x ¯ and b 1 = Σ ( x − x ¯ ) ( y − y – ) Σ ( x − x ¯ ) 2 = cov ( x , y ) s x 2 The sample means of the x values and the y values are x ¯ and y – , respectively. The best fit line always passes through the point ( x ¯ , y – ) called the points of means. The slope b can also be written as: b 1 = r y,x ( s y s x ) where s y = the standard deviation of the y values and s x = the standard deviation of the x values and r is the correlation coefficient between x and y . These equations are called the Normal Equations and come from another very important mathematical finding called the Gauss-Markov Theorem without which we could not do regression analysis. The Gauss-Markov Theorem tells us that the estimates we get from using the ordinary least squares (OLS) regression method will result in estimates that have some very important properties. In the Gauss-Markov Theorem it was proved that a least squares line is BLUE, which is, B est, L inear, U nbiased, E stimator. Best is the statistical property that an estimator is the one with the minimum variance. Linear refers to the property of the type of line being estimated. An unbiased estimator is one whose estimating function has an expected mean equal to the mean of the population. (You will remember that the expected value of µ x ¯ was equal to the population mean µ in accordance with the Central Limit Theorem. This is exactly the same concept here). Both Gauss and Markov were giants in the field of mathematics, and Gauss in physics too, in the 18 th century and early 19 th century. They barely overlapped chronologically and never in geography, but Markov’s work on this theorem was based extensively on the earlier work of Carl Gauss. The extensive applied value of this theorem had to wait until the middle of this last century. Using the OLS method we can now find the estimate of the error variance which is the variance of the squared errors, e 2 . This is sometimes called the standard error of the estimate . (Grammatically this is probably best said as the estimate of the error’s variance) The formula for the estimate of the error variance is: s e 2 = Σ ( y i − ŷ i ) 2 n − k = Σ e i 2 n − k where ŷ is the predicted value of y and y is the observed value, and thus the term ( y i − ŷ i ) 2 is the squared errors that are to be minimized to find the estimates of the regression line parameters. This is really just the variance of the error terms and follows our regular variance formula. One important note is that here we are dividing by ( n - k ) , which is the degrees of freedom. The degrees of freedom of a regression equation will be the number of observations, n, reduced by the number of estimated parameters, which includes the intercept as a parameter. The variance of the errors is fundamental in testing hypotheses for a regression. It tells us just how “tight” the dispersion is about the line. As we will see shortly, the greater the dispersion about the line, meaning the larger the variance of the errors, the less probable that the hypothesized independent variable will be found to have a significant effect on the dependent variable. In short, the theory being tested will more likely fail if the variance of the error term is high. Upon reflection this should not be a surprise. As we tested hypotheses about a mean we observed that large variances reduced the calculated test statistic and thus it failed to reach the tail of the distribution. In those cases, the null hypotheses could not be rejected. If we cannot reject the null hypothesis in a regression problem, we must conclude that the hypothesized independent variable has no effect on the dependent variable. A way to visualize this concept is to draw two scatter plots of x and y data along a predetermined line. The first will have little variance of the errors, meaning that all the data points will move close to the line. Now do the same except the data points will have a large estimate of the error variance, meaning that the data points are scattered widely along the line. Clearly the confidence about a relationship between x and y is effected by this difference between the estimate of the error variance. Residuals Plots A residuals plot can be used to help determine if a set of ( x , y ) data is linearly correlated. For each data point used to create the correlation line, a residual y - ŷ can be calculated, where y is the observed value of the response variable and ŷ is the value predicted by the correlation line. The difference between these values is called the residual. A residuals plot shows the explanatory variable x on the horizontal axis and the residual for that value on the vertical axis. The residuals plot is often shown together with a scatter plot of the data. While a scatter plot of the data should resemble a straight line, a residuals plot should appear random, with no pattern and no outliers. It should also show constant error variance, meaning the residuals should not consistently increase (or decrease) as the explanatory variable x increases. A residuals plot can be created using StatCrunch or a TI calculator. The plot should appear random. A box plot of the residuals is also helpful to verify that there are no outliers in the data. By observing the scatter plot of the data, the residuals plot, and the box plot of residuals, together with the linear correlation coefficient, we can usually determine if it is reasonable to conclude that the data are linearly correlated. EXAMPLE: A shop owner uses a straight-line regression to estimate the number of ice cream cones that would be sold in a day based on the temperature at noon. The owner has data for a 2-year period and chose nine days at random. A scatter plot of the data is shown, together with a residuals plot. Temperature ° F Ice cream cones sold 70 105 85 240 65 49 72 147 80 231 61 38 75 193 78 196 68 89 Table showing the number of ice cream cones sold on nine random days and the temperature at noon on those days. Scatter plot of the data looks like a straight line. Residuals plot appears random. Testing the Parameters of the Line The whole goal of the regression analysis was to test the hypothesis that the dependent variable, Y, was in fact dependent upon the values of the independent variables as asserted by some foundation theory, such as the consumption function example. Looking at the estimated equation under , we see that this amounts to determining the values of b 0 and b 1 . Notice that again we are using the convention of Greek letters for the population parameters and Roman letters for their estimates. The regression analysis output provided by the computer software will produce an estimate of b 0 and b 1 , and any other b's for other independent variables that were included in the estimated equation. The issue is how good are these estimates? In order to test a hypothesis concerning any estimate, we have found that we need to know the underlying sampling distribution. It should come as no surprise at his stage in the course that the answer is going to be the normal distribution. This can be seen by remembering the assumption that the error term in the population, ε, is normally distributed. If the error term is normally distributed and the variance of the estimates of the equation parameters, b 0 and b 1 , are determined by the variance of the error term, it follows that the variances of the parameter estimates are also normally distributed. And indeed this is just the case. We can see this by the creation of the test statistic for the test of hypothesis for the slope parameter, β 1 in our consumption function equation. To test whether or not Y does indeed depend upon X, or in our example, that consumption depends upon income, we need only test the hypothesis that β 1 equals zero. This hypothesis would be stated formally as: H 0 : β 1 = 0 H a : β 1 ≠ 0 If we cannot reject the null hypothesis, we must conclude that our theory has no validity. If we cannot reject the null hypothesis that β 1 = 0 then b 1 , the coefficient of Income, is zero and zero times anything is zero. Therefore the effect of Income on Consumption is zero. There is no relationship as our theory had suggested. Notice that we have set up the presumption, the null hypothesis, as \"no relationship\". This puts the burden of proof on the alternative hypothesis. In other words, if we are to validate our claim of finding a relationship, we must do so with a level of significance greater than 90, 95, or 99 percent. The status quo is ignorance, no relationship exists, and to be able to make the claim that we have actually added to our body of knowledge we must do so with significant probability of being correct. John Maynard Keynes got it right and thus was born Keynesian economics starting with this basic concept in 1936. The test statistic for this test comes directly from our old friend the standardizing formula: t c = b 1 − β 1 S b 1 where b 1 is the estimated value of the slope of the regression line, β 1 is the hypothesized value of beta, in this case zero, and S b 1 is the standard deviation of the estimate of b 1 . In this case we are asking how many standard deviations is the estimated slope away from the hypothesized slope. This is exactly the same question we asked before with respect to a hypothesis about a mean: how many standard deviations is the estimated mean, the sample mean, from the hypothesized mean? The test statistic is written as a student's t -distribution, but if the sample size is larger enough so that the degrees of freedom are greater than 30 we may again use the normal distribution. To see why we can use the student's t or normal distribution we have only to look at S b 1 ,the formula for the standard deviation of the estimate of b 1 : S b 1 = S e Σ ( x i − x – ) 2 or S b 1 = S e ( n − 1 ) S x 2 Where S e is the estimate of the error variance and S 2 x is the variance of x values of the coefficient of the independent variable being tested. We see that S e , the estimate of the error variance , is part of the computation. Because the estimate of the error variance is based on the assumption of normality of the error terms, we can conclude that the sampling distribution of the b's, the coefficients of our hypothesized regression line, are also normally distributed. One last note concerns the degrees of freedom of the test statistic, ν=n-k. Previously we subtracted 1 from the sample size to determine the degrees of freedom in a student's t problem. Here we must subtract one degree of freedom for each parameter estimated in the equation. For the example of the consumption function we lose 2 degrees of freedom, one for b 0 , the intercept, and one for b 1 , the slope of the consumption function. The degrees of freedom would be n - k - 1, where k is the number of independent variables and the extra one is lost because of the intercept. If we were estimating an equation with three independent variables, we would lose 4 degrees of freedom: three for the independent variables, k, and one more for the intercept. The decision rule for acceptance or rejection of the null hypothesis follows exactly the same form as in all our previous test of hypothesis. Namely, if the calculated value of t (or Z) falls into the tails of the distribution, where the tails are defined by α ,the required significance level in the test, we cannot accept the null hypothesis. If on the other hand, the calculated value of the test statistic is within the critical region, we cannot reject the null hypothesis. If we conclude that we cannot accept the null hypothesis, we are able to state with ( 1 - α ) level of confidence that the slope of the line is given by b 1 . This is an extremely important conclusion. Regression analysis not only allows us to test if a cause and effect relationship exists, we can also determine the magnitude of that relationship, if one is found to exist. It is this feature of regression analysis that makes it so valuable. If models can be developed that have statistical validity, we are then able to simulate the effects of changes in variables that may be under our control with some degree of probability , of course. For example, if advertising is demonstrated to effect sales, we can determine the effects of changing the advertising budget and decide if the increased sales are worth the added expense. Multicollinearity Our discussion earlier indicated that like all statistical models, the OLS regression model has important assumptions attached. Each assumption, if violated, has an effect on the ability of the model to provide useful and meaningful estimates. The Gauss-Markov Theorem has assured us that the OLS estimates are unbiased and minimum variance, but this is true only under the assumptions of the model. Here we will look at the effects on OLS estimates if the independent variables are correlated. The other assumptions and the methods to mitigate the difficulties they pose if they are found to be violated are examined in Econometrics courses. We take up multicollinearity because it is so often prevalent in Economic models and it often leads to frustrating results. The OLS model assumes that all the independent variables are independent of each other. This assumption is easy to test for a particular sample of data with simple correlation coefficients. Correlation, like much in statistics, is a matter of degree: a little is not good, and a lot is terrible. The goal of the regression technique is to tease out the independent impacts of each of a set of independent variables on some hypothesized dependent variable. If two 2 independent variables are interrelated, that is, correlated, then we cannot isolate the effects on Y of one from the other. In an extreme case where x 1 is a linear combination of x 2 , correlation equal to one, both variables move in identical ways with Y. In this case it is impossible to determine the variable that is the true cause of the effect on Y. (If the two variables were actually perfectly correlated, then mathematically no regression results could actually be calculated.) The normal equations for the coefficients show the effects of multicollinearity on the coefficients. b 1 = s y ( r x 1 y - r x 1 x 2 r x 2 y ) s x 1 ( 1 - r x 1 x 2 2 ) b 2 = s y ( r x 2 y - r x 1 x 2 r x 1 y ) s x 2 ( 1 - r x 1 x 2 2 ) b 0 = y - - b 1 x ¯ 1 - b 2 x ¯ 2 The correlation between x 1 and x 2 , r x 1 x 2 2 , appears in the denominator of both the estimating formula for b 1 and b 2 . If the assumption of independence holds, then this term is zero. This indicates that there is no effect of the correlation on the coefficient. On the other hand, as the correlation between the two independent variables increases the denominator decreases, and thus the estimate of the coefficient increases. The correlation has the same effect on both of the coefficients of these two variables. In essence, each variable is “taking” part of the effect on Y that should be attributed to the collinear variable. This results in biased estimates. Multicollinearity has a further deleterious impact on the OLS estimates. The correlation between the two independent variables also shows up in the formulas for the estimate of the variance for the coefficients. s b 1 2 = s e 2 ( n - 1 ) s x 1 2 ( 1 - r x 1 x 2 2 ) s b 2 2 = s e 2 ( n - 1 ) s x 2 2 ( 1 - r x 1 x 2 2 ) Here again we see the correlation between x 1 and x 2 in the denominator of the estimates of the variance for the coefficients for both variables. If the correlation is zero as assumed in the regression model, then the formula collapses to the familiar ratio of the variance of the errors to the variance of the relevant independent variable. If however the two independent variables are correlated, then the variance of the estimate of the coefficient increases. This results in a smaller t-value for the test of hypothesis of the coefficient. In short, multicollinearity results in failing to reject the null hypothesis that the X variable has no impact on Y when in fact X does have a statistically significant impact on Y. Said another way, the large standard errors of the estimated coefficient created by multicollinearity suggest statistical insignificance even when the hypothesized relationship is strong. How Good is the Equation? In the last section we concerned ourselves with testing the hypothesis that the dependent variable did indeed depend upon the hypothesized independent variable or variables. It may be that we find an independent variable that has some effect on the dependent variable, but it may not be the only one, and it may not even be the most important one. Remember that the error term was placed in the model to capture the effects of any missing independent variables. It follows that the error term may be used to give a measure of the \"goodness of fit\" of the equation taken as a whole in explaining the variation of the dependent variable, Y. The multiple correlation coefficient , also called the coefficient of multiple determination or the coefficient of determination , is given by the formula: R 2 = SSR SST where SSR is the regression sum of squares, the squared deviation of the predicted value of y from the mean value of y ( ŷ − y – ) , and SST is the total sum of squares which is the total squared deviation of the dependent variable, y, from its mean value, including the error term, SSE, the sum of squared errors. shows how the total deviation of the dependent variable, y, is partitioned into these two pieces. shows the estimated regression line and a single observation, x 1 . Regression analysis tries to explain the variation of the data about the mean value of the dependent variable, y. The question is, why do the observations of y vary from the average level of y? The value of y at observation x 1 varies from the mean of y by the difference ( y i − y – ). The sum of these differences squared is SST, the sum of squares total. The actual value of y at x 1 deviates from the estimated value, ŷ, by the difference between the estimated value and the actual value, ( y i − ŷ ). We recall that this is the error term, e, and the sum of these errors is SSE, sum of squared errors. The deviation of the predicted value of y, ŷ, from the mean value of y is ( ŷ − y – ) and is the SSR, sum of squares regression. It is called “regression” because it is the deviation explained by the regression. (Sometimes the SSR is called SSM for sum of squares mean because it measures the deviation from the mean value of the dependent variable, y, as shown on the graph.). Because the SST = SSR + SSE we see that the multiple correlation coefficient is the percent of the variance, or deviation in y from its mean value, that is explained by the equation when taken as a whole. R 2 will vary between zero and 1, with zero indicating that none of the variation in y was explained by the equation and a value of 1 indicating that 100% of the variation in y was explained by the equation. For time series studies expect a high R 2 and for cross-section data expect low R 2 . While a high R 2 is desirable, remember that it is the tests of the hypothesis concerning the existence of a relationship between a set of independent variables and a particular dependent variable that was the motivating factor in using the regression model. It is validating a cause and effect relationship developed by some theory that is the true reason that we chose the regression analysis. Increasing the number of independent variables will have the effect of increasing R 2 . To account for this effect the proper measure of the coefficient of determination is the R – 2 , adjusted for degrees of freedom, to keep down mindless addition of independent variables. There is no statistical test for the R 2 and thus little can be said about the model using R 2 with our characteristic confidence level. Two models that have the same size of SSE, that is sum of squared errors, may have very different R 2 if the competing models have different SST, total sum of squared deviations. The goodness of fit of the two models is the same; they both have the same sum of squares unexplained, errors squared, but because of the larger total sum of squares on one of the models the R 2 differs. Again, the real value of regression as a tool is to examine hypotheses developed from a model that predicts certain relationships among the variables. These are tests of hypotheses on the coefficients of the model and not a game of maximizing R 2 . Another way to test the general quality of the overall model is to test the coefficients as a group rather than independently. Because this is multiple regression (more than one X), we use the F-test to determine if our coefficients collectively affect Y. The hypothesis is: H o : β 1 = β 2 = … = β i = 0 H a : \"at least one of the βi is not equal to 0\" If the null hypothesis cannot be rejected, then we conclude that none of the independent variables contribute to explaining the variation in Y. Reviewing we see that SSR, the explained sum of squares, is a measure of just how much of the variation in Y is explained by all the variables in the model. SSE, the sum of the errors squared, measures just how much is unexplained. It follows that the ratio of these two can provide us with a statistical test of the model as a whole. Remembering that the F distribution is a ratio of Chi squared distributions and that variances are distributed according to Chi Squared, and the sum of squared errors and the sum of squares are both variances, we have the test statistic for this hypothesis as: F c = ( S S R k ) ( S S E n − k − 1 ) where n is the number of observations and k is the number of independent variables. It can be shown that this is equivalent to: F c = n − k − 1 k · R 2 1 − R 2 where R 2 is the coefficient of determination which is also a measure of the “goodness” of the model. As with all our tests of hypothesis, we reach a conclusion by comparing the calculated F statistic with the critical value given our desired level of confidence. If the calculated test statistic, an F statistic in this case, is in the tail of the distribution, then we cannot accept the null hypothesis. By not being able to accept the null hypotheses we conclude that this specification of this model has validity, because at least one of the estimated coefficients is significantly different from zero. An alternative way to reach this conclusion is to use the p-value comparison rule. The p-value is the area in the tail, given the calculated F statistic. In essence, the computer is finding the F value in the table for us. The computer regression output for the calculated F statistic is typically found in the ANOVA table section labeled “significance F\". How to read the output of an Excel regression is presented below. This is the probability of NOT accepting a false null hypothesis. If this probability is less than our pre-determined alpha error, then the conclusion is that we cannot accept the null hypothesis. Dummy Variables Thus far the analysis of the OLS regression technique assumed that the independent variables in the models tested were continuous random variables. There are, however, no restrictions in the regression model against independent variables that are binary. This opens the regression model for testing hypotheses concerning categorical variables such as gender, race, region of the country, before a certain data, after a certain date and innumerable others. These categorical variables take on only two values, 1 and 0, success or failure, from the binomial probability distribution. The form of the equation becomes: ŷ = b 0 + b 2 x 2 + b 1 x 1 where x 2 = 0 , 1 . X 2 is the dummy variable and X 1 is some continuous random variable. The constant, b 0 , is the y-intercept, the value where the line crosses the y-axis. When the value of X 2 = 0, the estimated line crosses at b 0 . When the value of X 2 = 1 then the estimated line crosses at b 0 + b 2 . In effect the dummy variable causes the estimated line to shift either up or down by the size of the effect of the characteristic captured by the dummy variable. Note that this is a simple parallel shift and does not affect the impact of the other independent variable; X 1 .This variable is a continuous random variable and predicts different values of y at different values of X 1 holding constant the condition of the dummy variable. An example of the use of a dummy variable is the work estimating the impact of gender on salaries. There is a full body of literature on this topic and dummy variables are used extensively. For this example the salaries of elementary and secondary school teachers for a particular state is examined. Using a homogeneous job category, school teachers, and for a single state reduces many of the variations that naturally effect salaries such as differential physical risk, cost of living in a particular state, and other working conditions. The estimating equation in its simplest form specifies salary as a function of various teacher characteristic that economic theory would suggest could affect salary. These would include education level as a measure of potential productivity, age and/or experience to capture on-the-job training, again as a measure of productivity. Because the data are for school teachers employed in a public school districts rather than workers in a for-profit company, the school district’s average revenue per average daily student attendance is included as a measure of ability to pay. The results of the regression analysis using data on 24,916 school teachers are presented below. Earnings Estimate for Elementary and Secondary School Teachers Variable Regression Coefficients (b) Standard Errors of the estimates for teacher's earnings function (s b ) Intercept 4269.9 Gender (man = 1) 632.38 13.39 Total Years of Experience 52.32 1.10 Years of Experience in Current District 29.97 1.52 Education 629.33 13.16 Total Revenue per ADA 90.24 3.76 R – 2 .725 n 24,916 The coefficients for all the independent variables are significantly different from zero as indicated by the standard errors. Dividing the standard errors of each coefficient results in a t-value greater than 1.96 which is the required level for 95% significance. The binary variable, our dummy variable of interest in this analysis, is gender where man is given a value of 1 and woman given a value of 0. The coefficient is significantly different from zero with a dramatic t-statistic of 47 standard deviations. We thus cannot accept the null hypothesis that the coefficient is equal to zero. Therefore we conclude that there is a premium paid teachers who are men of $632 after holding constant experience, education and the wealth of the school district in which the teacher is employed. It is important to note that these data are from some time ago and the $632 represents a six percent salary premium at that time. A graph of this example of dummy variables is presented below. In two dimensions, salary is the dependent variable on the vertical axis and total years of experience was chosen for the continuous independent variable on horizontal axis. Any of the other independent variables could have been chosen to illustrate the effect of the dummy variable. The relationship between total years of experience has a slope of $52.32 per year of experience and the estimated line has an intercept of $4,269 if the gender variable is equal to zero, for woman. If the gender variable is equal to 1, for man, the coefficient for the gender variable is added to the intercept and thus the relationship between total years of experience and salary is shifted upward parallel as indicated on the graph. Also marked on the graph are various points for reference. A woman school teacher who is a woman and has 10 years of experience receives a salary of $4,792 on the basis of her experience only, but this is still $109 less than a man with zero years of teaching experience. A more complex interaction between a dummy variable and the dependent variable can also be estimated. It may be that the dummy variable has more than a simple shift effect on the dependent variable, but also interacts with one or more of the other continuous independent variables. While not tested in the example above, it could be hypothesized that the impact of gender on salary was not a one-time shift, but impacted the value of additional years of experience on salary also. That is, school teacher’s salaries for women were discounted at the start, and further did not grow at the same rate from the effect of experience as for men. This would show up as a different slope for the relationship between total years of experience for men than for women. If this is so then women school teachers would not just start behind their colleagues who are men (as measured by the shift in the estimated regression line), but would fall further and further behind as time and experienced increased. The graph below shows how this hypothesis can be tested with the use of dummy variables and an interaction variable. The estimating equation shows how the slope of X 1 , the continuous random variable experience, contains two parts, b 1 and b 3 . This occurs because of the new variable X 2 X 1 , called the interaction variable, was created to allow for an effect on the slope of X 1 from changes in X 2 , the binary dummy variable. Note that when the dummy variable, X 2 = 0 the interaction variable has a value of 0, but when X 2 = 1 the interaction variable has a value of X 1 . The coefficient b 3 is an estimate of the difference in the coefficient of X 1 when X 2 = 1 compared to when X 2 = 0. In the example of teacher’s salaries, if there is a premium paid to teachers who are men that affects the rate of increase in salaries from experience, then the rate at which teachers’ salaries for men rises would be b 1 + b 3 and the rate at which teachers’ salaries for women rise would be simply b 1 . This hypothesis can be tested with the hypothesis: H 0 : β 3 = 0 | β 1 = 0 , β 2 = 0 H a : β 3 ≠ 0 | β 1 ≠ 0 , β 2 ≠ 0 This is a t-test using the test statistic for the parameter β 3 . If we cannot accept the null hypothesis that β 3 =0 we conclude there is a difference between the rate of increase for the group for whom the value of the binary variable is set to 1, males in this example. This estimating equation can be combined with our earlier one that tested only a parallel shift in the estimated line. The earnings/experience functions in are drawn for this case with a shift in the earnings function and a difference in the slope of the function with respect to total years of experience. A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a randomly selected student if you know the third exam score? x (third exam score) y (final exam score) 65 175 67 133 71 185 71 163 66 126 75 198 67 153 70 163 71 159 69 151 69 159 Table showing the scores on the final exam based on scores from the third exam. Scatter plot showing the scores on the final exam based on scores from the third exam. Try It SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet. X (depth in feet) Y (maximum dive time) 50 80 60 55 70 45 80 35 90 25 100 22 Suppose that you have at your disposal the information below for each of 30 drivers. Propose a model (including a very brief indication of symbols used to represent independent variables) to explain how miles per gallon vary from driver to driver on the basis of the factors measured. Information: miles driven per day weight of car number of cylinders in car average speed miles per gallon number of passengers Y j = b 0 + b 1 ⋅ X 1 + b 2 ⋅ X 2 + b 3 ⋅ X 3 + b 4 ⋅ X 4 + b 5 ⋅ X 6 + e j Consider a sample least squares regression analysis between a dependent variable (Y) and an independent variable (X). A sample correlation coefficient of −1 (minus one) tells us that there is no relationship between Y and X in the sample there is no relationship between Y and X in the population there is a perfect negative relationship between Y and X in the population there is a perfect negative relationship between Y and X in the sample. d. there is a perfect negative relationship between Y and X in the sample. In correlational analysis, when the points scatter widely about the regression line, this means that the correlation is negative. low. heterogeneous. between two measures that are unreliable. b. low Chapter Review It is hoped that this discussion of regression analysis has demonstrated the tremendous potential value it has as a tool for testing models and helping to better understand the world around us. The regression model has its limitations, especially the requirement that the underlying relationship be approximately linear. To the extent that the true relationship is nonlinear it may be approximated with a linear relationship or nonlinear forms of transformations that can be estimated with linear techniques. Double logarithmic transformation of the data will provide an easy way to test this particular shape of the relationship. A reasonably good quadratic form (the shape of the total cost curve from Microeconomics Principles) can be generated by the equation: Y = a + b 1 X + b 2 X 2 where the values of X are simply squared and put into the equation as a separate variable. There is much more in the way of econometric \"tricks\" that can bypass some of the more troublesome assumptions of the general regression model. This statistical technique is so valuable that further study would provide any student significant, statistically significant, dividends. Residual or “error” the value calculated from subtracting y 0 − y ^ 0 = e 0 . The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y that appears on the best-fit line. Sum of Squared Errors (SSE) the calculated value from adding up all the squared residual terms. The hope is that this value is very small when creating a model. R 2 – Coefficient of Determination This is a number between 0 and 1 that represents the percentage variation of the dependent variable that can be explained by the variation in the independent variable. Sometimes calculated by the equation R 2 = S S R S S T where SSR is the “Sum of Squares Regression” and SST is the “Sum of Squares Total.” The appropriate coefficient of determination to be reported should always be adjusted for degrees of freedom first.", "section": "The Regression Equation", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation As we have seen, the coefficient of an equation estimated using OLS regression analysis provides an estimate of the slope of a straight line that is assumed be the relationship between the dependent variable and at least one independent variable. From the calculus, the slope of the line is the first derivative and tells us the magnitude of the impact of a one unit change in the X variable upon the value of the Y variable measured in the units of the Y variable. As we saw in the case of dummy variables, this can show up as a parallel shift in the estimated line or even a change in the slope of the line through an interactive variable. Here we wish to explore the concept of elasticity and how we can use a regression analysis to estimate the various elasticities in which economists have an interest. The concept of elasticity is borrowed from engineering and physics where it is used to measure a material’s responsiveness to a force, typically a physical force such as a stretching/pulling force. It is from here that we get the term an “elastic” band. In economics, the force in question is some market force such as a change in price or income. Elasticity is measured as a percentage change/response in both engineering applications and in economics. The value of measuring in percentage terms is that the units of measurement do not play a role in the value of the measurement and thus allows direct comparison between elasticities. As an example, if the price of gasoline increased say 50 cents from an initial price of $3.00 and generated a decline in monthly consumption for a consumer from 50 gallons to 48 gallons we calculate the elasticity to be 0.25. The price elasticity is the percentage change in quantity resulting from some percentage change in price. A 16 percent increase in price has generated only a 4 percent decrease in demand: 16% price change → 4% quantity change or .04/.16 = .25. This is called an inelastic demand meaning a small response to the price change. This comes about because there are few if any real substitutes for gasoline; perhaps public transportation, a bicycle or walking. Technically, of course, the percentage change in demand from a price increase is a decline in demand thus price elasticity is a negative number. The common convention, however, is to talk about elasticity as the absolute value of the number. Some goods have many substitutes: pears for apples for plums, for grapes, etc. etc. The elasticity for such goods is larger than one and are called elastic in demand. Here a small percentage change in price will induce a large percentage change in quantity demanded. The consumer will easily shift the demand to the close substitute. While this discussion has been about price changes, any of the independent variables in a demand equation will have an associated elasticity. Thus, there is an income elasticity that measures the sensitivity of demand to changes in income: not much for the demand for food, but very sensitive for yachts. If the demand equation contains a term for substitute goods, say candy bars in a demand equation for cookies, then the responsiveness of demand for cookies from changes in prices of candy bars can be measured. This is called the cross-price elasticity of demand and to an extent can be thought of as brand loyalty from a marketing view. How responsive is the demand for Coca-Cola to changes in the price of Pepsi? Now imagine the demand for a product that is very expensive. Again, the measure of elasticity is in percentage terms thus the elasticity can be directly compared to that for gasoline: an elasticity of 0.25 for gasoline conveys the same information as an elasticity of 0.25 for $25,000 car. Both goods are considered by the consumer to have few substitutes and thus have inelastic demand curves, elasticities less than one. The mathematical formulae for various elasticities are: Price elasticity: η p = ( %∆Q ) ( %∆P ) Where η is the Greek small case letter eta used to designate elasticity. ∆ is read as “change”. Income elasticity: η Y = ( %∆Q ) ( %∆Y ) Where Y is used as the symbol for income. Cross-Price elasticity: η p1 = ( %∆ Q 1 ) ( %∆ P 2 ) Where P2 is the price of the substitute good. Examining closer the price elasticity we can write the formula as: η p = ( %∆Q ) ( %∆P ) = dQ dP ( P Q ) = b ( P Q ) Where b is the estimated coefficient for price in the OLS regression. The first form of the equation demonstrates the principle that elasticities are measured in percentage terms. Of course, the ordinary least squares coefficients provide an estimate of the impact of a unit change in the independent variable, X, on the dependent variable measured in units of Y. These coefficients are not elasticities, however, and are shown in the second way of writing the formula for elasticity as ( d Q d P ) , the derivative of the estimated demand function which is simply the slope of the regression line. Multiplying the slope times P Q provides an elasticity measured in percentage terms. Along a straight-line demand curve the percentage change, thus elasticity, changes continuously as the scale changes, while the slope, the estimated regression coefficient, remains constant. Going back to the demand for gasoline. A change in price from $3.00 to $3.50 was a 16 percent increase in price. If the beginning price were $5.00 then the same 50¢ increase would be only a 10 percent increase generating a different elasticity. Every straight-line demand curve has a range of elasticities starting at the top left, high prices, with large elasticity numbers, elastic demand, and decreasing as one goes down the demand curve, inelastic demand. In order to provide a meaningful estimate of the elasticity of demand the convention is to estimate the elasticity at the point of means. Remember that all OLS regression lines will go through the point of means. At this point is the greatest weight of the data used to estimate the coefficient. The formula to estimate an elasticity when an OLS demand curve has been estimated becomes: η p = b ( P − Q − ) Where P − and Q − are the mean values of these data used to estimate b , the price coefficient. The same method can be used to estimate the other elasticities for the demand function by using the appropriate mean values of the other variables; income and price of substitute goods for example. Logarithmic Transformation of the Data Ordinary least squares estimates typically assume that the population relationship among the variables is linear thus of the form presented in The Regression Equation . In this form the interpretation of the coefficients is as discussed above; quite simply the coefficient provides an estimate of the impact of a one unit change in X on Y measured in units of Y. It does not matter just where along the line one wishes to make the measurement because it is a straight line with a constant slope thus constant estimated level of impact per unit change. It may be, however, that the analyst wishes to estimate not the simple unit measured impact on the Y variable, but the magnitude of the percentage impact on Y of a one unit change in the X variable. Such a case might be how a unit change in experience, say one year, effects not the absolute amount of a worker’s wage, but the percentage impact on the worker’s wage. Alternatively, it may be that the question asked is the unit measured impact on Y of a specific percentage increase in X. An example may be “by how many dollars will sales increase if the firm spends X percent more on advertising?” The third possibility is the case of elasticity discussed above. Here we are interested in the percentage impact on quantity demanded for a given percentage change in price, or income or perhaps the price of a substitute good. All three of these cases can be estimated by transforming the data to logarithms before running the regression. The resulting coefficients will then provide a percentage change measurement of the relevant variable. To summarize, there are four cases: Unit ∆X → Unit ∆Y (Standard OLS case) Unit ∆X → %∆Y %∆X → Unit ∆Y %∆X → %∆Y (elasticity case) Case 1: The ordinary least squares case begins with the linear model developed above: Y = a + b X where the coefficient of the independent variable b = d Y d X is the slope of a straight line and thus measures the impact of a unit change in X on Y measured in units of Y. Case 2: The underlying estimated equation is: log ( Y ) = a + b X The equation is estimated by converting the Y values to logarithms and using OLS techniques to estimate the coefficient of the X variable, b. This is called a semi-log estimation. Again, differentiating both sides of the equation allows us to develop the interpretation of the X coefficient b: d ( log Y ) = b d X d Y Y = b d X Multiply by 100 to covert to percentages and rearranging terms gives: 100 b = %∆Y Unit ∆X 100 b is thus the percentage change in Y resulting from a unit change in X. Case 3: In this case the question is “what is the unit change in Y resulting from a percentage change in X?” What is the dollar loss in revenues of a five percent increase in price or what is the total dollar cost impact of a five percent increase in labor costs? The estimated equation for this case would be: Y = a + B log ( X ) Here the calculus differential of the estimated equation is: d Y = b d ( log X ) d Y = b d X X Divide by 100 to get percentage and rearranging terms gives: b 100 = d Y 100 d X X = Unit ∆Y %∆X Therefore, b 100 is the increase in Y measured in units from a one percent increase in X. Case 4: This is the elasticity case where both the dependent and independent variables are converted to logs before the OLS estimation. This is known as the log-log case or double log case, and provides us with direct estimates of the elasticities of the independent variables. The estimated equation is: log Y = a + b log X Differentiating we have: d ( log Y ) = b d ( log X ) d ( log Y ) = b 1 X d X thus: 1 Y d Y = b 1 X d X OR d Y Y = b d X X OR b = d Y d X ( X Y ) and b = %∆Y %∆X our definition of elasticity. We conclude that we can directly estimate the elasticity of a variable through double log transformation of the data. The estimated coefficient is the elasticity. It is common to use double log transformation of all variables in the estimation of demand functions to get estimates of all the various elasticities of the demand curve. In a linear regression, why do we need to be concerned with the range of the independent (X) variable? The precision of the estimate of the Y variable depends on the range of the independent (X) variable explored. If we explore a very small range of the X variable, we won't be able to make much use of the regression. Also, extrapolation is not recommended. Suppose one collected the following information where X is diameter of tree trunk and Y is tree height. X Y 4 8 2 4 8 18 6 22 10 30 6 8 Regression equation: y ^ i = −3.6 + 3.1 ⋅ X i What is your estimate of the average height of all trees having a trunk diameter of 7 inches? y ^ = −3.6 + ( 3.1 ⋅ 7 ) = 18.1 The manufacturers of a chemical used in flea collars claim that under standard test conditions each additional unit of the chemical will bring about a reduction of 5 fleas (i.e. where X j = amount of chemical and Y J = B 0 + B 1 ⋅ X J + E J , H 0 : B 1 = −5 Suppose that a test has been conducted and results from a computer include: Intercept = 60 Slope = −4 Standard error of the regression coefficient = 1.0 Degrees of Freedom for Error = 2000 95% Confidence Interval for the slope −2.04, −5.96 Is this evidence consistent with the claim that the number of fleas is reduced at a rate of 5 fleas per unit chemical? Most simply, since −5 is included in the confidence interval for the slope, we can conclude that the evidence is consistent with the claim at the 95% confidence level. Using a t test: H 0 : B 1 = −5 H A : B 1 ≠ −5 t calculated = −5 − ( −4 ) 1 = −1 t critical = −1.96 Since t calc < t crit we retain the null hypothesis that B 1 = −5 .", "section": "Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Predicting with a Regression Equation One important value of an estimated regression equation is its ability to predict the effects on Y of a change in one or more values of the independent variables. The value of this is obvious. Careful policy cannot be made without estimates of the effects that may result. Indeed, it is the desire for particular results that drive the formation of most policy. Regression models can be, and have been, invaluable aids in forming such policies. The Gauss-Markov theorem assures us that the point estimate of the impact on the dependent variable derived by putting in the equation the hypothetical values of the independent variables one wishes to simulate will result in an estimate of the dependent variable which is minimum variance and unbiased. That is to say that from this equation comes the best unbiased point estimate of y given the values of x. ŷ = b 0 + b , X 1 i + ⋯ + b k X k i Remember that point estimates do not carry a particular level of probability, or level of confidence, because points have no “width” above which there is an area to measure. This was why we developed confidence intervals for the mean and proportion earlier. The same concern arises here also. There are actually two different approaches to the issue of developing estimates of changes in the independent variable, or variables, on the dependent variable. The first approach wishes to measure the expected mean value of y from a specific change in the value of x: this specific value implies the expected value. Here the question is: what is the mean impact on y that would result from multiple hypothetical experiments on y at this specific value of x. Remember that there is a variance around the estimated parameter of x and thus each experiment will result in a bit of a different estimate of the predicted value of y. The second approach to estimate the effect of a specific value of x on y treats the event as a single experiment: you choose x and multiply it times the coefficient and that provides a single estimate of y. Because this approach acts as if there were a single experiment the variance that exists in the parameter estimate is larger than the variance associated with the expected value approach. The conclusion is that we have two different ways to predict the effect of values of the independent variable(s) on the dependent variable and thus we have two different intervals. Both are correct answers to the question being asked, but there are two different questions. To avoid confusion, the first case where we are asking for the expected value of the mean of the estimated y, is called a confidence interval as we have named this concept before. The second case, where we are asking for the estimate of the impact on the dependent variable y of a single experiment using a value of x, is called the prediction interval . The test statistics for these two interval measures within which the estimated value of y will fall are: Confidence Interval for Expected Value of Mean Value of y for x = x p ⏜ ¯ y = ± t α 2 s e 1 n + n ( x p - x ¯ ) 2 n ( ∑ x 2 ) - ( ∑ x ) 2 Prediction Interval for an Individual y for x = x p y ⏜ = ± t α 2 s e 1 + 1 n + n ( x p - x ¯ ) 2 n ( ∑ x 2 ) - ( ∑ x ) 2 Where s e is the standard deviation of the error term. The mathematical computations of these two test statistics are complex. Various computer regression software packages provide programs within the regression functions to provide answers to inquires of estimated predicted values of y given various values chosen for the x variable(s). It is important to know just which interval is being tested in the computer package because the difference in the size of the standard deviations will change the size of the interval estimated. This is shown in . Prediction and confidence intervals for regression equation; 95% confidence level. shows visually the difference the standard deviation makes in the size of the estimated intervals. The confidence interval, measuring the expected value of the dependent variable, is smaller than the prediction interval for the same level of confidence. The expected value method assumes that the experiment is conducted multiple times rather than just once as in the other method. The logic here is similar, although not identical, to that discussed when developing the relationship between the sample size and the confidence interval using the Central Limit Theorem. There, as the number of experiments increased, the distribution narrowed and the confidence interval became tighter around the expected value of the mean. It is also important to note that the intervals around a point estimate are highly dependent upon the range of data used to estimate the equation regardless of which approach is being used for prediction. Remember that all regression equations go through the point of means, that is, the mean value of y and the mean values of all independent variables in the equation. As the value of x chosen to estimate the associated value of y is further from the point of means the width of the estimated interval around the point estimate increases. Choosing values of x beyond the range of the data used to estimate the equation possess even greater danger of creating estimates with little use; very large intervals, and risk of error. shows this relationship. Confidence interval for an individual value of x, X p , at 95% level of confidence demonstrates the concern for the quality of the estimated interval whether it is a prediction interval or a confidence interval. As the value chosen to predict y, X p in the graph, is further from the central weight of the data, X ¯ , we see the interval expand in width even while holding constant the level of confidence. This shows that the precision of any estimate will diminish as one tries to predict beyond the largest weight of the data and most certainly will degrade rapidly for predictions beyond the range of the data. Unfortunately, this is just where most predictions are desired. They can be made, but the width of the confidence interval may be so large as to render the prediction useless. Only actual calculation and the particular application can determine this, however. Recall the third exam/final exam example . We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction. Assume the coefficient for X was determined to be significantly different from zero. Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores ( x -values) range from 65 to 75. Since 73 is between the x -values 65 and 75, we feel comfortable to substitute x = 73 into the equation. Then: y ^ = − 173.51 + 4.83 ( 73 ) = 179.08 We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average. a. What would you predict the final exam score to be for a student who scored a 66 on the third exam? a. 145.27 b. What would you predict the final exam score to be for a student who scored a 90 on the third exam? b. The x values in the data are between 65 and 75. Ninety is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will have a confidence interval that may not be meaningful.) To understand really how unreliable the prediction can be outside of the observed x values observed in the data, make the substitution x = 90 into the equation. y ^ = –173.51 + 4.83 ( 90 ) = 261.19 The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200. Try It Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8 x What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week? True or False? If False, correct it: Suppose you are performing a simple linear regression of Y on X and you test the hypothesis that the slope β is zero against a two-sided alternative. You have n = 25 observations and your computed test (t) statistic is 2.6. Then your P -value is given by .01 < P < .02, which gives borderline significance (i.e. you would reject H 0 at α = .02 but fail to reject H 0 at α = .01 ). True. t (critical, df = 23, two-tailed, α = .02) = ± 2.5 t critical, df = 23, two-tailed, α = .01 = ± 2.8 An economist is interested in the possible influence of \"Miracle Wheat\" on the average yield of wheat in a district. To do so they fit a linear regression of average yield per year against year after introduction of \"Miracle Wheat\" for a ten year period. The fitted trend line is y ^ j = 80 + 1.5 ⋅ X j ( Y j : Average yield in j year after introduction) ( X j : j year after introduction). What is the estimated average yield for the fourth year after introduction? Do you want to use this trend line to estimate yield for, say, 20 years after introduction? Why? What would your estimate be? 80 + 1.5 ⋅ 4 = 86 No. Most business statisticians would not want to extrapolate that far. If someone did, the estimate would be 110, but some other factors probably come into play with 20 years. An interpretation of r = 0.5 is that the following part of the Y-variation is associated with which variation in X: most half very little one quarter none of these d. one quarter Which of the following values of r indicates the most accurate prediction of one variable from another? r = 1.18 r = −.77 r = .68 b. r = −.77", "section": "Predicting with a Regression Equation", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "How to Use Microsoft Excel® for Regression Analysis This section of this chapter is here in recognition that what we are now asking requires much more than a quick calculation of a ratio or a square root. Indeed, the use of regression analysis was almost non- existent before the middle of the last century and did not really become a widely used tool until perhaps the late 1960’s and early 1970’s. Even then the computational ability of even the largest IBM machines is laughable by today’s standards. In the early days programs were developed by the researchers and shared. There was no market for something called “software” and certainly nothing called “apps”, an entrant into the market only a few years old. With the advent of the personal computer and the explosion of a vital software market we have a number of regression and statistical analysis packages to choose from. Each has their merits. We have chosen Microsoft Excel because of the wide-spread availability both on college campuses and in the post-college market place. Stata is an alternative and has features that will be important for more advanced econometrics study if you choose to follow this path. Even more advanced packages exist, but typically require the analyst to do some significant amount of programing to conduct their analysis. The goal of this section is to demonstrate how to use Excel to run a regression and then to do so with an example of a simple version of a demand curve. The first step to doing a regression using Excel is to load the program into your computer. If you have Excel you have the Analysis ToolPak although you may not have it activated. The program calls upon a significant amount of space so is not loaded automatically. To activate the Analysis ToolPak follow these steps: Click “File” > “Options” > “Add-ins” to bring up a menu of the add-in “ToolPaks”. Select “Analysis ToolPak” and click “GO” next to “Manage: excel add-ins” near the bottom of the window. This will open a new window where you click “Analysis ToolPak” (make sure there is a green check mark in the box) and then click “OK”. Now there should be an Analysis tab under the data menu. These steps are presented in the following screen shots. Click “Data” then “Data Analysis” and then click “Regression” and “OK”. Congratulations, you have made it to the regression window. The window asks for your inputs. Clicking the box next to the Y and X ranges will allow you to use the click and drag feature of Excel to select your input ranges. Excel has one odd quirk and that is the click and drop feature requires that the independent variables, the X variables, are all together, meaning that they form a single matrix. If your data are set up with the Y variable between two columns of X variables Excel will not allow you to use click and drag. As an example, say Column A and Column C are independent variables and Column B is the Y variable, the dependent variable. Excel will not allow you to click and drop the data ranges. The solution is to move the column with the Y variable to column A and then you can click and drag. The same problem arises again if you want to run the regression with only some of the X variables. You will need to set up the matrix so all the X variables you wish to regress are in a tightly formed matrix. These steps are presented in the following scene shots. Once you have selected the data for your regression analysis and told Excel which one is the dependent variable (Y) and which ones are the independent valuables (X‘s), you have several choices as to the parameters and how the output will be displayed. Refer to screen shot under “Input” section. If you check the “labels” box the program will place the entry in the first column of each variable as its name in the output. You can enter an actual name, such as price or income in a demand analysis, in row one of the Excel spreadsheet for each variable and it will be displayed in the output. The level of significance can also be set by the analyst. This will not change the calculated t statistic, called t stat, but will alter the p value for the calculated t statistic. It will also alter the boundaries of the confidence intervals for the coefficients. A 95 percent confidence interval is always presented, but with a change in this you will also get other levels of confidence for the intervals. Excel also will allow you to suppress the intercept. This forces the regression program to minimize the residual sum of squares under the condition that the estimated line must go through the origin. This is done in cases where there is no meaning in the model at some value other than zero, zero for the start of the line. An example is an economic production function that is a relationship between the number of units of an input, say hours of labor, and output. There is no meaning of positive output with zero workers. Once the data are entered and the choices are made click OK and the results will be sent to a separate new worksheet by default. The output from Excel is presented in a way typical of other regression package programs. The first block of information gives the overall statistics of the regression: Multiple R, R Squared, and the R squared adjusted for degrees of freedom, which is the one you want to report. You also get the Standard error (of the estimate) and the number of observations in the regression. The second block of information is titled ANOVA which stands for Analysis of Variance. Our interest in this section is the column marked F. This is the calculated F statistics for the null hypothesis that all of the coefficients are equal to zero verse the alternative that at least one of the coefficients are not equal to zero. This hypothesis test was presented in 13.4 under “How Good is the Equation?” The next column gives the p value for this test under the title “Significance F”. If the p value is less than say 0.05 (the calculated F statistic is in the tail) we can say with 90 % confidence that we cannot accept the null hypotheses that all the coefficients are equal to zero. This is a good thing: it means that at least one of the coefficients is significantly different from zero thus do have an effect on the value of Y. The last block of information contains the hypothesis tests for the individual coefficient. The estimated coefficients, the intercept and the slopes, are first listed and then each standard error (of the estimated coefficient) followed by the t stat (calculated student’s t statistic for the null hypothesis that the coefficient is equal to zero). We compare the t stat and the critical value of the student’s t, dependent on the degrees of freedom, and determine if we have enough evidence to reject the null that the variable has no effect on Y. Remember that we have set up the null hypothesis as the status quo and our claim that we know what caused the Y to change is in the alternative hypothesis. We want to reject the status quo and substitute our version of the world, the alternative hypothesis. The next column contains the p values for this hypothesis test followed by the estimated upper and lower bound of the confidence interval of the estimated slope parameter for various levels of confidence set by us at the beginning. Estimating the Demand for Roses Here is an example of using the Excel program to run a regression for a particular specific case: estimating the demand for roses. We are trying to estimate a demand curve, which from economic theory we expect certain variables affect how much of a good we buy. The relationship between the price of a good and the quantity demanded is the demand curve. Beyond that we have the demand function that includes other relevant variables: a person’s income, the price of substitute goods, and perhaps other variables such as season of the year or the price of complimentary goods. Quantity demanded will be our Y variable, and Price of roses, Price of carnations and Income will be our independent variables, the X variables. For all of these variables theory tells us the expected relationship. For the price of the good in question, roses, theory predicts an inverse relationship, the negatively sloped demand curve. Theory also predicts the relationship between the quantity demanded of one good, here roses, and the price of a substitute, carnations in this example. Theory predicts that this should be a positive or direct relationship; as the price of the substitute falls we substitute away from roses to the cheaper substitute, carnations. A reduction in the price of the substitute generates a reduction in demand for the good being analyzed, roses here. Reduction generates reduction is a positive relationship. For normal goods, theory also predicts a positive relationship; as our incomes rise we buy more of the good, roses. We expect these results because that is what is predicted by a hundred years of economic theory and research. Essentially we are testing these century-old hypotheses. The data gathered was determined by the model that is being tested. This should always be the case. One is not doing inferential statistics by throwing a mountain of data into a computer and asking the machine for a theory. Theory first, test follows. These data here are national average prices and income is the nation’s per capita personal income. Quantity demanded is total national annual sales of roses. These are annual time series data; we are tracking the rose market for the United States from 1984-2017, 33 observations. Because of the quirky way Excel requires how the data are entered into the regression package it is best to have the independent variables, price of roses, price of carnations and income next to each other on the spreadsheet. Once your data are entered into the spreadsheet it is always good to look at the data. Examine the range, the means and the standard deviations. Use your understanding of descriptive statistics from the very first part of this course. In large data sets you will not be able to “scan” the data. The Analysis ToolPac makes it easy to get the range, mean, standard deviations and other parameters of the distributions. You can also quickly get the correlations among the variables. Examine for outliers. Review the history. Did something happen? Was here a labor strike, change in import fees, something that makes these observations unusual? Do not take the data without question. There may have been a typo somewhere, who knows without review. Go to the regression window, enter the data and select 95% confidence level and click “OK”. You can include the labels in the input range if you have put a title at the top of each column, but be sure to click the “labels” box on the main regression page if you do. The regression output should show up automatically on a new worksheet. The first results presented is the R-Square, a measure of the strength of the correlation between Y and X 1 , X 2 , and X 3 taken as a group. Our R-square here of 0.699, adjusted for degrees of freedom, means that 70% of the variation in Y, demand for roses, can be explained by variations in X 1 , X 2 , and X 3 , Price of roses, Price of carnations and Income. There is no statistical test to determine the “significance” of an R 2 . Of course a higher R 2 is preferred, but it is really the significance of the coefficients that will determine the value of the theory being tested and which will become part of any policy discussion if they are demonstrated to be significantly different from zero. Looking at the third panel of output we can write the equation as: Y = b 0 + b 1 X 1 + b 2 X 2 + b 3 X 3 + e where b 0 is the intercept, b 1 is the estimated coefficient on price of roses, and b 2 is the estimated coefficient on price of carnations, b 3 is the estimated effect of income and e is the error term. The equation is written in Roman letters indicating that these are the estimated values and not the population parameters, β’s. Our estimated equation is: Quantity of roses sold = 183,475 − 1.76 Price of roses + 1.33 Price of carnations + 3.03 Income We first observe that the signs of the coefficients are as expected from theory. The demand curve is downward sloping with the negative sign for the price of roses. Further the signs of both the price of carnations and income coefficients are positive as would be expected from economic theory. Interpreting the coefficients can tell us the magnitude of the impact of a change in each variable on the demand for roses. It is the ability to do this which makes regression analysis such a valuable tool. The estimated coefficients tell us that an increase the price of roses by one dollar will lead to a 1.76 reduction in the number roses purchased. The price of carnations seems to play an important role in the demand for roses as we see that increasing the price of carnations by one dollar would increase the demand for roses by 1.33 units as consumers would substitute away from the now more expensive carnations. Similarly, increasing per capita income by one dollar will lead to a 3.03 unit increase in roses purchased. These results are in line with the predictions of economics theory with respect to all three variables included in this estimate of the demand for roses. It is important to have a theory first that predicts the significance or at least the direction of the coefficients. Without a theory to test, this research tool is not much more helpful than the correlation coefficients we learned about earlier. We cannot stop there, however. We need to first check whether our coefficients are statistically significant from zero. We set up a hypothesis of: H 0 : β 1 = 0 H a : β 1 ≠ 0 for all three coefficients in the regression. Recall from earlier that we will not be able to definitively say that our estimated b 1 is the actual real population of β 1 , but rather only that with (1-α)% level of confidence that we cannot reject the null hypothesis that our estimated β 1 is significantly different from zero. The analyst is making a claim that the price of roses causes an impact on quantity demanded. Indeed, that each of the included variables has an impact on the quantity of roses demanded. The claim is therefore in the alternative hypotheses. It will take a very large probability, 0.95 in this case, to overthrow the null hypothesis, the status quo, that β = 0. In all regression hypothesis tests the claim is in the alternative and the claim is that the theory has found a variable that has a significant impact on the Y variable. The test statistic for this hypothesis follows the familiar standardizing formula which counts the number of standard deviations, t, that the estimated value of the parameter, b 1 , is away from the hypothesized value, β 0 , which is zero in this case: t c = b 1 − β 0 S b 1 The computer calculates this test statistic and presents it as “t stat”. You can find this value to the right of the standard error of the coefficient estimate. The standard error of the coefficient for b 1 is S b 1 in the formula. To reach a conclusion we compare this test statistic with the critical value of the student’s t at degrees of freedom n-3-1 =29, and alpha = 0.025 (5% significance level for a two-tailed test). Our t stat for b 1 is approximately 5.90 which is greater than 1.96 (the critical value we looked up in the t-table), so we cannot accept our null hypotheses of no effect. We conclude that Price has a significant effect because the calculated t value is in the tail. We conduct the same test for b 2 and b 3 . For each variable, we find that we cannot accept the null hypothesis of no relationship because the calculated t-statistics are in the tail for each case, that is, greater than the critical value. All variables in this regression have been determined to have a significant effect on the demand for roses. These tests tell us whether or not an individual coefficient is significantly different from zero, but does not address the overall quality of the model. We have seen that the R squared adjusted for degrees of freedom indicates this model with these three variables explains 70% of the variation in quantity of roses demanded. We can also conduct a second test of the model taken as a whole. This is the F test presented in The Regression Equation of this chapter. Because this is a multiple regression (more than one X), we use the F-test to determine if our coefficients collectively affect Y. The hypothesis is: H 0 : β 1 = β 2 = ... = β i = 0 H a : \"at least one of the β i is not equal to 0\" Under the ANOVA section of the output we find the calculated F statistic for this hypotheses. For this example the F statistic is 21.9. Again, comparing the calculated F statistic with the critical value given our desired level of significance and the degrees of freedom will allow us to reach a conclusion. The best way to reach a conclusion for this statistical test is to use the p-value comparison rule. The p-value is the area in the tail, given the calculated F statistic. In essence the computer is finding the F value in the table for us and calculating the p-value. In the Summary Output under “significance F” is this probability. For this example, it is calculated to be 2.6 x 10 -5 , or 2.6 then moving the decimal five places to the left. (.000026) This is an almost infinitesimal level of probability and is certainly less than our alpha level of .05 for a 5 percent level of significance. By not being able to accept the null hypotheses we conclude that this specification of this model has validity because at least one of the estimated coefficients is significantly different from zero. Since F-calculated is greater than F-critical, we cannot accept H 0 , meaning that X 1 , X 2 and X 3 together has a significant effect on Y. The development of computing machinery and the software useful for academic and business research has made it possible to answer questions that just a few years ago we could not even formulate. Data is available in electronic format and can be moved into place for analysis in ways and at speeds that were unimaginable a decade ago. The sheer magnitude of data sets that can today be used for research and analysis gives us a higher quality of results than in days past. Even with only an Excel spreadsheet we can conduct very high level research. This section gives you the tools to conduct some of this very interesting research with the only limit being your imagination. A computer program for multiple regression has been used to fit y ^ j = b 0 + b 1 ⋅ X 1 j + b 2 ⋅ X 2 j + b 3 ⋅ X 3 j . Part of the computer output includes: i b i S b i 0 8 1.6 1 2.2 .24 2 -.72 .32 3 0.005 0.002 Calculation of confidence interval for b 2 consists of _______± (a student's t value) (_______) The confidence level for this interval is reflected in the value used for _______. The degrees of freedom available for estimating the variance are directly concerned with the value used for _______ −.72, .32 the t value the t value An investigator has used a multiple regression program on 20 data points to obtain a regression equation with 3 variables. Part of the computer output is: Variable Coefficient Standard Error of b i 1 0.45 0.21 2 0.80 0.10 3 3.10 0.86 0.80 is an estimate of ___________. 0.10 is an estimate of ___________. Assuming the responses satisfy the normality assumption, we can be 95% confident that the value of β 2 is in the interval,_______ ± [t .025 ⋅ _______], where t .025 is the critical value of the student's t -distribution with ____ degrees of freedom. The population value for β 2 , the change that occurs in Y with a unit change in X 2 , when the other variables are held constant. The population value for the standard error of the distribution of estimates of β 2 . .8, .1, 16 = 20 − 4.", "section": "How to Use Microsoft Excel® for Regression Analysis", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Statistical Tables F Distribution Table entry for p is the critical value F* with probability p lying to its right. F critical values Degrees of freedom in the numerator Degrees of freedom in the denominator p 1 2 3 4 5 6 7 8 9 1 .100 39.86 49.50 53.59 55.83 57.24 58.20 58.91 59.44 59.86 .050 161.45 199.50 215.71 224.58 230.16 233.99 236.77 238.88 240.54 .025 647.79 799.50 864.16 899.58 921.85 937.11 948.22 956.66 963.28 .010 4052.2 4999.5 5403.4 5624.6 5763.6 5859.0 5928.4 5981.1 6022.5 .001 405284 500000 540379 562500 576405 585937 592873 598144 602284 2 .100 8.53 9.00 9.16 9.24 9.29 9.33 9.35 9.37 9.38 .050 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38 .025 38.51 39.00 39.17 39.25 39.30 39.33 39.36 39.37 39.39 .010 98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 .001 998.50 999.00 999.17 999.25 999.30 999.33 999.36 999.37 999.39 3 .100 5.54 5.46 5.39 5.34 5.31 5.28 5.27 5.25 5.24 .050 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81 .025 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.47 .010 34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 .001 167.03 148.50 141.11 137.10 134.58 132.85 131.58 130.62 129.86 4 .100 4.54 4.32 4.19 4.11 4.05 4.01 3.98 3.95 3.94 .050 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 .025 12.22 10.65 9.98 9.60 9.36 9.20 9.07 8.98 8.90 .010 21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 .001 74.14 61.25 56.18 53.44 51.71 50.53 49.66 49.00 48.47 5 .100 4.06 3.78 3.62 3.52 3.45 3.40 3.37 3.34 3.32 .050 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 .025 10.01 8.43 7.76 7.39 7.15 6.98 6.85 6.76 6.68 .010 16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 .001 47.18 37.12 33.20 31.09 29.75 28.83 28.16 27.65 27.24 6 .100 3.78 3.46 3.29 3.18 3.11 3.05 3.01 2.98 2.96 .050 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 .025 8.81 7.26 6.60 6.23 5.99 5.82 5.70 5.60 5.52 .010 13.75 10.92 9.78 9.15 8.75 8.47 8.26 8.10 7.98 .001 35.51 27.00 23.70 21.92 20.80 20.03 19.46 19.03 18.69 7 .100 3.59 3.26 3.07 2.96 2.88 2.83 2.78 2.75 2.72 .050 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 .025 8.07 6.54 5.89 5.52 5.29 5.12 4.99 4.90 4.82 .010 12.25 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.72 .001 29.25 21.69 18.77 17.20 16.21 15.52 15.02 14.63 14.33 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 10 12 15 20 25 30 1 .100 60.19 60.71 61.22 61.74 62.05 62.26 .050 241.88 243.91 245.95 248.01 249.26 250.10 .025 968.63 976.71 984.87 993.10 998.08 1001.4 .010 6055.8 6106.3 6157.3 6208.7 6239.8 6260.6 .001 605621 610668 615764 620908 624017 626099 2 .100 9.39 9.41 9.42 9.44 9.45 9.46 .050 19.40 19.41 19.43 19.45 19.46 19.46 .025 39.40 39.41 39.43 39.45 39.46 39.46 .010 99.40 99.42 99.43 99.45 99.46 99.47 .001 999.40 999.42 999.43 999.45 999.46 999.47 3 .100 5.23 5.22 5.20 5.18 5.17 5.17 .050 8.79 8.74 8.70 8.66 8.63 8.62 .025 14.42 14.34 14.25 14.17 14.12 14.08 .010 27.23 27.05 26.87 26.69 26.58 26.50 .001 129.25 128.32 127.37 126.42 125.84 125.45 4 .100 3.92 3.90 3.87 3.84 3.83 3.82 .050 5.96 5.91 5.86 5.80 5.77 5.75 .025 8.84 8.75 8.66 8.56 8.50 8.46 .010 14.55 14.37 14.20 14.02 13.91 13.84 .001 48.05 47.41 46.76 46.10 45.70 45.43 5 .100 3.30 3.27 3.24 3.21 3.19 3.17 .050 4.74 4.68 4.62 4.56 4.52 4.50 .025 6.62 6.52 6.43 6.33 6.27 6.23 .010 10.05 9.89 9.72 9.55 9.45 9.38 .001 26.92 26.42 25.91 25.39 25.08 24.87 6 .100 2.94 2.90 2.87 2.84 2.81 2.80 .050 4.06 4.00 3.94 3.87 3.83 3.81 .025 5.46 5.37 5.27 5.17 5.11 5.07 .010 7.87 7.72 7.56 7.40 7.30 7.23 .001 18.41 17.99 17.56 17.12 16.85 16.67 7 .100 2.70 2.67 2.63 2.59 2.57 2.56 .050 3.64 3.57 3.51 3.44 3.40 3.38 .025 4.76 4.67 4.57 4.47 4.40 4.36 .010 6.62 6.47 6.31 6.16 6.06 5.99 .001 14.08 13.71 13.32 12.93 12.69 12.53 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 40 50 60 120 1000 1 .100 62.53 62.69 62.79 63.06 63.30 .050 251.14 251.77 252.20 253.25 254.19 .025 1005.6 1008.1 1009.8 1014.0 1017.7 .010 6286.8 6302.5 6313.0 6339.4 6362.7 .001 628712 630285 631337 633972 636301 2 .100 9.47 9.47 9.47 9.48 9.49 .050 19.47 19.48 19.48 19.49 19.49 .025 39.47 39.48 39.48 39.49 39.50 .010 99.47 99.48 99.48 99.49 99.50 .001 999.47 999.48 999.48 999.49 999.50 3 .100 5.16 5.15 5.15 5.14 5.13 .050 8.59 8.58 8.57 8.55 8.53 .025 14.04 14.01 13.99 13.95 13.91 .010 26.41 26.35 26.32 26.22 26.14 .001 124.96 124.66 124.47 123.97 123.53 4 .100 3.80 3.80 3.79 3.78 3.76 .050 5.72 5.70 5.69 5.66 5.63 .025 8.41 8.38 8.36 8.31 8.26 .010 13.75 13.69 13.65 13.56 13.47 .001 45.09 44.88 44.75 44.40 44.09 5 .100 3.16 3.15 3.14 3.12 3.11 .050 4.46 4.44 4.43 4.40 4.37 .025 6.18 6.14 6.12 6.07 6.02 .010 9.29 9.24 9.20 9.11 9.03 .001 24.60 24.44 24.33 24.06 23.82 6 .100 2.78 2.77 2.76 2.74 2.72 .050 3.77 3.75 3.74 3.70 3.67 .025 5.01 4.98 4.96 4.90 4.86 .010 7.14 7.09 7.06 6.97 6.89 .001 16.44 16.31 16.21 15.98 15.77 7 .100 2.54 2.52 2.51 2.49 2.47 .050 3.34 3.32 3.30 3.27 3.23 .025 4.31 4.28 4.25 4.20 4.15 .010 5.91 5.86 5.82 5.74 5.66 .001 12.33 12.20 12.12 11.91 11.72 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 1 2 3 4 5 6 7 8 9 8 .100 3.46 3.11 2.92 2.81 2.73 2.67 2.62 2.59 2.56 .050 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 .025 7.57 6.06 5.42 5.05 4.82 4.65 4.53 4.43 4.36 .010 11.26 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91 .001 25.41 18.49 15.83 14.39 13.48 12.86 12.40 12.05 11.77 9 .100 3.36 3.01 2.81 2.69 2.61 2.55 2.51 2.47 2.44 .050 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 .025 7.21 5.71 5.08 4.72 4.48 4.32 4.20 4.10 4.03 .010 10.56 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 .001 22.86 16.39 13.90 12.56 11.71 11.13 10.70 10.37 10.11 10 .100 3.29 2.92 2.73 2.61 2.52 2.46 2.41 2.38 2.35 .050 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 .025 6.94 5.46 4.83 4.47 4.24 4.07 3.95 3.85 3.78 .010 10.04 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 .001 21.04 14.91 12.55 11.28 10.48 9.93 9.52 9.20 8.96 11 .100 3.23 2.86 2.66 2.54 2.45 2.39 2.34 2.30 2.27 .050 4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 .025 6.72 5.26 4.63 4.28 4.04 3.88 3.76 3.66 3.59 .010 9.65 7.21 6.22 5.67 5.32 5.07 4.89 4.74 4.63 .001 19.69 13.81 11.56 10.35 9.58 9.05 8.66 8.35 8.12 12 .100 3.18 2.81 2.61 2.48 2.39 2.33 2.28 2.24 2.21 .050 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 .025 6.55 5.10 4.47 4.12 3.89 3.73 3.61 3.51 3.44 .010 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 .001 18.64 12.97 10.80 9.63 8.89 8.38 8.00 7.71 7.48 13 .100 3.14 2.76 2.56 2.43 2.35 2.28 2.23 2.20 2.16 .050 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 .025 6.41 4.97 4.35 4.00 3.77 3.60 3.48 3.39 3.31 .010 9.07 6.70 5.74 5.21 4.86 4.62 4.44 4.30 4.19 .001 17.82 12.31 10.21 9.07 8.35 7.86 7.49 7.21 6.98 14 .100 3.10 2.73 2.52 2.39 2.31 2.24 2.19 2.15 2.12 .050 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 .025 6.30 4.86 4.24 3.89 3.66 3.50 3.38 3.29 3.21 .010 8.86 6.51 5.56 5.04 4.69 4.46 4.28 4.14 4.03 .001 17.14 11.78 9.73 8.62 7.92 7.44 7.08 6.80 6.58 15 .100 3.07 2.70 2.49 2.36 2.27 2.21 2.16 2.12 2.09 .050 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 .025 6.20 4.77 4.15 3.80 3.58 3.41 3.29 3.20 3.12 .010 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 .001 16.59 11.34 9.34 8.25 7.57 7.09 6.74 6.47 6.26 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 10 12 15 20 25 30 40 50 60 120 1000 8 .100 2.54 2.50 2.46 2.42 2.40 2.38 2.36 2.35 2.34 2.32 2.30 .050 3.35 3.28 3.22 3.15 3.11 3.08 3.04 3.02 3.01 2.97 2.93 .025 4.30 4.20 4.10 4.00 3.94 3.89 3.84 3.81 3.78 3.73 3.68 .010 5.81 5.67 5.52 5.36 5.26 5.20 5.12 5.07 5.03 4.95 4.87 .001 11.54 11.19 10.84 10.48 10.26 10.11 9.92 9.80 9.73 9.53 9.36 9 .100 2.42 2.38 2.34 2.30 2.27 2.25 2.23 2.22 2.21 2.18 2.16 .050 3.14 3.07 3.01 2.94 2.89 2.86 2.83 2.80 2.79 2.75 2.71 .025 3.96 3.87 3.77 3.67 3.60 3.56 3.51 3.47 3.45 3.39 3.34 .010 5.26 5.11 4.96 4.81 4.71 4.65 4.57 4.52 4.48 4.40 4.32 .001 9.89 9.57 9.24 8.90 8.69 8.55 8.37 8.26 8.19 8.00 7.84 10 .100 2.32 2.28 2.24 2.20 2.17 2.16 2.13 2.12 2.11 2.08 2.06 .050 2.98 2.91 2.85 2.77 2.73 2.70 2.66 2.64 2.62 2.58 2.54 .025 3.72 3.62 3.52 3.42 3.35 3.31 3.26 3.22 3.20 3.14 3.09 .010 4.85 4.71 4.56 4.41 4.31 4.25 4.17 4.12 4.08 4.00 3.92 .001 8.75 8.45 8.13 7.80 7.60 7.47 7.30 7.19 7.12 6.94 6.78 11 .100 2.25 2.21 2.17 2.12 2.10 2.08 2.05 2.04 2.03 2.00 1.98 .050 2.85 2.79 2.72 2.65 2.60 2.57 2.53 2.51 2.49 2.45 2.41 .025 3.53 3.43 3.33 3.23 3.16 3.12 3.06 3.03 3.00 2.94 2.89 .010 4.54 4.40 4.25 4.10 4.01 3.94 3.86 3.81 3.78 3.69 3.61 .001 7.92 7.63 7.32 7.01 6.81 6.68 6.52 6.42 6.35 6.18 6.02 12 .100 2.19 2.15 2.10 2.06 2.03 2.01 1.99 1.97 1.96 1.93 1.91 .050 2.75 2.69 2.62 2.54 2.50 2.47 2.43 2.40 2.38 2.34 2.30 .025 3.37 3.28 3.18 3.07 3.01 2.96 2.91 2.87 2.85 2.79 2.73 .010 4.30 4.16 4.01 3.86 3.76 3.70 3.62 3.57 3.54 3.45 3.37 .001 7.29 7.00 6.71 6.40 6.22 6.09 5.93 5.83 5.76 5.59 5.44 13 .100 2.14 2.10 2.05 2.01 1.98 1.96 1.93 1.92 1.90 1.88 1.85 .050 2.67 2.60 2.53 2.46 2.41 2.38 2.34 2.31 2.30 2.25 2.21 .025 3.25 3.15 3.05 2.95 2.88 2.84 2.78 2.74 2.72 2.66 2.60 .010 4.10 3.96 3.82 3.66 3.57 3.51 3.43 3.38 3.34 3.25 3.18 .001 6.80 6.52 6.23 5.93 5.75 5.63 5.47 5.37 5.30 5.14 4.99 14 .100 2.10 2.05 2.01 1.96 1.93 1.91 1.89 1.87 1.86 1.83 1.80 .050 2.60 2.53 2.46 2.39 2.34 2.31 2.27 2.24 2.22 2.18 2.14 .025 3.15 3.05 2.95 2.84 2.78 2.73 2.67 2.64 2.61 2.55 2.50 .010 3.94 3.80 3.66 3.51 3.41 3.35 3.27 3.22 3.18 3.09 3.02 .001 6.40 6.13 5.85 5.56 5.38 5.25 5.10 5.00 4.94 4.77 4.62 15 .100 2.06 2.02 1.97 1.92 1.89 1.87 1.85 1.83 1.82 1.79 1.76 .050 2.54 2.48 2.40 2.33 2.28 2.25 2.20 2.18 2.16 2.11 2.07 .025 3.06 2.96 2.86 2.76 2.69 2.64 2.59 2.55 2.52 2.46 2.40 .010 3.80 3.67 3.52 3.37 3.28 3.21 3.13 3.08 3.05 2.96 2.88 .001 6.08 5.81 5.54 5.25 5.07 4.95 4.80 4.70 4.64 4.47 4.33 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 1 2 3 4 5 6 7 8 9 16 .100 3.05 2.67 2.46 2.33 2.24 2.18 2.13 2.09 2.06 .050 4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 .025 6.12 4.69 4.08 3.73 3.50 3.34 3.22 3.12 3.05 .010 8.53 6.23 5.29 4.77 4.44 4.20 4.03 3.89 3.78 .001 16.12 10.97 9.01 7.94 7.27 6.80 6.46 6.19 5.98 17 .100 3.03 2.64 2.44 2.31 2.22 2.15 2.10 2.06 2.03 .050 4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 .025 6.04 4.62 4.01 3.66 3.44 3.28 3.16 3.06 2.98 .010 8.40 6.11 5.19 4.67 4.34 4.10 3.93 3.79 3.68 .001 15.72 10.66 8.73 7.68 7.02 6.56 6.22 5.96 5.75 18 .100 3.01 2.62 2.42 2.29 2.20 2.13 2.08 2.04 2.00 .050 4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 .025 5.98 4.56 3.95 3.61 3.38 3.22 3.10 3.01 2.93 .010 8.29 6.01 5.09 4.58 4.25 4.01 3.84 3.71 3.60 .001 15.38 10.39 8.49 7.46 6.81 6.35 6.02 5.76 5.56 19 .100 3.36 3.01 2.81 2.69 2.61 2.55 2.51 2.47 2.44 .050 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 .025 7.21 5.71 5.08 4.72 4.48 4.32 4.20 4.10 4.03 .010 10.56 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 .001 22.86 16.39 13.90 12.56 11.71 11.13 10.70 10.37 10.11 20 .100 2.97 2.59 2.38 2.25 2.16 2.09 2.04 2.00 1.96 .050 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 .025 5.87 4.46 3.86 3.51 3.29 3.13 3.01 2.91 2.84 .010 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 .001 14.82 9.95 8.10 7.10 6.46 6.02 5.69 5.44 5.24 21 .100 2.96 2.57 2.36 2.23 2.14 2.08 2.02 1.98 1.95 .050 4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 .025 5.83 4.42 3.82 3.48 3.25 3.09 2.97 2.87 2.80 .010 8.02 5.78 4.87 4.37 4.04 3.81 3.64 3.51 3.40 .001 14.59 9.77 7.94 6.95 6.32 5.88 5.56 5.31 5.11 22 .100 2.95 2.56 2.35 2.22 2.13 2.06 2.01 1.97 1.93 .050 4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 .025 5.79 4.38 3.78 3.44 3.22 3.05 2.93 2.84 2.76 .010 7.95 5.72 4.82 4.31 3.99 3.76 3.59 3.45 3.35 .001 14.38 9.61 7.80 6.81 6.19 5.76 5.44 5.19 4.99 23 .100 2.94 2.55 2.34 2.21 2.11 2.05 1.99 1.95 1.92 .050 4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 .025 5.75 4.35 3.75 3.41 3.18 3.02 2.90 2.81 2.73 .010 7.88 5.66 4.76 4.26 3.94 3.71 3.54 3.41 3.30 .001 14.20 9.47 7.67 6.70 6.08 5.65 5.33 5.09 4.89 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 10 12 15 20 25 30 40 50 60 120 1000 16 .100 2.03 1.99 1.94 1.89 1.86 1.84 1.81 1.79 1.78 1.75 1.72 .050 2.49 2.42 2.35 2.28 2.23 2.19 2.15 2.12 2.11 2.06 2.02 .025 2.99 2.89 2.79 2.68 2.61 2.57 2.51 2.47 2.45 2.38 2.32 .010 3.69 3.55 3.41 3.26 3.16 3.10 3.02 2.97 2.93 2.84 2.76 .001 5.81 5.55 5.27 4.99 4.82 4.70 4.54 4.45 4.39 4.23 4.08 17 .100 2.00 1.96 1.91 1.86 1.83 1.81 1.78 1.76 1.75 1.72 1.69 .050 2.45 2.38 2.31 2.23 2.18 2.15 2.10 2.08 2.06 2.01 1.97 .025 2.92 2.82 2.72 2.62 2.55 2.50 2.44 2.41 2.38 2.32 2.26 .010 3.59 3.46 3.31 3.16 3.07 3.00 2.92 2.87 2.83 2.75 2.66 .001 5.58 5.32 5.05 4.78 4.60 4.48 4.33 4.24 4.18 4.02 3.87 18 .100 1.98 1.93 1.89 1.84 1.80 1.78 1.75 1.74 1.72 1.69 1.66 .050 2.41 2.34 2.27 2.19 2.14 2.11 2.06 2.04 2.02 1.97 1.92 .025 2.87 2.77 2.67 2.56 2.49 2.44 2.38 2.35 2.32 2.26 2.20 .010 3.51 3.37 3.23 3.08 2.98 2.92 2.84 2.78 2.75 2.66 2.58 .001 5.39 5.13 4.87 4.59 4.42 4.30 4.15 4.06 4.00 3.84 3.69 19 .100 1.96 1.91 1.86 1.81 1.78 1.76 1.73 1.71 1.70 1.67 1.64 .050 2.38 2.31 2.23 2.16 2.11 2.07 2.03 2.00 1.98 1.93 1.88 .025 2.82 2.72 2.62 2.51 2.44 2.39 2.33 2.30 2.27 2.20 2.14 .010 3.43 3.30 3.15 3.00 2.91 2.84 2.76 2.71 2.67 2.58 2.50 .001 5.22 4.97 4.70 4.43 4.26 4.14 3.99 3.90 3.84 3.68 3.53 20 .100 1.94 1.89 1.84 1.79 1.76 1.74 1.71 1.69 1.68 1.64 1.61 .050 2.35 2.28 2.20 2.12 2.07 2.04 1.99 1.97 1.95 1.90 1.85 .025 2.77 2.68 2.57 2.46 2.40 2.35 2.29 2.25 2.22 2.16 2.09 .010 3.37 3.23 3.09 2.94 2.84 2.78 2.69 2.64 2.61 2.52 2.43 .001 5.08 4.82 4.56 4.29 4.12 4.00 3.86 3.77 3.70 3.54 3.40 21 .100 1.92 1.87 1.83 1.78 1.74 1.72 1.69 1.67 1.66 1.62 1.59 .050 2.32 2.25 2.18 2.10 2.05 2.01 1.96 1.94 1.92 1.87 1.82 .025 2.73 2.64 2.53 2.42 2.36 2.31 2.25 2.21 2.18 2.11 2.05 .010 3.31 3.17 3.03 2.88 2.79 2.72 2.64 2.58 2.55 2.46 2.37 .001 4.95 4.70 4.44 4.17 4.00 3.88 3.74 3.64 3.58 3.42 3.28 22 .100 1.90 1.86 1.81 1.76 1.73 1.70 1.67 1.65 1.64 1.60 1.57 .050 2.30 2.23 2.15 2.07 2.02 1.98 1.94 1.91 1.89 1.84 1.79 .025 2.70 2.60 2.50 2.39 2.32 2.27 2.21 2.17 2.14 2.08 2.01 .010 3.26 3.12 2.98 2.83 2.73 2.67 2.58 2.53 2.50 2.40 2.32 .001 4.83 4.58 4.33 4.06 3.89 3.78 3.63 3.54 3.48 3.32 3.17 23 .100 1.89 1.84 1.80 1.74 1.71 1.69 1.66 1.64 1.62 1.59 1.55 .050 2.27 2.20 2.13 2.05 2.00 1.96 1.91 1.88 1.86 1.81 1.76 .025 2.67 2.57 2.47 2.36 2.29 2.24 2.18 2.14 2.11 2.04 1.98 .010 3.21 3.07 2.93 2.78 2.69 2.62 2.54 2.48 2.45 2.35 2.27 .001 4.73 4.48 4.23 3.96 3.79 3.68 3.53 3.44 3.38 3.22 3.08 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 1 2 3 4 5 6 7 8 9 24 .100 2.93 2.54 2.33 2.19 2.10 2.04 1.98 1.94 1.91 .050 4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 .025 5.72 4.32 3.72 3.38 3.15 2.99 2.87 2.78 2.70 .010 7.82 5.61 4.72 4.22 3.90 3.67 3.50 3.36 3.26 .001 14.03 9.34 7.55 6.59 5.98 5.55 5.23 4.99 4.80 25 .100 2.92 2.53 2.32 2.18 2.09 2.02 1.97 1.93 1.89 .050 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 .025 5.69 4.29 3.69 3.35 3.13 2.97 2.85 2.75 2.68 .010 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 .001 13.88 9.22 7.45 6.49 5.89 5.46 5.15 4.91 4.71 26 .100 2.91 2.52 2.31 2.17 2.08 2.01 1.96 1.92 1.88 .050 4.23 3.37 2.98 2.74 2.59 2.47 2.39 2.32 2.27 .025 5.66 4.27 3.67 3.33 3.10 2.94 2.82 2.73 2.65 .010 7.72 5.53 4.64 4.14 3.82 3.59 3.42 3.29 3.18 .001 13.74 9.12 7.36 6.41 5.80 5.38 5.07 4.83 4.64 27 .100 2.90 2.51 2.30 2.17 2.07 2.00 1.95 1.91 1.87 .050 4.21 3.35 2.96 2.73 2.57 2.46 2.37 2.31 2.25 .025 5.63 4.24 3.65 3.31 3.08 2.92 2.80 2.71 2.63 .010 7.68 5.49 4.60 4.11 3.78 3.56 3.39 3.26 3.15 .001 13.61 9.02 7.27 6.33 5.73 5.31 5.00 4.76 4.57 28 .100 2.89 2.50 2.29 2.16 2.06 2.00 1.94 1.90 1.87 .050 4.20 3.34 2.95 2.71 2.56 2.45 2.36 2.29 2.24 .025 5.61 4.22 3.63 3.29 3.06 2.90 2.78 2.69 2.61 .010 7.64 5.45 4.57 4.07 3.75 3.53 3.36 3.23 3.12 .001 13.50 8.93 7.19 6.25 5.66 5.24 4.93 4.69 4.50 29 .100 2.89 2.50 2.28 2.15 2.06 1.99 1.93 1.89 1.86 .050 4.18 3.33 2.93 2.70 2.55 2.43 2.35 2.28 2.22 .025 5.59 4.20 3.61 3.27 3.04 2.88 2.76 2.67 2.59 .010 7.60 5.42 4.54 4.04 3.73 3.50 3.33 3.20 3.09 .001 13.39 8.85 7.12 6.19 5.59 5.18 4.87 4.64 4.45 30 .100 2.88 2.49 2.28 2.14 2.05 1.98 1.93 1.88 1.85 .050 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 .025 5.57 4.18 3.59 3.25 3.03 2.87 2.75 2.65 2.57 .010 7.56 5.39 4.51 4.02 3.70 3.47 3.30 3.17 3.07 .001 13.29 8.77 7.05 6.12 5.53 5.12 4.82 4.58 4.39 40 .100 2.84 2.44 2.23 2.09 2.00 1.93 1.87 1.83 1.79 .050 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12 .025 5.42 4.05 3.46 3.13 2.90 2.74 2.62 2.53 2.45 .010 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89 .001 12.61 8.25 6.59 5.70 5.13 4.73 4.44 4.21 4.02 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 10 12 15 20 25 30 40 50 60 120 1000 24 .100 1.88 1.83 1.78 1.73 1.70 1.67 1.64 1.62 1.61 1.57 1.54 .050 2.25 2.18 2.11 2.03 1.97 1.94 1.89 1.86 1.84 1.79 1.74 .025 2.64 2.54 2.44 2.33 2.26 2.21 2.15 2.11 2.08 2.01 1.94 .010 3.17 3.03 2.89 2.74 2.64 2.58 2.49 2.44 2.40 2.31 2.22 .001 4.64 4.39 4.14 3.87 3.71 3.59 3.45 3.36 3.29 3.14 2.99 25 .100 1.87 1.82 1.77 1.72 1.68 1.66 1.63 1.61 1.59 1.56 1.52 .050 2.24 2.16 2.09 2.01 1.96 1.92 1.87 1.84 1.82 1.77 1.72 .025 2.61 2.51 2.41 2.30 2.23 2.18 2.12 2.08 2.05 1.98 1.91 .010 3.13 2.99 2.85 2.70 2.60 2.54 2.45 2.40 2.36 2.27 2.18 .001 4.56 4.31 4.06 3.79 3.63 3.52 3.37 3.28 3.22 3.06 2.91 26 .100 1.86 1.81 1.76 1.71 1.67 1.65 1.61 1.59 1.58 1.54 1.51 .050 2.22 2.15 2.07 1.99 1.94 1.90 1.85 1.82 1.80 1.75 1.70 .025 2.59 2.49 2.39 2.28 2.21 2.16 2.09 2.05 2.03 1.95 1.89 .010 3.09 2.96 2.81 2.66 2.57 2.50 2.42 2.36 2.33 2.23 2.14 .001 4.48 4.24 3.99 3.72 3.56 3.44 3.30 3.21 3.15 2.99 2.84 27 .100 1.85 1.80 1.75 1.70 1.66 1.64 1.60 1.58 1.57 1.53 1.50 .050 2.20 2.13 2.06 1.97 1.92 1.88 1.84 1.81 1.79 1.73 1.68 .025 2.57 2.47 2.36 2.25 2.18 2.13 2.07 2.03 2.00 1.93 1.86 .010 3.06 2.93 2.78 2.63 2.54 2.47 2.38 2.33 2.29 2.20 2.11 .001 4.41 4.17 3.92 3.66 3.49 3.38 3.23 3.14 3.08 2.92 2.78 28 .100 1.84 1.79 1.74 1.69 1.65 1.63 1.59 1.57 1.56 1.52 1.48 .050 2.19 2.12 2.04 1.96 1.91 1.87 1.82 1.79 1.77 1.71 1.66 .025 2.55 2.45 2.34 2.23 2.16 2.11 2.05 2.01 1.98 1.91 1.84 .010 3.03 2.90 2.75 2.60 2.51 2.44 2.35 2.30 2.26 2.17 2.08 .001 4.35 4.11 3.86 3.60 3.43 3.32 3.18 3.09 3.02 2.86 2.72 29 .100 1.83 1.78 1.73 1.68 1.64 1.62 1.58 1.56 1.55 1.51 1.47 .050 2.18 2.10 2.03 1.94 1.89 1.85 1.81 1.77 1.75 1.70 1.65 .025 2.53 2.43 2.32 2.21 2.14 2.09 2.03 1.99 1.96 1.89 1.82 .010 3.00 2.87 2.73 2.57 2.48 2.41 2.33 2.27 2.23 2.14 2.05 .001 4.29 4.05 3.80 3.54 3.38 3.27 3.12 3.03 2.97 2.81 2.66 30 .100 1.82 1.77 1.72 1.67 1.63 1.61 1.57 1.55 1.54 1.50 1.46 .050 2.16 2.09 2.01 1.93 1.88 1.84 1.79 1.76 1.74 1.68 1.63 .025 2.51 2.41 2.31 2.20 2.12 2.07 2.01 1.97 1.94 1.87 1.80 .010 2.98 2.84 2.70 2.55 2.45 2.39 2.30 2.25 2.21 2.11 2.02 .001 4.24 4.00 3.75 3.49 3.33 3.22 3.07 2.98 2.92 2.76 2.61 40 .100 1.76 1.71 1.66 1.61 1.57 1.54 1.51 1.48 1.47 1.42 1.38 .050 2.08 2.00 1.92 1.84 1.78 1.74 1.69 1.66 1.64 1.58 1.52 .025 2.39 2.29 2.18 2.07 1.99 1.94 1.88 1.83 1.80 1.72 1.65 .010 2.80 2.66 2.52 2.37 2.27 2.20 2.11 2.06 2.02 1.92 1.82 .001 3.87 3.64 3.40 3.14 2.98 2.87 2.73 2.64 2.57 2.41 2.25 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 1 2 3 4 5 6 7 8 9 50 .100 2.81 2.41 2.20 2.06 1.97 1.90 1.84 1.80 1.76 .050 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 2.07 .025 5.34 3.97 3.39 3.05 2.83 2.67 2.55 2.46 2.38 .010 7.17 5.06 4.20 3.72 3.41 3.19 3.02 2.89 2.78 .001 12.22 7.96 6.34 5.46 4.90 4.51 4.22 4.00 3.82 60 .100 2.79 2.39 2.18 2.04 1.95 1.87 1.82 1.77 1.74 .050 4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 .025 5.29 3.93 3.34 3.01 2.79 2.63 2.51 2.41 2.33 .010 7.08 4.98 4.13 3.65 3.34 3.12 2.95 2.82 2.72 .001 11.97 7.77 6.17 5.31 4.76 4.37 4.09 3.86 3.69 100 .100 2.76 2.36 2.14 2.00 1.91 1.83 1.78 1.73 1.69 .050 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 1.97 .025 5.18 3.83 3.25 2.92 2.70 2.54 2.42 2.32 2.24 .010 6.90 4.82 3.98 3.51 3.21 2.99 2.82 2.69 2.59 .001 11.50 7.41 5.86 5.02 4.48 4.11 3.83 3.61 3.44 200 .100 2.73 2.33 2.11 1.97 1.88 1.80 1.75 1.70 1.66 .050 3.89 3.04 2.65 2.42 2.26 2.14 2.06 1.98 1.93 .025 5.10 3.76 3.18 2.85 2.63 2.47 2.35 2.26 2.18 .010 6.76 4.71 3.88 3.41 3.11 2.89 2.73 2.60 2.50 .001 11.15 7.15 5.63 4.81 4.29 3.92 3.65 3.43 3.26 1000 .100 2.71 2.31 2.09 1.95 1.85 1.78 1.72 1.68 1.64 .050 3.85 3.00 2.61 2.38 2.22 2.11 2.02 1.95 1.89 .025 5.04 3.70 3.13 2.80 2.58 2.42 2.30 2.20 2.13 .010 6.66 4.63 3.80 3.34 3.04 2.82 2.66 2.53 2.43 .001 10.89 6.96 5.46 4.65 4.14 3.78 3.51 3.30 3.13 F critical values (continued) Degrees of freedom in the numerator Degrees of freedom in the denominator p 10 12 15 20 25 30 40 50 60 120 1000 50 .100 1.73 1.68 1.63 1.57 1.53 1.50 1.46 1.44 1.42 1.38 1.33 .050 2.03 1.95 1.87 1.78 1.73 1.69 1.63 1.60 1.58 1.51 1.45 .025 2.32 2.22 2.11 1.99 1.92 1.87 1.80 1.75 1.72 1.64 1.56 .010 2.70 2.56 2.42 2.27 2.17 2.10 2.01 1.95 1.91 1.80 1.70 .001 3.67 3.44 3.20 2.95 2.79 2.68 2.53 2.44 2.38 2.21 2.05 60 .100 1.71 1.66 1.60 1.54 1.50 1.48 1.44 1.41 1.40 1.35 1.30 .050 1.99 1.92 1.84 1.75 1.69 1.65 1.59 1.56 1.53 1.47 1.40 .025 2.27 2.17 2.06 1.94 1.87 1.82 1.74 1.70 1.67 1.58 1.49 .010 2.63 2.50 2.35 2.20 2.10 2.03 1.94 1.88 1.84 1.73 1.62 .001 3.54 3.32 3.08 2.83 2.67 2.55 2.41 2.32 2.25 2.08 1.92 100 .100 1.66 1.61 1.56 1.49 1.45 1.42 1.38 1.35 1.34 1.28 1.22 .050 1.93 1.85 1.77 1.68 1.62 1.57 1.52 1.48 1.45 1.38 1.30 .025 2.18 2.08 1.97 1.85 1.77 1.71 1.64 1.59 1.56 1.46 1.36 .010 2.50 2.37 2.22 2.07 1.97 1.89 1.80 1.74 1.69 1.57 1.45 .001 3.30 3.07 2.84 2.59 2.43 2.32 2.17 2.08 2.01 1.83 1.64 200 .100 1.63 1.58 1.52 1.46 1.41 1.38 1.34 1.31 1.29 1.23 1.16 .050 1.88 1.80 1.72 1.62 1.56 1.52 1.46 1.41 1.39 1.30 1.21 .025 2.11 2.01 1.90 1.78 1.70 1.64 1.56 1.51 1.47 1.37 1.25 .010 2.41 2.27 2.13 1.97 1.87 1.79 1.69 1.63 1.58 1.45 1.30 .001 3.12 2.90 2.67 2.42 2.26 2.15 2.00 1.90 1.83 1.64 1.43 1000 .100 1.61 1.55 1.49 1.43 1.38 1.35 1.30 1.27 1.25 1.18 1.08 .050 1.84 1.76 1.68 1.58 1.52 1.47 1.41 1.36 1.33 1.24 1.11 .025 2.06 1.96 1.85 1.72 1.64 1.58 1.50 1.45 1.41 1.29 1.13 .010 2.34 2.20 2.06 1.90 1.79 1.72 1.61 1.54 1.50 1.35 1.16 .001 2.99 2.77 2.54 2.30 2.14 2.02 1.87 1.77 1.69 1.49 1.22 Numerical entries represent the probability that a standard normal random variable is between 0 and z where z = x − μ σ . Standard Normal Probability Distribution: Z Table Standard Normal Distribution z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993 3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995 3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997 3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998 Student's t Distribution Upper critical values of Student's t Distribution with v Degrees of Freedom For selected probabilities, a , the table shows the values t v,a such that P ( t v > t v,a ) = a , where t v is a Student’s t random variable with v degrees of freedom. For example, the probability is .10 that a Student’s t random variable with 10 degrees of freedom exceeds 1.372. Probability of Exceeding the Critical Value v 0.10 0.05 0.025 0.01 0.005 0.001 1 3.078 6.314 12.706 31.821 63.657 318.313 2 1.886 2.920 4.303 6.965 9.925 22.327 3 1.638 2.353 3.182 4.541 5.841 10.215 4 1.533 2.132 2.776 3.747 4.604 7.173 5 1.476 2.015 2.571 3.365 4.032 5.893 6 1.440 1.943 2.447 3.143 3.707 5.208 7 1.415 1.895 2.365 2.998 3.499 4.782 8 1.397 1.860 2.306 2.896 3.355 4.499 9 1.383 1.833 2.262 2.821 3.250 4.296 10 1.372 1.812 2.228 2.764 3.169 4.143 11 1.363 1.796 2.201 2.718 3.106 4.024 12 1.356 1.782 2.179 2.681 3.055 3.929 13 1.350 1.771 2.160 2.650 3.012 3.852 14 1.345 1.761 2.145 2.624 2.977 3.787 15 1.341 1.753 2.131 2.602 2.947 3.733 16 1.337 1.746 2.120 2.583 2.921 3.686 17 1.333 1.740 2.110 2.567 2.898 3.646 18 1.330 1.734 2.101 2.552 2.878 3.610 19 1.328 1.729 2.093 2.539 2.861 3.579 20 1.325 1.725 2.086 2.528 2.845 3.552 21 1.323 1.721 2.080 2.518 2.831 3.527 22 1.321 1.717 2.074 2.508 2.819 3.505 23 1.319 1.714 2.069 2.500 2.807 3.485 24 1.318 1.711 2.064 2.492 2.797 3.467 25 1.316 1.708 2.060 2.485 2.787 3.450 26 1.315 1.706* 2.056 2.479 2.779 3.435 27 1.314 1.703 2.052 2.473 2.771 3.421 28 1.313 1.701 2.048 2.467 2.763 3.408 29 1.311 1.699 2.045 2.462 2.756 3.396 30 1.310 1.697 2.042 2.457 2.750 3.385 40 1.303 1.684 2.021 2.423 2.704 3.307 60 1.296 1.671 2.000 2.390 2.660 3.232 100 1.290 1.660 1.984 2.364 2.626 3.174 ∞ 1.282 1.645 1.960 2.326 2.576 3.090 NIST/SEMATECH e-Handbook of Statistical Methods, http://www.itl.nist.gov/div898/handbook/, September 2011. χ 2 Probability Distribution Area to the Right of the Critical Value of χ 2 df 0.995 0.990 0.975 0.950 0.900 0.100 0.050 0.025 0.010 0.005 1 0.000 0.000 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5 0.412 0.554 0.831 1.145 1.610 9.236 11.070 12.833 15.086 16.750 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188 11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757 12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.300 13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.819 14 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.319 15 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.801 16 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.267 17 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.718 18 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.156 19 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.582 20 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.997 21 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.401 22 8.643 9.542 10.982 12.338 14.041 30.813 33.924 36.781 40.289 42.796 23 9.260 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.181 24 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.559 25 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.928 26 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.290 27 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.195 46.963 49.645 28 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.993 29 13.121 14.256 16.047 17.708 19.768 39.087 42.557 45.722 49.588 52.336 30 13.787 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672 40 20.707 22.164 24.433 26.509 29.051 51.805 55.758 59.342 63.691 66.766 50 27.991 29.707 32.357 34.764 37.689 63.167 67.505 71.420 76.154 79.490 60 35.534 37.485 40.482 43.188 46.459 74.397 79.082 83.298 88.379 91.952 70 43.275 45.442 48.758 51.739 55.329 85.527 90.531 95.023 100.425 104.215 80 51.172 53.540 57.153 60.391 64.278 96.578 101.879 106.629 112.329 116.321 90 59.196 61.754 65.647 69.126 73.291 107.565 113.145 118.136 124.116 128.299 100 67.328 70.065 74.222 77.929 82.358 118.498 124.342 129.561 135.807 140.169 Data Sets Lap Times The following tables provide lap times from Terri Vogel's log book. Times are recorded in seconds for 2.5-mile laps completed in a series of races and practice runs. Race Lap Times (in seconds) Lap 1 Lap 2 Lap 3 Lap 4 Lap 5 Lap 6 Lap 7 Race 1 135 130 131 132 130 131 133 Race 2 134 131 131 129 128 128 129 Race 3 129 128 127 127 130 127 129 Race 4 125 125 126 125 124 125 125 Race 5 133 132 132 132 131 130 132 Race 6 130 130 130 129 129 130 129 Race 7 132 131 133 131 134 134 131 Race 8 127 128 127 130 128 126 128 Race 9 132 130 127 128 126 127 124 Race 10 135 131 131 132 130 131 130 Race 11 132 131 132 131 130 129 129 Race 12 134 130 130 130 131 130 130 Race 13 128 127 128 128 128 129 128 Race 14 132 131 131 131 132 130 130 Race 15 136 129 129 129 129 129 129 Race 16 129 129 129 128 128 129 129 Race 17 134 131 132 131 132 132 132 Race 18 129 129 130 130 133 133 127 Race 19 130 129 129 129 129 129 128 Race 20 131 128 130 128 129 130 130 Practice Lap Times (in seconds) Lap 1 Lap 2 Lap 3 Lap 4 Lap 5 Lap 6 Lap 7 Practice 1 142 143 180 137 134 134 172 Practice 2 140 135 134 133 128 128 131 Practice 3 130 133 130 128 135 133 133 Practice 4 141 136 137 136 136 136 145 Practice 5 140 138 136 137 135 134 134 Practice 6 142 142 139 138 129 129 127 Practice 7 139 137 135 135 137 134 135 Practice 8 143 136 134 133 134 133 132 Practice 9 135 134 133 133 132 132 133 Practice 10 131 130 128 129 127 128 127 Practice 11 143 139 139 138 138 137 138 Practice 12 132 133 131 129 128 127 126 Practice 13 149 144 144 139 138 138 137 Practice 14 133 132 137 133 134 130 131 Practice 15 138 136 133 133 132 131 131 Stock Prices The following table lists initial public offering (IPO) stock prices for all 1999 stocks that at least doubled in value during the first day of trading. IPO Offer Prices $17.00 $23.00 $14.00 $16.00 $12.00 $26.00 $20.00 $22.00 $14.00 $15.00 $22.00 $18.00 $18.00 $21.00 $21.00 $19.00 $15.00 $21.00 $18.00 $17.00 $15.00 $25.00 $14.00 $30.00 $16.00 $10.00 $20.00 $12.00 $16.00 $17.44 $16.00 $14.00 $15.00 $20.00 $20.00 $16.00 $17.00 $16.00 $15.00 $15.00 $19.00 $48.00 $16.00 $18.00 $9.00 $18.00 $18.00 $20.00 $8.00 $20.00 $17.00 $14.00 $11.00 $16.00 $19.00 $15.00 $21.00 $12.00 $8.00 $16.00 $13.00 $14.00 $15.00 $14.00 $13.41 $28.00 $21.00 $17.00 $28.00 $17.00 $19.00 $16.00 $17.00 $19.00 $18.00 $17.00 $15.00 $14.00 $21.00 $12.00 $18.00 $24.00 $15.00 $23.00 $14.00 $16.00 $12.00 $24.00 $20.00 $14.00 $14.00 $15.00 $14.00 $19.00 $16.00 $38.00 $20.00 $24.00 $16.00 $8.00 $18.00 $17.00 $16.00 $15.00 $7.00 $19.00 $12.00 $8.00 $23.00 $12.00 $18.00 $20.00 $21.00 $34.00 $16.00 $26.00 $14.00 References Data compiled by Jay R. Ritter of University of Florida using data from Securities Data Co. and Bloomberg .", "section": "Statistical Tables", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Mathematical Phrases, Symbols, and Formulas English Phrases Written Mathematically When the English says: Interpret this as: X is at least 4. X ≥ 4 The minimum of X is 4. X ≥ 4 X is no less than 4. X ≥ 4 X is greater than or equal to 4. X ≥ 4 X is at most 4. X ≤ 4 The maximum of X is 4. X ≤ 4 X is no more than 4. X ≤ 4 X is less than or equal to 4. X ≤ 4 X does not exceed 4. X ≤ 4 X is greater than 4. X > 4 X is more than 4. X > 4 X exceeds 4. X > 4 X is less than 4. X < 4 There are fewer X than 4. X < 4 X is 4. X = 4 X is equal to 4. X = 4 X is the same as 4. X = 4 X is not 4. X ≠ 4 X is not equal to 4. X ≠ 4 X is not the same as 4. X ≠ 4 X is different than 4. X ≠ 4 Symbols and Their Meanings Symbols and their Meanings Chapter (1st used) Symbol Spoken Meaning Sampling and Data The square root of same Sampling and Data π Pi 3.14159… (a specific number) Descriptive Statistics Q 1 Quartile one the first quartile Descriptive Statistics Q 2 Quartile two the second quartile Descriptive Statistics Q 3 Quartile three the third quartile Descriptive Statistics IQR interquartile range Q 3 – Q 1 = IQR Descriptive Statistics x – x-bar sample mean Descriptive Statistics μ mu population mean Descriptive Statistics s s sample standard deviation Descriptive Statistics s 2 s squared sample variance Descriptive Statistics σ sigma population standard deviation Descriptive Statistics σ 2 sigma squared population variance Descriptive Statistics Σ capital sigma sum Probability Topics { } brackets set notation Probability Topics S S sample space Probability Topics A Event A event A Probability Topics P ( A ) probability of A probability of A occurring Probability Topics P ( A | B ) probability of A given B prob. of A occurring given B has occurred Probability Topics P ( A ∪ B ) prob. of A or B prob. of A or B or both occurring Probability Topics P ( A ∩ B ) prob. of A and B prob. of both A and B occurring (same time) Probability Topics A ′ A-prime, complement of A complement of A, not A Probability Topics P ( A ') prob. of complement of A same Probability Topics G 1 green on first pick same Probability Topics P ( G 1 ) prob. of green on first pick same Discrete Random Variables PDF prob. density function same Discrete Random Variables X X the random variable X Discrete Random Variables X ~ the distribution of X same Discrete Random Variables ≥ greater than or equal to same Discrete Random Variables ≤ less than or equal to same Discrete Random Variables = equal to same Discrete Random Variables ≠ not equal to same Continuous Random Variables f ( x ) f of x function of x Continuous Random Variables pdf prob. density function same Continuous Random Variables U uniform distribution same Continuous Random Variables Exp exponential distribution same Continuous Random Variables f ( x ) = f of x equals same Continuous Random Variables m m decay rate (for exp. dist.) The Normal Distribution N normal distribution same The Normal Distribution z z -score same The Normal Distribution Z standard normal dist. same The Central Limit Theorem X – X -bar the random variable X -bar The Central Limit Theorem μ x – mean of X -bars the average of X -bars The Central Limit Theorem σ x – standard deviation of X -bars same Confidence Intervals CL confidence level same Confidence Intervals CI confidence interval same Confidence Intervals EBM error bound for a mean same Confidence Intervals EBP error bound for a proportion same Confidence Intervals t Student's t -distribution same Confidence Intervals df degrees of freedom same Confidence Intervals t α 2 student t with α /2 area in right tail same Confidence Intervals p ′ p -prime sample proportion of success Confidence Intervals q ′ q -prime sample proportion of failure Hypothesis Testing H 0 H -naught, H -sub 0 null hypothesis Hypothesis Testing H a H-a , H -sub a alternate hypothesis Hypothesis Testing H 1 H -1, H -sub 1 alternate hypothesis Hypothesis Testing α alpha probability of Type I error Hypothesis Testing β beta probability of Type II error Hypothesis Testing X 1 – – X 2 ¯ X 1-bar minus X 2-bar difference in sample means Hypothesis Testing μ 1 − μ 2 mu -1 minus mu -2 difference in population means Hypothesis Testing P ′ 1 − P ′ 2 P 1-prime minus P 2-prime difference in sample proportions Hypothesis Testing p 1 − p 2 p 1 minus p 2 difference in population proportions Chi-Square Distribution Χ 2 Ky -square Chi-square Chi-Square Distribution O Observed Observed frequency Chi-Square Distribution E Expected Expected frequency Linear Regression and Correlation y = a + bx y equals a plus b-x equation of a straight line Linear Regression and Correlation y ^ y -hat estimated value of y Linear Regression and Correlation r \"r\" same Linear Regression and Correlation ρ rho (\"row\") population correlation coefficient Linear Regression and Correlation ε error term for a regression line same Linear Regression and Correlation SSE Sum of Squared Errors same F -Distribution and ANOVA F F -ratio F -ratio Formulas Symbols you must know Population Sample N Size n μ Mean x _ σ 2 Variance s 2 σ Standard deviation s p Proportion p ′ Single data set formulae Population Sample μ = E ( x ) = 1 N ∑ i = 1 N ( x i ) Arithmetic mean x – = 1 n ∑ i = 1 n ( x i ) Geometric mean x ~ = ( ∏ i = 1 n X i ) 1 n Q 3 = 3 ( n + 1 ) 4 , Q 1 = ( n + 1 ) 4 Inter-quartile range I Q R = Q 3 − Q 1 Q 3 = 3 ( n + 1 ) 4 , Q 1 = ( n + 1 ) 4 σ 2 = 1 N ∑ i = 1 N ( x i − μ ) 2 Variance s 2 = 1 n ∑ i = 1 n ( x i − x _ ) 2 Single data set formulae Population Sample μ = E ( x ) = 1 N ∑ i = 1 N ( m i · f i ) Arithmetic mean x – = 1 n ∑ i = 1 n ( m i · f i ) Geometric mean x ~ = ( ∏ i = 1 n X i ) 1 n σ 2 = 1 N ∑ i = 1 N ( m i − μ ) 2 · f i Variance s 2 = 1 n ∑ i = 1 n ( m i − x _ ) 2 · f i C V = σ μ · 100 Coefficient of variation C V = s x _ · 100 Basic probability rules P ( A ∩ B ) = P ( A | B ) · P ( B ) Multiplication rule P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) Addition rule P ( A ∩ B ) = P ( A ) · P ( B ) or P ( A | B ) = P ( A ) Independence test Hypergeometric distribution formulae n C x = ( n x ) = n ! x ! ( n − x ) ! Combinatorial equation P ( x ) = ( A x ) ( N − A n − x ) ( N n ) Probability equation E ( X ) = μ = n p Mean σ 2 = ( N − n N − 1 ) n p ( q ) Variance Binomial distribution formulae P ( x ) = n ! x ! ( n − x ) ! p x ( q ) n − x Probability density function E ( X ) = μ = n p Arithmetic mean σ 2 = n p ( q ) Variance Geometric distribution formulae P ( X = x ) = ( 1 − p ) x − 1 ( p ) Probability when x is the first success. Probability when x is the number of failures before first success P ( X = x ) = ( 1 − p ) x ( p ) μ = 1 p Mean Mean μ = 1 − p p σ 2 = ( 1 − p ) p 2 Variance Variance σ 2 = ( 1 − p ) p 2 Poisson distribution formulae P ( x ) = e − μ μ x x ! Probability equation E ( X ) = μ Mean σ 2 = μ Variance Uniform distribution formulae f ( x ) = 1 b − a for a ≤ x ≤ b PDF E ( X ) = μ = a + b 2 Mean σ 2 = ( b − a ) 2 12 Variance Exponential distribution formulae P ( X ≤ x ) = 1 − e − m x Cumulative probability E ( X ) = μ = 1 m or m = 1 μ Mean and decay factor σ 2 = 1 m 2 = μ 2 Variance The following page of formulae requires the use of the \" Z \", \" t \", \" χ 2 \" or \" F \" tables. Z = x − μ σ Z-transformation for normal distribution Z = x − n p ′ n p ′ ( q ′ ) Normal approximation to the binomial Probability (ignores subscripts) Hypothesis testing Confidence intervals [bracketed symbols equal margin of error] (subscripts denote locations on respective distribution tables) Z c = x ¯ - μ 0 σ n Interval for the population mean when sigma is known x ¯ ± [ Z ( α / 2 ) σ n ] Z c = x ¯ - μ 0 s n Interval for the population mean when sigma is unknown but n > 30 x ¯ ± [ Z ( α / 2 ) s n ] t c = x ¯ - μ 0 s n Interval for the population mean when sigma is unknown but n < 30 x ¯ ± [ t ( n − 1 ) , ( α / 2 ) s n ] Z c = p ′ - p 0 p 0 q 0 n Interval for the population proportion p ′ ± [ Z ( α / 2 ) p ′ q ′ n ] t c = d – - δ 0 s d n Interval for difference between two means with matched pairs d – ± [ t ( n − 1 ) , ( α / 2 ) s d n ] where s d is the deviation of the differences Z c = ( x 1 – - x 2 – ) − δ 0 σ 1 2 n 1 + σ 2 2 n 2 Interval for difference between two means when sigmas are known ( x 1 – - x 2 – ) ± [ Z ( α / 2 ) σ 1 2 n 1 + σ 2 2 n 2 ] t c = ( x ¯ 1 - x ¯ 2 ) - δ 0 ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) Interval for difference between two means with equal variances when sigmas are unknown ( x ¯ 1 - x ¯ 2 ) ± [ t d f , ( α / 2 ) ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) ] where d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 − 1 ) ( ( s 1 ) 2 n 1 ) + ( 1 n 2 − 1 ) ( ( s 2 ) 2 n 2 ) Z c = ( p ′ 1 - p ′ 2 ) − δ 0 p ′ 1 ( q ′ 1 ) n 1 + p ′ 2 ( q ′ 2 ) n 2 Interval for difference between two population proportions ( p ′ 1 - p ′ 2 ) ± [ Z ( α / 2 ) p ′ 1 ( q ′ 1 ) n 1 + p ′ 2 ( q ′ 2 ) n 2 ] χ c 2 = ( n − 1 ) s 2 σ 0 2 Tests for GOF, Independence, and Homogeneity χ c 2 = Σ ( O − E ) 2 E where O = observed values and E = expected values F c = s 1 2 s 2 2 Where s 1 2 is the sample variance which is the larger of the two sample variances The next 3 formulae are for determining sample size with confidence intervals. (note: E represents the margin of error) n = Z ( a 2 ) 2 σ 2 E 2 Use when sigma is known E = x ¯ − μ n = Z ( a 2 ) 2 ( 0.25 ) E 2 Use when p ′ is unknown E = p ′ − p n = Z ( a 2 ) 2 [ p ′ ( q ′ ) ] E 2 Use when p ′ is unknown E = p ′ − p Simple linear regression formulae for y = a + b ( x ) r = Σ [ ( x − x ¯ ) ( y − y ¯ ) ] Σ ( x − x ¯ ) 2 * Σ ( y − y ¯ ) 2 = S x y S x S y = S S R S S T Correlation coefficient b = Σ [ ( x − x ¯ ) ( y − y ¯ ) ] Σ ( x − x ¯ ) 2 = S x y S S x = r y , x ( s y s x ) Coefficient b (slope) a = y ¯ − b ( x ¯ ) y-intercept s e 2 = Σ ( y i − y ^ i ) 2 n − k = Σ i = 1 n e i 2 n − k Estimate of the error variance S b = s e 2 ( x i − x ¯ ) 2 = s e 2 ( n − 1 ) s x 2 Standard error for coefficient b t c = b − β 0 s b Hypothesis test for coefficient β b ± [ t n − 2 , α / 2 S b ] Interval for coefficient β y ^ ± [ t α / 2 * s e ( 1 n + ( x p − x ¯ ) 2 s x ) ] Interval for expected value of y y ^ ± [ t α / 2 * s e ( 1 + 1 n + ( x p − x ¯ ) 2 s x ) ] Prediction interval for an individual y ANOVA formulae S S R = Σ i = 1 n ( y ^ i − y ¯ ) 2 Sum of squares regression S S E = Σ i = 1 n ( y ^ i − y ¯ i ) 2 Sum of squares error S S T = Σ i = 1 n ( y i − y ¯ ) 2 Sum of squares total R 2 = S S R S S T Coefficient of determination The following is the breakdown of a one-way ANOVA table for linear regression. Source of variation Sum of squares Degrees of freedom Mean squares F -ratio Regression S S R 1 or k − 1 M S R = S S R d f R F = M S R M S E Error S S E n − k M S E = S S E d f E Total S S T n − 1", "section": "Mathematical Phrases, Symbols, and Formulas", "book": "Introductory Business Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-business-statistics-2e"}
{"text": "Preface Welcome to Introductory Statistics 2e , an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost. The foundation of this textbook is Collaborative Statistics , by Barbara Illowsky and Susan Dean. Additional topics, examples, and innovations in terminology and practical applications have been added, all with a goal of increasing relevance and accessibility for students. About OpenStax OpenStax is part of Rice University, which is a 501(c)(3) nonprofit charitable corporation. Our mission is to make an amazing education accessible for all. Through our partnerships with philanthropic organizations and our alliance with other educational resource companies, we’re breaking down the most common barriers to learning. Because we believe that everyone should and can have access to knowledge. About OpenStax's resources Customization Introductory Statistics 2e is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY-NC-SA) license, which means that you can non-commercially distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors, and distribute all derivatives under the same license. Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book. Instructors also have the option of creating a customized version of their OpenStax book. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit your book page on OpenStax.org for more information. Art Attribution In Introductory Statistics 2e , most art contains attribution to its title, creator or rights holder, host platform, and license within the caption. For any art that is openly licensed, non-commercial users or organizations may reuse the art as long as they provide the same attribution to its original source. (Commercial entities should contact OpenStax to discuss reuse rights and permissions.) To maximize readability and content flow, some art does not include attribution in the text. If you reuse art from this text that does not have attribution provided, use the following attribution: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. Errata All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on OpenStax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on OpenStax.org. Format You can access this textbook for free in web view or PDF through OpenStax.org, and in low-cost print and iBooks editions. About Introductory Statistics 2e Introductory Statistics 2e follows scope and sequence requirements of a one-semester introduction to statistics course and is geared toward students majoring in fields other than math or engineering. The text assumes some knowledge of intermediate algebra and focuses on statistics application over theory. Introductory Statistics 2e includes innovative practical applications that make the text relevant and accessible, as well as collaborative exercises, technology integration problems, and statistics labs. Coverage and scope Chapter 1 Sampling and Data Chapter 2 Descriptive Statistics Chapter 3 Probability Topics Chapter 4 Discrete Random Variables Chapter 5 Continuous Random Variables Chapter 6 The Normal Distribution Chapter 7 The Central Limit Theorem Chapter 8 Confidence Intervals Chapter 9 Hypothesis Testing with One Sample Chapter 10 Hypothesis Testing with Two Samples Chapter 11 The Chi-Square Distribution Chapter 12 Linear Regression and Correlation Chapter 13 F Distribution and One-Way ANOVA Alternate sequencing Introductory Statistics 2e was conceived and written to fit a particular topical sequence, but it can be used flexibly to accommodate other course structures. One such potential structure, which fits reasonably well with the textbook content, is provided below. Please consider, however, that the chapters were not written to be completely independent, and that the proposed alternate sequence should be carefully considered for student preparation and textual consistency. Chapter 1 Sampling and Data Chapter 2 Descriptive Statistics Chapter 12 Linear Regression and Correlation Chapter 3 Probability Topics Chapter 4 Discrete Random Variables Chapter 5 Continuous Random Variables Chapter 6 The Normal Distribution Chapter 7 The Central Limit Theorem Chapter 8 Confidence Intervals Chapter 9 Hypothesis Testing with One Sample Chapter 10 Hypothesis Testing with Two Samples Chapter 11 The Chi-Square Distribution Chapter 13 F Distribution and One-Way ANOVA Pedagogical foundation and features Examples are placed strategically throughout the text to show students the step-by-step process of interpreting and solving statistical problems. To keep the text relevant for students, the examples are drawn from a broad spectrum of practical topics, including examples about college life and learning, health and medicine, retail and business, and sports and entertainment. Try It practice problems immediately follow many examples and give students the opportunity to practice as they read the text. They are usually based on practical and familiar topics, like the Examples themselves . Collaborative Exercises provide an in-class scenario for students to work together to explore presented concepts. Using the TI-83, 83+, 84, 84+ Calculator shows students step-by-step instructions to input problems into their calculator. The Technology Icon indicates where the use of a TI calculator or computer software is recommended. Practice, Homework, and Bringing It Together problems give the students problems at various degrees of difficulty while also including real-world scenarios to engage students. Changes to the Second Edition The revision of Introductory Statistics has been undertaken with the support of hundreds of faculty adopters and student users. Most of the edits are focused on currency, relevance, accuracy, and belonging. Dozens of examples and references have been changed in order to align with contemporary contexts and uses. Data and references have been extensively updated, as well. Accuracy checking and errata resolutions resulted in corrections and clarifications, both in the main text and in the answers/solutions. Finally, the second edition includes several hundred edits related to gender, race, ethnicity, age, academic status, and societal issues, designed to present the most informed and inclusive material. OpenStax only undertakes revisions when pedagogically necessary, and we work to ensure that the text maintains its organization in order to ease transition for instructors. Many of the above changes have been made within problems and examples. However, in nearly all cases, the problems retain their original numbering, so that instructors and online homework providers can move to the new edition with minimal impact on their assignment structures. A detailed transition guide is available on the instructor resource page for this textbook. Statistics labs These innovative activities were developed by Barbara Illowsky and Susan Dean in order to offer students the experience of designing, implementing, and interpreting statistical analyses. They are drawn from actual experiments and data-gathering processes and offer a unique hands-on and collaborative experience. The labs provide a foundation for further learning and classroom interaction that will produce a meaningful application of statistics. Statistics Labs appear at the end of each chapter and begin with student learning outcomes, general estimates for time on task, and any global implementation notes. Students are then provided with step-by-step guidance, including sample data tables and calculation prompts. The detailed assistance will help the students successfully apply the concepts in the text and lay the groundwork for future collaborative or individual work. Answers to Questions in the Book Answers to Examples are provided directly under the question. Answers to Try it Questions are provided at the Student Resources page. Answers to an assortment of Practice, Homework, and Bring it Together questions are provided at the end of each chapter for students. All answers to these questions are provided for instructors in the Instructor Answer Guide via the Instructor Resources page. Due to the high variability in responses, answers to Collaborative Exercises and Labs are not provided. Additional resources Student and instructor resources We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructor solution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which you can apply for when you log in or create your account on OpenStax.org. Take advantage of these resources to supplement your OpenStax book. Community Hubs OpenStax partners with the Institute for the Study of Knowledge Management in Education (ISKME) to offer Community Hubs on OER Commons – a platform for instructors to share community-created resources that support OpenStax books, free of charge. Through our Community Hubs, instructors can upload their own materials or download resources to use in their own courses, including additional ancillaries, teaching material, multimedia, and relevant course content. We encourage instructors to join the hubs for the subjects most relevant to your teaching and research as an opportunity both to enrich your courses and to engage with other faculty. To reach the Community Hubs, visit www.oercommons.org/hubs/OpenStax . Partner resources OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on OpenStax.org. About the authors Senior contributing authors Barbara Illowsky, De Anza College Susan Dean, De Anza College Contributing Authors Daniel Birmajer, Nazareth College Bryan Blount, Kentucky Wesleyan College Sheri Boyd, Rollins College Matthew Einsohn, Prescott College Nancy Foreman, West Los Angeles College James Helmreich, Marist College Lynette Kenyon, Collin County Community College Sheldon Lee, Viterbo University Jeff Taub, Maine Maritime Academy Reviewers Birgit Aquilonius, West Valley College Charles Ashbacher, Upper Iowa University, Cedar Rapids Abraham Biggs, Broward Community College Roberta Bloom, De Anza College Ernest Bonat, Portland Community College Sarah Boslaugh, Kennesaw State University David Bosworth, Hutchinson Community College George Bratton, University of Central Arkansas Jing Chang, College of Saint Mary Laurel Chiappetta, University of Pittsburgh Lenore Desilets, De Anza College Ann Flanigan, Kapiolani Community College David French, Tidewater Community College Mo Geraghty, De Anza College Larry Green, Lake Tahoe Community College Michael Greenwich, College of Southern Nevada Inna Grushko, De Anza College Valier Hauber, De Anza College Janice Hector, De Anza College Robert Henderson, Stephen F. Austin State University Mel Jacobsen, Snow College Mary Jo Kane, De Anza College Charles Klein, De Anza College Alexander Kolovos Sara Lenhart, Christopher Newport University Wendy Lightheart, Lane Community College Vladimir Logvenenko, De Anza College Jim Lucas, De Anza College Lisa Markus, De Anza College Miriam Masullo, SUNY Purchase Diane Mathios, De Anza College Robert McDevitt, Germanna Community College Mark Mills, Central College Cindy Moss, Skyline College Nydia Nelson, St. Petersburg College Benjamin Ngwudike, Jackson State University Jonathan Oaks, Macomb Community College Carol Olmstead, De Anza College Adam Pennell, Greensboro College Kathy Plum, De Anza College Lisa Rosenberg, Elon University Sudipta Roy, Kankakee Community College Javier Rueda, De Anza College Yvonne Sandoval, Pima Community College Rupinder Sekhon, De Anza College Travis Short, St. Petersburg College Frank Snow, De Anza College Abdulhamid Sukar, Cameron University Mary Teegarden, San Diego Mesa College John Thomas, College of Lake County Philip J. Verrecchia, York College of Pennsylvania Dennis Walsh, Middle Tennessee State University Cheryl Wartman, University of Prince Edward Island Carol Weideman, St. Petersburg College Andrew Wiesner, Pennsylvania State University", "section": "Preface", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction We encounter statistics in our daily lives more often than we probably realize and from many different sources, like the news. (credit: modification of work “New office” by Phil Whitehouse/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Recognize and differentiate between key terms. Apply various types of sampling methods to data collection. Create and interpret frequency tables. You are probably asking yourself the question, \"When and where will I use statistics?\" If you read any newspaper, watch television, or use the Internet, you will see statistical information. There are statistics about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or watch a television news program, you are given sample information. With this information, you may make a decision about the correctness of a statement, claim, or \"fact.\" Statistical methods can help you make the \"best educated guess.\" Since you will undoubtedly be given statistical information at some point in your life, you need to know some techniques for analyzing the information thoughtfully. Think about buying a house or managing a budget. Think about your chosen profession. The fields of economics, business, psychology, education, biology, law, computer science, police science, and early childhood development require at least one course in statistics. Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what \"good\" data can be distinguished from \"bad.\"", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Definitions of Statistics, Probability, and Key Terms The science of statistics deals with the collection, analysis, interpretation, and presentation of data . We see and use data in our everyday lives. In your classroom, try this exercise. Have class members write down the average time (in hours, to the nearest half-hour) they sleep per night. Your instructor will record the data. Then create a simple graph (called a dot plot ) of the data. A dot plot consists of a number line and dots (or points) positioned above the number line. For example, consider the following data: 5 5.5 6 6 6 6.5 6.5 6.5 6.5 7 7 8 8 9 The dot plot for this data would be as follows: Does your dot plot look the same as or different from the example? Why? If you did the same example in an English class with the same number of students, do you think the results would be the same? Why or why not? Where do your data appear to cluster? How might you interpret the clustering? The questions above ask you to analyze and interpret your data. With this example, you have begun your study of statistics. In this course, you will learn how to organize and summarize data. Organizing and summarizing data is called descriptive statistics . Two ways to summarize data are by graphing and by using numbers (for example, finding an average). After you have studied probability and probability distributions, you will use formal methods for drawing conclusions from \"good\" data. The formal methods are called inferential statistics . Statistical inference uses probability to determine how confident we can be that our conclusions are correct. Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful examination of the data. You will encounter what will seem to be too many mathematical formulas for interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to gain an understanding of your data. The calculations can be done using a calculator or a computer. The understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more confident in the decisions you make in life. Probability Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of an event occurring. For example, if you toss a fair coin four times, the outcomes may not be two heads and two tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails. The expected theoretical probability of heads in any one toss is 1 2 or 0.5. Even though the outcomes of a few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012 heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction 996 2000 is equal to 0.498 which is very close to 0.5, the expected probability. The theory of probability began with the study of games of chance such as poker. Predictions take the form of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of return on a client's investments. You might use probability to decide to buy a lottery ticket or not. In your study of statistics, you will use the power of mathematics through probability calculations to analyze and interpret your data. Key Terms In statistics, we generally want to study a population . You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample . The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population. Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16-ounce can contains 16 ounces of carbonated drink. From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter. A parameter is a numerical characteristic of the whole population that can be estimated by a statistic. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter. One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample . We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter. A variable , usually notated by capital letters such as X and Y , is a characteristic or measurement that can be determined for each member of a population. Variables may be numerical or categorical . Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category. If we let X equal the number of points earned by one math student at the end of a term, then X is a numerical variable. If we let Y be a person's party affiliation, then some examples of Y include Republican, Democrat, and Independent. Y is a categorical variable. We could do some math with values of X (calculate the average number of points earned, for example), but it makes no sense to do math with values of Y (calculating an average party affiliation makes no sense). Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value. Two words that come up often in statistics are mean and proportion . If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is 22 40 and the proportion of women students is 18 40 . Mean and proportion are discussed in more detail in later chapters. NOTE The words \" mean \" and \" average \" are often used interchangeably. The substitution of one word for the other is common practice. The technical term is \"arithmetic mean,\" and \"average\" is technically a center location. However, in practice among non-statisticians, \"average\" is commonly accepted for \"arithmetic mean.\" Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly surveyed 100 first year students at the college. Three of those students spent $150, $200, and $225, respectively. The population is all first year students attending ABC College this term. The sample could be all students enrolled in one section of a beginning statistics course at ABC College (although this sample may not represent the entire population). The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term. The statistic is the average (mean) amount of money spent (excluding books) by first year college students in the sample. The variable could be the amount of money spent (excluding books) by one first year student. Let X = the amount of money spent (excluding books) by one first year student attending ABC College. The data are the dollar amounts spent by the first year students. Examples of the data are $150, $200, and $225. Try It Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65, $75, and $95, respectively. Determine what the key terms refer to in the following study. A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below. 1. Population_____ 2. Statistic _____ 3. Parameter _____ 4. Sample _____ 5. Variable _____ 6. Data _____ all students who attended the college last year the cumulative GPA of one student who graduated from the college last year 3.65, 2.80, 1.50, 3.90 a group of students who graduated from the college last year, randomly selected the average cumulative GPA of students who graduated from the college last year all students who graduated from the college last year the average cumulative GPA of students in the study who graduated from the college last year 1. f 2. g 3. e 4. d 5. b 6. c Try It Determine what key terms refer to in the following study. A survey of athletes in a university was conducted to study the heights of athletes, in meters. Fill in the letter of the phrase that best describes each of the items below. Population ____ Statistics ____ Parameter ____ Sample ____ Variable ____ Data _____ the average height of athletes in the university the average height of athletes in the survey all athletes in the university all students in the university the height of one athlete a group of athletes randomly selected 1.82, 1.76, 1.69, 1.93 c b a f e g Determine what the key terms refer to in the following study. As part of a study designed to test the safety of electric automobiles, the National Transportation Safety Board collected and reviewed data about the effects of a crash on test dummies. Here is the criterion they used: Speed at which Cars Crashed Location of “drivers” (i.e. dummies) 35 miles/hour Front Seat Cars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars. The population is all cars containing dummies in the front seat. The sample is the 75 cars, selected by a simple random sample. The parameter is the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population. The statistic is proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample. The variable X = whether a dummy (if it had been a real person) who would have suffered head injuries. The data are either: yes, had head injury, or no, did not. Try It Determine what the key terms refer to in the following study. A survey is conducted to check the time taken by a mobile for charging of battery from 50% to 100%. The criteria used to collect the data are: Wattage of charger used Type of mobile used 30 W Android We want to know the proportion of Android mobiles that are charged to 100% within 30 minutes. We start with a simple random sample of 200 mobiles. The population is all Android mobiles. The sample is the 200 mobiles, selected by a simple random sample. The parameter is the proportion of Android mobiles that are charged to 100% within 30 minutes in the population. The statistic is the proportion of Android mobiles that are charged to 100% within 30 minutes in the survey. The variable X = the number of Android mobiles that are charged to 100% within 30 minutes. The data are either: yes, charged to 100% within 30 minutes; or no, charged to 100% taking more than 30 minutes. Determine what the key terms refer to in the following study. An insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit. The population is all medical doctors listed in the professional directory. The parameter is the proportion of medical doctors who have been involved in one or more malpractice suits in the population. The sample is the 500 doctors selected at random from the professional directory. The statistic is the proportion of medical doctors who have been involved in one or more malpractice suits in the sample. The variable X = whether an individual doctor has been involved in a malpractice suit. The data are either: yes, was involved in one or more malpractice lawsuits, or no, was not. Try It Determine what the key terms refer to in the following study. A study is to be conducted by a news agency to find the proportion of all truck drivers that have no points on their license. The agency selects 1000 truck drivers randomly from the directory of truck drivers and determines the number of truck drivers in the sample who have no points on their license. The population is all truck drivers present in the directory. The parameter is the proportion of truck drivers that have no points on their license in the population. The sample is the 1000 truck drivers selected randomly from the directory. The statistics is the proportion of truck drivers that have no points on their license in the sample. The variable X = the number of truck drivers that have no points on their license. The data are either: yes, they have no points on their license; or no, they have points on their license. Do the following exercise collaboratively with up to four people per group. Find a population, a sample, the parameter, the statistic, a variable, and data for the following study: You want to determine the average (mean) number of glasses of milk college students drink per day. Suppose yesterday, in your English class, you asked five students how many glasses of milk they drank the day before. The answers were 1, 0, 1, 3, and 4 glasses of milk. References The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/CrashTestDummies.html (accessed May 1, 2013). Chapter Review The mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text. Use the following information to answer the next five exercises . Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new drug is currently under study to address a respiratory virus. It is given to patients once the patient exhibits symptoms of the virus. Of interest is the average (mean) length of time in days from the time the patient starts the treatment until the symptoms are alleviated. Two researchers each follow a different set of 40 patients with the respiratory virus from the start of treatment until the symptoms are alleviated. The following data (in days) are collected. Researcher A: 3 4 11 15 16 17 22 44 37 16 14 24 25 15 26 27 33 29 35 44 13 21 22 10 12 8 40 32 26 27 31 34 29 17 8 24 18 47 33 34 Researcher B: 3 14 11 5 16 17 28 41 31 18 14 14 26 25 21 22 31 2 35 44 23 21 21 16 12 18 41 22 16 25 33 34 29 13 18 24 23 42 33 29 Determine what the key terms refer to in the example for Researcher A. population Patients with the virus sample parameter The average length of time (in months) patients with the virus live after treatment. statistic variable X = the length of time (in months) patients with the virus live after treatment HOMEWORK For each of the following eight exercises, identify: a. the population, b. the sample, c. the parameter, d. the statistic, e. the variable, and f. the data. Give examples where appropriate. A fitness center is interested in the mean amount of time a client exercises in the center each week. Ski resorts are interested in the mean age that children take their first ski and snowboard lessons. They need this information to plan their ski classes optimally. all children who take ski or snowboard lessons a group of these children the population mean age of children who take their first snowboard lesson the sample mean age of children who take their first snowboard lesson X = the age of one child who takes their first ski or snowboard lesson values for X , such as 3, 7, and so on A cardiologist is interested in the mean recovery period of their patients who have had heart attacks. Insurance companies are interested in the mean health costs each year of their clients, so that they can determine the costs of health insurance. the clients of the insurance companies a group of the clients the mean health costs of the clients the mean health costs of the sample X = the health costs of one client values for X , such as 34, 9, 82, and so on A politician is interested in the proportion of voters in their district who think the politician doing a good job. A marriage counselor is interested in the proportion of clients they counsel who stay married. all the clients of this counselor a group of clients of this marriage counselor the proportion of all her clients who stay married the proportion of the sample of the counselor’s clients who stay married X = whether a client stayed married yes, no Political pollsters may be interested in the proportion of people who will vote for a particular cause. A marketing company is interested in the proportion of people who will buy a particular product. all people (maybe in a certain geographic area, such as the United States) a group of the people the proportion of all people who will buy the product the proportion of the sample who will buy the product X = whether a person bought the product buy, not buy Use the following information to answer the next three exercises: A Lake Tahoe Community College instructor is interested in the mean number of days Lake Tahoe Community College math students are absent from class during a quarter. What is the population she is interested in? all Lake Tahoe Community College students all Lake Tahoe Community College English students all Lake Tahoe Community College students in her classes all Lake Tahoe Community College math students Consider the following: X = number of days a Lake Tahoe Community College math student is absent In this case, X is an example of a: variable. population. statistic. data. a The instructor’s sample produces a mean number of days absent of 3.5 days. This value is an example of a: parameter. data. statistic. variable. Average also called mean; a number that describes the central tendency of the data Categorical Variable variables that take on values that are names or labels Data a set of observations (a set of possible outcomes); most data can be put into two groups: qualitative (an attribute whose value is indicated by a label) or quantitative (an attribute whose value is indicated by a number). Quantitative data can be separated into two subgroups: discrete and continuous . Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage) Numerical Variable variables that take on values that are indicated by numbers Parameter a number that is used to represent a population characteristic and that generally cannot be determined easily Population all individuals, objects, or measurements whose properties are being studied Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur Proportion the number of successes divided by the total number in the sample Representative Sample a subset of the population that has the same characteristics as the population Sample a subset of the population studied Statistic a numerical characteristic of the sample; a statistic estimates the corresponding population parameter. Variable a characteristic of interest for each person or object in a population", "section": "Definitions of Statistics, Probability, and Key Terms", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Data, Sampling, and Variation in Data and Sampling Data may come from a population or from a sample. Lowercase letters like x or y generally are used to represent data values. Most data can be put into the following categories: Qualitative Quantitative Qualitative data are the result of categorizing or describing attributes of a population. Qualitative data are also often called categorical data . Hair color, blood type, ethnic group, the car a person drives, and the street a person lives on are examples of qualitative data. Qualitative data are generally described by words or letters. For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Researchers often prefer to use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or blood type. Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes of a population. Amount of money, pulse rate, weight, number of people living in your town, and number of students who take statistics are examples of quantitative data. Quantitative data may be either discrete or continuous . All data that are the result of counting are called quantitative discrete data . These data take on only certain numerical values. If you count the number of phone calls you receive for each day of the week, you might get values such as zero, one, two, or three. Data that are not only made up of counting numbers, but that may include fractions, decimals, or irrational numbers, are called quantitative continuous data . Continuous data are often the results of measurements like lengths, weights, or times. A list of the lengths in minutes for all the phone calls that you make in a week, with numbers like 2.4, 7.5, or 11.0, would be quantitative continuous data. Data Sample of Quantitative Discrete Data The data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data. Try It The data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this? Data Sample of Quantitative Continuous Data The data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data. Try It The data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this? You go to the supermarket and purchase three cans of soup (19 ounces tomato bisque, 14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts and peanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), and two desserts (16 ounces pistachio ice cream and 32 ounces chocolate chip cookies). Name data sets that are quantitative discrete, quantitative continuous, and qualitative. One Possible Solution: The three cans of soup, two packages of nuts, four kinds of vegetables and two desserts are quantitative discrete data because you count them. The weights of the soups (19 ounces, 14.1 ounces, 19 ounces) are quantitative continuous data because you measure weights as precisely as possible. Types of soups, nuts, vegetables and desserts are qualitative data because they are categorical. Try to identify additional data sets in this example. Try It The following list of materials was purchased by a purchase manager in a company: Two types of nails (2 kg box nails, 3 kg roofing nails) One type of oil (4 L machine oil) Four types of screws (3 kg wood screws, 5 kg machine screws, 1 kg set screws, 2 kg socket screws) Name data sets that are quantitative discrete, quantitative continuous, and qualitative. One possible solution: The two types of nails, one type of oil, and four types of screws are quantitative discrete data because you count them. The weights of materials are quantitative continuous data because you can measure weight as precisely as possible. Types of nails, oil, and screws are qualitative data because they are categorical. Try to identify additional data sets in the example. The data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative data. Try It The data are the colors of houses. You sample five houses. The colors of the houses are white, yellow, white, red, and white. What type of data is this? NOTE You may collect data as numbers and report it categorically. For example, the quiz scores for each student are recorded throughout the term. At the end of the term, the quiz scores are reported as A, B, C, D, or F. Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words \"the number of.\" the number of pairs of shoes you own the type of car you drive the distance it is from your home to the nearest grocery store the number of classes you take per school year. the type of calculator you use weights of dogs at an animal shelter number of correct answers on a quiz IQ scores (This may cause some discussion.) Items a, d, and g are quantitative discrete; items c, f, and h are quantitative continuous; items b and e are qualitative, or categorical. Try It Determine the correct data type (quantitative or qualitative) for the number of cars in a parking lot. Indicate whether quantitative data are continuous or discrete. A statistics professor collects information about the classification of her students as first-year students, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart . What type of data does this graph show? This pie chart shows the students in each year, which is qualitative (or categorical) data . Try It The registrar at State University keeps records of the number of credit hours students complete each semester. The data collected are summarized in the histogram. The class boundaries are 10 to less than 13, 13 to less than 16, 16 to less than 19, 19 to less than 22, and 22 to less than 25. What type of data does this graph show? Qualitative Data Discussion Below are tables comparing the number of part-time and full-time students at De Anza College and Foothill College enrolled for the most recent spring quarter. The tables display counts (frequencies) and percentages or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is compared to De Anza College. Most Recent Spring Quarter De Anza College Foothill College Number Percent Number Percent Full-time 9,200 40.9% Full-time 4,059 28.6% Part-time 13,296 59.1% Part-time 10,124 71.4% Total 22,496 100% Total 14,183 100% Tables are a good way of organizing and displaying data. But graphs can be even more helpful in understanding the data. There are no strict rules concerning which graphs to use. Two graphs that are used to display qualitative data are pie charts and bar graphs. In a pie chart , categories of data are represented by wedges in a circle and are proportional in size to the percent of individuals in each category. In a bar graph , the length of the bar for each category is proportional to the number or percent of individuals in each category. Bars may be vertical or horizontal. A Pareto chart consists of bars that are sorted into order by category size (largest to smallest). Look at and and determine which graph (pie or bar) you think displays the comparisons better. It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the data. We might make different choices of what we think is the “best” graph depending on the data and the context. Our choice also depends on what we are using the data for. Percentages That Add to More (or Less) Than 100% Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages add to more than 100% because students can be in more than one category. A bar graph is appropriate to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the percentages added to less than 100%. De Anza College Most Recent Spring Quarter Characteristic/Category Percent Full-Time Students 40.9% Students who intend to transfer to a 4-year educational institution 48.6% Students under age 25 61.0% TOTAL 150.5% Omitting Categories/Missing Data The table displays Ethnicity of Students but is missing the \"Other/Unknown\" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart. Ethnicity of Students at De Anza College in the Most Recent Fall Term Frequency Percent Asian 8,794 36.1% Black 1,412 5.8% Filipino 1,298 5.3% Hispanic/Latino 4,180 17.1% Native American 146 0.6% Pacific Islander 236 1.0% White 5,978 24.5% TOTAL 22,044 out of 24,382 90.4% out of 100% The following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us. This particular bar graph in can be difficult to understand visually. The graph in is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret. Bar Graph with Other/Unknown Category Pareto Chart With Bars Sorted by Size Pie Charts: No Missing Data The following pie charts have the “Other/Unknown” category included (since the percentages must add to 100%). The chart in (b) is organized by the size of each wedge, which makes it a more visually informative graph than the unsorted, alphabetical graph in (a). Sampling Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling . In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons. The easiest method to describe is called a simple random sample . Any group of n individuals is equally likely to be chosen as any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus class, which has 31 members not including Lisa. To choose a simple random sample of size three from the other members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out three names. A more technological way is for Lisa to first list the last names of the members of her class together with a two-digit number, as in : Class Roster ID Name ID Name ID Name 00 Anselmo 11 King 21 Roquero 01 Bautista 12 Legeny 22 Roth 02 Bayani 13 Lundquist 23 Rowell 03 Cheng 14 Macierz 24 Salangsang 04 Cuarismo 15 Motogawa 25 Slade 05 Cuningham 16 Okimoto 26 Stratcher 06 Fontecha 17 Patel 27 Tallai 07 Hong 18 Price 28 Tran 08 Hoobler 19 Quizon 29 Wai 09 Jiao 20 Reyes 30 Wood 10 Khan Lisa can use a table of random numbers (found in many statistics books and mathematical handbooks), a calculator, or a computer to generate random numbers. For this example, suppose Lisa chooses to generate random numbers from a calculator. The numbers generated are as follows: 0.94360 0.99832 0.14669 0.51470 0.40581 0.73381 0.04399 Lisa reads two-digit groups until she has chosen three class members (that is, she reads 0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could have generated more random numbers. The random numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However the third random number, 0.14669, contains 14 (the fourth random number also contains 14), the fifth random number contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist of Marcierz, Cuningham, and Cuarismo. To generate random numbers: Press MATH . Arrow over to PRB . Press 5:randInt(. Enter 0, 30) . Press ENTER for the first random number. Press ENTER two more times for the other 2 random numbers. If there is a repeat press ENTER again. Note: randInt(0, 30, 3) will generate 3 random numbers. Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster sample, and the systematic sample. To choose a stratified sample , divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample. To choose a cluster sample , divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample. To choose a systematic sample , randomly select a starting point and take every n th piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method. A type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others. Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents. Sampling with replacement is truly random sampling. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling without replacement is done. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low. Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors . A defective counting device can cause a nonsampling error. In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error. In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied. Critical Evaluation We need to evaluate the statistical studies we read about critically and analyze them before accepting the results of the studies. Common problems to be aware of include Problems with samples: A sample must be representative of the population. A sample that is not representative of the population is biased. Biased samples that are not representative of the population give results that are inaccurate and not valid. Self-selected samples: Responses only by people who choose to respond, such as call-in surveys, are often unreliable. Sample size issues: Samples that are too small may be unreliable. Larger samples are better, if possible. In some situations, having small samples is unavoidable and can still be used to draw conclusions. Examples: crash testing cars or medical testing for rare conditions Undue influence: collecting data or asking questions in a way that influences the response Non-response or refusal of subject to participate: The collected responses may no longer be representative of the population. Often, people with strong positive or negative opinions may answer surveys, which can affect the results. Causality: A relationship between two variables does not mean that one causes the other to occur. They may be related (correlated) because of their relationship through a different variable. Self-funded or self-interest studies: A study performed by a person or organization in order to support their claim. Is the study impartial? Read the study carefully to evaluate the work. Do not automatically assume that the study is good, but do not automatically assume the study is bad either. Evaluate it on its merits and the work done. Misleading use of data: improperly displayed graphs, incomplete data, or lack of context Confounding: When the effects of multiple factors on a response cannot be separated. Confounding makes it difficult or impossible to draw valid conclusions about the effect of each factor. As a class, determine whether or not the following samples are representative. If they are not, discuss the reasons. To find the average GPA of all students in a university, use all honor students at the university as the sample. To find out the most popular cereal among young people under the age of ten, stand outside a large supermarket for three hours and speak to every twentieth child under age ten who enters the supermarket. To find the average annual income of all adults in the United States, sample U.S. Representatives. Create a cluster sample by considering each state as a stratum (group). By using simple random sampling, select states to be part of the cluster. Then survey every U.S. Representative in the cluster. To determine the proportion of people taking public transportation to work, survey 20 people in New York City. Conduct the survey by sitting in Central Park on a bench and interviewing every person who sits next to you. To determine the average cost of a two-day stay in a hospital in Massachusetts, survey 100 hospitals across the state using simple random sampling. A study is done to determine the average tuition that San Jose State undergraduate students pay per semester. Each student in the following samples is asked how much tuition they paid for the Fall semester. What is the type of sampling in each case? A sample of 100 undergraduate San Jose State students is taken by organizing the students’ names by classification (first-year, sophomore, junior, or senior), and then selecting 25 students from each. A random number generator is used to select a student from the alphabetical listing of all undergraduate students in the Fall semester. Starting with that student, every 50th student is chosen until 75 students are included in the sample. A completely random method is used to select 75 students. Each undergraduate student in the fall semester has the same probability of being chosen at any stage of the sampling process. The first-year, sophomore, junior, and senior years are numbered one, two, three, and four, respectively. A random number generator is used to pick two of those years. All students in those two years are in the sample. An administrative assistant is asked to stand in front of the library one Wednesday and to ask the first 100 undergraduate students he encounters what they paid for tuition the Fall semester. Those 100 students are the sample. a. stratified; b. systematic; c. simple random; d. cluster; e. convenience Try It You are going to use the random number generator to generate different types of samples from the data. This table displays six sets of quiz scores (each quiz counts 10 points) for an elementary statistics class. #1 #2 #3 #4 #5 #6 5 7 10 9 8 3 10 5 9 8 7 6 9 10 8 6 7 9 9 10 10 9 8 9 7 8 9 5 7 4 9 9 9 10 8 7 7 7 10 9 8 8 8 8 9 10 8 8 9 7 8 7 7 8 8 8 10 9 8 7 Instructions: Use the Random Number Generator to pick samples. Create a stratified sample by column. Pick three quiz scores randomly from each column. Number each row one through ten. On your calculator, press Math and arrow over to PRB. For column 1, Press 5: randInt( and enter 1,10) . Press ENTER. Record the number. Press ENTER 2 more times (even the repeats). Record these numbers. Record the three quiz scores in column one that correspond to these three numbers. Repeat for columns two through six. These 18 quiz scores are a stratified sample. Create a cluster sample by picking two of the columns. Use the column numbers: one through six. Press MATH and arrow over to PRB. Press 5: randInt( and enter 1,6) . Press ENTER. Record the number. Press ENTER and record that number. The two numbers are for two of the columns. The quiz scores (20 of them) in these 2 columns are the cluster sample. Create a simple random sample of 15 quiz scores. Use the numbering one through 60. Press MATH . Arrow over to PRB. Press 5: randInt( and enter 1, 60) . Press ENTER 15 times and record the numbers. Record the quiz scores that correspond to these numbers. These 15 quiz scores are the random sample. Create a systematic sample of 12 quiz scores. Use the numbering one through 60. Press MATH . Arrow over to PRB. Press 5: randInt( and enter 1, 60) . Press ENTER. Record the number and the first quiz score. From that number, count ten quiz scores and record that quiz score. Keep counting ten quiz scores and recording the quiz score until you have a sample of 12 quiz scores. You may wrap around (go back to the beginning). Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A soccer coach selects six players from a group of boys aged eight to ten, seven players from a group of boys aged 11 to 12, and three players from a group of boys aged 13 to 14 to form a recreational soccer team. A pollster interviews all human resource personnel in five different high tech companies. A high school educational researcher interviews 50 public high school teachers and 50 private high school teachers. A medical researcher interviews every third cancer patient from a list of cancer patients at a local hospital. A high school counselor uses a computer to generate 50 random numbers and then picks students whose names correspond to the numbers. A student interviews classmates in their algebra class to determine how many pairs of jeans a student owns, on the average. a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f.convenience Try It Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A high school principal polls 50 first-year students, 50 sophomores, 50 juniors, and 50 seniors regarding policy changes for after school activities. If we were to examine two samples representing the same population, even if we used random sampling methods for the samples, they would not be exactly the same. Just as there is variation in data, there is variation in samples. As you become accustomed to sampling, the variability will begin to seem natural. NOTE In Confidence Intervals of the text, sample size formulas are provided which will determine sample sizes when sampling from a population. The sample size will be a function of the desired precision and not a function of the population size. It may be somewhat counterintuitive that the sample size does not depend on the population size. However, this implies that a sample size of 1,000 can be adequate to represent a population of 100,000 versus 1,000,000 given that the same level of precision is desired. When working in Confidence Intervals with sample size formulas, the student will notice that population size is not a factor in determining the sample size. Suppose ABC College has 10,000 part-time students (the population). We are interested in the average amount of money a part-time student spends on books in the fall term. Asking all 10,000 students is an almost impossible task. Suppose we take two different samples. First, we use convenience sampling and survey ten students from a first term organic chemistry class. Many of these students are taking first term calculus in addition to the organic chemistry class. The amount of money they spend on books is as follows: $128 $87 $173 $116 $130 $204 $147 $189 $93 $153 The second sample is taken using a list of senior citizens who take P.E. classes and taking every fifth senior citizen on the list, for a total of ten senior citizens. They spend: $50 $40 $36 $15 $50 $100 $40 $53 $22 $22 It is unlikely that any student is in both samples. a. Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population? a. No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are also taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average part-time student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample. b. Since these samples are not representative of the entire population, is it wise to use the results to describe the entire population? b. No. For these samples, each member of the population did not have an equally likely chance of being chosen. Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education, art, and early childhood development. (We assume that these are the only disciplines in which part-time students at ABC College are enrolled and that an equal number of part-time students are enrolled in each of the disciplines.) Each student is chosen using simple random sampling. Using a calculator, random numbers are generated and a student from a particular discipline is selected if they have a corresponding number. The students spend the following amounts: $180 $50 $150 $85 $260 $75 $180 $200 $200 $150 c. Is the sample biased? c. The sample is unbiased, but a larger sample would be recommended to increase the likelihood that the sample will be close to representative of the population. However, for a biased sampling technique, even a large sample runs the risk of not being representative of the population. Students often ask if it is \"good enough\" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes. Try It A local radio station has a fan base of 20,000 listeners. The station wants to know if its audience would prefer more music or more talk shows. Asking all 20,000 listeners is an almost impossible task. The station uses convenience sampling and surveys the first 200 people they meet at one of the station’s music concert events. 24 people said they’d prefer more talk shows, and 176 people said they’d prefer more music. Do you think that this sample is representative of (or is characteristic of) the entire 20,000 listener population? Variation in Data Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less than 16 ounces of liquid. In one study, eight 16-ounce cans were measured and produced the following amount (in ounces) of beverage: 15.8 16.1 15.2 14.8 15.8 15.9 16.0 15.5 Measurements of the amount of beverage in a 16-ounce can may vary because different people make the measurements or because the exact amount, 16 ounces of liquid, was not put into the cans. Manufacturers regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range. Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the same purpose. This is completely natural. However, if two or more of you are taking the same data and get very different results, it is time for you and the others to reevaluate your data-taking methods and your accuracy. Variation in Samples It was mentioned previously that two or more samples from the same population , taken randomly, and having close to the same characteristics of the population will likely be different from each other. Suppose Doreen and Jung both decide to study the average amount of time students at their college sleep each night. Doreen and Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample will be different from Jung's sample. Even if Doreen and Jung used the same sampling method, in all likelihood their samples would be different. Neither would be wrong, however. Think about what contributes to making Doreen’s and Jung’s samples different. If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results (the average amount of time a student sleeps) might be closer to the actual population average. But still, their samples would be, in all likelihood, different from each other. This variability in samples cannot be stressed enough. Size of a Sample The size of a sample (often called the number of observations) is important. The examples you have seen in this book so far have been small. Samples of only a few hundred observations, or even smaller, are sufficient for many purposes. In polling, samples that are from 1,200 to 1,500 observations are considered large enough and good enough if the survey is random and is well done. You will learn why when you study confidence intervals. Be aware that many large samples are biased. For example, call-in surveys are invariably biased, because people choose to respond or not. Divide into groups of two, three, or four. Your instructor will give each group one six-sided die. Try this experiment twice. Roll one fair die (six-sided) 20 times. Record the number of ones, twos, threes, fours, fives, and sixes you get in and (“frequency” is the number of times a particular face of the die occurs): First Experiment (20 rolls) Face on Die Frequency 1 2 3 4 5 6 Second Experiment (20 rolls) Face on Die Frequency 1 2 3 4 5 6 Did the two experiments have the same results? Probably not. If you did the experiment a third time, do you expect the results to be identical to the first or second experiment? Why or why not? Which experiment had the correct results? They both did. The job of the statistician is to see through the variability and draw appropriate conclusions. References Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/default.asp (accessed May 1, 2013). Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/methodology.asp (accessed May 1, 2013). Gallup-Healthways Well-Being Index. http://www.gallup.com/poll/146822/gallup-healthways-index-questions.aspx (accessed May 1, 2013). Data from http://www.bookofodds.com/Relationships-Society/Articles/A0374-How-George-Gallup-Picked-the-President Dominic Lusinchi, “’President’ Landon and the 1936 Literary Digest Poll: Were Automobile and Telephone Owners to Blame?” Social Science History 36, no. 1: 23-54 (2012), http://ssh.dukejournals.org/content/36/1/23.abstract (accessed May 1, 2013). “The Literary Digest Poll,” Virtual Laboratories in Probability and Statistics http://www.math.uah.edu/stat/data/LiteraryDigest.html (accessed May 1, 2013). “Gallup Presidential Election Trial-Heat Trends, 1936–2008,” Gallup Politics http://www.gallup.com/poll/110548/gallup-presidential-election-trialheat-trends-19362004.aspx#4 (accessed May 1, 2013). The Data and Story Library, http://lib.stat.cmu.edu/DASL/Datafiles/USCrime.html (accessed May 1, 2013). LBCC Distance Learning (DL) program data in 2010-2011, http://de.lbcc.edu/reports/2010-11/future/highlights.html#focus (accessed May 1, 2013). Data from San Jose Mercury News Chapter Review Data are individual items of information that come from a population or sample. Data may be classified as qualitative (categorical), quantitative continuous, or quantitative discrete. Because it is not practical to measure the entire population in a study, researchers use samples to represent the population. A random sample is a representative group from the population chosen by using a method that gives each individual in the population an equal chance of being included in the sample. Random sampling methods include simple random sampling, stratified sampling, cluster sampling, and systematic sampling. Convenience sampling is a nonrandom method of choosing a sample that often produces biased data. Samples that contain different individuals result in different data. This is true even when the samples are well-chosen and representative of the population. When properly selected, larger samples model the population more closely than smaller samples. There are many different potential problems that can affect the reliability of a sample. Statistical data needs to be critically analyzed, not simply accepted. Practice “Number of times per week” is what type of data? a. qualitative (categorical) b. quantitative discrete c. quantitative continuous Use the following information to answer the next four exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Antonio, Texas. The first house in the neighborhood around the park was selected randomly, and then the resident of every eighth house in the neighborhood around the park was interviewed. The sampling method was a. simple random b. systematic c. stratified d. cluster b “Duration (amount of time)” is what type of data? a. qualitative (categorical) b. quantitative discrete c. quantitative continuous The colors of the houses around the park are what kind of data? a. qualitative (categorical) b. quantitative discrete c. quantitative continuous a The population is ______________________ contains the total number of deaths worldwide as a result of earthquakes over a 13-year period. Year Total Number of Deaths 1 231 2 21,357 3 11,685 4 33,819 5 228,802 6 88,003 7 6,605 8 712 9 88,011 10 1,790 11 320,120 12 21,953 13 768 Total 823,856 Use to answer the following questions. What is the proportion of deaths between Year 8 and Year 13? What percent of deaths occurred before Year 2? What is the percent of deaths that occurred in Year 4 or after Year 11? What is the fraction of deaths that happened before Year 13? What kind of data is the number of deaths? Earthquakes are quantified according to the amount of energy they produce (examples are 2.1, 5.0, 6.7). What type of data is that? What contributed to the large number of deaths in Year 11? In Year 5? Explain. 0.5242 0.03% 6.86% 823,088 823,856 quantitative discrete quantitative continuous In both years, underwater earthquakes produced massive tsunamis. For the following four exercises, determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A group of test subjects is divided into twelve groups; then four of the groups are chosen at random. A market researcher polls every tenth person who walks into a store. systematic The first 50 people who walk into a sporting event are polled on their television preferences. A computer generates 100 random numbers, and 100 people whose names correspond with the numbers on the list are chosen. simple random Use the following information to answer the next seven exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new drug is currently under study to address a respiratory virus. It is given to patients once the patient exhibits symptoms of the virus. Of interest is the average (mean) length of time in days from the time the patient starts the treatment until the symptoms are alleviated. Two researchers each follow a different set of 40 patients with the respiratory virus from the start of treatment until the symptoms are alleviated. The following data (in days) are collected. Researcher A: 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34 Researcher B: 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29 Complete the tables using the data provided: Researcher A Survival Length (in days) Frequency Relative Frequency Cumulative Relative Frequency 0.5–6.5 6.5–12.5 12.5–18.5 18.5–24.5 24.5–30.5 30.5–36.5 36.5–42.5 42.5–48.5 Researcher B Survival Length (in days) Frequency Relative Frequency Cumulative Relative Frequency 0.5–6.5 6.5–12.5 12.5–18.5 18.5–24.5 24.5–30.5 30.5–36.5 36.5-45.5 Determine what the key term data refers to in the above example for Researcher A. values for X , such as 3, 4, 11, and so on List two reasons why the data may differ. Can you tell if one researcher is correct and the other one is incorrect? Why? No, we do not have enough information to make such a claim. Would you expect the data to be identical? Why or why not? Suggest at least two methods the researchers might use to gather random data. Take a simple random sample from each group. One way is by assigning a number to each patient and using a random number generator to randomly select patients. Suppose that the first researcher conducted his survey by randomly choosing one state in the nation and then randomly picking 40 patients from that state. What sampling method would that researcher have used? Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What sampling method would that researcher have used? What concerns would you have about this data set, based upon the data collection method? This would be convenience sampling and is not random. Use the following data to answer the next five exercises: Two researchers are gathering data on hours of video games played by school-aged children and young adults. They each randomly sample different groups of 150 students from the same school. They collect the following data. Researcher A Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency 0–2 26 0.17 0.17 2–4 30 0.20 0.37 4–6 49 0.33 0.70 6–8 25 0.17 0.87 8–10 12 0.08 0.95 10–12 8 0.05 1 Researcher B Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency 0–2 48 0.32 0.32 2–4 51 0.34 0.66 4–6 24 0.16 0.82 6–8 12 0.08 0.90 8–10 11 0.07 0.97 10–12 4 0.03 1 Give a reason why the data may differ. Would the sample size be large enough if the population is the students in the school? Yes, the sample size of 150 would be large enough to reflect a population of one school. Would the sample size be large enough if the population is school-aged children and young adults in the United States? Researcher A concludes that most students play video games between four and six hours each week. Researcher B concludes that most students play video games between two and four hours each week. Who is correct? Even though the specific data support each researcher’s conclusions, the different results suggest that more data need to be collected before the researchers can reach a conclusion. As part of a way to reward students for participating in the survey, the researchers gave each student a gift card to a video game store. Would this affect the data if students knew about the award before the study? Use the following data to answer the next five exercises: A pair of studies was performed to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The studies observed 200 stroke patients recovering over a period of several weeks. The first study collected the data in . The second study collected the data in . Group Showed improvement No improvement Deterioration Used program 142 43 15 Did not use program 72 110 18 Group Showed improvement No improvement Deterioration Used program 105 74 19 Did not use program 89 99 4 Given what you know, which study is correct? There is not enough information given to judge if either one is correct or incorrect. The first study was performed by the company that designed the software program. The second study was performed by the American Medical Association. Which study is more reliable? Both groups that performed the study concluded that the software works. Is this accurate? The software program seems to work because the second study shows that more patients improve while using the software than not. Even though the difference is not as large as that in the first study, the results from the second study are likely more reliable and still show improvement. The company takes the two studies as proof that their software causes mental improvement in stroke patients. Is this a fair statement? Patients who used the software were also a part of an exercise program whereas patients who did not use the software were not. Does this change the validity of the conclusions from ? Yes, because we cannot tell if the improvement was due to the software or the exercise; the data is confounded, and a reliable conclusion cannot be drawn. New studies should be performed. Is a sample size of 1,000 a reliable measure for a population of 5,000? Is a sample of 500 volunteers a reliable measure for a population of 2,500? No, even though the sample is large enough, the fact that the sample consists of volunteers makes it a self-selected sample, which is not reliable. A question on a survey reads: \"Do you prefer the delicious taste of Brand X or the taste of Brand Y?\" Is this a fair question? Is a sample size of two representative of a population of five? No, even though the sample is a large portion of the population, two responses are not enough to justify any conclusions. Because the population is so small, it would be better to include everyone in the population to get the most accurate data. Is it possible for two experiments to be well run with similar sample sizes to get different data? HOMEWORK For the following exercises, identify the type of data that would be used to describe a response (quantitative discrete, quantitative continuous, or qualitative), and give an example of the data. number of tickets sold to a concert quantitative discrete, 150 percent of body fat favorite baseball team qualitative, Oakland A’s time in line to buy groceries number of students enrolled at Evergreen Valley College quantitative discrete, 11,234 students most-watched television show brand of toothpaste qualitative, Crest distance to the closest movie theatre age of executives in Fortune 500 companies quantitative continuous, 47.3 years number of competing computer spreadsheet software packages Use the following information to answer the next two exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of resident use of a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every 8th house in the neighborhood around the park was interviewed. “Number of times per week” is what type of data? qualitative quantitative discrete quantitative continuous b “Duration (amount of time)” is what type of data? qualitative quantitative discrete quantitative continuous Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. Suppose an airline conducts a survey. Over Thanksgiving weekend, it surveys six flights from Boston to Salt Lake City to determine the number of babies on the flights. It determines the amount of safety equipment needed by the result of that study. Using complete sentences, list three things wrong with the way the survey was conducted. Using complete sentences, list three ways that you would improve the survey if it were to be repeated. The survey was conducted using six similar flights. The survey would not be a true representation of the entire population of air travelers. Conducting the survey on a holiday weekend will not produce representative results. Conduct the survey during different times of the year. Conduct the survey using flights to and from various locations. Conduct the survey on different days of the week. Suppose you want to determine the mean number of students per statistics class in your state. Describe a possible sampling method in three to five complete sentences. Make the description detailed. Suppose you want to determine the mean number of cans of soda drunk each month by students in their twenties at your school. Describe a possible sampling method in three to five complete sentences. Make the description detailed. Answers will vary. Sample Answer: You could use a systematic sampling method. Stop the tenth person as they leave one of the buildings on campus at 9:50 in the morning. Then stop the tenth person as they leave a different building on campus at 1:50 in the afternoon. List some practical difficulties involved in getting accurate results from a telephone survey. List some practical difficulties involved in getting accurate results from a mailed survey. Answers will vary. Sample Answer: Many people will not respond to mail surveys. If they do respond to the surveys, you can’t be sure who is responding. In addition, mailing lists can be incomplete. With your classmates, brainstorm some ways you could overcome these problems if you needed to conduct a phone or mail survey. An instructor takes a sample by gathering data on five randomly selected students from each Lake Tahoe Community College math class. The type of sampling used is cluster sampling stratified sampling simple random sampling convenience sampling b A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Jose. The first house in the neighborhood around the park was selected randomly and then every eighth house in the neighborhood around the park was interviewed. The sampling method was: simple random systematic stratified cluster Name the sampling method used in each of the following situations: A person in the airport is handing out questionnaires to travelers asking them to evaluate the airport’s service. The person does not ask travelers who are hurrying through the airport with their hands full of luggage but instead asks all travelers who are sitting near gates and not taking naps while they wait. A teacher wants to know if her students are doing homework, so they randomly select rows two and five and then call on all students in row two and all students in row five to present the solutions to homework problems to the class. The marketing manager for an electronics chain store wants information about the ages of its customers. Over the next two weeks, at each store location, 100 randomly selected customers are given questionnaires to fill out asking for information about age, as well as about other variables of interest. The librarian at a public library wants to determine what proportion of the library users are children. The librarian has a tally sheet on which they mark whether books are checked out by an adult or a child. The librarian records this data for every fourth patron who checks out books. A political party wants to know the reaction of voters to a debate between the candidates. The day after the debate, the party’s polling staff calls 1,200 randomly selected phone numbers. If a registered voter answers the phone or is available to come to the phone, that registered voter is asked whom they intend to vote for and whether the debate changed their opinion of the candidates. convenience cluster stratified systematic simple random A “random survey” was conducted of 3,274 people of the “microprocessor generation” (people born since 1971, the year the microprocessor was invented). It was reported that 48% of those individuals surveyed stated that if they had $2,000 to spend, they would use it for computer equipment. Also, 66% of those surveyed considered themselves relatively savvy computer users. Do you consider the sample size large enough for a study of this type? Why or why not? Based on your “gut feeling,” do you believe the percents accurately reflect the U.S. population for those individuals born since 1971? If not, do you think the percents of the population are actually higher or lower than the sample statistics? Why? Additional information: The survey, reported by Intel Corporation, was filled out by individuals who visited the Los Angeles Convention Center to see the Smithsonian Institute's road show called “America’s Smithsonian.” With this additional information, do you feel that all demographic and ethnic groups were equally represented at the event? Why or why not? With the additional information, comment on how accurately you think the sample statistics reflect the population parameters. The Well-Being Index is a survey that follows trends of U.S. residents on a regular basis. There are six areas of health and wellness covered in the survey: Life Evaluation, Emotional Health, Physical Health, Healthy Behavior, Work Environment, and Basic Access. Some of the questions used to measure the Index are listed below. Identify the type of data obtained from each question used in this survey: qualitative, quantitative discrete, or quantitative continuous. Do you have any health problems that prevent you from doing any of the things people your age can normally do? During the past 30 days, for about how many days did poor health keep you from doing your usual activities? In the last seven days, on how many days did you exercise for 30 minutes or more? Do you have health insurance coverage? qualitative quantitative discrete quantitative discrete qualitative In advance of the 1936 Presidential Election, a magazine titled Literary Digest released the results of an opinion poll predicting that the republican candidate Alf Landon would win by a large margin. The magazine sent post cards to approximately 10,000,000 prospective voters. These prospective voters were selected from the subscription list of the magazine, from automobile registration lists, from phone lists, and from club membership lists. Approximately 2,300,000 people returned the postcards. Think about the state of the United States in 1936. Explain why a sample chosen from magazine subscription lists, automobile registration lists, phone books, and club membership lists was not representative of the population of the United States at that time. What effect does the low response rate have on the reliability of the sample? Are these problems examples of sampling error or nonsampling error? During the same year, George Gallup conducted his own poll of 30,000 prospective voters. These researchers used a method they called \"quota sampling\" to obtain survey answers from specific subsets of the population. Quota sampling is an example of which sampling method described in this module? Crime-related and demographic statistics for 47 US states in 1960 were collected from government agencies, including the FBI's Uniform Crime Report . One analysis of this data found a strong connection between education and crime indicating that higher levels of education in a community correspond to higher crime rates. Which of the potential problems with samples discussed in could explain this connection? Causality: The fact that two variables are related does not guarantee that one variable is influencing the other. We cannot assume that crime rate impacts education level or that education level impacts crime rate. Confounding: There are many factors that define a community other than education level and crime rate. Communities with high crime rates and high education levels may have other lurking variables that distinguish them from communities with lower crime rates and lower education levels. Because we cannot isolate these variables of interest, we cannot draw valid conclusions about the connection between education and crime. Possible lurking variables include police expenditures, unemployment levels, region, average age, and size. Imagine you work for a polling company and a member of your team has proposed the following survey question: “Do you feel happy paying your taxes while some politicians are allowed to use loopholes and avoid paying their fair share of taxes?” As part of preliminary data collection, 11 people responded to this question. Each participant answered “NO!” Which of the potential problems with samples discussed in this module could explain this connection? A scholarly article about response rates begins with the following quote: “Declining contact and cooperation rates in random digit dial (RDD) national telephone surveys raise serious concerns about the validity of estimates drawn from such research.”(Scott Keeter et al., “Gauging the Impact of Growing Nonresponse on Estimates from a National RDD Telephone Survey,” Public Opinion Quarterly 70 no. 5 (2006), http://poq.oxfordjournals.org/content/70/5/759.full (accessed May 1, 2013).) The Pew Research Center for People and the Press admits: “The percentage of people we interview – out of all we try to interview – has been declining over the past decade or more.” (Frequently Asked Questions, Pew Research Center for the People & the Press, http://www.people-press.org/methodology/frequently-asked-questions/#dont-you-have-trouble-getting-people-to-answer-your-polls (accessed May 1, 2013).) What are some reasons for the decline in response rate over the past decade? Explain why researchers are concerned with the impact of the declining response rate on public opinion polls. Possible reasons: increased use of caller id, decreased use of landlines, increased use of private numbers, voice mail, privacy managers, hectic nature of personal schedules, decreased willingness to be interviewed When a large number of people refuse to participate, then the sample may not have the same characteristics of the population. Perhaps the majority of people willing to participate are doing so because they feel strongly about the subject of the survey. Bringing It Together Seven hundred and seventy-one distance learning students at Long Beach City College responded to surveys in a specific academic year. Highlights of the summary report are listed in . LBCC Distance Learning Survey Results Have computer at home 96% Unable to come to campus for classes 65% Age 41 or over 24% Would like LBCC to offer more DL courses 95% Took DL classes due to a disability 17% Live at least 16 miles from campus 13% Took DL courses to fulfill transfer requirements 71% What percent of the students surveyed do not have a computer at home? About how many students in the survey live at least 16 miles from campus? If the same survey were done at Great Basin College in Elko, Nevada, do you think the percentages would be the same? Why? Several online textbook retailers advertise that they have lower prices than on-campus bookstores. However, an important factor is whether the Internet retailers actually have the textbooks that students need in stock. Students need to be able to get textbooks promptly at the beginning of the college term. If the book is not available, then a student would not be able to get the textbook at all, or might get a delayed delivery if the book is back ordered. Martin, a college newspaper reporter, is investigating textbook availability at online retailers. He decides to investigate one textbook for each of the following seven subjects: calculus, biology, chemistry, physics, statistics, geology, and general engineering. He consults textbook industry sales data and selects the most popular nationally used textbook in each of these subjects. He visits websites for a random sample of major online textbook sellers and looks up each of these seven textbooks to see if they are available in stock for quick delivery through these retailers. Based on his investigation, Martin writes an article in which he draws conclusions about the overall availability of all college textbooks through online textbook retailers. Write an analysis of his study that addresses the following issues: Is his sample representative of the population of all college textbooks? Explain why or why not. Describe some possible sources of bias in this study, and how it might affect the results of the study. Give some suggestions about what could be done to improve the study. Answers will vary. Sample answer: The sample is not representative of the population of all college textbooks. Two reasons why it is not representative are that he only sampled seven subjects and he only investigated one textbook in each subject. There are several possible sources of bias in the study. The seven subjects that he investigated are all in mathematics and the sciences; there are many subjects in the humanities, social sciences, and other subject areas, (for example: literature, art, history, psychology, sociology, business) that he did not investigate at all. It may be that different subject areas exhibit different patterns of textbook availability, but his sample would not detect such results. He also looked only at the most popular textbook in each of the subjects he investigated. The availability of the most popular textbooks may differ from the availability of other textbooks in one of two ways: the most popular textbooks may be more readily available online, because more new copies are printed, and more students nationwide are selling back their used copies OR the most popular textbooks may be harder to find available online, because more student demand exhausts the supply more quickly. In reality, many college students do not use the most popular textbook in their subject, and this study gives no useful information about the situation for those less popular textbooks. He could improve this study by: expanding the selection of subjects he investigates so that it is more representative of all subjects studied by college students, and expanding the selection of textbooks he investigates within each subject to include a mixed representation of both the most popular and less popular textbooks. Cluster Sampling a method for selecting a random sample and dividing the population into groups (clusters); use simple random sampling to select a set of clusters. Every individual in the chosen clusters is included in the sample. Continuous Random Variable a random variable (RV) whose outcomes are measured; the height of trees in the forest is a continuous RV. Convenience Sampling a nonrandom method of selecting a sample; this method selects individuals that are easily accessible and may result in biased data. Nonsampling Error an issue that affects the reliability of sampling data other than natural variation; it includes a variety of human errors including poor study design, biased sampling methods, inaccurate information provided by study participants, data entry errors, and poor analysis. Qualitative Data a set of observations (a set of possible outcomes); qualitative data has an attribute whose value is indicated by a label Quantitative Data a set of observations (a set of possible outcomes); quantitative (an attribute whose value is indicated by a number) data can be separated into two subgroups: discrete and continuous . Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage). Random Sampling a method of selecting a sample that gives every member of the population an equal chance of being selected. Sampling Bias not all members of the population are equally likely to be selected Sampling Error the natural variation that results from selecting a sample to represent a larger population; this variation decreases as the sample size increases, so selecting larger samples reduces sampling error. Sampling with Replacement Once a member of the population is selected for inclusion in a sample, that member is returned to the population for the selection of the next individual. Sampling without Replacement A member of the population may be chosen for inclusion in a sample only once. If chosen, the member is not returned to the population before the next selection. Simple Random Sampling a straightforward method for selecting a random sample; give each member of the population a number. Use a random number generator to select a set of labels. These randomly selected labels identify the members of your sample. Stratified Sampling a method for selecting a random sample used to ensure that subgroups of the population are represented adequately; divide the population into groups (strata). Use simple random sampling to identify a proportionate number of individuals from each stratum. Systematic Sampling a method for selecting a random sample; list the members of the population. Use simple random sampling to select a starting point in the population. Let k = (number of individuals in the population)/(number of individuals needed in the sample). Choose every kth individual in the list starting with the one that was randomly selected. If necessary, return to the beginning of the population list to complete your sample.", "section": "Data, Sampling, and Variation in Data and Sampling", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Frequency, Frequency Tables, and Levels of Measurement Once you have a set of data, you will need to organize it so that you can analyze how frequently each datum occurs in the set. However, when calculating the frequency, you may need to round your answers so that they are as precise as possible. Answers and Rounding Off A simple way to round off answers is to carry your final answer one more decimal place than was present in the original data. Round off only the final answer. Do not round off any intermediate results, if possible. If it becomes necessary to round off intermediate results, carry them to at least twice as many decimal places as the final answer. For example, the average of the three quiz scores four, six, and nine is 6.3, rounded off to the nearest tenth, because the data are whole numbers. Most answers will be rounded off in this manner. It is not necessary to reduce most fractions in this course. Especially in Probability Topics , the chapter on probability, it is more helpful to leave an answer as an unreduced fraction. Levels of Measurement The way a set of data is measured is called its level of measurement . Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level): Nominal scale level Ordinal scale level Interval scale level Ratio scale level Data that is measured using a nominal scale is qualitative (categorical) . Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. For example, trying to classify people according to their favorite food does not make any sense. Putting pizza first and sushi second is not meaningful. Smartphone companies are another example of nominal scale data. The data are the names of the companies that make smartphones, but there is no agreed upon order of these brands, even though people may have personal preferences. Nominal scale data cannot be used in calculations. Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data. Another example of using the ordinal scale is a cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations. Like data measured on an ordinal scale, data that are measured on an interval scale have a definite ordering. While ordinal scale data are categorical, or qualitative, interval scale data are numerical, or quantitative. It is possible to calculate differences between values measured on an interval scale. There is no minimum or \"zero\" value, however. Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not represent a minimum value. In both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0. Data that are measured using the ratio scale give the most information. Ratio scale data is like interval scale data, but there is a minimum value and ratios can be calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points). The exams are machine-graded. The data can be put in order from lowest to highest: 20, 68, 80, 92. The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The minimum score is 0. So 80 is four times 20. The score of 80 is four times better than the score of 20. Frequency Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows: 5 6 3 3 2 4 7 5 2 3 5 6 5 4 4 3 5 2 5 3 . lists the different data values in ascending order and their frequencies. Frequency Table of Student Work Hours DATA VALUE FREQUENCY 2 3 3 5 4 3 5 6 6 2 7 1 A frequency is the number of times a value of the data occurs. According to , there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample. A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals. Frequency Table of Student Work Hours with Relative Frequencies DATA VALUE FREQUENCY RELATIVE FREQUENCY 2 3 3 20 or 0.15 3 5 5 20 or 0.25 4 3 3 20 or 0.15 5 6 6 20 or 0.30 6 2 2 20 or 0.10 7 1 1 20 or 0.05 The sum of the values in the relative frequency column of is 20 20 , or 1. Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in . Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies DATA VALUE FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 2 3 3 20 or 0.15 0.15 3 5 5 20 or 0.25 0.15 + 0.25 = 0.40 4 3 3 20 or 0.15 0.40 + 0.15 = 0.55 5 6 6 20 or 0.30 0.55 + 0.30 = 0.85 6 2 2 20 or 0.10 0.85 + 0.10 = 0.95 7 1 1 20 or 0.05 0.95 + 0.05 = 1.00 The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated. NOTE Because of rounding, the relative frequency column may not always sum to one, and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one. represents the heights, in inches, of a sample of 100 semiprofessional soccer players. Frequency Table of Soccer Player Height HEIGHTS (INCHES) FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 59.95–61.95 5 5 100 = 0.05 0.05 61.95–63.95 3 3 100 = 0.03 0.05 + 0.03 = 0.08 63.95–65.95 15 15 100 = 0.15 0.08 + 0.15 = 0.23 65.95–67.95 40 40 100 = 0.40 0.23 + 0.40 = 0.63 67.95–69.95 17 17 100 = 0.17 0.63 + 0.17 = 0.80 69.95–71.95 12 12 100 = 0.12 0.80 + 0.12 = 0.92 71.95–73.95 7 7 100 = 0.07 0.92 + 0.07 = 0.99 73.95–75.95 1 1 100 = 0.01 0.99 + 0.01 = 1.00 Total = 100 Total = 1.00 The data in this table have been grouped into the following intervals: 59.95 to 61.95 inches 61.95 to 63.95 inches 63.95 to 65.95 inches 65.95 to 67.95 inches 67.95 to 69.95 inches 69.95 to 71.95 inches 71.95 to 73.95 inches 73.95 to 75.95 inches NOTE This example is used again in Descriptive Statistics , where the method used to compute the intervals will be explained. In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints. From , find the percentage of heights that are less than 65.95 inches. If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are 5 + 3 + 15 = 23 players whose heights are less than 65.95 inches. The percentage of heights less than 65.95 inches is then 23 100 or 23%. This percentage is the cumulative relative frequency entry in the third row. Try It shows the amount, in inches, of annual rainfall in a sample of towns. Rainfall (Inches) Frequency Relative Frequency Cumulative Relative Frequency 2.95–4.97 6 6 50 = 0.12 0.12 4.97–6.99 7 7 50 = 0.14 0.12 + 0.14 = 0.26 6.99–9.01 15 15 50 = 0.30 0.26 + 0.30 = 0.56 9.01–11.03 8 8 50 = 0.16 0.56 + 0.16 = 0.72 11.03–13.05 9 9 50 = 0.18 0.72 + 0.18 = 0.90 13.05–15.07 5 5 50 = 0.10 0.90 + 0.10 = 1.00 Total = 50 Total = 1.00 From , find the percentage of rainfall that is less than 9.01 inches. From , find the percentage of heights that fall between 61.95 and 65.95 inches. Add the relative frequencies in the second and third rows: 0.03 + 0.15 = 0.18 or 18%. Try It From , find the percentage of rainfall that is between 6.99 and 13.05 inches. Use the heights of the 100 semiprofessional soccer players in . Fill in the blanks and check your answers. The percentage of heights that are from 67.95 to 71.95 inches is: ____. The percentage of heights that are from 67.95 to 73.95 inches is: ____. The percentage of heights that are more than 65.95 inches is: ____. The number of players in the sample who are between 61.95 and 71.95 inches tall is: ____. What kind of data are the heights? Describe how you could gather this data (the heights) so that the data are characteristic of all semiprofessional soccer players. Remember, you count frequencies . To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row. 29% 36% 77% 87 quantitative continuous get rosters from each team and choose a simple random sample from each Try It From , find the number of towns that have rainfall between 2.95 and 9.01 inches. In your class, have someone conduct a survey of the number of siblings each student has. Create a frequency table. Add to it a relative frequency column and a cumulative relative frequency column. Answer the following questions: What percentage of the students in your class have no siblings? What percentage of the students have from one to three siblings? What percentage of the students have fewer than three siblings? Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are as follows: 2 5 7 3 2 10 18 15 20 7 10 18 5 12 13 12 4 5 10 . was produced: Frequency of Commuting Distances DATA FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 2 2 2 19 2 19 3 1 1 19 3 19 4 1 1 19 4 19 5 3 3 19 7 19 7 2 2 19 9 19 10 3 3 19 12 19 12 2 2 19 14 19 13 1 1 19 15 19 15 1 1 19 16 19 18 1 1 19 17 19 20 1 1 19 19 19 Is the table correct? If it is not correct, what is wrong? True or False: Three percent of the people surveyed commute three miles or less. If the statement is not correct, what should it be? If the table is incorrect, make the corrections. What fraction of the people surveyed commute five or seven miles? What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between five and 13 miles (not including five and 13 miles)? The cumulative relative frequency column should read: 2 19 , 3 19 , 4 19 , 7 19 , 9 19 , 12 19 , 14 19 , 15 19 , 16 19 , 18 19 , 19 19 No. The frequency column sums to 18, not 19. Not all cumulative relative frequencies are correct. False. The frequency for three miles is one; for two miles, two. 5 19 7 19 , 12 19 , 7 19 Try It represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year? contains data for the number of years of service for 70 federal employees. Number of Years of Service Number of Federal Employees 24 2 25 1 26 3 27 0 28 4 29 6 30 11 31 12 32 7 33 8 34 6 35 10 Answer the following questions. What is the cumulative frequency for years of service between 30 and 35 (inclusive)? What is the relative frequency for 30 years of service? What is the relative frequency for 30 years of service or less? What is the relative frequency for 25 years of service or more? Cumulative frequency is 54. 11/70 or 0.157 or 15.7% 27/70 or 0.386 or 38.6% 68/70 or 0.971 or 97.1% Try It contains the total number of fatal motor vehicle traffic crashes in the United States for a period of 18 years. Year Total Number of Crashes Year Total Number of Crashes Year 1 36,254 Year 11 38,444 Year 2 37,241 Year 12 39,252 Year 3 37,494 Year 13 38,648 Year 4 37,324 Year 14 37,435 Year 5 37,107 Year 15 34,172 Year 6 37,140 Year 16 30,862 Year 7 37,526 Year 17 30,296 Year 8 37,862 Year 18 29,757 Year 9 38,491 Total 653,782 Year 10 38,477 Answer the following questions. What is the frequency of deaths measured from Year 7 through Year 11? What percentage of deaths occurred after Year 13? What is the relative frequency of deaths that occurred in Year 7 or before? What is the percentage of deaths that occurred in Year 18? What is the cumulative relative frequency for Year 13? Explain what this number tells you about the data. References “State & County QuickFacts,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/download_data.html (accessed May 1, 2013). “State & County QuickFacts: Quick, easy access to facts about people, business, and geography,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/index.html (accessed May 1, 2013). “Table 5: Direct hits by mainland United States Hurricanes (1851-2004),” National Hurricane Center, http://www.nhc.noaa.gov/gifs/table5.gif (accessed May 1, 2013). “Levels of Measurement,” http://infinity.cos.edu/faculty/woodbury/stats/tutorial/Data_Levels.htm (accessed May 1, 2013). Courtney Taylor, “Levels of Measurement,” about.com, http://statistics.about.com/od/HelpandTutorials/a/Levels-Of-Measurement.htm (accessed May 1, 2013). David Lane. “Levels of Measurement,” Connexions, http://cnx.org/content/m10809/latest/ (accessed May 1, 2013). Chapter Review Some calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth. In addition to rounding your answers, you can measure your data using the following four levels of measurement. Nominal scale level: data that cannot be ordered nor can it be used in calculations Ordinal scale level: data that can be ordered; the differences cannot be measured Interval scale level: data with a definite ordering but no starting point; the differences can be measured, but there is no such thing as a ratio. Ratio scale level: data with a starting point that can be ordered; the differences have meaning and ratios can be calculated. When organizing data, it is important to know how many times a value appears. How many statistics students study five hours or more for an exam? What percent of families on our block own two pets? Frequency, relative frequency, and cumulative relative frequency are measures that answer questions like these. What type of measure scale is being used? Nominal, ordinal, interval or ratio. High school soccer players classified by their athletic ability: Superior, Average, Above average Baking temperatures for various main dishes: 350, 400, 325, 250, 300 The colors of crayons in a 24-crayon box Social security numbers Incomes measured in dollars A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not satisfied Political outlook: extreme left, left-of-center, right-of-center, extreme right Time of day on an analog watch The distance in miles to the closest grocery store The dates 1066, 1492, 1644, 1947, and 1944 The heights of 21–65 year-old women Common letter grades: A, B, C, D, and F ordinal interval nominal nominal ratio ordinal nominal interval ratio interval ratio ordinal HOMEWORK Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below: Part-time Student Course Loads # of Courses Frequency Relative Frequency Cumulative Relative Frequency 1 30 0.6 2 15 3 Fill in the blanks in . What percent of students take exactly two courses? What percent of students take one or two courses? Sixty adults with gum disease were asked the number of times per week they used to floss before their diagnosis. The (incomplete) results are shown in . Flossing Frequency for Adults with Gum Disease # Flossing per Week Frequency Relative Frequency Cumulative Relative Freq. 0 27 0.4500 1 18 3 0.9333 6 3 0.0500 7 1 0.0167 Fill in the blanks in . What percent of adults flossed six times per week? What percent flossed at most three times per week? # Flossing per Week Frequency Relative Frequency Cumulative Relative Frequency 0 27 0.4500 0.4500 1 18 0.3000 0.7500 3 11 0.1833 0.9333 6 3 0.0500 0.9833 7 1 0.0167 1 5.00% 93.33% Nineteen immigrants to the U.S were asked how many years, to the nearest year, they have lived in the U.S. The data are as follows: 2 5 7 2 2 10 20 15 0 7 0 20 5 12 15 12 4 5 10 . was produced. Frequency of Immigrant Survey Responses Data Frequency Relative Frequency Cumulative Relative Frequency 0 2 2 19 0.1053 2 3 3 19 0.2632 4 1 1 19 0.3158 5 3 3 19 0.4737 7 2 2 19 0.5789 10 2 2 19 0.6842 12 2 2 19 0.7895 15 1 1 19 0.8421 20 1 1 19 1.0000 Fix the errors in . Also, explain how someone might have arrived at the incorrect number(s). Explain what is wrong with this statement: “47 percent of the people surveyed have lived in the U.S. for 5 years.” Fix the statement in b to make it correct. What fraction of the people surveyed have lived in the U.S. five or seven years? What fraction of the people surveyed have lived in the U.S. at most 12 years? What fraction of the people surveyed have lived in the U.S. fewer than 12 years? What fraction of the people surveyed have lived in the U.S. from five to 20 years, inclusive? How much time does it take to travel to work? shows the mean commute time by state for workers at least 16 years old who are not working at home. Find the mean travel time, and round off the answer properly. 24.0 24.3 25.9 18.9 27.5 17.9 21.8 20.9 16.7 27.3 18.2 24.7 20.0 22.6 23.9 18.0 31.4 22.3 24.0 25.5 24.7 24.6 28.1 24.9 22.6 23.6 23.4 25.7 24.8 25.5 21.2 25.7 23.1 23.0 23.9 26.0 16.3 23.1 21.4 21.5 27.0 27.0 18.6 31.7 23.3 30.1 22.9 23.3 21.7 18.6 The sum of the travel times is 1,173.1. Divide the sum by 50 to calculate the mean value: 23.462. Because each state’s travel time was measured to the nearest tenth, round this calculation to the nearest hundredth: 23.46. A leading business magazine publishes data on small businesses (defined as businesses that have been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion). shows the ages of the chief executive officers for the first 60 ranked small businesses. Age Frequency Relative Frequency Cumulative Relative Frequency 40–44 3 45–49 11 50–54 13 55–59 16 60–64 10 65–69 6 70–74 1 What is the frequency for CEO ages between 54 and 65? What percentage of CEOs are 65 years or older? What is the relative frequency of ages under 50? What is the cumulative relative frequency for CEOs younger than 55? Which graph shows the relative frequency and which shows the cumulative relative frequency? Use the following information to answer the next two exercises: contains data on hurricanes that have made direct hits on the U.S. Between 1851 and 2004. A hurricane is given a strength category rating based on the minimum wind speed generated by the storm. Frequency of Hurricane Direct Hits Category Number of Direct Hits Relative Frequency Cumulative Frequency 1 109 0.3993 0.3993 2 72 0.2637 0.6630 3 71 0.2601 4 18 0.9890 5 3 0.0110 1.0000 Total = 273 What is the relative frequency of direct hits that were category 4 hurricanes? 0.0768 0.0659 0.2601 Not enough information to calculate b What is the relative frequency of direct hits that were AT MOST a category 3 storm? 0.3480 0.9231 0.2601 0.3370 Cumulative Relative Frequency The term applies to an ordered set of observations from smallest to largest. The cumulative relative frequency is the sum of the relative frequencies for all values that are less than or equal to the given value. Frequency the number of times a value of the data occurs Relative Frequency the ratio of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes", "section": "Frequency, Frequency Tables, and Levels of Measurement", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Experimental Design and Ethics Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data. The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the explanatory variable . The affected variable is called the response variable . In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments . An experimental unit is a single object or individual to be measured. You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention. Additional variables that can cloud a study are called lurking variables . In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point, the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables. The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted: Results showed that believing one had taken the substance resulted in [ performance ] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment. McClung, M. Collins, D. “Because I know it will!”: placebo effects of an ergogenic aid on athletic performance. Journal of Sport & Exercise Psychology. 2007 Jun. 29(3):382-94. Web. April 30, 2013. When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group . This group is given a placebo treatment–an active treatment that cannot directly influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding or masking in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, they do not know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded. Researchers want to investigate whether taking aspirin regularly reduces the risk of heart attack. Four hundred people between the ages of 50 and 84 are recruited as participants. The people are divided randomly into two groups: one group will take aspirin, and the other group will take a placebo. Each person takes one pill each day for three years, but they don't know whether they are taking aspirin or the placebo. At the end of the study, researchers count the number of people in each group who have had heart attacks. Identify the following values for this study: population, sample, experimental units, explanatory variable, response variable, treatments. The population is people aged 50 to 84. The sample is the 400 people who participated. The experimental units are the individual people in the study. The explanatory variable is oral medication. The treatments are aspirin and a placebo. The response variable is whether a subject had a heart attack. Try It A study needs to be conducted of the effect of three medicines A, B, and C on the height of adults aged 30 to 45. 90 adults were selected randomly and divided into three equal groups. The first group was asked to take medicine A for 6 months. The second group was asked to take medicine B for 6 months. The third group was asked to take medicine C for 6 months. The average change in height in each group is calculated at the end of the study. Identify the following values for this study: population, sample, experimental units, explanatory variables, response variable, treatments. The population is adults aged 30 to 45. The sample is 90 adults that were selected randomly. The experimental units are the individual adults in the study. The explanatory variable is the medicines. The treatments are medicines A, B, and C. The response variable is the average change in height in the group. The Smell & Taste Treatment and Research Foundation conducted a study to investigate whether smell can affect learning. Subjects completed mazes multiple times while wearing masks. They completed the pencil and paper mazes three times wearing floral-scented masks, and three times with unscented masks. Participants were assigned at random to wear the floral mask during the first three trials or during the last three trials. For each trial, researchers recorded the time it took to complete the maze and the subject’s impression of the mask’s scent: positive, negative, or neutral. Describe the explanatory and response variables in this study. What are the treatments? Identify any lurking variables that could interfere with this study. Is it possible to use blinding in this study? The explanatory variable is scent, and the response variable is the time it takes to complete the maze. There are two treatments: a floral-scented mask and an unscented mask. All subjects experienced both treatments. The order of treatments was randomly assigned so there were no differences between the treatment groups. Random assignment eliminates the problem of lurking variables. Subjects will clearly know whether they can smell flowers or not, so subjects cannot be blinded in this study. Researchers timing the mazes can be blinded, though. The researcher who is observing a subject will not know which mask is being worn. Try It The Placebo Research Group conducted a study to find the extent of placebo effects. A group of men randomly selected were asked to take a test before and after taking a pill that induces a mild headache. The pill in half of the randomly selected men was replaced with a similar pill that has no effect. For each trial, researchers recorded the change in time men took to complete the tests before and after taking the pill. Describe the explanatory and response variable in this study. What are the treatments? Identify any lurking variables that could interfere with this study. Is it possible to use blinding in this study? The explanatory variable is the pill, and the response variable is the change in time taken between the two tests. There are two treatments: a pill that induces a mild headache and a pill that has no effect. All subjects have been given pills. The men for kind of pill intake are randomly selected. Random assignment eliminates the problem of lurking variables. Subjects do not know which pill they are taking, so they are blinded in this study. The researchers know which subject is taking which pill, so they cannot be blinded in this study. A researcher wants to study the effects of birth order on personality. Explain why this study could not be conducted as a randomized experiment. What is the main problem in a study that cannot be designed as a randomized experiment? The explanatory variable is birth order. You cannot randomly assign a person’s birth order. Random assignment eliminates the impact of lurking variables. When you cannot assign subjects to treatment groups at random, there will be differences between the groups other than the explanatory variable. Try It You are concerned about the effects of texting on driving performance. Design a study to test the response time of drivers while texting and while driving only. How many seconds does it take for a driver to respond when a leading car hits the brakes? Describe the explanatory and response variables in the study. What are the treatments? What should you consider when selecting participants? Your research partner wants to divide participants randomly into two groups: one to drive without distraction and one to text and drive simultaneously. Is this a good idea? Why or why not? Identify any lurking variables that could interfere with this study. How can blinding be used in this study? Ethics The widespread misuse and misrepresentation of statistical information often gives the field a bad name. Some say that “numbers don’t lie,” but the people who use numbers to support their claims often do. An investigation of famous social psychologist, Diederik Stapel, has led to the retraction of his articles from some of the world’s top journals including Journal of Experimental Social Psychology, Social Psychology, Basic and Applied Social Psychology, British Journal of Social Psychology, and the magazine Science . Diederik Stapel is a former professor at Tilburg University in the Netherlands. An extensive investigation involving three universities where Stapel has worked concluded that the psychologist is guilty of fraud on a colossal scale. Falsified data taints over 55 papers he authored and 10 Ph.D. dissertations that he supervised. Stapel did not deny that his deceit was driven by ambition. But it was more complicated than that, he told me. He insisted that he loved social psychology but had been frustrated by the messiness of experimental data, which rarely led to clear conclusions. His lifelong obsession with elegance and order, he said, led him to concoct sexy results that journals found attractive. “It was a quest for aesthetics, for beauty—instead of the truth,” he said. He described his behavior as an addiction that drove him to carry out acts of increasingly daring fraud, like a junkie seeking a bigger and better high. Yudhijit Bhattacharjee, “The Mind of a Con Man,” Magazine, New York Times, April 26, 2013. Available online at: http://www.nytimes.com/2013/04/28/magazine/diederik-stapels-audacious-academic-fraud.html?src=dayp&_r=2& (accessed May 1, 2013). The committee investigating Stapel concluded that he is guilty of several practices including: creating datasets, which largely confirmed the prior expectations, altering data in existing datasets, changing measuring instruments without reporting the change, and misrepresenting the number of experimental subjects. Clearly, it is never acceptable to falsify data the way this researcher did. Sometimes, however, violations of ethics are not as easy to spot. Researchers have a responsibility to verify that proper methods are being followed. The report describing the investigation of Stapel’s fraud states that, “statistical flaws frequently revealed a lack of familiarity with elementary statistics.” “Flawed Science: The Fraudulent Research Practices of Social Psychologist Diederik Stapel,” Tillburg University, November 28, 2012, http://www.tilburguniversity.edu/upload/064a10cd-bce5-4385-b9ff-05b840caeae6_120695_Rapp_nov_2012_UK_web.pdf (accessed May 1, 2013). Many of Stapel’s co-authors should have spotted irregularities in his data. Unfortunately, they did not know very much about statistical analysis, and they simply trusted that he was collecting and reporting data properly. Many types of statistical fraud are difficult to spot. Some researchers simply stop collecting data once they have just enough to prove what they had hoped to prove. They don’t want to take the chance that a more extensive study would complicate their lives by producing data contradicting their hypothesis. Professional organizations, like the American Statistical Association, clearly define expectations for researchers. There are even laws in the federal code about the use of research data. When a statistical study uses human participants, as in medical studies, both ethics and the law dictate that researchers should be mindful of the safety of their research subjects. The U.S. Department of Health and Human Services oversees federal regulations of research studies with the aim of protecting participants. When a university or other research institution engages in research, it must ensure the safety of all human subjects. For this reason, research institutions establish oversight committees known as Institutional Review Boards (IRB) . All planned studies must be approved in advance by the IRB. Key protections that are mandated by law include the following: Risks to participants must be minimized and reasonable with respect to projected benefits. Participants must give informed consent . This means that the risks of participation must be clearly explained to the subjects of the study. Subjects must consent in writing, and researchers are required to keep documentation of their consent. Data collected from individuals must be guarded carefully to protect their privacy. These ideas may seem fundamental, but they can be very difficult to verify in practice. Is removing a participant’s name from the data record sufficient to protect privacy? Perhaps the person’s identity could be discovered from the data that remains. What happens if the study does not proceed as planned and risks arise that were not anticipated? When is informed consent really necessary? Suppose your doctor wants a blood sample to check your cholesterol level. Once the sample has been tested, you expect the lab to dispose of the remaining blood. At that point the blood becomes biological waste. Does a researcher have the right to take it for use in a study? It is important that students of statistics take time to consider the ethical questions that arise in statistical studies. How prevalent is fraud in statistical studies? You might be surprised—and disappointed. There is a website dedicated to cataloging retractions of study articles that have been proven fraudulent. A quick glance will show that the misuse of statistics is a bigger problem than most people realize. Vigilance against fraud requires knowledge. Learning the basic theory of statistics will empower you to analyze statistical studies critically. Describe the unethical behavior in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected. A researcher is collecting data in a community. The researcher selects a block where they are comfortable walking because they know many of the people living on the street. No one seems to be home at four houses on the route. They do not record the addresses and do not return at a later time to try to find residents at home. The researcher skips four houses on the route because they are running late for an appointment. When they get home, they fill in the forms by selecting random answers from other residents in the neighborhood. By selecting a convenient sample, the researcher is intentionally selecting a sample that could be biased. Claiming that this sample represents the community is misleading. The researcher needs to select areas in the community at random. Intentionally omitting relevant data will create bias in the sample. Suppose the researcher is gathering information about jobs and child care. By ignoring people who are not home, They may be missing data from working families that are relevant to her study. They need to make every effort to interview all members of the target sample. It is never acceptable to fake data. Even though the responses the researcher are “real” responses provided by other participants, the duplication is fraudulent and can create bias in the data. They researcher needs to work diligently to interview everyone on their route. Try It Describe the unethical behavior, if any, in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected. A study is commissioned to determine the favorite brand of fruit juice among teens in California. The survey is commissioned by the seller of a popular brand of apple juice. There are only two types of juice included in the study: apple juice and cranberry juice. Researchers allow participants to see the brand of juice as samples are poured for a taste test. Twenty-five percent of participants prefer Brand X, 33% prefer Brand Y and 42% have no preference between the two brands. Brand X references the study in a commercial saying “Most teens like Brand X as much as or more than Brand Y.” References “Vitamin E and Health,” Nutrition Source, Harvard School of Public Health, http://www.hsph.harvard.edu/nutritionsource/vitamin-e/ (accessed May 1, 2013). Stan Reents. “Don’t Underestimate the Power of Suggestion,” athleteinme.com, http://www.athleteinme.com/ArticleView.aspx?id=1053 (accessed May 1, 2013). Ankita Mehta. “Daily Dose of Aspiring Helps Reduce Heart Attacks: Study,” International Business Times, July 21, 2011. Also available online at http://www.ibtimes.com/daily-dose-aspirin-helps-reduce-heart-attacks-study-300443 (accessed May 1, 2013). The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/ScentsandLearning.html (accessed May 1, 2013). M.L. Jacskon et al., “Cognitive Components of Simulated Driving Performance: Sleep Loss effect and Predictors,” Accident Analysis and Prevention Journal, Jan no. 50 (2013), http://www.ncbi.nlm.nih.gov/pubmed/22721550 (accessed May 1, 2013). “Earthquake Information by Year,” U.S. Geological Survey. http://earthquake.usgs.gov/earthquakes/eqarchives/year/ (accessed May 1, 2013). “Fatality Analysis Report Systems (FARS) Encyclopedia,” National Highway Traffic and Safety Administration. http://www-fars.nhtsa.dot.gov/Main/index.aspx (accessed May 1, 2013). Data from www.businessweek.com (accessed May 1, 2013). Data from www.forbes.com (accessed May 1, 2013). “America’s Best Small Companies,” http://www.forbes.com/best-small-companies/list/ (accessed May 1, 2013). U.S. Department of Health and Human Services, Code of Federal Regulations Title 45 Public Welfare Department of Health and Human Services Part 46 Protection of Human Subjects revised January 15, 2009. Section 46.111:Criteria for IRB Approval of Research. “April 2013 Air Travel Consumer Report,” U.S. Department of Transportation, April 11 (2013), http://www.dot.gov/airconsumer/april-2013-air-travel-consumer-report (accessed May 1, 2013). Lori Alden, “Statistics can be Misleading,” econoclass.com, http://www.econoclass.com/misleadingstats.html (accessed May 1, 2013). Maria de los A. Medina, “Ethics in Statistics,” Based on “Building an Ethics Module for Business, Science, and Engineering Students” by Jose A. Cruz-Cruz and William Frey, Connexions, http://cnx.org/content/m15555/latest/ (accessed May 1, 2013). Chapter Review A poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. To counter the influence of the power of suggestion in the study, participants in the control group receive an inactive placebo treatment that looks exactly like the active treatments, but cannot directly influence the response variable. To preserve the integrity of the placebo, both researchers and subjects may be blinded. When a study is designed properly, the only difference between treatment groups is the one imposed by the researcher. Therefore, when groups respond differently to different treatments, the difference must be due to the influence of the explanatory variable. “An ethics problem arises when you are considering an action that benefits you or some cause you support, hurts or reduces benefits to others, and violates some rule.” (Andrew Gelman, “Open Data and Open Methods,” Ethics and Statistics, http://www.stat.columbia.edu/~gelman/research/published/ChanceEthics1.pdf (accessed May 1, 2013).) Ethical violations in statistics are not always easy to spot. Professional associations and federal agencies post guidelines for proper conduct. It is important that you learn basic statistical procedures so that you can recognize proper data analysis. Design an experiment. Identify the explanatory and response variables. Describe the population being studied and the experimental units. Explain the treatments that will be used and how they will be assigned to the experimental units. Describe how blinding and placebos may be used to counter the power of suggestion. Discuss potential violations of the rule requiring informed consent. People in a correctional facility are offered good behavior credit in return for participation in a study. A research study is designed to investigate a new children’s allergy medication. Participants in a study are told that the new medication being tested is highly promising, but they are not told that only a small portion of participants will receive the new medication. Others will receive placebo treatments and traditional treatments. People who are incarcerated may not feel comfortable refusing participation, or may feel obligated to take advantage of the promised benefits. They may not feel truly free to refuse participation. Parents can provide consent on behalf of their children, but children are not competent to provide consent for themselves. All risks and benefits must be clearly outlined. Study participants must be informed of relevant aspects of the study in order to give appropriate consent. HOMEWORK How does sleep deprivation affect your ability to drive? A recent study measured the effects on 19 professional drivers. Each driver participated in two experimental sessions: one after normal sleep and one after 27 hours of total sleep deprivation. The treatments were assigned in random order. In each session, performance was measured on a variety of tasks including a driving simulation. Use key terms from this module to describe the design of this experiment. Explanatory variable: amount of sleep Response variable: performance measured in assigned tasks Treatments: normal sleep and 27 hours of total sleep deprivation Experimental Units: 19 professional drivers Lurking variables: none – all drivers participated in both treatments Random assignment: treatments were assigned in random order; this eliminated the effect of any “learning” that may take place during the first experimental session Control/Placebo: completing the experimental session under normal sleep conditions Blinding: researchers evaluating subjects’ performance must not know which treatment is being applied at the time An advertisement for Acme Investments displays the two graphs in to show the value of Acme’s product in comparison with the Other Guy’s product. Describe the potentially misleading visual effect of these comparison graphs. How can this be corrected? As the graphs show, Acme consistently outperforms the Other Guys! The graph in shows the number of complaints for six different airlines as reported to the US Department of Transportation in February 2013. Alaska, Pinnacle, and Airtran Airlines have far fewer complaints reported than American, Delta, and United. Can we conclude that American, Delta, and United are the worst airline carriers since they have the most complaints? You cannot assume that the numbers of complaints reflect the quality of the airlines. The airlines shown with the greatest number of complaints are the ones with the most passengers. You must consider the appropriateness of methods for presenting data; in this case displaying totals is misleading. Explanatory Variable the independent variable in an experiment; the value controlled by researchers Treatments different values or components of the explanatory variable applied in an experiment Response Variable the dependent variable in an experiment; the value that is measured for change at the end of an experiment Experimental Unit any individual or object to be measured Lurking Variable a variable that has an effect on a study even though it is neither an explanatory variable nor a response variable Random Assignment the act of organizing experimental units into treatment groups using random methods Control Group a group in a randomized experiment that receives an inactive treatment but is otherwise managed exactly as the other groups Informed Consent Any human subject in a research study must be cognizant of any risks or costs associated with the study. The subject has the right to know the nature of the treatments included in the study, their potential risks, and their potential benefits. Consent must be given freely by an informed, fit participant. Institutional Review Board a committee tasked with oversight of research programs that involve human subjects Placebo an inactive treatment that cannot directly affect the response variable, used to counter the power of suggestion Blinding not telling participants which treatment a subject is receiving Double-blind experiment an experiment in which both the subjects of an experiment and the researchers who work with the subjects are blinded", "section": "Experimental Design and Ethics", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Data Collection Experiment Data Collection Experiment Class Time: Names: Student Learning Outcomes The student will demonstrate the systematic sampling technique. The student will construct relative frequency tables. The student will interpret results and their differences from different data groupings. Movie Survey Ask five classmates from a different class how many movies they saw at the theater last month. Do not include rented movies. Record the data. In class, randomly pick one person. On the class list, mark that person’s name. Move down four names on the class list. Mark that person’s name. Continue doing this until you have marked 12 names. You may need to go back to the start of the list. For each marked name record the five data values. You now have a total of 60 data values. For each name marked, record the data. Order the Data Complete the two relative frequency tables below using your class data. Frequency of Number of Movies Viewed Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 0 1 2 3 4 5 6 7+ Frequency of Number of Movies Viewed Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 0–1 2–3 4–5 6–7+ Using the tables, find the percent of data that is at most two. Which table did you use and why? Using the tables, find the percent of data that is at most three. Which table did you use and why? Using the tables, find the percent of data that is more than two. Which table did you use and why? Using the tables, find the percent of data that is more than three. Which table did you use and why? Discussion Questions Is one of the tables “more correct” than the other? Why or why not? In general, how could you group the data differently? Are there any advantages to either way of grouping the data? Why did you switch between tables, if you did, when answering the question above?", "section": "Data Collection Experiment", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Sampling Experiment Sampling Experiment Class Time: Names: Student Learning Outcomes The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques. The student will explain the details of each procedure used. In this lab, you will be asked to pick several random samples of restaurants. In each case, describe your procedure briefly, including how you might have used the random number generator, and then list the restaurants in the sample you obtained. NOTE The following section contains restaurants stratified by city into columns and grouped horizontally by entree cost (clusters). Restaurants Used In Sample Restaurants Stratified by City and Entree Cost Entree Cost Under $10 $10 to under $15 $15 to under $20 Over $20 San Jose El Abuelo Taq, Pasta Mia, Emma’s Express, Bamboo Hut Emperor’s Guard, Creekside Inn Agenda, Gervais, Miro’s Blake’s, Eulipia, Hayes Mansion, Germania Palo Alto Senor Taco, Olive Garden, Taxi’s Ming’s, P.A. Joe’s, Stickney’s Scott’s Seafood, Poolside Grill, Fish Market Sundance Mine, Maddalena’s, Spago’s Los Gatos Mary’s Patio, Mount Everest, Sweet Pea’s, Andele Taqueria Lindsey’s, Willow Street Toll House Charter House, La Maison Du Cafe Mountain View Maharaja, New Ma’s, Thai-Rific, Garden Fresh Amber Indian, La Fiesta, Fiesta del Mar, Dawit Austin’s, Shiva’s, Mazeh Le Petit Bistro Cupertino Hobees, Hung Fu, Samrat, Panda Express Santa Barb. Grill, Mand. Gourmet, Bombay Oven, Kathmandu West Fontana’s, Blue Pheasant Hamasushi, Helios Sunnyvale Chekijababi, Taj India, Full Throttle, Tia Juana, Lemon Grass Pacific Fresh, Charley Brown’s, Cafe Cameroon, Faz, Aruba’s Lion & Compass, The Palace, Beau Sejour Santa Clara Rangoli, Armadillo Willy’s, Thai Pepper, Pasand Arthur’s, Katie’s Cafe, Pedro’s, La Galleria Birk’s, Truya Sushi, Valley Plaza Lakeside, Mariani’s A Simple Random Sample Pick a simple random sample of 15 restaurants. Describe your procedure. Complete the table with your sample. 1. __________ 6. __________ 11. __________ 2. __________ 7. __________ 12. __________ 3. __________ 8. __________ 13. __________ 4. __________ 9. __________ 14. __________ 5. __________ 10. __________ 15. __________ A Systematic Sample Pick a systematic sample of 15 restaurants. Describe your procedure. Complete the table with your sample. 1. __________ 6. __________ 11. __________ 2. __________ 7. __________ 12. __________ 3. __________ 8. __________ 13. __________ 4. __________ 9. __________ 14. __________ 5. __________ 10. __________ 15. __________ A Stratified Sample Pick a stratified sample , by city, of 20 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number. Describe your procedure. Complete the table with your sample. 1. __________ 6. __________ 11. __________ 16. __________ 2. __________ 7. __________ 12. __________ 17. __________ 3. __________ 8. __________ 13. __________ 18. __________ 4. __________ 9. __________ 14. __________ 19. __________ 5. __________ 10. __________ 15. __________ 20. __________ A Stratified Sample Pick a stratified sample , by entree cost, of 21 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number. Describe your procedure. Complete the table with your sample. 1. __________ 6. __________ 11. __________ 16. __________ 2. __________ 7. __________ 12. __________ 17. __________ 3. __________ 8. __________ 13. __________ 18. __________ 4. __________ 9. __________ 14. __________ 19. __________ 5. __________ 10. __________ 15. __________ 20. __________ 21. __________ A Cluster Sample Pick a cluster sample of restaurants from two cities. The number of restaurants will vary. Describe your procedure. Complete the table with your sample. 1. ________ 6. ________ 11. ________ 16. ________ 21. ________ 2. ________ 7. ________ 12. ________ 17. ________ 22. ________ 3. ________ 8. ________ 13. ________ 18. ________ 23. ________ 4. ________ 9. ________ 14. ________ 19. ________ 24. ________ 5. ________ 10. ________ 15. ________ 20. ________ 25. ________", "section": "Sampling Experiment", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction When you have large amounts of data, you will need to organize it in a way that makes sense. These ballots from an election are rolled together with similar ballots to keep them organized. (credit: modification of work “HELMAND PROVINCE, Afghanistan (Aug. 21, 2009) Voting ballots organized and arranged for counting by Afghan presidential election workers at a local school in the Nawa District.” by U.S. Marine Corps photo by Staff Sgt. William Greeson/ Wikimedia Commons, Public Domain) Chapter Objectives By the end of this chapter, the student should be able to: Display data graphically and interpret graphs: stemplots, histograms, and box plots. Recognize, describe, and calculate the measures of location of data: quartiles and percentiles. Recognize, describe, and calculate the measures of the center of data: mean, median, and mode. Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation, and range. Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data. In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called \"Descriptive Statistics.\" You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs. A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied. Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. NOTE This book contains instructions for constructing a histogram and a box plot for the TI-83+ and TI-84 calculators. The Texas Instruments (TI) website provides additional instructions for using these calculators.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs One simple graph, the stem-and-leaf graph or stemplot , comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit . For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem. For Professor Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest): 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Stem-and-Leaf Graph Stem Leaf 3 3 4 2 9 9 5 3 5 5 6 1 3 7 8 8 9 9 7 2 3 4 8 8 0 3 8 8 8 9 0 2 4 4 4 4 6 10 0 The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% ( 8 31 ) were in the 90s or 100, a fairly high number of As. Try It For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest): 32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61 Construct a stem plot for the data. The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later. The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data: 1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3 Do the data seem to have any concentration of values? NOTE The leaves are to the right of the decimal. The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers. Stem Leaf 1 1 5 2 3 5 7 3 2 3 3 5 8 4 0 2 5 5 7 8 5 5 6 6 5 7 7 8 9 10 11 12 3 Try It The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers: 0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0 A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. and show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data. Presidential Ages at Inauguration President Age President Age President Age Washington 57 Lincoln 52 Hoover 54 J. Adams 61 A. Johnson 56 F. Roosevelt 51 Jefferson 57 Grant 46 Truman 60 Madison 57 Hayes 54 Eisenhower 62 Monroe 58 Garfield 49 Kennedy 43 J. Q. Adams 57 Arthur 51 L. Johnson 55 Jackson 61 Cleveland 47 Nixon 56 Van Buren 54 B. Harrison 55 Ford 61 W. H. Harrison 68 Cleveland 55 Carter 52 Tyler 51 McKinley 54 Reagan 69 Polk 49 T. Roosevelt 42 G.H.W. Bush 64 Taylor 64 Taft 51 Clinton 47 Fillmore 50 Wilson 56 G. W. Bush 54 Pierce 48 Harding 55 Obama 47 Buchanan 65 Coolidge 51 Presidential Age at Death President Age President Age President Age Washington 67 Lincoln 56 Hoover 90 J. Adams 90 A. Johnson 66 F. Roosevelt 63 Jefferson 83 Grant 63 Truman 88 Madison 85 Hayes 70 Eisenhower 78 Monroe 73 Garfield 49 Kennedy 46 J. Q. Adams 80 Arthur 56 L. Johnson 64 Jackson 78 Cleveland 71 Nixon 81 Van Buren 79 B. Harrison 67 Ford 93 W. H. Harrison 68 Cleveland 71 Reagan 93 Tyler 71 McKinley 58 Polk 53 T. Roosevelt 60 Taylor 65 Taft 72 Fillmore 74 Wilson 67 Pierce 64 Harding 57 Buchanan 77 Coolidge 60 Ages at Inauguration Ages at Death 9 9 8 7 7 7 6 3 2 4 6 9 8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 5 3 6 6 7 7 8 9 8 5 4 4 2 1 1 1 0 6 0 0 3 3 4 4 5 6 7 7 7 8 7 0 1 1 1 2 3 4 7 8 8 9 8 0 1 3 5 8 9 0 0 3 3 The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stemand-leaf plot of these wins and losses. Losses Wins Season Losses Wins Season 34 48 1 41 41 22 34 48 2 39 43 23 46 36 3 44 38 24 46 36 4 39 43 25 36 46 5 25 57 26 47 35 6 40 42 27 51 31 7 36 46 28 53 29 8 26 56 29 51 31 9 32 50 30 41 41 10 19 31 31 36 46 11 54 28 32 32 50 12 57 25 33 51 31 13 49 33 34 40 42 14 47 35 35 39 43 15 54 28 36 42 40 16 69 13 37 48 34 17 56 26 38 32 50 18 52 30 39 25 57 19 45 37 40 32 50 20 35 47 41 30 52 21 29 53 42 Another type of graph that is useful for specific data values is a line graph . In the particular line graph shown in , the x -axis (horizontal axis) consists of data values and the y -axis (vertical axis) consists of frequency points . The frequency points are connected using line segments. In a survey, 40 parents were asked how many times per week a teenager must be reminded to do their chores. The results are shown in and in . Number of times teenager is reminded Frequency 0 2 1 5 2 8 3 14 4 7 5 4 Try It In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in . Construct a line graph. Number of times in shop Frequency 0 7 1 10 2 14 3 9 Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in has age groups represented on the x -axis and proportions on the y -axis . The percentage of U.S.-based TikTok users by age is shown in . Construct a bar graph using this data. Age groups Proportion (%) of TikTok users 10–19 32.5% 20–29 29.5% 30–39 16.4% 40–49 13.9% 50+ 7.1% Try It The population in Park City is made up of children, working-age adults, and retirees. shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions. Age groups Number of people Proportion of population Children 67,059 19% Working-age adults 152,198 43% Retirees 131,662 38% The columns in show the projected data for the year 2030 for the number and percentages of high school graduates by geographic region in the United States. Create a bar graph for this data with the geographic region (qualitative data) on the x -axis and the percentage of high school data (quantitative data) on the y -axis. Region Number of Graduates Percentage of Graduates Northeast 517,720 16.1% Midwest 695,170 21.6% South 1,253,540 39.0% West 749,400 23.3% Try It Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district. District Registered voter population Overall city population 1 15.5% 19.4% 2 12.2% 15.6% 3 9.8% 9.0% 4 17.4% 18.5% 5 22.8% 20.7% 6 22.3% 16.8% References Doyle, Brandon. \"TikTok Statistics – Updated Mar 2023.\" Wallaroo. March 21, 2023. https://wallaroomedia.com/blog/social-media/tiktok-statistics/. “Table 219.20. Public High School Graduates, by Region, State, and Jurisdiction: Selected Years, 1980-81 through 2030-31.” National Center for Education Statistics. Accessed May 24, 2023. https://nces.ed.gov/programs/digest/d21/tables/dt21_219.20.asp. “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). Chapter Review A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used. For each of the following data sets, create a stem plot and identify any outliers. The miles per gallon rating for 30 cars are shown below (lowest to highest). 19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43 Stem Leaf 1 9 9 9 2 0 1 1 5 5 5 6 6 8 9 3 1 1 2 2 3 4 5 6 7 7 8 8 8 8 4 1 3 3 The height in feet of 25 trees is shown below (lowest to highest). 25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54 The data are the prices of different laptops at an electronics store. Round each value to the nearest ten. 249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610 Stem Leaf 2 5 5 6 7 7 8 3 0 0 1 2 3 3 5 5 5 7 7 9 4 1 6 9 5 6 7 7 6 1 The data are daily high temperatures in a town for one month. 61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95 For the next three exercises, use the data to construct a line graph. In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in . Number of times in store Frequency 1 4 2 10 3 16 4 6 5 4 In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in . Years since last purchase Frequency 0 2 1 8 2 13 3 22 4 16 5 9 Several children were asked how many TV shows they watch each day. The results of the survey are shown in . Number of TV Shows Frequency 0 12 1 18 2 36 3 7 4 2 The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students. Seasons Number of students Proportion of population Spring 8 24% Summer 9 26% Autumn 11 32% Winter 6 18% Using the data from Ms. Ramirez’s math class supplied in , construct a bar graph showing the percentages. David County has six high schools. Each school sent students to participate in a county-wide science competition. shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school. High School Science competition population Overall student population Alabaster 28.9% 8.6% Concordia 7.6% 23.2% Genoa 12.1% 15.0% Mocksville 18.5% 14.3% Tynneson 24.2% 10.1% West End 8.7% 28.8% Use the data from the David County science competition supplied in . Construct a bar graph that shows the county-wide population percentage of students at each school. Homework Student grades on a chemistry exam were: 77, 78, 76, 81, 86, 51, 79, 82, 84, 99 Construct a stem-and-leaf plot of the data. Are there any potential outliers? If so, which scores are they? Why do you consider them outliers? contains the percentages of adults who were never married in the 50 U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Use a random number generator to randomly pick eight states. Construct a bar graph of the rates of those eight states. Construct a bar graph for all the states beginning with the letter \"A.\" Construct a bar graph for all the states beginning with the letter \"M.\" Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows. Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically) Press MATH Arrow over to PRB Press 5:randInt( Enter 51,1,8) Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}. Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.", "section": "Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Histograms, Frequency Polygons, and Time Series Graphs For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more. A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data. The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample. (Remember, frequency is defined as the number of times an answer occurs.) If: f = frequency n = total number of data values (or the sum of the individual frequencies), and RF = relative frequency, then: RF = f n For example, if three students in an English class of 40 students received from 90% to 100%, then, f = 3, n = 40, and RF = f n = 3 40 = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures. To construct a histogram , first decide how many bars or intervals , also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data. The following data are the heights (in inches to the nearest half inch) of 100 semiprofessional soccer players. The heights are continuous data, since height is measured. 60; 60.5; 61; 61; 61.5 63.5; 63.5; 63.5 64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5 66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5 68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5 70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71 72; 72; 72; 72.5; 72.5; 73; 73.5 74 The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point. 60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95. The largest value is 74, so 74 + 0.05 = 74.05 is the ending value. Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars. 74.05 - 59.95 = 14.1 14.1 ÷ 8 = 1.76 NOTE We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the number of bars or class intervals is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals. The boundaries are: 59.95 59.95 + 2 = 61.95 61.95 + 2 = 63.95 63.95 + 2 = 65.95 65.95 + 2 = 67.95 67.95 + 2 = 69.95 69.95 + 2 = 71.95 71.95 + 2 = 73.95 73.95 + 2 = 75.95 The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95. The following histogram displays the heights on the x -axis and relative frequency on the y -axis. Try It The following data are the shoe sizes of 50 students. The sizes are discrete data since shoe size is measured in whole and half units only. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars. 9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5 12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14 Create a histogram for the following data: the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data , since books are counted. 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 2; 2; 2; 2; 2; 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4 5; 5; 5; 5; 5 6; 6 Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books. Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5. Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ . 3.5 to 4.5 4.5 to 5.5 6 5.5 to 6.5 Calculate the number of bars as follows: 6.5 - 0.5 = 6 6 ÷ 1 = 6 where 1 is the width of a bar. Therefore, bars = 6. The following histogram displays the number of books on the x -axis and the frequency on the y -axis. Go to . There are calculator instructions for entering data and for creating a customized histogram. Create the histogram for . Press Y=. Press CLEAR to delete any equations. Press STAT 1:EDIT . If L1 has data in it, arrow up into the name L1 , press CLEAR and then arrow down. If necessary, do the same for L2 . Into L1 , enter 1, 2, 3, 4, 5, 6. Into L2 , enter 11, 10, 16, 6, 5, 2. Press WINDOW. Set Xmin = .5 , Xmax = 6.5 , Xscl = (6.5 – .5)/6 , Ymin = –1 , Ymax = 20 , Yscl = 1 , Xres = 1 . Press 2 nd Y=. Start by pressing 4:Plotsoff ENTER. Press 2 nd Y=. Press 1:Plot1 . Press ENTER. Arrow down to TYPE. Arrow to the 3 rd picture (histogram). Press ENTER. Arrow down to Xlist: Enter L1 (2 nd 1). Arrow down to Freq. Enter L2 (2 nd 2). Press GRAPH. Use the TRACE key and the arrow keys to examine the histogram. Try It The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted. 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3 20 student athletes play one sport. 22 student athletes play two sports. Eight student athletes play three sports. Fill in the blanks for the following sentence. Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to _____, the 2 in the middle of the interval from _____ to _____, and the 3 in the middle of the interval from _____ to _____. Using this data set, construct a histogram. Number of Hours My Classmates Spent Playing Video Games on Weekends 9.95 10 2.25 16.75 0 19.5 22.5 7.5 15 12.75 5.5 11 10 20.75 17.5 23 21.9 24 23.75 18 20 15 22.9 18.8 20.5 Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram. Try It The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram. 22 35 15 26 40 28 18 20 25 34 39 42 24 22 19 27 22 34 40 20 38 and 28 Use 10–19 as the first interval. Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals. Frequency Polygons Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons. To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x -axis and y -axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them. A frequency polygon was constructed from the frequency table below. Frequency Distribution for Calculus Final Test Scores Lower Bound Upper Bound Frequency Cumulative Frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 The first label on the x -axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x -axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x -axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side. Try It Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in . Age at Inauguration Frequency 41.5–46.5 4 46.5–51.5 11 51.5–56.5 14 56.5–61.5 9 61.5–66.5 4 66.5–71.5 2 Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets. We will construct an overlay frequency polygon comparing the scores from with the students’ final numeric grade. Frequency Distribution for Calculus Final Test Scores Lower Bound Upper Bound Frequency Cumulative Frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 Frequency Distribution for Calculus Final Grades Lower Bound Upper Bound Frequency Cumulative Frequency 49.5 59.5 10 10 59.5 69.5 10 20 69.5 79.5 30 50 79.5 89.5 45 95 89.5 99.5 5 100 Try It We will construct an overlay frequency polygon comparing the scores from with the students’ final test scores in algebra. Frequency Distribution for Algebra Final Test Scores Lower Bound Upper Bound Frequency Cumulative Frequency 49.5 59.5 10 10 59.5 69.5 5 15 69.5 79.5 40 55 79.5 89.5 35 90 89.5 99.5 10 100 Calculus Test Scores v Algebra Test Scores Image Insert Constructing a Time Series Graph Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with this data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected. One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph. To construct a time series graph, we must look at both pieces of our paired data set . We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur. The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only. Year Jan Feb Mar Apr May Jun Jul 1 181.7 183.1 184.2 183.8 183.5 183.7 183.9 2 185.2 186.2 187.4 188.0 189.1 189.7 189.4 3 190.7 191.8 193.3 194.6 194.4 194.5 195.4 4 198.3 198.7 199.8 201.5 202.5 202.9 203.5 5 202.416 203.499 205.352 206.686 207.949 208.352 208.299 6 211.080 211.693 213.528 214.823 216.632 218.815 219.964 7 211.143 212.193 212.709 213.240 213.856 215.693 215.351 8 216.687 216.741 217.631 218.009 218.178 217.965 218.011 9 220.223 221.309 223.467 224.906 225.964 225.722 225.922 10 226.665 227.663 229.392 230.085 229.815 229.478 229.104 Year Aug Sep Oct Nov Dec Annual 1 184.6 185.2 185.0 184.5 184.3 184.0 2 189.5 189.9 190.9 191.0 190.3 188.9 3 196.4 198.8 199.2 197.6 196.8 195.3 4 203.9 202.9 201.8 201.5 201.8 201.6 5 207.917 208.490 208.936 210.177 210.036 207.342 6 219.086 218.783 216.573 212.425 210.228 215.303 7 215.834 215.969 216.177 216.330 215.949 214.537 8 218.312 218.439 218.711 218.803 219.179 218.056 9 226.545 226.889 226.421 226.230 225.672 224.939 10 230.379 231.407 231.317 230.221 229.601 229.594 Try It The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO 2 emissions for the United States. CO2 Emissions Year Ukraine United Kingdom United States 1 352,259 540,640 5,681,664 2 343,121 540,409 5,790,761 3 339,029 541,990 5,826,394 4 327,797 542,045 5,737,615 5 328,357 528,631 5,828,697 6 323,657 522,247 5,656,839 7 272,176 474,579 5,299,563 Uses of a Time Series Graph Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot. How NOT to Lie with Statistics It is important to remember that the very reason we develop a variety of methods to present data is to develop insights into the subject of what the observations represent. We want to get a \"sense\" of the data. Are the observations all very much alike or are they spread across a wide range of values, are they bunched at one end of the spectrum or are they distributed evenly and so on. We are trying to get a visual picture of the numerical data. Shortly we will develop formal mathematical measures of the data, but our visual graphical presentation can say much. It can, unfortunately, also say much that is distracting, confusing and simply wrong in terms of the impression the visual leaves. Many years ago Darrell Huff wrote the book How to Lie with Statistics . It has been through 25 plus printings and sold more than one and one-half million copies. His perspective was a harsh one and used many actual examples that were designed to mislead. He wanted to make people aware of such deception, but perhaps more importantly to educate so that others do not make the same errors inadvertently. Again, the goal is to enlighten with visuals that tell the story of the data. Pie charts have a number of common problems when used to convey the message of the data. Too many pieces of the pie overwhelm the reader. More than perhaps five or six categories ought to give an idea of the relative importance of each piece. This is after all the goal of a pie chart, what subset matters most relative to the others. If there are more components than this then perhaps an alternative approach would be better or perhaps some can be consolidated into an \"other\" category. Pie charts cannot show changes over time, although we see this attempted all too often. In federal, state, and city finance documents pie charts are often presented to show the components of revenue available to the governing body for appropriation: income tax, sales tax motor vehicle taxes and so on. In and of itself this is interesting information and can be nicely done with a pie chart. The error occurs when two years are set side-by-side. Because the total revenues change year to year, but the size of the pie is fixed, no real information is provided and the relative size of each piece of the pie cannot be meaningfully compared. Histograms can be very helpful in understanding the data. Properly presented, they can be a quick visual way to present probabilities of different categories by the simple visual of comparing relative areas in each category. Here the error, purposeful or not, is to vary the width of the categories. This of course makes comparison to the other categories impossible. It does embellish the importance of the category with the expanded width because it has a greater area, inappropriately, and thus visually \"says\" that that category has a higher probability of occurrence. Time series graphs perhaps are the most abused. A plot of some variable across time should never be presented on axes that change part way across the page either in the vertical or horizontal dimension. Perhaps the time frame is changed from years to months. Perhaps this is to save space or because monthly data was not available for early years. In either case this confounds the presentation and destroys any value of the graph. If this is not done to purposefully confuse the reader, then it certainly is either lazy or sloppy work. Changing the units of measurement of the axis can smooth out a drop or accentuate one. If you want to show large changes, then measure the variable in small units, penny rather than thousands of dollars. And of course to continue the fraud, be sure that the axis does not begin at zero, zero. If it begins at zero, zero, then it becomes apparent that the axis has been manipulated. Perhaps you have a client that is concerned with the volatility of the portfolio you manage. An easy way to present the data is to use long time periods on the time series graph. Use months or better, quarters rather than daily or weekly data. If that doesn't get the volatility down then spread the time axis relative to the rate of return or portfolio valuation axis. If you want to show \"quick\" dramatic growth, then shrink the time axis. Any positive growth will show visually \"high\" growth rates. Do note that if the growth is negative then this trick will show the portfolio is collapsing at a dramatic rate. Again, the goal of descriptive statistics is to convey meaningful visuals that tell the story of the data. Purposeful manipulation is fraud and unethical at the worst, but even at its best, making these type of errors will lead to confusion on the part of the analysis. References Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker “Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013). “Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013). “Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013). “Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013). “CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013). “Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013). “Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013). Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach . CRC Press: 1980. “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). Chapter Review A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y -axis with the frequency being graphed on the x -axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table. Data Value (# cars) Frequency Relative Frequency Cumulative Relative Frequency What does the frequency column in sum to? Why? 65 What does the relative frequency column in sum to? Why? What is the difference between relative frequency and frequency for each data value in ? The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value. What is the difference between cumulative relative frequency and relative frequency for each data value? To construct the histogram for the data in , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling. Answers will vary. One possible histogram is shown: Construct a frequency polygon for the following: Pulse Rates for Females Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 Actual Speed in a 30 MPH Zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 Tar (mg) in Nonfiltered Cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2 Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger. Depth of Hunger Frequency 230–259 21 260–289 13 290–319 5 320–349 7 350–379 1 380–409 1 410–439 1 Find the midpoint for each class. These will be graphed on the x -axis. The frequency values will be graphed on the y -axis values. Use the two frequency tables to compare the life expectancy of males and females from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of females compared to males? Life Expectancy at Birth – Females Frequency 49–55 3 56–62 3 63–69 1 70–76 3 77–83 8 84–90 2 Life Expectancy at Birth – Males Frequency 49–55 3 56–62 3 63–69 1 70–76 1 77–83 7 84–90 5 Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births. Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009 Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354 Sex/Year 1870 1871 1872 1873 1874 1875 Female 56,431 56,099 57,472 58,233 60,109 60,146 Male 58,959 60,029 61,293 61,467 63,602 63,432 Total 115,390 116,128 118,765 119,700 123,711 123,578 The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for a city during the period from 1961 to 1973. Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36 Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33 Construct a double time series graph using a common x -axis for both sets of data. Which variable increased the fastest? Explain. Did the city's increase in police officers have an impact on the murder rate? Explain. Homework Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows: Publisher A # of books Freq. Rel. Freq. 0 10 1 12 2 16 3 12 4 8 5 6 6 2 8 2 Publisher B # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Publisher C # of books Freq. Rel. Freq. 0–1 20 2–3 35 4–5 12 6–7 2 8–9 1 Find the relative frequencies for each survey. Write them in the charts. Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? Make new histograms for Publisher A and Publisher B. This time, make bar widths of two. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group. Singles Amount($) Frequency Rel. Frequency 51–100 5 101–150 10 151–200 15 201–250 15 251–300 10 301–350 5 Couples Amount($) Frequency Rel. Frequency 100–150 5 201–250 5 251–300 5 301–350 5 351–400 10 401–450 10 451–500 10 501–550 10 551–600 5 601–650 5 Fill in the relative frequency for each group. Construct a histogram for the singles group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis. Construct a histogram for the couples group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis. Compare the two graphs: List two similarities between the graphs. List two differences between the graphs. Overall, are the graphs more similar or different? Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x -axis by $50, scale it by $100. Use relative frequency on the y -axis. Compare the graph for the singles with the new graph for the couples: List two similarities between the graphs. Overall, are the graphs more similar or different? How did scaling the couples graph differently change the way you compared it to the singles graph? Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences. Singles Amount($) Frequency Relative Frequency 51–100 5 0.08 101–150 10 0.17 151–200 15 0.25 201–250 15 0.25 251–300 10 0.17 301–350 5 0.08 Couples Amount($) Frequency Relative Frequency 100–150 5 0.07 201–250 5 0.07 251–300 5 0.07 301–350 5 0.07 351–400 10 0.14 401–450 10 0.14 451–500 10 0.14 501–550 10 0.14 551–600 5 0.07 601–650 5 0.07 See and . In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included). In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included). Compare the two graphs: Answers may vary. Possible answers include: Both graphs have a single peak. Both graphs use class intervals with width equal to $50. Answers may vary. Possible answers include: The couples graph has a class interval with no values. It takes almost twice as many class intervals to display the data for couples. Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same. Answers may vary. Compare the graph for the Singles with the new graph for the Couples: Both graphs have a single peak. Both graphs display 6 class intervals. Both graphs show the same general pattern. Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different. Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison. Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows. # of movies Frequency Relative Frequency Cumulative Relative Frequency 0 5 1 9 2 6 3 4 4 1 Construct a histogram of the data. Complete the columns of the chart. Use the following information to answer the next two exercises: Suppose 111 people who shopped in a special T-shirt store were asked the number of T-shirts they own costing more than $19 each. The percentage of people who own at most three T-shirts costing more than $19 each is approximately: 21 59 41 Cannot be determined c If the data were collected by asking the first 111 people who entered the store, then the type of sampling is: cluster simple random stratified convenience Following are the rates of unmarried adults for the 50 U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Construct a bar graph of the unmarried adult rates of your state and the four states closest to your state. Hint: Label the x -axis with the states. Answers will vary. Frequency the number of times a value of the data occurs Histogram a graphical representation in x - y form of the distribution of data in a data set; x represents the data and y represents the frequency, or relative frequency. The graph consists of contiguous rectangles. Relative Frequency the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes", "section": "Histograms, Frequency Polygons, and Time Series Graphs", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Measures of the Location of the Data The common measures of location are quartiles and percentiles . Quartiles divide an ordered data set into four equal parts. The three quartiles of a data set are labeled as Q 1 , Q 2 , and Q 3 . About one-fourth of the data falls on or below the first quartile Q 1 . About one-half of the data falls on or below the second quartile Q 2 . About three-fourths of the data falls on or below the third quartile Q 3 . In the same way, percentiles divide a data set into 100 equal parts. To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90 th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score. Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75 th percentile. That translates into a score of at least 1220. Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant. The median is a number that measures the \"center\" of the data. You can think of the median as the \"middle value,\" but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data. 1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1 Ordered from smallest to largest: 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5 Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two. 6.8 + 7.2 = 14 14 ÷ 2 = 7 The median is seven. Half of the values are smaller than seven and half of the values are larger than seven. Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q 1 , is the middle value of the lower half of the data, and the third quartile, Q 3 , is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set: 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5 The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two. 1; 1; 2; 2; 4; 6; 6.8 The number two, which is part of the data, is the first quartile . One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two. The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine. The third quartile , Q 3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example. The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile ( Q 3 ) and the first quartile ( Q 1 ). IQR = Q 3 – Q 1 The IQR can help to determine potential outliers . A value is suspected to be a potential outlier if it is less than (1.5)( IQR ) below the first quartile or more than (1.5)( IQR ) above the third quartile . Potential outliers always require further investigation. NOTE A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data. For the following 13 real estate prices, calculate the IQR and determine if any prices are potential outliers. Prices are in dollars. 389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000 Order the data from smallest to largest. 114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000 M = 488,800 Q 1 = 230,500 + 387,000 2 = 308,750 Q 3 = 639,000 + 659,000 2 = 649,000 IQR = 649,000 – 308,750 = 340,250 (1.5)( IQR ) = (1.5)(340,250) = 510,375 Q 1 – (1.5)( IQR ) = 308,750 – 510,375 = –201,625 Q 3 + (1.5)( IQR ) = 649,000 + 510,375 = 1,159,375 No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier . Try It For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars. $33,000 $64,500 $28,000 $54,000 $72,000 $68,500 $69,000 $42,000 $54,000 $120,000 $40,500 Test scores for a college statistics class held during the day are: 99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90 Test scores for a college statistics class held during the evening are: 98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5 For the two data sets, find the following: The interquartile range. Compare the two interquartile ranges. Any outliers in either set. The five number summary for the day and night classes is Minimum Q 1 Median Q 3 Maximum Day 32 56 74.5 82.5 99 Night 25.5 78 81 89 98 The IQR for the day group is Q 3 – Q 1 = 82.5 – 56 = 26.5 The IQR for the night group is Q 3 – Q 1 = 89 – 78 = 11 The interquartile range (the spread or variability) for the day class is larger than the night class IQR . This suggests more variation will be found in the day class’s class test scores. Day class outliers are found using the IQR times 1.5 rule. So, Q 1 - IQR (1.5) = 56 – 26.5(1.5) = 16.25 Q 3 + IQR (1.5) = 82.5 + 26.5(1.5) = 122.25 Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers. Night class outliers are calculated as: Q 1 – IQR (1.5) = 78 – 11(1.5) = 61.5 Q 3 + IQR(1.5) = 89 + 11(1.5) = 105.5 For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier. Try It Find the interquartile range for the following two data sets and compare them. Test Scores for Class A 69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94 Test Scores for Class B 90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100 Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were: AMOUNT OF SLEEP PER SCHOOL NIGHT (HOURS) FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 4 2 0.04 0.04 5 5 0.10 0.14 6 7 0.14 0.28 7 12 0.24 0.52 8 14 0.28 0.80 9 7 0.14 0.94 10 3 0.06 1.00 Find the 28 th percentile . Notice the 0.28 in the \"cumulative relative frequency\" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28 th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28 th percentile is between the last six and the first seven. The 28 th percentile is 6.5. Find the median . Look again at the \"cumulative relative frequency\" column and find 0.52. The median is the 50 th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50 th percentile is between the 25 th , or seven, and 26 th , or seven, values. The median is seven. Find the third quartile . The third quartile is the same as the 75 th percentile. You can \"eyeball\" this answer. If you look at the \"cumulative relative frequency\" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75 th percentile, then, must be an eight . Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q 3 , is the 38 th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.) Try it Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65 th percentile. Amount of time spent on route (hours) Frequency Relative Frequency Cumulative Relative Frequency 2 12 0.30 0.30 3 14 0.35 0.65 4 10 0.25 0.90 5 4 0.10 1.00 A Formula for Finding the k th Percentile If you were to do a little research, you would find several formulas for calculating the k th percentile. Here is one of them. k = the k th percentile. It may or may not be part of the data. i = the index (ranking or position of a data value) n = the total number of data Order the data from smallest to largest. Calculate i = k 100 ( n + 1 ) If i is an integer, then the k th percentile is the data value in the i th position in the ordered set of data. If i is not an integer, then round i up and round i down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example. Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the 70 th percentile. Find the 83 rd percentile. k = 70 i = the index n = 29 i = k 100 ( n + 1) = ( 70 100 )(29 + 1) = 21. Twenty-one is an integer, and the data value in the 21 st position in the ordered data set is 64. The 70 th percentile is 64 years. k = 83 rd percentile i = the index n = 29 i = k 100 ( n + 1) = ( 83 100 )(29 + 1) = 24.9, which is NOT an integer. Round it down to 24 and up to 25. The age in the 24 th position is 71 and the age in the 25 th position is 72. Average 71 and 72. The 83 rd percentile is 71.5 years. Try It Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Calculate the 20 th percentile and the 55 th percentile. NOTE You can calculate percentiles using calculators and computers. There are a variety of online calculators. Using : Find the 80 th percentile. Find the 90 th percentile. Find the first quartile. What is another name for the first quartile? Using the data from the frequency table, we have: Notice there are 50 data values in the table, so n = 50. Calculate the index i as follows: i = 80 100 ( 50 + 1 ) = 40 . 8 Since i = 40.8, calculate the mean of the 40 th and 41 st data values. The 40 th data value is 8, the 41 st data value is 9, and the mean of these two data values is 8.5. Thus, the 80 th percentile is 8.5. Calculate the index i as follows: i = 90 100 ( 50 + 1 ) = 45 . 9 Since i = 45.9, calculate the mean of the 45 th and 46 th data values. The 45 th data value is 9, the 46 th data value is 9, and the mean of these two data values is 9. Thus, the 90 th percentile is 9. Another name for the first quartile is the 25 th percentile. Proceed to calculate the 25 th percentile: Calculate the index i as follows: i = 25 100 ( 50 + 1 ) = 12 . 75 Since i = 12.75, calculate the mean of the 12 th and 13 th data values. The 12 th data value is 6, the 13 th data value is 6, and the mean of these two data values is 6. Thus, the first quartile is 6. Try It Refer to the . Find the third quartile. What is another name for the third quartile? Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions: How many students were surveyed? What kind of sampling did you do? Construct two different histograms. For each, starting value = _____ ending value = ____. Find the median, first quartile, and third quartile. Construct a table of the data to find the following: the 10 th percentile the 70 th percentile the percent of students who own less than four sweaters A Formula for Finding the Percentile of a Value in a Data Set Order the data from smallest to largest. x = the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile. y = the number of data values equal to the data value for which you want to find the percentile. n = the total number of data. Calculate x + 0.5 y n (100). Then round to the nearest integer. Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentile for 58. Find the percentile for 25. Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58. x = 18 and y = 1. x + 0.5 y n (100) = 18 + 0.5 ( 1 ) 29 (100) = 63.80. 58 is the 64 th percentile. Counting from the bottom of the list, there are three data values less than 25. There is one value of 25. x = 3 and y = 1. x + 0.5 y n (100) = 3 + 0.5 ( 1 ) 29 (100) = 12.07. Twenty-five is the 12 th percentile. Try It Listed are 30 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentiles for 47 and 31. Interpreting Percentiles, Quartiles, and Median A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15 th percentile. Low percentiles always correspond to lower data values. High percentiles always correspond to higher data values. A percentile may or may not correspond to a value judgment about whether it is \"good\" or \"bad.\" The interpretation of whether a certain percentile is \"good\" or \"bad\" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered \"good;\" in other contexts a high percentile might be considered \"good\". In many situations, there is no value judgment that applies. Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text. NOTE When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information. information about the context of the situation being considered the data value (value of the variable) that represents the percentile the percent of individuals or items with data values below the percentile the percent of individuals or items with data values above the percentile. On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation. Twenty-five percent of students finished the exam in 35 minutes or less. Seventy-five percent of students finished the exam in 35 minutes or more. A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.) Try It For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation. On a 20 question math test, the 70 th percentile for number of correct answers was 16. Interpret the 70 th percentile in the context of this situation. Try It On a 60 point written assignment, the 80 th percentile for the number of points earned was 49. Interpret the 80 th percentile in the context of this situation. At a community college, it was found that the 30 th percentile of credit units that students are enrolled for is seven units. Interpret the 30 th percentile in the context of this situation. Try It During a season, the 40 th percentile for points scored per player in a game is eight. Interpret the 40 th percentile in the context of this situation. Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown. 0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes 10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes; 30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes Determine the following five values. Min = 0 Q 1 = 20 Med = 40 Q 3 = 60 Max = 300 If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment. However, the principal needs to be careful. The value 300 appears to be a potential outlier. Q 3 + 1.5( IQR ) = 60 + (1.5)(40) = 120. The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values: Min = 0 Q 1 = 20 Q 3 = 60 Max = 120 We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results. Try It A college statistics instructor is investigating the amount of time students spend working on a final project in the course. The instructor would like students to spend approximately 3 to 4 hours as the typical amount of time to be spent on the project. The instructor collects data from a random sample of 10 students for the number of hours spent working on the final project. The results obtained are as follows: 2 hours; 3 hours; 5 hours; 3 hours; 4 hours; 4 hours; 3 hours; 11 hours; 3 hours; 2 hours. Determine the following five values: Min , Q 1 , Med , Q 3 , Max . Should the instructor modify the final project or leave as is? The five values are obtained as follows: Min = 2 Q 1 = 3 Med = 3 Q 3 = 4 Max = 11 Given that the 3 rd quartile is 4 hours, this indicates that 75% of the students spend 4 hours or less on the project. Also for the interquartile range: IQR = 4 – 3 = 1, we know that half the students spent between 3 and 4 hours on the project. This seems like a reasonable amount of time for students to spend on the final project. However, the value 11 hours appears to be a potential outlier. Q 3 + 1 . 5 ( I Q R ) = 4 + 1 . 5 ( 1 ) = 5 . 5 The value 11 is greater than 5.5, so it is an outlier. If we delete it and recalculate the five values, we get the following values: Min = 2 Q 1 = 2.5 Q 3 = 4 Max = 5 We still have 75% of students spending 4 hours or less on the project. Half of the students spend between 2.5 hours and 4 hours on the project. However, 10 students may be a relatively small sample size. The instructor might consider a larger sample size. References Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at http://usatoday30.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1 (accessed April 3, 2013). Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013). “1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013). Data from San Jose Mercury News . Data from Time Magazine ; survey by Yankelovich Partners, Inc. Chapter Review The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50 th percentile would be greater than 50 percent of the other observations in the set. Quartiles divide data into quarters. The first quartile ( Q 1 ) is the 25 th percentile,the second quartile ( Q 2 or median) is 50 th percentile, and the third quartile ( Q 3 ) is the 75 th percentile. The interquartile range, or IQR , is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q 1 from Q 3 , and can help determine outliers by using the following two expressions. Q 3 + IQR (1.5) Q 1 – IQR (1.5) Formula Review i = ( k 100 ) ( n + 1 ) where i = the ranking or position of a data value, k = the kth percentile, n = total number of data. Expression for finding the percentile of a data value: ( x + 0.5 y n ) (100) where x = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile, y = the number of data values equal to the data value for which you want to find the percentile, n = total number of data Listed are 29 ages for Academy Award winning best actors in order from smallest to largest. 18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the 40 th percentile. Find the 78 th percentile. The 40 th percentile is 37 years. The 78 th percentile is 70 years. Listed are 32 ages for Academy Award winning best actors in order from smallest to largest. 18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 Find the percentile of 37. Find the percentile of 72. Jesse was ranked 37 th in his graduating class of 180 students. At what percentile is Jesse’s ranking? Jesse graduated 37 th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37. x = 143 and y = 1. x + 0.5 y n (100) = 143 + 0.5 ( 1 ) 180 (100) = 79.72. Jesse’s rank of 37 puts him at the 80 th percentile. For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race? The 20 th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20 th percentile in the context of the situation. A bicyclist in the 90 th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90 th percentile in the context of the situation. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race? The 40 th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40 th percentile in the context of the situation. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed which is faster. 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster). On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain. Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85 th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85 th percentile in the context of this situation. When waiting in line at the DMV, the 85 th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer. In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78 th percentile. Should Li be pleased or upset by this result? Explain. In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1,700 in damage and was in the 90 th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90 th percentile in the context of this problem. The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of $1700 or more. The University of Wisconsin has two criteria used to set admission standards for students to be admitted to a college in the UW system: Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an \"admissions index\" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent? Students whose GPAs are at or above the 96 th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percentage of students from each high school are \"eligible in the local context\"? Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34 th percentile. The 34 th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses? You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more. Use the following information to answer the next six exercises. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. First quartile = _______ Second quartile = median = 50 th percentile = _______ 4 Third quartile = _______ Interquartile range ( IQR ) = _____ – _____ = _____ 6 – 4 = 2 10 th percentile = _______ 70 th percentile = _______ 6 Homework The median age for Black people in the U.S. is 30.9 years; for U.S. White people it is 42.3 years. Based upon this information, give two reasons why the median age for Black people could be lower than the median age for White people. Does the lower median age for Black people necessarily mean that Black people die younger than White people? Why or why not? How might it be possible for Black people and White people to die at approximately the same age, but for the median age for White people to be higher? Six hundred adult Americans were asked by telephone poll, \"What do you think constitutes a middle-class income?\" The results are in . Also, include left endpoint, but not the right endpoint. Salary ($) Relative Frequency < 20,000 0.02 20,000–25,000 0.09 25,000–30,000 0.19 30,000–40,000 0.26 40,000–50,000 0.18 50,000–75,000 0.17 75,000–99,999 0.02 100,000+ 0.01 What percentage of the survey answered \"not sure\"? What percentage think that middle-class is from $25,000 to $50,000? Construct a histogram of the data. Should all bars have the same width, based on the data? Why or why not? How should the <20,000 and the 100,000+ intervals be handled? Why? Find the 40 th and 80 th percentiles Construct a bar graph of the data 1 – (0.02+0.09+0.19+0.26+0.18+0.17+0.02+0.01) = 0.06 0.19+0.26+0.18 = 0.63 Answers may vary. 40 th percentile will fall between 30,000 and 40,000 80 th percentile will fall between 50,000 and 75,000 Answers may vary. Given the following box plot: which quarter has the smallest spread of data? What is that spread? which quarter has the largest spread of data? What is that spread? find the interquartile range ( IQR ). are there more data in the interval 5–10 or in the interval 10–13? How do you know this? which interval has the fewest data in it? How do you know this? 0–2 2–4 10–12 12–13 need more information The following box plot shows the U.S. population for a certain year. Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know? 12.6% are age 65 and over. Approximately what percentage of the population are working age adults (above age 17 to age 65)? more children; the left whisker shows that 25% of the population are children 17 and younger. The right whisker shows that 25% of the population are adults 50 and older, so adults 65 and over represent less than 25%. 62.4% Interquartile Range or IQR , is the range of the middle 50 percent of the data values; the IQR is found by subtracting the first quartile from the third quartile. Outlier an observation that does not fit the rest of the data Percentile a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50 th percentile. The first and third quartiles are the 25 th and the 75 th percentiles, respectively. Quartiles the numbers that separate the data into quarters; quartiles may or may not be part of the data. The second quartile is the median of the data.", "section": "Measures of the Location of the Data", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Box Plots Box plots (also called box-and-whisker plots or box-whisker plots ) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them. To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The \"whiskers\" extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data. NOTE You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values. Consider, again, this dataset. 1 1 2 2 4 6 6.8 7.2 8 8.3 9 10 10 11.5 The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot. NOTE See the calculator instructions on the TI web site or in the appendix. The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line. NOTE It is important to start a box plot with a scaled number line . Otherwise the box plot may not be useful. The following data are the heights of 40 students in a statistics class. 59 60 61 62 62 63 63 64 64 64 65 65 65 65 65 65 65 65 65 66 66 67 67 68 68 69 70 70 70 70 70 71 71 72 72 73 74 74 75 77 Construct a box plot with the following properties; the calculator intructions for the minimum and maximum values as well as the quartiles follow the example. Minimum value = 59 Maximum value = 77 Q 1: First quartile = 64.5 Q 2: Second quartile or median= 66 Q 3: Third quartile = 70 Each quarter has approximately 25% of the data. The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread. Range = maximum value – the minimum value = 77 – 59 = 18 Interquartile Range: IQR = Q 3 – Q 1 = 70 – 64.5 = 5.5. The interval 59–65 has more than 25% of the data so it has more data in it than the interval 66 through 70 which has 25% of the data. The middle 50% (middle half) of the data has a range of 5.5 inches. To find the minimum, maximum, and quartiles: Enter data into the list editor (Pres STAT 1:EDIT ). If you need to clear the list, arrow up to the name L1 , press CLEAR, and then arrow down. Put the data values into the list L1 . Press STAT and arrow to CALC. Press 1:1-VarStats . Enter L1 . Press ENTER. Use the down and up arrow keys to scroll. Smallest value = 59. Largest value = 77. Q 1 : First quartile = 64.5. Q 2 : Second quartile or median = 66. Q 3 : Third quartile = 70. To construct the box plot: Press 4:Plotsoff . Press ENTER. Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER. Arrow down to Xlist: Press 2nd 1 for L1 Arrow down to Freq: Press ALPHA . Press 1. Press Zoom. Press 9: ZoomStat . Press TRACE, and use the arrow keys to examine the box plot. Try It The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator, and state the interquartile range. 136 140 178 190 205 215 217 218 232 234 240 255 270 275 290 301 303 315 317 318 326 333 343 349 360 369 377 388 391 392 398 400 402 405 408 422 429 450 475 512 For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like: In this case, at least 25% of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive. Test scores for a college statistics class held during the day are: 99 56 78 55.5 32 90 80 81 56 59 45 77 84.5 84 70 72 68 32 79 90 Test scores for a college statistics class held during the evening are: 98 78 68 83 81 89 88 76 65 45 98 90 80 84.5 85 79 78 98 90 79 81 25.5 Find the smallest and largest values, the median, and the first and third quartile for the day class. Find the smallest and largest values, the median, and the first and third quartile for the night class. For each data set, what percentage of the data is between the smallest value and the first quartile? the first quartile and the median? the median and the third quartile? the third quartile and the largest value? What percentage of the data is between the first quartile and the largest value? Create a box plot for each set of data. Use one number line for both box plots. Which box plot has the widest spread for the middle 50% of the data (the data between the first and third quartiles)? What does this mean for that set of data in comparison to the other set of data? Min = 32 Q 1 = 56 M = 74.5 Q 3 = 82.5 Max = 99 Min = 25.5 Q 1 = 78 M = 81 Q 3 = 89 Max = 98 Day class: There are six data values ranging from 32 to 56: 30%. There are six data values ranging from 56 to 74.5: 30%. There are five data values ranging from 74.5 to 82.5: 25%. There are five data values ranging from 82.5 to 99: 25%. There are 16 data values between the first quartile, 56, and the largest value, 99: 75%. Night class: The first data set has the wider spread for the middle 50% of the data. The IQR for the first data set is greater than the IQR for the second set. This means that there is more variability in the middle 50% of the first data set. Try It The following data set shows the heights in inches for the boys in a class of 40 students. 66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74 The following data set shows the heights in inches for the girls in a class of 40 students. 61; 61; 62; 62; 63; 63; 63; 65; 65; 65; 66; 66; 66; 67; 68; 68; 68; 69; 69; 69 Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50% of the data. Graph a box-and-whisker plot for the data values shown. 10 10 10 15 35 75 90 95 100 175 420 490 515 515 790 The five numbers used to create a box-and-whisker plot are: Min: 10 Q 1 : 15 Med: 95 Q 3 : 490 Max: 790 The following graph shows the box-and-whisker plot. Try It Follow the steps you used to graph a box-and-whisker plot for the data values shown. 0 5 5 15 30 30 45 50 50 60 75 110 140 240 330 References Data from West Magazine . Chapter Review Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data. Use the following information to answer the next two exercises. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Construct a box plot below. Use a ruler to measure and scale accurately. Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas, but not in others? How can you tell? More than 25% of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25% and the bottom 25% are spread out evenly; the whiskers have the same length. Homework In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected. Have more U.S. citizens or more Germans surveyed been to over eight foreign countries? Compare the three box plots. What do they imply about the foreign travel of 20-year-old residents of the three countries when compared to each other? Given the following box plot, answer the questions. Think of an example (in words) where the data might fit into the above box plot. In 2–5 sentences, write down the example. What does it mean to have the first and second quartiles so close together, while the second to third quartiles are far apart? Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year. Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile. Given the following box plots, answer the questions. In complete sentences, explain why each statement is false. Data 1 has more data values above two than Data 2 has above two. The data sets cannot have the same mode. For Data 1 , there are more data values below four than there are above four. For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why in complete sentences. A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series. Which group is most likely to have an outlier? Explain how you determined that. Compare the three box plots. What do they imply about the age of purchasing a BMW from the series when compared to each other? Look at the BMW 5 series. Which quarter has the smallest spread of data? What is the spread? Look at the BMW 5 series. Which quarter has the largest spread of data? What is the spread? Look at the BMW 5 series. Estimate the interquartile range (IQR). Look at the BMW 5 series. Are there more data in the interval 31 to 38 or in the interval 45 to 55? How do you know this? Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this? 31–35 38–41 41–64 Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top 50% of buyers are more variable than the ages of the lower 50%. The BMW 3 series is most likely to have an outlier. It has the longest whisker. Comparing the median ages, younger people tend to buy the BMW 3 series, while older people tend to buy the BMW 7 series. However, this is not a rule, because there is so much variability in each data set. The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median. The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile. IQR ~ 17 years There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is concentrated. The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25% fall between 41 and 64. Since 25% of values fall between 31 and 38, we know that fewer than 25% fall between 31 and 35. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: # of movies Frequency 0 5 1 9 2 6 3 4 4 1 Construct a box plot of the data. Bringing It Together A certain small town in the United states has a population of 27,873 people. Their ages are as follows: Age Group Percent of Community 0–17 18.9 18–24 8.0 25–34 22.8 35–44 15.0 45–54 13.1 55–64 11.9 65+ 10.3 Construct a histogram of the age distribution for this small town. The bars will not be the same width for this example. Why not? What impact does this have on the reliability of the graph? What percentage of the community is under age 35? Which box plot most resembles the information above? For graph, answers may vary. 49.7% of the community is under the age of 35. Based on the information in the table, graph (a) most closely represents the data. Box plot a graph that gives a quick picture of the middle 50% of the data First Quartile the value that is the median of the of the lower half of the ordered data set Frequency Polygon looks like a line graph but uses intervals to display ranges of large amounts of data Interval also called a class interval; an interval represents a range of data and is used when displaying large data sets Paired Data Set two data sets that have a one to one relationship so that: both data sets are the same size, and each data point in one data set is matched with exactly one point from the other set. Skewed used to describe data that is not symmetrical; when the right side of a graph looks “chopped off” compared the left side, we say it is “skewed to the left.” When the left side of the graph looks “chopped off” compared to the right side, we say the data is “skewed to the right.” Alternatively: when the lower values of the data are more spread out, we say the data are skewed to the left. When the greater values are more spread out, the data are skewed to the right.", "section": "Box Plots", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Measures of the Center of the Data The \"center\" of a data set is also a way of describing location. The two most widely used measures of the \"center\" of the data are the mean (average) and the median . To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center. NOTE The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average\" is commonly accepted for “arithmetic mean.” When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an x with a bar over it (pronounced “ x bar”): x – . The Greek letter μ (pronounced \"mew\") represents the population mean . One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random. To see that both ways of calculating the mean are the same, consider the sample: 1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4 x ¯ = 1 + 1 + 1 + 2 + 2 + 3 + 4 + 4 + 4 + 4 + 4 11 = 2.7 x ¯ = 3 ( 1 ) + 2 ( 2 ) + 1 ( 3 ) + 5 ( 4 ) 11 = 2.7 In the second calculation, the frequencies are 3, 2, 1, and 5. You can quickly find the location of the median by using the expression n + 1 2 . The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is 97, then n + 1 2 = 97 + 1 2 = 49. The median is the 49 th value in the ordered data. If the total number of data values is 100, then n + 1 2 = 100 + 1 2 = 50.5. The median occurs midway between the 50 th and 51 st values. The location of the median and the value of the median are not the same. The upper case letter M is often used to represent the median. The next example illustrates the location of the median and the value of the median. A hospital administrator keeps track of the ages (in years) of patients visiting the emergency room over a one-week period (data are sorted from smallest to largest): 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47; Calculate the mean and the median. The calculation for the mean is: x ¯ = [ 3 + 4 + ( 8 ) ( 2 ) + 10 + 11 + 12 + 13 + 14 + ( 15 ) ( 2 ) + ( 16 ) ( 2 ) + ... + 35 + 37 + 40 + ( 44 ) ( 2 ) + 47 ] 40 = 23.6 To find the median, M , first use the formula for the location. The location is: n + 1 2 = 40 + 1 2 = 20.5 Starting at the smallest value, the median is located between the 20 th and 21 st values (the two 24s): 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47; M = 24 + 24 2 = 24 To find the mean and the median: Clear list L1. Pres STAT 4:ClrList . Enter 2nd 1 for list L1. Press ENTER. Enter data into the list editor. Press STAT 1:EDIT . Put the data values into list L1. Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER. Press the down and up arrow keys to scroll. x ¯ = 23.6, M = 24 Try It The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median. 3 4 5 7 7 7 7 8 8 9 9 10 10 10 10 10 11 12 12 13 14 14 15 15 17 17 18 19 19 19 21 21 22 22 23 24 24 24 24 Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the \"center\": the mean or the median? x ¯ = 5 , 000 , 000 + 49 ( 30 , 000 ) 50 = 129,400 M = 30,000 (There are 49 people who earn $30,000 and one person who earns $5,000,000.) The median is a better measure of the \"center\" than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data. Try It In a sample of 60 households, one house is worth $2,500,000. Twenty-nine houses are worth $280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median? Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal. Statistics exam scores for 20 students are as follows: 50 53 59 59 63 63 72 72 72 72 72 76 78 81 83 84 84 84 90 93 Find the mode. The most frequent score is 72, which occurs five times. Mode = 72. Try It The number of books checked out from the library from 25 students are as follows: 0 0 0 1 2 3 3 4 4 5 5 7 7 7 7 8 8 8 9 10 10 11 11 12 12 Find the mode. Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice. When is the mode the best measure of the \"center\"? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing. NOTE The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red. Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software. Try It Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is $50,000 and the mean is $47,500. What would be the best measure of the “center”? The Law of Large Numbers and the Mean The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean x ¯ of the sample is very likely to get closer and closer to µ . This is discussed in more detail later in the text. Sampling Distributions and Statistic of a Sampling Distribution You can think of a sampling distribution as a relative frequency distribution with a great many samples. (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below. # of movies Relative Frequency 0 5 30 1 15 30 2 6 30 3 3 30 4 1 30 If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution . A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean x ¯ is an example of a statistic which estimates the population mean μ . Calculating the Mean of Grouped Frequency Tables When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean: mean = data sum number of data values We simply need to modify the definition to fit within the restrictions of a frequency table. Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is lower boundary + upper boundary 2 . We can now modify the mean definition to be Mean of Frequency Table = ∑ f m ∑ f where f = the frequency of the interval and m = the midpoint of the interval. A frequency table displaying professor Blount’s last statistic test is shown. Find the best estimate of the class mean. Grade Interval Number of Students 50–56.5 1 56.5–62.5 0 62.5–68.5 4 68.5–74.5 4 74.5–80.5 2 80.5–86.5 3 86.5–92.5 4 92.5–98.5 1 Find the midpoints for all intervals Grade Interval Midpoint 50–56.5 53.25 56.5–62.5 59.5 62.5–68.5 65.5 68.5–74.5 71.5 74.5–80.5 77.5 80.5–86.5 83.5 86.5–92.5 89.5 92.5–98.5 95.5 Calculate the sum of the product of each interval frequency and midpoint. ∑ ​ f m 53.25 ( 1 ) + 59.5 ( 0 ) + 65.5 ( 4 ) + 71.5 ( 4 ) + 77.5 ( 2 ) + 83.5 ( 3 ) + 89.5 ( 4 ) + 95.5 ( 1 ) = 1460.25 μ = ∑ f m ∑ f = 1460.25 19 = 76.86 Try It A researcher conducted a study on the effect that playing video games has on memory recall. As part of the study, they compiled the following data: Hours Teenagers Spend on Video Games Number of Teenagers 0–3.5 3 3.5–7.5 7 7.5–11.5 12 11.5–15.5 7 15.5–19.5 9 What is the best estimate for the mean number of hours spent playing video games? References Data from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013). “Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013). Chapter Review The mean and the median can be calculated to help you find the \"center\" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set. Formula Review μ = ∑ f m ∑ f Where f = interval frequencies and m = interval midpoints. Find the mean for the following frequency tables. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Daily Low Temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 Points per Game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2 Use the following information to answer the next three exercises: The following data show the lengths of boats moored in a marina. The data are ordered from smallest to largest: 16 17 19 20 20 21 23 24 25 25 25 26 26 27 27 27 28 29 30 32 33 33 34 35 37 39 40 Calculate the mean. Mean: 16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 + 34 + 35 + 37 + 39 + 40 = 738; 738 27 = 27.33 Identify the median. Identify the mode. The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27 Use the following information to answer the next three exercises: Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Calculate the following: sample mean = x ¯ = _______ median = _______ 4 mode = _______ Homework The countries with the highest rates of obesity in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table. Percent of Population Number of Countries 11.4–20.45 29 20.45–29.45 13 29.45–38.45 4 38.45–47.45 0 47.45–56.45 2 56.45–65.45 1 65.45–74.45 0 74.45–83.45 1 What is the best estimate of the average obesity percentage for these countries? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How does the United States compare to other countries? gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children? Percent of Underweight Children Number of Countries 16–21.45 23 21.45–26.9 4 26.9–32.35 9 32.35–37.8 7 37.8–43.25 6 43.25–48.7 1 The mean percentage, x ¯ = 1328.65 50 = 26.57 Bringing It Together Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information. Javier Ercilia x ¯ 6.0 miles 6.0 miles s 4.0 miles 7.0 miles How can you determine which survey was correct ? Explain what the difference in the results of the surveys implies about the data. If the two histograms depict the distribution of values for each supervisor, which one depicts Ercilia's sample? How do you know? If the two box plots depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know? Use the following information to answer the next three exercises : We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section. Number of years Frequency Number of years Frequency 7 1 22 1 14 3 23 1 15 1 26 1 18 1 40 2 19 4 42 2 20 3 Total = 20 What is the IQR ? 8 11 15 35 a What is the mode? 19 19.5 14 and 20 22.65 Is this a sample or the entire population? sample entire population neither b Frequency Table a data representation in which grouped data is displayed along with the corresponding frequencies Mean a number that measures the central tendency of the data; a common name for mean is 'average.' The term 'mean' is a shortened form of 'arithmetic mean.' By definition, the mean for a sample (denoted by x ¯ ) is x ¯ = Sum of all values in the sample Number of values in the sample , and the mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population . Median a number that separates ordered data into halves; half the values are the same number or smaller than the median and half the values are the same number or larger than the median. The median may or may not be part of the data. Midpoint the mean of an interval in a frequency table Mode the value that appears most frequently in a set of data", "section": "Measures of the Center of the Data", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Skewness and the Mean, Median, and Mode Consider the following data set. 4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10 This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval. The histogram (shown in ) displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. The histogram for the data: 4 5 6 6 6 7 7 7 7 8 (shown in ) is not symmetrical. The right-hand side seems \"chopped off\" compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left. The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so. The histogram for the data: 6 7 7 7 7 8 8 8 9 10 , is also not symmetrical. It is skewed to the right . The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest . Again, the mean reflects the skewing the most. The mean is affected by outliers that do not influence the median. Therefore, when the distribution of data is skewed to the left, the mean is often less than the median. When the distribution is skewed to the right, the mean is often greater than the median. In symmetric distributions, we expect the mean and median to be approximately equal in value. This is an important connection between the shape of the distribution and the relationship of the mean and median. It is not, however, true for every data set. The most common exceptions occur in sets of discrete data. Skewness and symmetry become important when we discuss probability distributions in later chapters. Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors. Terry: 7; 9; 3; 3; 3; 4; 1; 3; 2; 2 Davis: 3; 3; 3; 4; 1; 4; 3; 2; 3; 1 Maris: 2; 3; 4; 4; 4; 6; 6; 6; 8; 3 Make a dot plot for the three authors and compare the shapes. Calculate the mean for each. Calculate the median for each. Describe any pattern you notice between the shape and the measures of center. Terry’s distribution has a right (positive) skew. Davis’ distribution has a left (negative) skew Maris’ distribution is symmetrically shaped. Terry’s mean is 3.7, Davis’ mean is 2.7, Maris’ mean is 4.6. Terry’s median is three, Davis’ median is three. Maris’ median is four. It appears that the median is always closest to the high point (the mode), while the mean tends to be farther out on the tail. In a symmetrical distribution, the mean and the median are both centrally located close to the high point of the distribution. Try It Discuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center? a. b. The Ages Former U.S Presidents Died 4 6 9 5 3 6 7 7 7 8 6 0 0 3 3 4 4 5 6 7 7 7 8 7 0 1 1 2 3 4 7 8 8 9 8 0 1 3 5 8 9 0 0 3 3 Key: 8|0 means 80. c. Chapter Review Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions . A left (or negative) skewed distribution has a shape like . A right (or positive) skewed distribution has a shape like . A symmetrical distribution looks like . Use the following information to answer the next three exercises: State whether the data are symmetrical, skewed to the left, or skewed to the right. 1 1 1 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 5 5 The data are symmetrical. The median is 3 and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical. 16 17 19 22 22 22 22 22 23 87 87 87 87 87 88 89 89 90 91 The data are skewed right. The median is 87.5 and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right. When the data are skewed left, what is the typical relationship between the mean and median? When the data are symmetrical, what is the typical relationship between the mean and median? When the data are symmetrical, the mean and median are close or the same. What word describes a distribution that has two modes? Describe the shape of this distribution. The distribution is skewed right because it looks pulled out to the right. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean is 4.1 and is slightly greater than the median, which is four. Describe the shape of this distribution. Describe the relationship between the mode and the median of this distribution. The mode and the median are the same. In this case, they are both five. Are the mean and the median the exact same in this distribution? Why or why not? Describe the shape of this distribution. The distribution is skewed left because it looks pulled out to the left. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean and the median are both six. The mean and median for the data are the same. 3 4 5 5 6 6 6 6 7 7 7 7 7 7 7 Is the data perfectly symmetrical? Why or why not? Which is the greatest, the mean, the mode, or the median of the data set? 11 11 12 12 12 12 13 15 17 22 22 22 The mode is 12, the median is 12.5, and the mean is 15.1. The mean is the largest. Which is the least, the mean, the mode, and the median of the data set? 56 56 56 58 59 60 62 64 64 65 67 Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why? The mean tends to reflect skewing the most because it is affected the most by outliers. In a perfectly symmetrical distribution, when would the mode be different from the mean and median? Homework The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years. What does it mean for the median age to rise? Give two reasons why the median age could rise. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?", "section": "Skewness and the Mean, Median, and Mode", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Measures of the Spread of the Data An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean. The standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean. The standard deviation provides a measure of the overall variation in a data set The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation. Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B . The average wait time at both supermarkets is five minutes. At supermarket A , the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes. Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B . Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average. The standard deviation can be used to determine whether a data value is close to or far from the mean. Suppose that Rosa and Binh both shop at supermarket A . Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A , the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean. Rosa waits for seven minutes: Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation. Rosa's wait time of seven minutes is two minutes longer than the average of five minutes. Rosa's wait time of seven minutes is one standard deviation above the average of five minutes. Binh waits for one minute. One is four minutes less than the average of five; four minutes is equal to two standard deviations. Binh's wait time of one minute is four minutes less than the average of five minutes. Binh's wait time of one minute is two standard deviations below the average of five minutes. A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate \"rule of thumb\" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.) The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because 5 + (1)(2) = 7. If one were also part of the data set, then one is two standard deviations to the left of five because 5 + (–2)(2) = 1. In general, a value = mean + (#ofSTDEV)(standard deviation) where #ofSTDEVs = the number of standard deviations #ofSTDEV does not need to be an integer One is two standard deviations less than the mean of five because: 1 = 5 + (–2)(2). The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population. sample: x = x ¯ + ( # o f S T D E V ) ( s ) Population: x = μ + ( # o f S T D E V ) ( σ ) The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. The symbol x ¯ is the sample mean and the Greek symbol μ is the population mean. Calculating the Standard Deviation If x is a number, then the difference \" x – mean\" is called its deviation . In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is x – μ . For sample data, in symbols a deviation is x – x ¯ . The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ . To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – x ¯ values for a sample, or the x – μ values for a population). The symbol σ 2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s 2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations. If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N , the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1 , one less than the number of items in the sample. Formulas for the Sample Standard Deviation s = Σ ( x − x ¯ ) 2 n − 1 or s = Σ f ( x − x ¯ ) 2 n − 1 For the sample standard deviation, the denominator is n - 1 , that is the sample size MINUS 1. Formulas for the Population Standard Deviation σ = Σ ( x − μ ) 2 N or σ = Σ f ( x – μ ) 2 N For the population standard deviation, the denominator is N , the number of items in the population. In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three. Sampling Variability of a Statistic The statistic of a sampling distribution was discussed in Descriptive Statistics: Measures of the Center of the Data . How much the statistic varies from one sample to another is known as the sampling variability of a statistic . You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is σ n where σ is the standard deviation of the population and n is the size of the sample. NOTE In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation σ x or s x from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.) In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of the students. The following data are the ages for a SAMPLE of n = 20 fifth grade students. The ages are rounded to the nearest half year: 9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; x ¯ = 9 + 9 .5(2) + 10(4) + 10 .5(4) + 11(6) + 11 .5(3) 20 = 10.525 The average age is 10.53 years, rounded to two places. The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s . Data Freq. Deviations Deviations 2 (Freq.)( Deviations 2 ) x f ( x – x ¯ ) ( x – x ¯ ) 2 ( f )( x – x ¯ ) 2 9 1 9 – 10.525 = –1.525 (–1.525) 2 = 2.325625 1 × 2.325625 = 2.325625 9.5 2 9.5 – 10.525 = –1.025 (–1.025) 2 = 1.050625 2 × 1.050625 = 2.101250 10 4 10 – 10.525 = –0.525 (–0.525) 2 = 0.275625 4 × 0.275625 = 1.1025 10.5 4 10.5 – 10.525 = –0.025 (–0.025) 2 = 0.000625 4 × 0.000625 = 0.0025 11 6 11 – 10.525 = 0.475 (0.475) 2 = 0.225625 6 × 0.225625 = 1.35375 11.5 3 11.5 – 10.525 = 0.975 (0.975) 2 = 0.950625 3 × 0.950625 = 2.851875 The total is 9.7375 The sample variance, s 2 , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s 2 = 9.7375 20 − 1 = 0.5125 The sample standard deviation s is equal to the square root of the sample variance: s = 0.5125 = 0.715891 , which is rounded to two decimal places, s = 0.72. Typically, you do the calculation for the standard deviation on your calculator or computer . The intermediate results are not rounded. This is done for accuracy. For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation) . Verify the mean and standard deviation or a calculator or computer. For a sample: x = x ¯ + (#ofSTDEVs)( s ) For a population: x = μ + (#ofSTDEVs)( σ ) For this example, use x = x ¯ + (#ofSTDEVs)( s ) because the data is from a sample Verify the mean and standard deviation on your calculator or computer. Find the value that is one standard deviation above the mean. Find ( x ¯ + 1s). Find the value that is two standard deviations below the mean. Find ( x ¯ – 2s). Find the values that are 1.5 standard deviations from (below and above) the mean. Clear lists L1 and L2. Press STAT 4:ClrList . Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2. Enter data into the list editor. Press STAT 1:EDIT . If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down. Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around. Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER. x ¯ = 10.525 Use Sx because this is sample data (not a population): Sx=0.715891 ( x ¯ + 1s) = 10.53 + (1)(0.72) = 11.25 ( x ¯ – 2 s ) = 10.53 – (2)(0.72) = 9.09 ( x ¯ – 1.5 s ) = 10.53 – (1.5)(0.72) = 9.45 ( x ¯ + 1.5 s ) = 10.53 + (1.5)(0.72) = 11.61 Try It On a baseball team, the ages of each of the players are as follows: 21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40 Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean. Explanation of the standard deviation calculation shown in the table The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero . (For , there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation. The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data. Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one ( n – 1). Why not divide by n ? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by ( n – 1) gives a better estimate of the population variance. NOTE Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic. The standard deviation, s or σ , is either zero or larger than zero. Describing the data with reference to the spread is called \"variability\". The variability in data depends upon the method by which the outcomes are obtained; for example, by measuring or by random sampling. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large. The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better \"feel\" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data . Display your data in a histogram or a box plot. Use the following data (first exam scores) from Professor Dean's spring pre-calculus class: 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator: The sample mean The sample standard deviation The median The first quartile The third quartile IQR Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart. See The sample mean = 73.5 The sample standard deviation = 17.9 The median = 73 The first quartile = 61 The third quartile = 90 IQR = 90 – 61 = 29 The x -axis goes from 32.5 to 100.5; y -axis goes from –2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5 – 32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5 + 13.6 = 46.1, 46.1 + 13.6 = 59.7, 59.7 + 13.6 = 73.3, 73.3 + 13.6 = 86.9, 86.9 + 13.6 = 100.5 = the ending value; No data values fall on an interval boundary. Data Frequency Relative Frequency Cumulative Relative Frequency 33 1 0.032 0.032 42 1 0.032 0.064 49 2 0.065 0.129 53 1 0.032 0.161 55 2 0.065 0.226 61 1 0.032 0.258 63 1 0.032 0.29 67 1 0.032 0.322 68 2 0.065 0.387 69 2 0.065 0.452 72 1 0.032 0.484 73 1 0.032 0.516 74 1 0.032 0.548 78 1 0.032 0.580 80 1 0.032 0.612 83 1 0.032 0.644 88 3 0.097 0.741 90 1 0.032 0.773 92 1 0.032 0.805 94 4 0.129 0.934 96 1 0.032 0.966 100 1 0.032 0.998 (Why isn't this value 1?) The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73 – 33 = 40) than the spread in the upper 50% (100 – 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores ( IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs. Try It The following data show the different types of pet food stores in the area carry. 6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12; Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator. Standard deviation of Grouped Frequency Tables Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: M e a n o f F r e q u e n c y T a b l e = ∑ f m ∑ f where f = interval frequencies and m = interval midpoints. Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean. Find the standard deviation for the data in . Class Frequency, f Midpoint, m f ⋅ m x ¯ m - x ¯ ( m - x ¯ ) 2 f ( m - x ¯ ) 2 0–2 1 1 1 7.58 -6.58 43.2964 43.2964 3–5 6 4 24 7.58 -3.58 12.8164 76.8984 6–8 10 7 70 7.58 -0.58 0.3364 3.364 9–11 7 10 70 7.58 2.42 5.8564 40.9948 12–14 0 13 0 7.58 5.42 29.3764 0 15–17 2 16 32 7.58 8.42 70.8964 141.7928 SUM ( Σ ) 26 197 306.3464 The values in the second, third, and fourth columns of are used to calculate the mean of the grouped frequency table, the value in the fifth column. x ¯ = Σ f m Σ f = 197 26 ≈ 7 . 58 . After calculating x ¯ , find the difference, m - x ¯ , for each midpoint, m . Next, square each difference. In the final column, calculate the product of the frequency and the squared difference for each class. The table makes it easy to use the formula for calculating the standard deviation of a grouped frequency table: s x = Σ f ( m - x ¯ ) 2 n - 1 = 43 . 2964 26 - 1 ≈ 3 . 50 . Although the formula is not complicated, these calculations are typically performed using technology. Try It Find the standard deviation for the data from the previous example Class Frequency, f 0–2 1 3–5 6 6–8 10 9–11 7 12–14 0 15–17 2 First, press the STAT key and select 1:Edit Input the midpoint values into L1 and the frequencies into L2 Select STAT , CALC , and 1: 1-Var Stats Select 2 nd then 1 then , 2 nd then 2 Enter You will see displayed both a population standard deviation, σ x , and the sample standard deviation, s x . Comparing Values from Different Data Sets The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading. For each data value, calculate how many standard deviations away from its mean the value is. Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs. # o f S T D E V s = value – mean standard deviation Compare the results of this calculation. #ofSTDEVs is often called a \" z -score\"; we can use the symbol z . In symbols, the formulas become: Sample x = x ¯ + zs z = x − x ¯ s Population x = μ + zσ z = x − μ σ Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school? Student GPA School Mean GPA School Standard Deviation John 2.85 3.0 0.7 Ali 77 80 10 For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer. z = # of STDEVs = value – mean standard deviation = x – μ σ For John, z = # o f S T D E V s = 2.85 – 3.0 0.7 = – 0.21 For Ali, z = # o f S T D E V s = 77 − 80 10 = − 0.3 John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean. John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. Try It Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team? Swimmer Time (seconds) Team Mean Time Team Standard Deviation Angie 26.2 27.2 0.8 Beth 27.3 30.1 1.4 The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data. For ANY data set, no matter what the distribution of the data is: At least 75% of the data is within two standard deviations of the mean. At least 89% of the data is within three standard deviations of the mean. At least 95% of the data is within 4.5 standard deviations of the mean. This is known as Chebyshev's Rule. For data having a distribution that is BELL-SHAPED and SYMMETRIC: Approximately 68% of the data is within one standard deviation of the mean. Approximately 95% of the data is within two standard deviations of the mean. More than 99% of the data is within three standard deviations of the mean. This is known as the Empirical Rule. It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the \"Normal\" or \"Gaussian\" probability distribution in later chapters. References Data from Microsoft Bookshelf. King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013). Chapter Review The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population. The Standard Deviation allows us to compare individual data or classes to the data set mean numerically. s = ∑ ​ ( x − x ¯ ) 2 n − 1 or s = ∑ ​ f ( x − x ¯ ) 2 n − 1 is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, μ , and the formula σ = ∑ ​ ( x − μ ) 2 N or σ = ∑ ​ f ( x − μ ) 2 N . Formula Review s x = ∑ f m 2 n − x ¯ 2 where s x = sample standard deviation x ¯ = sample mean Use the following information to answer the next two exercises : The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles. 29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150 Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth. s = 34.5 Find the value that is one standard deviation below the mean. Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team? Baseball Player Batting Average Team Batting Average Team Standard Deviation Fredo 0.158 0.166 0.012 Karl 0.177 0.189 0.015 For Fredo: z = 0.158 – 0.166 0.012 = –0.67 For Karl: z = 0.177 – 0.189 0.015 = –0.8 Fredo’s z -score of –0.67 is higher than Karl’s z -score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team. Use to find the value that is three standard deviations: above the mean below the mean Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84 . Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Daily Low Temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 Points per Game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2 s x = ∑ f m 2 n − x ¯ 2 = 193157.45 30 − 79.5 2 = 10.88 s x = ∑ f m 2 n − x ¯ 2 = 380945.3 101 − 60.94 2 = 7.62 s x = ∑ f m 2 n − x ¯ 2 = 440051.5 86 − 70.66 2 = 11.14 Homework Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at a local community college over a 29-year period. μ = 1000 FTES median = 1,014 FTES σ = 474 FTES first quartile = 528.5 FTES third quartile = 1,447.5 FTES n = 29 years A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer. The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or above the median. 75% of all years have an FTES: at or below: _____ at or above: _____ The population standard deviation = _____ 474 FTES What percent of the FTES were from 528.5 to 1447.5? How do you know? What is the IQR ? What does the IQR represent? 919 How many standard deviations away from the mean is the median? Additional Information: The population FTES for a specific six-year period was given in an updated report. The data are reported here. Year 1 2 3 4 5 6 Total FTES 1,585 1,690 1,735 1,935 2,021 1,890 Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR . Round to one decimal place. mean = 1,809.3 median = 1,812.5 standard deviation = 151.2 first quartile = 1,690 third quartile = 1,935 IQR = 245 What additional information is needed to construct a box plot for the FTES for Years 1 to 6 and a box plot for the 29-year period shown earlier? Compare the IQR for the previous 29-year period with the IQR for the six-year period shown in . Why do you suppose the IQR s are so different? Hint: Think about the number of years covered by each time period and what happened to higher education during those periods. Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer. Student GPA School Average GPA School Standard Deviation Thuy 2.7 3.2 0.8 Vichet 87 75 20 Kamala 8.6 8 0.4 A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer. For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type. An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he? Who is the fastest runner with respect to their class? Explain why. A selection of countries has poverty rates that range from 11.4% to 74.6%. This data is summarized in Table 2.76 . Percent of Population Number of Countries 11.4–20.45 29 20.45–29.45 13 29.45–38.45 4 38.45–47.45 0 47.45–56.45 2 56.45–65.45 1 65.45–74.45 0 74.45–83.45 1 What is the best estimate of the average poverty percentage for these countries? What is the standard deviation for the listed poverty rates? The United States has an average poverty rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ rate compared to the average rate? Explain. x ¯ = 23.32 Using the TI 83/84, we obtain a standard deviation of: s x = 12.95. The poverty rate of the United States is 10.58% higher than the average rate. Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the poverty percentage that is one standard deviation from the mean. The United States rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States does not have an unusually high percentage of people experiencing poverty. gives the percent of children under five considered to be underweight. Percent of Underweight Children Number of Countries 16–21.45 23 21.45–26.9 4 26.9–32.35 9 32.35–37.8 7 37.8–43.25 6 43.25–48.7 1 What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain. Bringing It Together Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: # of movies Frequency 0 5 1 9 2 6 3 4 4 1 Find the sample mean x ¯ . Find the approximate sample standard deviation, s . 1.48 1.12 Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows: X Frequency 1 2 2 5 3 8 4 12 5 12 6 0 7 1 Find the sample mean x – Find the sample standard deviation, s Construct a histogram of the data. Complete the columns of the chart. Find the first quartile. Find the median. Find the third quartile. Construct a box plot of the data. What percent of the students owned at least five pairs? Find the 40 th percentile. Find the 90 th percentile. Construct a line graph of the data Construct a stemplot of the data Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year. 177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265 Organize the data from smallest to largest value. Find the median. Find the first quartile. Find the third quartile. Construct a box plot of the data. The middle 50% of the weights are from _______ to _______. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? Assume the population was the San Francisco 49ers. Find: the population mean, μ . the population standard deviation, σ . the weight that is two standard deviations below the mean. When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he? That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer? 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302 241 205.5 272.5 205.5, 272.5 sample 236.34 37.50 161.34 0.84 std. dev. below the mean Young One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows: 3 8 –1 2 0 5 –3 1 –1 6 5 –2 What is the mean change score? What is the standard deviation for this population? What is the median change score? Find the change score that is 2.2 standard deviations below the mean. Refer to determine which of the following are true and which are false. Explain your solution to each part in complete sentences. The medians for all three graphs are the same. We cannot determine if any of the means for the three graphs is different. The standard deviation for graph b is larger than the standard deviation for graph a. We cannot determine if any of the third quartiles for the three graphs is different. True True True False In a recent issue of the IEEE Spectrum , 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference. Organize the data in a chart. Find the median, the first quartile, and the third quartile. Find the 65 th percentile. Find the 10 th percentile. Construct a box plot of the data. The middle 50% of the conferences last from _______ days to _______ days. Calculate the sample mean of days of engineering conferences. Calculate the sample standard deviation of days of engineering conferences. Find the mode. If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences. A survey of enrollment at 35 community colleges across the United States yielded the following figures: 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622 Organize the data into a chart with six intervals of equal width from 0 to 30,000. Label the two columns \"Enrollment\" and \"Frequency.\" Construct a histogram of the data. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean? Calculate the sample mean. Calculate the sample standard deviation. A school with an enrollment of 8000 would be how many standard deviations away from the mean? Enrollment Frequency 1000-5000 10 5000-10000 16 10000-15000 3 15000-20000 3 20000-25000 1 25000-30000 2 Answers may vary. mode 8628.74 6943.88 –0.09 Use the following information to answer the next two exercises. X = the number of days per week that 100 clients use a particular exercise facility. x Frequency 0 3 1 12 2 33 3 28 4 11 5 9 6 4 The 80 th percentile is _____ 5 80 3 4 The number that is 1.5 standard deviations BELOW the mean is approximately _____ 0.7 4.8 –2.8 Cannot be determined a Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the . # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion. If a data value is identified as an outlier, what should be done about it? Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.) Do parts a and c of this problem give the same answer? Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data? Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode? Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Variance mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as x – x ¯ where x is a value of the data and x ¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.", "section": "Measures of the Spread of the Data", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Descriptive Statistics Descriptive Statistics Class Time: Names: Student Learning Outcomes The student will construct a histogram and a box plot. The student will calculate univariate statistics. The student will examine the graphs to interpret what the data implies. Collect the Data Record the number of pairs of shoes you own. Randomly survey 30 classmates about the number of pairs of shoes they own. Record their values. Survey Results Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil and scale the axes. Calculate the following values. x ¯ = _____ s = _____ Are the data discrete or continuous? How do you know? In complete sentences, describe the shape of the histogram. Are there any potential outliers? List the value(s) that could be outliers. Use a formula to check the end values to determine if they are potential outliers. Analyze the Data Determine the following values. Min = _____ M = _____ Max = _____ Q 1 = _____ Q 3 = _____ IQR = _____ Construct a box plot of data What does the shape of the box plot imply about the concentration of data? Use complete sentences. Using the box plot, how can you determine if there are potential outliers? How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers? What does the IQR represent in this problem? Show your work to find the value that is 1.5 standard deviations: above the mean. below the mean.", "section": "Descriptive Statistics", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction Meteor showers are rare, but the probability of them occurring can be calculated. (credit: modification of work “A meteor during the peak of the 2009 Leonid Meteor Shower” by Navicore/ Wikimedia Commons, CC BY 3.0) Chapter Objectives By the end of this chapter, the student should be able to: Understand and use the terminology of probability. Determine whether two events are mutually exclusive and whether two events are independent. Calculate probabilities using the Addition Rules and Multiplication Rules. Construct and interpret Contingency Tables. Construct and interpret Venn Diagrams. Construct and interpret Tree Diagrams. It is often necessary to \"guess\" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs. You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach. Your instructor will survey your class. Count the number of students in the class today. Raise your hand if you have any change in your pocket or purse. Record the number of raised hands. Raise your hand if you rode a bus within the past month. Record the number of raised hands. Raise your hand if you answered \"yes\" to BOTH of the first two questions. Record the number of raised hands. Use the class data as estimates of the following probabilities. P (change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P (bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers. Find P (change). Find P (bus). Find P (change AND bus). Find the probability that a randomly chosen student in your class has change in their pocket or purse and rode a bus within the last month. Find P (change|bus). Find the probability that a randomly chosen student has change given that they rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Terminology Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment. A result of an experiment is called an outcome . The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = { H , T } where H = heads and T = tails are the outcomes. An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P ( A ). The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P ( A ) = 0 means the event A can never happen. P ( A ) = 1 means the event A always happens. P ( A ) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads). Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair , six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head ( H ) and a Tail ( T ) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer. To calculate the probability of an event A when all outcomes in the sample space are equally likely , count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is { HH , TH , HT , TT } where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition { HT , TH }, so P ( A ) = 2 4 = 0.5. Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P ( E ) = 2 6 . If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 2 6 of the rolls would result in an outcome of \"at least five\". You would not expect exactly 2 6 . The long-term relative frequency of obtaining this result would approach the theoretical probability of 2 6 as the number of repetitions grows larger and larger. This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.) It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair , or biased . Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely. \"OR\" Event : An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B . The event A OR B can also be written as A Union B, with notation as follows: A ∪ B For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. Then A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice. \"AND\" Event : An outcome is in the event A AND B if the outcome is in both A and B at the same time. The event A AND B can also be written as A Intersection B , with notation as follows: A ∩ B For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A ∩ B = {4, 5}. The complement of event A is denoted A′ (read \" A prime\"). A′ consists of all outcomes that are NOT in A . Notice that P ( A ) + P ( A′ ) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A′ = {5, 6}. P ( A ) = 4 6 , P ( A′ ) = 2 6 , and P ( A ) + P ( A′ ) = 4 6 + 2 6 = 1 The conditional probability of A given B is written P ( A | B ). P ( A | B ) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space . We calculate the probability of A from the reduced sample space B . The formula to calculate P ( A | B ) is P ( A | B ) = P ( A AND B ) P ( B ) where P ( B ) is greater than zero. For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P ( A | B ), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S ). We get the same result by using the formula. Remember that S has six outcomes. P ( A | B ) = P ( A AND B ) P ( B ) = ( the number of outcomes that are 2 or 3 and even in S ) 6 ( the number of outcomes that are even in S ) 6 = 1 6 3 6 = 1 3 Understanding Terminology and Symbols It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any. The sample space S is the whole numbers starting at one and less than 20. S = _____________________________ Let event A = the even numbers and event B = numbers greater than 13. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A AND B = ____________________, A OR B = ________________ P ( A AND B ) = _________, P ( A OR B ) = _____________ A′ = _____________, P ( A′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19} P ( A ) = 9 19 , P ( B ) = 6 19 A AND B = {14,16,18}, A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19} P ( A AND B ) = 3 19 , P ( A OR B ) = 12 19 A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P ( A′ ) = 10 19 P ( A ) + P ( A′ ) = 1 ( 9 19 + 10 19 = 1) P ( A | B ) = P ( A AND B ) P ( B ) = 3 6 , P ( B | A ) = P ( A AND B ) P ( A ) = 3 9 , No Try It The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)). S = _____________________________ Let event A = the sum is even and event B = the first number is prime. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A AND B = ____________________, A OR B = ________________ P ( A AND B ) = _________, P ( A OR B ) = _____________ B′ = _____________, P ( B′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? A fair, six-sided die is rolled. Describe the sample space S , identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up). Event T = the outcome is two. Event A = the outcome is an even number. Event B = the outcome is less than four. The complement of A . A GIVEN B B GIVEN A A AND B A OR B A OR B′ Event N = the outcome is a prime number. Event I = the outcome is seven. T = {2}, P ( T ) = 1 6 A = {2, 4, 6}, P ( A ) = 1 2 B = {1, 2, 3}, P ( B ) = 1 2 A′ = {1, 3, 5}, P ( A′ ) = 1 2 A | B = {2}, P ( A | B ) = 1 3 B | A = {2}, P ( B | A ) = 1 3 A AND B = {2}, P ( A AND B ) = 1 6 A OR B = {1, 2, 3, 4, 6}, P ( A OR B ) = 5 6 A OR B′ = {2, 4, 5, 6}, P ( A OR B′ ) = 2 3 N = {2, 3, 5}, P ( N ) = 1 2 A six-sided die does not have seven dots. P (7) = 0. Try It A number is randomly selected between 1 to 10. Describe the sample space S , identify each of the following events with a subset of S, and compute its probability (an outcome is the number selected). Event F = the outcome is 5. Event A = the number is more than 6. Event B = the number is odd. The complement of B . A GIVEN B . B GIVEN A . A OR B . A’ OR B . Event P = the outcome is a composite number. Event M = the number is a multiple of 3. F = 5 , P ( F ) = 1 10 A = 7 , 8 , 9 , P ( A ) = 3 10 B = 1 , 3 , 5 , 7 , 9 , P ( B ) = 5 10 B ' = 2 , 4 , 6 , 8 , 10 , P ( B ' ) = 5 10 A | B = 7 , 9 , P A | B = 2 5 B | A = 7 , 9 , P ( B | A ) = 1 2 A O R B = 1 , 3 , 5 , 7 , 8 , 9 , 10 , P ( A O R B ) = 7 10 A ' O R B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 9 , P ( A ' O R B ) = 4 5 P = 4 , 6 , 8 , 9 , 10 , P ( P ) = 1 2 M = 3 , 6 , 9 , P ( M ) = 3 10 describes the distribution of a random sample S of 100 individuals, organized by sex assigned at birth and whether they are right- or left-handed. Right-handed Left-handed Males 43 9 Females 44 4 Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities: P ( M ) P ( F ) P ( R ) P ( L ) P ( M AND R ) P ( F AND L ) P ( M OR F ) P ( M OR R ) P ( F OR L ) P ( M' ) P ( R | M ) P ( F | L ) P ( L | F ) P ( M ) = 0.52 P ( F ) = 0.48 P ( R ) = 0.87 P ( L ) = 0.13 P ( M AND R ) = 0.43 P ( F AND L ) = 0.04 P ( M OR F ) = 1 P ( M OR R ) = 0.96 P ( F OR L ) = 0.57 P ( M' ) = 0.48 P ( R | M ) = 0.8269 (rounded to four decimal places) P ( F | L ) = 0.3077 (rounded to four decimal places) P ( L | F ) = 0.0833 Try It The table describes the distribution of a random sample S of 100 individuals, organized by gender and whether they like tea or coffee. Tea Coffee Men 22 26 Women 16 36 Let's denote the events M = the subject is a man, W = the subject is a woman, T = the subject likes tea, and C = the subject likes coffee. Compute the following probabilities: P ( M ) P ( W ) P ( T ) P ( C ) P ( M AND T ) P ( W and C ) P ( M OR W ) P ( M OR T ) P ( W OR C ) P ( M ') P ( T | M ) P ( W | C ) P ( C | W ) P ( M ) = 0 . 48 P ( W ) = 0 . 52 P ( T ) = 0 . 38 P ( C ) = 0 . 62 P ( M A N D T ) = 0 . 22 P ( W a n d C ) = 0 . 36 P ( M O R W ) = 1 P ( M O R T ) = 0 . 64 P ( W O R C ) = 0 . 78 P ( M ' ) = 0 . 52 P ( T | M ) = 0 . 4583 (rounded to four decimal places) P ( W | C ) = 0 . 5806 (rounded to four decimal places) P ( C | W ) = 0 . 6923 (rounded to four decimal places) References “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013). Chapter Review In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive. Formula Review A and B are events P ( S ) = 1 where S is the sample space 0 ≤ P ( A ) ≤ 1 P ( A | B ) = P ( A AND B ) P ( B ) In a particular college class, there are men and women students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) Let W be the event that a student is a woman. Let M be the event that a student is a man. Let S be the event that a student has short hair. Let L be the event that a student has long hair. The probability that a student does not have long hair. The probability that a student is a man or has short hair. The probability that a student is a woman and has long hair. The probability that a student is a man, given that the student has long hair. The probability that a student has long hair, given that the student is a man. Of all the women students, the probability that a student has short hair. Of all students with long hair, the probability that a student is a woman. The probability that a student is a woman or has long hair. The probability that a randomly selected student is a man student with short hair. The probability that a student is a woman. P ( L′ ) = P ( S ) P ( M OR S ) P ( W AND L ) P ( M | L ) P ( L | M ) P ( S | W ) P ( W | L ) P ( W OR L ) P ( M AND S ) P ( W ) Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. One party favor is chosen from the box at random. Let H = the event of getting a hat. Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap. Let C = the event of getting a bag of confetti. Find P ( H ). Find P ( N ). P ( N ) = 15 42 = 5 14 = 0.36 Find P ( F ). Find P ( C ). P ( C ) = 5 42 = 0.12 Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. One jelly bean is chosen from the box at random. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. Find P ( B ). Find P ( G ). P ( G ) = 20 150 = 2 15 = 0.13 Find P ( P ). Find P ( R ). P ( R ) = 22 150 = 11 75 = 0.15 Find P ( Y ). Find P ( O ). P ( O ) = 150 - 22 - 38 - 20 - 28 - 26 150 = 16 150 = 8 75 = 0.11 Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. Find P ( A ). Find P ( E ). P ( E ) = 47 194 = 0.24 Find P ( F ). Find P ( N ). P ( N ) = 23 194 = 0.12 Find P ( O ). Find P ( S ). P ( S ) = 12 194 = 6 97 = 0.06 What is the probability of drawing a red card in a standard deck of 52 cards? What is the probability of drawing a club in a standard deck of 52 cards? 13 52 = 1 4 = 0.25 What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six? What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six? 3 6 = 1 2 = 0.5 Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area. Let B = the event of landing on blue. Let R = the event of landing on red. Let G = the event of landing on green. Let Y = the event of landing on yellow. If you land on Y , you get the biggest prize. Find P ( Y ). If you land on red, you don’t get a prize. What is P ( R )? P ( R ) = 4 8 = 0.5 Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. Write the symbols for the probability that a player is not an outfielder. Write the symbols for the probability that a player is an outfielder or is a great hitter. P ( O OR H ) Write the symbols for the probability that a player is an infielder and is not a great hitter. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder. P ( H | I ) Write the symbols for the probability that a player is an infielder, given that the player is a great hitter. Write the symbols for the probability that of all the outfielders, a player is not a great hitter. P ( N | O ) Write the symbols for the probability that of all the great hitters, a player is an outfielder. Write the symbols for the probability that a player is an infielder or is not a great hitter. P ( I OR N ) Write the symbols for the probability that a player is an outfielder and is a great hitter. Write the symbols for the probability that a player is an infielder. P ( I ) What is the word for the set of all possible outcomes? What is conditional probability? The likelihood that an event will occur given that another event has already occurred. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space? What is the sum of the probabilities of an event and its complement? 1 Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three. What does P ( E | M ) mean in words? What does P ( E OR M ) mean in words? the probability of landing on an even number or a multiple of three Homework The graph in displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045. Define three events in the graph. Describe in words what the entry 40 means. Describe in words the complement of the entry in question 2. Describe in words what the entry 30 means. Out of the men and women, what percent are men? Out of the women, what percent disapprove of Mayor Ford? Out of all the age groups, what percent approve of Mayor Ford? Find P (Approve|Men). Out of the age groups, what percent are more than 44 years old? Find P (Approve|Age < 35). Explain what is wrong with the following statements. Use complete sentences. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100% A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs. Conditional Probability the likelihood that an event will occur given that another event has already occurred Equally Likely Each outcome of an experiment has the same probability. Event a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S . An event is an arbitrary subset in S . It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A , B , C , and so on. Experiment a planned activity carried out under controlled conditions Outcome a particular result of an experiment Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S . Then: 0 ≤ P ( A ) ≤ 1 If A and B are any two mutually exclusive events, then P ( A OR B ) = P ( A ) + P (B). P ( S ) = 1 Sample Space the set of all possible outcomes of an experiment AND Event An outcome is in the event A AND B if the outcome is in both A AND B at the same time. Complement Event The complement of event A consists of all outcomes that are NOT in A . Conditional Probability of A GIVEN B P ( A | B ) is the probability that event A will occur given that the event B has already occurred. Or Event An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B .", "section": "Terminology", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Independent and Mutually Exclusive Events Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if the following are true: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A AND B ) = P ( A ) P ( B ) Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent events . Sampling may be done with replacement or without replacement . With replacement : If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. Without replacement : When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent. If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise . You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. a. Sampling with replacement: Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are { Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement: Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are { K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. Try It You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD . Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH . Which of a. or b. did you sample with replacement and which did you sample without replacement? a. Without replacement; b. With replacement Try It You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement. QS , 1 D , 1 C , QD KH , 7 D , 6 D , KH QS , 7 D , 6 D , KS Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P ( A AND B ) = 0. For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P ( A AND B ) = 2 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P ( A AND C ) = 0. Therefore, A and C are mutually exclusive. If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise . The following examples illustrate these definitions and terms. Flip two fair coins. (This is an experiment.) The sample space is { HH , HT , TH , TT } where T = tails and H = heads. The outcomes are HH , HT , TH , and TT . The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads. Let A = the event of getting at most one tail . (At most one tail means zero or one tail.) Then A can be written as { HH , HT , TH }. The outcome HH shows zero tails. HT and TH each show one tail. Let B = the event of getting all tails. B can be written as { TT }. B is the complement of A , so B = A′ . Also, P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1. The probabilities for A and for B are P ( A ) = 3 4 and P ( B ) = 1 4 . Let C = the event of getting all heads. C = { HH }. Since B = { TT }, P ( B AND C ) = 0. B and C are mutually exclusive. ( B and C have no members in common because you cannot have all tails and all heads at the same time.) Let D = event of getting more than one tail. D = { TT }. P ( D ) = 1 4 Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = { HT , HH }. P ( E ) = 2 4 Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = { HT , TH , TT }. P ( F ) = 3 4 Try It Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Flip two fair coins. Find the probabilities of the events. Let F = the event of getting at most one tail (zero or one tail). Let G = the event of getting two faces that are the same. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. Are F and G mutually exclusive? Let J = the event of getting all tails. Are J and H mutually exclusive? Look at the sample space in . Zero (0) or one (1) tails occur when the outcomes HH , TH , HT show up. P ( F ) = 3 4 Two faces are the same if HH or TT show up. P ( G ) = 2 4 A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P ( H ) = 2 4 F and G share HH so P ( F AND G ) is not equal to zero (0). F and G are not mutually exclusive. Getting all tails occurs when tails shows up on both coins ( TT ). H ’s outcomes are HH and HT . J and H have nothing in common so P ( J AND H ) = 0. J and H are mutually exclusive. Try It A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events: Let F = the event of getting the white ball twice. Let G = the event of getting two balls of different colors. Let H = the event of getting white on the first pick. Are F and G mutually exclusive? Are G and H mutually exclusive? Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}. Find the complement of A , A′ . The complement of A , A′ , is B because A and B together make up the sample space. P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1. Also, P ( A ) = 3 6 and P ( B ) = 3 6 . Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P ( C AND D ) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. Let event E = all faces less than five. E = {1, 2, 3, 4}. Are C and E mutually exclusive events? (Answer yes or no.) Why or why not? No. C = {3, 5} and E = {1, 2, 3, 4}. P ( C AND E ) = 1 6 . To be mutually exclusive, P ( C AND E ) must be zero. Find P ( C | A ). This is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P ( C | A ), find the probability of C using the sample space A . You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P ( C | A ) = 2 3 . Try It Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P ( A ) = 0.4 and P ( B ) = 0.2. P ( A AND B ) = 0.08. Are events A and B independent? Hint: You must show ONE of the following: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A AND B ) = P ( A ) P ( B ) P ( A | B ) = P ( A AND B ) P ( B ) = 0. 08 0.2 = 0 .4 = P ( A ) The events are independent because P ( A | B ) = P ( A ). Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P ( G ) = 0.6, P ( H ) = 0.5, and P ( G AND H ) = 0.3. Are G and H independent? If G and H are independent, then you must show ONE of the following: P ( G | H ) = P ( G ) P ( H | G ) = P ( H ) P ( G AND H ) = P ( G ) P ( H ) NOTE The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information. a. Show that P ( G | H ) = P ( G ). P ( G | H ) = P ( G AND H ) P ( H ) = 0 .3 0 .5 = 0.6 = P ( G ) b. Show P ( G AND H ) = P ( G ) P ( H ). P ( G ) P ( H ) = (0.6)(0.5) = 0.3 = P ( G AND H ) Since G and H are independent, knowing that a person is taking a science class does not change the chance that they are taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance they are taking math. For practice, show that P ( H | G ) = P ( H ) to show that G and H are independent events. Try It In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4. R = a red marble G = a green marble O = an odd-numbered marble The sample space is S = { R 1, R 2, R 3, R 4, R 5, R 6, G 1, G 2, G 3, G 4}. S has ten outcomes. What is P ( G AND O )? Let event C = taking an English class. Let event D = taking a speech class. Suppose P ( C ) = 0.75, P ( D ) = 0.3, P ( C | D ) = 0.75 and P ( C AND D ) = 0.225. Justify your answers to the following questions numerically. Are C and D independent? Are C and D mutually exclusive? What is P ( D | C )? Yes, because P ( C | D ) = P ( C ). No, because P ( C AND D ) is not equal to zero. P ( D | C ) = P ( C AND D ) P ( C ) = 0 .225 0.75 = 0.3 Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( B AND D ) = 0.20. Find P ( B | D ). Find P ( D | B ). Are B and D independent? Are B and D mutually exclusive? In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R 1, R 2, R 3, B 1, B 2, B 3, B 4, B 5. S has eight outcomes. P ( R ) = 3 8 . P ( B ) = 5 8 . P ( R AND B ) = 0. (You cannot draw one card that is both red and blue.) P ( E ) = 3 8 . (There are three even-numbered cards, R 2, B 2, and B 4.) P ( E | B ) = 2 5 . (There are five blue cards: B 1, B 2, B 3, B 4, and B 5. Out of the blue cards, there are two even cards; B 2 and B 4.) P ( B | E ) = 2 3 . (There are three even-numbered cards: R 2, B 2, and B 4. Out of the even-numbered cards, to are blue; B 2 and B 4.) The events R and B are mutually exclusive because P ( R AND B ) = 0. Let G = card with a number greater than 3. G = { B 4, B 5}. P ( G ) = 2 8 . Let H = blue card numbered between one and four, inclusive. H = { B 1, B 2, B 3, B 4}. P ( G | H ) = 1 4 . (The only card in H that has a number greater than three is B 4.) Since 2 8 = 1 4 , P ( G ) = P ( G | H ), which means that G and H are independent. Try It In a basketball arena, 70% of the fans are rooting for the home team. 25% of the fans are wearing blue. 20% of the fans are wearing blue and are rooting for the away team. Of the fans rooting for the away team, 67% are wearing blue. Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? In a particular college class, 60% of the students are women. Fifty percent of all students in the class have long hair. Forty-five percent of the students are women and have long hair. Of the women students, 75% have long hair. Let W be the event that a student is a women. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being a woman and having long hair independent? The following probabilities are given in this example: P ( W ) = 0.60; P ( L ) = 0.50 P ( W AND L ) = 0.45 P ( L | W ) = 0.75 NOTE The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P ( W | L ) yet, so you cannot use the second condition. >Check whether P ( W AND L ) = P ( W ) P ( L ). We are given that P ( W AND L ) = 0.45, but P ( W ) P ( L ) = (0.60)(0.50) = 0.30. The events of being a woman and having long hair are not independent because P ( W AND L ) does not equal P ( W ) P ( L ). Check whether P ( L | W ) equals P ( L ). We are given that P ( L | W ) = 0.75, but P ( L ) = 0.50; they are not equal. The events of being a woman and having long hair are not independent. The events of being a woman and having long hair are not independent; knowing that a student is a woman changes the probability that a student has long hair. Try It Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street. P ( I ) = 0.44 and P ( F ) = 0.56 P ( I AND F ) = 0 because Mark will take only one route to work. What is the probability of P ( I OR F )? Toss one fair coin (the coin has two sides, H and T ). The outcomes are ________. Count the outcomes. There are ____ outcomes. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes. Multiply the two numbers of outcomes. The answer is _______. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in part c. is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H 1 and T 6.) Event A = heads ( H ) on the coin followed by an even number (2, 4, 6) on the die. A = {_________________}. Find P ( A ). Event B = heads on the coin followed by a three on the die. B = {________}. Find P ( B ). Are A and B mutually exclusive? (Hint: What is P ( A AND B )? If P ( A AND B ) = 0, then A and B are mutually exclusive.) Are A and B independent? (Hint: Is P ( A AND B ) = P ( A ) P ( B )? If P ( A AND B ) = P ( A ) P ( B ), then A and B are independent. If not, then they are dependent). H and T ; 2 1, 2, 3, 4, 5, 6; 6 2(6) = 12 T 1, T 2, T 3, T 4, T 5, T 6, H 1, H 2, H 3, H 4, H 5, H 6 A = { H 2, H 4, H 6}; P ( A ) = 3 12 B = { H 3}; P ( B ) = 1 12 Yes, because P ( A AND B ) = 0 P ( A AND B ) = 0. P ( A ) P ( B ) = 3 12 · 1 12 = 1 48 . P ( A AND B ) does not equal P ( A ) P ( B ), so A and B are dependent. Try It A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing. Compute P ( T ). Compute P ( T | F ). Are T and F independent?. Are F and S mutually exclusive? Are F and S independent? References Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013). Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013). Chapter Review Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. Formula Review If A and B are independent, P ( A AND B ) = P ( A ) P ( B ), P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ). If A and B are mutually exclusive, P ( A OR B ) = P ( A ) + P ( B ) and P ( A AND B ) = 0. E and F are mutually exclusive events. P ( E ) = 0.4; P ( F ) = 0.5. Find P ( E ∣ F ). J and K are independent events. P ( J | K ) = 0.3. Find P ( J ). P ( J ) = 0.3 U and V are mutually exclusive events. P ( U ) = 0.26; P ( V ) = 0.37. Find: P ( U AND V ) = P ( U | V ) = P ( U OR V ) = Q and R are independent events. P ( Q ) = 0.4 and P ( Q AND R ) = 0.1. Find P ( R ). P ( Q AND R ) = P ( Q ) P ( R ) 0.1 = (0.4) P ( R ) P ( R ) = 0.25 Homework Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December in a certain year. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score. Find the probability that the selected occupation has an Emotional Health Index Score of 82.7. Find the probability that the selected occupation has an Emotional Health Index Score of 81.0. Find the probability that the selected occupation has an Emotional Health Index Score more than 81. Find the probability that the selected occupation has an Emotional Health Index Score between 80.5 and 82. If we know an occupation has an Emotional Health Index Score of 81.5 or more, what is the probability that it is 82.7? What is the probability that an occupation has an Emotional Health Index Score of 80.7 or 82.7? What is the probability that an occupation has an Emotional Health Index Score less than 80.2 given that it is already less than 81? What occupation has the highest Emotional Health Index Score? What occupation has the lowest Emotional Health Index Score? What is the range of the data? Compute the average EHIS. If all occupations are equally likely for a certain individual, what is the probability that they will have an occupation with lower than average EHIS? Bringing It Together A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data are compiled into . Shirt# ≤ 210 211–250 251–290 290≤ 1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P (Shirt# 1–33|≤ 210 pounds)? The probability that a person develops cancer is 0.4567. The probability that a person has at least one false positive test result (meaning the test comes back for cancer when the person does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a person develops cancer and P = a person has at least one false positive. P ( C ) = ______ P ( P | C ) = ______ P ( P | C' ) = ______ If a test comes up positive, based upon numerical values, can you assume that a male has cancer? Justify numerically and explain why or why not. P ( C ) = 0.4567 not enough information not enough information No, because over half (0.51) of men have at least one false positive text Given events G and H : P ( G ) = 0.43; P ( H ) = 0.26; P ( H AND G ) = 0.14 Find P ( H OR G ). Find the probability of the complement of event ( H AND G ). Find the probability of the complement of event ( H OR G ). Given events J and K : P ( J ) = 0.18; P ( K ) = 0.37; P ( J OR K ) = 0.45 Find P ( J AND K ). Find the probability of the complement of event ( J AND K ). Find the probability of the complement of event ( J OR K ). P ( J OR K ) = P ( J ) + P ( K ) − P ( J AND K ); 0.45 = 0.18 + 0.37 - P ( J AND K ); solve to find P ( J AND K ) = 0.10 P (NOT ( J AND K )) = 1 - P ( J AND K ) = 1 - 0.10 = 0.90 P (NOT ( J OR K )) = 1 - P ( J OR K ) = 1 - 0.45 = 0.55 Dependent Events If two events are NOT independent, then we say that they are dependent. Conditional Probability of One Event Given Another Event P ( A | B ) is the probability that event A will occur given that the event B has already occurred.", "section": "Independent and Mutually Exclusive Events", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. The Multiplication Rule If A and B are two events defined on a sample space , then: P ( A AND B ) = P ( B ) P ( A | B ). This rule may also be written as: P ( A | B ) = P ( A AND B ) P ( B ) (The probability of A given B equals the probability of A and B divided by the probability of B .) If A and B are independent , then P ( A | B ) = P ( A ). Then P ( A AND B ) = P ( A | B ) P ( B ) becomes P ( A AND B ) = P ( A ) P ( B ). The Addition Rule If A and B are defined on a sample space, then: P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ). If A and B are mutually exclusive , then P ( A AND B ) = 0. Then P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ) becomes P ( A OR B ) = P ( A ) + P ( B ). Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska Klaus can only afford one vacation. The probability that he chooses A is P ( A ) = 0.6 and the probability that he chooses B is P ( B ) = 0.35. P ( A AND B ) = 0 because Klaus can only afford to take one vacation Therefore, the probability that he chooses either New Zealand or Alaska is P ( A OR B ) = P ( A ) + P ( B ) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05. Try It Anna has to buy a new car. She has two choices, car A and car B. Anna can afford only one car. The probability that Anna will buy car A is P ( A ) = 0 . 25 , and the probability that Anna will buy car B is P ( B ) = 0 . 65 . Find: P ( A A N D B ) P ( A O R B ) P ( A A N D B ) = 0 because it is given that Anna can afford only one car. Since P ( A A N D B ) = 0 , then P ( A O R B ) = P ( A ) + P ( B ) = 0 . 25 + 0 . 65 = 0 . 9 . Note that the probability that Anna does not a car must be 0.1. Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P ( A ) = 0.65. B = the event Carlos is successful on his second attempt. P ( B ) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. a. What is the probability that he makes both goals? b. What is the probability that Carlos makes either the first goal or the second goal? c. Are A and B independent? d. Are A and B mutually exclusive? a. The problem is asking you to find P ( A AND B ) = P ( B AND A ). Since P ( B | A ) = 0.90: P ( B AND A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find P ( A OR B ). P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. No, they are not, because P ( B AND A ) = 0.585. P ( B ) P ( A ) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P ( B AND A ) So, P ( B AND A ) is not equal to P ( B ) P ( A ). d. No, they are not because P ( A and B ) = 0.585. To be mutually exclusive, P ( A AND B ) must equal zero. Try It Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = 0.75. D = the event Helen makes the second shot. P ( D ) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. a. What is the probability that the member is a novice swimmer? b. What is the probability that the member practices four times a week? c. What is the probability that the member is an advanced swimmer and practices four times a week? d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? a. 28 150 b. 80 150 c. 40 150 d. P (advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. e. No, these are not independent events. P (novice AND practices four times per week) = 0.0667 P (novice) P (practices four times per week) = 0.0996 0.0667 ≠ 0.0996 Try It A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M | S = math given speech What is the probability that Felicity enrolls in math and speech? Find P ( M AND S ) = P ( M | S ) P ( S ). What is the probability that Felicity enrolls in math or speech classes? Find P ( M OR S ) = P ( M ) + P ( S ) - P ( M AND S ). Are M and S independent? Is P ( M | S ) = P ( M )? Are M and S mutually exclusive? Is P ( M AND S ) = 0? a. 0.1625, b. 0.6875, c. No, d. No Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5. Find P ( B AND D ). Find P ( B OR D ). Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random. a. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? b. Given that the woman has breast cancer, what is the probability that she tests negative? c. What is the probability that the woman has breast cancer AND tests negative? d. What is the probability that the woman has breast cancer or tests negative? e. Are having breast cancer and testing negative independent events? f. Are having breast cancer and testing negative mutually exclusive? a. P ( B ) = 0.143; P ( N ) = 0.85 b. P ( N | B ) = 0.02 c. P ( B AND N ) = P ( B ) P ( N | B ) = (0.143)(0.02) = 0.0029 d. P ( B OR N ) = P ( B ) + P ( N ) - P ( B AND N ) = 0.143 + 0.85 - 0.0029 = 0.9901 e. No. P ( N ) = 0.85; P ( N | B ) = 0.02. So, P ( N | B ) does not equal P ( N ). f. No. P ( B AND N ) = 0.0029. For B and N to be mutually exclusive, P ( B AND N ) must be zero. Try It A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Refer to the information in . P = tests positive. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P ( P | B ) = 1 - P ( N | B ). What is the probability that a woman develops breast cancer and tests positive. Find P ( B AND P ) = P ( P | B ) P ( B ). What is the probability that a woman does not develop breast cancer. Find P ( B′ ) = 1 - P ( B ). What is the probability that a woman tests positive for breast cancer. Find P ( P ) = 1 - P ( N ). a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5. Find P ( B′ ). Find P ( D AND B ). Find P ( B | D ). Find P ( D AND B′ ). Find P ( D | B′ ). References DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013). “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). “Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013). Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013). Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). Chapter Review The multiplication rule and the addition rule are used for computing the probability of A and B , as well as the probability of A or B for two given events A , B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. Formula Review The multiplication rule: P ( A AND B ) = P ( A | B ) P ( B ) The addition rule: P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ) Use the following information to answer the next ten exercises. A local restaurant knows that the probability that a customer will order a pizza is 87%. The restaurant also knows that the probability that a customer will order a salad is 32%. Of the customers who order pizzas, 55% of them also order a salad. In this problem, let: Z = event that a customer orders a pizza S = event that a customer orders a salad Suppose that one customer is randomly selected. Find P ( Z ). Find P ( S ) : P ( S ) = 0.87 Find P ( S ). Find P ( S ) : P ( S ) = 0.32 Find P ( S | Z ). Find P ( S | Z ) : P ( S | Z ) = 0.55 In words, what is S | Z ? In words, what is S | Z ? : S | Z means, given that the customer has ordered pizza, the person also orders a salad. Find P ( Z A N D S ) . Find P ( Z A N D S ) : P ( Z A N D S ) = 0.4785 In words, what is Z A N D S ? In words, what is Z A N D S ? : Z A N D S represents the event that a customer orders a pizza and salad. Are P and S independent events? Show why or why not. Are P and S independent events? Show why or why not. No, because P ( S ) does not equal P ( S | Z ). Find P ( Z O R S ) . Find P ( Z O R S ) : P ( Z O R S ) = 0.7115 In words, what is Z O R S ? In words, what is Z O R S ? Z O R S represents the event that a customer orders a pizza or salad. Are Z and S mutually exclusive events? Show why or why not. Are Z and S mutually exclusive events? Show why or why not. No, because P ( S and Z ) does not equal 0. Homework Prior to the 2015 Supreme Court decision legalizing same-sex marriage nationwide, a survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them. In this problem, let: C = California registered voters who support same-sex marriage. B = California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them A = California registered voters who are 18 to 39 years old. Find P ( C ). Find P ( B ). Find P ( C | A ). Find P ( B | C ). In words, what is C | A ? In words, what is B | C ? Find P ( C AND B ). In words, what is C AND B ? Find P ( C OR B ). Are C and B mutually exclusive events? Show why or why not. A survey was conducted in a large city to measure the popularity of that city’s mayor. The survey was repeated every year for three years. The survey polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results the poll produced: In Year 1, 60% of the population approved of the mayor’s actions in office. In Year 2, 57% of the population approved of his actions. In Year 3, the percentage of popular approval was measured at 42%. What is the sample size for this study? What proportion in the poll disapproved of the mayor, according to the results from Year 3? How many people polled responded that they approved the mayor based on results from Year 3? What is the probability that a person supported the mayor, based on the data collected in Year 2? What is the probability that a person supported the mayor, based on the data collected in Year 1? The Forum Research surveyed 1,046 Torontonians. 58% 42% of 1,046 = 439 (rounding to the nearest integer) 0.57 0.60. Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range. (credit: film8ker/wikibooks) List the sample space of the 38 possible outcomes in roulette. You bet on red. Find P (red). You bet on -1st 12- (1st Dozen). Find P (-1st 12-). You bet on an even number. Find P (even number). Is getting an odd number the complement of getting an even number? Why? Find two mutually exclusive events. Are the events Even and 1st Dozen independent? Compute the probability of winning the following types of bets: Betting on two lines that touch each other on the table as in 1-2-3-4-5-6 Betting on three numbers in a line, as in 1-2-3 Betting on one number Betting on four numbers that touch each other to form a square, as in 10-11-13-14 Betting on two numbers that touch each other on the table, as in 10-11 or 10-13 Betting on 0-00-1-2-3 Betting on 0-1-2; or 0-00-2; or 00-2-3 P (Betting on two line that touch each other on the table) = 6 38 P (Betting on three numbers in a line) = 3 38 P (Bettting on one number) = 1 38 P (Betting on four number that touch each other to form a square) = 4 38 P (Betting on two number that touch each other on the table ) = 2 38 P (Betting on 0-00-1-2-3) = 5 38 P (Betting on 0-1-2; or 0-00-2; or 00-2-3) = 3 38 Compute the probability of winning the following types of bets: Betting on a color Betting on one of the dozen groups Betting on the range of numbers from 1 to 18 Betting on the range of numbers 19–36 Betting on one of the columns Betting on an even or odd number (excluding zero) Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card. G = card drawn is green E = card drawn is even-numbered List the sample space. P ( G ) = _____ P ( G | E ) = _____ P ( G AND E ) = _____ P ( G OR E ) = _____ Are G and E mutually exclusive? Justify your answer numerically. { G 1, G 2, G 3, G 4, G 5, Y 1, Y 2, Y 3} 5 8 2 3 2 8 6 8 No, because P ( G AND E ) does not equal 0. Roll two fair dice separately. Each die has six faces. List the sample space. Let A be the event that either a three or four is rolled first, followed by an even number. Find P ( A ). Let B be the event that the sum of the two rolls is at most seven. Find P ( B ). In words, explain what “ P ( A | B )” represents. Find P ( A | B ). Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification. A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin. List the sample space. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P ( A ). Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. NOTE The coin toss is independent of the card picked first. {( G , H ) ( G , T ) ( B , H ) ( B , T ) ( R , H ) ( R , T )} P ( A ) = P (blue) P (head) = ( 3 10 ) ( 1 2 ) = 3 20 Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P ( A AND B ) = 0 No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A ; if the card chosen is blue it is also (red or blue). P ( A AND C ) = P ( A ) = 3 20 An experiment consists of first rolling a die and then tossing a coin. List the sample space. Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P ( A ). Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on. List the sample space. Let A be the event that there are at least two tails. Find P ( A ). Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including justification. S = {( HHH ), ( HHT ), ( HTH ), ( HTT ), ( THH ), ( THT ), ( TTH ), ( TTT )} 4 8 Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P ( A AND B ) = 0. Consider the following scenario: Let P ( C ) = 0.4. Let P ( D ) = 0.5. Let P ( C | D ) = 0.6. Find P ( C AND D ). Are C and D mutually exclusive? Why or why not? Are C and D independent events? Why or why not? Find P ( C OR D ). Find P ( D | C ). Y and Z are independent events. Rewrite the basic Addition Rule P ( Y OR Z ) = P ( Y ) + P ( Z ) - P ( Y AND Z ) using the information that Y and Z are independent events. Use the rewritten rule to find P ( Z ) if P ( Y OR Z ) = 0.71 and P ( Y ) = 0.42. If Y and Z are independent, then P ( Y AND Z ) = P ( Y ) P ( Z ), so P ( Y OR Z ) = P ( Y ) + P ( Z ) - P ( Y ) P ( Z ). 0.5 G and H are mutually exclusive events. P ( G ) = 0.5 P ( H ) = 0.3 Explain why the following statement MUST be false: P ( H | G ) = 0.4. Find P ( H OR G ). Are G and H independent or dependent events? Explain in a complete sentence. According to the 2019 U.S. Census, approximately 331,449,281 people live in the United States. Of these people, 67,800,000 speak a language other than English at home. Of those who speak another language at home, 61.6% speak Spanish. Let: E = speaks English at home; E′ = speaks another language at home; S = speaks Spanish; Finish each probability statement by matching the correct answer. Probability Statements Answers a. P ( E ′ ) = i. 0.7954 b. P ( E ) = ii. 0.616 c. P ( S ∩ E ′ ) = iii. 0.2046 d. P ( S | E ′ ) = iv. 0.1260 iii i iv ii 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won green card. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let F = was a finalist. Are G and F independent or dependent events? Justify your answer numerically and also explain why. Are G and F mutually exclusive events? Justify your answer numerically and explain why. Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned. Let: R = money returned; E = economics classes; O = other classes Write a probability statement for the overall percent of money returned. Write a probability statement for the percent of money returned out of the economics classes. Write a probability statement for the percent of money returned out of the other classes. Is money being returned independent of the class? Justify your answer numerically and explain it. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer. P ( R ) = 0.44 P ( R | E ) = 0.56 P ( R | O ) = 0.31 No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P ( R | E ) ≠ P ( R ). No, this study definitely does not support that notion; in fact , it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P ( R | E ) > P ( R ). The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home Run Total Hits Babe Ruth 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,053 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 Total 7,918 2,127 583 1,723 12,351 Are \"the hit being made by Hank Aaron\" and \"the hit being a double\" independent events? Yes, because P (hit by Hank Aaron|hit is a double) = P (hit by Hank Aaron) No, because P (hit by Hank Aaron|hit is a double) ≠ P (hit is a double) No, because P (hit is by Hank Aaron|hit is a double) ≠ P (hit by Hank Aaron) Yes, because P (hit is by Hank Aaron|hit is a double) = P (hit is a double) P (type O OR Rh-) = P (type O) + P (Rh-) – P (type O AND Rh-) 0.38 = 0.45 + 0.07 – P (type O AND Rh-); Solve to find P (type O AND Rh-) = 0.14. 14% of people have type O, Rh- blood. P (NOT(type O AND Rh-)) = 1 – P (type O AND Rh-) = 1 – 0.14 = 0.86 86% of people do not have type O, Rh- blood. According to the American Red Cross, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Their data show that 45% of people have type O blood and 7% of people have Rh- factor; 38% of people have type O or Rh- factor. Find the probability that a person has both type O blood and the Rh- factor. Find the probability that a person does NOT have both type O blood and the Rh- factor. P (type O OR Rh-) = P (type O) + P (Rh-) - P (type O AND Rh-) 0.52 = 0.43 + 0.15 - P (type O AND Rh-); solve to find P (type O AND Rh-) = 0.06 6% of people have type O, Rh- blood P (NOT(type O AND Rh-)) = 1 - P (type O AND Rh-) = 1 - 0.06 = 0.94 94% of people do not have type O, Rh- blood At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper. Find the probability that a course has a final exam or a research project. Find the probability that a course has NEITHER of these two requirements. In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. In the box, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts. Find the probability that a cookie contains chocolate or nuts (he can't eat it). Find the probability that a cookie does not contain chocolate or nuts (he can eat it). Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts. P ( C OR N ) = P ( C ) + P ( N ) - P ( C AND N ) = 0.36 + 0.12 - 0.08 = 0.40 P (NEITHER chocolate NOR nuts) = 1 - P ( C OR N ) = 1 - 0.40 = 0.60 A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student Find P ( D AND E ). Find P ( E | D ). Find P ( D OR E ). Using an appropriate test, show whether D and E are independent. Using an appropriate test, show whether D and E are mutually exclusive. Independent Events The occurrence of one event has no effect on the probability of the occurrence of another event. Events A and B are independent if one of the following is true: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A AND B ) = P ( A ) P ( B ) Mutually Exclusive Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events A and B are mutually exclusive, then P ( A AND B ) = 0.", "section": "Two Basic Rules of Probability", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Contingency Tables A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner. Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: Speeding violation in the last year No speeding violation in the last year Total Uses cell phone while driving 25 280 305 Does not use cell phone while driving 45 405 450 Total 70 685 755 The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. Calculate the following probabilities using the table. a. Find P (Driver is a cell phone user). b. Find P (driver had no violation in the last year). c. Find P (Driver had no violation in the last year AND was a cell phone user). d. Find P (Driver is a cell phone user OR driver had no violation in the last year). e. Find P (Driver is a cell phone user GIVEN driver had a violation in the last year). f. Find P (Driver had no violation last year GIVEN driver was not a cell phone user) a. number of cell phone users total number in study = 305 755 b. number that had no violation total number in study = 685 755 c. 280 755 d. ( 305 755 + 685 755 ) − 280 755 = 710 755 e. 25 70 (The sample space is reduced to the number of drivers who had a violation.) f. 405 450 (The sample space is reduced to the number of drivers who were not cell phone users.) Try it shows the number of athletes who stretch before exercising and how many had injuries within the past year. Injury in last year No injury in last year Total Stretches 55 295 350 Does not stretch 231 219 450 Total 286 514 800 What is P (athlete stretches before exercising)? What is P (athlete stretches before exercising|no injury in the last year)? shows a random sample of 100 hikers and the areas of hiking they prefer. Hiking Area Preference Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Women 18 16 45 Men 14 55 Total 41 a. Complete the table. a. Hiking Area Preference Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Women 18 16 11 45 Men 16 25 14 55 Total 34 41 25 100 b. Are the events \"being a woman\" and \"preferring the coastline\" independent events? Let F = being a woman and let C = preferring the coastline. Find P ( F AND C ). Find P ( F ) P ( C ) Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent. b. P ( F AND C ) = 18 100 = 0.18 P ( F ) P ( C ) = ( 45 100 ) ( 34 100 ) = (0.45)(0.34) = 0.153 P ( F AND C ) ≠ P ( F ) P ( C ), so the events F and C are not independent. c. Find the probability that a person is a man given that the person prefers hiking near lakes and streams. Let M = being a man, and let L = prefers hiking near lakes and streams. What word tells you this is a conditional? Fill in the blanks and calculate the probability: P (___|___) = ___. Is the sample space for this problem all 100 hikers? If not, what is it? c. The word 'given' tells you that this is a conditional. P ( M | L ) = 25 41 No, the sample space for this problem is the 41 hikers who prefer lakes and streams. d. Find the probability that a person is a woman or prefers hiking on mountain peaks. Let F = being a woman, and let P = prefers mountain peaks. Find P ( F ). Find P ( P ). Find P ( F AND P ). Find P ( F OR P ). d. P ( F ) = 45 100 P ( P ) = 25 100 P ( F AND P ) = 11 100 P ( F OR P ) = 45 100 + 25 100 - 11 100 = 59 100 Try It shows a random sample of 200 cyclists and the routes they prefer. Let O = older and H = hilly path. Age Group Lake Path Hilly Path Wooded Path Total Younger 45 38 27 110 Older 26 52 12 90 Total 71 90 39 200 Out of the older group, what is the probability that the cyclist prefers a hilly path? Are the events “being older” and “preferring the hilly path” independent events? Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 and the probability he is not caught is 4 5 . If he goes out the second door, the probability he gets caught by Alissa is 1 4 and the probability he is not caught is 3 4 . The probability that Alissa catches Muddy coming out of the third door is 1 2 and the probability she does not catch Muddy is 1 2 . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1 3 . Door Choice Caught or Not Door One Door Two Door Three Total Caught 1 15 1 12 1 6 Not Caught 4 15 3 12 1 6 Total 1 The first entry 1 15 = ( 1 5 ) ( 1 3 ) is P (Door One AND Caught) The entry 4 15 = ( 4 5 ) ( 1 3 ) is P (Door One AND Not Caught) Verify the remaining entries. a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. b. What is the probability that Alissa does not catch Muddy? c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa? a. Door Choice Caught or Not Door One Door Two Door Three Total Caught 1 15 1 12 1 6 19 60 Not Caught 4 15 3 12 1 6 41 60 Total 5 15 4 12 2 6 1 b. 41 60 c. 9 19 Try It Anna has to buy a new car. She has two choices, car A and car B. Anna can afford only one car. The probability that Anna will buy car A is P ( A ) = 0 . 25 , and the probability that Anna will buy car B is P ( B ) = 0 . 65 . Find: P ( A A N D B ) P ( A O R B ) P ( A A N D B ) = 0 because it is given that Anna can afford only one car. Since P ( A A N D B ) = 0 , then P ( A O R B ) = P ( A ) + P ( B ) = 0 . 25 + 0 . 65 = 0 . 9 . Note that the probability that Anna does not a car must be 0.1. contains the number of crimes per 100,000 inhabitants in the United States over the span of several years. United States Crime Index Rates Per 100,000 Inhabitants Year Robbery Burglary Vandalism Vehicle Total 1 145.7 732.1 29.7 314.7 2 133.1 717.7 29.1 259.2 3 119.3 701 27.7 239.1 4 113.7 702.2 26.8 229.6 Total TOTAL each column and each row. Total data = 4,520.7 Find P (Year 2 AND Robbery). Find P (Year 3 AND Burglary). Find P (Year 3 OR Burglary). Find P (Year 4 | Vandalism). Find P (Vehicle | Year 1). a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575 Try It relates the weights and heights of a group of individuals participating in an observational study. Weight/Height Tall Medium Short Totals Overweight 18 28 14 Typical Weight Range 20 51 28 Underweight 12 25 9 Totals Find the total for each row and column Find the probability that a randomly chosen individual from this group is Tall. Find the probability that a randomly chosen individual from this group is Overweight and Tall. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Overweight. Find the probability that a randomly chosen individual from this group is Overweight given that the individual is Tall. Find the probability a randomly chosen individual from this group is Tall and Underweight. Are the events Overweight and Tall independent? References “Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (accessed May 3, 2013). Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services. Data from United States Senate. Available online at www.senate.gov (accessed May 2, 2013). Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013). “Human Blood Types.” Unite Blood Services, 2011. Available online at http://www.unitedbloodservices.org/learnMore.aspx (accessed May 2, 2013). Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013). “United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013). Chapter Review There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables. Use the following information to answer the next four exercises. shows a random sample of musicians and how they learned to play their instruments. Gender Self-taught Studied in School Private Instruction Total Woman 12 38 22 72 Man 19 24 15 58 Total 31 62 37 130 Find P (musician is a woman). Find P (musician is a man AND had private instruction). P (musician is a male AND had private instruction) = 15 130 = 3 26 = 0.12 Find P (musician is a woman OR is self taught). Are the events “being a woman musician” and “learning music in school” mutually exclusive events? The events are not mutually exclusive. It is possible to be a woman musician who learned music in school. Bringing It Together Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine , reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese people, and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Hispanic/Latino people, 2,305 Japanese people, and 3,970 White people. Complete the table using the data provided. Smoking Levels by Ethnicity Smoking Level Black Native Hawaiian Hispanic/Latino Japanese people White TOTALS 1–10 11–20 21–30 31+ TOTALS Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day. 35,065 100,450 Find the probability that the person was Hispanic/Latino. In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability. To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is 4,715 100,450 . In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability. In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability. To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is 4715 15,273 . Prove that smoking level/day and ethnicity are dependent events. Homework Use the information in the to answer the next eight exercises. The table shows the political party affiliation for various members of the U.S. Senate during two separate years when they are up for reelection. Up for reelection: Democratic Party Republican Party Other Total Year A 20 13 0 Year B 10 24 0 Total What is the probability that a randomly selected senator has an “Other” affiliation? 0 What is the probability that a randomly selected senator is up for reelection in Year B? What is the probability that a randomly selected senator is a Democrat and up for reelection in Year B? 10 67 What is the probability that a randomly selected senator is a Republican or is up for reelection in Year A? Suppose that a member of the U.S. Senate is randomly selected. Given that the randomly selected senator is up for reelection in Year B, what is the probability that this senator is a Democrat? 10 34 Suppose that a member of the U.S. Senate is randomly selected. What is the probability that the senator is up for reelection in Year A, knowing that this senator is a Republican? The events “Republican” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. d The events “Other” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. gives the number of participants in the recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted by age, race (Black or White), and sex. We are interested in possible relationships between age, race, and sex. Race and sex 15–24 25–40 41–65 Over 65 TOTALS White, male 1,165 2,036 3,703 8,395 White, female 1,076 2,242 4,060 9,129 Black, male 142 194 384 824 Black, female 131 290 486 1,061 All others TOTALS 2,792 5,279 9,354 21,081 Do not include \"all others\" for parts f and g. Fill in the column for cancer treatment for individuals over age 65. Fill in the row for all other races. Find the probability that a randomly selected individual was a White male. Find the probability that a randomly selected individual was a Black female. Find the probability that a randomly selected individual was Black Find the probability that a randomly selected individual was male. Out of the individuals over age 65, find the probability that a randomly selected individual was a Black or White male. Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 156 TOTALS 2,792 5,279 9,354 3,656 21,081 Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 278 517 721 156 1672 TOTALS 2,792 5,279 9,354 3,656 21,081 8,395 21,081 ≈ 0.3982 1,061 21,081 ≈ 0.0503 1,885 21,081 ≈ 0.0894 9,219 19,409 ≈ 0.475 1,595 3,500 ≈ 0.456 Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected. NAME Single Double Triple Home Run TOTAL HITS Babe Ruth 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,035 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 TOTAL 7,918 2,127 583 1,723 12,351 Find P (hit was made by Babe Ruth). 1518 2873 2873 12351 583 12351 4189 12351 Find P (hit was made by Ty Cobb|The hit was a Home Run). 4189 12351 117 1723 1723 4189 117 12351 b identifies a group of children by one of four hair colors, and by type of hair. Hair Type Brown Blond Black Red Totals Wavy 20 15 3 43 Straight 80 15 12 Totals 20 215 Complete the table. What is the probability that a randomly selected child will have wavy hair? What is the probability that a randomly selected child will have either brown or blond hair? What is the probability that a randomly selected child will have wavy brown hair? What is the probability that a randomly selected child will have red hair, given that they have straight hair? If B is the event of a child having brown hair, find the probability of the complement of B . In words, what does the complement of B represent? In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data were compiled into the following table. Shirt# ≤ 210 211–250 251–290 > 290 1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. Find the probability that his shirt number is from 1 to 33. Find the probability that he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds. 26 106 33 106 21 106 ( 26 106 ) + ( 33 106 ) - ( 21 106 ) = ( 38 106 ) 21 33 contingency table the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities.", "section": "Contingency Tables", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Tree and Venn Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities. Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of \"branches\" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram. In an urn, there are 11 balls. Three balls are red ( R ) and eight balls are blue ( B ). Draw two balls, one at a time, with replacement . \"With replacement\" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. Total = 64 + 24 + 24 + 9 = 121 The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R 1, R 2, and R 3 and each blue ball as B 1, B 2, B 3, B 4, B 5, B 6, B 7, and B 8. Then the nine RR outcomes can be written as: R 1 R 1 R 1 R 2 R 1 R 3 R 2 R 1 R 2 R 2 R 2 R 3 R 3 R 1 R 3 R 2 R 3 R 3 The other outcomes are similar. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space . a. List the 24 BR outcomes: B 1 R 1, B 1 R 2, B 1 R 3, ... b. Using the tree diagram, calculate P ( RR ). c. Using the tree diagram, calculate P ( RB OR BR ). d. Using the tree diagram, calculate P ( R on 1st draw AND B on 2nd draw). e. Using the tree diagram, calculate P ( R on 2nd draw GIVEN B on 1st draw). f. Using the tree diagram, calculate P ( BB ). g. Using the tree diagram, calculate P ( B on the 2nd draw given R on the first draw). a. B 1 R 1 B 1 R 2 B 1 R 3 B 2 R 1 B 2 R 2 B 2 R 3 B 3 R 1 B 3 R 2 B 3 R 3 B 4 R 1 B 4 R 2 B 4 R 3 B 5 R 1 B 5 R 2 B 5 R 3 B 6 R 1 B 6 R 2 B 6 R 3 B 7 R 1 B 7 R 2 B 7 R 3 B 8 R 1 B 8 R 2 B 8 R 3 b. P ( RR ) = ( 3 11 ) ( 3 11 ) = 9 121 c. P ( RB OR BR ) = ( 3 11 ) ( 8 11 ) + ( 8 11 ) ( 3 11 ) = 48 121 d. P ( R on 1st draw AND B on 2nd draw) = P ( RB ) = ( 3 11 ) ( 8 11 ) = 24 121 e. P ( R on 2nd draw GIVEN B on 1st draw) = P ( R on 2nd| B on 1st) = 24 88 = 3 11 This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB ). Twenty-four of the 88 possible outcomes are BR . 24 88 = 3 11 . f. P ( BB ) = 64 121 g. P ( B on 2nd draw| R on 1st draw) = 8 11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB ). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 24 33 . Try It In a standard deck, there are 52 cards. 12 cards are face cards (event F ) and 40 cards are not face cards (event N ). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P ( FF ). An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. \"Without replacement\" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, ( 3 11 ) ( 2 10 ) = 6 110 . Total = 56 + 24 + 24 + 6 110 = 110 110 = 1 NOTE If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement , so that on the second draw there are ten marbles left in the urn. Calculate the following probabilities using the tree diagram. a. P ( RR ) = ________ b. Fill in the blanks: P ( RB OR BR ) = ( 3 11 ) ( 8 10 ) + (___)(___) = 48 110 c. P ( R on 2nd| B on 1st) = d. Fill in the blanks. P ( R on 1st AND B on 2nd) = P ( RB ) = (___)(___) = 24 110 e. Find P ( BB ). f. Find P ( B on 2nd| R on 1st). a. P ( RR ) = ( 3 11 ) ( 2 10 ) = 6 110 b. P ( RB OR BR ) = ( 3 11 ) ( 8 10 ) + ( 8 11 ) ( 3 10 ) = 48 110 c. P ( R on 2nd| B on 1st) = 3 10 d. P ( R on 1st AND B on 2nd) = P ( RB ) = ( 3 11 ) ( 8 10 ) = 24 110 e. P ( BB ) = ( 8 11 ) ( 7 10 ) f. Using the tree diagram, P ( B on 2nd| R on 1st) = P ( R | B ) = 8 10 . If we are using probabilities, we can label the tree in the following general way. P ( R | R ) here means P ( R on 2nd| R on 1st) P ( B | R ) here means P ( B on 2nd| R on 1st) P ( R | B ) here means P ( R on 2nd| B on 1st) P ( B | B ) here means P ( B on 2nd| B on 1st) Try It In a standard deck, there are 52 cards. Twelve cards are face cards ( F ) and 40 cards are not face cards ( N ). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities. Find P ( FN OR NF ). Find P ( N | F ). Find P (at most one face card). Hint: \"At most one face card\" means zero or one face card. Find P (at least one face card). Hint: \"At least one face card\" means one or two face cards. A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption. What is the probability that both kittens are tabby? a. ( 1 2 ) ( 1 2 ) b. ( 4 9 ) ( 4 9 ) c. ( 4 9 ) ( 3 8 ) d. ( 4 9 ) ( 5 9 ) What is the probability that one kitten of each coloring is selected? a. ( 4 9 ) ( 5 9 ) b. ( 4 9 ) ( 5 8 ) c. ( 4 9 ) ( 5 9 ) + ( 5 9 ) ( 4 9 ) d. ( 4 9 ) ( 5 8 ) + ( 5 9 ) ( 4 8 ) What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? What is the probability of choosing two kittens of the same color? a. c, b. d, c. 4 8 , d. 32 72 Try It Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? Venn Diagram A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows: Try It Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation. Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = { TT , TH } and B = { TT , HT }. Therefore, A AND B = { TT }. A OR B = { TH , TT , HT }. The sample space when you flip two fair coins is X = { HH , HT , TH , TT }. The outcome HH is in NEITHER A NOR B . The Venn diagram is as follows: Try It Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time. If a student is selected at random, find the probability that the student belongs to a club. P ( C ) = 0.40 the probability that the student works part time. P ( PT ) = 0.50 the probability that the student belongs to a club AND works part time. P ( C AND PT ) = 0.05 the probability that the student belongs to a club given that the student works part time. P ( C | P T ) = P ( C AND P T ) P ( P T ) = 0.05 0.50 = 0.1 the probability that the student belongs to a club OR works part time. P ( C OR PT ) = P ( C ) + P ( PT ) - P ( C AND PT ) = 0.40 + 0.50 - 0.05 = 0.85 Try It Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works. A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood. The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor. We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor. P ( O ) = ___________ P ( R ) = ___________ P ( O AND R ) = ___________ P ( O OR R ) = ____________ In the Venn Diagram, describe the overlapping area using a complete sentence. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence. a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor. Try It In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2. Draw a Venn diagram representing the situation. Find the probability that the customer buys either a novel or a non-fiction book. In the Venn diagram, describe the overlapping area using a complete sentence. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. References Data from Clara County Public H.D. Data from the American Cancer Society. Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013). Data from the Federal Highway Administration, part of the United States Department of Transportation. Data from the United States Census Bureau, part of the United States Department of Commerce. Data from USA Today. “Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013). “Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html (accessed May 2, 2013). Chapter Review A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation. Homework Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red ( R ), four yellow ( Y ) and five blue ( B ) beads. For the coin, P ( H ) = 2 3 and P ( T ) = 1 3 where H is heads and T is tails. Find P (tossing a Head on the coin AND a Red bead) 2 3 5 15 6 36 5 36 Find P (Blue bead). 15 36 10 36 10 12 6 36 a A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?) Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain. For each complete path through the tree, write the event it represents and find the probabilities. Let S be the event that both cookies selected were the same flavor. Find P ( S ). Let T be the event that the cookies selected were different flavors. Find P ( T ) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods. Let U be the event that the second cookie selected is a butter cookie. Find P ( U ). Bringing It Together Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. Suppose that you randomly draw two cards, one at a time, with replacement . Let G 1 = first card is green Let G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 AND G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 independent events? Explain why or why not. P ( GG ) = ( 5 8 ) ( 5 8 ) = 25 64 P (at least one green) = P ( GG ) + P ( GY ) + P ( YG ) = 25 64 + 15 64 + 15 64 = 55 64 P ( G | G ) = 5 8 Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two. Suppose that you randomly draw two cards, one at a time, without replacement . G 1 = first card is green G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 AND G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 independent events? Explain why or why not. Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are women is 48.60. Of the women, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. men drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. Complete the following. Construct a table or a tree diagram of the situation. Find P (driver is a woman). Find P (driver is age 65 or over|driver is a woman). Find P (driver is age 65 or over AND a woman). In words, explain the difference between the probabilities in part c and part d. Find P (driver is age 65 or over). Are being age 65 or over and being a woman mutually exclusive events? How do you know? <20 20–64 >64 Totals Women 0.0244 0.3954 0.0661 0.486 Men 0.0259 0.4186 0.0695 0.514 Totals 0.0503 0.8140 0.1356 1 P ( F ) = 0.486 P (>64| F ) = 0.1361 P (>64 and F ) = P ( F ) P (>64| F ) = (0.486)(0.1361) = 0.0661 P (>64| F ) is the percentage of women drivers who are 65 or older and P (>64 and F ) is the percentage of drivers who are women and 65 or older. P (> 64 ) = P (>64 and F ) + P (>64 and M ) = 0.1356 No, being a woman and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F ) = 0.0661. Suppose that 10,000 U.S. licensed drivers are randomly selected. How many would you expect to be men? Using the table or tree diagram, construct a contingency table of gender versus age group. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is a woman. Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work. Assuming that the walkers walk alone, what percent of all commuters travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool? Car, Truck or Van Walk Public Transportation Other Totals Alone 0.7318 Not Alone 0.1332 Totals 0.8650 0.0390 0.0530 0.0430 1 If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P (Alone) = 0.7318 + 0.0390 = 0.7708. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771 (0.1332)(1,000) = 133 When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H ) while 110 showed a tail (event T ). On that basis, they claimed that it is not a fair coin. Based on the given data, find P ( H ) and P ( T ). Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin. Use the tree to find the probability of obtaining at least one head. Use the following information to answer the next two exercises. The following data represent the number of new vehicles purchased over a monthly time period in a certain county, within two age groups: people in their twenties and people in their thirties. Sedan SUV Minivan Other Totals Twenties 1,135 290 583 158 Thirties 1,246 463 241 190 Totals Suppose a person from this county is randomly selected. Find P (Person is in their twenties). Find P (Person purchases minivan). Find P (Person is in their twenties OR purchases an SUV). Find P (Person is in their twenties AND purchases a SEDAN). Find P (Person is in their thirties AND purchases an SUV). Find P (Person is in their twenties GIVEN person purchases a minivan). The completed contingency table is as follows: New Vehicles Purchased by Gender Sedan SUV Minivan Other Totals Twenties 1,135 290 583 158 2166 Thirties 1246 463 241 190 2140 Totals 2381 753 824 348 4306 Suppose a person from this county is randomly selected. P (Person is in their twenties) = 0.5030 P (Person purchases minivan) = 0.1914 P (Person is in their twenties OR has purchases an SUV) = 0.6105 P (Person is in their twenties AND purchases a SEDAN) = 0.0673 P (Person is in their thirties AND has purchases an SUV) = 0.1075 P (Person is in their twenties GIVEN person purchases a minivan) = 0.7075 Answer these questions using probability rules. Do NOT use the contingency table. A total of 6,481 new vehicles were purchased over a monthly time period in a certain county. These purchases will be our population. Of these purchases, 36% were made by people in their twenties, and 42% were purchases of sedans. 20% of purchases were for sedans purchased by people in their twenties. Find P (Person is in their twenties). Find P (Person purchased a sedan). Find P (Person is in their twenties GIVEN person purchased a sedan) P (Person is in their twenties) = 0.36. P (Person purchased a sedan) = 0.42 P (Person is in their twenties a woman GIVEN person purchased a sedan) = 0.4762 Tree Diagram the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies) Venn Diagram the visual representation of a sample space and events in the form of circles or ovals showing their intersections", "section": "Tree and Venn Diagrams", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Probability Topics Probability Topics Class time: Names: Student Learning Outcomes The student will use theoretical and empirical methods to estimate probabilities. The student will appraise the differences between the two estimates. The student will demonstrate an understanding of long-term relative frequencies. Do the Experiment Count out 40 mixed-color M&Ms®, which is approximately one small bag’s worth. Record the number of each color in . Use the information from this table to complete . Next, put the M&Ms in a cup. The experiment is to pick two M&Ms, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of . Do this 12 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of . After you record the pick, put both M&Ms back. Do this a total of 12 times, also. Use the data from to calculate the empirical probabilities shown in . Leave your answers in unreduced fractional form. Do not multiply out any fractions. Population Color Quantity Yellow ( Y ) Green ( G ) Blue ( BL ) Brown ( B ) Orange ( O ) Red ( R ) Theoretical Probabilities With Replacement Without Replacement P (2 reds) P ( R 1 B 2 OR B 1 R 2 ) P ( R 1 AND G 2 ) P ( G 2 | R 1 ) P (no yellows) P (doubles) P (no doubles) NOTE G 2 = green on second pick; R 1 = red on first pick; B 1 = brown on first pick; B 2 = brown on second pick; doubles = both picks are the same colour. Empirical Results With Replacement Without Replacement ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ ) Empirical Probabilities With Replacement Without Replacement P (2 reds) P ( R 1 B 2 OR B 1 R 2 ) P ( R 1 AND G 2 ) P ( G 2 | R 1 ) P (no yellows) P (doubles) P (no doubles) Discussion Questions Why are the “With Replacement” and “Without Replacement” probabilities different? Convert P (no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to four decimal places. Theoretical “With Replacement”: P (no yellows) = _______ Empirical “With Replacement”: P (no yellows) = _______ Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why? If you increased the number of times you picked two M&Ms to 240 times, why would empirical probability values change? Would this change (see part 3) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know? Explain the differences in what P ( G 1 AND R 2 ) and P ( R 1 | G 2 ) represent. Hint: Think about the sample space for each probability.", "section": "Probability Topics", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction You can use probability and discrete random variables to calculate the likelihood of lightning striking the ground five times during a half-hour thunderstorm. (credit: modification of work “CG lightning strike” by Axel Rouvin/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Recognize and understand discrete probability distribution functions, in general. Calculate and interpret expected values. Recognize the binomial probability distribution and apply it appropriately. Recognize the Poisson probability distribution and apply it appropriately. Recognize the geometric probability distribution and apply it appropriately. Recognize the hypergeometric probability distribution and apply it appropriately. Classify discrete word problems by their distributions. A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, they could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%? The manager of an auto dealership might be interested in the color preferences for new car buyers. Suppose on average the dealership sells 20 cars per month. What is the probability that a customer prefers red cars? These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment. Random Variable Notation Upper case letters such as X or Y denote a random variable. Lower case letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is given as a number. For example, let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT ; THH ; HTH ; HHT ; HTT ; THT ; TTH ; HHH . Then, x = 0, 1, 2, 3. X is in words and x is a number. Notice that for this example, the x values are countable outcomes. Because you can count the possible values that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is a discrete random variable. Toss a coin ten times and record the number of heads. After all members of the class have completed the experiment (tossed a coin ten times and counted the number of heads), fill in . Let X = the number of heads in ten tosses of the coin. x Frequency of x Relative Frequency of x Which value(s) of x occurred most frequently? If you tossed the coin 1,000 times, what values could x take on? Which value(s) of x do you think would occur most frequently? What does the relative frequency column sum to? Random Variable (RV) a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters X , Y , Z ,...; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters x, y, and z . For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in the two following ways. The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange}. We can tell what specific value x the random variable X takes only after performing the experiment.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Probability Distribution Function (PDF) for a Discrete Random Variable A discrete probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one. A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5. P ( x ) = probability that X takes on a value x . x P ( x ) 0 P ( x = 0) = 2 50 1 P ( x = 1) = 11 50 2 P ( x = 2) = 23 50 3 P ( x = 3) = 9 50 4 P ( x = 4) = 4 50 5 P ( x = 5) = 1 50 X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because: Each P ( x ) is between zero and one, inclusive. The sum of the probabilities is one, that is, 2 50 + 11 50 + 23 50 + 9 50 + 4 50 + 1 50 = 1 Try It A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P ( x ) = the probability that X takes on value x . Why is this a discrete probability distribution function (two reasons)? x P ( x ) 0 P ( x = 0) = 4 50 1 P ( x = 1) = 8 50 2 P ( x = 2) = 16 50 3 P ( x = 3) = 14 50 4 P ( x = 4) = 6 50 5 P ( x = 5) = 2 50 Suppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected. a. Let X = the number of days Nancy ____________________. b. X takes on what values? c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in . The table should have two columns labeled x and P ( x ). What does the P ( x ) column sum to? a. Let X = the number of days Nancy attends class per week. b. 0, 1, 2, and 3 c. x P ( x ) 0 0.01 1 0.04 2 0.15 3 0.80 Try It Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on? Chapter Review The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows: Each probability is between zero and one, inclusive ( inclusive means to include zero and one). The sum of the probabilities is one. Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution. Let X = the number of years a new hire will stay with the company. Let P ( x ) = the probability that a new hire will stay with the company x years. Complete using the data provided. x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 5 0.10 6 0.05 x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 0.10 5 0.10 6 0.05 P ( x = 4) = _______ P ( x ≥ 5) = _______ 0.10 + 0.05 = 0.15 On average, how long would you expect a new hire to stay with the company? What does the column “ P ( x )” sum to? 1 Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution. x P ( x ) 1 0.15 2 0.35 3 0.40 4 0.10 Define the random variable X . What is the probability the baker will sell more than one batch? P ( x > 1) = _______ 0.35 + 0.40 + 0.10 = 0.85 What is the probability the baker will sell exactly one batch? P ( x = 1) = _______ On average, how many batches should the baker make? 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45 Use the following information to answer the next four exercises: Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. Define the random variable X . Construct a probability distribution table for the data. x P ( x ) 0 0.03 1 0.04 2 0.08 3 0.85 We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic? Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time. Define the random variable X . Let X = the number of events Javier volunteers for each month. What values does x take on? Construct a PDF table. x P ( x ) 0 0.05 1 0.05 2 0.10 3 0.20 4 0.25 5 0.35 Find the probability that Javier volunteers for less than three events each month. P ( x < 3) = _______ Find the probability that Javier volunteers for at least one event each month. P ( x > 0) = _______ 1 – 0.05 = 0.95 Homework Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in . x P ( x ) 3 0.05 4 0.40 5 0.30 6 0.15 7 0.10 In words, define the random variable X . What does it mean that the values zero, one, and two are not included for x in the PDF? Probability Distribution Function (PDF) a mathematical description of a discrete random variable ( RV ), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome.", "section": "Probability Distribution Function (PDF) for a Discrete Random Variable", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Mean or Expected Value and Standard Deviation The expected value is often referred to as the \"long-term\" average or mean . This means that over the long term of doing an experiment over and over, you would expect this average. You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. As you learned in Probability Topics , probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers . The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together) . When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ . In other words, after conducting many trials of an experiment, you would expect this average value. The mean, μ , of a discrete probability function is the expected value. μ = ∑ ( x ∙ P x ) The standard deviation, σ , of the PDF is the square root of the variance. σ = ∑ [ x – μ 2 ∙ Ρ x ] When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes. NOTE To find the expected value or long term average, μ , simply multiply each value of the random variable by its probability and add the products. A soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, μ , of the number of days per week the soccer team plays soccer. To do the problem, first let the random variable X = the number of days the soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column x * P ( x ). In this column, you will multiply each x value by its probability. Expected Value Table This table is called an expected value table. The table helps you calculate the expected value or long-term average. x P ( x ) x * P ( x ) 0 0.2 (0)(0.2) = 0 1 0.5 (1)(0.5) = 0.5 2 0.3 (2)(0.3) = 0.6 Add the last column x * P ( x ) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1. The expected value is 1.1. The soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the soccer team plays soccer week after week after week. We say μ = 1.1. Try It A customer orders 1, 2, or 3 bottles of water. The probability of ordering 1 bottle is 0.6, the probability of ordering 2 bottles is 0.3, and the probability of ordering 3 bottles is 0.1. Find the long-term average or expected value, μ, of the number of bottles a customer will order. Let random variable X = the number of bottles of water a customer orders. X can take values 1, 2, 3. Construct a PDF table. x P ( x ) x * P ( x ) 1 0.6 ( 1 ) ( 0 . 6 ) = 0 . 6 2 0.3 ( 2 ) ( 0 . 3 ) = 0 . 6 3 0.1 ( 3 ) ( 0 . 1 ) = 0 . 3 The total of x * P ( x ) is obtained as 0 . 6 + 0 . 6 + 0 . 3 = 1 . 5 . So, the long-term average or expected value is obtained as μ = 1.5. Find the expected value of the number of times a newborn baby's crying wakes its parents after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its parents after midnight. Calculate the standard deviation of the variable as well. You expect a newborn to wake its parents after midnight 2.1 times per week, on the average. x P ( x ) x * P ( x ) ( x – μ ) 2 ⋅ P ( x ) 0 P ( x = 0) = 2 50 (0) ( 2 50 ) = 0 (0 – 2.1) 2 ⋅ 0.04 = 0.1764 1 P ( x = 1) = ( 11 50 ) (1) ( 11 50 ) = 11 50 (1 – 2.1) 2 ⋅ 0.22 = 0.2662 2 P ( x = 2) = 23 50 (2) ( 23 50 ) = 46 50 (2 – 2.1) 2 ⋅ 0.46 = 0.0046 3 P ( x = 3) = 9 50 (3) ( 9 50 ) = 27 50 (3 – 2.1) 2 ⋅ 0.18 = 0.1458 4 P ( x = 4) = 4 50 (4) ( 4 50 ) = 16 50 (4 – 2.1) 2 ⋅ 0.08 = 0.2888 5 P ( x = 5) = 1 50 (5) ( 1 50 ) = 5 50 (5 – 2.1) 2 ⋅ 0.02 = 0.1682 Add the values in the third column of the table to find the expected value of X : μ = Expected Value = 105 50 = 2.1 Use μ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value x , multiply the square of its deviation by its probability. (Each deviation has the format x – μ ). Add the values in the fourth column of the table: 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 The standard deviation of X is the square root of this sum: σ = 1.05 ≈ 1.0247 Try It A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value? x P ( x ) 0 P ( x = 0) = 4 50 1 P ( x = 1) = 8 50 2 P ( x = 2) = 16 50 3 P ( x = 3) = 14 50 4 P ( x = 4) = 6 50 5 P ( x = 5) = 2 50 Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game? To do this problem, set up an expected value table for the amount of money you can profit. Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of x are 100,000 dollars and −2 dollars. To win, you must get all five numbers correct, in order. The probability of choosing one correct number is 1 10 because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is ( 1 10 ) ( 1 10 ) ( 1 10 ) ( 1 10 ) ( 1 10 ) = ( 1 ) ( 10 − 5 ) = 0.00001. Therefore, the probability of winning is 0.00001 and the probability of losing is 1 − 0.00001 = 0.99999. The expected value table is as follows: Αdd the last column. –1.99998 + 1 = –0.99998 x P ( x ) x * P ( x ) Loss –2 0.99999 (–2)(0.99999) = –1.99998 Profit 100,000 0.00001 (100000)(0.00001) = 1 Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over. Try It You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term? Suppose you play a game with a biased coin. You play each game by tossing the coin once. P (heads) = 2 3 and P (tails) = 1 3 . If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, will you come out ahead? a. Define a random variable X . b. Complete the following expected value table. x WIN 10 1 3 LOSE –12 3 c. What is the expected value, μ ? Do you come out ahead? a. X = amount of profit b. x P ( x ) xP ( x ) WIN 10 1 3 10 3 LOSE –6 2 3 –12 3 c. Add the last column of the table. The expected value μ = – 2 3 . You lose, on average, about 67 cents each time you play the game so you do not come out ahead. Try It Suppose you play a game with a spinner. You play each game by spinning the spinner once. P (red) = 2 5 , P (blue) = 2 5 , and P (green) = 1 5 . If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win $10. Complete the following expected value table. x P ( x ) Red – 20 5 Blue 2 5 Green 10 Like data, probability distributions have standard deviations. To calculate the standard deviation ( σ ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled ( x – μ ) 2 P ( x ) and take the square root. x P ( x ) x * P ( x ) ( x – μ ) 2 P ( x ) 0 0.2 (0)(0.2) = 0 (0 – 1.1) 2 (0.2) = 0.242 1 0.5 (1)(0.5) = 0.5 (1 – 1.1) 2 (0.5) = 0.005 2 0.3 (2)(0.3) = 0.6 (2 – 1.1) 2 (0.3) = 0.243 Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root of 0.49, or σ = 0.49 = 0.7 Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating μ and σ . Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like and calculate the mean μ and standard deviation σ of X . Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Use the sample space to complete the following table: Calculating μ and σ . x P ( x ) x P ( x ) ( x – μ ) 2 ⋅ P ( x ) 0 9 36 0 (0 – 1) 2 ⋅ 9 36 = 9 36 1 18 36 18 36 (1 – 1) 2 ⋅ 18 36 = 0 2 9 36 18 36 (2 – 1) 2 ⋅ 9 36 = 9 36 Add the values in the third column to find the expected value: μ = 36 36 = 1. Use this value to complete the fourth column. Add the values in the fourth column and take the square root of the sum: σ = 18 36 ≈ 0.7071. Try It Two stacks of cards are placed on a table. Each stack has cards number 1 to 8. One card from each stack is picked up. Let X = number of cards showing an odd number. Construct a table like Table 4.11 and calculate the mean µ and standard deviation σ of X . The sample space has 64 outcomes: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (1,7) (1,8) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (2,7) (2,8) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (3,7) (3,8) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (4,7) (4,8) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (5,7) (5,8) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) (6,7) (6,8) (7,1) (7,2) (7,3) (7,4) (7,5) (7,6) (7,7) (7,8) (8,1) (8,2) (8,3) (8,4) (8,5) (8,6) (8,7) (8,8) The sample space is used to complete the following table: x P ( x ) x*P ( x ) x - μ 2 · P x 0 16 64 0 ( 0 - 1 ) 2 · 16 64 = 16 64 1 32 64 32 64 ( 1 - 1 ) 2 · 32 64 = 0 2 16 64 32 64 ( 2 - 1 ) 2 · 16 64 = 16 64 The mean is calculated by adding the values in the third column of P ( x ), so 64 64 = 1 . The square root of the sum of values obtained in the fourth column will give the standard deviation, σ = 32 64 = 0 . 7071 . On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay $20. Let X = the amount of profit from a bet. P (win) = P (one moderate earthquake will occur) = 21.42% P (loss) = P (one moderate earthquake will not occur) = 100% – 21.42% If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of X ? Construct a table similar to and to help you answer these questions. x P(x) x (Px) ( x – μ ) 2 P ( x ) win 50 0.2142 10.71 [50 – (–5.006)] 2 (0.2142) = 648.0964 loss –20 0.7858 –15.716 [–20 – (–5.006)] 2 (0.7858) = 176.6636 Mean = Expected Value = 10.71 + (–15.716) = –5.006. If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5.01 per bet. Standard Deviation = 648.0964 + 176.6636 ≈ 28.7186 Try It On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. As in , you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let X = the amount of profit from a bet. Find the mean and standard deviation of X . Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if they wish to cover these distributions. A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions. References Class Catalogue at the Florida State University. Available online at https://apps.oti.fsu.edu/RegistrarCourseLookup/SearchFormLegacy (accessed May 15, 2013). “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013). Chapter Review The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes. Formula Review Mean or Expected Value: μ = ∑ ​ x ∈ X x P ( x ) Standard Deviation: σ = ∑ ​ x ∈ X ( x − μ ) 2 P ( x ) Complete the expected value table. x P ( x ) x * P ( x ) 0 0.2 1 0.2 2 0.4 3 0.2 Find the expected value from the expected value table. x P ( x ) x * P ( x ) 2 0.1 2(0.1) = 0.2 4 0.3 4(0.3) = 1.2 6 0.4 6(0.4) = 2.4 8 0.2 8(0.2) = 1.6 0.2 + 1.2 + 2.4 + 1.6 = 5.4 Find the standard deviation. x P ( x ) x * P ( x ) ( x – μ ) 2 P ( x ) 2 0.1 2(0.1) = 0.2 (2–5.4) 2 (0.1) = 1.156 4 0.3 4(0.3) = 1.2 (4–5.4) 2 (0.3) = 0.588 6 0.4 6(0.4) = 2.4 (6–5.4) 2 (0.4) = 0.144 8 0.2 8(0.2) = 1.6 (8–5.4) 2 (0.2) = 1.352 Identify the mistake in the probability distribution table. x P ( x ) x * P ( x ) 1 0.15 0.15 2 0.25 0.50 3 0.30 0.90 4 0.20 0.80 5 0.15 0.75 The values of P ( x ) do not sum to one. Identify the mistake in the probability distribution table. x P ( x ) x * P ( x ) 1 0.15 0.15 2 0.25 0.40 3 0.25 0.65 4 0.20 0.85 5 0.15 1 Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. They have the following probability distribution. x P ( x ) x * P ( x ) 1 0.35 2 0.20 3 0.15 4 5 0.10 6 0.05 Define the random variable X . Let X = the number of years a physics major will spend doing post-graduate research. Define P ( x ), or the probability of x . Find the probability that a physics major will do post-graduate research for four years. P ( x = 4) = _______ 1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15 FInd the probability that a physics major will do post-graduate research for at most three years. P ( x ≤ 3) = _______ On average, how many years would you expect a physics major to spend doing post-graduate research? 1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution. Let X = the number of years a student will study ballet with the teacher. Let P ( x ) = the probability that a student will study ballet x years. Complete using the data provided. x P ( x ) x * P ( x ) 1 0.10 2 0.05 3 0.10 4 5 0.30 6 0.20 7 0.10 In words, define the random variable X . X is the number of years a student studies ballet with the teacher. P ( x = 4) = _______ P ( x < 4) = _______ 0.10 + 0.05 + 0.10 = 0.25 On average, how many years would you expect a child to study ballet with this teacher? What does the column \" P ( x )\" sum to and why? The sum of the probabilities sum to one because it is a probability distribution. What does the column \" x * P ( x )\" sum to and why? You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game? − 2 ( 40 52 ) + 30 ( 12 52 ) = − 1.54 + 6.92 = 5.38 You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game? HOMEWORK A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150. What are you interested in here? In words, define the random variable X . List the values that X may take on. Construct a PDF. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle? A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails. If the card is a face card, and the coin lands on Heads, you win $6 If the card is a face card, and the coin lands on Tails, you win $2 If the card is not a face card, you lose $2, no matter what the coin shows. Find the expected value for this game (expected net gain or loss). Explain what your calculations indicate about your long-term average profits and losses on this game. Should you play this game to win money? The variable of interest is X , or the gain or loss, in dollars. The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards. We first need to construct the probability distribution for X . We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value. Card Event X net gain/loss P ( X ) Face Card and Heads 6 ( 12 52 ) ( 1 2 ) = ( 6 52 ) Face Card and Tails 2 ( 12 52 ) ( 1 2 ) = ( 6 52 ) (Not Face Card) and (H or T) –2 ( 40 52 ) ( 1 ) = ( 40 52 ) Expected value = ( 6 ) ( 6 52 ) + ( 2 ) ( 6 52 ) + ( − 2 ) ( 40 52 ) = – 32 52 Expected value = –$0.62, rounded to the nearest cent If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average. You should not play this game to win money because the expected value indicates an expected average loss. You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss. Complete the PDF and answer the questions. x P ( x ) x P ( x ) 0 0.3 1 0.2 2 3 0.4 Find the probability that x = 2. Find the expected value. 0.1 1.6 Suppose that you are offered the following “deal.” You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6. What are you ultimately interested in here (the value of the roll or the money you win)? In words, define the Random Variable X . List the values that X may take on. Construct a PDF. Over the long run of playing this game, what are your expected average winnings per game? Based on numerical values, should you take the deal? Explain your decision in complete sentences. A venture capitalist, willing to invest $1,000,000, has three investments to choose from. The first investment, a software company, has a 10% chance of returning $5,000,000 profit, a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning $3,000,000 profit, a 40% chance of returning $1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning $6,000,000 profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars. Construct a PDF for each investment. Find the expected value for each investment. Which is the safest investment? Why do you think so? Which is the riskiest investment? Why do you think so? Which investment has the highest expected return, on average? Software Company x P ( x ) 5,000,000 0.10 1,000,000 0.30 –1,000,000 0.60 Hardware Company x P ( x ) 3,000,000 0.20 1,000,000 0.40 –1,000,00 0.40 Biotech Firm x P ( x ) 6,00,000 0.10 0 0.70 –1,000,000 0.20 $200,000; $600,000; $400,000 third investment because it has the lowest probability of loss first investment because it has the highest probability of loss second investment Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have. x P ( x ) x P ( x ) 0 0.10 1 0.20 2 0.30 3 4 0.10 5 0.05 6 (or more) 0.05 Find the probability that a married adult has three children. In words, what does the expected value in this example represent? Find the expected value. Is it more likely that a married adult will have two to three children or four to six children? How do you know? Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in . x P ( x ) 3 0.05 4 0.40 5 0.30 6 0.15 7 0.10 On average, how many years do you expect it to take for an individual to earn a B.S.? 4.85 years People visiting video game rental stores often rent more than one game at a time. The probability distribution for game rentals per customer at Game Stop is given in the following table. There is a five-game limit per customer at this store, so nobody ever rents more than five video games. x P ( x ) 0 0.03 1 0.50 2 0.24 3 4 0.07 5 0.04 Describe the random variable X in words. Find the probability that a customer rents three video games. Find the probability that a customer rents at least four video games. Find the probability that a customer rents at most two video games. Another shop, Games Galore, also rents video games. The probability distribution for video game rentals per customer at this shop is given as follows. Games Galore also has a five-game limit per customer. x P ( x ) 0 0.35 1 0.25 2 0.20 3 0.10 4 0.05 5 0.05 At which store is the expected number of video games rented per customer higher? If Game Stop estimates that they will have 300 customers next week, how many video games do they expect to rent next week? Answer in sentence form. If Game Stop expects 300 customers next week, and Games Galore projects that they will have 420 customers, for which store is the expected number of video game rentals for next week higher? Explain. Which of the two stores experiences more variation in the number of video game per customer? How do you know that? A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. Ten of the coupons are for a free gift worth $6. Eighty of the coupons are for a free gift worth $8. Six of the coupons are for a free gift worth $12. Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? Yes, I expect to come out ahead in money. No, I expect to come out behind in money. It doesn’t matter. I expect to break even. b Florida State University has 14 statistics classes scheduled for its summer term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students. What is the average class size assuming each class is filled to capacity? Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X . Find the mean of X . Find the standard deviation of X . In a lottery, there are 250 prizes of $5, 50 prizes of $25, and ten prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even? Let X = the amount of money to be won on a ticket. The following table shows the PDF for X . x P ( x ) 0 0.969 5 250 10,000 = 0.025 25 50 10,000 = 0.005 100 10 10,000 = 0.001 Calculate the expected value of X . 0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35 A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money. Expected Value expected arithmetic average when an experiment is repeated many times; also called the mean. Notations: μ . For a discrete random variable (RV) with probability distribution function P ( x ),the definition can also be written in the form μ = ∑ x P ( x ). Mean a number that measures the central tendency; a common name for mean is ‘average.’ The term ‘mean’ is a shortened form of ‘arithmetic mean.’ By definition, the mean for a sample (detonated by x ¯ ) is x ¯ = Sum of all values in the sample Number of values in the sample and the mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population . Mean of a Probability Distribution the long-term average of many trials of a statistical experiment Standard Deviation of a Probability Distribution a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution σ = ∑ [ x – μ 2 ∙ Ρ x ] The Law of Large Numbers As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero.", "section": "Mean or Expected Value and Standard Deviation", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Binomial Distribution There are three characteristics of a binomial experiment. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials. There are only two possible outcomes, called \"success\" and \"failure,\" for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1. The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p , of a success and probability, q , of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability p = 0.6. Then, q = 0.4. This means that for every true-false statistics question Joe answers, his probability of success ( p = 0.6) and his probability of failure ( q = 0.4) remain the same. The outcomes of a binomial experiment fit a binomial probability distribution . The random variable X = the number of successes obtained in the n independent trials. The mean, μ , and variance, σ 2 , for the binomial probability distribution are μ = np and σ 2 = npq . The standard deviation, σ , is then σ = n p q . Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials. At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A \"success\" could be defined as an individual who withdrew. The random variable X = the number of students who withdraw from the randomly selected elementary physics class. Try It The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a \"success\" be in this case? Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can be stated mathematically as P ( x = 15). Try It A trainer is teaching a rescued dolphin to catch live fish before returning it to the wild. The probability that the dolphin successfully catches a fish is 35%, and the probability that the dolphin does not successfully catch the fish is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically. A coin has been altered to weight the outcome from 0.5 to 0.25 and flipped 5 times. Each flip is independent. What is the probability of getting more than 3 heads? Let X = the number of heads in 5 flips of the fair coin. X takes on the values 0, 1, 2, 3, 4, 5. Since the coin is altered to result in p = 0.25, q is 0.75. The number of trials is n = 5. State the probability question mathematically. First develop fully the probability density function and graph the probability density function. With the fully developed probability density function we can simply read the solution to the question P x > 3 heads. P x > 3 = P x = 4 + P x = 5 = 0 . 0146 + 0 . 0007 = 0 . 0153 . We have added the two individual probabilities because of the addition rule from Probability Topics . also allows us to see the link between the probability density function and probability and area. We also see in the skew of the binomial distribution when p is not equal to 0.5. In the distribution is skewed right as a result of μ = n p = 1 . 25 because p = 0 . 25 . P x = x 0 = n x p x 1 - p n - x = 5 x 0 · 25 x 0 · 75 5 - x 0 etc . μ = np = 1 . 25 Try It A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically. Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly. a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial. b. If we are interested in the number of students who do their homework on time, then how do we define X ? c. What values does x take on? d. What is a \"failure,\" in words? e. If p + q = 1, then what is q ? f. The words \"at least\" translate as what kind of inequality for the probability question P ( x ____ 40). a. failure b. X = the number of statistics students who do their homework on time c. 0, 1, 2, …, 50 d. Failure is defined as a student who does not complete their homework on time. The probability of a success is p = 0.70. The number of trials is n = 50. e. q = 0.30 f. greater than or equal to (≥) The probability question is P ( x ≥ 40). Try It Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem. Notation for the Binomial: B = Binomial Probability Distribution Function X ~ B ( n , p ) Read this as \" X is a random variable with a binomial distribution.\" The parameters are n and p ; n = number of trials, p = probability of a success on each trial. It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education? Let X = the number of workers who have a high school diploma but do not pursue any further education. X takes on the values 0, 1, 2, ..., 20 where n = 20, p = 0.41, and q = 1 – 0.41 = 0.59. X ~ B (20, 0.41) Find P ( x ≤ 12). P ( x ≤ 12) = 0.9738. (calculator or computer) Go into 2 nd DISTR. The syntax for the instructions are as follows: To calculate ( x = value): binompdf( n , p , number) if \"number\" is left out, the result is the binomial probability table. To calculate P ( x ≤ value): binomcdf( n , p , number) if \"number\" is left out, the result is the cumulative binomial probability table. For this problem: After you are in 2 nd DISTR , arrow down to binomcdf . Press ENTER . Enter 20,0.41,12). The result is P ( x ≤ 12) = 0.9738. NOTE If you want to find P ( x = 12), use the pdf (binompdf). If you want to find P ( x > 12), use 1 - binomcdf(20,0.41,12). The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738. The graph of X ~ B (20, 0.41) is as follows: The y -axis contains the probability of x , where X = the number of workers who have only a high school diploma. The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(0.41) = 8.2. The formula for the variance is σ 2 = npq . The standard deviation is σ = n p q . σ = ( 20 ) ( 0.41 ) ( 0.59 ) = 2.20. Try It About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer. In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists. What values does x take on? What is the probability distribution? Find the following probabilities: the probability that two pages feature signature artists the probability that at most six pages feature signature artists the probability that more than three pages feature signature artists. Using the formulas, calculate the (i) mean and (ii) standard deviation. x = 0, 1, 2, 3, 4, 5, 6, 7, 8 X ~ B ( 100 , 8 560 ) P ( x = 2) = binompdf ( 100 , 8 560 , 2 ) = 0.2466 P ( x ≤ 6) = binomcdf ( 100 , 8 560 , 6 ) = 0.9994 P ( x > 3) = 1 – P ( x ≤ 3) = 1 – binomcdf ( 100 , 8 560 , 3 ) = 1 – 0.9443 = 0.0557 Mean = np = (100) ( 8 560 ) = 800 560 ≈ 1.4286 Standard Deviation = n p q = ( 100 ) ( 8 560 ) ( 552 560 ) ≈ 1.1867 Try It According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending. What is the probability distribution for X ? Use your calculator to find the following probabilities: the probability that 25 adults in the sample prefer saving over spending the probability that at most 20 adults prefer saving the probability that more than 30 adults prefer saving Using the formulas, calculate the (i) mean and (ii) standard deviation of X . The lifetime risk of developing cancer is about one in 67 (1.5%). Suppose we randomly sample 200 people. Let X = the number of people who will develop cancer. What is the probability distribution for X ? Using the formulas, calculate the (i) mean and (ii) standard deviation of X . Use your calculator to find the probability that at most eight people develop cancer Is it more likely that five or six people will develop cancer? Justify your answer numerically. X ~ B 200 , 0 . 015 Mean = n p = 200 0 . 015 = 3 Standard Deviation = n p q = 200 ( 0 . 015 ) ( 0 . 985 ) = 1 . 719 P x ≤ 8 = 0 . 9965 The probability that five people develop cancer is 0.1011. The probability that six people develop cancer is 0.0500. Try It During a certain NBA season, a player for the Los Angeles Clippers had the highest field goal completion rate in the league. This player scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by this player during the season. Let X = the number of shots that scored points. What is the probability distribution for X ? Using the formulas, calculate the (i) mean and (ii) standard deviation of X . Use your calculator to find the probability that this player scored with 60 of these shots. Find the probability that this player scored with more than 50 of these shots. The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement . The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is 6 16 . The probability of a student on the second draw is 5 15 , when the first draw selects a student. The probability is 6 15 , when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence. Try It A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why. References “Access to electricity (% of population),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc (accessed May 15, 2015). “Distance Education.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Distance_education (accessed May 15, 2013). “NBA Statistics – 2013,” ESPN NBA, 2013. Available online at http://espn.go.com/nba/statistics/_/seasontype/2 (accessed May 15, 2013). Newport, Frank. “Americans Still Enjoy Saving Rather than Spending: Few demographic differences seen in these views other than by income,” GALLUP® Economy, 2013. Available online at http://www.gallup.com/poll/162368/americans-enjoy-saving-rather-spending.aspx (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011 . Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013). “What are the key statistics about pancreatic cancer?” American Cancer Society, 2013. Available online at http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics (accessed May 15, 2013). Chapter Review A statistical experiment can be classified as a binomial experiment if the following conditions are met: There are a fixed number of trials, n . There are only two possible outcomes, called \"success\" and, \"failure\" for each trial. The letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial. The n trials are independent and are repeated using identical conditions. The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np , and the standard deviation is given by the formula σ = n p q . Formula Review X ~ B ( n , p ) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p . X = the number of successes in n independent trials n = the number of independent trials X takes on the values x = 0, 1, 2, 3, ..., n p = the probability of a success for any trial q = the probability of a failure for any trial p + q = 1 q = 1 – p The mean of X is μ = np . The standard deviation of X is σ = n p q . Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time first-year students from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time first-year students from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status. In words, define the random variable X . X = the number that reply “yes” X ~ _____(_____,_____) What values does the random variable X take on? 0, 1, 2, 3, 4, 5, 6, 7, 8 Construct the probability distribution function (PDF). x P ( x ) On average ( μ ), how many would you expect to answer yes? 5.7 What is the standard deviation ( σ )? What is the probability that at most five of the first-year students reply “yes”? 0.4151 What is the probability that at least two of the first-year students reply “yes”? HOMEWORK According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that they truly have the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Define the random variable and list its possible values. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, ...25 State the distribution of X . Find the probability that at least four of the 25 patients actually have the flu. 0.0165 On average, for every 25 patients calling in, how many do you expect to have the flu? People visiting a hot dog stand often purchase more than one hot dog. The probability distribution for purchases of number of hot dogs per customer at a hot dog stand is given . x P ( x ) 0 0.03 1 0.50 2 0.24 3 4 0.07 5 0.04 Describe the random variable X in words. Find the probability that a customer purchases three hot dogs. Find the probability that a customer purchases at least four hot dogs. Find the probability that a customer purchases at most two hot dogs. X = the number of hot dogs a person at a hot dog stand eats 0.12 0.11 0.77 A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many of the 12 students do we expect to attend the festivities? Find the probability that at most four students will attend. Find the probability that more than two students will attend. Use the following information to answer the next three exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. The expected number of wins for that upcoming month is: 1.67 12 382 1043 4.43 d. 4.43 Let X = the number of games won in that upcoming month. What is the probability that the San Jose Sharks win six games in that upcoming month? 0.1476 0.2336 0.7664 0.8903 What is the probability that the San Jose Sharks win at least five games in that upcoming month 0.3694 0.5266 0.4734 0.2305 c A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct. A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly. X = number of questions answered correctly X ~ B ( 32, 1 3 ) We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P ( x > 24). The event \"more than 24\" is the complement of \"less than or equal to 24.\" Using your calculator's distribution menu: 1 – binomcdf ( 32, 1 3 , 24 ) P ( x > 24) = 0 The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero. Six different colored dice are rolled. Of interest is the number of dice that show a one. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many dice would you expect to show a one? Find the probability that all six dice show a one. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically. More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many schools would you expect to offer such courses? Find the probability that at most ten offer such courses. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence. X = the number of college and universities that offer online offerings. 0, 1, 2, …, 13 X ~ B (13, 0.96) 12.48 0.0135 P ( x = 12) = 0.3186 P ( x = 13) = 0.5882 More likely to get 13. Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to attend their graduation? Find the probability that 17 or 18 attend. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically. At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to not to use the foil as their main weapon? Find the probability that six do not use the foil as their main weapon. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically. X = the number of fencers who do not use the foil as their main weapon 0, 1, 2, 3,... 25 X ~ B (25,0.40) 10 0.0442 The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising. Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many seniors are expected to have participated in after-school sports all four years of high school? Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many audits are expected in a 20-year period? Find the probability that a person is not audited at all. Find the probability that a person is audited more than twice. X = the number of audits in a 20-year period 0, 1, 2, …, 20 X ~ B (20, 0.02) 0.4 0.6676 0.0071 It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) What is the probability that at least eight have adequate earthquake supplies? Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? How many residents do you expect will have adequate earthquake supplies? There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The “house” rolls three dice. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back their $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back their $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back their $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) List the values that Y may take on. Then, construct one PDF table that includes both X and Y and their probabilities. Calculate the average expected matches over the long run of playing this game for the player. Calculate the average expected earnings over the long run of playing this game for the player. Determine who has the advantage, the player or the house. X = the number of matches 0, 1, 2, 3 X ~ B ( 3 , 1 6 ) In dollars: −1, 1, 2, 3 1 2 Multiply each Y value by the corresponding X probability from the PDF table. The answer is −0.0787. You lose about eight cents, on average, per game. The house has the advantage. Only 9% of the population of Niger has access to the internet. Suppose we randomly sample 150 people in Niger. Let X = the number of people who have access to the internet. What is the probability distribution for X ? Using the formulas, calculate the mean and standard deviation of X . Use your calculator to find the probability that 15 people in the sample have access to the internet. Find the probability that at most ten people in the sample have access to the internet. Find the probability that more than 25 people in the sample have access to the internet. The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in a certain country is 28.1%. Suppose you choose 15 people in that country at random. Let X = the number of people who are literate. Sketch a graph of the probability distribution of X . Using the formulas, calculate the (i) mean and (ii) standard deviation of X . Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate. X ~ B (15, 0.281) Mean = μ = np = 15(0.281) = 4.215 Standard Deviation = σ = n p q = 15 ( 0.281 ) ( 0.719 ) = 1.7409 P ( x > 5) = 1 – P ( x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246 P ( x = 3) = binompdf(15, 0.281, 3) = 0.1927 P ( x = 4) = binompdf(15, 0.281, 4) = 0.2259 It is more likely that four people are literate that three people are. Binomial Experiment a statistical experiment that satisfies the following three conditions: There are a fixed number of trials, n . There are only two possible outcomes, called \"success\" and, \"failure,\" for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. The n trials are independent and are repeated using identical conditions. Bernoulli Trials an experiment with the following characteristics: There are only two possible outcomes called “success” and “failure” for each trial. The probability p of a success is the same for any trial (so the probability q = 1 − p of a failure is the same for any trial). Binomial Probability Distribution a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B ( n , p ). The mean is μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = ( n x ) p x q n − x .", "section": "Binomial Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Geometric Distribution There are four main characteristics of a geometric experiment. A trial is repeated until a success occurs. Think of this as one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a \"success,\" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP. In theory, the number of trials could go on forever. The repeated trials are independent of each other. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 - p . For example, the probability of rolling a three when you throw one fair die is 1 / 6 . This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = 5 / 6 , the probability of a failure. The probability of getting a first three on the fifth roll is ( 5 / 6 ) ( 5 / 6 ) ( 5 / 6 ) ( 5 / 6 ) ( 1 / 6 ) = 0 . 0804 The random variable X represents the number of the trial in which the first success occurs. That is, X = the number of independent trials until the first success. The following are additional attributes of the geometric distribution: The random variable is discrete. The random variable may be defined in two ways depending upon the analyst’s interest. In the example above for throwing a die, the question was “What is the probability that the first success will be on the fifth throw?” Alternatively, the question could be asked as “What is the probability that it takes four failures before a success?” We will see that each way of asking the question will alter the form of the geometric probability density function slightly and will change the mean and standard deviation of the geometric pdf. Implicit in the random variable is that the probability of a success is constant and therefore so is the probability of a failure. For flipping a coin this is obvious, but in experiments that require skill, such as hitting a baseball or throwing a dart, one might consider that learning during the experiment would alter the probability of a success. The geometric distribution cannot capture “learning” thus the historical probability of success is assumed to be constant. The geometric distribution is “memoryless.” There are very few probability density functions that are what is known as “memoryless,” and the Geometric distribution is the only one with a discrete random variable that is memoryless. As an example, historically Major League Baseball player Jones has a record of hitting the ball for at least an advance to first base with a probability of 0.20. Jones has not had a hit in his last 10 times at bat. What is the probability that Jones will get a hit in his third time at bat? The answer ignores his 10 previous failures. All events prior to the events in current time are irrelevant and thus are considered “memoryless.” Formally: P x = n + k | x ≥ k + 1 = P ( x = n ) , where k = number of previous failures Jones’s probability of a hit begins anew each time he comes to bat. This feature of the geometric distribution results in a curious result: Drawing parts from a manufacturing process to test for parts that are defective, the geometric distribution begins with a clean slate each time the tests begin with no consideration of previous test results. More on this when we get to the exponential probability density function. You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P ( x = 5). Try It You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? A safety engineer feels that 35% of all industrial accidents in the plant are caused by failure of employees to follow instructions. They decide to look at the accident reports (selected randomly and replaced in the pile after reading) until they find one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until they find a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions? Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P ( x ≥ 3). (\"At least\" translates to a \"greater than or equal to\" symbol). On average, how many reports would the safety engineer expect to review until they find a report showing an accident caused by something other than failure to follow instructions? This question asks you to find the expected value or the mean. This is the average number of failures from something other than following instructions. The formula for the mean of this geometric distribution is: μ = E x = 1 p ≈ 2 . 8 Note that the mean does not need to be a whole number although the random variable is discrete and must be a counting number What is the probability that the safety engineer will have to examine at least three reports until they find a report showing an accident caused by employee failure to follow instructions? This question is answered by first defining the random variable. Let X = the number of accidents the safety engineer must examine until they find a report showing an accident caused by employee failure to follow instructions. X can take on the values 1, 2, 3, 4, … ∞. Unlike the binomial distribution with a fixed number of trials, the geometric distribution may have an infinite number of failed trials before a success occurs. This second question asks you to find P ( x ≥ 3 ) . (\"At least\" translates to a \"greater than or equal to\" symbol.) In a binomial distribution, we saw the solution to questions of “more than” or “less than” would be to calculate each probability individually and add from zero to three. If the probability of interest is “greater than or equal to” (≥) the individual probabilities are added from zero to three and then subtracted from one. P x ≥ 3 = 1 - P ( x < 3 ) Because we cannot add the full range of the geometric distribution random variable because it goes through infinity, an alternative solution is developed. An alternative geometric distribution pair of formulas provides solutions for questions asking for probabilities “more than” and “less than.” If the question is: What is the probability it takes MORE THAN n events to get first success? P x > n = ( 1 - p ) n What is the probability it takes LESS THAN n event for success? P x < n = 1 - ( 1 - p ) n Try It An instructor feels that 15% of students get below a C on their final exam. They decide to look at final exams (selected randomly and replaced in the pile after reading) until they find one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until they find one with a grade below a C. What is the probability question stated mathematically? Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says they live within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if they live within five miles of you. There is no definite number of trials (number of times you ask a student). a. Let X = the number of ____________ you must ask ____________ one says yes. b. What values does X take on? c. What are p and q ? d. The probability question is P (_______). a. Let X = the number of students you must ask until one says yes. b. X can be any positive integer: 1, 2, 3, 4, …, c. p = 0.55; q = 0.45 d. P ( x = 4) Try It You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q ? Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G ( p ) Read this as \" X is a random variable with a geometric distribution .\" The parameter is p ; p = the probability of a success for each trial. CASE I: Random Variable X Is Event of First Success In this case we ask, “What is the probability that we will have some number x of events of interest to us of failures before a success?” The geometric pdf tells us the probability that the first occurrence of success requires x number of failure independent trials, each with probability ( 1 - p ) . If the probability of success on each trial is p , then the probability that the x th trial (out of x trials) is the first success is: P X = x = ( 1 - p ) x - 1 p for x = 1 , 2 , 3 , . . . . Like the binomial distribution, the geometric distribution has the parameters of the mean and standard deviation. The expected value of X, the mean for Case I, is μ = 1 p . This tells us how many failed trials to expect until we get the first success. This count includes in the count of trials the trial that results in success. The above form of the geometric distribution is used for modeling the number of trials until the first success. The number of trials includes the one that is a success: x = all trials including the one that is a success. This can be seen in the form of the formula. If X = number of trials including the success, then we must multiply the probability of failure, ( 1 - p ) , times the number of failures, that is x - 1 . The standard deviation of Case I of the geometric distribution is: σ = 1 p 1 p - 1 CASE II: Random Variable X Is Number of Failures BEFORE a Success By contrast to Case I, the following form of the geometric distribution used for modeling number of failures until the first success is: P X = x = ( 1 - p ) x p for x = 0 , 1 , 2 , 3 , . . . . In this case the trial that is the success is not counted as a trial in the formula: x = number of failures. The expected value, the mean, of this distribution is μ = 1 - p p . This tells us how many failures to expect before we have a success. In either case, the sequence of probabilities is a geometric sequence. In Case II, the standard deviation parameter is: σ = 1 - p p 2 . GEOMETRIC: P = . 02 , Common ratio = r = . 98 The y-axis in contains the probability of x , and the x -axis is the random number components tested. For example, at x = 1 the probability it will be found to be defective is 0.0196. With two components tested, the probability the second component is defective is graphed at a probability of 0.0196 at x = 2 on the x axis. For the probability that the third component is defective we find P X = 3 = 0 . 019208 . (The first two are the same because of rounding in the computations.) Notice on that the probabilities decline by the same step down with each change in the value of x. This increment is called the common ratio. This exists for the geometric probability distribution uniquely. The common ratio, called r, can be calculated by dividing any value by the previous value, e.g., P ( x = 5 ) P ( x = 4 ) = 0 . 018447 0 . 018823 = 0 . 98 . For this set of data, therefore, the common ratio is 0.98. The common ratio then multiplied by any other probability value will provide the next probability value in the sequence. For example, the probability that the sixth component tested is a failure is 0.018078. Check this using the formula from Case I. Now we have the P x = 6 . Knowing this and the common ratio we can calculate the P x = 7 by simple multiplication: P x = 7 = 0 . 018078 * 0 . 98 = 0 . 017716 , , the same value we found earlier by using the geometric probability distribution. This common ratio increment is the same ratio between every number and is called a geometric progression and thus the name for this probability density function. Once the common ratio is calculated, any P x = x a one desires to know can be easily found. The number of components that you would expect to test until you find the first defective component is the mean, μ = 50 for this case of defective components. The formula for the mean of the geometric distribution for the random variable defined as number of failures before first success is μ = 1 p = 1 0 . 02 = 50 See for an example where the geometric random variable is defined as number of trials until first success. The expected value of this formula for the geometric distribution will be different from this version of the distribution. Case II also has a variance, but this is changed from the Case I formula. This formula for the variance is: P X = x = ( 1 - p ) x - 1 p for x = 1 , 2 , 3 , . . . . The formula for the variance is σ 2 = 1 p 1 p - 1 = 1 0 . 02 1 0 . 02 - 1 = 2 , 450 The standard deviation is σ = 1 p 1 p - 1 = 1 0 . 02 1 0 . 02 - 1 = 49 . 5 Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3, ... where p = 0.02. Find P ( x = 7). P ( x = 7) = 0.0177. To find the probability that x = 7, Enter 2 nd , DISTR Scroll down and select geometpdf ( Press ENTER Enter 0.02, 7); press ENTER to see the result: P ( x = 7) = 0.0177 To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function. The probability that the seventh component is the first defect is 0.0177. The formula for the variance is σ 2 = ( 1 p ) ( 1 p − 1 ) = ( 1 0.02 ) ( 1 0.02 − 1 ) = 2,450 The standard deviation is σ = ( 1 p ) ( 1 p − 1 ) = ( 1 0. 02 ) ( 1 0. 02 − 1 ) = 49.5 Try It The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. The lifetime risk of developing cancer is about one in 67 (1.5%). Let X = the number of people you ask until one says they have cancer. Then X is a discrete random variable with a geometric distribution: X ~ G 1 67 or X ~ G ( 0 . 015 ) What is the probability of that you ask ten people before one says they have cancer? What is the probability that you must ask 20 people? Find the (i) mean and (ii) standard deviation of X . P ( x = 10 ) = g e o m e t r i c p d f ( 0 . 015 , 10 ) = 0 . 0131 P ( x = 20 ) = g e o m e t r i c p d f ( 0 . 015 , 20 ) = 0 . 0113 Mean = μ = 1 p = 1 0 . 015 = 66 . 67 Standard Deviation = σ = 1 - p p 2 = 1 - 0 . 015 [ 0 . 015 ] 2 = 66 . 16 Try It The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the number of Afghani women you ask until one says that she is literate. What is the probability distribution of X ? What is the probability that you ask five women before one says she is literate? What is the probability that you must ask ten women? Find the (i) mean and (ii) standard deviation of X . References “Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at http://www.pewsocialtrends.org/files/2010/10/millennials-confident-connected-open-to-change.pdf (accessed May 15, 2013). “Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-to-change/ (accessed May 15, 2013). “Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013). “Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at http://ec.europa.eu/europeaid/where/asia/documents/afgh_brochure_summary_en.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013). “UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/video/afghan-female-literacy-centers.html (accessed May 15, 2013). Chapter Review There are three characteristics of a geometric experiment: There are one or more Bernoulli trials with all failures except the last one, which is a success. In theory, the number of trials could go on forever. There must be at least one trial. The probability, p , of a success and the probability, q , of a failure are the same for each trial. In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G ( p ) where p is the probability of success in a single trial. The mean of the geometric distribution X ~ G ( p ) is μ = 1 p and the standard deviation is σ ( 1 − p ) p 2 = 1 p ( 1 p − 1 ) . Formula Review X ~ G( p ) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p . X = the number of independent trials until the first success X takes on the values x = 1, 2, 3, ... p = the probability of a success for any trial q = the probability of a failure for any trial p + q = 1 q = 1 – p The mean is μ = 1 p . The standard deviation is σ = 1 – p p 2 = 1 p ( 1 p − 1 ) . Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time first-year students from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select first-year students from the study until you find one who replies “yes.” You are interested in the number of first-year students you must ask. In words, define the random variable X . X = the number of first-year students selected from the study until one replied \"yes\" that same-sex couples should have the right to legal marital status. X ~ _____(_____,_____) What values does the random variable X take on? 1,2,… Construct the probability distribution function (PDF). Stop at x = 6. x P ( x ) 1 2 3 4 5 6 On average ( μ ), how many first-year students would you expect to have to ask until you found one who replies \"yes?\" 1.4 What is the probability that you will need to ask fewer than three first-year students? HOMEWORK A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many dealerships would we expect her to have to call until she finds one that has the car? Find the probability that she must call at most four dealerships. Find the probability that she must call three or four dealerships. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many adults in America do you expect to survey until you find one who will watch the Super Bowl? Find the probability that you must ask seven people. Find the probability that you must ask three or four people. X = the number of adults in America who are surveyed until one says they will watch the Super Bowl. X ~ G (0.40) 2.5 0.0187 0.2304 It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies? What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies? How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies? How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies? In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many pages do you expect to advertise footwear on them? Is it probable that all twenty will advertise footwear on them? Why or why not? What is the probability that fewer than ten will advertise footwear on them? Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it? How many pages do you expect to need to survey in order to find one that advertises footwear? X = the number of pages that advertise footwear X takes on the values 0, 1, 2, ..., 20 X ~ B (20, 29 192 ) 3.02 No 0.9997 X = the number of pages we must survey until we find one that advertises footwear. X ~ G ( 29 192 ) 0.3881 6.6207 pages Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. p = probability of success (event F occurs) q = probability of failure (event F does not occur) Write the description of the random variable X . What are the values that X can take on? Find the values of p and q . Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. Emery has music practice three days a week. They practice for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on? 0, 1, 2, and 3 Medical researchers in a certain country have documented that the percentage of people ages 15 to 49 who are infected with a certain respiratory virus is 17.3%. Let X = the number of people you test until you find a person infected with the virus. Sketch a graph of the distribution of the discrete random variable X . What is the probability that you must test 30 people to find one with the virus? What is the probability that you must ask ten people? Find the (i) mean and (ii) standard deviation of the distribution of X . According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millenials you ask until you find a person without a profile on a social networking site. Describe the distribution of X . Find the (i) mean and (ii) standard deviation of X . What is the probability that you must ask ten people to find one person without a social networking site? What is the probability that you must ask 20 people to find one person without a social networking site? What is the probability that you must ask at most five people? X ~ G (0.25) Mean = μ = 1 p = 1 0.25 = 4 Standard Deviation = σ = 1 − p p 2 = 1 − 0 .25 0.25 2 ≈ 3.4641 P ( x = 10) = geometpdf(0.25, 10) = 0.0188 P ( x = 20) = geometpdf(0.25, 20) = 0.0011 P ( x ≤ 5) = geometcdf(0.25, 5) = 0.7627 Geometric Distribution a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable X is defined as the number of trials until the first success. Notation: X ~ G ( p ). The mean is μ = 1 p and the standard deviation is σ = 1 p ( 1 p − 1 ) . The probability of exactly x failures before the first success is given by the formula: P ( X = x ) = p (1 – p ) x – 1 . Geometric Experiment a statistical experiment with the following properties: There are one or more Bernoulli trials with all failures except the last one, which is a success. In theory, the number of trials could go on forever. There must be at least one trial. The probability, p , of a success and the probability, q , of a failure do not change from trial to trial.", "section": "Geometric Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Hypergeometric Distribution There are five characteristics of a hypergeometric experiment. You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. For example, you want to choose a softball team from a combined group of 11 men and 13 women. The team consists of ten players. Each pick is not independent, since sampling is without replacement. In the softball example, the probability of picking a woman first is 13 24 . The probability of picking a man second is 11 23 if a woman was picked first. It is 10 23 if a man was picked first. The probability of the second pick depends on what happened in the first pick. You are not dealing with Bernoulli Trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let X = the number of gumdrops in the sample of 50. X takes on the values x = 0, 1, 2, ..., 50. What is the probability statement written mathematically? P ( x = 35) Try It A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample? Suppose a shipment of 100 laptops is known to have ten defective laptops. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 laptops, at most two are defective. The two groups are the 90 non-defective laptops and the 10 defective laptops. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective laptops. The size of the sample is 12 laptops. (They may be non-defective or defective.) Let X = the number of defective laptops in the sample of 12. X takes on the values 0, 1, 2, ..., 10. X may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective laptops. Write the probability statement mathematically. P ( x ≤ 2) Try It A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is X , and what values does it take on? You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men? This is a hypergeometric problem because you are choosing your committee from two groups (men and women). a. Are you choosing with or without replacement? b. What is the group of interest? c. How many are in the group of interest? d. How many are in the other group? e. Let X = _________ on the committee. What values does X take on? f. The probability question is P (_______). a. without b. the men c. 15 men d. 18 women e. Let X = the number of men on the committee. x = 0, 1, 2, …, 7. f. P ( x > 4) Try It A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem. Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function X ~ H ( r , b , n ) Read this as \" X is a random variable with a hypergeometric distribution.\" The parameters are r , b , and n ; r = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen sample. A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee? Let X = the number of men on the committee of four. The men are the group of interest (first group). X takes on the values 0, 1, 2, 3, 4, where r = 6 , b = 5 , and n = 4 . X ~ H (6, 5, 4) Find P ( x = 2). P ( x = 2) = 0.4545 (calculator or computer) NOTE Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do. The probability that there are two men on the committee is about 0.45. The graph of X ~ H (6, 5, 4) is: The y -axis contains the probability of X , where X = the number of men on the committee. You would expect m = 2.18 (about two) men on the committee. The formula for the mean is μ = n r r + b = ( 4 ) ( 6 ) 6 + 5 = 2.18 Try It An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has ten slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample? Chapter Review A hypergeometric experiment is a statistical experiment with the following properties: You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. Each pick is not independent, since sampling is without replacement. You are not dealing with Bernoulli Trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. The distribution of X is denoted X ~ H ( r , b , n ), where r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. It follows that n ≤ r + b . The mean of X is μ = n r r + b and the standard deviation is σ = r b n ( r + b − n ) ( r + b ) 2 ( r + b − 1) . Formula Review X ~ H ( r , b , n ) means that the discrete random variable X has a hypergeometric probability distribution with r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. X = the number of items from the group of interest that are in the chosen sample, and X may take on the values x = 0, 1, ..., up to the size of the group of interest. (The minimum value for X may be larger than zero in some instances.) n ≤ r + b The mean of X is given by the formula μ = n r r + b and the standard deviation is = r b n ( r + b − n ) ( r + b ) 2 ( r + b − 1) . Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. In words, define the random variable X . X = the number of business majors in the sample. X ~ _____(_____,_____) What values does X take on? 2, 3, 4, 5, 6, 7, 8, 9 Find the standard deviation. On average ( μ ), how many would you expect to be business majors? 6.26 HOMEWORK A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many Shotokan Karate students do we expect to be in that first demonstration? In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many pages do you expect to advertise footwear on them? Calculate the standard deviation. X = the number of pages that advertise footwear 0, 1, 2, 3, ..., 20 X ~ H (29, 163, 20); r = 29, b = 163, n = 20 3.03 1.5197 Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many instructors do you expect on the committee who are not technically proficient? Find the probability that at least five on the committee are not technically proficient. Find the probability that at most three on the committee are not technically proficient. Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) Are you choosing the nine athletes with or without replacement? X = the number of Patriots picked 0, 1, 2, 3, 4 X ~ H (4, 8, 9) Without replacement A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart? What is the group of interest? How many are in the group of interest? How many are in the other group? Let X = _________. What values does X take on? The probability question is P (_______). Find the probability in question. Find the (i) mean and (ii) standard deviation of X . Hypergeometric Experiment a statistical experiment with the following properties: You take samples from two groups. You are concerned with a group of interest, called the first group. You sample without replacement from the combined groups. Each pick is not independent, since sampling is without replacement. You are not dealing with Bernoulli Trials. Hypergeometric Probability a discrete random variable (RV) that is characterized by: A fixed number of trials. The probability of success is not the same from trial to trial. We sample from two groups of items when we are interested in only one group. X is defined as the number of successes out of the total number of items chosen. Notation: X ~ H ( r , b , n ), where r = the number of items in the group of interest, b = the number of items in the group not of interest, and n = the number of items chosen.", "section": "Hypergeometric Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Poisson Distribution There are two main characteristics of a Poisson experiment. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages. The Poisson distribution may be used to approximate the binomial if the probability of success is \"small\" (such as 0.01) and the number of trials is \"large\" (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a \"success.\" The random variable X = the number of occurrences in the interval of interest. The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three? Let X = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is ( 5 30 ) (12) = 2 loaves of bread. The probability question asks you to find P ( x = 3). Try It The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes? A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question. P ( x < 5) Try It An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically. You notice that a news reporter says \"uh,\" on average, two times per broadcast. What is the probability that the news reporter says \"uh\" more than two times per broadcast. This is a Poisson problem because you are interested in knowing the number of times the news reporter says \"uh\" during a broadcast. a. What is the interval of interest? b. What is the average number of times the news reporter says \"uh\" during one broadcast? c. Let X = ____________. What values does X take on? d. The probability question is P (______). a. one broadcast b. 2 c. Let X = the number of times the news reporter says \"uh\" during one broadcast. x = 0, 1, 2, 3, ... d. P ( x > 2) Try It An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution. Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P ( μ ) Read this as \" X is a random variable with a Poisson distribution.\" The parameter is μ (or λ ); μ (or λ ) = the mean for the interval of interest. The standard deviation of the Poisson distribution with mean µ is σ =√ μ Leah receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes? Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or 1 4 hour.) x = 0, 1, 2, 3, ... If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives ( 1 8 ) (6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem. Press 1 – and then press 2 nd DISTR . Arrow down to poissoncdf . Press ENTER . Enter (.75,1). The result is P ( x > 1) = 0.1734. NOTE The TI calculators use λ (lambda) for the mean. The Poisson distribution is discrete, and thus > 1 includes all whole numbers through infinity. The solution is to subtract the probability less than 1 thus 1 - [ P ( x = 0 ) + P ( x = 1 ) ] P ( x ) = μ x e - μ x ! P ( x > 1 ) = 1 - P ( x ≤ 1 ) = 1 - [ P ( x = 0 ) + P ( x = 1 ) ] = 1 - [ 0 . 75 0 e - 0 . 75 0 ! + 0 . 75 1 e - 0 . 75 1 ! ] = 1 - [ ( 1 ) ( 0 . 4724 ) 1 + ( 0 . 75 ) ( 0 . 4724 ) 1 ] = 1 - [ 0 . 4724 + 0 . 3543 ] = 0 . 1733 The y -axis in contains the probability of x where X = the number of calls in 15 minutes. Try It A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (147). The mean is 147 emails. What is the probability that an email user receives exactly 160 emails per day? What is the probability that an email user receives at most 160 emails per day? What is the standard deviation? P ( x = 160) = poissonpdf(147, 160) ≈ 0.0180 P ( x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666 Standard Deviation = σ = μ = 147 ≈ 12.1244 Try It According to a recent poll by the Pew Internet Project, people between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (187). The mean is 187 text messages. What is the probability that a person sends exactly 175 texts per day? What is the probability that a person sends at most 150 texts per day? What is the standard deviation? Text message users receive or send an average of 41.5 text messages per day. How many text messages does a text message user receive or send per hour? What is the probability that a text message user receives or sends two messages per hour? What is the probability that a text message user receives or sends more than two messages per hour? Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is 41.5 24 ≈ 1.7292. X ~ P (1.7292), so P ( x = 2) = poissonpdf(1.7292, 2) ≈ 0.2653 P ( x > 2) = 1 – P ( x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505 Try It Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,700 arrivals and departures each day. How many airplanes arrive and depart the airport per hour? What is the probability that there are exactly 100 arrivals and departures in one hour? What is the probability that there are at most 100 arrivals and departures in one hour? On a certain day in May starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? Let X = the number of days with low seismic activity. Using the binomial distribution: P ( x = 10) = binompdf(200, .0102, 10) ≈ 0.000039 Using the Poisson distribution: Calculate μ = np = 200(0.0102) ≈ 2.04 P ( x = 10) = poissonpdf(2.04, 10) ≈ 0.000045 We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0. Try It On a certain day in May starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? References “ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atl.com/about-atl/atl-factsheet/ (accessed February 18, 2019). Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013). “Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html (accessed May 15, 2013). “Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www.state.sc.us/dmh/anorexia/statistics.htm (accessed May 15, 2013). “Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where there is an average of 60 births a day,” theguardian, 2013. Available online at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013). “How Americans Use Text Messaging,” Pew Internet, 2013. Available online at http://pewinternet.org/Reports/2011/Cell-Phone-Texting-2011/Main-Report.aspx (accessed May 15, 2013). Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volum is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013). “One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline. Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-day-mothers-bed.html (accessed May 15, 2013). Vanderkam, Laura. “Stop Checking Your Email, Now.” CNNMoney, 2013. Available online at http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15, 2013). “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013). Chapter Review A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is \"small\" (less than or equal to 0.05) and the number of trials is \"large\" (greater than or equal to 20). Formula Review X ~ P ( μ ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest. X takes on the values x = 0, 1, 2, 3, ... The mean μ is typically given. The variance is σ 2 = μ , and the standard deviation is σ = μ . The probability of having exactly x successes in r trials is P X = x = e - μ μ x x ! . When P ( μ ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial. Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day. Assume the event occurs independently in any given day. Define the random variable X . What values does X take on? 0, 1, 2, 3, 4, … What is the probability of getting 150 customers in one day? What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day. 0.0485 What is the probability that the store will have more than 12 customers in the first hour? What is the probability that the store will have fewer than 12 customers in the first two hours? 0.0214 Which type of distribution can the Poisson model be used to approximate? When would you do this? Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age. Assume the event occurs independently in any given day. In words, define the random variable X . X = the number of U.S. teens who die from motor vehicle injuries per day. X ~ _____(_____,_____) What values does X take on? 0, 1, 2, 3, 4, ... For the given values of the random variable X , fill in the corresponding probabilities. Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically. No Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically. HOMEWORK The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon. Find the mean and standard deviation of X . What is the probability that the office receives at most six calls at noon on Monday? Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon? What is the probability that the office receives more than eight calls at noon? X ~ P (5.5); μ = 5.5; σ = 5.5 ≈ 2.3452 P ( x ≤ 6) = poissoncdf(5.5, 6) ≈ 0.6860 There is a 15.7% probability that the law staff will receive more calls than they can handle. P ( x > 8) = 1 – P ( x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – 0.8944 = 0.1056 The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour. Find the mean and standard deviation of X . Sketch a graph of the probability distribution of X . What is the probability that the maternity ward will deliver three babies in one hour? What is the probability that the maternity ward will deliver at most three babies in one hour? What is the probability that the maternity ward will deliver more than five babies in one hour? A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions. Let X = the number of defective bulbs in a string. Using the Poisson distribution: μ = np = 100(0.03) = 3 X ~ P (3) P ( x ≤ 4) = poissoncdf(3, 4) ≈ 0.8153 Using the binomial distribution: X ~ B (100, 0.03) P ( x ≤ 4) = binomcdf(100, 0.03, 4) ≈ 0.8179 The Poisson approximation is very good—the difference between the probabilities is only 0.0026. The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) Find the probability that she has no children. Find the probability that she has fewer children than the Japanese average. Find the probability that she has more children than the Japanese average. The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen. In words, define the Random Variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) Find the probability that she has no children. Find the probability that she has fewer children than the Spanish average. Find the probability that she has more children than the Spanish average . X = the number of children for a Spanish woman 0, 1, 2, 3,... X ~ P (1.47) 0.2299 0.5679 0.4321 Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces: In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _______ Find the probability that she has no litters in one year. Find the probability that she has at least two litters in one year. Find the probability that she has exactly three litters in one year. The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many cookies do we expect to have an extra fortune? Find the probability that none of the cookies have an extra fortune. Find the probability that more than three have an extra fortune. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences. X = the number of fortune cookies that have an extra fortune 0, 1, 2, 3,... 144 X ~ B (144, 0.03) or P (4.32) 4.32 0.0124 or 0.0133 0.6300 or 0.6264 As n gets larger, the probabilities get closer together. For every 200 U.S. kids and teens, the average number who suffer from obsessive compulsive disorder is one. Out of a randomly chosen group of 600 U.S. kids and teens determine the following. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to suffer from obsessive compulsive disorder? Find the probability that no one suffers from obsessive compulsive disorder. Find the probability that more than four suffer from obsessive compulsive disorder. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many are expected to be audited? Find the probability that no one was audited. Find the probability that at least three were audited. X = the number of people audited in one year 0, 1, 2, ..., 100 X ~ P (2) 2 0.1353 0.3233 Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) How many seniors are expected to have participated in after-school sports all four years of high school? Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes. In words, define the random variable X . List the values that X may take on. Give the distribution of X . X ~ _____(_____,_____) On average, how many pieces of egg shell do you expect to be in the cake? What is the probability that there will not be any pieces of egg shell in the cake? Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes? Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why? X = the number of shell pieces in one cake 0, 1, 2, 3,... X ~ P (1.5) 1.5 0.2231 0.0001 Yes Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week. In words, the random variable X = _________________ the number of times Mrs. Plum’s cats wake her up each week. the number of times Mrs. Plum’s cats wake her up each hour. the number of times Mrs. Plum’s cats wake her up each night. the number of times Mrs. Plum’s cats wake her up. Find the probability that her cats will wake her up no more than five times next week. 0.5000 0.9329 0.0378 0.0671 d Poisson Probability Distribution a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable: The probability that the event occurs in a given interval is the same for all intervals. The events occur with a known mean and independently of the time since the last event. The distribution is defined by the mean μ of the event in the interval. Notation: X ~ P ( μ ). The mean is μ = np . The standard deviation is σ = μ . The probability of having exactly x successes in r trials is P ( X = x ) = ( e − μ ) μ x x ! . The Poisson distribution is often used to approximate the binomial distribution, when n is “large” and p is “small” (a general rule is that n should be greater than or equal to 20 and p should be less than or equal to 0.05).", "section": "Poisson Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Discrete Distribution (Playing Card Experiment) Discrete Distribution (Playing Card Experiment) Class Time: Names: Student Learning Outcomes The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution. The student will compare technology-generated simulation and a theoretical distribution. The student will demonstrate an understanding of long-term probabilities. Supplies One full deck of playing cards One programming calculator Procedure The experimental procedure for empirical data is to pick one card from a deck of shuffled cards. The theoretical probability of picking a diamond from a deck is _________. Shuffle a deck of cards. Pick one card from it. Record whether it was a diamond or not a diamond. Put the card back and reshuffle. Do this a total of ten times. Record the number of diamonds picked. Let X = number of diamonds. Theoretically, X ~ B (_____,_____) Organize the Data Record the number of diamonds picked for your class with playing cards in . Then calculate the relative frequency. x Frequency Relative Frequency 0 __________ 1 __________ 2 __________ 3 __________ 4 __________ 5 __________ 6 __________ 7 __________ 8 __________ 9 __________ 10 __________ Calculate the following: x ¯ = ________ s = ________ Construct a histogram of the empirical data. Theoretical Distribution Build the theoretical PDF chart based on the distribution in the Procedure section. x P ( x ) 0 1 2 3 4 5 6 7 8 9 10 Calculate the following: μ = ____________ σ = ____________ Construct a histogram of the theoretical distribution. Using the Data NOTE RF = relative frequency Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. P ( x = 3) = _______________________ P (1 < x < 4) = _______________________ P ( x ≥ 8) = _______________________ Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. RF ( x = 3) = _______________________ RF (1 < x < 4) = _______________________ RF ( x ≥ 8) = _______________________ Discussion Questions For questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical, empirical, and simulation distributions. Use complete sentences. Describe the three most significant differences between the graphs or distributions of the theoretical, empirical, and simulation distributions. Using your answers from questions 1 and 2, does it appear that the two sets of data fit the theoretical distribution? In complete sentences, explain why or why not. Suppose that the experiment had been repeated 500 times. Would you expect or to change, and how would it change? Why? Why wouldn’t the other table(s) change?", "section": "Discrete Distribution (Playing Card Experiment)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Discrete Distribution (Dice Experiment Using Three Regular Dice) Discrete Distribution (Dice Experiment Using Three Regular Dice) Class Time: Names: Student Learning Outcomes The student will compare empirical data and a theoretical distribution to determine if a dice experiment game fits a discrete distribution. The student will demonstrate an understanding of long-term probabilities. Supplies Three regular six-sided dice Procedure Round answers to relative frequency and probability problems to four decimal places. The experimental procedure is to bet on a specific number appearing on the dice. Then, roll three dice and count the number of matches. The number of matches will decide your profit. What is the theoretical probability of one die matching the the specific number? Choose one number to place a bet on. Roll the three dice. Count the number of matches. Let X = number of matches. Theoretically, X ~ B (______,______) Let Y = profit per game. Organize the Data In , fill in the y value that corresponds to each x value. Next, record the number of matches picked for your class. Then, calculate the relative frequency. Complete the table. x y Frequency Relative Frequency 0 1 2 3 Calculate the following: x ¯ = _______ s x = ________ y ¯ = _______ s y = _______ Explain what x ¯ represents. Explain what y ¯ represents. Based upon the experiment: What was the average profit per game? Did this represent an average win or loss per game? How do you know? Answer in complete sentences. Construct a histogram of the empirical data. Theoretical Distribution Build the theoretical PDF chart for x and y based on the distribution from the Procedure section. x y P ( x ) = P ( y ) 0 1 2 3 Calculate the following: μ x = _______ σ x = _______ μ x = _______ Explain what μ x represents. Explain what μ y represents. Based upon theory: What was the expected profit per game? Did the expected profit represent an average win or loss per game? How do you know? Answer in complete sentences. Construct a histogram of the theoretical distribution. Use the Data NOTE RF = relative frequency Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. P ( x = 3) = _________________ P (0 < x < 3) = _________________ P ( x ≥ 2) = _________________ Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. RF (x = 3) = _________________ RF (0 < x < 3) = _________________ RF ( x ≥ 2) = _________________ Discussion Question For questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions. Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not. Suppose that the experiment had been repeated 500 times. Would you expect or to change, and how would it change? Why? Why wouldn’t the other table change?", "section": "Discrete Distribution (Dice Experiment Using Three Regular Dice)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction The heights of these radish plants are continuous random variables. (credit: modification of work “Radish (Raphanus sativus): White rust caused by Albugo candida” by Scot Nelson/ Flickr, Public Domain) Chapter Objectives By the end of this chapter, the student should be able to: Recognize and understand continuous probability density functions in general. Recognize the uniform probability distribution and apply it appropriately. Recognize the exponential probability distribution and apply it appropriately. Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables. NOTE The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important. Properties of Continuous Probability Distributions The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. The curve is called the probability density function (abbreviated as pdf ). We use the symbol f ( x ) to represent the curve. f ( x ) is the function that corresponds to the graph; we use the density function f ( x ) to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf ). The cumulative distribution function is used to evaluate probability as area. The outcomes are measured, not counted. The entire area under the curve and above the x-axis is equal to one. Probability is found for intervals of x values rather than for individual x values. P(c < x < d) is the probability that the random variable X is in the interval between the values c and d . P(c < x < d) is the area under the curve, above the x -axis, to the right of c and the left of d . P(x = c) = 0 The probability that x takes on any single individual value is zero. The area below the curve, above the x -axis, and between x = c and x = c has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero. P(c < x < d) is the same as P(c ≤ x ≤ d) because probability is equal to area. We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus. However, because most students taking this course have not studied calculus, we will not be using calculus in this textbook. There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way. In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions. The graph shows a Uniform Distribution with the area between x = 3 and x = 6 shaded to represent the probability that the value of the random variable X is in the interval between three and six. The graph shows an Exponential Distribution with the area between x = 2 and x = 4 shaded to represent the probability that the value of the random variable X is in the interval between two and four. The graph shows the Standard Normal Distribution with the area between x = 1 and x = 2 shaded to represent the probability that the value of the random variable X is in the interval between one and two. Uniform Distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b . Notation: X ~ U ( a , b ). The mean is μ = a + b 2 and the standard deviation is σ = ( b − a ) 2 12 . The probability density function is f ( x ) = 1 b − a for a < x < b or a ≤ x ≤ b . The cumulative distribution is P ( X ≤ x ) = x − a b − a . Exponential Distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital; the notation is X ~ Exp ( m ). The mean is μ = 1 m and the standard deviation is σ = 1 m . The probability density function is f ( x ) = me −mx , x ≥ 0 and the cumulative distribution function is P ( X ≤ x ) = 1 − e −mx .", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Continuous Probability Functions We begin by defining a continuous probability density function. We use the function notation f ( x ). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f ( x ) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA. Consider the function f ( x ) = 1 20 for 0 ≤ x ≤ 20. x = a real number. The graph of f ( x ) = 1 20 is a horizontal line. However, since 0 ≤ x ≤ 20, f ( x ) is restricted to the portion between x = 0 and x = 20, inclusive. f ( x ) = 1 20 for 0 ≤ x ≤ 20. The graph of f ( x ) = 1 20 is a horizontal line segment when 0 ≤ x ≤ 20. The area between f ( x ) = 1 20 where 0 ≤ x ≤ 20 and the x -axis is the area of a rectangle with base = 20 and height = 1 20 . AREA = 20 ( 1 20 ) = 1 Suppose we want to find the area between f( x ) = 1 20 and the x -axis where 0 < x < 2. AREA = ( 2 – 0 ) ( 1 20 ) = 0.1 ( 2 – 0 ) = 2 = base of a rectangle REMINDER area of a rectangle = (base)(height). The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as P (0 < x < 2) = P ( x < 2) = 0.1. Suppose we want to find the area between f ( x ) = 1 20 and the x -axis where 4 < x < 15. AREA = ( 15 – 4 ) ( 1 20 ) = 0.55 ( 15 – 4 ) = 11 = the base of a rectangle The area corresponds to the probability P (4 < x < 15) = 0.55. Suppose we want to find P ( x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P ( x = 15) = (base)(height) = (0) ( 1 20 ) = 0 P ( X ≤ x ), which can also be written as P ( X < x ) for continuous distributions, is called the cumulative distribution function or CDF. Notice the \"less than or equal to\" symbol. We can also use the CDF to calculate P ( X > x ). The CDF gives \"area to the left\" and P ( X > x ) gives \"area to the right.\" We calculate P ( X > x ) for continuous distributions as follows: P ( X > x ) = 1 – P ( X < x ). Label the graph with f ( x ) and x . Scale the x and y axes with the maximum x and y values. f ( x ) = 1 20 , 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. P ( 2.3 < x < 12.7 ) = ( base ) ( height ) = ( 12.7 − 2.3 ) ( 1 20 ) = 0.52 Try It Consider the function f ( x ) = 1 8 for 0 ≤ x ≤ 8. Draw the graph of f ( x ) and find P (2.5 < x < 7.5). Chapter Review The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P ( a < x < b ). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f ( x ), is used to draw the graph of the probability distribution. The total area under the graph of f ( x ) is one. The area under the graph of f ( x ) and between values a and b gives the probability P ( a < x < b ). The cumulative distribution function (cdf) of X is defined by P ( X ≤ x ). It is a function of x that gives the probability that the random variable is less than or equal to x . Formula Review Probability density function (pdf) f ( x ): f ( x ) ≥ 0 The total area under the curve f ( x ) is one. Cumulative distribution function (cdf): P ( X ≤ x ) Which type of distribution does the graph illustrate? Uniform Distribution Which type of distribution does the graph illustrate? Which type of distribution does the graph illustrate? Normal Distribution What does the shaded area represent? P (___< x < ___) What does the shaded area represent? P (___< x < ___) P (6 < x < 7) For a continuous probablity distribution, 0 ≤ x ≤ 15. What is P ( x > 15)? What is the area under f ( x ) if the function is a continuous probability density function? one For a continuous probability distribution, 0 ≤ x ≤ 10. What is P ( x = 7)? A continuous probability function is restricted to the portion between x = 0 and 7. What is P ( x = 10)? zero f ( x ) for a continuous probability function is 1 5 , and the function is restricted to 0 ≤ x ≤ 5. What is P ( x < 0)? f ( x ), a continuous probability function, is equal to 1 12 , and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)? one Find the probability that x falls in the shaded area. Find the probability that x falls in the shaded area. 0.625 Find the probability that x falls in the shaded area. f ( x ), a continuous probability function, is equal to 1 3 and the function is restricted to 1 ≤ x ≤ 4. Describe P ( x > 3 2 ) . The probability is equal to the area from x = 3 2 to x = 4 above the x-axis and up to f ( x ) = 1 3 . Homework For each probability and percentile problem, draw the picture. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor. What part of the experiment will yield discrete data? What part of the experiment will yield continuous data? When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why? Age is a measurement, regardless of the accuracy used.", "section": "Continuous Probability Functions", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints. The data in are 55 smiling times, in seconds, of an eight-week-old baby. 10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9 12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8 1.3 0.7 8.9 11.9 10.9 7.3 5.9 3.7 17.9 19.2 9.8 5.8 6.9 2.6 5.8 21.7 11.8 3.4 2.1 4.5 6.3 10.7 8.9 9.4 9.4 7.6 10.0 3.3 6.7 7.8 11.6 13.8 18.6 The sample mean = 11.65 and the sample standard deviation = 6.08. We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely . The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let X = length, in seconds, of an eight-week-old baby's smile. The notation for the uniform distribution is X ~ U ( a , b ) where a = the lowest value of x and b = the highest value of x . The probability density function is f ( x ) = 1 b − a for a ≤ x ≤ b . For this example, x ~ U (0, 23) and f ( x ) = 1 23 − 0 for 0 ≤ X ≤ 23. Formulas for the theoretical mean and standard deviation are μ = a + b 2 and σ = ( b − a ) 2 12 For this problem, the theoretical mean and standard deviation are μ = 0 + 23 2 = 11.50 seconds and σ = ( 23 − 0 ) 2 12 = 6.64 seconds. Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. Try It The data that follow record the total weight, to the nearest pound, of fish caught by passengers on 35 different charter fishing boats on one summer day. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and b . Write the distribution in proper notation, and calculate the theoretical mean and standard deviation. 1 12 4 10 4 14 11 7 11 4 13 2 4 6 3 10 0 12 6 9 10 5 13 4 10 14 12 11 6 10 11 0 11 13 2 a. Refer to . What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds? P (2 < x < 18) = (base)(height) = (18 – 2) ( 1 23 ) = 16 23 . b. Find the 90 th percentile for an eight-week-old baby's smiling time. b. Ninety percent of the smiling times fall below the 90 th percentile, k , so P ( x < k ) = 0.90. P ( x < k ) = 0.90 ( base ) ( height ) = 0.90 ( k − 0 ) ( 1 23 ) = 0.90 k = ( 23 ) ( 0.90 ) = 20.7 c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS . c. This probability question is a conditional . You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds. Find P ( x > 12| x > 8) There are two ways to do the problem. For the first way , use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. Write a new f ( x ): f ( x ) = 1 23 − 8 = 1 15 for 8 < x < 23 P ( x > 12| x > 8) = (23 − 12) ( 1 15 ) = 11 15 For the second way , use the conditional formula from Probability Topics with the original distribution X ~ U (0, 23): P ( A | B ) = P ( A AND B ) P ( B ) For this problem, A is ( x > 12) and B is ( x > 8). So, P ( x > 12 | x > 8) = P ( x > 12 AND x > 8 ) P ( x > 8 ) = P ( x > 12 ) P ( x > 8 ) = 11 23 15 23 = 11 15 Darker shaded area represents P(x > 12). Entire shaded area shows P(x > 8). Try It A distribution is given as X ~ U (0, 20). What is P (2 < x < 18)? Find the 90 th percentile. The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. a. What is the probability that a person waits fewer than 12.5 minutes? b. On the average, how long must a person wait? Find the mean, μ , and the standard deviation, σ . c. Ninety percent of the time, the time a person must wait falls below what value? This asks for the 90 th percentile. a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U (0, 15). Write the probability density function. f ( x ) = 1 15 − 0 = 1 15 for 0 ≤ x ≤ 15. Find P ( x < 12.5). Draw a graph. P ( x < k ) = ( base ) ( height ) = ( 12.5 - 0 ) ( 1 15 ) = 0.8333 The probability a person waits less than 12.5 minutes is 0.8333. b. μ = a + b 2 = 15 + 0 2 = 7.5. On the average, a person must wait 7.5 minutes. σ = ( b - a ) 2 12 = ( 15 - 0 ) 2 12 = 4.3. The Standard deviation is 4.3 minutes. c. Find the 90 th percentile. Draw a graph. Let k = the 90 th percentile. P ( x < k ) = ( base ) ( height ) = ( k − 0 ) ( 1 15 ) 0.90 = ( k ) ( 1 15 ) k = ( 0.90 ) ( 15 ) = 13.5 k is sometimes called a critical value. The 90 th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. Try It The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive. Find a and b and describe what they represent. Write the distribution. Find the mean and the standard deviation. What is the probability that the duration of games for a team for a single season is between 480 and 500 hours? What is the 65 th percentile for the duration of games for a team in a single season? Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X = the time, in minutes, it takes a nine-year old child to eat a donut. Then X ~ U (0.5, 4). a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______. b. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. The second question has a conditional probability . You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see ). You must reduce the sample space. First way : Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a new f ( x ): f ( x ) = 1 4 − 1.5 = 2 5 for 1.5 ≤ x ≤ 4. Find P ( x > 2| x > 1.5). Draw a graph. P ( x > 2 | x > 1.5) = (base)(new height) = (4 − 2) ( 2 5 ) = 4 5 a. 0.5714 b. 4 5 The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is 4 5 . Second way: Draw the original graph for X ~ U (0.5, 4). Use the conditional formula P ( x > 2| x > 1.5) = P ( x > 2 AND x > 1.5 ) P ( x > 1 .5 ) = P ( x > 2 ) P ( x > 1.5 ) = 2 3.5 2.5 3.5 = 0 .8 = 4 5 Try It Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. Then X ~ U (6, 15). Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes. Ace Heating and Air Conditioning Service finds that the amount of time a repair tech needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x = the time needed to fix a furnace. Then x ~ U (1.5, 4). Find the probability that a randomly selected furnace repair requires more than two hours. Find the probability that a randomly selected furnace repair requires less than three hours. Find the 30 th percentile of furnace repair times. The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent? Find the mean and standard deviation a. To find f ( x ): f ( x ) = 1 4 − 1.5 = 1 2.5 so f ( x ) = 0.4 P ( x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8 Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time x is greater than two b. P ( x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6 The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0; since X ~ U (1.5, 4), x can not be less than 1.5. Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three c. Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times. P ( x < k ) = 0.30 P ( x < k ) = (base)(height) = ( k – 1.5)(0.4) 0.3 = ( k – 1.5) (0.4) ; Solve to find k : 0.75 = k – 1.5, obtained by dividing both sides by 0.4 k = 2.25 , obtained by adding 1.5 to both sides The 30 th percentile of repair times is 2.25 hours. 30% of repair times are 2.25 hours or less. d. Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. P ( x > k ) = 0.25 P ( x > k ) = (base)(height) = (4 – k )(0.4) 0.25 = (4 – k )(0.4) ; Solve for k : 0.625 = 4 − k , obtained by dividing both sides by 0.4 −3.375 = − k , obtained by subtracting four from both sides: k = 3.375 The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75 th percentile of furnace repair times. e. μ = a + b 2 and σ = ( b − a ) 2 12 μ = 1.5 + 4 2 = 2.75 hours and σ = ( 4 – 1.5 ) 2 12 = 0.7217 hours Try It The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car. Write the random variable X in words. X = __________________. Write the distribution. Graph the distribution. Find P ( x > 19). Find the 50 th percentile. Chapter Review If X has a uniform distribution where a < x < b or a ≤ x ≤ b , then X takes on values between a and b (may include a and b ). All values x are equally likely. We write X ∼ U ( a , b ). The mean of X is μ = a + b 2 . The standard deviation of X is σ = ( b − a ) 2 12 . The probability density function of X is f ( x ) = 1 b − a for a ≤ x ≤ b . The cumulative distribution function of X is P ( X ≤ x ) = x − a b − a . X is continuous. The probability P ( c < X < d ) may be found by computing the area under f ( x ), between c and d . Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. Formula Review X = a real number between a and b (in some instances, X can take on the values a and b ). a = smallest X ; b = largest X X ~ U (a, b) The mean is μ = a + b 2 The standard deviation is σ = ( b – a ) 2 12 Probability density function: f ( x ) = 1 b − a for a ≤ X ≤ b Area to the Left of x : P ( X < x ) = ( x – a ) ( 1 b − a ) Area to the Right of x : P ( X > x ) = ( b – x ) ( 1 b − a ) Area Between c and d : P ( c < x < d ) = (base)(height) = ( d – c ) ( 1 b − a ) Uniform: X ~ U ( a , b ) where a < x < b pdf: f ( x ) = 1 b − a for a ≤ x ≤ b cdf: P ( X ≤ x ) = x − a b − a mean µ = a + b 2 standard deviation σ = ( b − a ) 2 12 P ( c < X < d ) = ( d – c ) ( 1 b – a ) References McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 1.5 2.4 3.6 2.6 1.6 2.4 2.0 3.5 2.5 1.8 2.4 2.5 3.5 4.0 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as X ~ U (1.5, 4.5). What type of distribution is this? In this distribution, outcomes are equally likely. What does this mean? It means that the value of x is just as likely to be any number between 1.5 and 4.5. What is the height of f ( x ) for the continuous probability distribution? What are the constraints for the values of x ? 1.5 ≤ x ≤ 4.5 Graph P (2 < x < 3). What is P (2 < x < 3)? 0.3333 What is P (x < 3.5| x < 4)? What is P ( x = 1.5)? zero What is the 90 th percentile of square footage for homes? Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. 0.6 Use the following information to answer the next eight exercises. A distribution is given as X ~ U (0, 12). What is a ? What does it represent? What is b ? What does it represent? b is 12, and it represents the highest value of x . What is the probability density function? What is the theoretical mean? six What is the theoretical standard deviation? Draw the graph of the distribution for P ( x > 9). Find P ( x > 9). Find the 40 th percentile. 4.8 Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. What is being measured here? In words, define the random variable X . X = The age (in years) of cars in the staff parking lot Are the data discrete or continuous? The interval of values for x is ______. 0.5 to 9.5 The distribution for X is ______. Write the probability density function. f ( x ) = 1 9 where x is between 0.5 and 9.5, inclusive. Graph the probability distribution. Sketch the graph of the probability distribution. Identify the following values: Lowest value for x ¯ : _______ Highest value for x ¯ : _______ Height of the rectangle: _______ Label for x -axis (words): _______ Label for y -axis (words): _______ Find the average age of the cars in the lot. μ = 5 Find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, and shade the area of interest. Find the probability. P ( x < 4) = _______ Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, shade the area of interest. Find the probability. P ( x < 4| x < 7.5) = _______ Answers may vary. 3.5 7 What has changed in the previous two problems that made the solutions different? Find the third quartile of ages of cars in the lot. This means you will have to find the value such that 3 4 , or 75%, of the cars are at most (less than or equal to) that age. Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.75. The third quartile is _______ Answers may vary. k = 7.25 7.25 Homework For each probability and percentile problem, draw the picture. Births are approximately uniformly distributed throughout the year. They can be said to follow a uniform distribution from 0 to 52 (spread of 52 weeks). X ~ _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that a person is born at the exact moment week 19 ends. That is, find P ( x = 19) = _________ P (2 < x < 31) = _________ Find the probability that a person is born after week 40. P (12 < x | x < 28) = _________ Find the 70 th percentile. Find the minimum for the upper quarter. A random number generator picks a number from one to nine in a uniform manner. X ~ _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ P (3.5 < x < 7.25) = _________ P ( x > 5.67) P ( x > 5| x > 3) = _________ Find the 90 th percentile. X ~ U (1, 9) Answers may vary. f ( x ) = 1 8 where 1 ≤ x ≤ 9 five 2.3 15 32 333 800 2 3 8.2 According to a study by Dr. John McDougall of his live-in weight loss program, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. Define the random variable. X = _________ X ~ _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that the individual lost more than ten pounds in a month. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month. P (7 < x < 13| x > 9) = __________. State this in a probability question, similarly to parts g and h, draw the picture, and find the probability. A subway train arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. Define the random variable. X = _______ X ~ _______ Graph the probability distribution. f ( x ) = _______ μ = _______ σ = _______ Find the probability that the commuter waits less than one minute. Find the probability that the commuter waits between three and four minutes. Sixty percent of commuters wait more than how long for the train? State this in a probability question, similarly to parts g and h, draw the picture, and find the probability. X represents the length of time a commuter must wait for a train to arrive on the Red Line. X ~ U (0, 8) Graph the probability distribution. f ( x ) = 1 8 where 0 ≤ x ≤ 8 four 2.31 1 8 1 8 3.2 The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class. Define the random variable. X = _________ X ~ _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that she is over 6.5 years old. Find the probability that she is between four and six years old. Find the 70 th percentile for the age of first graders on September 1 at Garden Elementary School. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. What is the average waiting time (in minutes)? zero two three four d Find the 30 th percentile for the waiting times (in minutes). two 2.4 2.75 three The probability of waiting more than seven minutes given a person has waited more than four minutes is? 0.125 0.25 0.5 0.75 b The time (in minutes) until the next bus departs a major bus depot follows a distribution with f ( x ) = 1 20 where x goes from 25 to 45 minutes. Define the random variable. X = ________ X ~ ________ Graph the probability distribution. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous). μ = ________ σ = ________ Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. P (25 < x < 55) = _________. State this in a probability statement, similarly to parts g and h, draw the picture, and find the probability. Find the 90 th percentile. This means that 90% of the time, the time is less than _____ minutes. Find the 75 th percentile. In a complete sentence, state what this means. (See part j.) Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution. Find the probability that the value of the stock is more than $19. Find the probability that the value of the stock is between $19 and $22. Find the upper quartile - 25% of all days the stock is above what value? Draw the graph. Given that the stock is greater than $18, find the probability that the stock is more than $21. The probability density function of X is 1 25 − 16 = 1 9 . P ( X > 19) = (25 – 19) ( 1 9 ) = 6 9 = 2 3 . P (19 < X < 22) = (22 – 19) ( 1 9 ) = 3 9 = 1 3 . The area must be 0.25, and 0.25 = (width) ( 1 9 ) , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75. This is a conditional probability question. P(x > 21| x > 18). You can do this two ways: Draw the graph where a is now 18 and b is still 25. The height is 1 ( 25 − 18 ) = 1 7 So, P ( x > 21| x > 18) = (25 – 21) ( 1 7 ) = 4/7. Use the formula: P ( x > 21| x > 18) = P ( x > 21 AND x > 18 ) P ( x > 18 ) = P ( x > 21 ) P ( x > 18 ) = ( 25 − 21 ) ( 25 − 18 ) = 4 7 . A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution. Find the average time between fireworks. Find probability that the time between fireworks is greater than four seconds. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. Find the probability that the truck driver goes more than 650 miles in a day. Find the probability that the truck drivers goes between 400 and 650 miles in a day. At least how many miles does the truck driver travel on the furthest 10% of days? P ( X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 = 0.125 P (400 < X < 650) = 650 − 400 700 − 300 = 250 400 = 0.625 0.10 = width 700 − 300 , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days. Conditional Probability the likelihood that an event will occur given that another event has already occurred.", "section": "The Uniform Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Exponential Distribution The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money. Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts. Let X = amount of time (in minutes) a postal clerk spends with their customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes. X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m , the decay parameter. m = 1 μ . Therefore, m = 1 4 = 0.25. The standard deviation, σ , is the same as the mean. μ = σ The distribution notation is X ~ Exp ( m ). Therefore, X ~ Exp (0.25). The probability density function is f ( x ) = me - mx . The number e = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key \" e x .\" If you enter one for x , the calculator will display the value e . The curve is: f ( x ) = 0.25 e –0.25 x where x is at least zero and m = 0.25. For example, f (5) = 0.25 e (-0.25)(5) = 0.072. The value 0.072 is the height of the curve when x = 5. In below, you will learn how to find probabilities using the decay parameter. The graph is as follows: Notice the graph is a declining curve. When x = 0, f ( x ) = 0.25 e (−0.25)(0) = (0.25)(1) = 0.25 = m . The maximum value on the y -axis is m . Try It The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution. a. Using the information in , find the probability that a clerk spends four to five minutes with a randomly selected customer. a. Find P (4 < x < 5). The cumulative distribution function (CDF) gives the area to the left. P ( x < x ) = 1 – e –mx P ( x < 5) = 1 – e (−0.25)(5) = 0.7135 and P ( x < 4) = 1 – e (–0.25)(4) = 0.6321 NOTE You can do these calculations easily on a calculator. The probability that a postal clerk spends four to five minutes with a randomly selected customer is P (4 < x < 5) = P ( x < 5) – P ( x < 4) = 0.7135 − 0.6321 = 0.0814. On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5). b. Half of all customers are finished within how long? (Find the 50 th percentile) b. Find the 50 th percentile. P ( x < k ) = 0.50, k = 2.8 minutes (calculator or computer) Half of all customers are finished within 2.8 minutes. You can also do the calculation as follows: P ( x < k ) = 0.50 and P ( x < k ) = 1 – e –0.25 k Therefore, 0.50 = 1 − e −0.25 k and e −0.25 k = 1 − 0.50 = 0.5 Take natural logs: ln ( e –0.25 k ) = ln (0.50). So, –0.25 k = ln (0.50) Solve for k : k = l n ( 0.50 ) - 0.25 = 2.8 minutes. The calculator simplifies the calculation for percentile k . See the following two notes. NOTE A formula for the percentile k is k = l n ( 1 − A r e a T o T h e L e f t ) − m where ln is the natural log. On the home screen, enter ln(1 – 0.50)/–0.25. Press the (-) for the negative. c. Which is larger, the mean or the median? c. From part b, the median or 50 th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger. Try It The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait? Have each class member count the change they have in their pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean. Let X = the amount of money a student in your class has in their pocket or purse. The distribution for X is approximately exponential with mean, μ = _______ and m = _______. The standard deviation, σ = ________. Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $.40 in their pocket or purse. (Shade P ( x < 0.40)). On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than 7 years? a. Let x = the amount of time (in years) a computer part lasts. μ = 10 so m = 1 μ = 1 10 = 0.1 Find P ( x > 7). Draw the graph. P ( x > 7) = 1 – P ( x < 7). Since P ( X < x ) = 1 – e –mx then P ( X > x ) = 1 –(1 – e –mx ) = e -mx P ( x > 7) = e (–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966. On the home screen, enter e^(-.1*7). b. On the average, how long would five computer parts last if they are used one after another? b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years. c. Eighty percent of computer parts last at most how long? c. Find the 80 th percentile. Draw the graph. Let k = the 80 th percentile. Solve for k : k = l n ( 1 – 0.80 ) – 0.1 = 16.1 years Eighty percent of the computer parts last at most 16.1 years. On the home screen, enter ln ( 1 – 0.80 ) – 0.1 d. What is the probability that a computer part lasts between nine and 11 years? d. Find P (9 < x < 11). Draw the graph. P (9 < x < 11) = P ( x < 11) – P ( x < 9) = (1 – e (–0.1)(11) ) – (1 – e (–0.1)(9) ) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737. On the home screen, enter e ^(–0.1*9) – e ^(–0.1*11). Try It On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day? Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12 . If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes. What is m , μ , and σ ? The probability that you must wait more than five minutes is _______ . m = 1 12 μ = 12 σ = 12 P ( x > 5) = 0.6592 Try It Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1 20 . Let X = the distance people are willing to commute in miles. What is m , μ , and σ ? What is the probability that a person is willing to commute more than 25 miles? The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed. On average, how many minutes elapse between two successive arrivals? When the store first opens, how long on average does it take for three customers to arrive? After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive. Seventy percent of the customers arrive within how many minutes of the previous customer? Is an exponential distribution reasonable for this situation? Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive. Let X = the time between arrivals, in minutes. By part a, μ = 2, so m = 1 2 = 0.5. Therefore, X ∼ Exp (0.5). The cumulative distribution function is P ( X < x ) = 1 – e (-0.5)(x) . Therefore P ( X < 1) = 1 – e (–0.5)(1) ≈ 0.3935. 1 - e ^(–0.5) ≈ 0.3935 P ( X > 5) = 1 – P ( X < 5) = 1 – (1 – e (-0.50)(5) ) = e –2.5 ≈ 0.0821. 1 – (1 – e^((-0.50)(5))) or e^( – 5*0.5) We want to solve 0.70 = P ( X < x ) for x . Substituting in the cumulative distribution function gives 0.70 = 1 – e –0.5 x , so that e –0.5x = 0.30. Converting this to logarithmic form gives –0.5 x = ln (0.30), or x = l n ( 0.30 ) – 0.5 ≈ 2.41 minutes. Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer. You are finding the 70 th percentile k so you can use the formula k = l n ( 1 – A r e a _ T o _ T h e _ L e f t _ O f _ k ) ( – m ) k = l n ( 1 – 0.70 ) ( – 0.5 ) ≈ 2.41 minutes This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others. Try It Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution. On average, how many seconds elapse between two successive cars? After a car passes by, how long on average will it take for another seven cars to pass by? Find the probability that after a car passes by, the next car will pass within the next 20 seconds. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds. Memorylessness of the Exponential Distribution In recall that the amount of time between customers is exponentially distributed with a mean of two minutes ( X ~ Exp (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property . Specifically, the memoryless property says that P ( X > r + t | X > r ) = P ( X > t ) for all r ≥ 0 and t ≥ 0 For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation. P ( X > 5 + 1 | X > 5) = P ( X > 1) = e ( – 0.5 ) ( 1 ) ≈ 0.6065. This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival. The exponential distribution is often used to model the longevity of an electrical or mechanical device. In , the lifetime of a certain computer part has the exponential distribution with a mean of ten years ( X ~ Exp (0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P ( X > 17| X > 10) = P ( X > 7) = 0.4966. Refer to where the time a postal clerk spends with a customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that the customer will spend at least an additional three minutes with the postal clerk? The decay parameter of X is m = 1 4 = 0.25, so X ∼ Exp (0.25). The cumulative distribution function is P ( X < x ) = 1 – e –0.25 x . We want to find P ( X > 7| X > 4). The memoryless property says that P ( X > 7| X > 4) = P ( X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk. This is P ( X > 3) = 1 – P ( X < 3) = 1 – (1 – e –0.25⋅3 ) = e –0.75 ≈ 0.4724. 1–(1–e^(–0.25*3)) = e^(–0.25*3). Try It Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years. Relationship between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ , then P ( X = k ) = λ k e − λ k ! . Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. ( k ! = k *( k –1*)( k –2)*( k –3)*…3*2*1) Suppose X has the Poisson distribution with mean λ . Compute P ( X = k ) by entering 2 nd , VARS(DISTR) , C: poissonpdf( λ , k ) . To compute P ( X ≤ k ), enter 2 nd , VARS (DISTR) , D:poissoncdf( λ , k ) . At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution. Find the average time between two successive calls. Find the probability that after a call is received, the next call occurs in less than ten seconds. Find the probability that exactly five calls occur within a minute. Find the probability that less than five calls occur within a minute. Find the probability that more than 40 calls occur in an eight-minute period. On average there are four calls occur per minute, so 15 seconds, or 15 60 = 0.25 minutes occur between successive calls on average. Let T = time elapsed between calls. From part a, μ = 0.25, so m = 1 0.25 = 4. Thus, T ∼ Exp (4). The cumulative distribution function is P ( T < t ) = 1 – e –4 t . The probability that the next call occurs in less than ten seconds (ten seconds = 1/6 minute) is P ( T < 1 6 ) = 1 – e ( – 4 ) ( 1 6 ) ≈ 0.4866. Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. Therefore, X ∼ Poisson (4), and so P ( X = 5) = 4 5 e − 4 5 ! ≈ 0.1563. (5! = (5)(4)(3)(2)(1)) poissonpdf(4, 5) = 0.1563. Keep in mind that X must be a whole number, so P ( X < 5) = P ( X ≤ 4). To compute this, we could take P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4). Using technology, we see that P ( X ≤ 4) = 0.6288. poisssoncdf(4, 4) = 0.6288 Let Y = the number of calls that occur during an eight minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period. Hence, Y ∼ Poisson (32). Therefore, P ( Y > 40) = 1 – P ( Y ≤ 40) = 1 – 0.9294 = 0.0707. 1 – poissoncdf(32, 40). = 0.0707 Try It In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. Calculate the probability that there are at most 2 accidents occur in any given week. What is the probability that there is at least two weeks between any 2 accidents? Chapter Review If X has an exponential distribution with mean μ , then the decay parameter is m = 1 μ , and we write X ∼ Exp ( m ) where x ≥ 0 and m > 0 . The probability density function of X is f ( x ) = me -mx (or equivalently f ( x ) = 1 μ e − x / μ . The cumulative distribution function of X is P ( X ≤ x ) = 1 – e – mx . The exponential distribution has the memoryless property , which says that future probabilities do not depend on any past information. Mathematically, it says that P ( X > x + k | X > x ) = P ( X > k ). If T represents the waiting time between events, and if T ∼ Exp ( λ ), then the number of events X per unit time follows the Poisson distribution with mean λ . The probability density function of X is P ( X = k ) = λ k e − k k ! . This may be computed using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf( λ , k ). The cumulative distribution function P ( X ≤ k ) may be computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf( λ , k ). Formula Review Exponential: X ~ Exp ( m ) where m = the decay parameter pdf: f ( x ) = me (– mx ) where x ≥ 0 and m > 0 cdf: P ( X ≤ x ) = 1 – e (– mx ) mean µ = 1 m standard deviation σ = µ percentile k : k = l n ( 1 − A r e a T o T h e L e f t O f k ) ( − m ) Additionally P ( X > x ) = e (– mx ) P ( a < X < b ) = e (– ma ) – e (– mb ) Memoryless Property: P ( X > x + k | X > x ) = P ( X > k ) Poisson probability: P ( X = k ) = λ k e − λ k ! with mean λ k ! = k *( k -1)*( k -2)*( k -3)*…3*2*1 References Data from the United States Census Bureau. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013). “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013). Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf‎ (accessed June 11, 2013). Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp (0.2) What type of distribution is this? Are outcomes equally likely in this distribution? Why or why not? No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. What is m ? What does it represent? What is the mean? five What is the standard deviation? State the probability density function. f ( x ) = 0.2e -0.2 x Graph the distribution. Find P (2 < x < 10). 0.5350 Find P ( x > 6). Find the 70 th percentile. 6.02 Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp (0.75). What is m ? What is the probability density function? f ( x ) = 0.75 e -0.75 x What is the cumulative distribution function? Draw the distribution. Find P ( x < 4). Find the 30 th percentile. 0.4756 Find the median. Which is larger, the mean or the median? The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33 , which is greater than 0.9242. Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. What is being measured here? Are the data discrete or continuous? continuous In words, define the random variable X . What is the decay rate ( m )? m = 0.000121 The distribution for X is ______. Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P ( x < 5,730). Sketch the graph, and shade the area of interest. Find the probability. P ( x < 5,730) = __________ Answers may vary. P ( x < 5,730) = 0.5001 Find the percentage of carbon-14 lasting longer than 10,000 years. Sketch the graph, and shade the area of interest. Find the probability. P ( x > 10,000) = ________ Thirty percent (30%) of carbon-14 will decay within how many years? Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.30. Answers may vary. k = 2947.73 Homework Suppose that the length of phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. Define the random variable. X = ________________. Is X continuous or discrete? X ~ ________ μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a phone call lasts less than nine minutes. Find the probability that a phone call lasts more than nine minutes. Find the probability that a phone call lasts between seven and nine minutes. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. Define the random variable. X = _________________________________. Is X continuous or discrete? X ~ ________ On average, how long would you expect one car battery to last? On average, how long would you expect nine car batteries to last, if they are used one after another? Find the probability that a car battery lasts more than 36 months. Seventy percent of the batteries last at least how long? X = the useful life of a particular car battery, measured in months. X is continuous. X ~ Exp (0.025) 40 months 360 months 0.4066 14.27 At one point in time, the percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. Define the random variable. X = _________________________________. Is X continuous or discrete? X ~ ________ μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the percent is less than 12. Find the probability that the percent is between eight and 14. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. Why is this number different from 9.848%? What would make this number higher than 9.848%? The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. Define the random variable. X = _________________________________. Is X continuous or discrete? X ~ = ________ μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the person retired after age 70. Do more people retire before age 65 or after age 65? In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet? X = the time (in years) after reaching age 60 that it takes an individual to retire X is continuous. X ~ Exp ( 1 5 ) five five Answers may vary. 0.1353 before 18.3 The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150. Define the random variable. X = _________________________________. X ~ = ________ μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a car required over $300 for maintenance during its first year. Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphones is known to follow an exponential distribution. The decay rate is: 0.3333 0.5000 2 3 a What is the probability that a smartphone will fail within two years of the date of purchase? 0.8647 0.4866 0.2212 0.9997 What is the median lifetime of these phones (in years)? 0.1941 1.3863 2.0794 5.5452 c Let X ~ Exp (0.1). decay rate = ________ μ = ________ Graph the probability distribution function. On the graph, shade the area corresponding to P ( x < 6) and find the probability. Sketch a new graph, shade the area corresponding to P (3 < x < 6) and find the probability. Sketch a new graph, shade the area corresponding to P ( x < 7) and find the probability. Sketch a new graph, shade the area corresponding to the 40 th percentile and find the value. Find the average value of x . Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. Find the probability that a light bulb lasts less than one year. Find the probability that a light bulb lasts between six and ten years. Seventy percent of all light bulbs last at least how long? A company decides to offer a warranty to give refunds to light bulbs whose lifetime is among the lowest two percent of all bulbs. To the nearest month, what should be the cutoff lifetime for the warranty to take place? If a light bulb has lasted seven years, what is the probability that it fails within the 8 th year. Let T = the life time of a light bulb. The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is P ( T < t ) = 1 − e − t 8 Therefore, P ( T < 1) = 1 – e – 1 8 ≈ 0.1175. We want to find P (6 < t < 10). To do this, P (6 < t < 10) – P ( t < 6) = ( 1 – e – 1 8 * 10 ) – ( 1 – e – 1 8 * 6 ) ≈ 0.7135 – 0.5276 = 0.1859 We want to find 0.70 = P ( T > t ) = 1 – ( 1 – e − t 8 ) = e − t 8 . Solving for t , e – t 8 = 0.70, so – t 8 = ln (0.70), and t = –8 ln (0.70) ≈ 2.85 years. Or use t = l n (area_to_the_right) ( – m ) = l n (0 .70) – 1 8 ≈ 2. 85 years . We want to find 0.02 = P ( T < t ) = 1 – e – t 8 . Solving for t , e – t 8 = 0.98, so – t 8 = ln (0.98), and t = –8 ln (0.98) ≈ 0.1616 years, or roughly two months. The warranty should cover light bulbs that last less than 2 months. Or use ln (area_to_the_right) ( – m) = ln (1 – 0.2 ) – 1 8 = 0.1616. We must find P ( T < 8| T > 7). Notice that by the rule of complement events, P ( T < 8| T > 7) = 1 – P ( T > 8| T > 7). By the memoryless property ( P ( X > r + t | X > r ) = P ( X > t )). So P ( T > 8| T > 7) = P ( T > 1) = 1 – ( 1 – e – 1 8 ) = e – 1 8 ≈ 0.8825 Therefore, P ( T < 8| T > 7) = 1 – 0.8825 = 0.1175. At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution. On average, how much time occurs between five consecutive calls? Find the probability that after a call is received, it takes more than three minutes for the next call to occur. Ninety-percent of all calls occur within how many minutes of the previous call? Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute. Find the probability that less than 20 calls occur within an hour. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. What is the probability that an entire season elapses with a single no-hitter? If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? What is the probability that there are more than 3 no-hitters in a single season? Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, ( X = 0) = 3 0 e – 3 0 ! = e –3 ≈ 0.0498 NOTE You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 1 3 season. For the exponential, µ = 1 3 . Therefore, m = 1 μ = 3 and T ∼ Exp (3). The desired probability is P ( T > 1) = 1 – P ( T < 1) = 1 – (1 – e –3 ) = e –3 ≈ 0.0498. Let T = duration of time between no-hitters. We find P ( T > 2| T > 1), and by the memoryless property this is simply P ( T > 1), which we found to be 0.0498 in part a. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P ( X > 3) = 1 – P ( X ≤ 3) = 0.3528. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential. What is the probability that the next earthquake occurs within the next three months? Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes? What is the probability of zero earthquakes occurring in 2014? What is the probability that at least two earthquakes will occur in 2014? According to the American Red Cross, about one out of nine people in the U.S. have Type B positive blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B positive blood types that arrive roughly follows the Poisson distribution. If 100 people arrive, how many on average would be expected to have Type B positive blood? What is the probability that over 10 people out of these 100 have type B positive blood? What is the probability that more than 20 people arrive before a person with type B positive blood is found? 100 9 = 11.11 P ( X > 10) = 1 – P ( X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532. The number of people with Type B positive blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B positive arrivals is roughly exponential with mean μ = 9 and m = 1 9 . The cumulative distribution function of X is P ( X < x ) = 1 − e − x 9 . Thus, P ( X > 20) = 1 - P ( X ≤ 20) = 1 − ( 1 − e − 20 9 ) ≈ 0.1084. NOTE We could also deduce that each person arriving has a 8/9 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive blood is ( 8 9 ) 20 ≈ 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between Type B positive people is discrete instead of continuous.) A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution. Find the probability that the duration between two successive visits to the web site is more than ten minutes. The top 25% of durations between visits are at least how long? Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes? Find the probability that less than 7 visits occur within a one-hour period. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed. Find the probability that the time between two successive visits to the urgent care facility is less than 2 minutes. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes? Find the probability that more than eight patients arrive during a half-hour period. Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 . The cdf is P ( T < t ) = 1 − e t 7 P ( T < 2) = 1 - 1 − e − 2 7 ≈ 0.2485. P ( T > 15) = 1 − P ( T < 15 ) = 1 − ( 1 − e − 15 7 ) ≈ e − 15 7 ≈ 0.1173 . P ( T > 15| T > 10) = P ( T > 5) = 1 − ( 1 − e − 5 7 ) = e − 5 7 ≈ 0.4895 . Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 , X ∼ Poisson ( 30 7 ) . Find P ( X > 8) = 1 – P ( X ≤ 8) ≈ 0.0311. decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x . It is the value m in the probability density function f ( x ) = me (- mx ) of an exponential random variable. It is also equal to m = 1 μ , where μ is the mean of the random variable. memoryless property For an exponential random variable X , the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that X exceeds x + k , given that it has exceeded x , is the same as the probability that X would exceed k if we had no knowledge about it. In symbols we say that P ( X > x + k | X > x ) = P ( X > k ). Poisson distribution If there is a known average of λ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to P ( X = k ) = λ k e − λ k ! .", "section": "The Exponential Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Continuous Distribution Continuous Distribution Class Time: Names: Student Learning Outcomes The student will compare and contrast empirical data from a random number generator with the uniform distribution. Collect the Data Use a random number generator to generate 50 values between zero and one (inclusive). List them in . Round the numbers to four decimal places or set the calculator MODE to four places. Complete the table. Calculate the following: x ¯ = _______ s = _______ first quartile = _______ third quartile = _______ median = _______ Organize the Data Construct a histogram of the empirical data. Make eight bars. Construct a histogram of the empirical data. Make five bars. Describe the Data In two to three complete sentences, describe the shape of each graph. (Keep it simple. Does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end, and so on. One way to help you determine a shape is to draw a smooth curve roughly through the top of the bars.) Describe how changing the number of bars might change the shape. Theoretical Distribution In words, X = _____________________________________. The theoretical distribution of X is X ~ U (0,1). In theory, based upon the distribution X ~ U (0,1), complete the following. μ = ______ σ = ______ first quartile = ______ third quartile = ______ median = __________ Are the empirical values (the data) in the section titled Collect the Data close to the corresponding theoretical values? Why or why not? Plot the Data Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges. Do you notice any potential outliers? If so, which values are they? Either way, justify your answer numerically. (Recall that any DATA that are less than Q 1 – 1.5( IQR ) or more than Q 3 + 1.5( IQR ) are potential outliers. IQR means interquartile range.) Compare the Data For each of the following parts, use a complete sentence to comment on how the value obtained from the data compares to the theoretical value you expected from the distribution in the section titled Theoretical Distribution . minimum value: _______ first quartile: _______ median: _______ third quartile: _______ maximum value: _______ width of IQR : _______ overall shape: _______ Based on your comments in the section titled Collect the Data , how does the box plot fit or not fit what you would expect of the distribution in the section titled Theoretical Distribution ? Discussion Question Suppose that the number of values generated was 500, not 50. How would that affect what you would expect the empirical data to be and the shape of its graph to look like?", "section": "Continuous Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction If you ask enough people about their shoe size, you will find that your graphed data is shaped like a bell curve and can be described as normally distributed. (credit: modification of work “Shoe shop” by Lo/ Flickr, Public Domain) Chapter Objectives By the end of this chapter, the student should be able to: Recognize the normal probability distribution and apply it appropriately. Recognize the standard normal probability distribution and apply it appropriately. Compare normal probabilities by converting to the standard normal distribution. The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures): the mean ( μ ) and the standard deviation ( σ ). If X is a quantity to be measured that has a normal distribution with mean ( μ ) and standard deviation ( σ ), we designate this by writing The probability density function is a rather complicated mathematical function. Do not memorize it . It is not necessary. f ( x ) = 1 σ ⋅ 2 ⋅ π ⋅ e − 1 2 ⋅ ( x − μ σ ) 2 The cumulative distribution function is P ( X < x ). It is calculated either by a calculator or a computer, or it is looked up in a table. Technology has made the tables virtually obsolete. For that reason, as well as the fact that there are various table formats, we are not including table instructions. The curve is symmetric about a vertical line drawn through the mean, μ . In theory, the mean is the same as the median, because the graph is symmetric about μ . As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Since the area under the curve must equal one, a change in the standard deviation, σ , causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ . A change in μ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution . Collaborative Classroom Activity Your instructor will record the heights of both men and women in your class, separately. Draw histograms of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do you think that if you had recorded 200 data values for men and 200 for women that the curves would look bell-shaped? Calculate the mean for each data set. Write the means on the x -axis of the appropriate graph below the peak. Shade the approximate area that represents the probability that one randomly chosen male is taller than 72 inches. Shade the approximate area that represents the probability that one randomly chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability appear to be more than 0.5? Formula Review X ∼ N ( μ , σ ) μ = the mean σ = the standard deviation Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 σ 2 2 , where μ is the mean of the distribution and σ is the standard deviation; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution .", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Standard Normal Distribution The standard normal distribution is a normal distribution of standardized values called z -scores . A z -score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows: x = μ + ( z )( σ ) = 5 + (3)(2) = 11 The z -score is three. The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = x − μ σ produces the distribution Z ~ N (0, 1). The value x in the given equation comes from a normal distribution with mean μ and standard deviation σ . Z -Scores If X is a normally distributed random variable and X ~ N(μ, σ) , then the z -score is: z = x – μ σ The z -score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ . Values of x that are larger than the mean have positive z -scores, and values of x that are smaller than the mean have negative z -scores. If x equals the mean, then x has a z -score of zero. Suppose X ~ N(5, 6) . This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then: z = x – μ σ = 17 – 5 6 = 2 This means that x = 17 is two standard deviations (2 σ ) above or to the right of the mean μ = 5. Notice that: 5 + (2)(6) = 17 (The pattern is μ + zσ = x ) Now suppose x = 1. Then: z = x – μ σ = 1 – 5 6 = –0.67 (rounded to two decimal places) This means that x = 1 is 0.67 standard deviations (–0.67 σ ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1) Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ . Or, when z is positive, x is greater than μ , and when z is negative x is less than μ . Try It What is the z -score of x , when x = 1 and X ~ N (12,3)? Some doctors believe that a person can lose five pounds, on the average, in a month by reducing their fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N (5, 2). Fill in the blanks. a. Suppose a person lost ten pounds in a month. The z -score when x = 10 pounds is z = 2.5 (verify). This z -score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). a. This z -score tells you that x = 10 is 2.5 standard deviations to the right of the mean five . b. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z -score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean. b. z = –4 . This z -score tells you that x = –3 is four standard deviations to the left of the mean. c. Suppose the random variables X and Y have the following normal distributions: X ~ N (5, 6) and Y ~ N (2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z ? c. z = y − μ σ = 4 − 2 1 = 2 where µ = 2 and σ = 1. The z -score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own ) standard deviations to the right of their respective means. The z -score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N (5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N (2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means . Try It Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. X ~ N (16,4). Suppose Jerome scores ten points in a game. The z –score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?). The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ , then the Empirical Rule states the following: About 68% of the x values lie between –1 σ and +1 σ of the mean µ (within one standard deviation of the mean). About 95% of the x values lie between –2 σ and +2 σ of the mean µ (within two standard deviations of the mean). About 99.7% of the x values lie between –3 σ and +3 σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. The z -scores for +1 σ and –1 σ are +1 and –1, respectively. The z -scores for +2 σ and –2 σ are +2 and –2, respectively. The z -scores for +3 σ and –3 σ are +3 and –3 respectively. The empirical rule is also known as the 68-95-99.7 rule. The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N (170, 6.28). a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z -score when x = 168 cm is z = _______. This z -score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z -score of z = 1.27. What is the male’s height? The z -score ( z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. a. –0.32, 0.32, left, 170 b. 177.98 cm, 1.27, right Try It Use the information in to answer the following questions. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The z -score when x = 176 cm is z = _______. This z -score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z -score of z = –2. What is the male’s height? The z -score ( z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males from 1984 to 1985. Then Y ~ N (172.36, 6.34). The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N (170, 6.28). Find the z -scores for x = 160.58 cm and y = 162.85 cm. Interpret each z -score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations? The z -score for x = -160.58 is z = –1.5. The z -score for y = 162.85 is z = –1.5. Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction. Try It In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N (496, 114). Find the z -scores for x 1 = 325 and x 2 = 366.21. Interpret each z -score. What can you say about x 1 = 325 and x 2 = 366.21 as they compare to their respective means and standard deviations? Suppose x has a normal distribution with mean 50 and standard deviation 6. About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1 σ = (–1)(6) = –6 and 1 σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z -scores are –1 and +1 for 44 and 56, respectively. About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2 σ = (–2)(6) = –12 and 2 σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z -scores are –2 and +2 for 38 and 62, respectively. About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 99.7% of the x values lie between –3 σ = (–3)(6) = –18 and 3 σ = (3)(6) = 18 from the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z -scores are –3 and +3 for 32 and 68, respectively. Try It Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie? From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N (172.36, 6.34). About 68% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively. About 95% of the y values lie between what two values? These values are ________________. The z -scores are ________________ respectively. About 99.7% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively. About 68% of the values lie between 166.02 cm and 178.7 cm. The z -scores are –1 and 1. About 95% of the values lie between 159.68 cm and 185.04 cm. The z -scores are –2 and 2. About 99.7% of the values lie between 153.34 cm and 191.38 cm. The z -scores are –3 and 3. Try It The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points. About 68% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively. About 95% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively. About 99.7% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively. References “Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013). “The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013). “2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013). “Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013). Data from the San Jose Mercury News . Data from The World Almanac and Book of Facts . “List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013). Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013). Chapter Review A z -score is a standardized value. Its distribution is the standard normal, Z ~ N (0, 1). The mean of the z -scores is zero and the standard deviation is one. If z is the z -score for a value x from the normal distribution N ( µ , σ ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ . Formula Review z = a standardized value ( z -score) mean = 0; standard deviation = 1 To find the observed value, x , when the z -scores is known: x = μ + ( z ) σ z -score: z = x – μ σ Z = the random variable for z -scores A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________. ounces of water in a bottle A normal distribution has a mean of 61 and a standard deviation of 15. What is the median? X ~ N (1, 2) σ = _______ 2 A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________. X ~ N (–4, 1) What is the median? –4 X ~ N (3, 5) σ = _______ X ~ N (–2, 1) μ = _______ –2 What does a z -score measure? What does standardizing a normal distribution do to the mean? The mean becomes zero. Is X ~ N (0, 1) a standardized normal distribution? Why or why not? What is the z -score of x = 12, if it is two standard deviations to the right of the mean? z = 2 What is the z -score of x = 9, if it is 1.5 standard deviations to the left of the mean? What is the z -score of x = –2, if it is 2.78 standard deviations to the right of the mean? z = 2.78 What is the z -score of x = 7, if it is 0.133 standard deviations to the left of the mean? Suppose X ~ N (2, 6). What value of x has a z -score of three? x = 20 Suppose X ~ N (8, 1). What value of x has a z -score of –2.25? Suppose X ~ N (9, 5). What value of x has a z -score of –0.5? x = 6.5 Suppose X ~ N (2, 3). What value of x has a z -score of –0.67? Suppose X ~ N (4, 2). What value of x is 1.5 standard deviations to the left of the mean? x = 1 Suppose X ~ N (4, 2). What value of x is two standard deviations to the right of the mean? Suppose X ~ N (8, 9). What value of x is 0.67 standard deviations to the left of the mean? x = 1.97 Suppose X ~ N (–1, 2). What is the z -score of x = 2? Suppose X ~ N (12, 6). What is the z -score of x = 2? z = –1.67 Suppose X ~ N (9, 3). What is the z -score of x = 9? Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z -score of x = 5.5? z ≈ –0.33 In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean. 0.67, right In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean. 3.14, left In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean. About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution? about 68% About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution? About what percent of x values lie between the second and third standard deviations (both sides)? about 4% Suppose X ~ N (15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15). Suppose X ~ N (–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3). between –5 and –1 Suppose X ~ N (–3, 1). Between what x values does 34.14% of the data lie? About what percent of x values lie between the mean and three standard deviations? about 50% About what percent of x values lie between the mean and one standard deviation? About what percent of x values lie between the first and second standard deviations from the mean (both sides)? about 27% About what percent of x values lie between the first and third standard deviations(both sides)? Use the following information to answer the next two exercises: The life of wearable fitness devices is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Define the random variable X in words. X = _______________. The lifetime of a wearable fitness device measured in years. X ~ _____(_____,_____) Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the median recovery time? 2.7 5.3 7.4 2.1 What is the z -score for a patient who takes ten days to recover? 1.5 0.2 2.2 7.3 c The length of time to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? The data cannot follow the uniform distribution. The data cannot follow the exponential distribution.. The data cannot follow the normal distribution. I only II only III only I, II, and III The heights of the 430 National Basketball Association players were listed on team rosters at the start of a recent season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z -score and interpret it using complete sentences. 77 inches 85 inches If an NBA player reported his height had a z -score of 3.5, would you believe him? Explain your answer. Use the z -score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. Use the z -score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely. The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution. Calculate the z -scores for the male systolic blood pressures 100 and 150 millimeters. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him? Kyle’s doctor told him that the z -score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14). Which answer(s) is/are correct? Kyle’s systolic blood pressure is 175. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. Calculate Kyle’s blood pressure. iv Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N (10.2, 0.8). Calculate the z -scores that correspond to the following weights and interpret them. 11 kg 7.9 kg 12.2 kg During a certain year, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115. Calculate the z -score for an SAT score of 720. Interpret it using a complete sentence. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? During a different year, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took? Let X = an SAT math score and Y = an ACT math score. Use the z-score formula. X = 720 : z = 720 - 520 115 = 1 . 74 . The exam score of 720 is 1.74 standard deviations above the mean of 520. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. X – μ σ = 700 – 514 117 ≈ 1.59, the z -score for the SAT. Y – μ σ = 30 – 21 5.3 ≈ 1.70, the z -scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z -score). Standard Normal Distribution a continuous random variable (RV) X ~ N (0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N (0, 1). z-score the linear transformation of the form z = x – μ σ ; if this transformation is applied to any normal distribution X ~ N ( μ , σ ) the result is the standard normal distribution Z ~ N (0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ , the result is called the z -score of x . The z -score allows us to compare data that are normally distributed but scaled differently.", "section": "The Standard Normal Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Using the Normal Distribution The shaded area in the following graph indicates the area to the left of x . This area is represented by the probability P ( X < x ). Normal tables, computers, and calculators provide or calculate the probability P ( X < x ). The area to the right is then P ( X > x ) = 1 – P ( X < x ). Remember, P ( X < x ) = Area to the left of the vertical line through x . P ( X > x ) = 1 – P ( X < x ) = Area to the right of the vertical line through x . P ( X < x ) is the same as P ( X ≤ x ) and P ( X > x ) is the same as P ( X ≥ x ) for continuous distributions. Calculations of Probabilities Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. NOTE To calculate the probability, use the probability tables provided in without the use of technology. The tables include instructions for how to use them. If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772. Try It If the area to the left of x is 0.012, then what is the area to the right? The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. a. Find the probability that a randomly selected student scored more than 65 on the exam. a. Let X = a score on the final exam. X ~ N (63, 5), where μ = 63 and σ = 5. Draw a graph. Then, find P ( x > 65). P ( x > 65) = 0.3446 The probability that any student selected at random scores more than 65 is 0.3446. Go into 2nd DISTR . After pressing 2nd DISTR , press 2:normalcdf . The syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10 99 ) by pressing 1 , the EE key (a 2nd key) and then 99 . Or, you can enter 10^99 instead. The number 10 99 is way out in the right tail of the normal curve. We are calculating the area between 65 and 10 99 . In some instances, the lower number of the area might be –1E99 (= –10 99 ). The number –10 99 is way out in the left tail of the normal curve. Historical Note The TI probability program calculates a z -score and then the probability from the z -score. Before technology, the z -score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the z -score was used. You calculate the z -score and look up the area to the left. The probability is the area to the right. z = 65 – 63 5 = 0.4 Area to the left is 0.6554. P ( x > 65) = P ( z > 0.4) = 1 – 0.6554 = 0.3446 Find the percentile for a student scoring 65: *Press 2nd Distr *Press 2:normalcdf ( *Enter lower bound, upper bound, mean, standard deviation followed by ) *Press ENTER . For this Example, the steps are 2nd Distr 2:normalcdf (65,1,2nd EE,99,63,5) ENTER The probability that a selected student scored more than 65 is 0.3446. To find the probability that a selected student scored more than 65, subtract the percentile from 1. b. Find the probability that a randomly selected student scored less than 85. b. Draw a graph. Then find P ( x < 85), and shade the graph. Using a computer or calculator, find P ( x < 85) = 1. normalcdf(0,85,63,5) = 1 (rounds to one) The probability that one student scores less than 85 is approximately one (or 100%). c. Find the 90 th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k ). c. Find the 90 th percentile. For each problem or part of a problem, draw a new graph. Draw the x -axis. Shade the area that corresponds to the 90 th percentile. Let k = the 90 th percentile. The variable k is located on the x -axis. P ( x < k ) is the area to the left of k . The 90 th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k , and ten percent are the same or higher. The variable k is often called a critical value . k = 69.4 The 90 th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step: invNorm in 2nd DISTR . invNorm(area to the left, mean, standard deviation) For this problem, invNorm(0.90,63,5) = 69.4 d. Find the 70 th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k ). d. Find the 70 th percentile. Draw a new graph and label it appropriately. k = 65.6 The 70 th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above. invNorm(0.70,63,5) = 65.6 Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65. A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day. a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N (2, 0.5) where μ = 2 and σ = 0.5. Find P (1.8 < x < 2.75). The probability for which you are looking is the area between x = 1.8 and x = 2.75. P (1.8 < x < 2.75) = 0.5886 normalcdf(1.8,2.75,2,0.5) = 0.5886 The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment. b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25 th percentile, k , where P ( x < k ) = 0.25. invNorm(0.25,2,0.5) = 1.66 The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours. Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old. b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. c. Find the 80 th percentile of this distribution, and interpret it in a complete sentence. a. normalcdf(23,64.7,36.9,13.9) = 0.8186 b. normalcdf(–10 99 ,50.8,36.9,13.9) = 0.8413 c. invNorm(0.80,36.9,13.9) = 48.6 The 80 th percentile is 48.6 years. 80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less. Try It Use the information in to answer the following questions. Find the 30 th percentile, and interpret it in a complete sentence. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place). a. Calculate the interquartile range ( IQR ). b. Forty percent of the smartphone users from 13 to 55+ are at least what age? a. IQR = Q 3 – Q 1 Calculate Q 3 = 75 th percentile and Q 1 = 25 th percentile. invNorm(0.75,36.9,13.9) = Q 3 = 46.2754 invNorm(0.25,36.9,13.9) = Q 1 = 27.5246 IQR = Q 3 – Q 1 = 18.8 b. Find k where P ( x ≥ k ) = 0.40 (\"At least\" translates to \"greater than or equal to.\") 0.40 = the area to the right. Area to the left = 1 – 0.40 = 0.60. The area to the left of k = 0.60. invNorm(0.60,36.9,13.9) = 40.4215. k = 40.4. Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years. Try It Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points. Calculate the first- and third-quartile scores for this exam. The middle 50% of the exam scores are between what two values? A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph. b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______. c. Find the 90 th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence. a. normalcdf(6,10^99,5.85,0.24) = 0.2660 b. 1 – 0.20 = 0.80 The tails of the graph of the normal distribution each have an area of 0.40. Find k1 , the 40 th percentile, and k2 , the 60 th percentile (0.40 + 0.20 = 0.60). k1 = invNorm(0.40,5.85,0.24) = 5.79 cm k2 = invNorm(0.60,5.85,0.24) = 5.91 cm c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm. Try It Using the information from , answer the following: The middle 40% of mandarin oranges from this farm are between ______ and ______. Find the 16 th percentile and interpret it in a complete sentence. References “Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013). “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013). “Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013). “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013). “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013). Chapter Review The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one. Formula Review Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation. Standard Normal Distribution: Z ~ N (0, 1). Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation) Calculator function for the k th percentile: k = invNorm (area to the left of k , mean, standard deviation) How would you represent the area to the left of one in a probability statement? P ( x < 1) What is the area to the right of one? Is P ( x < 1) equal to P ( x ≤ 1)? Why? Yes, because they are the same in a continuous distribution: P ( x = 1) = 0 How would you represent the area to the left of three in a probability statement? What is the area to the right of three? 1 – P ( x < 3) or P ( x > 3) If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x ? If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x ? 1 – 0.543 = 0.457 Use the following information to answer the next four exercises: X ~ N (54, 8) Find the probability that x > 56. Find the probability that x < 30. 0.0013 Find the 80 th percentile. Find the 60 th percentile. 56.03 X ~ N (6, 2) Find the probability that x is between three and nine. X ~ N (–3, 4) Find the probability that x is between one and four. 0.1186 X ~ N (4, 5) Find the maximum of x in the bottom quartile. Use the following information to answer the next three exercise: The life of wearable fitness devices is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Find the probability that a wearable fitness device will break down during the guarantee period. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (0 < x < ____________) = ___________ (Use zero for the minimum value of x .) Answers may vary. 3, 0.1979 Find the probability that a wearable fitness device will last between 2.8 and six years. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (__________ < x < __________) = __________ Find the 70 th percentile of the distribution for the time a wearable fitness device lasts. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%. P ( x < k ) = __________ Therefore, k = _________ Answers may vary. 0.70, 4.78 years Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the probability of spending more than two days in recovery? 0.0580 0.8447 0.0553 0.9420 The 90 th percentile for recovery times is? 8.89 7.07 7.99 4.32 c Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? Yes No Unable to determine Find the probability that it takes at least eight minutes to find a parking space. 0.0001 0.9270 0.1862 0.0668 d Seventy percent of the time, it takes more than how many minutes to find a parking space? 1.24 2.41 3.95 6.05 According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual. X ~ _____(_____,_____) Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically. X ~ N (66, 2.5) 0.5404 No, the probability that an Asian male is over 72 inches tall is 0.0082 IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. X ~ _____(_____,_____) Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement. The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories. X ~ _____(_____,_____) Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement. X ~ N (36, 10) The probability that a person consumes more than 40% of their calories as fat is 0.3446. Approximately 25% of people consume less than 29.26% of their calories as fat. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X . Shade the region corresponding to the probability. Find the probability. Find the 80 th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. In words, define the random variable X . X ~ _____(_____,_____) Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. What percent of the children spend over ten hours per day unsupervised? Seventy percent of the children spend at least how long per day unsupervised? X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day. X ~ N (3, 1.5) The probability that the child spends less than one hour a day unsupervised is 0.0918. The probability that a child spends over ten hours a day unsupervised is less than 0.0001. 2.21 hours In a certain presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for Candidate A. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for Candidate A was bell-shaped. Let X = number of votes for Candidate A for an election district. State the approximate distribution of X . Is 1,956.8 a population mean or a sample mean? How do you know? Find the probability that a randomly selected district had fewer than 1,600 votes for Candidate A. Sketch the graph and write the probability statement. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for Candidate A. Find the third quartile for votes for Candidate A. Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days. In words, define the random variable X . X ~ _____(_____,_____) If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement. Sixty percent of all trials of this type are completed within how many days? X = the distribution of the number of days a particular type of criminal trial will take X ~ N (21, 7) The probability that a randomly selected trial will last more than 24 days is 0.3336. 22.77 A motorcycle racer averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of their race times is normally distributed. We are interested in one of their randomly selected laps. In words, define the random variable X . X ~ _____(_____,_____) Find the percent of the racer's laps that are completed in less than 130 seconds. The fastest 3% of the racer's laps are under _____. The middle 80% of the racer's laps are from _______ seconds to _______ seconds. Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. displays the ordered real data (in minutes): 0.50 4.25 5 6 7.25 1.75 4.25 5.25 6 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 6.5 8 2.5 4.75 5.5 6.5 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 6 6.75 9.75 3.75 5 6 6.75 10.75 Calculate the sample mean and the sample standard deviation. Construct a histogram. Draw a smooth curve through the midpoints of the tops of the bars. In words, describe the shape of your histogram and smooth curve. Let the sample mean approximate μ and the sample standard deviation approximate σ . The distribution of X can then be approximated by X ~ _____(_____,_____) Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes. Determine the cumulative relative frequency for waiting less than 6.1 minutes. Why aren’t the answers to part f and part g exactly the same? Why are the answers to part f and part g as close as they are? If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion. mean = 5.51, s = 2.15 Answers may vary. Answers may vary. Answers may vary. X ~ N (5.51, 2.15) 0.6029 The cumulative frequency for less than 6.1 minutes is 0.64. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well. Suppose that Ric and Anita attend different colleges. Ric’s GPA is the same as the average GPA at their school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false. Ric’s actual GPA is lower than Anita’s actual GPA. Ric is not passing because their z -score is zero. Anita is in the 70 th percentile of students at her college. shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums. 40,000 40,000 45,050 45,500 46,249 48,134 49,133 50,071 50,096 50,466 50,832 51,100 51,500 51,900 52,000 52,132 52,200 52,530 52,692 53,864 54,000 55,000 55,000 55,000 55,000 55,000 55,000 55,082 57,000 58,008 59,680 60,000 60,000 60,492 60,580 62,380 62,872 64,035 65,000 65,050 65,647 66,000 66,161 67,428 68,349 68,976 69,372 70,107 70,585 71,594 72,000 72,922 73,379 74,500 75,025 76,212 78,000 80,000 80,000 82,300 Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data). Construct a histogram. Draw a smooth curve through the midpoints of the tops of the bars of the histogram. In words, describe the shape of your histogram and smooth curve. Let the sample mean approximate μ and the sample standard deviation approximate σ . The distribution of X can then be approximated by X ~ _____(_____,_____). Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample. Why aren’t the answers to part f and part g exactly the same? mean = 60,136 s = 10,468 Answers will vary. Answers will vary. Answers will vary. X ~ N (60136, 10468) 0.7440 The cumulative relative frequency is 43/60 = 0.717. The answers for part f and part g are not the same, because the normal distribution is only an approximation. An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. The person who is being sued for lack of child support was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the biological parent. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the parent? What is the probability that he could be the parent? Calculate the z -scores first, and then use those to calculate the probability. An automotive factory can build an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample. n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = n p q = (100)(0 .1)(0 .9) = 3 z = ±1: x 1 = µ + zσ = 10 + 1(3) = 13 and x 2 = µ – zσ = 10 – 1(3) = 7. 68% of the defective cars will fall between seven and 13. z = ±2: x 1 = µ + zσ = 10 + 2(3) = 16 and x 2 = µ – zσ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16 z = ±3: x 1 = µ + zσ = 10 + 3(3) = 19 and x 2 = µ – zσ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19. We flip a coin 100 times ( n = 100) and note that it only comes up heads 20% ( p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following: There is about a 68% chance that the number of heads will be somewhere between ___ and ___. There is about a ____chance that the number of heads will be somewhere between 12 and 28. There is about a ____ chance that the number of heads will be somewhere between eight and 32. A $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are somewhere between 34 and 54 prizes. somewhere between 54 and 64 prizes. more than 64 prizes. n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = n p q = (190)(0 .2)(0 .8) = 5.5136 For this problem: P (34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641 For this problem: P (54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018 For this problem: P ( x > 64) = normalcdf(64,10 99 ,48,5.5136) = 0.0000012 (approximately 0) On average, 28% of 18 to 34 year olds check social media before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of 5%. Find the probability that the percent of 18 to 34-year-olds who check social media before getting out of bed in the morning is at least 30. Find the 95 th percentile, and express it in a sentence.", "section": "Using the Normal Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Normal Distribution (Lap Times) Normal Distribution (Lap Times) Class Time: Names: Student Learning Outcome The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution. Directions Round the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places. Collect the Data Use the data from Appendix C . Use a stratified sampling method by lap (races 1 to 20) and a random number generator to pick six lap times from each stratum. Record the lap times below for laps two to seven. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Calculate the following: x ¯ = _______ s = _______ Draw a smooth curve through the tops of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end, and so on?) Analyze the Distribution Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data? X ~ _____(_____,_____) How does the histogram help you arrive at the approximate distribution? Describe the Data Use the data you collected to complete the following statements. The IQR goes from __________ to __________. IQR = __________. ( IQR = Q 3 – Q 1 ) The 15 th percentile is _______. The 85 th percentile is _______. The median is _______. The empirical probability that a randomly chosen lap time is more than 130 seconds is _______. Explain the meaning of the 85 th percentile of this data. Theoretical Distribution Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data. The IQR goes from __________ to __________. IQR = _______. The 15 th percentile is _______. The 85 th percentile is _______. The median is _______. The probability that a randomly chosen lap time is more than 130 seconds is _______. Explain the meaning of the 85 th percentile of this distribution. Discussion Questions Do the data from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution ? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution , explain why or why not.", "section": "Normal Distribution (Lap Times)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Normal Distribution (Pinkie Length) Normal Distribution (Pinkie Length) Class Time: Names: Student Learning Outcomes The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution. Collect the Data Measure the length of your pinky finger (in centimeters). Randomly survey 30 adults for their pinky finger lengths. Round the lengths to the nearest 0.5 cm. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Calculate the following. x ¯ = _______ s = _______ Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end, and so on?) Analyze the Distribution Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution of the data you collected? X ~ _____(_____,_____) How does the histogram help you arrive at the approximate distribution? Describe the Data Using the data you collected complete the following statements. (Hint: order the data) Remember ( IQR = Q 3 – Q 1 ) IQR = _______ The 15 th percentile is _______. The 85 th percentile is _______. Median is _______. What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm? Explain the meaning of the 85 th percentile of this data. Theoretical Distribution Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation. IQR = _______ The 15 th percentile is _______. The 85 th percentile is _______. Median is _______. What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm? Explain the meaning of the 85 th percentile of this data. Discussion Questions Do the data you collected give a close approximation to the theoretical distribution? In complete sentences and comparing the results in the sections titled Describe the Data and Theoretical Distribution , explain why or why not.", "section": "Normal Distribution (Pinkie Length)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction If you want to figure out the distribution of the change people carry in their pockets, using the central limit theorem and assuming your sample is large enough, you will find that the distribution is normal and bell-shaped. (credit: modification of work “american coins” by Paula R. Lively/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Recognize central limit theorem problems. Classify continuous word problems by their distributions. Apply and interpret the central limit theorem for means. Apply and interpret the central limit theorem for sums. Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem . The central limit theorem (clt for short) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, μ , and a known standard deviation, σ . The first alternative says that if we collect samples of size n with a \"large enough n ,\" calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are \"large enough,\" calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape. The size of the sample, n , that is required in order to be \"large enough\" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement. It would be difficult to overstate the importance of the central limit theorem in statistical theory. Knowing that data, even if its distribution is not normal, behaves in a predictable way is a powerful tool. Collaborative Classroom Activity Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times. Each time a person rolls more than one die, they calculate the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, 6 on one roll. The mean is 2 + 2 + 3 + 4 + 6 5 = 3.4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means. Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5. Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Since the \"mean\" when you roll one die is just the face on the die, what distribution do these means appear to be representing? Draw the graph for the means using two dice. Do the sample means show any kind of pattern? Draw the graph for the means using five dice. Do you see any pattern emerging? Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice? As the number of dice rolled increases from one to two to five to ten, the following is happening: The mean of the sample means remains approximately the same. The spread of the sample means (the standard deviation of the sample means) gets smaller. The graph appears steeper and thinner. You have just demonstrated the central limit theorem (clt). The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution). Sampling Distribution Given simple random samples of size n from a given population with a measured characteristic such as mean, proportion, or standard deviation for each sample, the probability distribution of all the measured characteristics is called a sampling distribution.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Central Limit Theorem for Sample Means (Averages) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose: μ X = the mean of X σ X = the standard deviation of X If you draw random samples of size n , then as n increases, the random variable x ¯ which consists of sample means, tends to be normally distributed and X ¯ ~ N ( μ x , σ X n ) The central limit theorem for sample means says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate their means, those means tend to follow a normal distribution (the sampling distribution). As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n . The variable n is the number of values that are averaged together, not the number of times the experiment is done. To put it more formally, if you draw random samples of size n , the distribution of the random variable X ¯ , which consists of sample means, is called the sampling distribution of the mean . The sampling distribution of the mean approaches a normal distribution as n , the sample size , increases. The random variable X ¯ has a different z -score associated with it from that of the random variable X . The mean x ¯ is the value of X ¯ in one sample. z = x ¯ − μ x ( σ X n ) μ X is the average of both X and X ¯ . σ x = σ X n = standard deviation of X ¯ and is called the standard error of the mean. To find probabilities for means on the calculator, follow these steps. 2nd DISTR 2:normalcdf n o r m a l c d f ( l o w e r v a l u e o f t h e a r e a , u p p e r v a l u e o f t h e a r e a , m e a n , s t a n d a r d d e v i a t i o n s a m p l e s i z e ) where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population. a. Find the probability that the sample mean is between 85 and 92. a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean . Let x ¯ = the mean of a sample of size 25. Since μ X = 90, σ X = 15, and n = 25, x ¯ ~ N ( 90 , 15 25 ) . Find P (85 < x ¯ < 92). Draw a graph. P (85 < x ¯ < 92) = 0.6997 The probability that the sample mean is between 85 and 92 is 0.6997. normalcdf (lower value, upper value, mean, standard error of the mean) The parameter list is abbreviated (lower value, upper value, μ , σ n ) normalcdf (85,92,90, 15 25 ) = 0.6997 b. Find the value that is two standard deviations above the expected value, 90, of the sample mean. b. To find the value that is two standard deviations above the expected value 90, use the formula: value = μ x + (#ofTSDEVs) ( σ x n ) value = 90 + 2 ( 15 25 ) = 96 The value that is two standard deviations above the expected value is 96. The standard error of the mean is σ x n = 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n . Try It An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50. The length of time, in hours, it takes an \"over 40\" group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours . A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. Let X = the time, in hours, it takes to play one soccer match. The probability question asks you to find a probability for the sample mean time, in hours , it takes to play one soccer match. Let x ¯ = the mean time, in hours, it takes to play one soccer match. If μ X = _________, σ X = __________, and n = ___________, then X ¯ ~ N (______, ______) by the central limit theorem for means . μ X = 2, σ X = 0.5, n = 50, and X ~ N ( 2, 0.5 50 ) Find P (1.8 < x ¯ < 2.3). Draw a graph. P (1.8 < x ¯ < 2.3) = 0.9977 normalcdf ( 1. 8,2 .3,2, .5 50 ) = 0.9977 The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977. Try It The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours. To find percentiles for means on the calculator, follow these steps. 2 nd DIStR 3:invNorm k = invNorm ( area to the left of k , mean, s t a n d a r d d e v i a t i o n s a m p l e s i z e ) where: k = the k th percentile mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n In a recent study, it was reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100. What are the mean and standard deviation for the sample mean ages of iPad users? What does the distribution look like? Find the probability that the sample mean age is more than 30 years (the reported mean age of iPad users in this particular study). Find the 95 th percentile for the sample mean age (to one decimal place). Since the sample mean tends to target the population mean, we have μ χ = μ = 34. The sample standard deviation is given by σ χ = σ n = 15 100 = 15 10 = 1.5 The central limit theorem states that for large sample sizes( n ), the sampling distribution will be approximately normal. The probability that the sample mean age is more than 30 is given by P ( X ¯ > 30 ) = normalcdf (30,E99,34,1.5) = 0.9962 Let k = the 95 th percentile. k = invNorm ( 0. 95,34, 15 100 ) = 36.5 Try It In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy? The mean number of minutes for app engagement by an iPad user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60. What are the mean and standard deviation for the sample mean number of minutes for app engagement by an iPad user? What is the standard error of the mean? Find the 90 th percentile for the sample mean time of minutes for app engagement by an iPad user. Interpret this value in a complete sentence. Find the probability that the sample mean is between eight minutes and 8.5 minutes. μ x ¯ = μ = 8.2 σ x ¯ = σ n = 1 60 = 0.13 This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60. Let k = the 90 th percentile k = invNorm ( 0. 90,8 .2, 1 60 ) = 8.37. This values indicates that 90 percent of the average app engagement time for iPad users is less than 8.37 minutes. P (8 < x ¯ < 8.5) = normalcdf ( 8,8 .5,8 .2, 1 60 ) = 0.9293 Try It Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x ¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces? References Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013). Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013). Data from the United States Department of Agriculture. Chapter Review In a population whose distribution may be known or unknown, if the size ( n ) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ( n ). Formula Review The Central Limit Theorem for Sample Means: x ¯ ~ N ( μ x , σ x n ) The Mean X ¯ : μ x Central Limit Theorem for Sample Means z-score and standard error of the mean: z = x ¯ − μ x ( σ x n ) Standard Error of the Mean (Standard Deviation ( X ¯ )): σ x n Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let x ¯ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. What is the mean, standard deviation, and sample size? mean = 4 hours; standard deviation = 1.2 hours; sample size = 16 Complete the distributions. X ~ _____(_____,_____) X ¯ ~ _____(_____,_____) Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. P (________ < x < ________) = _______ a. Answers may vary. b. 3.5, 4.25, 0.2441 Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. P (________________) = _______ What causes the probabilities in and to be different? The fact that the two distributions are different accounts for the different probabilities. Find the 95 th percentile for the mean time to complete one month's reviews. Sketch the graph. The 95 th Percentile =____________ Homework Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. In words, Χ = ____________ Χ ~ _____(_____,_____) In words, X ¯ = ____________ x ¯ ~ ______ (______, ______) Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Explain why there is a difference in part e and part f. Χ = amount of change a student carries E x p ( 1 0 . 88 ) o r Χ ~ E x p ( 1 . 1364 ) x ¯ = average amount of change carried by a sample of 25 sstudents. x ¯ ~ N (0.88, 0.176) 0.0819 0.4276 The distributions are different. Part e asks about the probability for an individual and part f asks about the probability of the mean value for a sample of 25. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. If x ¯ = average distance in feet for 49 fly balls, then x ¯ ~ _______(_______,_______) What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for x ¯ . Shade the region corresponding to the probability. Find the probability. Find the 80 th percentile of the distribution of the average of 49 fly balls. According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. In words, Χ = _____________ In words, X ¯ = _____________ X ¯ ~ _____(_____,_____) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. Would you be surprised if one taxpayer finished their Form 1040 in more than 12 hours? In a complete sentence, explain why. length of time for an individual to complete IRS form 1040, in hours. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours. N ( 10 .53, 1 3 ) Yes. I would be surprised, because the probability is almost 0. No. I would not be totally surprised because the probability is 0.2312 Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let x ¯ the average of the 49 races. X ¯ ~ _____(_____,_____) Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. Find the 80 th percentile for the average of these 49 marathons. Find the median of the average running times. The length of songs in a collector’s Apple Music album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. In words, Χ = _________ Χ ~ _____________ In words, X ¯ = _____________ X ¯ ~ _____(_____,_____) Find the first quartile for the average song length, X ¯ . The IQR (interquartile range) for the average song length, X ¯ , is from ___ - ___. the length of a song, in minutes, in the collection U (2, 3.5) the average length, in minutes, of the songs from a sample of five albums from the collection N (2.75, 0.0660) 2.71 minutes 0.09 minutes In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. In words, Χ = _____________ In words, X ¯ = _____________ X ¯ ~ _____(_____,_____) The IQR for x ¯ is from _______ acres to _______ acres. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. When the sample size is large, the mean of X ¯ is approximately equal to the mean of Χ . When the sample size is large, x ¯ is approximately normally distributed. When the sample size is large, the standard deviation of x ¯ is approximately the same as the standard deviation of Χ . True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. False. The standard deviation of the sample distribution of the means will decrease as the sample size increases; however, the standard deviation of the sample distribution of the means will not equal the standard deviation of X. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let x ¯ = average percent of fat calories. x ¯ ~ ______(______, ______) For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. Find the first quartile for the average percent of fat calories. The distribution of income in some developing countries is considered wedge shaped (many low income people, very few middle income people, and even fewer high income people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country. In words, Χ = _____________ In words, X ¯ = _____________ X ¯ ~ _____(_____,_____) How is it possible for the standard deviation to be greater than the average? Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200? X = the yearly income of someone in a developing country the average salary from samples of 1,000 residents of a developing country X ¯ ∼ N ( 2000, 8000 1000 ) Very wide differences in data values can have averages smaller than standard deviations. The distribution of the sample mean will have higher probabilities closer to the population mean. P (2000 < x ¯ < 2100) = 0.1537 P (2100 < x ¯ < 2200) = 0.1317 Which of the following is NOT TRUE about the distribution for averages? The mean, median, and mode are equal. The area under the curve is one. The curve never touches the x -axis. The curve is skewed to the right. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is: X ¯ ~ N (4.59, 0.10) X ¯ ~ N ( 4 .59, 0.10 16 ) X ¯ ~ N ( 4 .59, 16 0.10 ) X ¯ ~ N ( 4 .59, 16 0.10 ) b Average a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean. Central Limit Theorem Given a random variable (RV) with known mean μ and known standard deviation, σ , we are sampling with size n , and we are interested in two new RVs: the sample mean, X ¯ , and the sample sum, ΣΧ . If the size ( n ) of the sample is sufficiently large, then X ¯ ~ N ( μ , σ n ) and ΣΧ ~ N ( nμ , ( n )( σ )). If the size ( n ) of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean, and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean. Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation; notation: Χ ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called a standard normal distribution . Standard Error of the Mean the standard deviation of the distribution of the sample means, or σ n .", "section": "The Central Limit Theorem for Sample Means (Averages)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Central Limit Theorem for Sums Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose: μ X = the mean of Χ σ Χ = the standard deviation of X If you draw random samples of size n , then as n increases, the random variable Σ X consisting of sums tends to be normally distributed and Σ Χ ~ N (( n )( μ Χ ), ( n )( σ Χ )). The central limit theorem for sums says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate the sum of each sample, these sums tend to follow a normal distribution. As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size. The random variable Σ X has the following z -score associated with it: Σ x is one sum. z = Σ x – ( n ) ( μ X ) ( n ) ( σ X ) ( n )( μ X ) = the mean of Σ X ( n ) ( σ X ) = standard deviation of Σ X To find probabilities for sums on the calculator, follow these steps. 2 nd DISTR 2: normalcdf normalcdf (lower value of the area, upper value of the area, ( n )(mean), ( n )(standard deviation)) where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500. Find the sum that is 1.5 standard deviations above the mean of the sums. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values. Σ X = the sum or total of 80 values. Since μ X = 90, σ X = 15, and n = 80, Σ X ~ N ((80)(90), ( 80 )(15)) mean of the sums = ( n )( μ X ) = (80)(90) = 7,200 standard deviation of the sums = ( n )( σ X ) = ( 80 ) (15) sum of 80 values = Σx = 7,500 a. Find P (Σ x > 7,500) P (Σ x > 7,500) = 0.0127 normalcdf (lower value, upper value, mean of sums, stdev of sums) The parameter list is abbreviated(lower, upper, ( n )( μ X , ( n ) ( σ X )) normalcdf (7500,1E99,(80)(90), ( 80 ) (15)) = 0.0127 REMINDER 1E99 = 10 99 . Press the EE key for E. b. Find Σ x where z = 1.5. Σ x = ( n )( μ X ) + ( z ) ( n ) ( σ Χ ) = (80)(90) + (1.5)( 80 )(15) = 7,401.2 Try It An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400. To find percentiles for sums on the calculator, follow these steps. 2 nd DIStR 3:invNorm k = invNorm (area to the left of k , ( n )(mean), ( n ) (standard deviation)) where: k is the k th percentile mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n In a recent study, it was reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50. What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution? Find the probability that the sum of the ages is between 1,500 and 1,800 years. Find the 80 th percentile for the sum of the 50 ages. μ Σx = nμ x = 50(34) = 1,700 and σ Σx = n σ x = ( 50 ) (15) = 106.07 The distribution is normal for sums by the central limit theorem. P (1500 < Σ x < 1800) = normalcdf (1,500, 1,800, (50)(34), ( 50 ) (15)) = 0.7974 Let k = the 80 th percentile. k = invNorm (0.80,(50)(34), ( 50 ) (15)) = 1,789.3 Try It In a recent study, it was reported that the mean age of iPad users is 35 years. Suppose the standard deviation is ten years. The sample size is 39. What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution? Find the probability that the sum of the ages is between 1,400 and 1,500 years. Find the 90 th percentile for the sum of the 39 ages. The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70. What are the mean and standard deviation for the sums? Find the 95 th percentile for the sum of the sample. Interpret this value in a complete sentence. Find the probability that the sum of the sample is at least ten hours. μ Σx = nμ x = 70(8.2) = 574 minutes and σ Σx = ( n ) ( σ x ) = ( 70 ) (1) = 8.37 minutes Let k = the 95 th percentile. k = invNorm (0.95,(70)(8.2), ( 70 ) (1)) = 587.76 minutes Ninety five percent of the sums of app engagement times are at most 587.76 minutes. ten hours = 600 minutes P (Σ x ≥ 600) = normalcdf (600,E99,(70)(8.2), ( 70 ) (1)) = 0.0009 The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem? Find the 84 th and 16 th percentiles for the sum of the sample. Interpret these values in context. References Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May 17, 2013). Chapter Review The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of μ X and a standard deviation of σ x , the mean of the sums is nμ x and the standard deviation is ( n ) ( σ x ) where n is the sample size. Formula Review The Central Limit Theorem for Sums: ∑X ~ N [( n )( μ x ),( n )( σ x )] Mean for Sums ( ∑X ): ( n )( μ x ) The Central Limit Theorem for Sums z -score and standard deviation for sums: z for the sample mean = Σ x – ( n ) ( μ X ) ( n ) ( σ X ) Standard deviation for Sums ( ∑X ): ( n ) ( σ x ) Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population. Find the probability that the sum of the 95 values is greater than 7,650. 0.3345 Find the probability that the sum of the 95 values is less than 7,400. Find the sum that is two standard deviations above the mean of the sums. 7,833.92 Find the sum that is 1.5 standard deviations below the mean of the sums. Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly. Find the probability that the sum of the 40 values is greater than 7,500. 0.0089 Find the probability that the sum of the 40 values is less than 7,000. Find the sum that is one standard deviation above the mean of the sums. 7,326.49 Find the sum that is 1.5 standard deviations below the mean of the sums. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums. 77.45% Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100. Find the probability that the sum of the 100 values is greater than 3,910. Find the probability that the sum of the 100 values is less than 3,900. 0.4207 Find the probability that the sum of the 100 values falls between the numbers you found in and . Find the sum with a z –score of –2.5. 3,888.5 Find the sum with a z –score of 0.5. Find the probability that the sums will fall between the z -scores –2 and 1. 0.8186 Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest. What is the mean of ΣX ? What is the standard deviation of ΣX ? 5 What is P ( Σx = 290)? What is P ( Σx > 290)? 0.9772 True or False: only the sums of normal distributions are also normal distributions. In order for the sums of a distribution to approach a normal distribution, what must be true? The sample size, n , gets larger. What three things must you know about a distribution to find the probability of sums? An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42? 49 An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200? Use the following information to answer the next three exercises. A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. They sample 400 customers. What is the z -score for Σx = 840? 26.00 What is the z -score for Σx = 1,186? What is P ( Σx < 1,186)? 0.1587 Use the following information to answer the next three exercises: An unknown distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest. What is the mean of ΣX ? What is the standard deviation of ΣX ? 1,000 What is P ( Σx > 9,000)? Homework Which of the following is NOT TRUE about the theoretical distribution of sums? The mean, median and mode are equal. The area under the curve is one. The curve never touches the x -axis. The curve is skewed to the right. Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials. In words, ΣX = ______________ ΣX ~ _____(_____,_____) Find the probability that the total length of the nine trials is at least 225 days. Ninety percent of the total of nine of these types of trials will last at least how long? the total length of time for nine criminal trials N (189, 21) 0.0432 162.09; ninety percent of the total nine trials of this type will last 162 days or more. Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children. In words, X = _____________ The distribution is _______. In words, ΣX = _______________ ΣX ~ _____(_____,_____) Find the probability that the total weight of open boxes is less than 250 pounds. Find the 35 th percentile for the total weight of open boxes of cereal. Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district. In words, X = ______________ X ~ _____(_____,_____) In words, ΣX = _____________ ΣX ~ _____(_____,_____) Find the probability that the teachers earn a total of over $400,000. Find the 90 th percentile for an individual teacher's salary. Find the 90 th percentile for the sum of ten teachers' salary. If we surveyed 70 teachers instead of ten, graphically, how would that change the distribution in part d? If each of the 70 teachers received a $3,000 raise, graphically, how would that change the distribution in part b? X = the salary of one elementary school teacher in the district X ~ N (44,000, 6,500) ΣX ~ sum of the salaries of ten elementary school teachers in the sample ΣX ~ N (44000, 20554.80) 0.9742 $52,330.09 466,342.04 Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve. If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.", "section": "The Central Limit Theorem for Sums", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Using the Central Limit Theorem It is important for you to understand when to use the central limit theorem . If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable. Examples of the Central Limit Theorem Law of Large Numbers The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x ¯ of the sample tends to get closer and closer to μ . From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for X ¯ is σ n .) This means that the sample mean x ¯ must be close to the population mean μ . We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. Central Limit Theorem for the Mean and Sum Examples A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find: The probability that the mean stress score for the 75 students is less than two. The 90 th percentile for the mean stress score for the 75 students. The probability that the total of the 75 stress scores is less than 200. The 90 th percentile for the total stress score for the 75 students. Let X = one stress score. Problems a and b ask you to find a probability or a percentile for a mean . Problems c and d ask you to find a probability or a percentile for a total or sum . The sample size, n , is equal to 75. Since the individual stress scores follow a uniform distribution, X ~ U (1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution). μ X = a + b 2 = 1 + 5 2 = 3 σ X = ( b – a ) 2 12 = ( 5 – 1) 2 12 = 16 12 = 4 3 , or approximately 1 . 15 . For problems a. and b., let X ¯ = the mean stress score for the 75 students. Then, X ¯ ∼ N ( 3, 1 .15 75 ) a. Find P ( x ¯ < 2). Draw the graph. a. P ( x ¯ < 2) = 0 The probability that the mean stress score is less than two is about zero. normalcdf ( 1,2,3, 1 .15 75 ) = 0 REMINDER The smallest stress score is one. b. Find the 90 th percentile for the mean of 75 stress scores. Draw a graph. b. Let k = the 90 th precentile. Find k , where P ( x ¯ < k ) = 0.90. k = 3.2 The 90 th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2. invNorm ( 0 .90,3, 1.15 75 ) = 3.2 For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N [(75)(3), ( 75 ) (1.15)] c. Find P ( Σx < 200). Draw the graph. c. The mean of the sum of 75 stress scores is (75)(3) = 225 The standard deviation of the sum of 75 stress scores is 75 4 3 = 75 × 4 3 = 100 = 10 P ( Σx < 200) = –0.0062 The probability that the total of 75 scores is less than 200 is about 0.0062. normalcdf (75,200,225,10). REMINDER The smallest total of 75 stress scores is 75, because the smallest single score is one. d. Find the 90 th percentile for the total of 75 stress scores. Draw a graph. d. Let k = the 90 th percentile. Find k where P ( Σx < k ) = 0.90. k = 237.8 The 90 th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8. invNorm (0.90,(75)(3), ( 75 ) (1.15)) = 237.8 Try It Use the information in , but use a sample size of 55 to answer the following questions. Find P ( x ¯ < 7). Find P ( Σx > 170). Find the 80 th percentile for the mean of 55 scores. Find the 85 th percentile for the sum of 55 scores. Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. X ∼ Exp ( 1 22 ) . From previous chapters, we know that μ = 22 and σ = 22. Let X ¯ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance. X ¯ ~ N ( 22, 22 80 ) by the central limit theorem for sample means Using the clt to find probability Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P( x ¯ > 20). Draw the graph. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P ( x > 20). Explain why the probabilities in parts a and b are different. Find: P ( x ¯ > 20) P ( x ¯ > 20) = 0.79199 using normalcdf ( 20,1E99,22, 22 80 ) The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance. REMINDER 1E99 = 10 99 and –1E99 = –10 99 . Press the EE key for E. Or just use 10 99 instead of 1E99. Find P (x > 20). Remember to use the exponential distribution for an individual: X ~ E x p ( 1 22 ) . P ( x > 20 ) = e ( − ( 1 22 ) ( 20 ) ) or e (–0.04545(20)) = 0.4029 P ( x > 20) = 0.4029 but P ( x ¯ > 20) = 0.7919 The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean. Using the clt to find percentiles Find the 95 th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph. Let k = the 95 th percentile. Find k where P ( x ¯ < k ) = 0.95 k = 26.0 using invNorm ( 0 .95,22 , 22 80 ) = 26.0 The 95 th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time. Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes. Try It Use the information in , but change the sample size to 144. Find P (20 < x ¯ < 30). Find P ( Σx is at least 3,000). Find the 75 th percentile for the sample mean excess time of 144 customers. Find the 85 th percentile for the sum of 144 excess times used by customers. In the United States, a robbery occurs every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100. Find the median, the first quartile, and the third quartile for the sample mean time of robberies in the United States. Find the median, the first quartile, and the third quartile for the sum of sample times of robberies in the United States. Find the probability that a robbery occurs on the average between 1.75 and 1.85 minutes. Find the value that is two standard deviations above the sample mean. Find the IQR for the sum of the sample times. We have, μ x = μ = 2 and σ x = σ n = 0.5 10 = 0.05. Therefore: 50 th percentile = μ x = μ = 2 25 th percentile = invNorm (0.25,2,0.05) = 1.97 75 th percentile = invNorm (0.75,2,0.05) = 2.03 We have μ Σx = n ( μ x ) = 100(2) = 200 and σ μx = n ( σ x ) = 10(0.5) = 5. Therefore 50 th percentile = μ Σx = n ( μ x ) = 100(2) = 200 25 th percentile = invNorm(0.25,200,5) = 196.63 75 th percentile = invNorm(0.75,200,5) = 203.37 P (1.75 < x ¯ < 1.85) = normalcdf (1.75,1.85,2,0.05) = 0.0013 Using the z -score equation, z = x ¯ – μ x ¯ σ x ¯ , and solving for x , we have x = 2(0.05) + 2 = 2.1 The IQR is 75 th percentile – 25 th percentile = 203.37 – 196.63 = 6.74 Try It Based on data from the National Health Survey, females between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for females between the ages of 18 to 24 follow a normal distribution. If one female from this population is randomly selected, find the probability that their systolic blood pressure is greater than 120. If 40 females from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120. If the sample were four females between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used? A study was done regarding attendance at Broadway shows in New York City. The age range of the attendees was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years. In a sample of 25 attendees, what is the probability that the mean age is less than 35? Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results. In a sample of 49 attendees, what is the probability that the sum of the ages is no less than 1,600? Is it likely that the sum of the ages of the 49 attendees is at most 1,595? Interpret the results. Find the 95th percentile for the sample mean age of 65 attendees. Interpret the results. Find the 90th percentile for the sum of the ages of 65 attendees. Interpret the results. P ( x ¯ < 35) = normalcdf (- E 99,35,30.9,1.8) = 0.9886 P ( x ¯ > 50) = normalcdf (50, E 99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50. P ( Σx ≥ 1,600) = normalcdf (1600,E99,1514.10,63) = 0.0864 P ( Σx ≤ 1,595) = normalcdf (-E99,1595,1514.10,63) = 0.9005. This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1595. The 95th percentile = invNorm (0.95,30.9,1.1) = 32.7. This indicates that 95% of the attendees in the sample of 65 are younger than 32.7 years, on average. The 90th percentile = invNorm (0.90,2008.5,72.56) = 2101.5. This indicates that 90% of the attendees in the sample of 65 have a sum of ages less than 2,101.5 years. Try It According to Boeing data, the 757 airliner carries 200 passengers and has doors with a height of 72 inches. Assume for a certain population of men we have a mean height of 69.0 inches and a standard deviation of 2.8 inches. What doorway height would allow 95% of men to enter the aircraft without bending? Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men? For engineers designing the 757, which result is more relevant: the height from part a or part b? Why? HISTORICAL NOTE Normal Approximation to the Binomial Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n , you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution : there are a certain number n of independent trials the outcomes of any trial are success or failure each trial has the same probability of a success p Recall that if X is the binomial random variable, then X ~ B ( n, p ). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = n p q . Remember that q = 1 – p . In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example. Suppose in a local Kindergarten through 12 th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. Find the probability that at least 150 favor a charter school. Find the probability that at most 160 favor a charter school. Find the probability that more than 155 favor a charter school. Find the probability that fewer than 147 favor a charter school. Find the probability that exactly 175 favor a charter school. Let X = the number that favor a charter school for grades K trough 5. X ~ B ( n, p ) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = n p q . The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y . Y ~ N (159, 8.6447). See The Normal Distribution for help with calculator instructions. For part a, you include 150 so P ( X ≥ 150) has normal approximation P ( Y ≥ 149.5) = 0.8641. normalcdf (149.5,10^99,159,8.6447) = 0.8641. For part b, you include 160 so P ( X ≤ 160) has normal appraximation P ( Y ≤ 160.5) = 0.5689. normalcdf (0,160.5,159,8.6447) = 0.5689 For part c, you exclude 155 so P ( X > 155) has normal approximation P ( y > 155.5) = 0.6572. normalcdf (155.5,10^99,159,8.6447) = 0.6572. For part d, you exclude 147 so P ( X < 147) has normal approximation P ( Y < 146.5) = 0.0741. normalcdf (0,146.5,159,8.6447) = 0.0741 For part e, P ( X = 175) has normal approximation P (174.5 < Y < 175.5) = 0.0083. normalcdf (174.5,175.5,159,8.6447) = 0.0083 Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in \"binomial probability distribution calculation\" in an Internet browser, you can find at least one online calculator for the binomial. For , the probabilities are calculated using the following binomial distribution: ( n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial. P ( X ≥ 150) : 1 - binomialcdf (300,0.53,149) = 0.8641 P ( X ≤ 160) : binomialcdf (300,0.53,160) = 0.5684 P ( X > 155) : 1 - binomialcdf (300,0.53,155) = 0.6576 P ( X < 147) : binomialcdf (300,0.53,146) = 0.0742 P ( X = 175) :(You use the binomial pdf.) binomialpdf (300,0.53,175) = 0.0083 Try It In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor. References Data from the Wall Street Journal. “National Health and Nutrition Examination Survey.” Center for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013). Chapter Review The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean x ¯ gets to μ . Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deivation? What is the distribution for the mean weight of 100 25-pound lifting weights? Find the probability that the mean actual weight for the 100 weights is less than 24.9. U (24, 26), 25, 0.5774 N (25, 0.0577) 0.0416 Draw the graph from Find the probability that the mean actual weight for the 100 weights is greater than 25.2. 0.0003 Draw the graph from Find the 90 th percentile for the mean weight for the 100 weights. 25.07 Draw the graph from What is the distribution for the sum of the weights of 100 25-pound lifting weights? Find P ( Σx < 2,450). N (2,500, 5.7735) 0 Draw the graph from Find the 90 th percentile for the total weight of the 100 weights. 2,507.40 Draw the graph from Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken. What is the standard deviation? What is the parameter m ? 10 1 10 What is the distribution for the length of time one battery lasts? What is the distribution for the mean length of time 64 batteries last? N ( 10, 10 8 ) What is the distribution for the total length of time 64 batteries last? Find the probability that the sample mean is between seven and 11. 0.7799 Find the 80 th percentile for the total length of time 64 batteries last. Find the IQR for the mean amount of time 64 batteries last. 1.69 Find the middle 80% for the total amount of time 64 batteries last. Use the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken. Find P ( Σx > 420). 0.0072 Find the 90 th percentile for the sums. Find the 15 th percentile for the sums. 391.54 Find the first quartile for the sums. Find the third quartile for the sums. 405.51 Find the 80 th percentile for the sums. Homework The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds. In words, Χ = _______ Χ ~ _____(_____,_____) In words, X ¯ = ____________ X ¯ ~ _____(_____,_____) Before doing any calculations, which do you think will be higher? Explain why. The probability that an individual attention span is less than ten minutes. The probability that the average attention span for the 60 children is less than ten minutes? Calculate the probabilities in part e. Explain why the distribution for X ¯ is not exponential. The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows. 8.625 30.25 27.625 46.75 32.875 18.25 5 0.125 2.9375 6.875 28.25 24.25 21 1.5 30.25 71 43.5 49.25 2.5625 31 16.5 9.5 18.5 18 9 10.5 16.625 1.25 18 12.87 7 12.875 2.875 60.25 29.25 In words, Χ = ______________ x ¯ = _____ s x = _____ n = _____ Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of ten. In words, describe the distribution of stock prices. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five pieces together until you have ten averages. List those ten averages. Use the ten averages from part e to calculate the following. x ¯ = _____ s x = _____ Construct a histogram of the distribution of the averages. Start at x = -0.0005. Use bar widths of ten. Does this histogram look like the graph in part c? In one or two complete sentences, explain why the graphs either look the same or look different? Based upon the theory of the central limit theorem , X ¯ ~ _____(_____,____) X = the closing stock prices for U.S. semiconductor manufacturers i. $20.71; ii. $17.31; iii. 35 Exponential distribution, Χ ~ Exp ( 1 20.71 ) Answers will vary. i. $20.71; ii. $11.14 Answers will vary. Answers will vary. Answers will vary. N ( 20 .71, 17.31 5 ) Use the following information to answer the next three exercises: Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. Χ ~ _____(_____,_____) U (0,4) U (10,2) Eχp (2) N (2,1) The average wait time is: one hour. two hours. two and a half hours. four hours. b Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is: 1 4 1 2 3 4 3 8 Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited. The 90 th percentile sample average wait time (in minutes) for a sample of 100 riders is: 315.0 40.3 38.5 65.2 b Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes? yes no There is not enough information. Use the following to answer the next two exercises: The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. What's the approximate probability that the average price for 16 gas stations is over $4.69? almost zero 0.1587 0.0943 unknown a Find the probability that the average price for 30 gas stations is less than $4.55. 0.6554 0.3446 0.0142 0.9858 0 Suppose in a local Kindergarten through 12 th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate following using the normal approximation to the binomial distribtion. Find the probability that less than 100 favor a charter school for grades K through 5. Find the probability that 170 or more favor a charter school for grades K through 5. Find the probability that no more than 140 favor a charter school for grades K through 5. Find the probability that there are fewer than 130 that favor a charter school for grades K through 5. Find the probability that exactly 150 favor a charter school for grades K through 5. If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology. 0 0.1123 0.0162 0.0003 0.0268 Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places. Find the probability that Janice is the driver at most 20 days. Find the probability that Roberta is the driver more than 16 days. Find the probability that Barbara drives exactly 24 of those 96 days. X ~ N (60, 9). Suppose that you form random samples of 25 from this distribution. Let X ¯ be the random variable of averages. Let ΣX be the random variable of sums. For parts c through f, sketch the graph, shade the region, label and scale the horizontal axis for X ¯ , and find the probability. Sketch the distributions of X and X ¯ on the same graph. X ¯ ~ _____(_____,_____) P ( x ¯ < 60) = _____ Find the 30 th percentile for the mean. P (56 < x ¯ < 62) = _____ P (18 < x ¯ < 58) = _____ Σx ~ _____(_____,_____) Find the minimum value for the upper quartile for the sum. P (1,400 < Σx < 1,550) = _____ Answers may vary. X ¯ ~ N ( 60, 9 25 ) 0.5000 59.06 0.8536 0.1333 N (1500, 45) 1530.35 0.6877 Suppose that the length of research papers is uniformly distributed from ten to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers. In words, X = _____________ X ~ _____(_____,_____) μ x = _____ σ x = _____ In words, X ¯ = ______________ X ¯ ~ _____(_____,_____) In words, ΣX = _____________ ΣX ~ _____(_____,_____) Without doing any calculations, do you think that it’s likely that the professor will need to read a total of more than 1,050 pages? Why? Calculate the probability that the professor will need to read a total of more than 1,050 pages. Why is it so unlikely that the average length of the papers will be less than 12 pages? Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district. Find the 90 th percentile for an individual teacher’s salary. Find the 90 th percentile for the average teacher’s salary. $52,330 $46,634 The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital. In words, X = _____________ In words, X ¯ = ___________________ X ¯ ~ _____(_____,_____) In words, ΣX = _______________ ΣX ~ _____(_____,_____) Is it likely that an individual stayed more than five days in the hospital? Why or why not? Is it likely that the average stay for the 80 women was more than five days? Why or why not? Which is more likely: An individual stayed more than five days. the average stay of 80 women was more than five days. If we were to sum up the women’s stays, is it likely that, collectively they spent more than a year in the hospital? Why or why not? For each problem, wherever possible, provide graphs and use the calculator. NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company’s claim reasonable? We have μ = 17, σ = 0.8, x ¯ = 16.7, and n = 30. To calculate the probability, we use normalcdf (lower, upper, μ , σ n ) = normalcdf ( E – 99,16 .7,17, 0. 8 30 ) = 0.0200. If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim. Men have an average weight of 172 pounds with a standard deviation of 29 pounds. Find the probability that 20 randomly selected men will have a sum weight greater than 3600 lbs. If 20 men have a sum weight greater than 3500 lbs, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain. M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class. Red Orange Yellow Brown Blue Green 0.751 0.735 0.883 0.696 0.881 0.925 0.841 0.895 0.769 0.876 0.863 0.914 0.856 0.865 0.859 0.855 0.775 0.881 0.799 0.864 0.784 0.806 0.854 0.865 0.966 0.852 0.824 0.840 0.810 0.865 0.859 0.866 0.858 0.868 0.858 1.015 0.857 0.859 0.848 0.859 0.818 0.876 0.942 0.838 0.851 0.982 0.868 0.809 0.873 0.863 0.803 0.865 0.809 0.888 0.932 0.848 0.890 0.925 0.842 0.940 0.878 0.793 0.832 0.833 0.905 0.977 0.807 0.845 0.850 0.841 0.852 0.830 0.932 0.778 0.856 0.833 0.814 0.842 0.881 0.791 0.778 0.818 0.810 0.786 0.864 0.881 0.853 0.825 0.864 0.855 0.873 0.942 0.880 0.825 0.882 0.869 0.931 0.912 0.887 The bag contained 465 candies and he listed weights in the table came from randomly selected candies. Count the weights. Find the mean sample weight and the standard deviation of the sample weights of candies in the table. Find the sum of the sample weights in the table and the standard deviation of the sum of the weights. If 465 M&Ms are randomly selected, find the probability that their weights sum to at least 396.9. Is the Mars Company’s M&M labeling accurate? For the sample, we have n = 100, x ¯ = 0.862, s = 0.05 Σ x ¯ = 85.65, Σs = 5.18 normalcdf (396.9, E 99,(465)(0.8565),(0.05)( 465 )) ≈ 1 Since the probability of a sample of size 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that Mars is correctly labeling their M&M packages. The Screw Right Company claims their 3 4 inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded. 0.757 0.723 0.754 0.737 0.757 0.741 0.722 0.741 0.743 0.742 0.740 0.758 0.724 0.739 0.736 0.735 0.760 0.750 0.759 0.754 0.744 0.758 0.765 0.756 0.738 0.742 0.758 0.757 0.724 0.757 0.744 0.738 0.763 0.756 0.760 0.768 0.761 0.742 0.734 0.754 0.758 0.735 0.740 0.743 0.737 0.737 0.725 0.761 0.758 0.756 The screws were randomly selected from the local home repair store. Find the mean diameter and standard deviation for the sample Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible? Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time? Use normalcdf ( E – 99,1 .1,1, 1 70 ) = 0.7986. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points? Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine? We assume that the weights of coins are normally distributed in the population. Since we have normalcdf ( 5. 111,5 .291,5 .201, 0. 065 280 ) ≈ 0.8338, we expect (1 – 0.8338)280 ≈ 47 coins to be rejected. Exponential Distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital, notation: X ~ Exp ( m ). The mean is μ = 1 m and the standard deviation is σ = 1 m . The probability density function is f ( x ) = me –mx , x ≥ 0 and the cumulative distribution function is P ( X ≤ x ) = 1 – e –mx . Mean a number that measures the central tendency; a common name for mean is \"average.\" The term \"mean\" is a shortened form of \"arithmetic mean.\" By definition, the mean for a sample (denoted by x ¯ ) is x ¯ = Sum of all values in the sample Number of values in the sample , and the mean for a population (denoted by μ ) is μ = Sum of all values in the population Number of values in the population . Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation.; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Uniform Distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b ; often referred as the Rectangular Distribution because the graph of the pdf has the form of a rectangle. Notation: X ~ U ( a , b ). The mean is μ = a + b 2 and the standard deviation is σ = ( b – a) 2 12 . The probability density function is f ( x ) = 1 b – a for a < x < b or a ≤ x ≤ b . The cumulative distribution is P ( X ≤ x ) = x – a b – a .", "section": "Using the Central Limit Theorem", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Central Limit Theorem (Pocket Change) Central Limit Theorem (Pocket Change) Class Time: Names: Student Learning Outcomes The student will demonstrate and compare properties of the central limit theorem. NOTE This lab works best when sampling from several classes and combining data. Collect the Data Count the change in your pocket. (Do not include bills.) Randomly survey 30 classmates. Record the values of the change in . Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Calculate the following ( n = 1; surveying one person at a time): x ¯ = _______ s = _______ Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Collecting Averages of Pairs Repeat steps one through five of the section Collect the Data. with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs. Randomly survey 30 pairs of classmates. Record the values of the average of their change in . Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data . Sketch the graph using a ruler and a pencil. Calculate the following ( n = 2; surveying two people at a time): x ¯ = _______ s = _______ Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Collecting Averages of Groups of Five Repeat steps one through five (of the section titled Collect the Data ) with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of five. Randomly survey 30 groups of five classmates. Record the values of the average of their change. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data . Sketch the graph using a ruler and a pencil. Calculate the following ( n = 5; surveying five people at a time): x ¯ = _______ s = _______ Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions Why did the shape of the distribution of the data change, as n changed? Use one to two complete sentences to explain what happened. In the section titled Collect the Data , what was the approximate distribution of the data? X ~ _____(_____,_____) In the section titled Collecting Averages of Groups of Five , what was the approximate distribution of the averages? X ¯ ~ _____(_____,_____) In one to two complete sentences, explain any differences in your answers to the previous two questions.", "section": "Central Limit Theorem (Pocket Change)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Central Limit Theorem (Cookie Recipes) Central Limit Theorem (Cookie Recipes) Class Time: Names: Student Learning Outcomes The student will demonstrate and compare properties of the central limit theorem. Given X = length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of the different recipes makes the same quantity of cookies.) Recipe # X Recipe # X Recipe # X Recipe # X 1 1 16 2 31 3 46 2 2 5 17 2 32 4 47 2 3 2 18 4 33 5 48 11 4 5 19 6 34 6 49 5 5 6 20 1 35 6 50 5 6 1 21 6 36 1 51 4 7 2 22 5 37 1 52 6 8 6 23 2 38 2 53 5 9 5 24 5 39 1 54 1 10 2 25 1 40 6 55 1 11 5 26 6 41 1 56 2 12 1 27 4 42 6 57 4 13 1 28 1 43 2 58 3 14 3 29 6 44 6 59 6 15 2 30 2 45 2 60 5 Calculate the following: μ x = _______ σ x = _______ Collect the Data Use a random number generator to randomly select four samples of size n = 5 from the given population. Record your samples in . Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. Complete the table: Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups: Means: x ¯ = ____ x ¯ = ____ x ¯ = ____ x ¯ = ____ Calculate the following: x ¯ = _______ s x ¯ = _______ Again, use a random number generator to randomly select four samples from the population. This time, make the samples of size n = 10. Record the samples in . As before, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups Means: x ¯ = ____ x ¯ = ____ x ¯ = ____ x ¯ = ____ Calculate the following: x ¯ = ______ s x ¯ = ______ For the original population, construct a histogram. Make intervals with a bar width of one day. Sketch the graph using a ruler and pencil. Scale the axes. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the Procedure for n = 5 For the sample of n = 5 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 a day. Sketch the graph using a ruler and pencil. Scale the axes. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the Procedure for n = 10 For the sample of n = 10 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 a day. Sketch the graph using a ruler and pencil. Scale the axes. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions Compare the three histograms you have made, the one for the population and the two for the sample means. In three to five sentences, describe the similarities and differences. State the theoretical (according to the clt) distributions for the sample means. n = 5: x ¯ ~ _____(_____,_____) n = 10: x ¯ ~ _____(_____,_____) Are the sample means for n = 5 and n = 10 “close” to the theoretical mean, μ x ? Explain why or why not. Which of the two distributions of sample means has the smaller standard deviation? Why? As n changed, why did the shape of the distribution of the data change? Use one to two complete sentences to explain what happened.", "section": "Central Limit Theorem (Cookie Recipes)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction Have you ever wondered what the average number of M&Ms in a bag at the grocery store is? You can use confidence intervals to answer this question. (credit: modification of work “sweet, orange, food, green, red, color, brown, blue, colorful, yellow, chocolate, snack, dessert, toy, plain, candy, sweetness, treat, confectionery, coated, m ms, hard shell, snack food, jelly bean”/ Pxhere, Public Domain) Chapter Objectives By the end of this chapter, the student should be able to: Calculate and interpret confidence intervals for estimating a population mean and a population proportion. Interpret the Student's t probability distribution as the sample size changes. Discriminate between problems applying the normal and the Student's t distributions. Calculate the sample size required to estimate a population mean and a population proportion given a desired confidence level and margin of error. Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion. We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics . The sample data help us to make an estimate of a population parameter . We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals. In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed. If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from Apple Music. If so, you could conduct a survey and calculate the sample mean, x ¯ , and the sample standard deviation, s . You would use x ¯ to estimate the population mean and s to estimate the population standard deviation. The sample mean, x ¯ , is the point estimate for the population mean, μ . The sample standard deviation, s , is the point estimate for the population standard deviation, σ . Each of x ¯ and s is called a statistic. A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. It provides a range of reasonable values in which we expect the population parameter to fall. There is no guarantee that a given confidence interval does capture the parameter, but there is a predictable probability of success. Suppose, for the Apple Music example, we do not know the population mean μ , but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is σ n = 1 100 = 0.1 The Empirical Rule , which applies to bell-shaped distributions, says that in approximately 95% of the samples, the sample mean, x ¯ , will be within two standard deviations of the population mean μ . For our Apple Music example, two standard deviations is (2)(0.1) = 0.2. The sample mean x ¯ is likely to be within 0.2 units of μ . Because x ¯ is within 0.2 units of μ , which is unknown, then μ is likely to be within 0.2 units of x ¯ in 95% of the samples. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between x ¯ − 0 .2 and x ¯ + 0 .2 in 95% of all the samples. For the Apple Music example, suppose that a sample produced a sample mean x ¯ = 2 . Then the unknown population mean μ is between x ¯ − 0.2 = 2 − 0.2 = 1.8 and x ¯ + 0.2 = 2 + 0.2 = 2.2 We say that we are 95% confident that the unknown population mean number of songs downloaded from Apple Music per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2). The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ or our sample produced an x ¯ that is not within 0.2 units of the true mean μ . The first possibility happens for 95% of well-chosen samples. It is important to remember that the second possibility happens for 5% of samples, even though correct procedures are followed. Remember that a confidence interval is created for an unknown population parameter like the population mean, μ . Confidence intervals for some parameters have the form: (point estimate – margin of error, point estimate + margin of error ) The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean. When you read newspapers and journals, some reports will use the phrase \"margin of error.\" Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. These are two ways of expressing the same concept. NOTE Although the text only covers symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation). Have your instructor record the number of meals each student in your class eats out in a week. Assume that the standard deviation is known to be three meals. Construct an approximate 95% confidence interval for the true mean number of meals students eat out each week. Calculate the sample mean. Let σ = 3 and n = the number of students surveyed. Construct the interval ( x ¯ − 2 ⋅ σ n , x ¯ + 2 ⋅ σ n ) . We say we are approximately 95% confident that the true mean number of meals that students eat out in a week is between __________ and ___________. Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: the desired confidence level, information that is known about the distribution (for example, known standard deviation), the sample and its size. Inferential Statistics also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if four out of the 100 calculators sampled are defective we might infer that four percent of the production is defective. Parameter a numerical characteristic of a population Point Estimate a single number computed from a sample and used to estimate a population parameter", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "A Single Population Mean using the Normal Distribution A confidence interval for a population mean, when the population standard deviation is known, is based on the conclusion of the Central Limit Theorem that the sampling distribution of the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x ¯ = 10 and we have constructed the 90% confidence interval (5, 15) where EBM = 5. Calculating the Confidence Interval To construct a confidence interval for a single unknown population mean μ , where the population standard deviation is known , we need x ¯ as an estimate for μ and we need the margin of error. Here, the margin of error ( EBM ) is called the error bound for a population mean (abbreviated EBM ). The sample mean x ¯ is the point estimate of the unknown population mean μ . The confidence interval estimate will have the form: (point estimate - error bound, point estimate + error bound) or, in symbols,( x ¯ – E B M , x ¯ + E B M ) The margin of error ( EBM ) depends on the confidence level (abbreviated CL ). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of their conclusions. There is another probability called alpha ( α ). α is related to the confidence level, CL . α is the probability that the interval does not contain the unknown population parameter. Mathematically, α + CL = 1. Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5. x ¯ = 7 and EBM = 2.5 The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5). If the confidence level ( CL ) is 95%, then we say that, \"We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5.\" Try It Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean? A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x ¯ = 2 , and we have constructed the 90% confidence interval (1.8, 2.2) where EBM = 0.1. To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the normal distribution. X ¯ ± Z a / 2 σ n = 2 ± 1 . 645 ( 0 . 1 ) = 2 ± 0 . 1645 1 . 8355 ≤ μ ≤ 2 . 1645 To capture the central 90%, we must go out 1.645 \"standard deviations\" on either side of the calculated sample mean. The value 1.645 is the z -score from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail. It is important that the \"standard deviation\" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is σ n . The fraction σ n , is commonly called the \"standard error of the mean\" in order to distinguish clearly the standard deviation for a mean from the population standard deviation σ . In summary, as a result of the central limit theorem: X ¯ is normally distributed, that is, X ¯ ~ N ( μ X , σ n ) . When the population standard deviation σ is known, we use a normal distribution to calculate the error bound. Calculating the Confidence Interval To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are: Calculate the sample mean x ¯ from the sample data. Remember, in this section we already know the population standard deviation σ . Find the z -score that corresponds to the confidence level. Calculate the error bound EBM . Construct the confidence interval. Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.) We will first examine each step in more detail, and then illustrate the process with some examples. Finding the z -score for the Stated Confidence Level When we know the population standard deviation σ , we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N (0, 1). The confidence level, CL , is the area in the middle of the standard normal distribution. CL = 1 – α , so α is the area that is split equally between the two tails. Each of the tails contains an area equal to α 2 . The z-score that has an area to the right of α 2 is denoted by z α 2 . For example, when CL = 0.95, α = 0.05 and α 2 = 0.025; we write z α 2 = z 0.025 . The area to the right of z 0.025 is 0.025 and the area to the left of z 0.025 is 1 – 0.025 = 0.975. z α 2 = z 0. 025 = 1 .96 , using a calculator, computer or a standard normal probability table. invNorm (0.975, 0, 1) = 1.96 NOTE Remember to use the area to the LEFT of z α 2 ; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z ~ N (0, 1). Writing the Interpretation The interpretation should clearly state the confidence level ( CL ), explain what population parameter is being estimated (here, a population mean ), and state the confidence interval (both endpoints). \"We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units).\" Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90% confidence interval for the true (population) mean of statistics exam scores. You can use technology to calculate the confidence interval directly. The first solution is shown step-by-step. The second solution uses the TI-83, 83+, and 84+ calculators. The solution is shown step by step: The formula for a confidence interval for an unknown population mean assuming we know the population standard deviation is: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) For a 90% confidence interval, visualize an area of 0.90 centered under the normal curve (See ). The remaining area for the two tails of the normal distribution is then 0.10, which indicates that the area in the left tail is one-half of 0.10, which is 0.05. The corresponding z-score that cuts off an area of 0.05 in the left tail is 1.645. In this example we are given that the population standard deviation σ = 3 . We are also given that the sample size n = 36 and the sample mean X ¯ = 68 . Substituting these values in the confidence interval formula results in the following: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) 68 - 1 . 645 3 36 ≤ μ ≤ 68 + 1 . 645 3 36 68 - 0 . 8225 ≤ μ ≤ 68 + 0 . 8225 67 . 1775 ≤ μ ≤ 68 . 8225 We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Press STAT and arrow over to TESTS . Arrow down to 7:ZInterval . Press ENTER . Arrow to Stats and press ENTER . Arrow down and enter three for σ , 68 for x ¯ , 36 for n , and .90 for C-level . Arrow down to Calculate and press ENTER . The confidence interval is (to three decimal places)(67.178, 68.822). Explanation of 90% Confidence Level: Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score. Try It Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time. (34.1347, 37.8653) The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. shows the highest SAR level for a random selection of cell phone models as measured by the FCC. SAR Data for a sample of 30 cell phones 1.11 1.36 0.74 1.48 1.34 0.5 1.43 1.18 0.4 1.3 1.3 0.867 1.09 1.26 0.68 0.455 1.29 0.51 1.41 0.36 1.13 0.82 0.52 0.3 0.78 1.6 1.48 1.25 1.39 1.38 Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. Assume that the population standard deviation is σ = 0.337. To find the confidence interval, start by finding the point estimate: the sample mean. x ¯ = 1.024 Next, find the EBM . Because you are creating a 98% confidence interval, CL = 0.98. You need to find z 0.01 having the property that the area under the normal density curve to the right of z 0.01 is 0.01 and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find z 0.01 = 2.326. E B M = ( z 0.01 ) σ n = ( 2.326 ) 0.337 30 = 0.1431 To find the 98% confidence interval, find x ¯ ± E B M . x ¯ – EBM = 1.024 – 0.1431 = 0.8809 x ¯ – EBM = 1.024 – 0.1431 = 1.1671 We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram. Press STAT and arrow over to TESTS . Arrow down to 7:ZInterval . Press ENTER . Arrow to Stats and press ENTER . Arrow down and enter the following values: σ : 0.337 x ¯ : 1.024 n : 30 C -level: 0.98 Arrow down to Calculate and press ENTER. The confidence interval is (to three decimal places) (0.881, 1.167). Try It shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337. SAR Data for a sample of 20 cell phones 1.48 1.53 0.8 0.68 1.15 1.4 1.36 1.24 0.77 0.57 0.462 0.2 1.36 0.51 1.39 0.3 1.3 0.73 0.7 0.869 Notice the difference in the confidence intervals calculated in and the following Try It exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter. Changing the Confidence Level or Sample Size Suppose we change the original problem in by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score. To find the confidence interval, you need the sample mean, x ¯ , and the EBM . x ¯ = 68 EBM = ( z α 2 ) ( σ n ) σ = 3; n = 36; The confidence level is 95% ( CL = 0.95). CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025 z α 2 = z 0.025 The area to the right of z 0.025 is 0.025 and the area to the left of z 0.025 is 1 – 0.025 = 0.975. z α 2 = z 0.025 = 1.96 when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.) EBM = (1.96) ( 3 36 ) = 0.98 x ¯ – EBM = 68 – 0.98 = 67.02 x ¯ + EBM = 68 + 0.98 = 68.98 Notice that the EBM is larger for a 95% confidence level in the original problem. We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Explanation of 95% Confidence Level: Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score. Comparing the results: The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. Summary: Effect of Changing the Confidence Level Increasing the confidence level increases the error bound, making the confidence interval wider. Decreasing the confidence level decreases the error bound, making the confidence interval narrower. Try It Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time. Suppose we change the original problem in to see what happens to the error bound if the sample size is changed. Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36? x ¯ = 68 EBM = ( z α 2 ) ( σ n ) σ = 3; The confidence level is 90% ( CL =0.90); z α 2 = z 0.05 = 1.645. If we increase the sample size n to 100, we decrease the error bound. When n = 100: EBM = ( z α 2 ) ( σ n ) = (1.645) ( 3 100 ) = 0.4935. If we decrease the sample size n to 25, we increase the error bound. When n = 25: EBM = ( z α 2 ) ( σ n ) = (1.645) ( 3 25 ) = 0.987. Summary: Effect of Changing the Sample Size Increasing the sample size causes the error bound to decrease, making the confidence interval narrower. Decreasing the sample size causes the error bound to increase, making the confidence interval wider. Try It Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time. Working Backwards to Find the Error Bound or Sample Mean When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean. Finding the Error Bound From the upper value for the interval, subtract the sample mean, OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two. Finding the Sample Mean Subtract the error bound from the upper value of the confidence interval, OR, average the upper and lower endpoints of the confidence interval. Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know. Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean. Calculate the Error Bound: If we know that the sample mean is 68: EBM = 68.82 – 68 = 0.82. If we don't know the sample mean: EBM = ( 68.82 − 67.18 ) 2 = 0.82. Calculate the Sample Mean: If we know the error bound: x ¯ = 68.82 – 0.82 = 68 If we don't know the error bound: x ¯ = ( 67.18 + 68.82 ) 2 = 68. Try It Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population mean when the population standard deviation is known is EBM = ( z α 2 ) ( σ n ) . The formula for sample size is n = z 2 σ 2 E B M 2 , found by solving the error bound formula for n . In this formula, z is z α 2 , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study. The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed? From the problem, we know that σ = 15 and EBM = 2. z = z 0.025 = 1.96, because the confidence level is 95%. n = z 2 σ 2 E B M 2 = ( 1.96 ) 2 ( 15 ) 2 2 2 = 216.09 using the sample size equation. Use n = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough. Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students. Try It The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed? References “American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t (accessed July 2, 2013). “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013). “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm (accessed September 30,2013). Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/2000growthchart-us.pdf (accessed July 2, 2013). La, Lynn, Kent German. \"Cell Phone Radiation Levels.\" c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013). “Mean Income in the Past 12 Months (in 2011 Inflaction-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013). “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013). “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013). Chapter Review In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean ( EBM ). A confidence interval has the general form: (lower bound, upper bound) = (point estimate – EBM , point estimate + EBM ) The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem, E B M = z σ n Given a confidence interval, you can work backwards to find the error bound ( EBM ) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean. Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal: n = z 2 σ 2 E B M 2 Formula Review X ¯ ~ N ( μ X , σ n ) The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size. The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by (lower bound, upper bound) = (point estimate – EBM , point estimate + EBM ) = ( x ¯ − E B M , x ¯ + E B M ) = ( x ¯ − z σ n , x ¯ + z σ n ) EBM = z σ n = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known. CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter z α 2 = the z -score with the property that the area to the right of the z-score is ∝ 2 this is the z -score used in the calculation of \"EBM where α = 1 – CL . n = z 2 σ 2 E B M 2 = the formula used to determine the sample size ( n ) needed to achieve a desired margin of error at a given level of confidence General form of a confidence interval (lower value, upper value) = (point estimate−error bound, point estimate + error bound) To find the error bound when you know the confidence interval error bound = upper value−point estimate OR error bound = upper value − lower value 2 Single Population Mean, Known Standard Deviation, Normal Distribution Use the Normal Distribution for Means, Population Standard Deviation is Known EBM = z α 2 ⋅ σ n The confidence interval has the format ( x ¯ − EBM , x ¯ + EBM ). Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. Identify the following: x ¯ = _____ σ = _____ n = _____ 244 15 50 In words, define the random variables X and X ¯ . Which distribution should you use for this problem? N ( 244 , 15 50 ) Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why? As the sample size increases, there will be less variability in the mean, so the interval size decreases. Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Identify the following: x ¯ = _____ σ = _____ n = _____ In words, define the random variables X and X ¯ . X is the time in minutes it takes to complete the U.S. Census short form. X ¯ is the mean time it took a sample of 200 people to complete the U.S. Census short form. Which distribution should you use for this problem? Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound. CI: (7.9441, 8.4559) EBM = 0.26 If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why? The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases. Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not? Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. Identify the following: x ¯ = ______ σ = ______ n = ______ x ¯ = 2.2 σ = 0.2 n = 20 In words, define the random variable X . In words, define the random variable X ¯ . X ¯ is the mean weight of a sample of 20 heads of lettuce. Which distribution should you use for this problem? Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. EBM = 0.07 CI: (2.1264, 2.2736) Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. In complete sentences, explain why the confidence interval in is wider than in . The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals. In complete sentences, give an interpretation of what the interval in means. What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same? The confidence level would increase. What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same? Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student. x ¯ = _____ 30.4 n = _____ ________ = 15 σ In words, define the random variable X ¯ . What is x ¯ estimating? μ Is σ x known? As a result of your answer to , state the exact distribution to use when calculating the confidence interval. normal Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises . How much area is in both tails (combined)? α =________ How much area is in each tail? α 2 =________ 0.025 Identify the following specifications: lower limit upper limit error bound The 95% confidence interval is:__________________. (24.52,36.28) Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. In one complete sentence, explain what the interval means. We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28. Using the same mean, standard deviation, and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know? Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why? The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean. Homework Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches. x ¯ =________ σ =________ n =________ In words, define the random variables X and X ¯ . Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean height of male Swedes. State the confidence interval. Sketch the graph. Calculate the error bound. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why? 71 3 48 X is the height of a Swiss male, and is the mean height from a sample of 48 Swiss males. Normal. We know the standard deviation for the population, and the sample size is greater than 30. CI: (70.151, 71.49) EBM = 0.849 The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean. Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal. In words, define the random variables X and X ¯ . Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean length of engineering conferences. State the confidence interval. Sketch the graph. Calculate the error bound. Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. x ¯ =________ σ =________ n =________ In words, define the random variables X and X ¯ . Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean time to complete the tax forms. State the confidence interval. Sketch the graph. Calculate the error bound. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why? Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why? x ¯ = 23.6 σ = 7 n = 100 X is the time needed to complete an individual tax form. X ¯ is the mean time to complete tax forms from a sample of 100 customers. N ( 23.6 , 7 100 ) because we know sigma. (22.228, 24.972) EBM = 1.372 It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce. x ¯ =________ σ =________ s x =________ In words, define the random variable X . In words, define the random variable X ¯ . Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean weight of the candies. State the confidence interval. Sketch the graph. Calculate the error bound. Construct a 98% confidence interval for the population mean weight of the candies. State the confidence interval. Sketch the graph. Calculate the error bound. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e. In complete sentences, give an interpretation of what the interval in part f means. A camp director is interested in the mean number of letters each child sends during their camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8. x ¯ =________ σ =________ n =________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population mean number of letters campers send home. State the confidence interval. Sketch the graph. Calculate the error bound. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why? 7.9 2.5 20 X is the number of letters a single camper will send home. X ¯ is the mean number of letters sent home from a sample of 20 campers. N 7.9 ( 2.5 20 ) CI: (6.98, 8.82) EBM : 0.92 The error bound and confidence interval will decrease. What is meant by the term “90% confident” when constructing a confidence interval for a mean? If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During a certain campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200. $3,600 $1,243,900 $10,900 $385,200 $581,500 $7,400 $2,900 $400 $3,714,500 $632,500 $391,000 $467,400 $56,800 $5,800 $405,200 $733,200 $8,000 $468,700 $75,200 $41,000 $13,300 $9,500 $953,800 $1,113,500 $1,109,300 $353,900 $986,100 $88,600 $378,200 $13,200 $3,800 $745,100 $5,800 $3,072,100 $1,626,700 $512,900 $2,309,200 $6,600 $202,400 $15,800 Find the point estimate for the population mean. Using 95% confidence, calculate the error bound. Create a 95% confidence interval for the mean total individual contributions. Interpret the confidence interval in the context of the problem. x ¯ = $568,873 CL = 0.95 α = 1 – 0.95 = 0.05 z α 2 = 1.96 EBM = z 0.025 σ n = 1.96 909200 40 = $281,764 x ¯ − EBM = 568,873 − 281,764 = 287,109 x ¯ + EBM = 568,873 + 281,764 = 850,637 Alternate solution: Press STAT and arrow over to TESTS . Arrow down to 7:ZInterval . Press ENTER . Arrow to Stats and press ENTER . Arrow down and enter the following values: σ : 909,200 x ¯ : 568,873 n : 40 CL : 0.95 Arrow down to Calculate and press ENTER . The confidence interval is ($287,114, $850,632). Notice the small difference between the two solutions–these differences are simply due to rounding error in the hand calculations. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income. The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure? Use the formula for EBM , solved for n : n = z 2 σ 2 E B M 2 From the statement of the problem, you know that σ = 2.5, and you need EBM = 1. z = z 0.035 = 1.812 (This is the value of z for which the area under the density curve to the right of z is 0.035.) n = z 2 σ 2 E B M 2 = 1.812 2 2.5 2 1 2 ≈ 20.52 You need to measure at least 21 male students to achieve your goal. Confidence Level (CL) the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter. Error Bound for a Population Mean ( EBM ) the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation.", "section": "A Single Population Mean using the Normal Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "A Single Population Mean using the Student t Distribution In practice, we rarely know the population standard deviation . In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval. William S. Gosset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to \"discover\" what is called the Student's t-distribution . The name comes from the fact that Gosset wrote under the pen name \"Student.\" Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ . If you draw a simple random sample of size n from a population that has an approximately normal distribution with mean μ and unknown population standard deviation σ and calculate the t -score t = x ¯ – μ ( s n ) , then the t -scores follow a Student's t-distribution with n – 1 degrees of freedom . The t -score has the same interpretation as the z -score . It measures how far x ¯ is from its mean μ . For each sample size n , there is a different Student's t-distribution. The degrees of freedom , n – 1 , come from the calculation of the sample standard deviation s . This calculation requires n deviations ( x – x ¯ values ) . Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df). Properties of the Student's t-Distribution The graph for the Student's t-distribution is similar to the standard normal curve. The mean for the Student's t-distribution is zero and the distribution is symmetric about zero. The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution. The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution. The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality. Calculators and computers can easily calculate any Student's t-probabilities. The TI-83,83+, and 84+ have a tcdf function to find the probability for given values of t . The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability. For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified. The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.) A probability table for the Student's t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-Distribution.) When using a t -table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. A Student's t table (See ) gives t -scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's t-probabilities. The notation for the Student's t-distribution (using T as the random variable) is: T ~ t df where df = n – 1. For example, if we have a sample of size n = 20 items, then we calculate the degrees of freedom as df = n - 1 = 20 - 1 = 19 and we write the distribution as T ~ t 19 . If the population standard deviation is not known , the error bound for a population mean is: E B M = ( t α 2 ) ( s n ) , t σ 2 is the t -score with area to the right equal to α 2 , use df = n – 1 degrees of freedom, and s = sample standard deviation. The format for the confidence interval is: ( x ¯ − E B M , x ¯ + E B M ) . To calculate the confidence interval directly: Press STAT . Arrow over to TESTS . Arrow down to 8:TInterval and press ENTER (or just press 8). Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data. The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators. 8.6 9.4 7.9 6.8 8.3 7.3 9.2 9.6 8.7 11.4 10.3 5.4 8.1 5.5 6.9 The first solution is step-by-step. The second solution uses the TI-83+ and TI-84 calculators. To find the confidence interval, you need the sample mean, x ¯ , and the EBM . x ¯ = 8.2267 s = 1.6722 n = 15 df = 15 – 1 = 14 CL so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025 t α 2 = t 0.025 The area to the right of t 0.025 is 0.025, and the area to the left of t 0.025 is 1 – 0.025 = 0.975 t α 2 = t 0.025 = 2.14 using invT(.975,14) on the TI-84+ calculator. E B M = ( t α 2 ) ( s n ) E B M = ( 2.14 ) ( 1.6722 15 ) = 0.924 x ¯ – EBM = 8.2267 – 0.9240 = 7.3 x ¯ + EBM = 8.2267 + 0.9240 = 9.15 The 95% confidence interval is (7.30, 9.15). We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15. Press STAT and arrow over to TESTS . Arrow down to 8:TInterval and press ENTER (or you can just press 8 ). Arrow to Data and press ENTER . Arrow down to List and enter the list name where you put the data. There should be a 1 after Freq . Arrow down to C-level and enter 0.95 Arrow down to Calculate and press ENTER . The 95% confidence interval is (7.3006, 9.1527) NOTE When calculating the error bound, a probability table for the Student's t-distribution can also be used to find the value of t . The table gives t -scores that correspond to the confidence level (column) and degrees of freedom (row); the t -score is found where the row and column intersect in the table. Try It You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5 The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. The scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the \"In utero/newborn\" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. shows how many of the targeted chemicals were found in each infant’s cord blood. 79 145 147 160 116 100 159 151 156 126 137 83 156 94 121 144 123 114 139 99 Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood. From the sample, you can calculate x ¯ = 127.45 and s = 25.965. There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19. You are asked to calculate a 90% confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10 α 2 = 0.05, t α 2 = t 0.05 By definition, the area to the right of t 0.05 is 0.05 and so the area to the left of t 0.05 is 1 – 0.05 = 0.95. Use a table, calculator, or computer to find that t 0.05 = 1.729. E B M = t α 2 ( s n ) = 1.729 ( 25.965 20 ) ≈ 10.038 x ¯ – EBM = 127.45 – 10.038 = 117.412 x ¯ + EBM = 127.45 + 10.038 = 137.488 We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488. Enter the data as a list. Press STAT and arrow over to TESTS . Arrow down to 8:TInterval and press ENTER (or you can just press 8 ). Arrow to Data and press ENTER . Arrow down to List and enter the list name where you put the data. Arrow down to Freq and enter 1. Arrow down to C-level and enter 0.90 Arrow down to Calculate and press ENTER . The 90% confidence interval is (117.41, 137.49). Try It A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in . Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week. 0 3 1 20 9 5 10 1 10 4 14 2 4 4 5 References “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013). Data from Microsoft Bookshelf . Data from http://www.businessweek.com/. Data from http://www.forbes.com/. “Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013). “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013). “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013). Chapter Review In many cases, the researcher does not know the population standard deviation, σ , of the measure being studied. In these cases, it is common to use the sample standard deviation, s , as an estimate of σ . The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: t = x ¯ − μ s n The t -score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = ( t α 2 ) s n where t α 2 is the t -score with area to the right equal to α 2 , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 for a given α . Formula Review s = the standard deviation of sample values. t = x ¯ − μ s n is the formula for the t -score which measures how far away a measure is from the population mean in the Student’s t-distribution df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample T ~ t df the random variable, T , has a Student’s t-distribution with df degrees of freedom E B M = t α 2 s n = the error bound for the population mean when the population standard deviation is unknown t α 2 is the t -score in the Student’s t-distribution with area to the right equal to α 2 The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound) = (point estimate – EBM , point estimate + EBM ) = ( x ¯ – t s n , x ¯ + t s n ) Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. Identify the following: x ¯ =_______ s x =_______ n =_______ n – 1 =_______ Define the random variables X and X ¯ in words. X is the number of hours a patient waits in the emergency room before being called back to be examined. X ¯ is the mean wait time of 70 patients in the emergency room. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound. CI: (1.3808, 1.6192) EBM = 0.12 Explain in complete sentences what the confidence interval means. Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Identify the following: x ¯ =_______ s x =_______ n =_______ n – 1 =_______ x ¯ = 151 s x = 32 n = 108 n – 1 = 107 Define the random variable X in words. Define the random variable X ¯ in words. X ¯ is the mean number of hours spent watching television per month from a sample of 108 Americans. Which distribution should you use for this problem? Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound. CI: (142.92, 159.08) EBM = 8.08 Why would the error bound change if the confidence level were lowered to 95%? Use the following information to answer the next 13 exercises: The data in are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 1 2 7 3 18 4 7 5 6 Calculate the following: x ¯ =______ s x =______ n =______ 3.26 1.02 39 Define the random variable X ¯ in words. What is x ¯ estimating? μ Is σ x known? As a result of your answer to , state the exact distribution to use when calculating the confidence interval. t 38 Construct a 95% confidence interval for the true mean number of colors on national flags. How much area is in both tails (combined)? How much area is in each tail? 0.025 Calculate the following: lower limit upper limit error bound The 95% confidence interval is_____. (2.93, 3.59) Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean. In one complete sentence, explain what the interval means. We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. Using the same x ¯ , s x , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know? The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean. Using the same x ¯ , s x , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why? Homework In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice Calculate p ′. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag. State the confidence interval. Sketch the graph. Calculate the error bound. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. State the confidence interval. Sketch the graph. Calculate the error bound. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why? 8629 6944 35 34 t 34 CI: (6244, 11,014) EB = 2385 It will become smaller Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean time wasted. State the confidence interval. Sketch the graph. Calculate the error bound. Explain in a complete sentence what the confidence interval means. A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean length of time. State the confidence interval. Sketch the graph. Calculate the error bound. What does it mean to be “95% confident” in this problem? x ¯ = 2.51 s x = 0.318 n = 9 n - 1 = 8 the effective length of time for a tranquilizer the mean effective length of time of tranquilizers from a sample of nine patients We need to use a Student’s-t distribution, because we do not know the population standard deviation. CI: (2.27, 2.76) Answers may vary. EBM : 0.25 If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 99% confidence interval for the population mean length of time using training wheels. State the confidence interval. Sketch the graph. Calculate the error bound. Why would the error bound change if the confidence level were lowered to 90%? The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns. The FEC has reported financial information for 556 Leadership PACs that operating during a certain election cycle. The following table shows the total receipts during this cycle for a random selection of 30 Leadership PACs. $46,500.00 $0 $40,966.50 $105,887.20 $5,175.00 $29,050.00 $19,500.00 $181,557.20 $31,500.00 $149,970.80 $2,555,363.20 $12,025.00 $409,000.00 $60,521.70 $18,000.00 $61,810.20 $76,530.80 $119,459.20 $0 $63,520.00 $6,500.00 $502,578.00 $705,061.10 $708,258.90 $135,810.00 $2,000.00 $2,000.00 $0 $1,287,933.80 $219,148.30 x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the specific election cycle. Use the Student's t-distribution. x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29 CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04 α 2 = 0.02 t α 2 = t 0.02 = 2.150 E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 ) ~ $ 204 , 561.66 x ¯ - EBM = $251,854.23 - $204,561.66 = $47,292.57 x ¯ + EBM = $251,854.23+ $204,561.66 = $456,415.89 We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the specific election cycle lies between $47,292.57 and $456,415.89. Alternate Solution Enter the data as a list. Press STAT and arrow over to TESTS . Arrow down to 8:TInterval . Press ENTER . Arrow to Data and press ENTER . Arrow down and enter the name of the list where the data is stored. Enter Freq : 1 Enter C-Level : 0.96 Arrow down to Calculate and press Enter . The 96% confidence interval is ($47,262, $456,447). The difference between solutions arises from rounding differences. Forbes magazine published data on the best small firms in a certain year. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 49 Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. x ¯ = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. State the confidence interval. Sketch the graph. Calculate the error bound. x ¯ = 11.6 s x = 4.1 n = 225 n - 1 = 224 X is the number of unoccupied seats on a single flight. X ¯ is the mean number of unoccupied seats from a sample of 225 flights. We will use a Student’s-t distribution, because we do not know the population standard deviation. CI: (11.12 , 12.08) Answers may vary. EBM : 0.48 In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Explain your choice. Define the random variable X ¯ in words. Construct a 95% confidence interval for the population mean cost of a used car. State the confidence interval. Sketch the graph. Calculate the error bound. Explain what a “95% confidence interval” means for this study. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. State the confidence interval. Sketch the graph. Calculate the error bound. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies. Calculate the mean. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not? CI: (7.64 , 9.36) EBM : 0.86 The sample should have been increased. Answers will vary. Answers will vary. Answers will vary. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. x ¯ = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean worth of coupons. State the confidence interval. Sketch the graph. Calculate the error bound. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. Find the 95% Confidence Interval for the true population mean for the amount of soda served. (12.42, 14.18) (12.32, 14.29) (12.50, 14.10) Impossible to determine b What is the error bound? 0.87 1.98 0.99 1.74 Degrees of Freedom ( df ) the number of objects in a sample that are free to vary Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 / 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation, notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t -Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student; the major characteristics of the random variable (RV) are: It is continuous and assumes any real values. The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. It approaches the standard normal distribution as n get larger. There is a \"family\" of t–distributions: each representative of the family is completely defined by the number of degrees of freedom, which is one less than the number of data.", "section": "A Single Population Mean using the Student t Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "A Population Proportion During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03). Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval, the sample size, the error bound , and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution . (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B ( n , p ) where n is the number of trials and p is the probability of a success. To form a proportion, take X , the random variable for the number of successes and divide it by n , the number of trials (or the sample size). The random variable p ′ (read \"P prime\") is that proportion, P ′ = X n (Sometimes the random variable is denoted as P ^ , read \"P hat\".) When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial. X ~ N ( n p , n p q ) If we divide the random variable, the mean, and the standard deviation by n , we get a normal distribution of proportions with p ′ , called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by n .) X n = P ′ ~ N ( n p n , n p q n ) Using algebra to simplify : n p q n = p q n p ′ follows a normal distribution for proportions : X n = P ′ ~ N ( p , p q n ) The confidence interval has the form ( p ′ – EBP , p ′ + EBP ). EBP is error bound for the proportion. p ′ = x n p ′ = the estimated proportion of successes ( p ′ is a point estimate for p , the true proportion.) x = the number of successes n = the size of the sample The error bound for a proportion is E B P = ( z α 2 ) ( p ′ q ′ n ) where q ′ = 1 – p ′ This formula is similar to the error bound formula for a mean, except that the \"appropriate standard deviation\" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is σ n . For a proportion, the appropriate standard deviation is p q n . However, in the error bound formula, we use p ′ q ′ n as the standard deviation, instead of p q n . In the error bound formula, the sample proportions p ′ and q ′ are estimates of the unknown population proportions p and q . The estimated proportions p ′ and q ′ are used because p and q are not known. The sample proportions p ′ and q ′ are calculated from the data: p ′ is the estimated proportion of successes, and q ′ is the estimated proportion of failures. The confidence interval can be used only if the number of successes np ′ and the number of failures nq ′ are both greater than five. NOTE For the normal distribution of proportions, the z -score formula is as follows. If P ′ ~ N ( p , p q n ) then the z -score formula is z = p ′ − p p q n Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have smartphones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have smartphones. Of the 500 people surveyed, 421 responded yes - they own smartphones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have smartphones. The first solution is step-by-step. The second solution uses a function of the TI-83, 83+ or 84 calculators. Let X = the number of people in the sample who have cell phones. X is binomial. X ~ B ( 500 , 421 500 ) . To calculate the confidence interval, you must find p ′ , q ′ , and EBP . n = 500 x = the number of successes = 421 p ′ = x n = 421 500 = 0.842 p ′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. q ′ = 1 - p ′ = 1 - 0.842 = 0.158 Since CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ( α 2 ) = 0.025. Then z α 2 = z 0.025 = 1.96 Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z 0.025 . Remember that the area to the right of z 0.025 is 0.025 and the area to the left of z 0.025 is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table. E B P = ( z α 2 ) p ′ q ′ n = ( 1.96 ) ( 0.842 ) ( 0.158 ) 500 = 0.032 p ′ – E B P = 0.842 – 0.032 = 0.81 p ′ + E B P = 0.842 + 0.032 = 0.874 The confidence interval for the true binomial population proportion is ( p ′ – E B P , p ′ + E B P ) = ( 0 . 810 , 0 . 874 ) . Interpretation: We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have smartphones. Explanation of 95% Confidence Level: Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have smartphones. Press STAT and arrow over to TESTS . Arrow down to A:1-PropZint . Press ENTER . Arrow down to x and enter 421. Arrow down to n and enter 500. Arrow down to C-Level and enter .95. Arrow down to Calculate and press ENTER . The confidence interval is (0.81003, 0.87397). Try It Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval. The first solution is step-by-step. The second solution uses a function of the TI-83, 83+, or 84 calculators. x = 300 and n = 500 p ′ = x n = 300 500 = 0.600 q ′ = 1 - p ′ = 1 - 0.600 = 0.400 Since CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10 ( α 2 ) = 0.05 z α 2 = z 0.05 = 1.645 Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z 0.05 . Remember that the area to the right of z 0.05 is 0.05 and the area to the left of z 0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. E B P = ( z α 2 ) p ′ q ′ n = ( 1.645 ) ( 0.60 ) ( 0.40 ) 500 = 0.036 p ′ – E B P = 0.60 − 0.036 = 0.564 p ′ + E B P = 0.60 + 0.036 = 0.636 The confidence interval for the true binomial population proportion is ( p ′ – EBP , p ′ + EBP ) = (0.564,0.636). Interpretation: We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%. Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters. Explanation of 90% Confidence Level: Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters. Press STAT and arrow over to TESTS . Arrow down to A:1-PropZint . Press ENTER . Arrow down to x and enter 300. Arrow down to n and enter 500. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER . The confidence interval is (0.564, 0.636). Try It A student polls their school to see if students in the school district are for or against the new legislation regarding school uniforms. They survey 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. “Plus Four” Confidence Interval for p There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed. Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2. Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten. A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke. Six students out of 25 reported smoking within the past week, so x = 6 and n = 25. Because we are using the “plus-four” method, we will use x = 6 + 2 = 8 and n = 25 + 4 = 29. p ′ = x n = 8 29 ≈ 0.276 q ′ = 1 – p ′ = 1 – 0.276 = 0.724 Since CL = 0.95, we know α = 1 – 0.95 = 0.05 and α 2 = 0.025. z 0.025 = 1.96 E B P = ( z α 2 ) p ′ q ′ n = ( 1.96 ) 0.276 ( 0.724 ) 29 ≈ 0.163 p ′ – EBP = 0.276 – 0.163 = 0.113 p ′ + EBP = 0.276 + 0.163 = 0.439 We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439. Press STAT and arrow over to TESTS . Arrow down to A:1-PropZint . Press ENTER . REMINDER Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials. Arrow down to x and enter eight. Arrow down to n and enter 29. Arrow down to C-Level and enter 0.95. Arrow down to Calculate and press ENTER. The confidence interval is (0.113, 0.439). Try It Out of a random sample of 65 first-year students at State University, 31 students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of first-year students at State University who have declared a major. A marketing research firm conducted a survey regarding the potential purchase of electric vehicles among adults aged 18–29. In a group of 50 adults aged 18–29, 13 of them reported they would consider the purchase of an electric vehicle. Use the “plus four” method to find a 90% confidence interval for the true proportion of adults aged 18–29 who would consider the purchase of an electric vehicle. Using “plus-four,” we have x = 13 + 2 = 15 and n = 50 + 4 = 54. p ′ = 15 54 ≈ 0.278 q ′ = 1 – p ′ = 1 - 0.241 = 0.722 Since CL = 0.90, we know α = 1 – 0.90 = 0.10 and α 2 = 0.05. z 0.05 = 1.645 E B P = ( z α 2 ) ( p ′ q ′ n ) = ( 1.645 ) ( ( 0.278 ) ( 0.722 ) 54 ) ≈ 0.100 p ′ – EBP = 0.278 – 0.100 = 0.178 p ′ + EBP = 0.278 + 0.100 = 0.378 We are 90% confident that between 17.8% and 37.8% of adults aged 18–29 would consider the purchase of an electric vehicle. Press STAT and arrow over to TESTS . Arrow down to A:1-PropZint . Press ENTER . Arrow down to x and enter 15. Arrow down to n and enter 54. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER . The confidence interval is (0.178, 0.378). Try It The survey referenced in talked to adults aged 18–29 in smaller focus groups but also interviewed additional individuals over the phone. When the study was complete, 588 adults aged 18–29 had answered the question about their potential purchase of an electric vehicle with 159 saying that they would consider such a purchase. Use the “plus-four” method to find a 90% confidence interval for the true proportion of adults aged 18–29 who would consider the purchase of an electric vehicle based on this larger sample. Compare the results to those in . Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is E B P = ( z α 2 ) ( p ′ q ′ n ) Solving for n gives you an equation for the sample size. n = ( z α 2 ) 2 ( p ′ q ′ ) E B P 2 Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones. From the problem, we know that EBP = 0.03 (3%=0.03) and z α 2 z 0.05 = 1.645 because the confidence level is 90%. However, in order to find n , we need to know the estimated (sample) proportion p ′. Remember that q ′ = 1 – p ′. But, we do not know p ′ yet. Since we multiply p ′ and q ′ together, we make them both equal to 0.5 because p ′ q ′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n . This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size n , use the formula and make the substitutions. n = z 2 p ′ q ′ E B P 2 gives n = 1.645 2 ( 0.5 ) ( 0.5 ) 0.03 2 = 751.7 Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones. Try It Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones? References Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling. Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013). Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, and Meredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx (accessed July 2, 2013). Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.” Pew Research Center: Internet and American Life Project. Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013). Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at http://www.gallup.com/poll/162758/three-four-workers-plan-work-past-retirement-age.aspx (accessed July 2, 2013). The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2, 2013). Zogby. “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. Available online at http://www.zogbyanalytics.com/news/299-americans-neither-worried-nor-prepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). “52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process (accessed July 2, 2013). Chapter Review Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning. Let p ′ represent the sample proportion, x/n , where x represents the number of successes and n represents the sample size. Let q ′ = 1 – p ′ . Then the confidence interval for a population proportion is given by the following formula: (lower bound, upper bound) = ( p ′ – E B P , p ′ + E B P ) = ( p ′ – z p ′ q ′ n , p ′ + z p ′ q ′ n ) The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate p ′ = x + 2 n + 4 , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples. Formula Review p′ = x / n where x represents the number of successes and n represents the sample size. The variable p ′ is the sample proportion and serves as the point estimate for the true population proportion. q ′ = 1 – p ′ p ′ ~ N ( p , p q n ) The variable p ′ has a binomial distribution that can be approximated with the normal distribution shown here. EBP = the error bound for a proportion = z α 2 p ′ q ′ n Confidence interval for a proportion: (lower bound, upper bound) = ( p ′ – E B P , p ′ + E B P ) = ( p ′ – z p ′ q ′ n , p ′ + z p ′ q ′ n ) n = z α 2 2 p ′ q ′ E B P 2 provides the number of participants needed to estimate the population proportion with confidence 1 - α and margin of error EBP . Use the normal distribution for a single population proportion p ′ = x n E B P = ( z α 2 ) p ′ q ′ n p ′ + q ′ = 1 The confidence interval has the format ( p ′ – EBP , p ′ + EBP ). x ¯ is a point estimate for μ p ′ is a point estimate for ρ s is a point estimate for σ Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05? If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why? The sample size needed would increase. As the confidence level increases, α decreases and z ( a 2 ) increases. To maintain the same error bound, the size of the sample needs to increase. Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. Identify the following: x = ______ n = ______ p ′ = ______ Define the random variables X and p ′ in words. X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P ′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. Which distribution should you use for this problem? Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.5321, 0.6679) EBM : 0.0679 List two difficulties the company might have in obtaining random results, if this survey were done by email. Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P ′ in words. X is the number of “successes” where an executive prefers a truck. P ′ is the percentage of executives sampled who prefer a truck. Which distribution should you use for this problem? Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.19432, 0.33068) EBM : 0.0707 Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given in the survey is ±2%. Explain what the ±2% means. The sampling error means that the true mean can be 2% above or below the sample mean. Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important. Define the random variable X in words. Define the random variable P ′ in words. P ′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. Which distribution should you use for this problem? Construct a 90% confidence interval, and state the confidence interval and the error bound. CI: (0.62735, 0.67265) EBM : 0.02265 What would happen to the confidence interval if the level of confidence were 95%? Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. What is being counted? The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class. In words, define the random variable X . Calculate the following: x = _______ n = _______ p ′ = _______ x = 64 n = 80 p ′ = 0.8 State the estimated distribution of X . X ~________ Define a new random variable P ′. What is p ′ estimating? p In words, define the random variable P ′. State the estimated distribution of P ′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet. P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 ) . (0.72171, 0.87829). How much area is in both tails (combined)? How much area is in each tail? 0.04 Calculate the following: lower limit upper limit error bound The 92% confidence interval is _______. (0.72; 0.88) Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. In one complete sentence, explain what the interval means. With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%. Using the same p ′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know? Using the same p ′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why? The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? Homework Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03? If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 1,068 The sample size would need to be increased since the critical value increases as the confidence level increases. Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. x = __________ n = __________ p ′ = __________ Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion who claim they always buckle up. State the confidence interval. Sketch the graph. Calculate the error bound. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job. State the confidence interval. Sketch the graph. Calculate the error bound. X = the number of people who feel that the president is doing an acceptable job; P ′ = the proportion of people in a sample who feel that the president is doing an acceptable job. N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 ) CI: (0.59, 0.63) Answers may vary. EBM : 0.02 An article regarding interracial dating and marriage recently appeared in the Washington Post . Of the 1,709 randomly selected adults, 315 identified themselves as Hispanic/Latino, 323 identified themselves as Black, 254 identified themselves as Asian, and 779 identified themselves as White. In this survey, 86% of Black people said that they would welcome a White person into their families. Among Asian people, 77% would welcome a White person into their families, 71% would welcome a Hispanic/Latino person, and 66% would welcome a Black person. We are interested in finding the 95% confidence interval for the percent of all Black adults who would welcome a White person into their families. Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval. State the confidence interval. Sketch the graph. Calculate the error bound. Refer to the information in . Construct three 95% confidence intervals. percent of all Asians who would welcome a White person into their families. percent of all Asians who would welcome a Hispanic/Latino into their families. percent of all Asians who would welcome a Black person into their families. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? (0.72, 0.82) (0.65, 0.76) (0.60, 0.72) Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Hispanic/Latino person into their families. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families. Stanford University conducted a study of whether running is healthy for people over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period. State the confidence interval. Sketch the graph. Calculate the error bound. Explain what a “97% confidence interval” means for this study. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine . One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem. State the confidence interval. Sketch the graph. Calculate the error bound. Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. X = the number of adult Americans who feel that crime is the main problem; p ′ = the proportion of adult Americans who feel that crime is the main problem Since we are estimating a proportion, given p ′ = 0.2 and n = 1000, the distribution we should use is N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) . CI: (0.18, 0.22) Answers may vary. EBM : 0.02 One way to lower the sampling error is to increase the sample size. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%. Refer to . Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools. State the confidence interval. Sketch the graph. Calculate the error bound. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. Use the following information to answer the next three exercises: According to a Field Poll, 79% of Georgia adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing the state. We wish to construct a 90% confidence interval for the true proportion of Georgia adults who feel that education and the schools is one of the top issues facing Georgia. A point estimate for the true population proportion is: 0.90 1.27 0.79 400 c A 90% confidence interval for the population proportion is _______. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) The error bound is approximately _____. 1.581 0.791 0.059 0.030 d Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not. Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. 0.6614 0.3386 173 338 a In a specific year, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error. Determine the estimated proportion from the sample. Determine the sample size. Identify CL and α . Calculate the error bound based on the information provided. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values. Create a confidence interval for the results of this study. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? A national survey of 1,000 adults was conducted by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. Find the point estimate and the error bound for this confidence interval. Can we (with 95% confidence) conclude that more than half of all American adults believe this? Use the point estimate from part a and n = 1,000 to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education. Can we (with 75% confidence) conclude that at least half of all American adults believe this? p ′ = (0 .55 + 0 .49) 2 = 0.52; EBP = 0.55 - 0.52 = 0.03 No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125 z α 2 = 1.150 . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) E B P = ( 1.150 ) 0.52 ( 0.48 ) 1 , 000 ≈ 0.018 ( p ′ - EBP , p ′ + EBP ) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Alternate Solution STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75. Answer is (0.502, 0.538) Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence. Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. This survey was conducted through automated telephone interviews. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2020 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview? CL = 0.95 α = 1 – 0.95 = 0.05 α 2 = 0.025 z α 2 = 1.96. Use p ′ = q ′ = 0.5. n = z α 2 2 p ′ q ′ E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.05 2 = 384.16 You need to interview at least 385 students to estimate the proportion to within 5% at 95% confidence. In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem. Binomial Distribution a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B ( n , p ). The mean is μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = n x p x q n − x .", "section": "A Population Proportion", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Confidence Interval (Home Costs) Confidence Interval (Home Costs) Class Time: Names: Student Learning Outcomes The student will calculate the 90% confidence interval for the mean cost of a home in the area in which this school is located. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval. Collect the Data Check the Real Estate section in your local newspaper. Record the sale prices for 35 randomly selected homes recently listed in the county. NOTE Many newspapers list them only one day per week. Also, we will assume that homes come up for sale randomly. Complete the table: Describe the Data Compute the following: x ¯ = _____ s x = _____ n = _____ In words, define the random variable X ¯ . State the estimated distribution to use. Use both words and symbols. Find the Confidence Interval Calculate the confidence interval and the error bound. Confidence Interval: _____ Error Bound: _____ How much area is in both tails (combined)? α = _____ How much area is in each tail? α 2 = _____ Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample mean. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data on the first page and count how many of the data values lie within the confidence interval. What percent is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%. Describe the Confidence Interval In two to three complete sentences, explain what a confidence interval means (in general), as if you were talking to someone who has not taken statistics. In one to two complete sentences, explain what this confidence interval means for this particular study. Use the Data to Construct Confidence Intervals Using the given information, construct a confidence interval for each confidence level given. Confidence level EBM/Error Bound Confidence Interval 50% 80% 95% 99% What happens to the EBM as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens.", "section": "Confidence Interval (Home Costs)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Confidence Interval (Place of Birth) Confidence Interval (Place of Birth) Class Time: Names: Student Learning Outcomes The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state. The student will interpret confidence intervals. The student will determine the effects of changing conditions on the confidence interval. Collect the Data Survey the students in your class, asking them if they were born in this state. Let X = the number that were born in this state. n = ____________ x = ____________ In words, define the random variable P′ . State the estimated distribution to use. Find the Confidence Interval and Error Bound Calculate the confidence interval and the error bound. Confidence Interval: _____ Error Bound: _____ How much area is in both tails (combined)? α = _____ How much area is in each tail? α 2 = _____ Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample proportion. Describe the Confidence Interval In two to three complete sentences, explain what a confidence interval means (in general), as though you were talking to someone who has not taken statistics. In one to two complete sentences, explain what this confidence interval means for this particular study. Construct a confidence interval for each confidence level given. Confidence level EBP/Error Bound Confidence Interval 50% 80% 95% 99% What happens to the EBP as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens.", "section": "Confidence Interval (Place of Birth)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Confidence Interval (Women's Heights) Confidence Interval (Women's Heights) Class Time: Names: Student Learning Outcomes The student will calculate a 90% confidence interval using the given data. The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean. Given: Heights of 100 Women (in Inches) 59.4 71.6 69.3 65.0 62.9 66.5 61.7 55.2 67.5 67.2 63.8 62.9 63.0 63.9 68.7 65.5 61.9 69.6 58.7 63.4 61.8 60.6 69.8 60.0 64.9 66.1 66.8 60.6 65.6 63.8 61.3 59.2 64.1 59.3 64.9 62.4 63.5 60.9 63.3 66.3 61.5 64.3 62.9 60.6 63.8 58.8 64.9 65.7 62.5 70.9 62.9 63.1 62.2 58.7 64.7 66.0 60.5 64.7 65.4 60.2 65.0 64.1 61.1 65.3 64.6 59.2 61.4 62.0 63.5 61.4 65.5 62.3 65.5 64.7 58.8 66.1 64.9 66.9 57.9 69.8 58.5 63.4 69.2 65.9 62.2 60.0 58.1 62.5 62.4 59.1 66.4 61.2 60.4 58.7 66.7 67.5 63.2 56.6 67.7 62.5 lists the heights of 100 women. Use a random number generator to select ten data values randomly. Calculate the sample mean and the sample standard deviation. Assume that the population standard deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your sample of ten values. Write the confidence interval you obtained in the first space of . Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into . 90% Confidence Intervals __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Discussion Questions The actual population mean for the 100 heights given is μ = 63.4. Using the class listing of confidence intervals, count how many of them contain the population mean μ ; i.e., for how many intervals does the value of μ lie between the endpoints of the confidence interval? Divide this number by the total number of confidence intervals generated by the class to determine the percent of confidence intervals that contains the mean μ . Write this percent here: _____________. Is the percent of confidence intervals that contain the population mean μ close to 90%? Suppose we had generated 100 confidence intervals. What do you think would happen to the percent of confidence intervals that contained the population mean? When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated based on that data. How many of the 100 data values lie within your confidence interval? What percent is this? Is this percent close to 90%? Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem. Suppose you obtained the heights of ten women and calculated a confidence interval from this information. Without knowing the population mean μ , would you have any way of knowing for certain if your interval actually contained the value of μ ? Explain.", "section": "Confidence Interval (Women's Heights)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (credit: modification of work “black and white, view, dog, animal, fur, mammal, garden, head, vertebrate, beautiful, elegant, rush, attention, dalmatian, dog breed, dog head, noble, stains, animal portrait, head drawing, spotty, hundeportrait, dalmatians”/ Pxhere, Public Domain) Chapter Objectives By the end of this chapter, the student should be able to: Differentiate between Type I and Type II Errors Describe hypothesis testing in general and in practice Conduct and interpret hypothesis tests for a single population mean, population standard deviation known. Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown. Conduct and interpret hypothesis tests for a single population proportion. One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year. A statistician will make a decision about these claims. This process is called \" hypothesis testing .\" A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests. Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will: Set up two contradictory hypotheses. Collect sample data (in homework problems, the data or summary statistics will be given to you). Determine the correct distribution to perform the hypothesis test. Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis. Make a decision and write a meaningful conclusion. NOTE To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Appendix E . Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: The desired confidence level. Information that is known about the distribution (for example, known standard deviation). The sample and its size. Hypothesis Testing Based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Null and Alternative Hypotheses The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints. H 0 : The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action. H a : The alternative hypothesis: It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 . This is usually what the researcher is trying to prove. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are \"reject H 0 \" if the sample information favors the alternative hypothesis or \"do not reject H 0 \" or \"decline to reject H 0 \" if the sample information is insufficient to reject the null hypothesis. Mathematical Symbols Used in H 0 and H a : H 0 H a equal (=) not equal (≠) or greater than (>) or less than (<) greater than or equal to (≥) less than (<) less than or equal to (≤) more than (>) NOTE H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis. H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ .30 H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30 Try It A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses. We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are: H 0 : μ = 2.0 H a : μ ≠ 2.0 Try It We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66 We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are: H 0 : μ ≥ 5 H a : μ < 5 Try It We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45 In an issue of U. S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066 Try It On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : p __ 0.40 H a : p __ 0.40 Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class. Chapter Review In a hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties. Formula Review H 0 and H a are contradictory. If H o has: equal (=) greater than or equal to (≥) less than or equal to (≤) then H a has: not equal (≠) or greater than (>) or less than (<) less than (<) greater than (>) If α ≤ p -value, then do not reject H 0 . If α > p -value, then reject H 0 . α is preconceived. Its value is set before the hypothesis test starts. The p -value is calculated from the data. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words. The random variable is the mean Internet speed in Megabits per second. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses. The American family has an average of two children. What is the random variable? Describe in words. The random variable is the mean number of children an American family has. The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words. The random variable is the proportion of people picked at random in Times Square visiting the city. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses. In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses. H 0 : p = 0.42 H a : p < 0.42 Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal. H 0 : ________ H a : ________ A random survey of 75 student loan recipients revealed that the mean length of time in repayment is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time in repayment could likely be 15 years, what would the null and alternative hypotheses be? H 0 : __________ H a : __________ H 0 : μ = 15 H a : μ ≠ 15 The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5% of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population, what would the null and alternative hypotheses be? H 0 : ________ H a : ________ Homework Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, H 0 , and the alternative hypothesis. H a , in terms of the appropriate parameter ( μ or p ). The mean number of years Americans work before retiring is 34. At most 60% of Americans vote in presidential elections. The mean starting salary for San Jose State University graduates is at least $100,000 per year. Twenty-nine percent of high school seniors get drunk each month. Fewer than 5% of adults ride the bus to work in Los Angeles. The mean number of cars a person owns in her lifetime is not more than ten. About half of Americans prefer to live away from cities, given the choice. Europeans have a mean paid vacation each year of six weeks. The chance of developing breast cancer is under 11% for females. Private universities' mean tuition cost is more than $20,000 per year. H 0 : μ = 34; H a : μ ≠ 34 H 0 : p ≤ 0.60; H a : p > 0.60 H 0 : μ ≥ 100,000; H a : μ < 100,000 H 0 : p = 0.29; H a : p ≠ 0.29 H 0 : p = 0.05; H a : p < 0.05 H 0 : μ ≤ 10; H a : μ > 10 H 0 : p = 0.50; H a : p ≠ 0.50 H 0 : μ = 6; H a : μ ≠ 6 H 0 : p ≥ 0.11; H a : p < 0.11 H 0 : μ ≤ 20,000; H a : μ > 20,000 Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin? The alternative hypothesis is: p < 0.30 p ≤ 0.30 p ≥ 0.30 p > 0.30 A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is: p = 0.20 p > 0.20 p < 0.20 p ≤ 0.20 c Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are: H o : x ¯ = 4.5, H a : x ¯ > 4.5 H o : μ ≥ 4.5, H a : μ < 4.5 H o : μ = 4.75, H a : μ > 4.75 H o : μ = 4.5, H a : μ > 4.5 References Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm. Hypothesis a statement about the value of a population parameter, in case of two hypotheses, the statement assumed to be true is called the null hypothesis (notation H 0 ) and the contradictory statement is called the alternative hypothesis (notation H a ).", "section": "Null and Alternative Hypotheses", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Outcomes and the Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis H 0 and the decision to reject or not. The outcomes are summarized in the following table: ACTION H 0 IS ACTUALLY ... True False Do not reject H 0 Correct Outcome Type II error Reject H 0 Type I Error Correct Outcome The four possible outcomes in the table are: The decision is not to reject H 0 when H 0 is true (correct decision). The decision is to reject H 0 when H 0 is true (incorrect decision known as a Type I error ). The decision is not to reject H 0 when, in fact, H 0 is false (incorrect decision known as a Type II error ). The decision is to reject H 0 when H 0 is false ( correct decision whose probability is called the Power of the Test ). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. α = probability of a Type I error = P (Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P (Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. α and β should be as small as possible because they are probabilities of errors. They are rarely zero. The Power of the Test is 1 – β . Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors. Suppose the null hypothesis, H 0 , is: Navah's rock climbing equipment is safe. Type I error : Navah thinks that her rock climbing equipment may not be safe when, in fact, it really is safe. Type II error : Navah thinks that her rock climbing equipment may be safe when, in fact, it is not safe. α = probability that Navah thinks her rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Navah thinks her rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Navah thinks her rock climbing equipment is safe, she will go ahead and use it.) Try It Suppose the null hypothesis, H 0 , is: the blood cultures contain no traces of pathogen X . State the Type I and Type II errors. Suppose the null hypothesis, H 0 , is: The victim of an automobile accident is alive when they arrive at the emergency room of a hospital. Type I error : The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error : The emergency crew does not know if the victim is alive when, in fact, the victim is dead. α = probability that the emergency crew thinks the victim is dead when, in fact, the victim is really alive = P (Type I error). β = probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P (Type II error). The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat them.) Try It Suppose the null hypothesis, H 0 , is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II? A company called Genetic Labs claims to be able to increase the likelihood that a pregnancy will result in a male being born. Statisticians want to test the claim. Suppose that the null hypothesis, H 0 , is: Genetic Labs has no effect on sex outcome. Type I error : This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that Genetic Labs influences the sex outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α . Type II error : This results when we fail to reject a false null hypothesis. In context, we would state that Genetic Labs does not influence the sex outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β . The error of greater consequence would be the Type I error since people would use the Genetic Labs product in hopes of increasing the chances of having a male. Try It “Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence. A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious? Type I : A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%. Type II : A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%. In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. Try It Determine both Type I and Type II errors for the following scenario: Assume a null hypothesis, H 0 , that states the percentage of adults with jobs is at least 88%. Identify the Type I and Type II errors from these four statements. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88% Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%. Chapter Review In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters α and β , for a Type I and a Type II error respectively. The power of the test, 1 – β , quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable. Formula Review α = probability of a Type I error = P (Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P (Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000. A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences. For Exercise 9.12 , what are α and β in words? α = the probability that you think the bag cannot withstand -15 degrees F, when in fact it can β = the probability that you think the bag can withstand -15 degrees F, when in fact it cannot In words, describe 1 – β For Exercise 9.12 . A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H 0 , is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences. Type I: The procedure will go well, but the doctors think it will not. Type II: The procedure will not go well, but the doctors think it will. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H 0 , is: the surgical procedure will go well. Which is the error with the greater consequence? The power of a test is 0.981. What is the probability of a Type II error? 0.019 A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H 0 , is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H 0 , is: the sample does not contain E-coli. The probability that the sample does not contain E-coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E-coli, but the microbiologist thinks it does not is 0.002. What is the power of this test? 0.998 A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H 0 , is: the sample contains E-coli. Which is the error with the greater consequence? Homework State the Type I and Type II errors in complete sentences given the following statements. The mean number of years Americans work before retiring is 34. At most 60% of Americans vote in presidential elections. The mean starting salary for San Jose State University graduates is at least $100,000 per year. Twenty-nine percent of high school seniors get drunk each month. Fewer than 5% of adults ride the bus to work in Los Angeles. The mean number of cars a person owns in their lifetime is not more than ten. About half of Americans prefer to live away from cities, given the choice. Europeans have a mean paid vacation each year of six weeks. The chance of developing breast cancer is under 11% for females. Private universities mean tuition cost is more than $20,000 per year. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years. Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do. Type I error: We conclude that the mean number of cars a person owns in their lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in their lifetime is not more than 10 when, in fact, it is more than 10. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of females who develop breast cancer is at least 11%, when in fact it is less than 11%. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000. For statements a-j in Exercise 9.66 , answer the following in complete sentences. State a consequence of committing a Type I error. State a consequence of committing a Type II error. When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error? To conclude the drug is safe when in, fact, it is unsafe. Not to conclude the drug is safe when, in fact, it is safe. To conclude the drug is safe when, in fact, it is safe. Not to conclude the drug is unsafe when, in fact, it is unsafe. b A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________. at least 20%, when in fact, it is less than 20%. 20%, when in fact, it is 20%. less than 20%, when in fact, it is at least 20%. less than 20%, when in fact, it is less than 20%. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours is more than seven hours. is at most seven hours. is at least seven hours. is less than seven hours. d Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is: to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same to conclude that the mean hours per week currently is 4.5, when in fact, it is higher to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher Type I Error The decision is to reject the null hypothesis when, in fact, the null hypothesis is true. Type II Error The decision is not to reject the null hypothesis when, in fact, the null hypothesis is false.", "section": "Outcomes and the Type I and Type II Errors", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Probability Distribution Needed for Hypothesis Testing Earlier in the course, we discussed sampling distributions. Particular distributions are associated with various types of hypothesis testing. The following table summarizes various hypothesis tests and corresponding probability distributions that will be used to conduct the test (based on the assumptions shown below): Type of Hypothesis Test Population Parameter Estimated value (point estimate) Probability Distribution Used Hypothesis test for the mean, when the population standard deviation is known Population mean μ Sample mean x ¯ Normal distribution, X ¯ ~ N ( μ X , σ X n ) Hypothesis test for the mean, when the population standard deviation is unknown and the distribution of the sample mean is approximately normal Population mean μ Sample mean x ¯ Student’s t-distribution, t d f Hypothesis test for proportions Population proportion p Sample proportion p ' Normal distribution, P ' ~ N ( p , p · q n ) Assumptions When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed , or your sample size is sufficiently large. You know the value of the population standard deviation , which, in reality, is rarely known. When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t -test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t -test will work even if the population is not approximately normally distributed). When you perform a hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution : there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( n p > 5 and n q > 5 ). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and σ = p q n . Remember that q = 1 - p . Chapter Review In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied. When testing for a single population mean: A Student's t -test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation. When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of successes and the mean number of failures satisfy the conditions: np > 5 and nq > 5 where n is the sample size, p is the probability of a success, and q is the probability of a failure. Formula Review If there is no given preconceived α , then use α = 0.05. Types of Hypothesis Tests Single population mean, known population variance (or standard deviation): Normal test . Single population mean, unknown population variance (or standard deviation): Student's t -test . Single population proportion: Normal test . For a single population mean , we may use a normal distribution with the following mean and standard deviation. Means: μ = μ x ¯ and σ x ¯ = σ x n A single population proportion , we may use a normal distribution with the following mean and standard deviation. Proportions: µ = p and σ = p q n . Which two distributions can you use for hypothesis testing for this chapter? A normal distribution or a Student’s t -distribution Which distribution do you use when you are testing a population mean and the population standard deviation is known? Assume a normal distribution, with n ≥ 30. Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large. Use a Student’s t -distribution A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal. A population has a mean of 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test? a normal distribution for a single population mean It is thought that 42% of respondents in a taste test would prefer Brand A . In a particular test of 100 people, 39% preferred Brand A . What distribution should you use to perform a hypothesis test? You are performing a hypothesis test of a single population mean using a Student’s t -distribution. What must you assume about the distribution of the data? It must be approximately normally distributed. You are performing a hypothesis test of a single population mean using a Student’s t -distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq ? They must both be greater than five. You are performing a hypothesis test of a single population proportion. You find out that np is less than five. What must you do to be able to perform a valid hypothesis test? You are performing a hypothesis test of a single population proportion. The data come from which distribution? binomial distribution Homework It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X ¯ ~ ________________ N ( 7.24 , 1.93 22 ) N ( 7.24 , 1.93 ) t 22 t 21 d Binomial Distribution a discrete random variable (RV) that arises from Bernoulli trials. There are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in n trials. The notation is: X ~ B(n, p) μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = ( n x ) p x q n − x . Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 , where μ is the mean of the distribution, and σ is the standard deviation, notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Student's t -Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. The major characteristics of the random variable (RV) are: It is continuous and assumes any real values. The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. It approaches the standard normal distribution as n gets larger. There is a \"family\" of t distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items.", "section": "Probability Distribution Needed for Hypothesis Testing", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Rare Events, the Sample, Decision and Conclusion Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test. Rare Events Suppose you make an assumption about a property of the population (this assumption is the null hypothesis ). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption —it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.) For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is 1 200 = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A \"rare event\" has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket. Using the Sample to Test the Null Hypothesis Use the sample data to calculate the actual probability of getting the test result, called the p -value . The p -value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample. A large p -value calculated from the data indicates that we should not reject the null hypothesis . The smaller the p -value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. Draw a graph that shows the p -value. The hypothesis test is easier to perform if you use a graph because you see the problem more clearly. Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal. The null hypothesis could be H 0 : μ ≤ 15 The alternate hypothesis is H a : μ > 15 The words \"is more than\" translates as a \">\" so \" μ > 15\" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis. By the Central Limit Theorem , the distribution of sample means, for samples of 10 loaves, will be normal with mean μ = 15 and standard deviation σ n = 0.5 10 = 0.16 . Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p -value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm. The p -value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means. p -value= P ( x ¯ > 17) which is approximately zero. A p -value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone) , we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population of the baker's loaves of bread is greater than 15 cm. Try It A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5. H 0 : μ ≤ 12 H a : μ > 12 The p -value is 0.0013 Draw a graph that shows the p -value. Decision and Conclusion A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p -value and a preset or preconceived α (also called a \"significance level\") . A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. When you make a decision to reject or not reject H 0 , do as follows: If α > p -value, reject H 0 . The results of the sample data are significant. There is sufficient evidence to conclude that H 0 is an incorrect belief and that the alternative hypothesis , H a , may be correct. If α ≤ p -value, do not reject H 0 . The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis, H a , may be correct. When you \"do not reject H 0 \", it does not mean that you should believe that H 0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H o . Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem. When using the p -value to evaluate a hypothesis test, it is sometimes useful to use the following memory device If the p -value is low, the null must go. If the p -value is high, the null must fly. This memory aid relates a p -value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p -value higher than the established alpha (the p is high) as not rejecting the null hypothesis. Fill in the blanks. Reject the null hypothesis when ______________________________________. The results of the sample data _____________________________________. Do not reject the null when hypothesis when __________________________________________. The results of the sample data ____________________________________________. Reject the null hypothesis when the p -value is less than the established alpha value . The results of the sample data support the alternative hypothesis . Do not reject the null hypothesis when the p -value is greater than the established alpha value . The results of the sample data do not support the alternative hypothesis . Try It It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows: H 0 : p = 0.50, H a : p > 0.50 α = 0.01 p -value = 0.025 Interpret the results and state a conclusion in simple, non-technical terms. Chapter Review When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p -value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind: α > p -value, reject the null hypothesis α ≤ p -value, do not reject the null hypothesis When do you reject the null hypothesis? The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely? The outcome of winning is very unlikely. The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why? It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 inches. Do the data support the claim that the mean height is less than 73 inches? The p -value is almost zero. State the null and alternative hypotheses and interpret the p -value. H 0 : μ > = 73 H a : μ < 73 The p -value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim. The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The p -value is 0.0264. State the null and alternative hypotheses and interpret the p -value. Does the shaded region represent a low or a high p -value compared to a level of significance of 1%? The shaded region shows a low p -value. What should you do when α > p -value? What should you do if α = p -value? Do not reject H 0 . If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences. Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal. Is this a test of means or proportions? means What symbol represents the random variable for this test? In words, define the random variable for this test. the mean time spent in jail for 26 first time convicted burglars Is σ known and, if so, what is it? Calculate the following: x ¯ _______ σ _______ s x _______ n _______ 3 1.5 1.8 26 Since both σ and s x are given, which should be used? In one to two complete sentences, explain why. State the distribution to use for the hypothesis test. X ¯ ~ N ( 2.5 , 1.5 26 ) A random survey of 75 student loan recipients revealed that the mean length of time in repayment is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time in repayment could likely be 15 years. Is this a test of one mean or proportion? State the null and alternative hypotheses. H 0 : ____________________ H a : ____________________ Is this a right-tailed, left-tailed, or two-tailed test? What symbol represents the random variable for this test? In words, define the random variable for this test. Is the population standard deviation known and, if so, what is it? Calculate the following: x ¯ = _____________ s = ____________ n = ____________ Which test should be used? State the distribution to use for the hypothesis test. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Homework The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5% of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. Is this a test of one mean or proportion? State the null and alternative hypotheses. H 0 : ____________________ H a : ____________________ Is this a right-tailed, left-tailed, or two-tailed test? What symbol represents the random variable for this test? In words, define the random variable for this test. Calculate the following: x = ________________ n = ________________ p ′ = _____________ Calculate σ x = __________. Show the formula set-up. State the distribution to use for the hypothesis test. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Level of Significance of the Test probability of a Type I error (reject the null hypothesis when it is true). Notation: α. In hypothesis testing, the Level of Significance is called the preconceived α or the preset α. p -value the probability that an event will happen purely by chance assuming the null hypothesis is true. The smaller the p -value, the stronger the evidence is against the null hypothesis.", "section": "Rare Events, the Sample, Decision and Conclusion", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Additional Information and Full Hypothesis Test Examples In a hypothesis test problem, you may see words such as \"the level of significance is 1%.\" The \"1%\" is the preconceived or preset α . The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. If no level of significance is given, a common standard to use is α = 0.05. When you calculate the p -value and draw the picture, the p -value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed. The alternative hypothesis , H a , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test. H a never has a symbol that contains an equal sign. Thinking about the meaning of the p -value : A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules. The following examples illustrate a left-, right-, and two-tailed test. H o : μ = 5, H a : μ < 5 Test of a single population mean. H a tells you the test is left-tailed. The picture of the p -value is as follows: Try It H 0 : μ = 10, H a : μ < 10 Assume the p -value is 0.0935. What type of test is this? Draw the picture of the p -value. H 0 : p ≤ 0.2 H a : p > 0.2 This is a test of a single population proportion. H a tells you the test is right-tailed . The picture of the p -value is as follows: Try It H 0 : μ ≤ 1, H a : μ > 1 Assume the p -value is 0.1243. What type of test is this? Draw the picture of the p -value. H 0 : p = 50 H a : p ≠ 50 This is a test of a single population mean. H a tells you the test is two-tailed . The picture of the p -value is as follows. Try It H 0 : p = 0.5, H a : p ≠ 0.5 Assume the p -value is 0.2564. What type of test is this? Draw the picture of the p -value. Full Hypothesis Test Examples Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean . H 0 : μ = 16.43 H a : μ < 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The \"<\" tells you this is left-tailed. Determine the distribution needed: Random variable: X ¯ = the mean time to swim the 25-yard freestyle. Distribution for the test: X ¯ is normal (population standard deviation is known: σ = 0.8) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15. Calculate the p -value using the normal distribution for a mean: p -value = P ( x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16. Graph: μ = 16.43 comes from H 0 . Our assumption is μ = 16.43. Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p -value: α = 0.05 p -value = 0.0187 α > p -value Make a decision: Since α > p -value, reject H 0 . This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds. Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles. The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Stats and press ENTER . Arrow down and enter 16.43 for μ 0 (null hypothesis), .8 for σ , 16 for the sample mean, and 15 for n . Arrow down to μ : (alternate hypothesis) and arrow over to < μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value ( p = 0.0187) but it also calculates the test statistic ( z -score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 ( p -value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Try It The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal. First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion. Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Stats and press ENTER . Arrow down and enter 40 for μ 0 (null hypothesis), 2 for σ , 45 for the sample mean, and 20 for n . Arrow down to μ : (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value but it also calculates the test statistic ( z -score) for the sample mean. Select <, ≠, or > for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER . A shaded graph appears with test statistic and p -value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Historical Note ( ) The traditional way to compare the two probabilities, α and the p -value, is to compare the critical value ( z -score from α ) to the test statistic ( z -score from data). The calculated test statistic for the p -value is –2.08. (From the Central Limit Theorem, the test statistic formula is z = x ¯ − μ X ( σ X n ) . For this problem, x ¯ = 16, μ X = 16.43 from the null hypothes is, σ X = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see 15.Tables in the Table of Contents). The z -score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z -score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p -value), reject H 0 . Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p -value is very common. For this problem, the p -value, 0.0187 is considerably smaller than α , 0.05. You can be confident about your decision to reject. The graph shows α , the p -value, and the test statistic and the critical value. A college football coach records the mean weight that the players can bench press as 275 pounds , with a standard deviation of 55 pounds . Three of the players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3) 215(3) 225(1) 241(2) 252(2) 265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1) . Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds . Set up the Hypothesis Test: Since the problem is about a mean weight, this is a test of a single population mean . H 0 : μ = 275 H a : μ > 275 This is a right-tailed test. Calculating the distribution needed: Random variable: X ¯ = the mean weight, in pounds, lifted by the football players. Distribution for the test: It is normal because σ is known. X ¯ ~ N ( 275 , 55 30 ) x ¯ = 286.2 pounds (from the data). σ = 55 pounds (Always use σ if you know it.) We assume μ = 275 pounds unless our data shows us otherwise. Calculate the p -value using the normal distribution for a mean and using the sample mean as input (see for using the data as input): p -value = P ( x ¯ > 286.2 ) = 0.1323 . Interpretation of the p -value: If H 0 is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event. Compare α and the p -value: α = 0.025 p -value = 0.1323 Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. Put the data and frequencies into lists. Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Data and press ENTER . Arrow down and enter 275 for μ 0 , 55 for σ , the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value ( p = 0.1331, a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic ( z -score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 ( p -value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Try It A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2). Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 . Set up the Hypothesis Test: H 0 : μ = 475 H 0 : μ > 475 This is a right-tailed test. The random variable X is the mean time of employees working in a day. Distribution for the test: σ is known. X ~ N 475 , 45 20 x ¯ = 475 . 5 minutes (from the data). σ = 45 minutes. We assume μ = 475 minutes unless our data shows otherwise. The p -value is calculated: p -value = P ( x ¯ > 475 . 5 ) = 0 . 5044 Interpretation of p -value: If H 0 is true, then there is a 0.5044, or 50.44%, chance that the mean time of employees working is 475.5 minutes or more. Since a 50.44% chance is large enough, the mean time of 475.5 is not a rare event. Compare α and p values: α = 0 . 025 and p = 0 . 5044 Make a decision: Since α < p value, do not reject H 0 . Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean time of employees is more than 475 minutes. Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71 . He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution. Set up the hypothesis test: A 5% level of significance means that α = 0.05. This is a test of a single population mean . H 0 : μ = 65 H a : μ > 65 Since the instructor thinks the average score is higher, use a \">\". The \">\" means the test is right-tailed. Determine the distribution needed: Random variable: X ¯ = average score on the first statistics test. Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's t . Use t df . Therefore, the distribution for the test is t 9 where n = 10 and df = 10 - 1 = 9. Calculate the p -value using the Student's t -distribution: p -value = P ( x ¯ > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data. Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more. Compare α and the p -value: Since α = 0.05 and p -value = 0.0396. α > p -value. Make a decision: Since α > p -value, reject H 0 . This means you reject μ = 65. In other words, you believe the average test score is greater than 65. Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is greater than 65, just as the math instructor thinks. Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for μ 0 , the name of the list where you put the data, and 1 for Freq: . Arrow down to μ : and arrow over to > μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value ( p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 ( p -value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Try It It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors. Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance. Set up the hypothesis test: The 1% level of significance means that α = 0.01. This is a test of a single population proportion . H 0 : p = 0.50 H a : p ≠ 0.50 The words \"is the same or different from\" tell you this is a two-tailed test. Calculate the distribution needed: Random variable: P′ = the percent of of first-time brides who are younger than their grooms. Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion. P ′ ~ N ( p , p ⋅ q n ) Therefore, P ′ ~ N ( 0.5 , 0.5 ⋅ 0.5 100 ) where p = 0.50, q = 1− p = 0.50, and n = 100 Calculate the p -value using the normal distribution for proportions: p -value = P ( p′ < 0.47 or p′ > 0.53) = 0.5485 where x = 53, p′ = x n = 53 100 = 0.53. Interpretation of the p -value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion p ' is 0.53 or more OR 0.47 or less (see the graph in ). μ = p = 0.50 comes from H 0 , the null hypothesis. p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.) Compare α and the p -value: Since α = 0.01 and p -value = 0.5485. α < p -value. Make a decision: Since α < p -value, you cannot reject H 0 . Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%. Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for p 0 , 53 for x and 100 for n . Arrow down to Prop and arrow to not equals p 0 . Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the p -value ( p = 0.5485) and the test statistic ( z -score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 ( p -value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.) Try It A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance. First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion. Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones. a. The value that helps determine the p -value is p′ . Calculate p′ . b. What is a success for this problem? c. What is the level of significance? d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Calculate the p -value. e. Make a decision. _____________(Reject/Do not reject) H 0 because____________. Set up the Hypothesis Test: H 0 : p = 0.30 H a : p ≠ 0.30 Determine the distribution needed: The random variable is P′ = proportion of households that have three cell phones. The distribution for the hypothesis test is P ' ~ N ( 0.30 , ( 0.30 ) ⋅ ( 0.70 ) 150 ) a. p′ = x n where x is the number of successes and n is the total number in the sample. x = 43, n = 150 p′ = 43 150 b. A success is having three cell phones in a household. c. The level of significance is the preset α . Since α is not given, assume that α = 0.05. d. p -value = 0.7216 e. Assuming that α = 0.05, α < p -value. The decision is do not reject H 0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%. Try It Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors. The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p . The distribution for the test is normal. The estimated proportion p ′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it! My dog has so many fleas, They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42. At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased. Now it is time for you to have some fun With the level of significance being .01, You must help me figure out Use the new shampoo or go without? Set up the hypothesis test: H 0 : p ≤ 0.25 H a : p > 0.25 Determine the distribution needed: In words, CLEARLY state what your random variable X ¯ or P′ represents. P′ = The proportion of fleas that are killed by the new shampoo State the distribution to use for the test. Normal: N ( 0.25 , ( 0.25 ) ( 1 − 0.25 ) 42 ) Test Statistic: z = 2.3163 Calculate the p -value using the normal distribution for proportions: p -value = 0.0103 In one to two complete sentences, explain what the p -value means for this problem. If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 ( 17 42 ) or more. Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p -value. Compare α and the p -value: Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences. alpha decision reason for decision 0.01 Do not reject H 0 α < p -value Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%. Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%. NOTE This test result is not very definitive since the p -value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return. Try It A car soap gets rid of 30% of stains on the car. After adding a new compound to the soap, the soap is used on a car and found to wash 20 stains out of the 50 stains on the car. With the level of significance being 0.01, find out if adding the new compound to soap is beneficial. Set up hypothesis test: H 0 : p ≤ 0 . 3 H 0 : p > 0 . 3 The distribution to use for the test is: Normal: N 0 . 3 , ( 0 . 3 ) ( 1 - 0 . 3 ) 50 The test statistic is obtained to be z = 1.543. The p -value is obtained to be p = 0.0614. If the null hypothesis is true, then there is a 0.0614 probability that the sample proportion is 0.4 20 50 or more. At 1% level of significance, the value of α is less than the p -value. So, the data does not show sufficient evidence that the percentage of stains removed is more than 30%. The confidence interval is obtained as (0.273, 0.527). The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass. 1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal. Let’s follow a four-step process to answer this statistical question. State the Question : We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be H 0 : μ ≤ 1 H a : μ > 1 Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's t-distribution. Assume the underlying population is normal. Based on the sample of 11 data values shown above, sample mean, sample standard deviation, and test statistic are calculated as follows: x ¯ = 1 . 04 s = 0 . 0659 t = 2 . 014 To calculate the p -value, note that this is a right-tailed test. Then, find the area under the t -distribution to the right of the test statistic 2.014 (using 10 degrees of freedom). This area in the right tail is 0.036, and thus the p -value = 0.036. State the Conclusions : Since the p -value ( p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one. Try It The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows: 205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205 Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal. The hypotheses, at a significance level of 0.1, will be as follows: H 0 : μ ≤ 200 H 0 : μ > 200 Since the standard deviation is not known, Student’s t -distribution is to be used. The data is input into the TI-83 as follows: Since the p -value is less than alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average boiling point is greater than 200. In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. We will follow the four-step process. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be H 0 : p ≤ 0.00034 H a : p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population. The associated TI results are Since the p -value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users. Try It In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. The test hypotheses are as follows: H 0 : p ≤ 0 . 00041 H 0 : p > 0 . 00041 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes disease-causing environments, we want to minimize the chances of incorrectly identifying causes of diseases. The testing will be done with x = 138 and n = 390,000. The sample is sufficiently large because we have np = 390,000(0.00041) = 159.9, nq = 390,000(0.99959) = 389,840.1, two independent outcomes, and a fixed probability of success p = 0.00041. Thus, we will be able to generalize our results to the population. The associated TI results are: Since the p -value = 0.0832 is greater than alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher skin diseases rates for moisturizer users. Statistical data indicates that in a certain country there are approximately 268,608,618 residents aged 12 and older. For a certain period of time, statistical data also indicates that the percentage of residents with blood type AB negative (AB-) is 207,754 individuals. This translates into a percentage of 0.078% with this rather rare blood type. In a certain province of the country, there were 11 people with blood type AB- out of the population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the percentage of residents in the entire country with blood type AB- versus the percentage in the local province. Use a significance level of 0.01. We will follow the four-step plan. We need to test whether the proportion of residents with AB- blood type in the local province is statistically different as compared to the proportion in the entire country. Since we are presented with proportions, we will use a one-proportion z -test. The hypotheses for the test will be: H 0 ∶ p = 0 . 00078 H a ∶ p ≠ 0 . 00078 Note the sample proportion is p ' = 11 37 , 937 = 0 . 00029 The test statistic is calculated as z = -3.4189. To calculate the p -value, note that this is a two-tailed test. Find the area under the normal distribution to the left of the test statistic and then double this area. The area to the left of the test statistic is 0.000314, and this area doubled results in the p -value of 0.00063. Since the p -value, p = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of individuals with blood type AB- in the local province is different from the proportion of individuals in the entire country. Try It According to the U.S. Census, there are approximately 201,456,463 residents 20 and older. Statistics from the Criminal National Network indicate that, on average, 104,354 murders occur each year for people aged 20 and older. This translates into a percentage of murder of 0.052%. In Ohio, there were reported 127 murders for a population of 427,648. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local murder percentage and the national murder percentage. Use a significance level of 0.01. The hypotheses for the one-proportion z -test are as follow: H 0 : p ≤ 0 . 00052 H 0 : p > 0 . 00052 The following screenshots display the summary statistics from the hypothesis test: Since the p -value = 0 is less than alpha value of 0.01, the sample data includes that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of murders in Ohio is different from the national average population. Chapter Review The hypothesis test itself has an established process. This can be summarized as follows: Determine H 0 and H a . Remember, they are contradictory. Determine the random variable. Determine the distribution for the test. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p -value. (A z -score and a t -score are examples of test statistics.) Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences. Notice that in performing the hypothesis test, you use α and not β . β is needed to help determine the sample size of the data that is used in calculating the p -value. Remember that the quantity 1 – β is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when it should be. Assume H 0 : μ = 9 and H a : μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume H 0 : μ ≤ 6 and H a : μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? Assume H 0 : p = 0.25 and H a : p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? This is a two-tailed test. Draw the general graph of a left-tailed test. Draw the graph of a two-tailed test. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use? Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use? a right-tailed test A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use? You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use? a left-tailed test If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test? This is a two-tailed test. Homework For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's- t distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim? H 0 : μ ≥ 50,000 H a : μ < 50,000 Let X ¯ = the average lifespan of a brand of tires. normal distribution z = -2.315 p -value = 0.0103 Answers may vary. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. (43,537, 49,463) From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level? H 0 : μ = $1.00 H a : μ ≠ $1.00 Let X ¯ = the average cost of a daily newspaper. normal distribution z = –0.866 p -value = 0.3865 Answers may vary. Alpha: 0.01 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.01. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. ($0.84, $1.06) An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten? H 0 : μ = 10 H a : μ ≠ 10 Let X ¯ the mean number of sick days an employee takes per year. Student’s t -distribution t = –1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten. (4.9443, 11.806) In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level? Your statistics instructor claims that 60% of the students who take their Elementary Statistics class go through life feeling more enriched. For some reason that the instructor can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of their past Elementary Statistics students and find that 34 feel more enriched as a result of taking the class. Now, what do you think? H 0 : p ≥ 0.6 H a : p < 0.6 Let P′ = the proportion of students who feel more enriched as a result of taking Elementary Statistics. normal for a single proportion 1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: There is insufficient evidence to conclude that less than 60 percent of the students feel more enriched. Confidence Interval: (0.409, 0.654) The “plus-4s” confidence interval is (0.411, 0.648) A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief. Refer to Exercise 9.81 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. H 0 : μ = 4 H a : μ ≠ 4 Let X ¯ the average I.Q. of a set of brown trout. two-tailed Student's t-test t = 1.95 p -value = 0.076 Check student’s solution. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is greater than 0.05 Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. (3.8865,5.9468) According to an article in Newsweek , the birth ratio of females to males is 100:105. In China, the birth ratio is 100: 114 (46.7% females). Suppose you want to explore the reported figures of the percent of females born in China. You conduct a study. In this study, you count the number of females and males born in 150 randomly chosen recent births. There are 60 females and 90 males born of the 150. Based on your study, do you believe that the percent of females born in China is 46.7? A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. H 0 : p ≥ 0.13 H a : p < 0.13 Let P′ = the proportion of Americans who have seen or sensed angels normal for a single proportion –2.688 p -value = 0.0036 Check student’s solution. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%. (0, 0.0623). The“plus-4s” confidence interval is (0.0022, 0.0978) The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. They ask ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should the engineer count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55. Use the “Lap time” data for Lap 4 (see ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given. H 0 : μ ≥ 129 H a : μ < 129 Let X ¯ = the average time in seconds that Terri finishes Lap 4. Student's t -distribution t = 1.209 0.8792 Check student’s solution. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds. (128.63, 130.37) Use the “Initial Public Offering” data (see ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices. NOTE The following questions were written by past students. They are excellent problems! \"Asian Family Reunion,\" by Chau Nguyen Every two years it comes around. We all get together from different towns. In my honest opinion, It's not a typical family reunion. Not forty, or fifty, or sixty, But how about seventy companions! The kids would play, scream, and shout One minute they're happy, another they'll pout. The teenagers would look, stare, and compare From how they look to what they wear. The men would chat about their business That they make more, but never less. Money is always their subject And there's always talk of more new projects. The women get tired from all of the chats They head to the kitchen to set out the mats. Some would sit and some would stand Eating and talking with plates in their hands. Then come the games and the songs And suddenly, everyone gets along! With all that laughter, it's sad to say That it always ends in the same old way. They hug and kiss and say \"good-bye\" And then they all begin to cry! I say that 60 percent shed their tears But my mom counted 35 people this year. She said that boys and men will always have their pride, So we won't ever see them cry. I myself don't think she's correct, So could you please try this problem to see if you object? H 0 : p = 0.60 H a : p < 0.60 Let P′ = the proportion of family members who shed tears at a reunion. normal for a single proportion –1.71 0.0438 Check student’s solution. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p -value and alpha are quite close, so other tests should be done. We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The“plus-4s” confidence interval (see Confidence Intervals ) is (0.3861, 0.6139) Note that here the “large-sample” 1 – PropZTest provides the approximate p -value of 0.0438. Whenever a p -value based on a normal approximation is close to the level of significance, the exact p -value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course. \"The Problem with Angels,\" by Cyndy Dowling Although this problem is wholly mine, The catalyst came from the magazine, Time. On the magazine cover I did find The realm of angels tickling my mind. Inside, 69% I found to be In angels, Americans do believe. Then, it was time to rise to the task, Ninety-five high school and college students I did ask. Viewing all as one group, Random sampling to get the scoop. So, I asked each to be true, \"Do you believe in angels?\" Tell me, do! Hypothesizing at the start, Totally believing in my heart That the proportion who said yes Would be equal on this test. Lo and behold, seventy-three did arrive, Out of the sample of ninety-five. Now your job has just begun, Solve this problem and have some fun. \"Blowing Bubbles,\" by Sondra Prull Studying stats just made me tense, I had to find some sane defense. Some light and lifting simple play To float my math anxiety away. Blowing bubbles lifts me high Takes my troubles to the sky. POIK! They're gone, with all my stress Bubble therapy is the best. The label said each time I blew The average number of bubbles would be at least 22. I blew and blew and this I found From 64 blows, they all are round! But the number of bubbles in 64 blows Varied widely, this I know. 20 per blow became the mean They deviated by 6, and not 16. From counting bubbles, I sure did relax But now I give to you your task. Was 22 a reasonable guess? Find the answer and pass this test! H 0 : μ ≥ 22 H a : μ < 22 Let X ¯ = the mean number of bubbles per blow. Student's t -distribution –2.667 p -value = 0.00486 Check student’s solution. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22. (18.501, 21.499) \"Dalmatian Darnation,\" by Kathy Sparling A greedy dog breeder named Spreckles Bred puppies with numerous freckles The Dalmatians he sought Possessed spot upon spot The more spots, he thought, the more shekels. His competitors did not agree That freckles would increase the fee. They said, “Spots are quite nice But they don't affect price; One should breed for improved pedigree.” The breeders decided to prove This strategy was a wrong move. Breeding only for spots Would wreak havoc, they thought. His theory they want to disprove. They proposed a contest to Spreckles Comparing dog prices to freckles. In records they looked up One hundred one pups: Dalmatians that fetched the most shekels. They asked Mr. Spreckles to name An average spot count he'd claim To bring in big bucks. Said Spreckles, “Well, shucks, It's for one hundred one that I aim.” Said an amateur statistician Who wanted to help with this mission. “Twenty-one for the sample Standard deviation's ample: They examined one hundred and one Dalmatians that fetched a good sum. They counted each spot, Mark, freckle and dot And tallied up every one. Instead of one hundred one spots They averaged ninety six dots Can they muzzle Spreckles’ Obsession with freckles Based on all the dog data they've got? \"Macaroni and Cheese, please!!\" by Nedda Misherghi and Rachelle Hall As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value. One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook: Price per box of Mac and Cheese: 5 stores @ $2.02 15 stores @ $0.25 3 stores @ $1.29 6 stores @ $0.35 4 stores @ $2.27 7 stores @ $1.50 5 stores @ $1.89 8 stores @ 0.75. I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!) H 0 : μ ≤ 1 H a : μ > 1 Let X ¯ = the mean cost in dollars of macaroni and cheese in a certain town. Student's t -distribution t = 0.340 p -value = 0.36756 Check student’s solution. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05 Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1. (0.8291, 1.241) \"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark,\" by Jacqueline Ghodsi THE CHARACTERS (in order of appearance): HAMLET, Prince of Denmark and student of Statistics POLONIUS, Hamlet’s tutor HOROTIO, friend to Hamlet and fellow student Scene: The great library of the castle, in which Hamlet does his lessons Act I (The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.) POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination! HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable. POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true. HORATIO (to Hamlet): What should we do, my Lord? HAMLET: Go to thy purpose, Horatio. HORATIO: To what end, my Lord? HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no. (Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.) Act II (The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.) POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations? HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes. POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.) HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.) (Curtain falls) \"Untitled,\" by Stephen Chen I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%. So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right? H 0 : p = 0.01 H a : p > 0.01 Let P′ = the proportion of errors generated Normal for a single proportion 2.13 0.0165 Check student’s solution. Alpha: 0.05 Decision: Reject the null hypothesis Reason for decision: The p -value is less than 0.05. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01. Confidence interval: (0, 0.094). The“plus-4s” confidence interval is (0.004, 0.144). \"Japanese Girls’ Names\" by Kumi Furuichi It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on. However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children. I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation. Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko. \"Phillip’s Wish,\" by Suzanne Osorio My nephew likes to play Chasing the girls makes his day. He asked his mother If it is okay To get his ear pierced. She said, “No way!” To poke a hole through your ear, Is not what I want for you, dear. He argued his point quite well, Says even my macho pal, Mel, Has gotten this done. It’s all just for fun. C’mon please, mom, please, what the hell. Again Phillip complained to his mother, Saying half his friends (including their brothers) Are piercing their ears And they have no fears He wants to be like the others. She said, “I think it’s much less. We must do a hypothesis test. And if you are right, I won’t put up a fight. But, if not, then my case will rest.” We proceeded to call fifty guys To see whose prediction would fly. Nineteen of the fifty Said piercing was nifty And earrings they’d occasionally buy. Then there’s the other thirty-one, Who said they’d never have this done. So now this poem’s finished. Will his hopes be diminished, Or will my nephew have his fun? H 0 : p = 0.50 H a : p < 0.50 Let P′ = the proportion of friends that has a pierced ear. normal for a single proportion –1.70 p -value = 0.0448 Check student’s solution. Alpha: 0.05 Decision: Reject the null hypothesis Reason for decision: The p -value is less than 0.05. (However, they are very close.) Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears. Confidence Interval: (0.245, 0.515): The “plus-4s” confidence interval is (0.259, 0.519). \"The Craven,\" by Mark Salangsang Once upon a morning dreary In stats class I was weak and weary. Pondering over last night’s homework Whose answers were now on the board This I did and nothing more. While I nodded nearly napping Suddenly, there came a tapping. As someone gently rapping, Rapping my head as I snore. Quoth the teacher, “Sleep no more.” “In every class you fall asleep,” The teacher said, his voice was deep. “So a tally I’ve begun to keep Of every class you nap and snore. The percentage being forty-four.” “My dear teacher I must confess, While sleeping is what I do best. The percentage, I think, must be less, A percentage less than forty-four.” This I said and nothing more. “We’ll see,” he said and walked away, And fifty classes from that day He counted till the month of May The classes in which I napped and snored. The number he found was twenty-four. At a significance level of 0.05, Please tell me am I still alive? Or did my grade just take a dive Plunging down beneath the floor? Upon thee I hereby implore. Toastmasters International cites a report by Gallup Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at their school fear public speaking. The student randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at their school is less than 40%. H 0 : p = 0.40 H a : p < 0.40 Let P′ = the proportion of schoolmates who fear public speaking. normal for a single proportion –1.01 p -value = 0.1563 Check student’s solution. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking. Confidence Interval: (0.3241, 0.4240): The “plus-4s” confidence interval is (0.3257, 0.4250). Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data. According to New York City's Community Health Survey, the most recent adult smoking rate is 12%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 12% or if it has decreased. H 0 : p = 0.12 H a : p < 0.12 Let P ' = the proportion of NYC residents that smoke. Normal for a single proportion 0.2207 p -value = 0.5873 Answers may vary alpha 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: The p -value is greater than 0.05. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14. Confidence interval: Confidence Interval: (0.0502, 0.2070): The “plus-4s” confidence interval (see Confidence Intervals ) is (0.0676, 0.2297). The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. They randomly survey 56 online students and find that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test. H 0 : p = 0.12 H a : p < 0.12 Let P ' = the proportion of NYC residents that smoke. Normal for a single proportion 0.2207 p -value = 0.5873 Answers may vary alpha 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. ($68,757, $73,485) La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old. Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin? After conducting the test, your decision and conclusion are Reject H 0 : There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Reject H 0 : There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. c A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is: There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Marvel Universe is less than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Marvel Universe is more than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Marvel Universe is less than 20%. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Marvel Universe is at least 20%. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion? There is enough evidence to conclude that the mean number of hours is more than 4.75 There is enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.75 c Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question. State the null and alternate hypothesis. State the p -value. State alpha. What is your decision? Write a conclusion. Answer any other questions asked in the problem. According to the Center for Disease Control website, in 2022 at least 2% of high school students reported they had smoked a cigarette. An Introduction to Statistics class conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. They surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? H 0 : p = 0.488 H a : p ≠ 0.488 p -value = 0.0114 alpha = 0.05 Reject the null hypothesis. There is enough evidence to conclude that the percent of families owning stocks is not 48.8%. The survey does not appear to be accurate. Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate? The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level in Kentucky? Are the results applicable across the country? Why? H 0 : p = 0.517 H a : p ≠ 0.517 p -value = 0.9203. alpha = 0.05. Do not reject the null hypothesis. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is different from the proportion of homes heated by natural gas across the country. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation. For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so they asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library? The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. H 0 : µ ≥ 11.52 H a : µ < 11.52 p -value = 0.000002 which is almost 0. alpha = 0.05. Reject the null hypothesis. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average. We would make the same conclusion if alpha was 1% because the p -value is almost 0. A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities? A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals 3 2 1 3 7 2 9 4 6 6 8 0 5 6 4 2 1 3 4 1 At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year? H 0 : µ ≤ 5.8 H a : µ > 5.8 p -value = 0.9987 alpha = 0.05 Do not reject the null hypothesis. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year. According to the U.S. Census Bureau , the mean family size in the U.S. is 3.13. A sample of a college math class resulted in the following family sizes: 5 4 5 4 4 3 6 4 3 3 5 5 6 3 3 2 7 4 5 2 2 2 3 2 At α = 0.05 level, is the class’ mean family size greater than the national average? Does the census result remain valid? Why? The student academic group on a college campus claims that first-year students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 first-year students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? H 0 : µ ≥ 150 H a : µ < 150 p -value = 0.0622 alpha = 0.01 Do not reject the null hypothesis. At the 1% significance level, there is not enough evidence to conclude that first-year students study less than 2.5 hours per day, on average. The student academic group’s claim appears to be correct. References Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC. Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013). Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013). Data from Growing by Degrees by Allen and Seaman. Data from La Leche League International. Available online at http://www.lalecheleague.org/Law/BAFeb01.html. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013). Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013) Data from the U.S. Census Bureau, available online at http://quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013). Data from the United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/. Data from Toastmasters International. Available online at http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1. Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013). Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013). “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf. Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013). Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at http://www.rainn.org/get-information/statistics/frequency-of-sexual-assault (accessed June 27, 2013). Central Limit Theorem Given a random variable (RV) with known mean μ and known standard deviation σ. We are sampling with size n and we are interested in two new RVs - the sample mean, X ¯ , and the sample sum, Σ X . If the size n of the sample is sufficiently large, then X ¯ ~ N ( μ , σ n ) and Σ X ~ N ( n μ , n σ ) . If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean.", "section": "Additional Information and Full Hypothesis Test Examples", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Hypothesis Testing of a Single Mean and Single Proportion Hypothesis Testing of a Single Mean and Single Proportion Class Time: Names: Student Learning Outcomes The student will select the appropriate distributions to use in each case. The student will conduct hypothesis tests and interpret the results. Television Survey In a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower. H 0 : _____________ H a : _____________ In words, define the random variable. __________ = ______________________ The distribution to use for the test is _______________________. Determine the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Determine the p -value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence. Language Survey About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%. H 0 : ___________ H a : ___________ In words, define the random variable. __________ = _______________ The distribution to use for the test is ________________ Determine the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Determine the p -value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence. Jeans Survey Suppose that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal. H 0 : _____________ H a : _____________ In words, define the random variable. __________ = ______________________ The distribution to use for the test is _______________________. Determine the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Determine the p -value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.", "section": "Hypothesis Testing of a Single Mean and Single Proportion", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River) you can use a slightly different technique when conducting a hypothesis test. (credit: modification of work “scrambled egg, toast and home-made berry jam (no added sugar) for breakfast” by Chloe Lim/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Classify hypothesis tests by type. Conduct and interpret hypothesis tests for two population means, population standard deviations known. Conduct and interpret hypothesis tests for two population means, population standard deviations unknown. Conduct and interpret hypothesis tests for two population proportions. Conduct and interpret hypothesis tests for matched or paired samples. Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs . Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p -values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests. This chapter deals with the following hypothesis tests: Independent groups (samples are independent) Test of two population means. Test of two population proportions. Matched or paired samples (samples are dependent) Test of the population mean of differences of measures for paired individuals or objects.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Two Population Means with Unknown Standard Deviations The two independent samples are simple random samples from two distinct populations. For the two distinct populations: if the sample sizes are small, the distributions are important (should be normal) if the sample sizes are large, the distributions are not important (need not be normal) NOTE The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, X ¯ 1 – X ¯ 2 , and divide by the standard error in order to standardize the difference. The result is a t-score test statistic. Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means , X ¯ 1 – X ¯ 2 . The standard error is: ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 The test statistic ( t -score) is calculated as follows: ( x ¯ 1 – x ¯ 2 ) – ( μ 1 – μ 2 ) ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 where: s 1 and s 2 , the sample standard deviations, are estimates of σ 1 and σ 2 , respectively. σ 1 and σ 2 are the unknown population standard deviations. x ¯ 1 and x ¯ 2 are the sample means. μ 1 and μ 2 are the population means. The number of degrees of freedom ( df ) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student's t -distribution with df as follows: Degrees of freedom d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 – 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 – 1 ) ( ( s 2 ) 2 n 2 ) 2 When both sample sizes n 1 and n 2 are five or larger, the Student's t approximation is very good. Notice that the sample variances ( s 1 ) 2 and ( s 2 ) 2 are not pooled. (If the question comes up, do not pool the variances.) NOTE It is not necessary to compute this by hand. A calculator or computer easily computes it. Independent groups The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in . Each populations has a normal distribution. Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation Girls 9 2 0.866 Boys 16 3.2 1.00 Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance. The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μ g is the population mean for girls and μ b is the population mean for boys. This is a test of two independent groups , two population means . Random variable : X ¯ g − X ¯ b = difference in the sample mean amount of time girls and boys play sports each day. H 0 : μ g = μ b H 0 : μ g – μ b = 0 H a : μ g ≠ μ b H a : μ g – μ b ≠ 0 The words \"the same\" tell you H 0 has an \"=\". Since there are no other words to indicate H a , assume it says \"is different.\" This is a two-tailed test. Distribution for the test: Use t df where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances. Calculate the p -value using a Student's t -distribution: p -value = 0.0054 Graph: s g = 0.866 s b = 1 So, x ¯ g – x ¯ b = 2 – 3.2 = –1.2 Half the p -value is below –1.2 and half is above 1.2. Make a decision: Since α > p -value, reject H 0 . This means you reject μ g = μ b . The means are different. Press STAT . Arrow over to TESTS and press 4:2-SampTTest . Arrow over to Stats and press ENTER . Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER . Arrow down to Pooled: and No . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw. Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys). Try It Two samples are shown in . Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 5 1 Population B 16 4.7 1.2 NOTE When the sum of the sample sizes is larger than 30 ( n 1 + n 2 > 30) you can use the normal distribution to approximate the Student's t . Use the Student's t distribution, however, whenever possible. A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions. a. Is this a test of two means or two proportions? a. two means b. Are the populations standard deviations known or unknown? b. unknown c. Which distribution do you use to perform the test? c. Student's t d. What is the random variable? d. X ¯ A - X ¯ B e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols. e. H o : μ A ≤ μ B H a : μ A > μ B f. Is this test right-, left-, or two-tailed? f. right g. What is the p -value? g. 0.1928 h. Do you reject or not reject the null hypothesis? h. Do not reject. i. Conclusion: i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B. Try It A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed. Are the population standard deviations known? Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion? A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in and . Online Class 67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4 Face-to-face Class 77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right, left, or two tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. (See the conclusion in , and write yours in a similar fashion) two means unknown Student's t X ¯ 1 – X ¯ 2 H 0 : μ 1 = μ 2 Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes. H a : μ 1 < μ 2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class. left-tailed p -value = 0.0011 Reject the null hypothesis The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class. First put the data for each group into two lists (such as L1 and L2). Press STAT . Arrow over to TESTS and press 4:2SampTTest . Make sure Data is highlighted and press ENTER . Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ 1 : and arrow to < μ 2 (less than). Press ENTER . Arrow down to Pooled: No. Press ENTER . Arrow down to Calculate and press ENTER . NOTE Be careful not to mix up the information for Group 1 and Group 2! Try It Two professors, A and B, teach classes on the same subjects. Scores of 10 students from each class are selected randomly. The final exam scores of the students are as follows: Professor A: 97 62 73 58 84 74 66 93 73 85 Professor B: 85 64 74 55 76 67 72 84 71 98 Professor A says that the mean score of their class is more than the mean of professor B’s class. Is professor A correct? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right-, left-, or two-tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ___ (is/is not) sufficient evidence to conclude that ___. two means unknown Student’s t X 1 - X 2 H 0 : μ 1 = μ 2 Null hypothesis: the means of final exam scores are equal for both professors. H 0 : μ 2 < μ 1 Alternative hypothesis: the mean of final exam scores for class of professor B is less than mean of final exam scores for class of professor A. left-tailed p -value = 0.03 Reject the null hypothesis. Professor A is correct. The evidence shows that the mean of final exam scores of the class of professor A is more than that of the class of professor B. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of final exam scores of the class of professor A is more than that of the class of professor B. Cohen's Standards for Small, Medium, and Large Effect Sizes Cohen's d is a measure of effect size based on the differences between two means. Cohen’s d , named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes. Cohen's Standard Effect Sizes Size of effect d Small 0.2 medium 0.5 Large 0.8 Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: d = x ¯ 1 – x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 – 1 ) s 1 2 + ( n 2 – 1 ) s 2 2 n 1 + n 2 – 2 Calculate Cohen’s d for . Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. μ 1 = 4 s 1 = 1.5 n 1 = 11 μ 2 = 3.5 s 2 = 1 n 2 = 9 d = 0.384 The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them. Try It Calculate Cohen’s d for . Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. μ 1 = 5 s 1 = 1 . 2 n 1 = 15 μ 2 = 4 . 5 s 2 = 0 . 8 n 2 = 20 d = 0.49 The effect is medium because 0.49 is very close to Cohen’s value of 0.5 for medium effect size. The size of the differences of the means for the two companies is small, indicating that there is not a significant difference between them. Calculate Cohen’s d for . Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem. d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference. Try It Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in and , respectively. Northeast 94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6 77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5 76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0 West 126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6 116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6 78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5 Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right, left, or two tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. Calculate Cohen’s d and interpret it. References Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com Data from Microsoft Bookshelf . Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013). “List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age (accessed June 17, 2013). “Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013). “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013). Chapter Review Two population means from independent samples where the population standard deviations are not known Random Variable: X ¯ 1 − X ¯ 2 = the difference of the sampling means Distribution: Student's t -distribution with degrees of freedom (variances not pooled) Formula Review Standard error: SE = ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Test statistic ( t -score): t = ( x ¯ 1 − x ¯ 2 ) − ( μ 1 − μ 2 ) ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Degrees of freedom: d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 − 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 − 1 ) ( ( s 2 ) 2 n 2 ) 2 where: s 1 and s 2 are the sample standard deviations, and n 1 and n 2 are the sample sizes. x ¯ 1 and x ¯ 2 are the sample means. Cohen’s d is the measure of effect size: d = x ¯ 1 − x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2 Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for independent group means, population standard deviations, and/or variances known independent group means, population standard deviations, and/or variances unknown matched or paired samples single mean two proportions single proportion It is believed that 70% of men pass their drivers test in the first attempt, while 65% of women pass the test in the first attempt. Of interest is whether the proportions are in fact equal. two proportions A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. matched or paired samples The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their mid-level professionals are paid differently, on average. The average worker in Germany gets eight weeks of paid vacation. single mean According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market. It is believed that the average grade on an English essay in a particular school system for women is higher than for men. A random sample of 31 women had a mean score of 82 with a standard deviation of three, and a random sample of 25 men had a mean score of 76 with a standard deviation of four. independent group means, population standard deviations and/or variances unknown The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? two proportions A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6. independent group means, population standard deviations and/or variances unknown Varsity athletes practice five times a week, on average. A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? independent group means, population standard deviations and/or variances unknown A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? two proportions Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. Are standard deviations known or unknown? What is the random variable? The random variable is the difference between the mean amounts of sugar in the two soft drinks. Is this a one-tailed or two-tailed test? Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for White people born in 1900 and 33.0 years for non-White people. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 White people, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 non-White people, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for White and non-White people. Is this a test of means or proportions? means State the null and alternative hypotheses. H 0 : __________ H a : __________ Is this a right-tailed, left-tailed, or two-tailed test? two-tailed In symbols, what is the random variable of interest for this test? In words, define the random variable of interest for this test. the difference between the mean life spans of White and non-White people Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for . This is a comparison of two population means with unknown population standard deviations. Calculate the test statistic and p -value. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p -value. Answers may vary. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Reject the null hypothesis p -value < 0.05 At the 5% level of significance, the evidence supports the claim that life expectancy in the 1900s was different between White people and non-White people. Does it appear that the means are the same? Why or why not? Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from 29 men and 16 women. The men took an average of three English courses with a standard deviation of 0.8. The women took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. Subscripts: 1: two-year colleges; 2: four-year colleges H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 X ¯ 1 – X ¯ 2 is the difference between the mean enrollments of the two-year colleges and the four-year colleges. Student’s- t test statistic: -0.2480 p -value: 0.4019 Answers may vary. Alpha: 0.05 Decision: Do not reject Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. At Rachel’s 11 th birthday party, eight guests were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The children thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 21 47 40 30 28 22 21 23 25 45 43 37 35 29 32 Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. Subscripts: 1: mechanical engineering; 2: electrical engineering H 0 : µ 1 ≥ µ 2 H a : µ 1 < µ 2 X ¯ 1 − X ¯ 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers. t 108 test statistic: t = –0.82 p -value: 0.2061 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers. Marketing companies have collected data implying that teenage girls use more social media apps on their smartphones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with smartphones, the mean number of social media apps for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from to answer the next four exercises. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 X ¯ 1 − X ¯ 2 is the difference between the mean times for completing a lap in races and in practices. t 20.32 test statistic: –4.70 p -value: 0.0001 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. Repeat the test in Exercise 10.83 , but use Lap 5 data this time. Repeat the test in Exercise 10.83 , but this time combine the data from Laps 1 and 5. H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 is the difference between the mean times for completing a lap in races and in practices. t 40.94 test statistic: –5.08 p -value: zero Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?” Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. Western Eastern Los Angeles 9 D.C. United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Conduct a hypothesis test to answer the next two exercises. The exact distribution for the hypothesis test is: the normal distribution the Student's t -distribution the uniform distribution the exponential distribution If the level of significance is 0.05, the conclusion is: There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score. Unable to determine c Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. They take random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is: There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. Elijah wants to know whether textbook costs are different for different courses of study. He selects a random sample of 33 sociology textbooks offered on a popular online site. The mean price of his sample is $74.64 with a standard deviation of $49.36. He then selects a random sample of 33 math and science textbooks from the same site. The mean price of this sample is $111.56 with a standard deviation of $66.90. Is the mean price of a sociology textbook lower than the mean price of a math or science textbook? Test at a 1% significance level. Test: two independent sample means, population standard deviations unknown. μ 1 = the the mean price of a sociology text on the selected site. μ 2 = the mean price of a math/science text on the selected site. Random variable: X 1 ¯ - X 1 ¯ = the difference in the sample mean textbook price between sociology texts and math/science texts. Hypotheses: H 0 : μ 1 - μ 2 = 0 , H a : μ 1 - μ 2 < μ 2 which can be expressed as H 0 s : μ 1 - μ 2 , Ha μ 1 < μ 2 . Distribution for the test: Use t d f ; because each sample has more than 30 observations, d f = n 1 + n 2 - 2 = 33 + 33 - 2 = 64 . Estimate the critical value on the t -table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column. Calculate the test statistic: t c = ( X ¯ 1 - X ¯ 2 ) - 0 s 1 2 n 2 + s 2 2 n 2 = ( 74 . 64 - 111 . 56 ) - 0 49 . 36 2 33 + 66 . 90 2 33 = - 2 . 55 . Using a calculator with t c = - 2 . 55 and d f = 64 , the left-tailed p -value: Decision: Reject H 0 . Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. They take random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: μ day > μ night μ day < μ night μ day = μ night μ day ≠ μ night d Degrees of Freedom ( df ) the number of objects in a sample that are free to vary. Standard Deviation A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Variable (Random Variable) a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters X , Y , Z ,... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters x , y , z ,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in the two following ways. The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. We can tell what specific value x of the random variable X takes only after performing the experiment.", "section": "Two Population Means with Unknown Standard Deviations", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is X 1 ¯ – X 2 ¯ . The normal distribution has the following format: Normal distribution is: X ¯ 1 – X ¯ 2 ~ N [ μ 1 – μ 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ] The standard deviation is: ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 The test statistic ( z -score) is: z = ( x ¯ 1 – x ¯ 2 ) – ( μ 1 – μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax . Both populations have a normal distributions. The data are recorded in . Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation 1 3 0.33 2 2.9 0.36 Does the data indicate that wax 1 is more effective than wax 2 ? Test at a 5% level of significance. This is a test of two independent groups, two population means, population standard deviations known. Random Variable : X ¯ 1 – X ¯ 2 = difference in the mean number of months the competing floor waxes last. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The words \"is more effective\" says that wax 1 lasts longer than wax 2 , on average. \"Longer\" is a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is: X ¯ 1 – X ¯ 2 ~ N ( 0 , 0.33 2 20 + 0.36 2 20 ) Since μ 1 ≤ μ 2 then μ 1 – μ 2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p -value using the normal distribution: p -value = 0.1799 Graph: X ¯ 1 – X ¯ 2 = 3 – 2.9 = 0.1 Compare α and the p -value: α = 0.05 and p -value = 0.1799. Therefore, α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts. Press STAT . Arrow over to TESTS and press 3:2-SampZTest . Arrow over to Stats and press ENTER . Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ 1: and arrow to > μ 2 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw . Try It The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines of each type are randomly assigned to be tested. Both populations have normal distributions. shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine Sample Mean Number of RPM Population Standard Deviation 1 1,500 50 2 1,600 60 An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. During a certain year, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance. This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators Random variable: X ¯ 1 – X ¯ 2 = difference in the mean age of Democratic and Republican U.S. senators. H 0 : µ 1 ≤ µ 2 H 0 : µ 1 – µ 2 ≤ 0 H a : µ 1 > µ 2 H a : µ 1 – µ 2 > 0 The words \"older than\" translates as a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is: X ¯ 1 – X ¯ 2 ∼ N [ 0 , ( 9.55 ) 2 30 + ( 10.17 ) 2 30 ] Since µ 1 ≤ µ 2 , µ 1 – µ 2 ≤ 0 and the mean for the normal distribution is zero. (Calculating the p -value using the normal distribution gives p -value = 0.4955) Graph: Compare α and the p -value: α = 0.05 and p -value = 0.4955. Therefore, α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. Try It The average age of 10 professors selected randomly in university A is 46.672 with a standard deviation of 8.53. The average age of 10 professors selected randomly in university B is 47.531 with a standard deviation of 7.83. Does the data indicate that university A has older professors than university B, on average? Test at a 5% level of significance. The total sample size is 10 + 10 = 20., which is greater than 10, so we can use the Student’s t -distribution. The random variable is X 1 - X 2 = difference between the mean age of universities A and B. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H 0 : μ 1 > μ 2 H 0 : μ 1 - μ 2 > 0 The phrase “older than” translates as a “>” symbol and goes into H 0 . Therefore, this is a right-tailed test. The distribution is: X 1 - X 2 ~ N 0 , ( 8 . 53 ) 2 10 + ( 7 . 83 ) 2 10 The mean for the normal distribution is 0 since μ 1 ≤ μ 2 and μ 1 - μ 2 ≤ 0 . The p -value is obtained as 0.4258. Since alpha is less than the p -value, do not reject the H 0 . At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of professors in university A is more than the mean age of professors in university B. References Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). “State-Specific Prevalence of Obesity AmongAdults—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013). Chapter Review A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics: Random variable: X ¯ 1 − X ¯ 2 = the difference of the means Distribution: normal distribution Formula Review Normal Distribution: X ¯ 1 − X ¯ 2 ∼ N [ μ 1 − μ 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ] . Generally µ 1 – µ 2 = 0. Test Statistic ( z -score): z = ( x ¯ 1 − x ¯ 2 ) − ( μ 1 − μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Generally µ 1 - µ 2 = 0. where: σ 1 and σ 2 are the known population standard deviations. n 1 and n 2 are the sample sizes. x ¯ 1 and x ¯ 2 are the sample means. μ 1 and μ 2 are the population means. Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation Wesley 86 3 Rodriguez 91 7 What is the random variable? The difference in mean speeds of the fastball pitches of the two pitchers State the null and alternative hypotheses. What is the test statistic? –2.46 What is the p -value? At the 1% significance level, what is your conclusion? At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food 16 2.5 No food 14 1.5 Is the population standard deviation known or unknown? State the null and alternative hypotheses. Subscripts: 1 = Food, 2 = No Food H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 What is the p -value? Draw the graph of the p -value. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 95 Alloy Zeta 900 105 State the null and alternative hypotheses. Subscripts: 1 = Gamma, 2 = Zeta H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 Is this a right-, left-, or two-tailed test? What is the p -value? 0.0062 Draw the graph of the p -value. At the 1% significance level, what is your conclusion? There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. Subscripts: 1 = boys, 2 = girls H 0 : µ 1 ≤ µ 2 H a : µ 1 > µ 2 The random variable is the difference in the mean auto insurance costs for boys and girls. normal test statistic: z = 2.50 p -value: 0.0062 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans H 0 : µ 1 ≥ µ 2 H a : µ 1 < µ 2 The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. normal test statistic: 6.36 p -value: 0 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score 2 2 3 3 4 2 1 1 2 4 Husband’s Score 2 2 1 3 2 1 1 1 2 4 H 0 : µ d = 0 H a : µ d < 0 The random variable X d is the average difference between husband’s and wife’s satisfaction level. t 9 test statistic: t = –1.86 p -value: 0.0479 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis, but run another test. Reason for Decision: p -value < alpha Conclusion: This is a weak test because alpha and the p -value are close. However, there is insufficient evidence to conclude that the mean difference is negative.", "section": "Two Population Means with Known Standard Deviations", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Comparing Two Independent Population Proportions When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: The two independent samples are simple random samples that are independent. The number of successes is at least five, and the number of failures is at least five, for each of the samples. Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. Like the case of differences in sample means, we construct a sampling distribution for differences in sample proportions: p ' A = X A n A and p ' B = X B n B are the sample proportions for the two sets of data in question X A and X B . The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H 0 : p A = p B . To conduct the test, we use a pooled proportion , p c . The pooled proportion is calculated as follows: p c = x A + x B n A + n B The distribution for the differences is: P ′ A − P ′ B ~ N [ 0 , p c ( 1 − p c ) ( 1 n A + 1 n B ) ] The test statistic ( z -score) is: z = ( p ′ A − p ′ B ) − ( p A − p B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance. The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for medication A and medication B, respectively. Then p A and p B are the desired population proportions. Random Variable: P′ A – P′ B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B. H 0 : p A = p B p A – p B = 0 H a : p A ≠ p B p A – p B ≠ 0 The words \"is a difference\" tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: p c = x A + x B n A + n B = 20 + 12 200 + 200 = 0.08 1 – p c = 0.92 P ′ A – P ′ B ~ N [ 0 , ( 0.08 ) ( 0.92 ) ( 1 200 + 1 200 ) ] P′ A – P′ B follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.1404. Estimated proportion for group A: p ′ A = x A n A = 20 200 = 0.1 Estimated proportion for group B: p ′ B = x B n B = 12 200 = 0.06 Graph: P′ A – P′ B = 0.1 – 0.06 = 0.04. Half the p -value is below –0.04, and half is above 0.04. Compare α and the p -value: α = 0.01 and the p -value = 0.1404. α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B . Press STAT . Arrow over to TESTS and press 6:2-PropZTest . Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1 : and arrow to not equal p2 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw . Try It Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. A research study was conducted about gender differences regarding the use of seat belts in motor vehicles. The researcher believed that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. The data collected represents a random sample of U.S. adults and is summarized in . Is the proportion of women not wearing seat belts less than the proportion of men not wearing seat belts? Test at a 1% level of significance. Men Women Does not wear seat belts 183 156 Total number surveyed 2231 2169 This is a test of two population proportions. Let M and F be the subscripts for men and women. Then p M and p F are the desired population proportions. Random Variable: p′ F − p′ M = difference in the proportions of men and women who do not wear seat belts. H 0 : p F = p M H 0 : p F – p M = 0 H a : p F < p M H a : p F – p M < 0 The words \"less than\" tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: p c = x F + x M n F + n M = 156 + 183 2169 + 2231 = 0 .077 1 − p c = 0.923 Therefore, p ′ F – p ′ M ∼ N ( 0 , ( 0.077 ) ( 0.923 ) ( 1 2169 + 1 2231 ) ) p′ F – p′ M follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.1045 Estimated proportion for women: 0.0719 Estimated proportion for men: 0.082 Graph: Decision: Since α < p -value, Do not reject H 0 Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. Try It A survey was conducted about the favorable beverage as tea. The data collected is summarized in the table. Is the proportion of men favoring tea more than women favoring tea? Test at a 1% level of significance. Men Women Favor tea 16 18 Total surveyed 230 218 Let p M and p F be desired population proportions. H 0 : p F = p M H 0 : p F - p M = 0 H 0 : p F < p M H 0 : p F - p M < 0 The test is for “less than,” so it is left-tailed. Distribution for the test: p C = x F + x M n F + n M = 18 + 16 218 + 230 = 0 . 076 1 - p C = 0 . 924 p ' F - p ' M ~ N 0 , ( 0 . 076 ) ( 0 . 924 ) 1 218 + 1 230 This follows an approximate normal distribution. Calculate the p-value using the normal distribution: p -value = 0.1023 Decision: Since α < p - v a l u e , do not reject H 0 . Conclusion: At 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of females favoring tea is less than the proportion of males favoring tea. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER . Arrow down to Calculate and press ENTER. The p -value is P = 0.1045 and the test statistic is z = -1.256. A marketing firm claims that the proportion of younger adults who own electric vehicles is greater than the proportion of older adults who own electric vehicles. A random sample of U.S. adults was taken, and the results of the survey indicate the following: Out of a sample of 232 older adults (aged 35 or older), 5% own electric vehicles. Out of a sample of 1,343 young adults (aged 34 or younger), 10% own electric vehicles. Test at the 5% level of significance. Is the proportion of younger adults greater than the proportion of older adults with respect to owning electric vehicles? This is a test of two population proportions. Let Y and O be the subscripts for younger adults and older adults, respectively. Then pY and pO are the desired population proportions. Random Variable: p’Y and p’O = difference in the proportions of younger and older adults who own electric vehicles. H 0 : p Y = p O H 0 : p Y − p O = 0 H a : p Y > p O H a : p Y - p O > 0 The words \"greater than\" indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal: p c = x y + x o n y + n o = 134 + 12 1343 + 232 = 0 . 0927 1 - p c = 0 . 9073 Therefore, p ' y - p ' O ~ N 0 , 0 . 0927 ) ( 0 . 9073 ) ( 1 1343 + 1 232 ) p ' y - p ' O follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.0077 Estimated proportion for group Y: 0.10 Estimated proportion for group O: 0.05 Graph: Decision: Since  > p -value, reject the H 0 . Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of younger adults own electric vehicles as compared to older adults. TI-83+ and TI-84: Press STAT . Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER . Arrow down to Calculate and press ENTER . The P-value is P = 0.0092 and the test statistic is Z = 2.33. Try It A government researcher is investigating whether there is a difference in the use of helmets by motorcyclists in different geographic regions for those states where use of helmets is required by law. The research shows that for motorcyclists in the northeast U.S., 7622 out of 113,231 motorcyclists did not wear helmets. In the southeast U.S., 7439 out of 104,873 motorcyclists did not wear helmets. Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? two proportions b. Which distribution do you use to perform the test? normal for two proportions c. What is the random variable? Subscripts: 1 = Northeast, 2 = Southeast P ′ 2 - P ′ 2 d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols. Subscripts: 1 = Northeast, 2 = Southeast H 0 : p 1 = p 2 H 0 : p 1 − p 2 = 0 H a : p 1 ≠ p 2 H a : p 1 − p 2 ≠ 0 e. Is this test right-, left-, or two-tailed? two-tailed f. What is the p-value? p -value = 0.00086 g. Do you reject or not reject the null hypothesis? Reject the H 0 . h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________. At the 5% significance level, from the sample data, there is sufficient evidence to conclude that there is a difference between the proportion of motorcyclists who did not wear helmets for northeast and southeast regions of the U.S. References Data from Educational Resources , December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994. “State of the States.” Gallup, 2013. Available online at http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive (accessed June 17, 2013). “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/westnile/index.htm (accessed June 17, 2013). Chapter Review Test of two population proportions from independent samples. Random variable: p ^ A – p ^ B = difference between the two estimated proportions Distribution: normal distribution Formula Review Pooled Proportion: p c = x F + x M n F + n M Distribution for the differences: p ′ A − p ′ B ∼ N [ 0 , p c ( 1 − p c ) ( 1 n A + 1 n B ) ] where the null hypothesis is H 0 : p A = p B or H 0 : p A – p B = 0. Test Statistic ( z -score): z = ( p ′ A − p ′ B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) where the null hypothesis is H 0 : p A = p B or H 0 : p A − p B = 0. where p′ A and p′ B are the sample proportions, p A and p B are the population proportions, P c is the pooled proportion, and n A and n B are the sample sizes. Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS 1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS 2 had system failures within the first eight hours of operation. OS 2 is believed to be more stable (have fewer crashes) than OS 1 . Is this a test of means or proportions? What is the random variable? P ′ OS1 – P ′ OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS 1 and OS 2 . State the null and alternative hypotheses. What is the p -value? 0.1018 What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. Is this a test of means or proportions? proportions State the null and alternative hypotheses. H 0 : _________ H a : _________ Is this a right-tailed, left-tailed, or two-tailed test? How do you know? right-tailed What is the random variable of interest for this test? In words, define the random variable for this test. The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for the Exercise 10.56 . Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. Calculate the test statistic. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p -value. Answers may vary. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Reject the null hypothesis. p -value < alpha At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. We are interested in whether the proportions of conferred physical science degrees for people aged 21 to 24 are the same for White and Black people in the United States. The number of conferred degrees for White people in a given year is 4,930. Five hundred eighty were aged 21 to 24. The estimate for Black people is 330. Forty were aged 21 to 24. We will let people with conferred physical science degrees be our population. H 0 : P W = P B H a : P W ≠ P B The random variable is the difference in the proportions of White and Black people with conferred degrees in a given year, aged 21 to 24. normal for two proportions test statistic: –0.1944 p -value: 0.8458 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of White and Black people with conferred physical science degrees, aged 21 to 24, are different. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic/Latino students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic/Latino students out of a total of 2,441 students. In general, do you think that the percent of Hispanic/Latino students at the two colleges is basically the same or different? Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The random variable is the difference between the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College. normal for two proportions test statistic: 4.29 p -value: 0.00002 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College are different. Use the following information to answer the next three exercises. Some individuals who were exposed to the COVID-19 virus experienced a loss of taste. In a sample of COVID patients during 2022, there were 629 reported cases of loss of taste out of a total of 1,021 reported cases. In 2021, there were 486 reported cases of loss of taste out of a sample of 712 cases. Is the 2022 proportion of loss of taste more than the 2021 proportion? Using a 1% level of significance, conduct an appropriate hypothesis test. “2022” subscript: 2022 group. “2021” subscript: 2021 group This is: a test of two proportions a test of two independent means a test of a single mean a test of matched pairs. An appropriate null hypothesis is: p 2022 ≤ p 2021 p 2022 ≥ p 2021 μ 2022 ≤ μ 2021 p 2022 > p 2021 a The p -value is 0.0022. At a 1% level of significance, the appropriate conclusion is There is sufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is less than the proportion of people in the United States in 2021 who experienced loss of taste. There is insufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is more than the proportion of people in the United States in 2021 who experienced loss of taste. There is insufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is less than the proportion of people in the United States in 2021 who experienced loss of taste. There is sufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is more than the proportion of people in the United States in 2021 who experienced loss of taste. Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. Test: two independent sample proportions. Random variable: p ′ 1 - p ′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed p -value : 0.0033 Decision: Reject the null hypothesis. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older. Adults aged 18 years old and older were randomly selected for a survey on working from home. The researchers wanted to determine if the proportion of women who work from home is less than the proportion of men who work from home. The results are shown in . Test at the 1% level of significance. Number who work from home Sample size Men 42,769 155,525 Women 67,169 248,775 Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a Tablet 69 231 Sample Size 628 2,309 Test: two independent sample proportions Random variable: p′ 1 − p′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 > p 2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: right-tailed p -value: 0.2354 Decision: Do not reject the H 0 . Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. A group of friends debated whether more people aged 21-30 use wearable fitness devices than people aged 31-40. They consulted a research study of wearable fitness devices use among adults. The results of the survey indicate that of the 973 randomly sampled people in their twenties, 379 use wearable fitness devices. For people in their thirties, 404 of the 1,304 who were randomly sampled use wearable fitness devices. Test at the 5% level of significance. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Subscripts: 1: men; 2: women H 0 : p 1 ≤ p 2 H a : p 1 > p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. normal for two proportions test statistic: 0.22 p -value: 0.4133 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. Joan Nguyen recently claimed that the proportion of college-age men with at least one pierced ear is as high as the proportion of college-age women. She conducted a survey in her classes. Out of 107 men, 20 had at least one pierced ear. Out of 92 women, 47 had at least one pierced ear. Do you believe that the proportion of men has reached the proportion of women? H 0 : p 1 = p 2 H a : p 1 ≠ p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women that have at least one pierced ear. normal for two proportions test statistic: –4.82 p -value: zero Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of men and women with at least one pierced ear is different. Use the data sets found in to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? \"To Breakfast or Not to Breakfast?\" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what they have achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in , solve our problem. Work hours with breakfast Work hours without breakfast 8 6 7 5 9 5 5 4 9 7 8 7 10 7 7 5 6 6 9 5 H 0 : µ d = 0 H a : µ d > 0 The random variable X d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t 9 test statistic: 4.8963 p -value: 0.0004 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased. Pooled Proportion estimate of the common value of p 1 and p 2 .", "section": "Comparing Two Independent Population Proportions", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Matched or Paired Samples When using a hypothesis test for matched or paired samples, the following characteristics should be present: Simple random sampling is used. Sample sizes are often small. Two measurements (samples) are drawn from the same pair of individuals or objects. Differences are calculated from the matched or paired samples. The differences form the sample that is used for the hypothesis test. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences. The test statistic ( t -score) is: t = x ¯ d − μ d ( s d n ) A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in . A lower score indicates less pain. The \"before\" value is matched to an \"after\" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Corresponding \"before\" and \"after\" values form matched pairs. (Calculate \"after\" – \"before.\") After Data Before Data Difference 6.8 6.6 0.2 2.4 6.5 -4.1 7.4 9 -1.6 8.5 10.3 -1.8 8.1 11.3 -3.2 6.1 8.1 -2 3.4 6.3 -2.9 2 11.6 -9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: x – d = –3.13 and s d = 2.91 Verify these values. Let μ d be the population mean for the differences. We use the subscript d to denote \"differences.\" Random variable: X ¯ d = the mean difference of the sensory measurements H 0 : μ d ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ d is the population mean of the differences.) H a : μ d < 0 The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t 7 . (Notice that the test is for a single population mean.) Calculate the test statistic and p -value using the Student's t -distribution: test statistic, t = –3.036, p -value = 0.0095 Graph: X ¯ d is the random variable for the differences. The sample mean and sample standard deviation of the differences are: x ¯ d = –3.13 s ¯ d = 2.91 Compare α and the p -value: α = 0.05 and p -value = 0.0095. α > p -value. Make a decision: Since α > p -value, reject H 0 . This means that μ d < 0 and there is improvement. Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain. NOTE For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time ( after - before ) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1 st list name - 2 nd list name. The calculator will do the subtraction, and you will have the differences in the third list. Use your list of differences as the data. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 0 for μ 0 , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ : and arrow over to < μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate ). Press ENTER . Try It A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level. Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 A college softball coach was interested in whether the college's strength development class increased their players' maximum lift (in pounds). Four players participated in the study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 241 338 368 Amount of weight lifted after the class 295 252 330 360 The coach wants to know if the strength development class makes the players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x ¯ d = 21.3 , s d = 46.7 NOTE The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative. Using the difference data, this becomes a test of a single __________ (fill in the blank). Define the random variable: X ¯ d mean difference in the maximum lift per player. The distribution for the hypothesis test is t 3 . H 0 : μ d ≤ 0, H a : μ d > 0 Graph: Calculate the p -value: The p -value is 0.2150 Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p -value. What is the conclusion? At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. It should be noted that this data set consists of a very small sample size, with the possibility of an outlier data value. It is important to confirm that the matched pairs have differences that come from a population that is normal. The coach may want to consider increasing the sample size of players in the study. Try It A new prep class was designed to improve SAT test scores. Four students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in . Are the scores, on average, higher after the class? Test at a 5% level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 1960 1920 2150 Score after class 1920 2160 2200 2100 Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in . Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand 30 26 34 17 19 26 20 Weaker Hand 28 14 27 18 17 26 16 Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant. Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x ¯ d = 3.71, s d = 4.5. Random variable: X ¯ d = mean difference in the distances between the hands. Distribution for the hypothesis test: t 6 H 0 : μ d = 0 H a : μ d ≠ 0 Graph: Calculate the p -value: The p -value is 0.0716 (using the data directly). (test statistic = 2.18. p -value = 0.0719 using ( x ¯ d = 3.71 , s d = 4.5. ) Decision: Assume α = 0.05. Since α < p -value, Do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put. Try-It Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 111 135 140 125 Off-hand 105 109 98 111 99 Chapter Review A hypothesis test for matched or paired samples (t-test) has these characteristics: Test the differences by subtracting one measurement from the other measurement Random Variable: x ¯ d = mean of the differences Distribution: Student’s-t distribution with n – 1 degrees of freedom If the number of differences is small (less than 30), the differences must follow a normal distribution. Two samples are drawn from the same set of objects. Samples are dependent. Formula Review Test Statistic ( t -score): t = x ¯ d − μ d ( s d n ) where: x ¯ d is the mean of the sample differences. μ d is the mean of the population differences. s d is the sample standard deviation of the differences. n is the sample size. Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in . The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. Installation A B C D E F G H Before 3 6 4 2 5 8 2 6 After 1 5 2 0 1 0 2 2 What is the random variable? the mean difference of the system failures State the null and alternative hypotheses. What is the p -value? 0.0067 Draw the graph of the p -value. What conclusion can you draw about the software patch? With a p -value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. Subject A B C D E F Before 3 4 3 2 4 5 After 4 5 6 4 5 7 State the null and alternative hypotheses. What is the p -value? 0.0021 What is the sample mean difference? Draw the graph of the p -value. What conclusion can you draw about the juggling class? Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 State the null and alternative hypotheses. H 0 : μ d ≥ 0 H a : μ d < 0 What is the test statistic? What is the p -value? 0.0699 What is the sample mean difference? What is the conclusion? We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in . Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 140 220 230 110 120 240 220 200 190 180 150 190 200 360 300 280 300 260 240 p -value = 0.1494 At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. Use the following information to answer the next two exercises. A new heart disease prevention drug was tried on a group of 224 patients. Forty-five patients developed heart disease after four years. In a control group of 224 patients, 68 developed heart disease after four years. We want to test whether the method of treatment reduces the proportion of patients that develop heart disease after four years or if the proportions of the treated group and the untreated group stay the same. Let the subscript t = treated patient and ut = untreated patient. The appropriate hypotheses are: H 0 : p t < p ut and H a : p t ≥ p ut H 0 : p t ≤ p ut and H a : p t > p ut H 0 : p t = p ut and H a : p t ≠ p ut H 0 : p t = p ut and H a : p t < p ut If the p -value is 0.0062 what is the conclusion (use α = 0.05)? The method has no effect. There is sufficient evidence to conclude that the method reduces the proportion of patients who develop heart disease after four years. There is sufficient evidence to conclude that the method increases the proportion of patients who develop heart disease after four years. There is insufficient evidence to conclude that the method reduces the proportion of patients who develop heart disease after four years. b Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: x ¯ d = −10.2 s d = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training. The distribution for the test is: t 5 t 6 N (−10.2, 8.4) N(−10.2, 8.4 6 ) If α = 0.05, the p -value and the conclusion are 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training. c A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 The correct decision is: Reject H 0 . Do not reject the H 0 . A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. The group compared the estimates of new female breast cancer cases by southern state in Year 1 and in Year 2. The results are in . Southern States Year 1 Year 2 Alabama 3,450 3,720 Arkansas 2,150 2,280 Florida 15,540 15,710 Georgia 6,970 7,310 Kentucky 3,160 3,300 Louisiana 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas 15,050 14,980 Virginia 6,190 6,280 Test: two matched pairs or paired samples ( t -test) Random variable: X ¯ d Distribution: t 12 H 0 : μ d = 0 H a : μ d > 0 The mean of the differences of new female breast cancer cases in the south between Year 2 and Year 1 is greater than zero. The estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. Graph: right-tailed p -value: 0.0004 Decision: Reject H 0 Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in Year 2 than in Year 1. A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in . Test at the 1% level of significance. Cities Hyatt Regency prices in dollars Hilton prices in dollars Atlanta 107 169 Boston 358 289 Chicago 209 299 Dallas 209 198 Denver 167 169 Indianapolis 179 214 Los Angeles 179 169 New York City 625 459 Philadelphia 179 159 Washington, DC 245 239 A politician asked their staff to determine whether the underemployment rate in the northeast decreased year over year. The results are in . Northeastern States Year 1 Year 2 Connecticut 17.3 16.4 Delaware 17.4 13.7 Maine 19.3 16.1 Maryland 16.0 15.5 Massachusetts 17.6 18.2 New Hampshire 15.4 13.5 New Jersey 19.2 18.7 New York 18.5 18.7 Ohio 18.2 18.8 Pennsylvania 16.5 16.9 Rhode Island 20.7 22.4 Vermont 14.7 12.3 West Virginia 15.5 17.3 Test: matched or paired samples ( t -test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8} Random Variable: X ¯ d Distribution: H 0 : μ d = 0 H a : μ d < 0 The mean of the differences of the rate of underemployment in the northeastern states between Year 2 and Year 1 is less than zero. The underemployment rate went down from Year 1 to Year 2. Graph: left-tailed. p -value: 0.1207 Decision: Do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from Year 1 to Year 2. Bringing It Together Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. independent group means, population standard deviations and/or variances known independent group means, population standard deviations and/or variances unknown matched or paired samples single mean two proportions single proportion A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. e The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. d A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively. According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this. f According to a recent study, U.S. companies have a mean maternity-leave of six weeks. A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. e A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 300 960 920 1010 1100 840 880 1100 1070 1250 1320 860 860 1330 1370 790 770 990 1040 1110 1200 740 850 University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked. f Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown . Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size 41 41 Sample mean 97.5 98.1 Sample standard deviation 17.5 19.2 Two independent means, normal distribution Two independent means, Student’s-t distribution Matched or paired samples, Student’s-t distribution Two population proportions, normal distribution A golf instructor is interested in determining if a new technique for improving players’ golf scores is effective. The instructor takes four (4) new students and records their 18-hole scores before learning the technique and then after having taken the class. The instructor conducts a hypothesis test. The data are as . Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 This is: a test of two independent means. a test of two proportions. a test of a single mean. a test of a single proportion. a", "section": "Matched or Paired Samples", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Hypothesis Testing for Two Means and Two Proportions Hypothesis Testing for Two Means and Two Proportions Class Time: Names: Student Learning Outcomes The student will select the appropriate distributions to use in each case. The student will conduct hypothesis tests and interpret the results. Supplies: the business section from two consecutive days’ newspapers three small packages of M&Ms® five small packages of Reese's Pieces® Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements. H 0 : _________ H a : _________ In words, define the random variable. The distribution to use for the test is _____________. Calculate the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Calculate the p -value. Do you reject or not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence. Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day. H 0 : ________ H a : ________ In words, define the random variable. The distribution to use for the test is _____________. Calculate the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Calculate the p -value: Do you reject or not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence. Candy Survey Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands. H 0 : ________ H a : ________ In words, define the random variable. What distribution should be used for this test? Calculate the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Calculate the p -value. Do you reject or not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence. Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample. H 0 : ________ H a : ________ In words, define the random variable. The distribution to use for the test is ________________. Calculate the test statistic using your data. Draw a graph and label it appropriately. Shade the actual level of significance. Graph: Calculate the p -value. Do you reject or not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.", "section": "Hypothesis Testing for Two Means and Two Proportions", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: modification of work “The first few times I used my Asda 'bag for life' I left the receipts in the bottom of the bag” by Pete/ Flickr, Public domain) Chapter Objectives By the end of this chapter, the student should be able to: Interpret the chi-square probability distribution as the sample size changes. Conduct and interpret chi-square goodness-of-fit hypothesis tests. Conduct and interpret chi-square test of independence hypothesis tests. Conduct and interpret chi-square homogeneity hypothesis tests. Conduct and interpret chi-square single variance hypothesis tests. Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example the test of independence, which determines if events are independent, such as in the movie example the test of a single variance, which tests variability, such as in the coffee example NOTE Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see ). TI-83+ and TI-84 calculator instructions are included in the text. Collaborative Classroom Exercise Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like). Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Facts About the Chi-Square Distribution The notation for the chi-square distribution is: X ∼ χ d f 2 where df = degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chi-square probabilities then use df = n - 1. The degrees of freedom for the three major uses are each calculated differently.) For the χ 2 distribution, the population mean is μ = df and the population standard deviation is σ = 2 ( d f ) . The random variable is shown as χ 2 , but may be any upper case letter. The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables. χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + ... + ( Z k ) 2 The curve is nonsymmetrical and skewed to the right. There is a different chi-square curve for each df . The test statistic for any test is always greater than or equal to zero. When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ 1,000 2 the mean, μ = df = 1,000 and the standard deviation, σ = 2 ( 1,000 ) = 44.7. Therefore, X ~ N (1,000, 44.7), approximately. The mean, μ , is located just to the right of the peak. References Data from Parade Magazine . “HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011. Chapter Review The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df . For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. Formula Review χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + … ( Z df ) 2 chi-square distribution random variable μ χ 2 = df chi-square distribution population mean σ χ 2 = 2 ( d f ) Chi-Square distribution population standard deviation If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? mean = 25 and standard deviation = 7.0711 If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. When does the chi-square curve approximate a normal distribution? when the number of degrees of freedom is greater than 90 Where is μ located on a chi-square curve? Is it more likely the df is 90, 20, or two in the graph? df = 2 Homework Decide whether the following statements are true or false. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. true The standard deviation of the chi-square distribution is twice the mean. The mean and the median of the chi-square distribution are the same if df = 24. false", "section": "Facts About the Chi-Square Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Goodness-of-Fit Test In this type of hypothesis test, you determine whether the data \"fit\" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: Σ k ( O − E ) 2 E where: O = observed values (data) E = expected values (from theory) k = the number of different data cells or categories The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form ( O − E ) 2 E . The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The expected value for each cell needs to be at least five in order for you to use this test. Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to . Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9–11 6 12+ 2 A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in displays the results of that survey. Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9–11 1 12+ 4 Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test. H 0 : Student absenteeism fits faculty perception. The alternative hypothesis is the opposite of the null hypothesis. H a : Student absenteeism does not fit faculty perception. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test? a. No. Notice that the expected number of absences for the \"12+\" entry is less than five (it is two). Combine that group with the \"9–11\" group to create new tables where the number of students for each entry are at least five. The new results are in and . Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9+ 8 Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9+ 5 b. What is the number of degrees of freedom ( df )? b. There are four \"cells\" or categories in each of the new tables. df = number of cells – 1 = 4 – 1 = 3 Try It A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in . Number produced Number defective 0–100 5 101–200 6 201–300 7 301–400 8 401–500 10 A random sample was taken to determine the actual number of defects. shows the results of the survey. Number produced Number defective 0–100 5 101–200 7 201–300 8 301–400 9 401–500 11 State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom. Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in . For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level. Day of the Week Employees were Most Absent Monday Tuesday Wednesday Thursday Friday Number of Absences 15 12 9 9 15 The null and alternative hypotheses are: H 0 : The absent days occur with equal frequencies, that is, they fit a uniform distribution. H a : The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected ( E ) values. The values in the table are the observed ( O ) values or data. This time, calculate the χ 2 test statistic by hand. Make a chart with the following headings and fill in the columns: Expected ( E ) values (12, 12, 12, 12, 12) Observed ( O ) values (15, 12, 9, 9, 15) ( O – E ) ( O – E ) 2 ( O – E ) 2 E Now add (sum) the last column. The sum is three. This is the χ 2 test statistic. To find the p -value, calculate P ( χ 2 > 3). This test is right-tailed. (Use a computer or calculator to find the p -value. You should get p -value = 0.5578.) The dfs are the number of cells – 1 = 5 – 1 = 4 Press 2nd DISTR . Arrow down to χ 2 cdf . Press ENTER . Enter (3,10^99,4) . Rounded to four decimal places, you should see 0.5578, which is the p-value. Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.) The decision is not to reject the null hypothesis. Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies. TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList ). Enter the list name and press ENTER . Try It Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in . Sunday Monday Tuesday Wednesday Thursday Friday Saturday Number of Students 11 8 10 7 10 5 5 From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use? One study indicates that the number of streaming services that American families have is distributed (this is the given distribution for the American population) as in . Number of Streaming Services Percent 0 10 1 16 2 55 3 11 4+ 8 The table contains expected ( E ) percents. A random sample of 600 families in the far western United States resulted in the data in . Number of Streaming Services Frequency 0 66 1 119 2 340 3 60 4+ 15 Total = 600 The table contains observed ( O ) frequency values. At the 1% significance level, does it appear that the distribution \"number of streaming services\" of far western United States families is different from the distribution for the American population as a whole? This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected ( E ) frequencies, multiply the percentage by 600. The expected frequencies are shown in . Number of Streaming Services Percent Expected Frequency 0 10 (0.10)(600) = 60 1 16 (0.16)(600) = 96 2 55 (0.55)(600) = 330 3 11 (0.11)(600) = 66 over 3 8 (0.08)(600) = 48 Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600. H 0 : The \"number of streaming services\" distribution of far western United States families is the same as the \"number of streaming services\" distribution of the American population. H a : The \"number of streaming services\" distribution of far western United States families is different from the \"number of streaming services\" distribution of the American population. Distribution for the test: χ 4 2 where df = (the number of cells) – 1 = 5 – 1 = 4. NOTE df ≠ 600 – 1 Calculate the test statistic: χ 2 = 29.65 Graph: Probability statement: p -value = P ( χ 2 > 29.65) = 0.000006 Compare α and the p -value: α = 0.01 p -value = 0.000006 So, α > p -value. Make a decision: Since α > p -value, reject H o . This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole. Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the \"number of streaming services\" distribution for the far western United States is different from the \"number of streaming services\" distribution for the American population as a whole. Press STAT and ENTER . Make sure to clear lists L1 , L2 , and L3 if they have data in them (see the note at the end of ). Into L1 , put the observed frequencies 66 , 119 , 340 , 60 , 15 . Into L2 , put the expected frequencies .10*600, .16*600 , .55*600 , .11*600 , .08*600 . Arrow over to list L3 and up to the name area \"L3\" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see \"sum\" (Enter L3) . Rounded to 2 decimal places, you should see 29.65 . Press 2nd DISTR . Press 7 or Arrow down to 7:χ2cdf and press ENTER . Enter (29.65,1E99,4) . Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start. Try It The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in . Number of Pets Percent 0 18 1 25 2 30 3 18 4+ 9 A random sample of 1,000 students from the Eastern United States resulted in the data in . Number of Pets Frequency 0 210 1 240 2 320 3 140 4+ 90 At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p -value? Suppose you flip two coins 100 times. The results are 20 HH , 27 HT , 30 TH , and 23 TT . Are the coins fair? Test at a 5% significance level. This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is { HH , HT , TH , TT }. Out of 100 flips, you would expect 25 HH , 25 HT , 25 TH , and 25 TT . This is the expected distribution. The question, \"Are the coins fair?\" is the same as saying, \"Does the distribution of the coins (20 HH , 27 HT , 30 TH , 23 TT ) fit the expected distribution?\" Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three . Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed. H 0 : The coins are fair. H a : The coins are not fair. Distribution for the test: χ 2 2 where df = 3 – 1 = 2. Calculate the test statistic: χ 2 = 2.14 Graph: Probability statement: p -value = P ( χ 2 > 2.14) = 0.3430 Compare α and the p -value: α = 0.05 p -value = 0.3430 α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: There is insufficient evidence to conclude that the coins are not fair. Press STAT and ENTER . Make sure you clear lists L1 , L2 , and L3 if they have data in them. Into L1 , put the observed frequencies 20 , 57 , 23 . Into L2 , put the expected frequencies 25 , 50 , 25 . Arrow over to list L3 and up to the name area \"L3\" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see \"sum\" . Enter L3 . Rounded to two decimal places, you should see 2.14 . Press 2nd DISTR . Arrow down to 7:χ2cdf (or press 7 ). Press ENTER . Enter 2.14,1E99,2) . Rounded to four places, you should see .3430 , which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start. Try It Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom? MDG Region Adult Literacy Rate (%) Developed Regions 99.0 Commonwealth of Independent States 99.5 Northern Africa 67.3 Sub-Saharan Africa 62.5 Latin America and the Caribbean 91.0 Eastern Asia 93.8 Southern Asia 61.9 South-Eastern Asia 91.9 Western Asia 84.5 Oceania 66.4 References Data from the U.S. Census Bureau Data from the College Board. Available online at http://www.collegeboard.com. Data from the U.S. Census Bureau, Current Population Reports. Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92. Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013). Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013). Chapter Review To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five. Formula Review ∑ k ( O − E ) 2 E goodness-of-fit test statistic where: O : observed values E : expected values k : number of different data cells or categories df = k − 1 degrees of freedom Determine the appropriate test to be used in the next three exercises. An archeologist is calculating the distribution of the frequency of the number of artifacts they find in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, they compare the actual number of artifacts found in each grid section to see if their expectation was accurate. An economist is deriving a model to predict outcomes on the stock market. They create a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, The economist records the actual points on the index. They want to see how well the model matched what actually happened. a goodness-of-fit test A personal trainer is putting together a weight-lifting program for their clients. For a 90-day program, they expect each client to lift a specific maximum weight each week. As the program goes along, the trainer records the actual maximum weights their clients lifted. They want to know how well their expectations met with what was observed. Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in . Grade Proportion A 0.25 B 0.30 C 0.35 D 0.10 The actual distribution for a class of 20 is in . Grade Frequency A 7 B 7 C 5 D 1 d f = ______ 3 State the null and alternative hypotheses. χ 2 test statistic = ______ 2.04 p -value = ______ At the 5% significance level, what can you conclude? We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores. Use the following information to answer the next nine exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in . Ethnicity Number of Cases White 2,229 Hispanic/Latino 1,157 Black/African-American 457 Asian, Pacific Islander 232 Total = 4,075 The percentage of each ethnic group in Santa Clara County is as in . Ethnicity Percentage of total county population Number expected (round to two decimal places) White 42.9% 1748.18 Hispanic/Latino 26.7% Black/African-American 2.6% Asian, Pacific Islander 27.8% Total = 100% If the ethnicities of COVID-19 cases victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group. Perform a goodness-of-fit test to determine whether the occurrence of COVID-19 cases follows the ethnicities of the general population of Santa Clara County. H 0 : _______ H 0 : the distribution of COVID-19 cases follows the ethnicities of the general population of Santa Clara County. H a : _______ Is this a right-tailed, left-tailed, or two-tailed test? right-tailed degrees of freedom = _______ χ 2 test statistic = _______ 2016.136 p -value = _______ Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p -value. Let α = 0.05 Decision: ________________ Reason for the Decision: ________________ Conclusion (write out in complete sentences): ________________ Graph: Answers may vary. Decision: Reject the null hypothesis. Reason for the Decision: p -value < alpha Conclusion (write out in complete sentences): The make-up of COVID-19 cases does not fit the ethnicities of the general population of Santa Clara County. Does it appear that the pattern of COVID-19 cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? Homework For each problem, use a solution sheet to solve the hypothesis test problem. Go to for the chi-square solution sheet. Round expected frequency to two decimal places. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in are the result of the 120 rolls. Face Value Frequency Expected Frequency 1 15 2 29 3 16 4 15 5 30 6 15 The marital status distribution of the population of U.S. men, ages 15 and older, is as shown in . Marital Status Percent Expected Frequency never married 31.3 married 56.1 widowed 2.5 divorced/separated 10.1 Suppose that a random sample of 400 U.S. young adults, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in , rounding to two decimal places. Marital Status Frequency never married 140 married 238 widowed 2 divorced/separated 20 Marital Status Percent Expected Frequency never married 31.3 125.2 married 56.1 224.4 widowed 2.5 10 divorced/separated 10.1 40.4 The data fits the distribution. The data does not fit the distribution. 3 chi-square distribution with df = 3 19.27 0.0002 Check student’s solution. Alpha = 0.05 Decision: Reject null Reason for decision: p -value < alpha Conclusion: Data does not fit the distribution. Use the following information to answer the next two exercises: The columns in contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for students using mass transit to get to school, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who used mass transit to get to school. Race/Ethnicity Percentage Who Use Mass Transit to Get to School Overall Student Population Survey Frequency Asian, Asian American, or Pacific Islander 10.2% 5.4% 113 Black or African-American 8.2% 14.5% 94 Hispanic or Latino 15.5% 15.9% 136 American Indian or Alaska Native 0.6% 1.2% 10 White 59.4% 61.6% 604 Not reported/other 6.1% 1.4% 43 Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity. Perform a goodness-of-fit test to determine whether the local results follow the distribution of those that use mass transit, based on ethnicity. H 0 : The local results follow the distribution of the percentage of students who use mass transit to get to school H a : The local results do not follow the distribution of the percentage of students who use mass transit to get to school df = 5 chi-square distribution with df = 5 chi-square test statistic = 13.4 p -value = 0.0199 Answers may vary. Alpha = 0.05 Decision: Reject null when a = 0.05 Reason for Decision: p -value < alpha Conclusion: Local data do not fit the mass transit Distribution. Decision: Do not reject null when a = 0.01 Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the of students who use mass transit. The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in . Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area. Race Lake Tahoe Frequency Manhattan Frequency Asian Indian 131 174 Chinese 118 557 Filipino 1,045 518 Japanese 80 54 Korean 12 29 Vietnamese 9 21 Other 24 66 Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 first-year students from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006) . Suppose a survey of 5,000 graduating women and 5,000 graduating men was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for and . The second column in each table does not add to 100% because of rounding. Conduct a goodness-of-fit test to determine if the actual college majors of graduating women fit the distribution of their expected majors. Major Women - Expected Major Women - Actual Major Arts & Humanities 14.0% 670 Biological Sciences 8.4% 410 Business 13.1% 685 Education 13.0% 650 Engineering 2.6% 145 Physical Sciences 2.6% 125 Professional 18.9% 975 Social Sciences 13.0% 605 Technical 0.4% 15 Other 5.8% 300 Undecided 8.0% 420 H 0 : The actual college majors of graduating women fit the distribution of their expected majors H a : The actual college majors of graduating women do not fit the distribution of their expected majors df = 10 chi-square distribution with df = 10 test statistic = 11.48 p -value = 0.3211 Answers may vary. Alpha = 0.05 Decision: Do not reject null when a = 0.05 and a = 0.01 Reason for decision: p -value > alpha Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating women do not fit the distribution of their expected majors. Conduct a goodness-of-fit test to determine if the actual college majors of graduating men fit the distribution of their expected majors. Major Men - Expected Major Men - Actual Major Arts & Humanities 11.0% 600 Biological Sciences 6.7% 330 Business 22.7% 1130 Education 5.8% 305 Engineering 15.6% 800 Physical Sciences 3.6% 175 Professional 9.3% 460 Social Sciences 7.6% 370 Technical 1.8% 90 Other 8.2% 400 Undecided 6.6% 340 Read the statement and decide whether it is true or false. In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true. true In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not. true The test to use to determine if a six-sided die is fair is a goodness-of-fit test. In a goodness-of fit test, if the p -value is 0.0113, in general, do not reject the null hypothesis. false A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values. Business Type Number in class Observed Number that recycle one commodity Expected number that recycle one commodity Office 35 19 17.5 Retail/Wholesale 48 27 24 Food/Restaurants 53 35 26.5 Manufacturing/Medical 52 21 26 Hotel/Mixed 24 9 12 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population. Age class (years) Obese (percentage) Expected USA average (percentage) 20–30 15.0 32.6 31–40 26.5 32.6 41–50 13.6 36.6 51–60 21.9 36.6 61–70 21.0 39.7 The hypotheses for the goodness-of-fit test are: H 0 : Surveyed obese fit the distribution of expected obese H a : Surveyed obese do not fit the distribution of expected obese Use a chi-square distribution with df = 4 to evaluate the data. The test statistic is X 2 = 9.85 The P -value = 0.0431 At 5% significance level, α = 0.05. For this data, P < α. Reject the null hypothesis. At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.", "section": "Goodness-of-Fit Test", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Test of Independence Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test: Σ ( i ⋅ j ) ( O – E ) 2 E where: O = observed values E = expected values i = the number of rows in the table j = the number of columns in the table There are i ⋅ j terms of the form ( O – E ) 2 E . A test of independence determines whether two factors are independent or not. You first encountered the term independence in Probability Topics . As a review, consider the following example. NOTE The expected value for each cell needs to be at least five in order for you to use this test. Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P ( A AND B ) = P ( A ) P ( B ). A AND B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not. Let y = expected number of drivers who used a cell phone while driving and received speeding violations. If A and B are independent, then P ( A AND B ) = P ( A ) P ( B ). By substitution, y 755 = ( 70 755 ) ( 305 755 ) Solve for y : y = ( 70 ) ( 305 ) 755 = 28.3 About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations. In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors , the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent) . If we do a test of independence using the example, then the null hypothesis is: H 0 : Being a cell phone user while driving and receiving a speeding violation are independent events. If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit. The number of degrees of freedom for the test of independence is: df = (number of columns - 1)(number of rows - 1) The following formula calculates the expected number ( E ): E = (row total)(column total) total number surveyed Try It A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll? In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In is a sample of the adult volunteers and the number of hours they volunteer per week. Number of Hours Worked Per Week by Volunteer Type (Observed) The table contains observed (O) values (data). Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total Community College Students 111 96 48 255 Four-Year College Students 96 133 61 290 Nonstudents 91 150 53 294 Column Total 298 379 162 839 Is the number of hours volunteered independent of the type of volunteer? The observed table and the question at the end of the problem, \"Is the number of hours volunteered independent of the type of volunteer?\" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer . This test is always right-tailed. H 0 : The number of hours volunteered is independent of the type of volunteer. H a : The number of hours volunteered is dependent on the type of volunteer. The expected results are in . Number of Hours Worked Per Week by Volunteer Type (Expected) Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours Community College Students 90.57 115.19 49.24 Four-Year College Students 103.00 131.00 56.00 Nonstudents 104.42 132.81 56.77 The table contains expected ( E ) values (data). For example, the calculation for the expected frequency for the top left cell is E = ( row total ) ( column total ) total number surveyed = ( 255 ) ( 298 ) 839 = 90.57 Calculate the test statistic: χ 2 = 12.99 (calculator or computer) Distribution for the test: χ 4 2 df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4 Graph: Probability statement: p -value= P ( χ 2 > 12.99) = 0.0113 Compare α and the p -value: Since no α is given, assume α = 0.05. p -value = 0.0113. α > p -value. Make a decision: Since α > p -value, reject H 0 . This means that the factors are not independent. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another. For the example in , if there had been another type of volunteer, teenagers, what would the degrees of freedom be? Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 3 ENTER 3 ENTER . Enter the table values by row from . Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . If necessary, use the arrow keys to move the cursor after Observed: and press 2nd MATRX . Press 1:[A] to select matrix A. It is not necessary to enter expected values. The matrix listed after Expected: can be blank. Arrow down to Calculate . Press ENTER . The test statistic is 12.9909 and the p -value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate . Try It The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. shows the results: Industry Sector 2000 2010 2020 Total Nonagriculture wage and salary 13,243 13,044 15,018 41,305 Goods-producing, excluding agriculture 2,457 1,771 1,950 6,178 Services-providing 10,786 11,273 13,068 35,127 Agriculture, forestry, fishing, and hunting 240 214 201 655 Nonagriculture self-employed and unpaid family worker 931 894 972 2,797 Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36 Secondary jobs as a self-employed or unpaid family worker 196 144 152 492 Total 27,867 27,351 31,372 86,590 We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom. De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. Need to Succeed in School vs. Anxiety Level Need to Succeed in School High Anxiety Med-high Anxiety Medium Anxiety Med-low Anxiety Low Anxiety Row Total High Need 35 42 53 15 10 155 Medium Need 18 48 63 33 31 193 Low Need 4 5 11 15 17 52 Column Total 57 95 127 63 58 400 a. How many high anxiety level students are expected to have a high need to succeed in school? a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 155 ⋅ 57 400 = 22.09 The expected number of students who have a high anxiety level and a high need to succeed in school is about 22. b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety? b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 8.19 The expected number of students who have a low need to succeed in school and a med-low level of anxiety is 8. Try It Refer back to the information in . How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020? References DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/Rls2436.pdf (accessed May 24, 2013). Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-ice-cream (accessed May 24, 2013) “Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013). Chapter Review To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5. Formula Review Test of Independence The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1). The test statistic is Σ ( i ⋅ j ) ( O – E ) 2 E where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table. If the null hypothesis is true, the expected number E = (row total)(column total) total surveyed . Determine the appropriate test to be used in the next three exercises. A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups. a test of independence The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. They take a random sample of 100 players from different organizations. A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. They take a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing. a test of independence Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel. Traveling Distance Third class Second class First class Total 1–100 miles 21 14 6 41 101–200 miles 18 16 8 42 201–300 miles 16 17 15 48 301–400 miles 12 14 21 47 401–500 miles 6 6 10 22 Total 73 67 60 200 State the hypotheses. H 0 : _______ H a : _______ df = _______ 8 How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets? How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets? 6.6 What is the test statistic? What is the p -value? 0.0435 What can you conclude at the 5% level of significance? Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people. Complete the table. Smoking Levels by Ethnicity (Observed) Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese American White TOTALS 1-10 11-20 21-30 31+ TOTALS Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White Totals 1-10 9,886 2,745 12,831 8,378 7,650 41,490 11-20 6,514 3,062 4,932 10,680 9,877 35,065 21-30 1,671 1,419 1,406 4,715 6,062 15,273 31+ 759 788 800 2,305 3,970 8,622 Totals 18,830 8,014 19,969 26,078 27,559 10,0450 State the hypotheses. H 0 : _______ H a : _______ Enter expected values in . Round to two decimal places. Calculate the following values: Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White 1-10 7777.57 3310.11 8248.02 10771.29 11383.01 11-20 6573.16 2797.52 6970.76 9103.29 9620.27 21-30 2863.02 1218.49 3036.20 3965.05 4190.23 31+ 1616.25 687.87 1714.01 2238.37 2365.49 df = _______ χ 2 test statistic = ______ 10,301.8 p -value = ______ Is this a right-tailed, left-tailed, or two-tailed test? Explain why. right Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p -value. State the decision and conclusion (in a complete sentence) for the following preconceived levels of α . α = 0.05 Decision: ___________________ Reason for the decision: ___________________ Conclusion (write out in a complete sentence): ___________________ Reject the null hypothesis. p -value < alpha There is sufficient evidence to conclude that smoking level is dependent on ethnic group. α = 0.01 Decision: ___________________ Reason for the decision: ___________________ Conclusion (write out in a complete sentence): ___________________ Homework For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier. U.S. Ski Area Beginner Intermediate Advanced Tahoe 20 30 40 Utah 10 30 60 Colorado 10 40 50 Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in . Conduct a test of independence. Family Size Sub & Compact Mid-size Full-size Van & Truck 1 20 35 40 35 2 20 50 70 80 3–4 20 50 100 90 5+ 20 30 70 70 H 0 : Car size is independent of family size. H a : Car size is dependent on family size. df = 9 chi-square distribution with df = 9 test statistic = 15.8284 p -value = 0.0706 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent. College students may be interested in whether or not their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. shows the data. Conduct a test of independence. Major < $50,000 $50,000 – $68,999 $69,000 + English 5 20 5 Engineering 10 30 60 Nursing 10 15 15 Business 10 20 30 Psychology 20 30 20 Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in . Conduct a test of independence. Location 20–29 30–39 40–49 50 and over Niagara Falls 15 25 25 20 Poconos 15 25 25 10 Europe 10 25 15 5 Virgin Islands 20 25 15 5 H 0 : Honeymoon locations are independent of bride’s age. H a : Honeymoon locations are dependent on bride’s age. df = 9 chi-square distribution with df = 9 test statistic = 15.7027 p -value = 0.0734 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent. A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and their choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence. Sport 18 - 25 26 - 30 31 - 40 41 and over racquetball 42 58 30 46 tennis 58 76 38 65 swimming 72 60 65 33 A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in . Conduct a test of independence. Type of Fries Northeast South Central West skinny fries 70 50 20 25 curly fries 100 60 15 30 steak fries 20 40 10 10 H 0 : The types of fries sold are independent of the location. H a : The types of fries sold are dependent on the location. df = 6 chi-square distribution with df = 6 test statistic =18.8369 p -value = 0.0044 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent. According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by men in the following age groups. He is interested in whether the age of the man and the amount of life insurance purchased are independent events. Conduct a test for independence. Age of Men None < $200,000 $200,000–$400,000 $401,001–$1,000,000 $1,000,001+ 20–29 40 15 40 0 5 30–39 35 5 20 20 10 40–49 20 0 30 0 30 50+ 40 30 15 15 10 Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence. Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate < $30,000 15 25 10 5 $30,000–$40,000 20 40 70 30 $40,000–$50,000 10 20 40 55 $50,000–$60,000 5 10 20 60 $60,000+ 0 5 10 150 H 0 : Salary is independent of level of education. H a : Salary is dependent on level of education. df = 12 chi-square distribution with df = 12 test statistic = 255.7704 p -value = 0 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent. Read the statement and decide whether it is true or false. The number of degrees of freedom for a test of independence is equal to the sample size minus one. The test for independence uses tables of observed and expected data values. true The test to use when determining if the college or university a student chooses to attend is related to their socioeconomic status is a test for independence. In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed. true An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on , do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5% significance level. U.S. region/Flavor Strawberry Chocolate Vanilla Rocky Road Mint Chocolate Chip Pistachio Row total West 12 21 22 19 15 8 97 Midwest 10 32 22 11 15 6 96 East 8 31 27 8 15 7 96 South 15 28 30 8 15 6 102 Column Total 45 112 101 46 60 27 391 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level. Age Group\\ Net Worth Value (in millions of US dollars) 1–5 6–24 ≥25 Row Total 17–25 8 7 5 20 26–30 6 5 9 20 Column Total 14 12 14 40 H 0 : Age is independent of the youngest online entrepreneurs’ net worth. H a : Age is dependent on the net worth of the youngest online entrepreneurs. df = 2 chi-square distribution with df = 2 test statistic = 1.76 p -value 0.4144 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent. A 2013 poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in , and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5% significance level. Opinion/Ethnicity Asian-American White/Non-Hispanic Black Hispanic/Latino Row Total Against tax 48 433 41 160 682 In Favor of tax 54 234 24 147 459 No opinion 16 43 16 19 94 Column Total 118 710 81 326 1235 Contingency Table a table that displays sample values for two different factors that may be dependent or contingent on one another; it facilitates determining conditional probabilities.", "section": "Test of Independence", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Test for Homogeneity The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity , can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. NOTE The expected value for each cell needs to be at least five in order for you to use this test. Hypotheses H 0 : The distributions of the two populations are the same. H a : The distributions of the two populations are not the same. Test Statistic Use a χ 2 test statistic. It is computed in the same way as the test for independence. Degrees of Freedom ( df ) df = number of columns - 1 Requirements All values in the table must be greater than or equal to five. Common Uses Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values. Do men and women college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected men college students and 300 randomly selected women college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in . Do male and female college students have the same distribution of living arrangements? Distribution of Living Arragements for College Men and College Women Dormitory Apartment With Parents Other Men 72 84 49 45 Women 91 86 88 35 H 0 : The distribution of living arrangements for men college students is the same as the distribution of living arrangements for women college students. H a : The distribution of living arrangements for men college students is not the same as the distribution of living arrangements for women college students. Degrees of Freedom ( df ): df = number of columns – 1 = 4 – 1 = 3 Distribution for the test: χ 3 2 Calculate the test statistic: χ 2 = 10.1287 (calculator or computer) Probability statement: p -value = P ( χ 2 >10.1287) = 0.0175 Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 2 ENTER 4 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 10.1287 and the p -value = 0.0175. Do the procedure a second time but arrow down to Draw instead of calculate . Compare α and the p -value: Since no α is given, assume α = 0.05. p -value = 0.0175. α > p -value. Make a decision: Since α > p -value, reject H 0 . This means that the distributions are not the same. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for men and women college students are not the same. Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. Try It Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in . Do families and singles have the same distribution of cars? Test at a level of significance of 0.05. Sport Sedan Hatchback Truck Van/SUV Family 5 15 35 17 28 Single 45 65 37 46 7 Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake? Perez Chung Stevens Before 167 128 135 After 214 197 225 H 0 : The distribution of voter preferences was the same before and after the earthquake. H a : The distribution of voter preferences was not the same before and after the earthquake. Degrees of Freedom ( df ): df = number of columns – 1 = 3 – 1 = 2 Distribution for the test: χ 2 2 Calculate the test statistic : χ 2 = 3.2603 (calculator or computer) Probability statement: p -value= P ( χ 2 > 3.2603) = 0.1959 Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 2 ENTER 3 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 3.2603 and the p -value = 0.1959. Do the procedure a second time but arrow down to Draw instead of calculate . Compare α and the p -value: α = 0.05 and the p -value = 0.1959. α < p -value. Make a decision: Since α < p -value, do not reject H o . Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake. Try It Ivy League schools receive many applications, but only some can be accepted. At the schools listed in , two types of applications are accepted: regular and early decision. Application Type Accepted Brown Columbia Cornell Dartmouth Penn Yale Regular 2,115 1,792 5,306 1,734 2,685 1,245 Early Decision 577 627 1,228 444 1,195 761 We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p -value, and draw a conclusion about the test of homogeneity. References Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013). “Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/pubsinfo.asp?pubid=2009030 (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/2009030_sup.pdf (accessed May 24, 2013). Chapter Review To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five. Formula Review ∑ i ⋅ j ( O − E ) 2 E Homogeneity test statistic where: O = observed values E = expected values i = number of rows in data contingency table j = number of columns in data contingency table df = ( i −1)( j −1) Degrees of freedom A math teacher wants to see if two of their classes have the same distribution of test scores. What test should they use? test for homogeneity What are the null and alternative hypotheses for ? A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should they use? test for homogeneity A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should they use? What condition must be met to use the test for homogeneity? All values in the table must be greater than or equal to five. Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in . 20–30 30–40 40–50 50–60 Private Practice 16 40 38 6 Hospital 8 44 59 39 State the null and alternative hypotheses. df = _______ 3 What is the test statistic? What is the p -value? 0.00005 What can you conclude at the 5% significance level? Homework For each word problem, use a solution sheet to solve the hypothesis test problem. Go to for the chi-square solution sheet. Round expected frequency to two decimal places. A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in . Conduct a test of homogeneity. Test at a 5% level of significance. Open Conscientious Extrovert Agreeable Neurotic Business 41 52 46 61 58 Social Science 72 75 63 80 65 H 0 : The distribution for personality types is the same for both majors H a : The distribution for personality types is not the same for both majors df = 4 chi-square with df = 4 test statistic = 3.01 p -value = 0.5568 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in . Conduct a test for homogeneity at a 5% level of significance. French Toast Pancakes Waffles Omelettes Men 47 35 28 53 Women 65 59 55 60 A fish and wildlife technician is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance. H 0 : The distribution for fish caught is the same in Green Valley Lake and in Echo Lake. H a : The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake. 3 chi-square with df = 3 11.75 p -value = 0.0083 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake In a recent year, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5% significance level. Did you expect the result you obtained? Reasons for Homeschooling Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) Row Total Concern about the environment of other schools 1,321 309 1,630 Dissatisfaction with academic instruction at other schools 1,096 258 1,354 To provide religious or moral instruction 1,257 540 1,797 Child has special needs, other than physical or mental 315 55 370 Nontraditional approach to child’s education 984 99 1,083 Other reasons (e.g., finances, travel, family time, etc.) 485 216 701 Column Total 5,458 1,477 6,935 When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for a specific six-year period. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level. Year European Union United States Row Total 6 3,413 7,164 10,557 5 3,302 7,057 10,359 4 3,505 7,488 10,993 3 3,537 7,758 11,295 2 3,595 7,697 11,292 1 3,613 7,847 11,460 Column Total 20,965 45,011 65,976 H 0 : The distribution of average energy use in the USA is the same as in Europe between Year 1 and Year 6. H a : The distribution of average energy use in the USA is not the same as in Europe between Year 1 and Year 6. df = 4 chi-square with df = 4 test statistic = 2.7434 p -value = 0.7395 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from Year 1 to Year 6. The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. presents the number of Top Safety Picks in six car categories for two specific years. Analyze the table data to conclude Year 1 and Year 2. Derive your results at the 5% significance level. Year \\ Car Type Small Mid-Size Large Small SUV Mid-Size SUV Large SUV Row Total Year 1 12 22 10 10 27 6 87 Year 2 31 30 19 11 29 4 124 Column Total 43 52 29 21 56 10 211", "section": "Test for Homogeneity", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Comparison of the Chi-Square Tests You have seen the χ 2 test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which χ 2 test is the appropriate one to use. Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an unknown distribution \"fits\" a known distribution. In this case there will be a single qualitative survey question or a single outcome of an experiment from a single population. Goodness-of-Fit is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution. The null and alternative hypotheses are: H 0 : The population fits the given distribution. H a : The population does not fit the given distribution. Independence: Use the test for independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a contingency table will be constructed. The goal is to see if the two variables are unrelated (independent) or related (dependent). The null and alternative hypotheses are: H 0 : The two variables (factors) are independent. H a : The two variables (factors) are dependent. Homogeneity: Use the test for homogeneity to decide if two populations with unknown distributions have the same distribution as each other. In this case there will be a single qualitative survey question or experiment given to two different populations. The null and alternative hypotheses are: H 0 : The two populations follow the same distribution. H a : The two populations have different distributions. Chapter Review The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown. Which test do you use to decide whether an observed distribution is the same as an expected distribution? a goodness-of-fit test What is the null hypothesis for the type of test from ? Which test would you use to decide whether two factors have a relationship? a test for independence Which test would you use to decide if two populations have the same distribution? How are tests of independence similar to tests for homogeneity? Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ ( i j ) ( O - E ) 2 E . In addition, all values must be greater than or equal to five. How are tests of independence different from tests for homogeneity? Homework For each word problem, use a solution sheet to solve the hypothesis test problem. Go to for the chi-square solution sheet. Round expected frequency to two decimal places. Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance. H 0 : The distribution for technology use is the same for community college students and university students. H a : The distribution for technology use is not the same for community college students and university students. 2 chi-square with df = 2 7.05 p -value = 0.0294 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities. Read the statement and decide whether it is true or false. If df = 2, the chi-square distribution has a shape that reminds us of the exponential. Bringing It Together Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests. If you did a left-tailed test, what would you be testing? The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected. Testing to see if the data fits the distribution “too well” or is too perfect.", "section": "Comparison of the Chi-Square Tests", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Test of a Single Variance A test of a single variance assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is: ( n - 1 ) s 2 σ 2 where: n = the total number of data s 2 = sample variance σ 2 = population variance You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Even though we are given the population standard deviation, we can set up the test using the population variance as follows. H 0 : σ 2 = 5 2 H a : σ 2 > 5 2 Try It A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes. With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers . Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ 2 , or the population standard deviation, σ . Random Variable: The sample standard deviation, s , is the random variable. Let s = standard deviation for the waiting times. H 0 : σ 2 = 7.2 2 H a : σ 2 < 7.2 2 The word \"less\" tells you this is a left-tailed test. Distribution for the test: χ 24 2 , where: n = the number of customers sampled df = n – 1 = 25 – 1 = 24 Calculate the test statistic: χ 2 = ( n − 1 ) s 2 σ 2 = ( 25 − 1 ) ( 3.5 ) 2 7.2 2 = 5.67 where n = 25, s = 3.5, and σ = 7.2. Graph: Probability statement: p -value = P ( χ 2 < 5.67) = 0.000042 Compare α and the p -value: α = 0.05 p -value = 0.000042 α > p -value Make a decision: Since α > p -value, reject H 0 . This means that you reject σ 2 = 7.2 2 . In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. In 2nd DISTR , use 7:χ2cdf . The syntax is (lower, upper, df) for the parameter list. For , χ2cdf(-1E99,5.67,24) . The p -value = 0.000042. Try It The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p -value, and draw a conclusion. Test at the 1% significance level. References “AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013). Data from the World Bank, June 5, 2012. Chapter Review To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation). Formula Review χ 2 = ( n − 1 ) ⋅ s 2 σ 2 Test of a single variance statistic where: n : sample size s : sample standard deviation σ : population standard deviation df = n – 1 Degrees of freedom Test of a Single Variance Use the test to determine variation. The degrees of freedom is the number of samples – 1. The test statistic is ( n – 1 ) ⋅ s 2 σ 2 , where n = the total number of data, s 2 = sample variance, and σ 2 = population variance. The test may be left-, right-, or two-tailed. Use the following information to answer the next three exercises: An archer’s standard deviation for hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less. What type of test should be used? a test of a single variance State the null and alternative hypotheses. Is this a right-tailed, left-tailed, or two-tailed test? a left-tailed test Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. What type of test should be used? State the null and alternative hypotheses. H 0 : σ 2 = 0.81 2 ; H a : σ 2 > 0.81 2 df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. What type of test should be used? a test of a single variance What is the test statistic? What is the p -value? 0.0542 What can you conclude at the 5% significance level? Homework Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. Is the traveler disputing the claim about the average or about the variance? A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 225 Is this a right-tailed, left-tailed, or two-tailed test? H 0 : __________ H 0 : σ 2 ≤ 150 df = ________ chi-square test statistic = ________ 36 p -value = ________ Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p -value. Answers may vary. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________ How did you know to test the variance instead of the mean? The claim is that the variance is no more than 150 minutes. If an additional test were done on the claim of the average delay, which distribution would you use? If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? a Student's t - or normal distribution For each word problem, use a solution sheet to solve the hypothesis test problem. Go to for the chi-square solution sheet. Round expected frequency to two decimal places. A plant manager is concerned their equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. H 0 : σ = 15 H a : σ > 15 df = 42 chi-square with df = 42 test statistic = 26.88 p -value = 0.9663 Answers may vary. Alpha = 0.05 Decision: Do not reject null hypothesis. Reason for decision: p -value > alpha Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. H 0 : σ ≤ 3 H a : σ > 3 df = 17 chi-square distribution with df = 17 test statistic = 28.73 p -value = 0.0371 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in . Does the students’ survey indicate that the standard deviation is greater than 0.75? # of births Frequency 0 5 1 30 2 10 3 5 According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. They count the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11 H 0 : σ = 2 H a : σ ≠ 2 df = 14 chi-square distiribution with df = 14 chi-square test statistic = 5.2094 p -value = 0.0346 Answers may vary. Alpha = 0.05 Decision: Reject the null hypothesis Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2. The manager of \"Frenchies\" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis? The sample standard deviation is $34.29. H 0 : σ 2 = 25 2 H a : σ 2 > 25 2 df = n – 1 = 7. test statistic: x 2 = x 7 2 = ( n – 1 ) s 2 25 2 = ( 8 – 1 ) ( 34.29 ) 2 25 2 = 13.169 ; p -value: P ( x 7 2 > 13.169 ) = 1 – P ( x 7 2 ≤ 13.169 ) = 0.0681 Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test. (a) at the 5% significance level (b) at the 1% significance level Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172;", "section": "Test of a Single Variance", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Lab 1: Chi-Square Goodness-of-Fit Lab 1: Chi-Square Goodness-of-Fit Class Time: Names: Student Learning Outcome The student will evaluate data collected to determine if they fit either the uniform or exponential distributions. Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. (Or, ask three cashiers for the last ten amounts. Be sure to include the express lane, if it is open.) NOTE You may need to combine two categories so that each cell has an expected value of at least five. Record the values. Construct a histogram of the data. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Calculate the following: x ¯ = ________ s = ________ s 2 = ________ Uniform Distribution Test to see if grocery receipts follow the uniform distribution. Using your lowest and highest values, X ~ U (_______, _______) Divide the distribution into fifths. Calculate the following: lowest value = _________ 20 th percentile = _________ 40 th percentile = _________ 60 th percentile = _________ 80 th percentile = _________ highest value = _________ For each fifth, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Fifth Observed Expected 1 st 2 nd 3 rd 4 th 5 th H 0 : ________ H a : ________ What distribution should you use for a hypothesis test? Why did you choose this distribution? Calculate the test statistic. Find the p -value. Sketch a graph of the situation. Label and scale the x -axis. Shade the area corresponding to the p -value. State your decision. State your conclusion in a complete sentence. Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter 1 x ¯ . Using 1 x ¯ as the decay parameter, X ~ Exp (_________). Calculate the following: lowest value = ________ first quartile = ________ 37 th percentile = ________ median = ________ 63 rd percentile = ________ 3 rd quartile = ________ highest value = ________ For each cell, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Cell Observed Expected 1 st 2 nd 3 rd 4 th 5 th 6 th H 0 : ________ H a : ________ What distribution should you use for a hypothesis test? Why did you choose this distribution? Calculate the test statistic. Find the p -value. Sketch a graph of the situation. Label and scale the x -axis. Shade the area corresponding to the p -value. State your decision. State your conclusion in a complete sentence. Discussion Questions Did your data fit either distribution? If so, which? In general, do you think it’s likely that data could fit more than one distribution? In complete sentences, explain why or why not.", "section": "Lab 1: Chi-Square Goodness-of-Fit", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Lab 2: Chi-Square Test of Independence Lab 2: Chi-Square Test of Independence Class Time: Names: Student Learning Outcome The student will evaluate if there is a significant relationship between favorite type of snack and gender. Collect the Data Using your class as a sample, complete the following chart. Ask each other what your favorite snack is, then total the results. Note You may need to combine two food categories so that each cell has an expected value of at least five. Favorite type of snack sweets (candy & baked goods) ice cream chips & pretzels fruits & vegetables Total men women Total Looking at , does it appear to you that there is a dependence between gender and favorite type of snack food? Why or why not? Hypothesis Test Conduct a hypothesis test to determine if the factors are independent: H 0 : ________ H a : ________ What distribution should you use for a hypothesis test? Why did you choose this distribution? Calculate the test statistic. Find the p -value. Sketch a graph of the situation. Label and scale the x -axis. Shade the area corresponding to the p -value. State your decision. State your conclusion in a complete sentence. Discussion Questions Is the conclusion of your study the same as or different from your answer to answer to question two under Collect the Data ? Why do you think that occurred?", "section": "Lab 2: Chi-Square Test of Independence", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction Linear regression and correlation can help you determine if an auto mechanic’s salary is related to his work experience. (credit: modification of work “USPS commissions local repair-shop for some needed work on its older trucks” by Joshua Rothhaas/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Discuss basic ideas of linear regression and correlation. Create and interpret a line of best fit. Calculate and interpret the correlation coefficient. Calculate and interpret outliers. Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee. The type of data described in the examples is bivariate data — \"bi\" for two variables. In reality, statisticians use multivariate data, meaning many variables. In this chapter, you will be studying the simplest form of regression, \"linear regression\" with one independent variable ( x ). This involves data that fits a line in two dimensions. You will also study correlation which measures how strong the relationship is.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Linear Equations Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form: y = a + bx where a and b are constant numbers. The variable x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable. The following examples are linear equations. y = 3 + 2x y = –0.01 + 1.2x Try It Is the following an example of a linear equation? y = –0.125 – 3.5 x The graph of a linear equation of the form y = a + bx is a straight line . Any line that is not vertical can be described by this equation. Graph the equation y = –1 + 2 x . Try It Is the following an example of a linear equation? Why or why not? A local small business completes federal tax returns for customers. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job. Find the equation that expresses the total cost in terms of the number of hours required to complete the job. Let x = the number of hours it takes to get the job done. Let y = the total cost to the customer. The $31.50 is a fixed cost. If it takes x hours to complete the tax return, then (32)( x ) is the cost of the tax return processing only. The total cost is: y = 31.50 + 32 x Try It Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class. Slope and Y -Intercept of a Linear Equation For the linear equation y = a + bx , b = slope and a = y -intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the y -intercept is the y coordinate of the point (0, a ) where the line crosses the y -axis. Three possible graphs of y = a + bx . (a) If b > 0, the line slopes upward to the right. (b) If b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right. Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15 x . What are the independent and dependent variables? What is the y -intercept and what is the slope? Interpret them using complete sentences. The independent variable ( x ) is the number of hours Svetlana tutors each session. The dependent variable ( y ) is the amount, in dollars, Svetlana earns for each session. The y -intercept is 25 ( a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 ( b = 15). For each session, Svetlana earns $15 for each hour she tutors. Try It Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20 x . What are the independent and dependent variables? What is the y -intercept and what is the slope? Interpret them using complete sentences. References Data from the Centers for Disease Control and Prevention. Data from the National Center for agency reporting flu cases and TB Prevention. Chapter Review The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form y = mx + b , where m and b are constants, x is the independent variable, y is the dependent variable. In a statistical context, a linear equation is written in the form y = a + bx , where a and b are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation y = a + bx , the constant b , called a coefficient, represents the slope . The constant a is called the y -intercept. The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable ( y ) changes for every one unit increase in the independent ( x ) variable, on average. The y -intercept is used to describe the dependent variable when the independent variable equals zero. Formula Review y = a + bx where a is the y -intercept and b is the slope. The variable x is the independent variable and y is the dependent variable. Use the following information to answer the next three exercises . A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour. What are the dependent and independent variables? dependent variable: fee amount; independent variable: time Find the equation that expresses the total fee in terms of the number of hours the equipment is rented. Graph the equation from . Use the following information to answer the next two exercises . A credit card company charges $10 when a payment is late, and $5 a day each day the payment remains unpaid. Find the equation that expresses the total fee in terms of the number of days the payment is late. Graph the equation from . Is the equation y = 10 + 5 x – 3 x 2 linear? Why or why not? Which of the following equations are linear? a. y = 6 x + 8 b. y + 7 = 3 x c. y – x = 8 x 2 d. 4 y = 8 y = 6 x + 8, 4 y = 8, and y + 7 = 3 x are all linear equations. Does the graph show a linear equation? Why or why not? contains real data for the first two decades of flu reporting. Adults and Adolescents only, United States Year # flu cases diagnosed # flu deaths Pre-1981 91 29 1981 319 121 1982 1,170 453 1983 3,076 1,482 1984 6,240 3,466 1985 11,776 6,878 1986 19,032 11,987 1987 28,564 16,162 1988 35,447 20,868 1989 42,674 27,591 1990 48,634 31,335 1991 59,660 36,560 1992 78,530 41,055 1993 78,834 44,730 1994 71,874 49,095 1995 68,505 49,456 1996 59,347 38,510 1997 47,149 20,736 1998 38,393 19,005 1999 25,174 18,454 2000 25,522 17,347 2001 25,643 17,402 2002 26,464 16,371 Total 802,118 489,093 Use the columns \"year\" and \"# flu cases diagnosed. Why is “year” the independent variable and “# flu cases diagnosed.” the dependent variable (instead of the reverse)? The number of flu cases depends on the year. Therefore, year becomes the independent variable and the number of flu cases is the dependent variable. Use the following information to answer the next two exercises . A specialty cleaning company charges an equipment fee and an hourly labor fee. A linear equation that expresses the total amount of the fee the company charges for each session is y = 50 + 100 x . What are the independent and dependent variables? What is the y -intercept and what is the slope? Interpret them using complete sentences. The y -intercept is 50 ( a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 ( b = 100). For each session, the company charges $100 for each hour they clean. Use the following information to answer the next three questions . Due to erosion, a river shoreline is losing several thousand pounds of soil each year. A linear equation that expresses the total amount of soil lost per year is y = 12,000 x . What are the independent and dependent variables? How many pounds of soil does the shoreline lose in a year? 12,000 pounds of soil What is the y -intercept? Interpret its meaning. Use the following information to answer the next two exercises . The price of a single issue of stock can fluctuate throughout the day. A linear equation that represents the price of stock for Shipment Express is y = 15 – 1.5 x where x is the number of hours passed in an eight-hour day of trading. What are the slope and y -intercept? Interpret their meaning. The slope is –1.5 ( b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y -intercept is $15 ( a = 15). This means the price of stock before the trading day was $15. If you owned this stock, would you want a positive or negative slope? Why? Homework For each of the following situations, state the independent variable and the dependent variable. A study is done to determine if elderly drivers are involved in more motor vehicle fatalities than other drivers. The number of fatalities per 100,000 drivers is compared to the age of drivers. A study is done to determine if the weekly grocery bill changes based on the number of family members. Insurance companies base life insurance premiums partially on the age of the applicant. Utility bills vary according to power consumption. A study is done to determine if a higher education reduces the crime rate in a population. independent variable: age; dependent variable: fatalities independent variable: # of family members; dependent variable: grocery bill independent variable: age of applicant; dependent variable: insurance premium independent variable: power consumption; dependent variable: utility independent variable: higher education (years); dependent variable: crime rates Piece-rate systems are widely debated incentive payment plans. In a recent study of loan officer effectiveness, the following piece-rate system was examined: % of goal reached < 80 80 100 120 Incentive n/a $4,000 with an additional $125 added per percentage point from 81–99% $6,500 with an additional $125 added per percentage point from 101–119% $9,500 with an additional $125 added per percentage point starting at 121% If a loan officer makes 95% of their goal, write the linear function that applies based on the incentive plan table. In context, explain the y -intercept and slope.", "section": "Linear Equations", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Scatter Plots Before we take up the discussion of linear regression and correlation, we need to examine a way to display the relation between two variables x and y . The most common and easiest way is a scatter plot . The following example illustrates a scatter plot. An educational researcher collects data on the vocabulary size of children as a function of age. The data is shown in Table 12.1. Is there a relationship between age and vocabulary size for young children? Construct a scatter plot. Let x = Child’s Age, and let y = Vocabulary Size. Age (years) Vocabulary Size (number of words) 3 655 4 1098 6 2463 7 3195 To create a scatter plot: Enter your X data into list L1 and your Y data into list L2. Press 2nd STATPLOT ENTER to use Plot 1. On the input screen for PLOT 1, highlight On and press ENTER . (Make sure the other plots are OFF.) For TYPE: highlight the very first icon, which is the scatter plot, and press ENTER . For Xlist:, enter L1 ENTER and for Ylist: L2 ENTER. For Mark: it does not matter which symbol you highlight, but the square is the easiest to see. Press ENTER . Make sure there are no other equations that could be plotted. Press Y = and clear any equations out. Press the ZOOM key and then the number 9 (for menu item \"ZoomStat\") ; the calculator will fit the window to the data. You can press WINDOW to see the scaling of the axes. Try It Amelia plays basketball for her high school. She wants to improve to play at the college level. She notices that the number of points she scores in a game goes up in response to the number of hours she practices her jump shot each week. She records the following data: X (hours practicing jump shot) Y (points scored in a game) 5 15 7 22 9 28 10 31 11 33 12 36 Construct a scatter plot and state if what Amelia thinks appears to be true. A scatter plot shows the direction of a relationship between the variables. A clear direction happens when there is either: High values of one variable occurring with high values of the other variable or low values of one variable occurring with low values of the other variable. High values of one variable occurring with low values of the other variable. You can determine the strength of the relationship by looking at the scatter plot and seeing how close the points are to a line, a power function, an exponential function, or to some other type of function. For a linear relationship there is an exception. Consider a scatter plot where all the points fall on a horizontal line providing a \"perfect fit.\" The horizontal line would in fact show no relationship. When you look at a scatterplot, you want to notice the overall pattern and any deviations from the pattern. The following scatterplot examples illustrate these concepts. In this chapter, we are interested in scatter plots that show a linear pattern. Linear patterns are quite common. The linear relationship is strong if the points are close to a straight line, except in the case of a horizontal line where there is no relationship. If we think that the points show a linear relationship, we would like to draw a line on the scatter plot. This line can be calculated through a process called linear regression . However, we only calculate a regression line if one of the variables helps to explain or predict the other variable. If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x Chapter Review Scatter plots are particularly helpful graphs when we want to see if there is a linear relationship among data points. They indicate both the direction of the relationship between the x variables and the y variables, and the strength of the relationship. We calculate the strength of the relationship between an independent variable and a dependent variable using linear regression. Does the scatter plot appear linear? Strong or weak? Positive or negative? The data appear to be linear with a strong, positive correlation. Does the scatter plot appear linear? Strong or weak? Positive or negative? Does the scatter plot appear linear? Strong or weak? Positive or negative? The data appear to have no correlation. Homework The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data. Year Cuba’s PPP Year Cuba’s PPP 1 1,700 8 4,000 2 1,700 9 11,000 4 2,300 10 9,500 5 2,900 11 9,700 6 3,000 12 9,900 7 3,500 Answers may vary. The following table shows the number of faculty and number of students at several colleges. Construct a scatter plot of the data College Faculty Students College A 47 940 College B 58 1102 College C 26 533 College D 63 1244 Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data. School Mid-Career Salary (in thousands) Yearly Tuition Princeton 137 57,410 Harvey Mudd 135 62,817 CalTech 127 60,864 US Naval Academy 122 0 West Point 120 0 MIT 118 57,986 Lehigh University 118 59,930 NYU-Poly 117 58,168 Babson College 117 56,576 Stanford 114 56,169 For graph: Answers may vary. Note that tuition is the independent variable and salary is the dependent variable. If the level of significance is 0.05 and the p -value is 0.06, what conclusion can you draw? If there are 15 data points in a set of data, what is the number of degree of freedom? 13", "section": "Scatter Plots", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The Regression Equation Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to \"fit\" a straight line. This is called a Line of Best Fit or Least-Squares Line . If you know a person's pinky (smallest) finger length, do you think you could predict that person's height? Collect data from your class (pinky finger length, in inches). The independent variable, x , is pinky finger length and the dependent variable, y , is height. For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler . Then \"by eye\" draw a line that appears to \"fit\" the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y -intercept of the line by extending your line so it crosses the y -axis. Using the slopes and the y -intercepts, write your equation of \"best fit.\" Do you think everyone will have the same equation? Why or why not? According to your equation, what is the predicted height for a pinky length of 2.5 inches? A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score? x (third exam score) y (final exam score) 65 175 67 133 71 185 71 163 66 126 75 198 67 153 70 163 71 159 69 151 69 159 Table showing the scores on the final exam based on scores from the third exam. Scatter plot showing the scores on the final exam based on scores from the third exam. Try It SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet. X (depth in feet) Y (maximum dive time) 50 80 60 55 70 45 80 35 90 25 100 22 The third exam score, x , is the independent variable and the final exam score, y , is the dependent variable. We will plot a regression line that best \"fits\" the data. If each of you were to fit a line \"by eye,\" you would draw different lines. We can use what is called a least-squares regression line to obtain the best fit line. Consider the following diagram. Each point of data is of the the form ( x , y ) and each point of the line of best fit using least-squares linear regression has the form ( x , ŷ ). The ŷ is read \" y hat\" and is the estimated value of y . It is the value of y obtained using the regression line. It is not generally equal to y from data. The term y 0 – ŷ 0 = ε 0 is called the \"error\" or residual . It is not an error in the sense of a mistake. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y . In other words, it measures the vertical distance between the actual data point and the predicted point on the line. If the observed data point lies above the line, the residual is positive, and the line underestimates the actual data value for y . If the observed data point lies below the line, the residual is negative, and the line overestimates that actual data value for y . In the diagram in , y 0 – ŷ 0 = ε 0 is the residual for the point shown. Here the point lies above the line and the residual is positive. ε = the Greek letter epsilon For each data point, you can calculate the residuals or errors, y i - ŷ i = ε i for i = 1, 2, 3, ..., 11. Each | ε | is a vertical distance. For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 ε values. If you square each ε and add, you get ( ε 1 ) 2 + ( ε 2 ) 2 + ... + ( ε 11 ) 2 = Σ i = 1 11 ε 2 This is called the Sum of Squared Errors (SSE) . Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation: y ^ = a + b x where a = y ¯ − b x ¯ and b = Σ ( x − x ¯ ) ( y − y ¯ ) Σ ( x − x ¯ ) 2 . The sample means of the x values and the y values are x ¯ and y ¯ , respectively. The best fit line always passes through the point ( x ¯ , y ¯ ) . The slope b can be written as b = r ( s y s x ) where s y = the standard deviation of the y values and s x = the standard deviation of the x values. r is the correlation coefficient, which is discussed in the next section. Residuals Plots A residuals plot can be used to help determine if a set of ( x , y ) data is linearly correlated. For each data point used to create the correlation line, a residual y - ŷ can be calculated, where y is the observed value of the response variable and ŷ is the value predicted by the correlation line. The difference between these values is called the residual. A residuals plot shows the explanatory variable x on the horizontal axis and the residual for that value on the vertical axis. The residuals plot is often shown together with a scatter plot of the data. While a scatter plot of the data should resemble a straight line, a residuals plot should appear random, with no pattern and no outliers. It should also show constant error variance, meaning the residuals should not consistently increase (or decrease) as the explanatory variable x increases. A residuals plot can be created using StatCrunch or a TI calculator. The plot should appear random. A box plot of the residuals is also helpful to verify that there are no outliers in the data. By observing the scatter plot of the data, the residuals plot, and the box plot of residuals, together with the linear correlation coefficient, we can usually determine if it is reasonable to conclude that the data are linearly correlated. EXAMPLE: A shop owner uses a straight-line regression to estimate the number of ice cream cones that would be sold in a day based on the temperature at noon. The owner has data for a 2-year period and chose nine days at random. A scatter plot of the data is shown, together with a residuals plot. Temperature ° F Ice cream cones sold 70 105 85 240 65 49 72 147 80 231 61 38 75 193 78 196 68 89 Table showing the number of ice cream cones sold on nine random days and the temperature at noon on those days. Scatter plot of the data looks like a straight line. Residuals plot appears random. Least Squares Criteria for Best Fit The process of fitting the best-fit line is called linear regression . The idea behind finding the best-fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is, made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least-squares regression line . NOTE Computer spreadsheets, statistical software, and many calculators can quickly calculate the best-fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best-fit line and create a scatterplot are shown at the end of this section. THIRD EXAM vs FINAL EXAM EXAMPLE: The graph of the line of best fit for the third-exam/final-exam example is as follows: The least squares regression line (best-fit line) for the third-exam/final-exam example has the equation: y ^ = − 173.51 + 4.83 x REMINDER Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship between the variables, then it is reasonable to use a best fit line to make predictions for y given x within the domain of x -values in the sample data, but not necessarily for x -values outside that domain. You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam. You should NOT use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x -values in the sample data, which are between 65 and 75. UNDERSTANDING SLOPE The slope of the line, b , describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English. INTERPRETATION OF THE SLOPE: The slope of the best-fit line tells us how the dependent variable ( y ) changes for every one unit increase in the independent ( x ) variable, on average. THIRD EXAM vs FINAL EXAM EXAMPLE Slope: The slope of the line is b = 4.83. Interpretation: For a one-point increase in the score on the third exam, the final exam score increases by 4.83 points, on average. Using the Linear Regression T Test: LinRegTTest In the STAT list editor, enter the X data in list L1 and the Y data in list L2 , paired so that the corresponding ( x , y ) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.) On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest . (Be careful to select LinRegTTest , as some calculators may also have a different item called LinRegTInt.) On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1 On the next line, at the prompt β or ρ , highlight \"≠ 0\" and press ENTER Leave the line for \"RegEq:\" blank Highlight Calculate and press ENTER. The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items. The second line says y = a + bx . Scroll down to find the values a = –173.513, and b = 4.8273; the equation of the best fit line is ŷ = –173.51 + 4.83 x The two items at the bottom are r 2 = 0.43969 and r = 0.663. For now, just note where to find these values; we will discuss them in the next two sections. Graphing the Scatterplot and Regression Line We are assuming your X data is already entered in list L1 and your Y data is in list L2 Press 2nd STATPLOT ENTER to use Plot 1 On the input screen for PLOT 1, highlight On , and press ENTER For TYPE: highlight the very first icon which is the scatterplot and press ENTER Indicate Xlist: L1 and Ylist: L2 For Mark: it does not matter which symbol you highlight. Press the ZOOM key and then the number 9 (for menu item \"ZoomStat\") ; the calculator will fit the window to the data To graph the best-fit line, press the \"Y=\" key and type the equation –173.5 + 4.83X into equation Y1. (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, Ymax NOTE Another way to graph the line after you create a scatter plot is to use LinRegTTest. Make sure you have done the scatter plot. Check it on your screen. Go to LinRegTTest and enter the lists. At RegEq: press VARS and arrow over to Y-VARS. Press 1 for 1:Function. Press 1 for 1:Y1. Then arrow down to Calculate and do the calculation for the line of best fit. Press Y = (you will see the regression equation). Press GRAPH. The line will be drawn.\" The Correlation Coefficient r Besides looking at the scatter plot and seeing that a line seems reasonable, how can you tell if the line is a good predictor? Use the correlation coefficient as another indicator (besides the scatterplot) of the strength of the relationship between x and y . The correlation coefficient, r , developed by Karl Pearson in the early 1900s, is numerical and provides a measure of strength and direction of the linear association between the independent variable x and the dependent variable y . The correlation coefficient is calculated as r = n Σ ( x y ) − ( Σ x ) ( Σ y ) [ n Σ x 2 − ( Σ x ) 2 ] [ n Σ y 2 − ( Σ y ) 2 ] where n = the number of data points. If you suspect a linear relationship between x and y , then r can measure how strong the linear relationship is. What the VALUE of r tells us: The value of r is always between –1 and +1: –1 ≤ r ≤ 1. The size of the correlation r indicates the strength of the linear relationship between x and y . Values of r close to –1 or to +1 indicate a stronger linear relationship between x and y . If r = 0 there is likely no linear correlation. It is important to view the scatterplot, however, because data that exhibit a curved or horizontal pattern may have a correlation of 0. If r = 1, there is perfect positive correlation. If r = –1, there is perfect negative correlation. In both these cases, all of the original data points lie on a straight line. Of course,in the real world, this will not generally happen. What the SIGN of r tells us A positive value of r means that when x increases, y tends to increase and when x decreases, y tends to decrease (positive correlation) . A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends to increase (negative correlation) . The sign of r is the same as the sign of the slope, b , of the best-fit line. NOTE Strong correlation does not suggest that x causes y or y causes x . We say \"correlation does not imply causation.\" (a) A scatter plot showing data with a positive correlation. 0 < r < 1 (b) A scatter plot showing data with a negative correlation. –1 < r < 0 (c) A scatter plot showing data with zero correlation. r = 0 The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can quickly calculate r . The correlation coefficient r is the bottom item in the output screens for the LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions). The Coefficient of Determination The variable r 2 is called the coefficient of determination and is the square of the correlation coefficient, but is usually stated as a percent, rather than in decimal form. It has an interpretation in the context of the data: r 2 , when expressed as a percent, represents the percent of variation in the dependent (predicted) variable y that can be explained by variation in the independent (explanatory) variable x using the regression (best-fit) line. 1 – r 2 , when expressed as a percentage, represents the percent of variation in y that is NOT explained by variation in x using the regression line. This can be seen as the scattering of the observed data points about the regression line. Consider the third exam/final exam example introduced in the previous section The line of best fit is: ŷ = –173.51 + 4.83x The correlation coefficient is r = 0.6631 The coefficient of determination is r 2 = 0.6631 2 = 0.4397 Interpretation of r 2 in the context of this example: Approximately 44% of the variation (0.4397 is approximately 0.44) in the final-exam grades can be explained by the variation in the grades on the third exam, using the best-fit regression line. Therefore, approximately 56% of the variation (1 – 0.44 = 0.56) in the final exam grades can NOT be explained by the variation in the grades on the third exam, using the best-fit regression line. (This is seen as the scattering of the points about the line.) Chapter Review A regression line, or a line of best fit, can be drawn on a scatter plot and used to predict outcomes for the x and y variables in a given data set or sample data. There are several ways to find a regression line, but usually the least-squares regression line is used because it creates a uniform line. Residuals, also called “errors,” measure the distance from the actual value of y and the estimated value of y . The Sum of Squared Errors, when set to its minimum, calculates the points on the line of best fit. Regression lines can be used to predict values within the given set of data, but should not be used to make predictions for values outside the set of data. The correlation coefficient r measures the strength of the linear association between x and y . The variable r has to be between –1 and +1. When r is positive, the x and y will tend to increase and decrease together. When r is negative, x will increase and y will decrease, or the opposite, x will decrease and y will increase. The coefficient of determination r 2 , is equal to the square of the correlation coefficient. When expressed as a percent, r 2 represents the percent of variation in the dependent variable y that can be explained by variation in the independent variable x using the regression line. Use the following information to answer the next five exercises . A random sample of ten professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars). x y x y 0 2 5 12 3 8 4 9 2 7 3 9 1 3 0 3 5 13 4 10 Draw a scatter plot of the data. Use regression to find the equation for the line of best fit. ŷ = 2.23 + 1.99 x Draw the line of best fit on the scatter plot. What is the slope of the line of best fit? What does it represent? The slope is 1.99 ( b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year. What is the y -intercept of the line of best fit? What does it represent? What does an r value of zero mean? It means that there is no correlation between the data sets. When n = 2 and r = 1, are the data significant? Explain. When n = 100 and r = -0.89, is there a significant correlation? Explain. Yes, there are enough data points and the value of r is strong enough to show that there is a strong negative correlation between the data sets. Homework What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern? Explain what it means when a correlation has an r 2 of 0.72. It means that 72% of the variation in the dependent variable ( y ) can be explained by the variation in the independent variable ( x ). Can a coefficient of determination be negative? Why or why not?", "section": "The Regression Equation", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Testing the Significance of the Correlation Coefficient The correlation coefficient, r , tells us about the strength and direction of the linear relationship between x and y . However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient r and the sample size n , together. We perform a hypothesis test of the \"significance of the correlation coefficient\" to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population. The sample data are used to compute r , the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r , is our estimate of the unknown population correlation coefficient. The symbol for the population correlation coefficient is ρ , the Greek letter \"rho.\" ρ = population correlation coefficient (unknown) r = sample correlation coefficient (known; calculated from sample data) The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is \"close to zero\" or \"significantly different from zero\". We decide this based on the sample correlation coefficient r and the sample size n . If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is \"significant.\" Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. What the conclusion means: There is a significant linear relationship between x and y . We can use the regression line to model the linear relationship between x and y in the population. If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is \"not significant\". Conclusion: \"There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero.\" What the conclusion means: There is not a significant linear relationship between x and y . Therefore, we CANNOT use the regression line to model a linear relationship between x and y in the population. NOTE If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values. If r is not significant OR if the scatter plot does not show a linear trend, the line should not be used for prediction. If r is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate or reliable for prediction OUTSIDE the domain of observed x values in the data. PERFORMING THE HYPOTHESIS TEST Null Hypothesis: H 0 : ρ = 0 Alternate Hypothesis: H a : ρ ≠ 0 WHAT THE HYPOTHESES MEAN IN WORDS: Null Hypothesis H 0 : The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship (correlation) between x and y in the population. Alternate Hypothesis H a : The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population. DRAWING A CONCLUSION: There are two methods of making the decision. The two methods are equivalent and give the same result. Method 1: Using the p -value Method 2: Using a table of critical values In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05 NOTE Using the p -value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.) METHOD 1: Using a p -value to make a decision To calculate the p -value using LinRegTTEST : On the LinRegTTEST input screen, on the line prompt for β or ρ , highlight \" ≠ 0 \" The output screen shows the p-value on the line that reads \"p =\". (Most computer statistical software can calculate the p -value.) If the p -value is less than the significance level ( α = 0.05): Decision: Reject the null hypothesis. Conclusion: \"There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero.\" If the p -value is NOT less than the significance level ( α = 0.05) Decision: DO NOT REJECT the null hypothesis. Conclusion: \"There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is NOT significantly different from zero.\" Calculation Notes: You will use technology to calculate the p -value. The following describes the calculations to compute the test statistics and the p -value: The p -value is calculated using a t -distribution with n - 2 degrees of freedom. The formula for the test statistic is t = r n − 2 1 − r 2 . The value of the test statistic, t , is shown in the computer or calculator output along with the p -value. The test statistic t has the same sign as the correlation coefficient r . The p -value is the combined area in both tails. An alternative way to calculate the p -value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR. THIRD-EXAM vs FINAL-EXAM EXAMPLE: p -value method Consider the third exam/final exam example . The line of best fit is: ŷ = -173.51 + 4.83 x with r = 0.6631 and there are n = 11 data points. Can the regression line be used for prediction? Given a third exam score ( x value), can we use the line to predict the final exam score (predicted y value)? H 0 : ρ = 0 H a : ρ ≠ 0 α = 0.05 The p -value is 0.026 (from LinRegTTest on your calculator or from computer software). The p -value, 0.026, is less than the significance level of α = 0.05. Decision: Reject the Null Hypothesis H 0 Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score ( x ) and the final exam score ( y ) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. METHOD 2: Using a table of Critical Values to make a decision The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of r is significant or not . Compare r to the appropriate critical value in the table. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may want to use the line for prediction. Suppose you computed r = 0.801 using n = 10 data points. df = n - 2 = 10 - 2 = 8. The critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r > positive critical value, then r is significant. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be used for prediction. If you view this example on a number line, it will help you. r is not significant between -0.632 and +0.632. r = 0.801 > +0.632. Therefore, r is significant. Try It For a given line of best fit, you computed that r = 0.6501 using n = 12 data points and the critical value is 0.576. Can the line be used for prediction? Why or why not? Suppose you computed r = –0.624 with 14 data points. df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction r = –0.624 < -0.532. Therefore, r is significant. Try It For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical value is 0.666. Can the line be used for prediction? Why or why not? Suppose you computed r = 0.776 and n = 6. df = 6 – 2 = 4. The critical values are –0.811 and 0.811. Since –0.811 < 0.776 < 0.811, r is not significant, and the line should not be used for prediction. -0.811 < r = 0.776 < 0.811. Therefore, r is not significant. Try It For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is = 0.707. Can the line be used for prediction? Why or why not? THIRD-EXAM vs FINAL-EXAM EXAMPLE: critical value method Consider the third exam/final exam example . The line of best fit is: ŷ = –173.51+4.83 x with r = 0.6631 and there are n = 11 data points. Can the regression line be used for prediction? Given a third-exam score ( x value), can we use the line to predict the final exam score (predicted y value)? H 0 : ρ = 0 H a : ρ ≠ 0 α = 0.05 Use the \"95% Critical Value\" table for r with df = n – 2 = 11 – 2 = 9. The critical values are –0.602 and +0.602 Since 0.6631 > 0.602, r is significant. Decision: Reject the null hypothesis. Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score ( x ) and the final exam score ( y ) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if r is significant and the line of best fit associated with each r can be used to predict a y value. If it helps, draw a number line. r = –0.567 and the sample size, n , is 19. The df = n – 2 = 17. The critical value is –0.456. –0.567 < –0.456 so r is significant. r = 0.708 and the sample size, n , is nine. The df = n – 2 = 7. The critical value is 0.666. 0.708 > 0.666 so r is significant. r = 0.134 and the sample size, n , is 14. The df = 14 – 2 = 12. The critical value is 0.532. 0.134 is between –0.532 and 0.532 so r is not significant. r = 0 and the sample size, n , is five. No matter what the dfs are, r = 0 is between the two critical values so r is not significant. Try It For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not? Assumptions in Testing the Significance of the Correlation Coefficient Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between x and y in the population. The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this. The assumptions underlying the test of significance are: The relationship between the variables being correlated should be linear. The data points should fall along an approximate straight-line pattern when plotted as ( x , y ) data points on a scatter plot. The y values for any particular x value are normally distributed about the line. This implies that there are more y values scattered closer to the line than are scattered farther away. Assumption (1) implies that these normal distributions are centered on the line: the means of these normal distributions of y values lie on the line. The standard deviations of the population y values about the line are equal for each value of x . In other words, each of these normal distributions of y values has the same shape and spread about the line. The residual errors are mutually independent (no pattern). The data are produced from a well-designed, random sample or randomized experiment. The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered further away from the line. Chapter Review Linear regression is a procedure for fitting a straight line of the form ŷ = a + bx to data. The conditions for regression are: Linear In the population, there is a linear relationship that models the average value of y for different values of x . Independent The residuals are assumed to be independent. Normal The y values are distributed normally for any value of x . Equal variance The standard deviation of the y values is equal for each x value. Random The data are produced from a well-designed random sample or randomized experiment. The slope b and intercept a of the least-squares line estimate the slope β and intercept α of the population (true) regression line. To estimate the population standard deviation of y , σ , use the standard deviation of the residuals, s . s = S E E n − 2 . The variable ρ (rho) is the population correlation coefficient. To test the null hypothesis H 0 : ρ = hypothesized value , use a linear regression t-test. The most common null hypothesis is H 0 : ρ = 0 which indicates there is no linear relationship between x and y in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest). Formula Review Least Squares Line or Line of Best Fit: y ^ = a + b x where a = y -intercept b = slope Standard deviation of the residuals: s = S E E n − 2 . where SSE = sum of squared errors n = the number of data points When testing the significance of the correlation coefficient, what is the null hypothesis? When testing the significance of the correlation coefficient, what is the alternative hypothesis? H a : ρ ≠ 0 If the level of significance is 0.05 and the p -value is 0.04, what conclusion can you draw?", "section": "Testing the Significance of the Correlation Coefficient", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Prediction Recall the third exam/final exam example . We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction. Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores ( x -values) range from 65 to 75. Since 73 is between the x -values 65 and 75 , substitute x = 73 into the equation. Then: y ^ = − 173.51 + 4.83 ( 73 ) = 179.08 We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average. Recall the third exam/final exam example . a. What would you predict the final exam score to be for a student who scored a 66 on the third exam? b. What would you predict the final exam score to be for a student who scored a 90 on the third exam? a. 145.27 b. The x values in the data are between 65 and 75. Ninety is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable.) To understand really how unreliable the prediction can be outside of the observed x values observed in the data, make the substitution x = 90 into the equation. y ^ = –173.51 + 4.83 ( 90 ) = 261.19 The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200. NOTE The process of predicting inside of the observed x values observed in the data is called interpolation . The process of predicting outside of the observed x values observed in the data is called extrapolation . Try It Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8 x What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week? References Data from the Centers for Disease Control and Prevention. Data from the National Center for agency reporting flu cases and TB Prevention. Data from the United States Census Bureau. Available online at http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html Data from the National Center for Health Statistics. Chapter Review After determining the presence of a strong correlation coefficient and calculating the line of best fit, you can use the least squares regression line to make predictions about your data. Use the following information to answer the next two exercises . An electronics retailer used regression to find a simple model to predict sales growth in the first quarter of the new year (January through March). The model is good for 90 days, where x is the day. The model can be written as follows: ŷ = 101.32 + 2.48 x where ŷ is in thousands of dollars. What would you predict the sales to be on day 60? $250,120 What would you predict the sales to be on day 90? Use the following information to answer the next three exercises . A landscaping company is hired to mow the grass for several large properties. The total area of the properties combined is 1,350 acres. The rate at which one person can mow is as follows: ŷ = 1350 – 1.2 x where x is the number of hours and ŷ represents the number of acres left to mow. How many acres will be left to mow after 20 hours of work? 1,326 acres How many acres will be left to mow after 100 hours of work? How many hours will it take to mow all of the lawns? (When is ŷ = 0?) 1,125 hours, or when x = 1,125 contains real data for the first two decades of flu cases reporting. Adults and Adolescents only, United States Year # flu cases diagnosed # flu deaths Pre-1981 91 29 1981 319 121 1982 1,170 453 1983 3,076 1,482 1984 6,240 3,466 1985 11,776 6,878 1986 19,032 11,987 1987 28,564 16,162 1988 35,447 20,868 1989 42,674 27,591 1990 48,634 31,335 1991 59,660 36,560 1992 78,530 41,055 1993 78,834 44,730 1994 71,874 49,095 1995 68,505 49,456 1996 59,347 38,510 1997 47,149 20,736 1998 38,393 19,005 1999 25,174 18,454 2000 25,522 17,347 2001 25,643 17,402 2002 26,464 16,371 Total 802,118 489,093 Graph “year” versus “# flu cases diagnosed” (plot the scatter plot). Do not include pre-1981 data. Perform linear regression. What is the linear equation? Round to the nearest whole number. Answers may vary. Find the correlation coefficient. r = ________ Solve. When x = 1985, ŷ = _____ When x = 1990, ŷ =_____ When x = 1970, ŷ =______ Why doesn’t this answer make sense? When x = 1985, ŷ = 25,52 When x = 1990, ŷ = 34,275 When x = 1970, ŷ = –725 Why doesn’t this answer make sense? The range of x values was 1981 to 2002; the year 1970 is not in this range. The regression equation does not apply, because predicting for the year 1970 is extrapolation, which requires a different process. Also, a negative number does not make sense in this context, where we are predicting flu cases diagnosed. Does the line seem to fit the data? Why or why not? What does the correlation imply about the relationship between time (years) and the number of diagnosed flu cases reported in the U.S.? Also, the correlation r = 0.4526. If r is compared to the value in the 95% Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But the scatter plot indicates otherwise. Plot the two given points on the following graph. Then, connect the two points to form the regression line. Obtain the graph on your calculator or computer. Write the equation: ŷ = ____________ y ^ = –3,448,225 + 1750 x Hand draw a smooth curve on the graph that shows the flow of the data. Does the line seem to fit the data? Why or why not? There was an increase in flu cases diagnosed until 1993. From 1993 through 2002, the number of flu cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data. Do you think a linear fit is best? Why or why not? What does the correlation imply about the relationship between time (years) and the number of diagnosed flu cases reported in the U.S.? Since there is no linear association between year and # of flu cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable “causes” the other variable. Graph “year” vs. “# flu cases diagnosed.” Do not include pre-1981. Label both axes with words. Scale both axes. Enter your data into your calculator or computer. The pre-1981 data should not be included. Why is that so? Write the linear equation, rounding to four decimal places: We don’t know if the pre-1981 data was collected from a single year. So we don’t have an accurate x value for this figure. Regression equation: ŷ (#Flu Cases) = –3,448,225 + 1749.777 (year) Coefficients Intercept –3,448,225 X Variable 1 1,749.777 Find the correlation coefficient. correlation = _____ Homework Recently, the annual number of driver deaths per 100,000 for the selected age groups was as follows: Age Number of Driver Deaths per 100,000 16–19 38 20–24 36 25–34 24 35–54 20 55–74 18 75+ 28 For each age group, pick the midpoint of the interval for the x value. (For the 75+ group, use 80.) Using “ages” as the independent variable and “Number of driver deaths per 100,000” as the dependent variable, make a scatter plot of the data. Calculate the least squares (best–fit) line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Predict the number of deaths for ages 40 and 60. Based on the given data, is there a linear relationship between age of a driver and driver fatality rate? What is the slope of the least squares (best-fit) line? Interpret the slope. Age Number of Driver Deaths per 100,000 16–19 38 20–24 36 25–34 24 35–54 20 55–74 18 75+ 28 Check student’s solution. ŷ = 35.5818045 – 0.19182491 x r = –0.57874 For four df and alpha = 0.05, the LinRegTTest gives p -value = 0.2288 so we do not reject the null hypothesis; there is not a significant linear relationship between deaths and age. Using the table of critical values for the correlation coefficient, with four df , the critical value is 0.811. The correlation coefficient r = –0.57874 is not less than –0.811, so we do not reject the null hypothesis. There is not a linear relationship between the two variables, as evidenced by a p-value greater than 0.05. shows the life expectancy for an individual born in the United States in certain years. Year of Birth Life Expectancy 1930 59.7 1940 62.9 1950 70.2 1965 69.7 1973 71.4 1982 74.5 1987 75 1992 75.7 2010 78.7 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the ordered pairs. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated life expectancy for an individual born in 1950 and for one born in 1982. Why aren’t the answers to part e the same as the values in that correspond to those years? Use the two points in part e to plot the least squares line on your graph from part b. Based on the data, is there a linear relationship between the year of birth and life expectancy? Are there any outliers in the data? Using the least squares line, find the estimated life expectancy for an individual born in 1850. Does the least squares line give an accurate estimate for that year? Explain why or why not. What is the slope of the least-squares (best-fit) line? Interpret the slope. The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition ten, for various pages is given in Page number Maximum value ($) 4 16 14 19 25 15 32 17 43 19 57 15 72 16 85 15 90 17 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the ordered pairs. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated maximum values for the restaurants on page ten and on page 70. Does it appear that the restaurants giving the maximum value are placed in the beginning of the “Fine Dining” section? How did you arrive at your answer? Suppose that there were 200 pages of restaurants. What do you estimate to be the maximum value for a restaurant listed on page 200? Is the least squares line valid for page 200? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. We wonder if the better discounts appear earlier in the book so we select page as X and discount as Y . Check student’s solution. ŷ = 17.21757 – 0.01412 x r = – 0.2752 For seven df and alpha = 0.05, using LinRegTTest p -value = 0.4736 so we do not reject; there is a not a significant linear relationship between page and discount. Using the table of critical values for the correlation coefficient, with seven df , the critical value is 0.666. The correlation coefficient xi = –0.2752 is not less than 0.666 so we do not reject. There is not a significant linear correlation so it appears there is no relationship between the page and the amount of the discount. As the page number increases by one page, the discount decreases by $0.01412 gives the gold medal times for every other Summer Olympics for the women’s 100-meter freestyle (swimming). Year Time (seconds) 1912 82.2 1924 72.4 1932 66.8 1952 66.8 1960 61.2 1968 60.0 1976 55.65 1984 55.92 1992 54.64 2000 53.8 2008 53.1 2016 52.7 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least squares line. Put the equation in the form of: ŷ = a + bx . Find the correlation coefficient. Is the decrease in times significant? Find the estimated gold medal time for 1932. Find the estimated time for 1984. Why are the answers from part f different from the chart values? Does it appear that a line is the best way to fit the data? Why or why not? Use the least-squares line to estimate the gold medal time for the next Summer Olympics. Do you think that your answer is reasonable? Why or why not? State # letters in name Year entered the Union Rank for entering the Union Area (square miles) Alabama 7 1819 22 52,423 Colorado 8 1876 38 104,100 Hawaii 6 1959 50 10,932 Iowa 4 1846 29 56,276 Maryland 8 1788 7 12,407 Missouri 8 1821 24 69,709 New Jersey 9 1787 3 8,722 Ohio 4 1803 17 44,828 South Carolina 13 1788 8 32,008 Utah 4 1896 45 84,904 Wisconsin 9 1848 30 65,499 We are interested in whether or not the number of letters in a state name depends upon the year the state entered the Union. Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx . Find the correlation coefficient. What does it imply about the significance of the relationship? Find the estimated number of letters (to the nearest integer) a state would have if it entered the Union in 1900. Find the estimated number of letters a state would have if it entered the Union in 1940. Does it appear that a line is the best way to fit the data? Why or why not? Use the least-squares line to estimate the number of letters a new state that enters the Union this year would have. Can the least squares line be used to predict it? Why or why not? Year is the independent or x variable; the number of letters is the dependent or y variable. Check student’s solution. no ŷ = 47.03 – 0.0216 x –0.4280 The r-value indicates that there is not a significant correlation between the year the state entered the union and the number of letters in the name. No, the relationship does not appear to be linear; the correlation is not significant.", "section": "Prediction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Outliers In some data sets, there are values (observed data points) called outliers . Outliers are observed data points that are far from the least squares line. They have large \"errors\", where the \"error\" or residual is the vertical distance from the line to the point. Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier. Besides outliers, a sample may contain one or a few points that are called influential points . Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly. Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them. Identifying Outliers We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier . The standard deviation used is the standard deviation of the residuals or errors. We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods. In the third exam/final exam example , you can determine if there is an outlier or not. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or –1. Graphical Identification of Outliers With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2 s or more, then we would consider the data point to be \"too far\" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3: As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412 . Line Y2 = –173.5 + 4.83 x –2(16.4) and line Y3 = –173.5 + 4.83 x + 2(16.4) where ŷ = –173.5 + 4.83 x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit. Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the \"Y=\"equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line. Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers. Try It Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6. Numerical Identification of Outliers In , the first two columns are the third-exam and final-exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83 x . The residuals, or errors, have been calculated in the fourth column of the table: observed y value−predicted y value = y − ŷ . s is the standard deviation of all the y − ŷ = ε values where n = the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as: s = S S E n − 2 NOTE We divide by ( n – 2) because the regression model involves two estimates. Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35 –17 16 –6 –19 9 3 –1 –10 –9 –1 . x y ŷ y – ŷ 65 175 140 175 – 140 = 35 67 133 150 133 – 150= –17 71 185 169 185 – 169 = 16 71 163 169 163 – 169 = –6 66 126 145 126 – 145 = –19 75 198 189 198 – 189 = 9 67 153 150 153 – 150 = 3 70 163 164 163 – 164 = –1 71 159 169 159 – 169 = –10 69 151 160 151 – 160 = –9 69 159 160 159 – 160 = –1 We are looking for all data points for which the residual is greater than 2 s = 2(16.4) = 32.8 or less than –32.8. Compare these values to the residuals in column four of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35. How does the outlier affect the best fit line? Numerically and graphically, we have identified the point (65, 175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on and delete the outlier, so that we can explore how it affects the results, as a learning experience. Compute a new best-fit line and correlation coefficient using the ten remaining points: On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39 x and r = 0.9121 The new line with r = 0.9121 is a stronger correlation than the original ( r = 0.6631) because r = 0.9121 is closer to one. This means that the new line is a better fit to the ten remaining data values. The line can better predict the final exam score given the third exam score. Numerical Identification of Outliers: Calculating s and Finding Outliers Manually If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following. First, square each | y – ŷ | The squares are 35 2 17 2 16 2 6 2 19 2 9 2 3 2 1 2 10 2 9 2 1 2 Then, add (sum) all the | y – ŷ | squared terms using the formula Σ i = 1 11 ( | y i − y ^ i | ) 2 = Σ i = 1 11 ε i 2 (Recall that y i – ŷ i = ε i .) = 35 2 + 17 2 + 16 2 + 6 2 + 19 2 + 9 2 + 3 2 + 1 2 + 10 2 + 9 2 + 1 2 = 2440 = SSE . The result, SSE is the Sum of Squared Errors. Next, calculate s , the standard deviation of all the y – ŷ = ε values where n = the total number of data points. The calculation is s = SSE n – 2 . For the third exam/final exam problem, s = 2440 11 – 2 = 16.47 . Next, multiply s by 2: (2)(16.47) = 32.94 32.94 is 2 standard deviations away from the mean of the y – ŷ values. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance is at least 2 s , then we would consider the data point to be \"too far\" from the line of best fit. We call that point a potential outlier . For the example, if any of the | y – ŷ | values are at least 32.94, the corresponding ( x , y ) data point is a potential outlier. For the third exam/final exam problem, all the | y – ŷ |'s are less than 31.29 except for the first one which is 35. 35 > 31.29 That is, | y – ŷ | ≥ (2)(s) The point which corresponds to | y – ŷ | = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.) NOTE When outliers are deleted, the researcher should either record that data was deleted, and why, or the researcher should provide results both with and without the deleted data. If data is erroneous and the correct values are known (e.g., student one actually scored a 70 instead of a 65), then this correction can be made to the data. The next step is to compute a new best-fit line using the ten remaining points. The new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39 x and r = 0.9121 Using this new line of best fit (based on the remaining ten data points in the third exam/final exam example ), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line? Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam. The original line predicted ŷ = –173.51 + 4.83(73) = 179.08 so the prediction using the new line with the outlier eliminated differs from the original prediction. Try It The data points for the graph from the third exam/final exam example are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10. The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. In the following table, x is the year and y is the CPI. Data x y x y 1915 10.1 1979 72.6 1926 17.7 1980 82.4 1935 13.7 1986 109.6 1940 14.7 1991 130.7 1947 24.1 1999 166.6 1952 26.5 2010 219.2 1964 31.0 2020 258 1969 36.7 2023 299.2 1975 49.3 Draw a scatterplot of the data. Calculate the least squares line. Write the equation in the form ŷ = a + bx . Draw the line on the scatterplot. Find the correlation coefficient. Is it significant? What is the average CPI for the year 1990? See . ŷ = –3204 + 1.662 x is the equation of the line of best fit. r = 0.8694 The number of data points is n = 14. Use the 95% Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12. n – 2 = 12. The corresponding critical value is 0.532. Since 0.8694 > 0.532, r is significant. ŷ = –3204 + 1.662(1990) = 103.4 CPI Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2 = –3204 + 1.662X – 2(25.4) and Y3 = –3204 + 1.662X + 2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside it.) NOTE In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician should prefer to use other methods to fit a curve to this data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate. If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt; our data is taken from the column entitled \"Annual Avg.\" (third column from the right). For example you could add more current years of data. Try adding the more recent years: 2010: CPI = 219.2; 2020: CPI = 258.0; 2023: CPI = 299.2. See how it affects the model. (Check: y ⏜ = - 5030 + 2 . 598 x ; r = 0 . 9067 . Is r significant? Is the fit better with the addition of the new points?) Try It The following table shows economic development measured in per capita income PCINC. Year PCINC Year PCINC 1870 340 1920 1050 1880 499 1930 1170 1890 592 1940 1364 1900 757 1950 1836 1910 927 1960 2132 What are the independent and dependent variables? Draw a scatter plot. Use regression to find the line of best fit and the correlation coefficient. Interpret the significance of the correlation coefficient. Is there a linear relationship between the variables? Find the coefficient of determination and interpret it. What is the slope of the regression equation? What does it mean? Use the line of best fit to estimate PCINC for 1900, for 2000. Determine if there are any outliers. 95% Critical Values of the Sample Correlation Coefficient Table Degrees of Freedom: n – 2 Critical Values: (+ and –) 1 0.997 2 0.950 3 0.878 4 0.811 5 0.754 6 0.707 7 0.666 8 0.632 9 0.602 10 0.576 11 0.555 12 0.532 13 0.514 14 0.497 15 0.482 16 0.468 17 0.456 18 0.444 19 0.433 20 0.423 21 0.413 22 0.404 23 0.396 24 0.388 25 0.381 26 0.374 27 0.367 28 0.361 29 0.355 30 0.349 40 0.304 50 0.273 60 0.250 70 0.232 80 0.217 90 0.205 100 0.195 References Data from the House Ways and Means Committee, the Health and Human Services Department. Data from Microsoft Bookshelf. Data from the United States Department of Labor, the Bureau of Labor Statistics. Data from the Physician’s Handbook, 1990. Data from the United States Department of Labor, the Bureau of Labor Statistics. Chapter Review To determine if a point is an outlier, do one of the following: Input the following equations into the TI 83, 83+,84, 84+: y 1 = a + b x y 2 = a + b x + 2 s y 3 = a + b x − 2 s where s is the standard deviation of the residuals If any point is above y 2 or below y 3 then the point is considered to be an outlier. Use the residuals and compare their absolute values to 2 s where s is the standard deviation of the residuals. If the absolute value of any residual is greater than or equal to 2 s , then the corresponding point is an outlier. Note: The calculator function LinRegTTest (STATS TESTS LinRegTTest) calculates s . Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69. Do there appear to be any outliers? Yes, there appears to be an outlier at (6, 58). A point is removed, and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why? What effect did the potential outlier have on the line of best fit? The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data. Are you more or less confident in the predictive ability of the new line of best fit? The Sum of Squared Errors for a data set of 18 numbers is 49. What is the standard deviation? s = 1.75 The Standard Deviation for the Sum of Squared Errors for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier? Homework The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level). Height (in feet) Stories 1,050 57 428 28 362 26 529 40 790 60 401 22 380 38 1,454 110 1,127 100 700 46 Using “stories” as the independent variable and “height” as the dependent variable, make a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Calculate the least squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated heights for 32 stories and for 94 stories. Based on the data in , is there a linear relationship between the number of stories in tall buildings and the height of the buildings? Are there any outliers in the data? If so, which point(s)? What is the estimated height of a building with six stories? Does the least squares line give an accurate estimate of height? Explain why or why not. Based on the least squares line, adding an extra story is predicted to add about how many feet to a building? What is the slope of the least squares (best-fit) line? Interpret the slope. Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration. Percent return: 74; 66; 81; 52; 73; 62; 52; 45; 62; 46; 60; 46; 38 Percent new: 5; 6; 8; 11; 12; 15; 16; 17; 18; 18; 19; 20; 20 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70% of the adults from the previous year have returned. What is the prediction? a. and b. Check student’s solution. c. The slope of the regression line is -0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32% new sparrow hawks, which doesn’t make sense since if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3%. d. If we examine r2, we see that only 57.52% of the variation in the percent of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages. e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66%, our observed new percentage, 6%, is almost 6% less than the predicted new value of 11.98%. If we remove this data pair, we see only an adjusted slope of -.2789 and an adjusted intercept of 30.9816. In other words, even though this data generates the largest residual, it is not an outlier, nor is the data pair an influential point. f. If there are 70% returning birds, we would expect to see y = –.2789(70) + 30.9816 = 0.114 or 11.4% new birds in the colony. The following table shows data on average per capita coffee consumption and heart disease rate in a random sample of 10 countries. Yearly coffee consumption in liters 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7 Death from heart diseases 221 167 131 191 220 297 71 172 211 300 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. Do the data provide convincing evidence that there is a linear relationship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days: Swim Time Heart Rate 34.12 144 35.72 152 34.72 124 34.05 140 34.13 152 35.73 146 36.17 128 35.57 136 35.37 144 35.57 148 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. Check student’s solution. Check student’s solution. We have a slope of –1.4946 with a y -intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate will decrease by 1.5 beats per minute. If the student is not swimming at all, the y -intercept indicates that his heart rate will be 193.88 beats per minute. While the slope has meaning (the longer it takes to swim 2,000 meters, the less effort the heart puts out), the y -intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low. Since only 1.5% of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship. The point (34.72, 124) generates the largest residual of –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953 with the y-intercept changing to 247.1616. While the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y -intercept becomes more meaningful. A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are White men. Population Size Homicide rate per 100,000 people 558,724 8.6 538,584 8.9 519,171 8.52 500,457 8.89 482,418 13.07 465,029 14.57 448,267 21.36 432,109 28.03 416,533 31.49 401,518 37.39 387,046 46.26 373,095 47.24 359,647 52.33 Use your calculator to construct a scatter plot of the data. What should the independent variable be? Why? Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot. Discuss what the following mean in context. The slope of the regression equation The y-intercept of the regression equation The correlation r The coefficient of determination r2. Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. School Mid-Career Salary (in thousands) Yearly Tuition Princeton 137 57,410 Harvey Mudd 135 62,817 CalTech 127 60,864 US Naval Academy 122 0 West Point 120 0 MIT 118 57,986 Lehigh University 118 59,930 NYU-Poly 117 58,168 Babson College 117 56,576 Stanford 114 56,169 Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer. If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = 0.002 x + 0.309 with a correlation coefficient of 0.533 and a coefficient of determination of 0.284. There is not enough evidence to state that there is a strong linear correlation between tuition costs and future salaries, even when the service academies are removed. Bring It Together The average number of people in a family that attended college for various years is given in . Year Number of Family Members Attending College 1969 4.0 1973 3.6 1975 3.2 1979 3.0 1983 3.0 1988 3.0 1991 2.9 Using “year” as the independent variable and “Number of Family Members Attending College” as the dependent variable, draw a scatter plot of the data. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Pick two years between 1969 and 1991 and find the estimated number of family members attending college. Based on the data in , is there a linear relationship between the year and the average number of family members attending college? Using the least-squares line, estimate the number of family members attending college for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not. Are there any outliers in the data? What is the estimated average number of family members attending college for 1986? Does the least squares line give an accurate estimate for that year? Explain why or why not. What is the slope of the least squares (best-fit) line? Interpret the slope. The percent of women who are wage and salary workers who are paid hourly rates is given in for a 14-year period. Year Percent of workers paid hourly rates 1 61.2 2 60.7 3 61.3 4 61.3 5 61.8 6 61.7 7 61.8 8 62.0 9 62.7 12 62.8 14 62.9 Using “year” as the independent variable and “percent” as the dependent variable, draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated percents for year 13 and year 10. Based on the data, is there a linear relationship between the year and the percent of female wage and salary earners who are paid hourly rates? Are there any outliers in the data? What is the estimated percent for the year 72? Does the least-squares line give an accurate estimate for that year? Explain why or why not. What is the slope of the least-squares (best-fit) line? Interpret the slope. Check student's solution. yes ŷ = −266.8863+0.1656x 0.9448; Yes 62.8233; 62.3265 yes no; (Year 9, 62.7) 72.5937; no slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. Use the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in . Size (ounces) Cost ($) Cost per ounce 16 3.99 32 4.99 64 5.99 200 10.99 Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? If the laundry detergent were sold in a 40-ounce size, find the estimated cost. If the laundry detergent were sold in a 90-ounce size, find the estimated cost. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the given data? Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. Complete for the cost per ounce of the different sizes. Using “size” as the independent variable and “cost per ounce” as the dependent variable, draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? If the laundry detergent were sold in a 40-ounce size, find the estimated cost per ounce. If the laundry detergent were sold in a 90-ounce size, find the estimated cost per ounce. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the the data? Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would cost per ounce? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. Size (ounces) Cost ($) cents/oz 16 3.99 24.94 32 4.99 15.59 64 5.99 9.36 200 10.99 5.50 Check student’s solution. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. ŷ = 20.2368 – 0.0819x r = –0.8086 40-oz: 16.96 cents/oz 90-oz: 12.87 cents/oz The relationship is not linear; the least squares line is not appropriate. no outliers No, you would be extrapolating. The 300-oz size is outside the range of x . slope = –0.08194; for each additional ounce in size, the cost per ounce decreases by 0.082 cents. According to a flyer by a Prudential Insurance Company representative, the costs of approximate probate fees and taxes for selected net taxable estates are as follows: Net Taxable Estate ($) Approximate Probate Fees and Taxes ($) 600,000 30,000 750,000 92,500 1,000,000 203,000 1,500,000 438,000 2,000,000 688,000 2,500,000 1,037,000 3,000,000 1,350,000 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx . Find the correlation coefficient. Is it significant? Find the estimated total cost for a next taxable estate of $1,000,000. Find the cost for $2,500,000. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? Based on these results, what would be the probate fees and taxes for an estate that does not have any assets? What is the slope of the least-squares (best-fit) line? Interpret the slope. The following are advertised sale prices of hard disk drives on Amazon’s website for various capacity hard drives. Hard Disk Drive Capacity (in terabytes) Sale Price ($) 9 147 20 197 27 297 31 447 35 1177 40 2177 60 2497 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated sale price for a 32 terabyte hard drive. Find the cost for a 50 terabyte hard drive. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? What is the slope of the least-squares (best-fit) line? Interpret the slope. Size is x , the independent variable, price is y , the dependent variable. Answers may vary. The relationship does not appear to be linear. ŷ = –745.252 + 54.75569 x r = 0.8944, yes it is significant 32-terabyte: $1006.93, 50-terabyte: $1992.53 No, the relationship does not appear to be linear. However, r is significant. no, the 60-terabyte hard drive For each additional terabyte, the price increases by $54.76 shows the average heights for American boys in a specific year. Age (years) Height (cm) birth 50.8 2 83.8 3 91.4 5 106.6 7 119.3 10 137.1 14 157.5 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated average height for a one-year-old. Find the estimated average height for an eleven-year-old. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? Use the least squares line to estimate the average height for a sixty-two-year-old man. Do you think that your answer is reasonable? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. State # letters in name Year entered the Union Ranks for entering the Union Area (square miles) Alabama 7 1819 22 52,423 Colorado 8 1876 38 104,100 Hawaii 6 1959 50 10,932 Iowa 4 1846 29 56,276 Maryland 8 1788 7 12,407 Missouri 8 1821 24 69,709 New Jersey 9 1787 3 8,722 Ohio 4 1803 17 44,828 South Carolina 13 1788 8 32,008 Utah 4 1896 45 84,904 Wisconsin 9 1848 30 65,499 We are interested in whether there is a relationship between the ranking of a state and the area of the state. What are the independent and dependent variables? What do you think the scatter plot will look like? Make a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. What does it imply about the significance of the relationship? Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas? Use the two points in part f to plot the least-squares line on your graph from part b. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers? Use the least squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not? Delete “Hawaii” and substitute “Alaska” for it. Alaska is the forty-ninth, state with an area of 656,424 square miles. Calculate the new least-squares line. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case? Do you think that, in general, newer states are larger than the original states? Let rank be the independent variable and area be the dependent variable. Answers may vary. There appears to be a linear relationship, with one outlier. ŷ (area) = 24177.06 + 1010.478 x r = 0.50047, r is not significant so there is no relationship between the variables. Alabama: 46407.576 Colorado: 62575.224 Alabama estimate is closer than Colorado estimate. If the outlier is removed, there is a linear relationship. There is one outlier (Hawaii). rank 51: 75711.4; no Alabama 7 1819 22 52,423 Colorado 8 1876 38 104,100 Hawaii 6 1959 50 10,932 Iowa 4 1846 29 56,276 Maryland 8 1788 7 12,407 Missouri 8 1821 24 69,709 New Jersey 9 1787 3 8,722 Ohio 4 1803 17 44,828 South Carolina 13 1788 8 32,008 Utah 4 1896 45 84,904 Wisconsin 9 1848 30 65,499 ŷ = –87065.3 + 7828.532x Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. yes, with the exception of Hawaii Outlier an observation that does not fit the rest of the data", "section": "Outliers", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Regression (Distance from School) Regression (Distance from School) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Use eight members of your class for the sample. Collect bivariate data (distance an individual lives from school, the cost of supplies for the current term). Complete the table. Distance from school Cost of supplies this term Which variable should be the dependent variable and which should be the independent variable? Why? Graph “distance” vs. “cost.” Plot the points on the graph. Label both axes with words. Scale both axes. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: ŷ = ______ Is the correlation significant? Why or why not? (Answer in one to three complete sentences.) Supply an answer for the following senarios: For a person who lives eight miles from campus, predict the total cost of supplies this term: For a person who lives eighty miles from campus, predict the total cost of supplies this term: Obtain the graph on your calculator or computer. Sketch the regression line. Discussion Questions Answer each question in complete sentences. Does the line seem to fit the data? Why? What does the correlation imply about the relationship between the distance and the cost? Are there any outliers? If so, which point is an outlier? Should the outlier, if it exists, be removed? Why or why not?", "section": "Regression (Distance from School)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Regression (Textbook Cost) Regression (Textbook Cost) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Survey ten textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook). Complete the table. Number of pages Cost of textbook Which variable should be the dependent variable and which should be the independent variable? Why? Graph “pages” vs. “cost.” Plot the points on the graph in Analyze the Data . Label both axes with words. Scale both axes. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: y = ______ Is the correlation significant? Why or why not? (Answer in complete sentences.) Supply an answer for the following senarios: For a textbook with 400 pages, predict the cost. For a textbook with 600 pages, predict the cost. Obtain the graph on your calculator or computer. Sketch the regression line. Discussion Questions Answer each question in complete sentences. Does the line seem to fit the data? Why? What does the correlation imply about the relationship between the number of pages and the cost? Are there any outliers? If so, which point(s) is an outlier? Should the outlier, if it exists, be removed? Why or why not?", "section": "Regression (Textbook Cost)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Regression (Fuel Efficiency) Regression (Fuel Efficiency) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Find a reputable source that provides information on total fuel efficiency (in miles per gallon) and weight (in pounds) of new model cars with automatic transmissions. We will use this data to determine the relationship, if any, between the fuel efficiency of a car and its weight. Using your random number generator, randomly select 20 cars from the list and record their weights and fuel efficiency into . Weight Fuel Efficiency Which variable should be the dependent variable and which should be the independent variable? Why? By hand, do a scatterplot of “weight” vs. “fuel efficiency”. Plot the points on graph paper. Label both axes with words. Scale both axes accurately. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to 4 decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: ŷ = ______ Obtain the graph of the regression line on your calculator. Sketch the regression line on the same axes as your scatter plot. Discussion Questions Is the correlation significant? Explain how you determined this in complete sentences. Is the relationship a positive one or a negative one? Explain how you can tell and what this means in terms of weight and fuel efficiency. In one or two complete sentences, what is the practical interpretation of the slope of the least squares line in terms of fuel efficiency and weight? For a car that weighs 4,000 pounds, predict its fuel efficiency. Include units. Can we predict the fuel efficiency of a car that weighs 10,000 pounds using the least squares line? Explain why or why not. Answer each question in complete sentences. Does the line seem to fit the data? Why or why not? What does the correlation imply about the relationship between fuel efficiency and weight of a car? Is this what you expected? Are there any outliers? If so, which point is an outlier?", "section": "Regression (Fuel Efficiency)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Introduction One-way ANOVA is used to measure information from several groups. (credit: modification of work “Magazine Stack” by thebittenword.com/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Interpret the F probability distribution as the number of groups and the sample size change. Discuss two uses for the F distribution: one-way ANOVA and the test of two variances. Conduct and interpret one-way ANOVA. Conduct and interpret hypothesis tests of two variances. Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to their upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests comparing averages between more than two groups, statisticians have developed a method called \" Analysis of Variance \" (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or one-way ANOVA . You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is just a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer.", "section": "Introduction", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a one-way ANOVA test, there are five basic assumptions to be fulfilled: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances) . The factor is a categorical variable. The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups: H 0 : μ 1 = μ 2 = μ 3 = ... = μ k H a : At least two of the group means μ 1 , μ 2 , μ 3 , ..., μ k are not equal. That is, μ i ≠ μ j for some i ≠ j . The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (blue box plots), H 0 : μ 1 = μ 2 = μ 3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots). (a) H 0 is true. All means are the same; the differences are due to random variation. (b) H 0 is not true. The means are not all the same; the differences are too large to be due to random variation. Chapter Review Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances). Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they? Write one assumption. Each population from which a sample is taken is assumed to be normal. Write another assumption. Write a third assumption. The populations are assumed to have equal standard deviations (or variances). Write a fourth assumption. Write the final assumption. The response is a numerical value. State the null hypothesis for a one-way ANOVA test if there are four groups. State the alternative hypothesis for a one-way ANOVA test if there are three groups. H a : At least two of the group means μ 1 , μ 2 , μ 3 are not equal. When do you use an ANOVA test? Homework Three different traffic routes are tested for mean driving time. The entries in the are the driving times in minutes on the three different routes. Route 1 Route 2 Route 3 30 27 16 32 29 41 27 28 22 35 36 31 State SS between , SS within , and the F statistic. SS between = 26 SS within = 441 F = 0.2653 Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = State the hypotheses. H 0 : ____________ H a : ____________ Analysis of Variance also referred to as ANOVA, is a method of testing whether or not the means of three or more populations are equal. The method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F -ratio. One-Way ANOVA a method of testing whether or not the means of three or more populations are equal; the method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. there is one independent variable and one dependent variable. The test statistic for analysis of variance is the F -ratio. Variance mean of the squared deviations from the mean; the square of the standard deviation. For a set of data, a deviation can be represented as x – x ¯ where x is a value of the data and x ¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.", "section": "One-Way ANOVA", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "The F Distribution and the F-Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution , named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F 4,10 . NOTE The F distribution is derived from the Student's t-distribution. The values of the F distribution are squares of the corresponding values of the t -distribution. One-Way ANOVA expands the t -test for comparing more than two groups. The scope of that derivation is beyond the level of this course. It is preferable to use ANOVA when there are more than two groups instead of performing pairwise t -tests because performing multiple tests introduces the likelihood of making a Type 1 error. To calculate the F ratio , two estimates of the variance are made. Variance between samples : An estimate of σ 2 that is the variance of the sample means multiplied by n (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. Variance within samples : An estimate of σ 2 that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. SS between = the sum of squares that represents the variation among the different samples SS within = the sum of squares that represents the variation within samples that is due to chance. To find a \"sum of squares\" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics . MS means \" mean square .\" MS between is the variance between groups, and MS within is the variance within groups. Calculation of Sum of Squares and Mean Square k = the number of different groups n j = the size of the j th group s j = the sum of the values in the j th group n = total number of all the values combined (total sample size: ∑ n j ) x = one value: ∑ x = ∑ s j Sum of squares of all values from every group combined: ∑ x 2 Total sum of squares: S S total = ∑ x 2 ( ∑ x ) 2 n Explained variation: sum of squares representing variation among the different samples: SS between = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n Unexplained variation: sum of squares representing variation within samples due to chance: S S within = S S total – S S between df 's for different groups ( df 's for the numerator): d f between = k – 1 Equation for errors within samples ( df 's for the denominator): df within = n – k Mean square (variance estimate) explained by the different groups: MS between = S S between d f between Mean square (variance estimate) that is due to chance (unexplained): MS within = S S within d f within MS between and MS within can be written as follows: M S between = S S between d f between = S S between k − 1 M S w i t h i n = S S w i t h i n d f w i t h i n = S S w i t h i n n − k The one-way ANOVA test depends on the fact that MS between can be influenced by population differences among means of the several groups. Since MS within compares values of each group to its own group mean, the fact that group means might be different does not affect MS within . The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MS between and MS within should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances. F -Ratio or F Statistic F = M S between M S within If MS between and MS within estimate the same value (following the belief that H 0 is true), then the F -ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MS between consists of the population variance plus a variance produced from the differences between the samples. MS within is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MS between will generally be larger than MS within .Then the F -ratio will be larger than one. However, if the population effect is small, it is not unlikely that MS within will be larger in a given sample. The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F -ratio can be written as: F -Ratio Formula when the groups are the same size F = n ⋅ s x ¯ 2 s 2 pooled where ... n = the sample size df numerator = k – 1 df denominator = n – k s 2 pooled = the mean of the sample variances (pooled variance) s x ¯ 2 = the variance of the sample means Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software. Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) SS (Factor) k – 1 MS (Factor) = SS (Factor)/( k – 1) F = MS (Factor)/ MS (Error) Error (Within) SS (Error) n – k MS (Error) = SS (Error)/( n – k ) Total SS (Total) n – 1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in . Plan 1: n 1 = 4 Plan 2: n 2 = 3 Plan 3: n 3 = 3 5 3.5 8 4.5 7 4 4 3.5 3 4.5 s 1 = 16.5, s 2 =15, s 3 = 15.5 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. S S ( b e t w e e n ) = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n = s 1 2 4 + s 2 2 3 + s 3 2 3 − ( s 1 + s 2 + s 3 ) 2 10 where n 1 = 4, n 2 = 3, n 3 = 3 and n = n 1 + n 2 + n 3 = 10 = ( 16.5 ) 2 4 + ( 15 ) 2 3 + ( 15.5 ) 2 3 − ( 16.5 + 15 + 15.5 ) 2 10 S S ( b e t w e e n ) = 2.2458 S ( t o t a l ) = ∑ x 2 − ( ∑ x ) 2 n = ( 5 2 + 4.5 2 + 4 2 + 3 2 + 3.5 2 + 7 2 + 4.5 2 + 8 2 + 4 2 + 3.5 2 ) − ( 5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5 ) 2 10 = 244 − 47 2 10 = 244 − 220.9 S S ( t o t a l ) = 23.1 S S ( w i t h i n ) = S S ( t o t a l ) − S S ( b e t w e e n ) = 23.1 − 2.2458 S S ( w i t h i n ) = 20.8542 One-Way ANOVA Table: The formulas for SS (Total) , SS (Factor) = SS (Between) and SS (Error) = SS (Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively). Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) SS (Factor) = SS (Between) = 2.2458 k – 1 = 3 groups – 1 = 2 MS (Factor) = SS (Factor)/( k – 1) = 2.2458/2 = 1.1229 F = MS (Factor)/ MS (Error) = 1.1229/2.9792 = 0.3769 Error (Within) SS (Error) = SS (Within) = 20.8542 n – k = 10 total data – 3 groups = 7 MS (Error) = SS (Error)/( n – k ) = 20.8542/7 = 2.9792 Total SS (Total) = 2.2458 + 20.8542 = 23.1 n – 1 = 10 total data – 1 = 9 Try It As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments bare soil a commercial ground cover black plastic straw compost All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants: Bare: n 1 = 3 Ground Cover: n 2 = 3 Plastic: n 3 = 3 Straw: n 4 = 3 Compost: n 5 = 3 2,625 5,348 6,583 7,285 6,277 2,997 5,682 8,560 6,897 7,818 4,915 5,482 3,830 9,230 8,677 Create the one-way ANOVA table. The one-way ANOVA hypothesis test is always right-tailed because larger F -values are way out in the right tail of the F -distribution curve and tend to make us reject H 0 . Notation The notation for the F distribution is F ~ F df ( num ), df ( denom ) where df ( num ) = df between and df ( denom ) = df within The mean for the F distribution is μ = d f ( d e n o m ) d f ( d e n o m ) – 2 References Tomato Data, Marist College School of Science (unpublished student research) Chapter Review Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. Formula Review S S between = ∑ ​ [ ( s j ) 2 n j ] − ( ∑ ​ s j ) 2 n S S total = ∑ ​ x 2 − ( ∑ ​ x ) 2 n S S within = S S total − S S between df between = df ( num ) = k – 1 df within = df(denom) = n – k MS between = S S between d f between MS within = S S within d f within F = M S between M S within F ratio when the groups are the same size: F = n s x ¯ 2 s 2 p o o l e d Mean of the F distribution: µ = d f ( n u m ) d f ( d e n o m ) − 2 where: k = the number of groups n j = the size of the j th group s j = the sum of the values in the j th group n = the total number of all values (observations) combined x = one value (one observation) from the data s x ¯ 2 = the variance of the sample means s 2 p o o l e d = the mean of the sample variances (pooled variance) Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in are the weights for the different groups. Group 1 Group 2 Group 3 216 202 170 198 213 165 240 284 182 187 228 197 176 210 201 What is the Sum of Squares Factor? 4,939.2 What is the Sum of Squares Error? What is the df for the numerator? 2 What is the df for the denominator? What is the Mean Square Factor? 2,469.6 What is the Mean Square Error? What is the F statistic? 3.7416 Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in . Team 1 Team 2 Team 3 Team 4 1 2 0 3 2 3 1 4 0 2 1 4 3 4 0 3 2 4 0 2 What is SS between ? What is the df for the numerator? 3 What is MS between ? What is SS within ? 13.2 What is the df for the denominator? What is MS within ? 0.825 What is the F statistic? Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p -value, so it is likely that we will reject the null hypothesis. Homework Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = H 0 : µ 1 = µ 2 = µ 3 = µ 4 = µ 5 Hα : At least any two of the group means µ 1 , µ 2 , …, µ 5 are not equal. degrees of freedom – numerator: df ( num ) = _________ degrees of freedom – denominator: df ( denom ) = ________ df ( denom ) = 15 F statistic = ________", "section": "The F Distribution and the F-Ratio", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Facts About the F Distribution Here are some facts about the F distribution. The curve is not symmetrical but skewed to the right. There is a different curve for each set of df s. The F statistic is greater than or equal to zero. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal. Other uses for the F distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter. Let’s return to the slicing tomato exercise in . The means of the tomato yields under the five mulching conditions are represented by μ 1 , μ 2 , μ 3 , μ 4 , μ 5 . We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. The null and alternative hypotheses are: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j some i ≠ j The one-way ANOVA results are shown in Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) 36,648,561 5 – 1 = 4 36,648,561 4 = 9,162,140 9,162,140 2,044,672 .6 = 4 .4810 Error (Within) 20,446,726 15 – 5 = 10 20,446,726 10 = 2,044,672 .6 Total 57,095,287 15 – 1 = 14 Distribution for the test: F 4,10 df ( num ) = 5 – 1 = 4 df ( denom ) = 15 – 5 = 10 Test statistic: F = 4.4810 Probability Statement: p -value = P ( F > 4.481) = 0.0248. Compare α and the p -value: α = 0.05, p -value = 0.0248 Make a decision: Since α > p -value, we reject H 0 . Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields. To find these results on the calculator: Press STAT. Press 1:EDIT . Put the data into the lists L 1 , L 2 , L 3 , L 4 , L 5 . Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter L 1 , L 2 , L 3 , L 4 , L 5 ). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p -value of the test. The calculator displays: F = 4.4810 p = 0.0248 ( p -value) Factor df = 4 SS = 36648560.9 MS = 9162140.23 Error df = 10 SS = 20446726 MS = 2044672.6 Try It There are multiple variants of the virus that causes COVID-19. The length of hospital stays for patients afflicted with various strains of COVID-19 is shown in . Delta Strain Omicron Strain Alpha Strain Gamma Strain Beta Strain 13.9 11.7 18.2 16.9 9.3 14.9 15.1 14.6 12.8 15.8 16.8 9.9 10.1 11.2 16.4 Test whether the mean length of hospital stay is the same or different for the various strains of COVID-19. Construct the ANOVA table, find the p -value, and state your conclusion. Use a 5% significance level. We test for equality of mean length of hospital stay: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j , some i ≠ j The one-way ANOVA table results are shown: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F P-value Factor (Between) 14.056 5 – 1 = 4 3.514 0.353095 0.8362 Error (Within) 99.52 15 – 5 = 10 9.952 Total 113.576 15 – 1 = 14 Distribution for the test: F 4 , 10 Probability Statement: p - v a l u e = P ( F > 0 . 3531 ) = 0 . 8362 Compare  and the p -value: α = 0 . 05 , p - v a l u e = 08362 , α > p - v a l u e Make a decision: Since  > p -value, we do not reject H 0 . Conclusion: At the 5% significance level, there is insufficient evidence from these data that different strains of COVID-19 virus will cause a significant difference in the length of a hospital stay. Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in . MEAN GRADES FOR FOUR SORORITIES Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 2.63 2.63 3.79 1.85 1.77 3.78 3.45 2.83 3.25 4.00 3.08 1.69 1.86 2.55 2.26 3.33 2.21 2.45 3.18 Using a significance level of 1%, is there a difference in mean grades among the sororities? Let μ 1 , μ 2 , μ 3 , μ 4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design , because each factor (i.e., sorority) has the same number of observations. H 0 : μ 1 = μ 2 = μ 3 = μ 4 H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 are equal. Distribution for the test: F 3,16 where k = 4 groups and n = 20 samples in total df ( num )= k – 1 = 4 – 1 = 3 df ( denom ) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph: Probability statement: p -value = P ( F > 2.23) = 0.1241 Compare α and the p -value: α = 0.01 p -value = 0.1241 α < p -value Make a decision: Since α < p -value, you cannot reject H 0 . Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Put the data into lists L 1 , L 2 , L 3 , and L 4 . Press STAT and arrow over to TESTS . Arrow down to F:ANOVA . Press ENTER and Enter ( L1,L2,L3,L4 ). The calculator displays the F statistic, the p -value and the values for the one-way ANOVA table: F = 2.2303 p = 0.1241 ( p -value) Factor df = 3 SS = 2.88732 MS = 0.96244 Error df = 16 SS = 6.9044 MS = 0.431525 Try It Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in . GPAs FOR FOUR SPORTS TEAMS Basketball Baseball Hockey Lacrosse 3.6 2.1 4.0 2.0 2.9 2.6 2.0 3.6 2.5 3.9 2.6 3.9 3.3 3.1 3.2 2.7 3.8 3.4 3.2 2.5 Use a significance level of 5%, and determine if there is a difference in GPA among the teams. A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in . Tommy's Plants Tara's Plants Nick's Plants 24 25 23 21 31 27 23 23 22 30 20 30 23 28 20 Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance. This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = n ⋅ s x ¯ 2 s 2 pooled . First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants Sample Mean 24.2 25.4 24.4 Sample Variance 11.7 18.3 16.3 Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = s x ¯ 2 Then MS between = n s x ¯ 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s 2 pooled Then MS within = s 2 pooled = 15.433. The F statistic (or F ratio) is F = M S between M S within = n s x ¯ 2 s 2 p o o l e d = ( 5 ) ( 0.413 ) 15.433 = 0.134 The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12 The distribution for the test is F 2,12 and the F statistic is F = 0.134 The p -value is P ( F > 0.134) = 0.8759. Decision: Since α = 0.03 and the p -value = 0.8759, do not reject H 0 . (Why?) Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. To calculate the p -value: *Press 2nd DISTR *Arrow down to Fcdf (and press ENTER . *Enter 0.134, E99 , 2, 12) *Press ENTER The p -value is 0.8759. Try It Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in . From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States they have visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets in . References Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall, 1994. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 50. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118. “MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012. Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), \"A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich,\" Journal of the American Medical Association , 268, 1578-1580. Chapter Review The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p -value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p -value) to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from need to be used for hand calculations. In the case of balanced data (the groups are the same size) however, simplified calculations based on group means and variances may be used. In practice, of course, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look of your data! An F statistic can have what values? What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? The curves approximate the normal distribution. Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in . Team 1 Team 2 Team 3 Team 4 Team 5 36 32 48 38 41 42 35 50 44 39 51 38 39 46 40 What is the df(num) ? What is the df(denom) ? ten What are the Sum of Squares and Mean Squares Factors? What are the Sum of Squares and Mean Squares Errors? SS = 237.33; MS = 23.73 What is the F statistic? What is the p -value? 0.1614 At the 5% significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Group A Group B Group C 101 151 101 108 149 109 98 160 198 107 112 186 111 126 160 What is the df(num) ? two What is the df(denom) ? What are the SS between and MS between ? SS = 5,700.4; MS = 2,850.2 What are the SS within and MS within ? What is the F Statistic? 3.6101 What is the p -value? At the 10% significance level, are the scores among the different groups different? Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level. Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = Enter the data into your calculator or computer. p -value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α . α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ Homework DIRECTIONS Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in . Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. Weights of Student Lab Rats Linda's rats Tuan's rats Javier's rats 43.5 47.0 51.2 39.4 40.5 40.9 41.3 38.9 37.9 46.0 46.3 45.0 38.2 44.2 48.6 H 0 : µ L = µ T = µ J H a : at least any two of the means are different df ( num ) = 2; df ( denom ) = 12 F distribution 0.67 0.5305 Answers may vary. Decision: Do not reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in . Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. working-class professional (middle incomes) professional (wealthy) 17.8 16.5 8.5 26.7 17.4 6.3 49.4 22.0 4.6 9.4 7.4 12.6 65.4 9.4 11.0 47.1 2.1 28.6 19.5 6.4 15.4 51.2 13.9 9.3 Use the following information to answer the next two exercises. lists the number of pages in four different types of magazines. home decorating news health computer 172 87 82 104 286 94 153 136 163 123 87 98 205 106 103 207 197 101 96 146 Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? H a : µ c = µ n = µ h At least any two of the magazines have different mean lengths. df ( num ) = 2, df ( denom ) = 12 F distribtuion F = 15.28 p -value = 0.0005 Answers may vary. Alpha: 0.05 Decision: Reject the Null Hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that shows the results of a study. CNN FOX Local 45 15 72 12 43 37 18 68 56 38 50 60 23 31 51 35 22 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the means for the final exams the same for all statistics class delivery types? shows the scores on final exams from several randomly selected classes that used the different delivery types. Online Hybrid Face-to-Face 72 83 80 84 73 78 77 84 84 80 81 81 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ o = μ h = μ f At least two of the means are different. df ( n ) = 2, df ( d ) = 13 F 2,13 0.64 0.5437 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: The mean scores of different class delivery are not different. Are the mean number of times a month a person eats out the same for White, Black, Hispanic/Latino and Asian people? Suppose that shows the results of a study. White Black Hispanic/Latino Asian 6 4 7 8 8 1 3 3 2 5 5 5 4 2 4 1 6 6 7 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that shows the results of a study. Powder Machine Made Hard Packed 1,210 2,107 2,846 1,080 1,149 1,638 1,537 862 2,019 941 1,870 1,178 1,528 2,233 1,382 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ p = μ m = μ h At least any two of the means are different. df ( n ) = 2, df ( d ) = 12 F 2,12 3.13 0.0807 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew. Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy 5.1 meters 3.1 meters 4.7 meters 5.3 meters Medium 4 meters 3.5 meters 4.5 meters 6.1 meters Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? Yes or No. Why is this a balanced design? Calculate the sample mean and sample standard deviation for each group. Does the weight of the paper have an effect on how far the plane will travel? Use a 1% level of significance. Complete the test using the method shown in the bean plant example in . variance of the group means __________ MS between = ___________ mean of the three sample variances ___________ MS within = _____________ F statistic = ____________ df(num) = __________, df(denom) = ___________ number of groups _______ number of observations _______ p -value = __________ ( P ( F > _______) = __________) Graph the p -value. decision: _______________________ conclusion: _______________________________________________________________ DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 16.4 26.7 28.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 14.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three groups: The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H 0 : μ 1 = μ 2 = μ 3 ; H a : μ i ≠ μ j some i ≠ j . Define μ 1 , μ 2 , μ 3 , as the population mean number of eggs laid by the three groups of fruit flies. F statistic = 8.6657; p -value = 0.0004 Decision: Since the p -value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t -test to compare the RS and NS groups they are significantly different ( p = 0.0013). Similarly, SS and NS are significantly different ( p = 0.0006). However, the two selected groups, RS and SS are not significantly different ( p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 98 97.3 97.4 98.7 98.7 98.2 98.8 97.6 98 97.4 97.5 98.8 98.8 98.2 98.8 97.7 98 97.4 97.6 98.8 98.8 98.3 98.9 97.8 98 97.4 97.7 98.8 98.8 98.4 99 97.8 98.1 97.5 97.8 98.8 98.9 98.4 99 97.9 98.3 97.6 97.9 99.2 99 98.5 99 97.9 98.3 97.6 98 99.3 99 98.5 99.2 98 98.3 97.8 98 99.1 98.6 99.5 98.2 98.4 97.8 98 99.1 98.6 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98 98.4 99.9 99.3 98.2 98.5 98 98.6 100 99.4 98.2 98.6 98 98.6 100.8", "section": "Facts About the F Distribution", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Test of Two Variances Another of the uses of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers. In order to perform a F test of two variances, it is important that the following are true: The populations from which the two samples are drawn are normally distributed. The two populations are independent of each other. Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p -values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here. Suppose we sample randomly from two independent normal populations. Let σ 1 2 and σ 2 2 be the population variances and s 1 2 and s 2 2 be the sample variances. Let the sample sizes be n 1 and n 2 . Since we are interested in comparing the two sample variances, we use the F ratio: F = [ ( s 1 ) 2 ( σ 1 ) 2 ] [ ( s 2 ) 2 ( σ 2 ) 2 ] F has the distribution F ~ F ( n 1 – 1, n 2 – 1) where n 1 – 1 are the degrees of freedom for the numerator and n 2 – 1 are the degrees of freedom for the denominator. If the null hypothesis is σ 1 2 = σ 2 2 , then the F Ratio becomes F = [ ( s 1 ) 2 ( σ 1 ) 2 ] [ ( s 2 ) 2 ( σ 2 ) 2 ] = ( s 1 ) 2 ( s 2 ) 2 . NOTE The F ratio could also be ( s 2 ) 2 ( s 1 ) 2 . It depends on H a and on which sample variance is larger. If the two populations have equal variances, then s 1 2 and s 2 2 are close in value and F = ( s 1 ) 2 ( s 2 ) 2 is close to one. But if the two population variances are very different, s 1 2 and s 2 2 tend to be very different, too. Choosing s 1 2 as the larger sample variance causes the ratio ( s 1 ) 2 ( s 2 ) 2 to be greater than one. If s 1 2 and s 2 2 are far apart, then F = ( s 1 ) 2 ( s 2 ) 2 is a large number. Therefore, if F is close to one, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than one, then the evidence is against the null hypothesis. A test of two variances may be left, right, or two-tailed. Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor's grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%. Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively. n 1 = n 2 = 30. H 0 : σ 1 2 = σ 2 2 and H a : σ 1 2 < σ 2 2 Calculate the test statistic: By the null hypothesis ( σ 1 2 = σ 2 2 ) , the F statistic is: F = [ ( s 1 ) 2 ( σ 1 ) 2 ] [ ( s 2 ) 2 ( σ 2 ) 2 ] = ( s 1 ) 2 ( s 2 ) 2 = 52.3 89.9 = 0.5818 Distribution for the test: F 29,29 where n 1 – 1 = 29 and n 2 – 1 = 29. Graph: This test is left tailed. Draw the graph labeling and shading appropriately. Probability statement: p -value = P ( F < 0.5818) = 0.0753 Compare α and the p -value: α = 0.10 α > p -value. Make a decision: Since α > p -value, reject H 0 . Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller. Press STAT and arrow over to TESTS . Arrow down to D:2-SampFTest . Press ENTER . Arrow to Stats and press ENTER . For Sx1 , n1 , Sx2 , and n2 , enter ( 52.3 ) , 30 , ( 89.9 ) , and 30 . Press ENTER after each. Arrow to σ1: and σ2 . Press ENTER . Arrow down to Calculate and press ENTER . F = 0.5818 and p -value = 0.0753. Do the procedure again and try Draw instead of Calculate . Try It The New York Choral Society divides its singers who are men up into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different? Tenor1 Bass2 Tenor 1 Bass 2 Tenor 1 Bass 2 69 72 67 72 68 67 72 75 70 74 67 70 71 67 65 70 64 70 66 75 72 66 69 76 74 70 68 72 74 72 68 75 71 71 72 64 68 74 66 74 73 70 75 68 72 66 72 References “MLB Vs. Division Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/order/true. Chapter Review The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom. Assumptions: The populations from which the two samples are drawn are normally distributed. The two populations are independent of each other. Formula Review F has the distribution F ~ F ( n 1 – 1, n 2 – 1) F = s 1 2 σ 1 2 s 2 2 σ 2 2 If σ 1 = σ 2 , then F = s 1 2 s 2 2 Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an F test of two variances. Name one assumption that must be true. The populations from which the two samples are drawn are normally distributed. What is the other assumption that must be true? Use the following information to answer the next five exercises. Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times. Test the claim at the 10% level. Assume that commute times are normally distributed. State the null and alternative hypotheses. H 0 : σ 1 = σ 2 H a : σ 1 < σ 2 or H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 < σ 2 2 What is s 1 in this problem? What is s 2 in this problem? 4.11 What is n ? What is the F statistic? 0.7159 What is the p -value? Is the claim accurate? No, at the 10% level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker. Use the following information to answer the next four exercises. Two students are interested in whether or not there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are more consistent. State the null and alternative hypotheses. What is the F Statistic? 2.8674 What is the p -value? At the 5% significance level, do we reject the null hypothesis? Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student. Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records their speeds going up 35 hills. The first cyclist has a variance of 23.8 and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. Assume that commute times are normally distributed. State the null and alternative hypotheses. What is the F Statistic? 0.7414 At the 5% significance level, what can we say about the cyclists’ variances? Homework Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Linda's rats Tuan's rats Javier's rats 43.5 47.0 51.2 39.4 40.5 40.9 41.3 38.9 37.9 46.0 46.3 45.0 38.2 44.2 48.6 Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%. H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 ≠ σ 1 2 df ( num ) = 4; df ( denom ) = 4 F 4, 4 3.00 2(0.1563) = 0.3126. Using the TI-83+/84+ function 2-SampFtest, you get the test statistic as 2.9986 and p -value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335 but the p -value is the same because this is a two-tailed test. Answers may vary. Decision: Do not reject the null hypothesis; Conclusion: There is insufficient evidence to conclude that the variances are different. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows. working-class professional (middle incomes) professional (wealthy) 17.8 16.5 8.5 26.7 17.4 6.3 49.4 22.0 4.6 9.4 7.4 12.6 65.4 9.4 11.0 47.1 2.1 28.6 19.5 6.4 15.4 51.2 13.9 9.3 Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups. Use a 5% significance level. Use the following information to answer the next two exercises. lists the number of pages in four different types of magazines. home decorating news health computer 172 87 82 104 286 94 153 136 163 123 87 98 205 106 103 207 197 101 96 146 Which two magazine types do you think have the same variance in length? Which two magazine types do you think have different variances in length? The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances. Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that the shows the results of a study. Saturday Sunday Saturday Sunday 75 44 62 137 18 58 0 82 150 61 124 39 94 19 50 127 62 99 31 141 73 60 118 73 89 Are the variances for incomes on the East Coast and the West Coast the same? Suppose that shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05. East West 38 71 47 126 30 42 82 51 75 44 52 90 115 88 67 H 0 : = σ 1 2 = σ 2 2 H a : σ 1 2 ≠ σ 1 2 df ( n ) = 7, df ( d ) = 6 F 7,6 0.8117 0.7825 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: There is not sufficient evidence to conclude that the variances are different. Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100 mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how? Here are the data: 0 mg 100 mg 200 mg 0 mg 100 mg 200 mg 242 248 246 245 246 248 244 245 250 248 247 252 247 248 248 248 250 250 242 247 246 244 246 248 246 243 245 242 244 250 King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul) during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but declined significantly afterwards. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from a fourth. These spanned most of his reign. We have data on the silver content of the coins: First Coinage Second Coinage Third Coinage Fourth Coinage 5.9 6.9 4.9 5.3 6.8 9.0 5.5 5.6 6.4 6.6 4.6 5.5 7.0 8.1 4.5 5.1 6.6 9.3 6.2 7.7 9.2 5.8 7.2 8.6 5.8 6.9 6.2 Did the silver content of the coins change over the course of Manuel’s reign? Here are the means and variances of each coinage. The data are unbalanced. First Second Third Fourth Mean 6.7444 8.2429 4.875 5.6143 Variance 0.2953 1.2095 0.2025 0.1314 Here is a strip chart of the silver content of the coins: While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table: Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) 37.748 4 – 1 = 3 12.5825 26.272 Error (Within) 11.015 27 – 4 = 23 0.4789 Total 48.763 27 – 1 = 26 P ( F > 26.272) = 0; Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth. The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in a certain year Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in this year, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same or not. Note that the data are not balanced, as two divisions had five teams, while one had only four. Division Team Wins East NY Yankees 95 East Baltimore 93 East Tampa Bay 90 East Toronto 73 East Boston 69 Division Team Wins Central Detroit 88 Central Chicago Sox 85 Central Kansas City 72 Central Cleveland 68 Central Minnesota 66 Division Team Wins West Oakland 94 West Texas 93 West LA Angels 89 West Seattle 75 Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season. While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust. Here is the ANOVA table for the data: Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) 344.16 3 – 1 = 2 172.08 Error (Within) 1,219.55 14 – 3 = 11 110.87 1.5521 Total 1,563.71 14 – 1 = 13 P ( F > 1.5521) = 0.2548 Since the p -value is so large, there is not good evidence against the null hypothesis of equal means. We decline to reject the null hypothesis. Thus, for 2012, there is not any have any good evidence of a significant difference in mean number of wins between the divisions of the American League.", "section": "Test of Two Variances", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Lab: One-Way ANOVA One-Way ANOVA Class Time: Names: Student Learning Outcome The student will conduct a simple one-way ANOVA test involving three variables. Collect the Data Record the price per pound of eight fruits, eight vegetables, and eight breads in your local supermarket. Fruits Vegetables Breads Explain how you could try to collect the data randomly. Analyze the Data and Conduct a Hypothesis Test State the null hypothesis and the alternative hypothesis. Compute the following: Fruit: x ¯ = ______ s x = ______ n = ______ Vegetables: x ¯ = ______ s x = ______ n = ______ Bread: x ¯ = ______ s x = ______ n = ______ Find the following: df ( num ) = ______ df ( denom ) = ______ State the approximate distribution for the test. Test statistic: F = ______ Sketch a graph of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p -value. p -value = ______ Test at α = 0.05. State your decision and conclusion. Decision: Why did you make this decision? Conclusion (write a complete sentence). Based on the results of your study, is there a need to investigate any of the food groups’ prices? Why or why not?", "section": "Lab: One-Way ANOVA", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Review Exercises (Ch 3-13) These review exercises are designed to provide extra practice on concepts learned before a particular chapter. For example, the review exercises for Chapter 3, cover material learned in chapters 1 and 2. Chapter 3 Use the following information to answer the next six exercises: In a survey of 100 stocks on NASDAQ, the average percent increase for the past year was 9% for NASDAQ stocks. 1 . The “average increase” for all NASDAQ stocks is the: population statistic parameter sample variable 2. All of the NASDAQ stocks are the: population statistics parameter sample variable 3. Nine percent is the: population statistics parameter sample variable 4. The 100 NASDAQ stocks in the survey are the: population statistic parameter sample variable 5. The percent increase for one stock in the survey is the: population statistic parameter sample variable 6. Would the data collected by qualitative, quantitative discrete, or quantitative continuous? Use the following information to answer the next two exercises: Thirty people spent two weeks around Mardi Gras in New Orleans. Their two-week weight gain is below. (Note: a loss is shown by a negative weight gain.) Weight Gain Frequency –2 3 –1 5 0 2 1 4 4 13 6 2 11 1 7. Calculate the following values: the average weight gain for the two weeks the standard deviation the first, second, and third quartiles 8. Construct a histogram and box plot of the data. Chapter 4 Use the following information to answer the next two exercises: A recent poll concerning credit cards found that 35 percent of respondents use a credit card that gives them a mile of air travel for every dollar they charge. Thirty percent of the respondents charge more than $2,000 per month. Of those respondents who charge more than $2,000, 80 percent use a credit card that gives them a mile of air travel for every dollar they charge. 9. What is the probability that a randomly selected respondent will spend more than $2,000 AND use a credit card that gives them a mile of air travel for every dollar they charge? (0.30)(0.35) (0.80)(0.35) (0.80)(0.30) (0.80) 10. Are using a credit card that gives a mile of air travel for each dollar spent AND charging more than $2,000 per month independent events? Yes No, and they are not mutually exclusive either. No, but they are mutually exclusive. Not enough information given to determine the answer 11. A sociologist wants to know the opinions of employed adult women about government funding for day care. She obtains a list of 520 members of a local business and professional women’s club and mails a questionnaire to 100 of these women selected at random. Sixty-eight questionnaires are returned. What is the population in this study? all employed adult women all the members of a local business and professional women’s club the 100 women who received the questionnaire all employed women with children Use the following information to answer the next two exercises: An article from The San Jose Mercury News was concerned with the racial mix of the 1500 students at Prospect High School in Saratoga, CA. The table summarizes the results. (Male and female values are approximate.) Suppose one Prospect High School student is randomly selected. Gender/Ethnic group White Asian Hispanic Black American Indian Male 400 168 115 35 16 Female 440 132 140 40 14 12. Find the probability that a student is Asian or Male. 13. Find the probability that a student is Black given that the student is female. 14. A sample of pounds lost, in a certain month, by individual members of a weight reducing clinic produced the following statistics: Mean = 5 lbs. Median = 4.5 lbs. Mode = 4 lbs. Standard deviation = 3.8 lbs. First quartile = 2 lbs. Third quartile = 8.5 lbs. The correct statement is: One fourth of the members lost exactly two pounds. The middle fifty percent of the members lost from two to 8.5 lbs. Most people lost 3.5 to 4.5 lbs. All of the choices above are correct. 15. What does it mean when a data set has a standard deviation equal to zero? All values of the data appear with the same frequency. The mean of the data is also zero. All of the data have the same value. There are no data to begin with. 16. The statement that describe the illustration is: the mean is equal to the median. There is no first quartile. The lowest data value is the median. The median equals Q 1 + Q 3 2 . 17. According to a recent article in the San Jose Mercury News the average number of babies born with significant hearing loss (deafness) is approximately 2 per 1000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. 18. A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. Ten of the coupons are for a free gift worth $6. Eighty of the coupons are for a free gift worth $8. Six of the coupons are for a free gift worth $12. Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? Yes, I expect to come out ahead in money. No, I expect to come out behind in money. It doesn’t matter. I expect to break even. Use the following information to answer the next four exercises: Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that they truly have the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. 19. Define the random variable and list its possible values. 20. State the distribution of X . 21. Find the probability that at least four of the 25 patients actually have the flu. 22. On average, for every 25 patients calling in, how many do you expect to have the flu? Use the following information to answer the next two exercises: Different types of writing can sometimes be distinguished by the number of letters in the words used. A student interested in this fact wants to study the number of letters of words used by Jhumpa Lahiri in her novels. She opens a novel at random and records the number of letters of the first 250 words on the page. 23. What kind of data was collected? qualitative quantitative continuous quantitative discrete 24. What is the population under study? Chapter 5 Use the following information to answer the next seven exercises: A recent study of mothers of junior high school children in Santa Clara County reported that 76% of the mothers are employed in paid positions. Of those mothers who are employed, 64% work full-time (over 35 hours per week), and 36% work part-time. However, out of all of the mothers in the population, 49% work full-time. The population under study is made up of mothers of junior high school children in Santa Clara County. Let E = employed and F = full-time employment. 25. Find the percent of all mothers in the population that are NOT employed. Find the percent of mothers in the population that are employed part-time. 26. The “type of employment” is considered to be what type of data? 27. Find the probability that a randomly selected mother works part-time given that she is employed. 28. Find the probability that a randomly selected person from the population will be employed or work full-time. 29. Being employed and working part-time: mutually exclusive events? Why or why not? independent events? Why or why not? Use the following additional information to answer the next two exercises: We randomly pick ten mothers from the above population. We are interested in the number of the mothers that are employed. Let X = number of mothers that are employed. 30. State the distribution for X . 31. Find the probability that at least six are employed. 32. We expect the statistics discussion board to have, on average, 14 questions posted to it per week. We are interested in the number of questions posted to it per day. Define X . What are the values that the random variable may take on? State the distribution for X . Find the probability that from ten to 14 (inclusive) questions are posted to the listserv on a randomly picked day. 33. A person invests $1,000 into stock of a company that hopes to go public in one year. The probability that the person will lose all his money after one year (i.e. his stock will be worthless) is 35%. The probability that the person’s stock will still have a value of $1,000 after one year (i.e. no profit and no loss) is 60%. The probability that the person’s stock will increase in value by $10,000 after one year (i.e. will be worth $11,000) is 5%. Find the expected profit after one year. 34. Rachel’s piano cost $3,000. The average cost for a piano is $4,000 with a standard deviation of $2,500. Becca’s guitar cost $550. The average cost for a guitar is $500 with a standard deviation of $200. Matt’s drums cost $600. The average cost for drums is $700 with a standard deviation of $100. Whose cost was lowest when compared to their own instrument? 35. Explain why each statement is either true or false given the box plot in . Twenty-five percent of the data re at most five. There is the same amount of data from 4–5 as there is from 5–7. There are no data values of three. Fifty percent of the data are four. Using the following information to answer the next two exercises: 64 faculty members were asked the number of cars they owned (including spouse and children’s cars). The results are given in the following graph: 36. Find the approximate number of responses that were three. 37. Find the first, second and third quartiles. Use them to construct a box plot of the data. Use the following information to answer the next three exercises: shows data gathered from 15 girls on the Snow Leopard soccer team when they were asked how they liked to wear their hair. Supposed one girl from the team is randomly selected. Hair Style/Hair Color Blond Brown Black Ponytail 3 2 5 Plain 2 2 1 38. Find the probability that the girl has black hair GIVEN that she wears a ponytail. 39. Find the probability that the girl wears her hair plain OR has brown hair. 40. Find the probability that the girl has blond hair AND that she wears her hair plain. Chapter 6 Use the following information to answer the next two exercises: X ~ U (3, 13) 41. Explain which of the following are false and which are true. f ( x ) = 1 10 , 3 ≤ x ≤ 13 There is no mode The median is less than the mean. P ( x > 10) = P ( x ≤ 6) 42. Calculate: the mean. the median. the 65 th percentile. 43. Which of the following is true for the box plot in ? Twenty-five percent of the data are at most five. There is about the same amount of data from 4–5 as there is from 5–7. There are no data values of three. Fifty percent of the data are four. 44. If P ( G | H ) = P ( G ), then which of the following is correct? G and H are mutually exclusive events. P ( G ) = P ( H ) Knowing that H has occurred will affect the chance that G will happen. G and H are independent events. 45. If P ( J ) = 0.3, P ( K ) = 0.63, and J and K are independent events, then explain which are correct and which are incorrect. P ( J AND K ) = 0 P ( J OR K ) = 0.9 P ( J OR K ) = 0.72 P ( J ) ≠ P ( J | K ) 46. On average, five students from each high school class get full scholarships to four-year colleges. Assume that most high school classes have about 500 students. X = the number of students from a high school class that get full scholarships to four-year schools. Which of the following is the distribution of X ? P (5) B (500, 5) Exp ( 1 5 ) N ( 5 , ( 0.01 ) ( 0.99 ) 500 ) Chapter 7 Use the following information to answer the next three exercises: Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. 47. X ~ _________ U (0, 4) U (10, 20) Exp (2) N (2, 1) 48. The average wait time is: 1 hour. 2 hours. 2.5 hours. 4 hours. 49. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least 1.5 more hours is: 1 4 1 2 3 4 3 8 50. Given: X ~ Exp ( 1 3 ) Find P ( x > 1). Calculate the minimum value for the upper quartile. Find P ( x = 1 3 ) 51. 40% of full-time students took 4 years to graduate 30% of full-time students took 5 years to graduate 20% of full-time students took 6 years to graduate 10% of full-time students took 7 years to graduate The expected time for full-time students to graduate is: 4 years 4.5 years 5 years 5.5 years 52. Which of the following distributions is described by the following example? Many people can run a short distance of under two miles, but as the distance increases, fewer people can run that far. binomial uniform exponential normal 53. The length of time to brush one’s teeth is generally thought to be exponentially distributed with a mean of 3 4 minutes. Find the probability that a randomly selected person brushes their teeth less than 3 4 minutes. 0.5 3 4 0.43 0.63 54. Which distribution accurately describes the following situation? The chance that a teenage boy regularly gives his mother a kiss goodnight is about 20%. Fourteen teenage boys are randomly surveyed. Let X = the number of teenage boys that regularly give their mother a kiss goodnight. B (14,0.20) P (2.8) N (2.8,2.24) Exp ( 1 0.20 ) 55. A 2008 report on technology use states that approximately 20% of U.S. households have never sent an e-mail. Suppose that we select a random sample of fourteen U.S. households. Let X = the number of households in a 2008 sample of 14 households that have never sent an email B (14,0.20) P (2.8) N (2.8,2.24) Exp ( 1 0.20 ) Chapter 8 Use the following information to answer the next three exercises: Suppose that a sample of 15 randomly chosen people were put on a special weight loss diet. The amount of weight lost, in pounds, follows an unknown distribution with mean equal to 12 pounds and standard deviation equal to three pounds. Assume that the distribution for the weight loss is normal. 56. To find the probability that the mean amount of weight lost by 15 people is no more than 14 pounds, the random variable should be: number of people who lost weight on the special weight loss diet. the number of people who were on the diet. the mean amount of weight lost by 15 people on the special weight loss diet. the total amount of weight lost by 15 people on the special weight loss diet. 57. Find the probability asked for in Question 56 . 58. Find the 90 th percentile for the mean amount of weight lost by 15 people. Using the following information to answer the next three exercises: The time of occurrence of the first accident during rush-hour traffic at a major intersection is uniformly distributed between the three hour interval 4 p.m. to 7 p.m. Let X = the amount of time (hours) it takes for the first accident to occur. 59. What is the probability that the time of occurrence is within the first half-hour or the last hour of the period from 4 to 7 p.m.? cannot be determined from the information given 1 6 1 2 1 3 60. The 20 th percentile occurs after how many hours? 0.20 0.60 0.50 1 61. Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let C = the total cumulative time. Then C follows which distribution? U (0,3) Exp (13) N (60, 5.477) N (1.5, 0.01875) 62. Using the information in Question 61 , find the probability that the total time for all first accidents to occur is more than 43 hours. Use the following information to answer the next two exercises: The length of time a parent must wait for his children to clean their rooms is uniformly distributed in the time interval from one to 15 days. 63. How long must a parent expect to wait for his children to clean their rooms? eight days three days 14 days six days 64. What is the probability that a parent will wait more than six days given that the parent has already waited more than three days? 0.5174 0.0174 0.7500 0.2143 Use the following information to answer the next five exercises: Twenty percent of the students at a local community college live in within five miles of the campus. Thirty percent of the students at the same community college receive some kind of financial aid. Of those who live within five miles of the campus, 75% receive some kind of financial aid. 65. Find the probability that a randomly chosen student at the local community college does not live within five miles of the campus. 80% 20% 30% cannot be determined 66. Find the probability that a randomly chosen student at the local community college lives within five miles of the campus or receives some kind of financial aid. 50% 35% 27.5% 75% 67. Are living in student housing within five miles of the campus and receiving some kind of financial aid mutually exclusive? yes no cannot be determined 68. The interest rate charged on the financial aid is _______ data. quantitative discrete quantitative continuous qualitative discrete qualitative 69. The following information is about the students who receive financial aid at the local community college. 1st quartile = $250 2nd quartile = $700 3rd quartile = $1200 These amounts are for the school year. If a sample of 200 students is taken, how many are expected to receive $250 or more? 50 250 150 cannot be determined Use the following information to answer the next two exercises: P ( A ) = 0.2, P ( B ) = 0.3; A and B are independent events. 70. P ( A AND B ) = ______ 0.5 0.6 0 0.06 71. P ( A OR B ) = _______ 0.56 0.5 0.44 1 72. If H and D are mutually exclusive events, P ( H ) = 0.25, P ( D ) = 0.15, then P ( H | D ). 1 0 0.40 0.0375 Chapter 9 73. Rebecca and Matt are 14 year old twins. Matt’s height is two standard deviations below the mean for 14 year old boys’ height. Rebecca’s height is 0.10 standard deviations above the mean for 14 year old girls’ height. Interpret this. Matt is 2.1 inches shorter than Rebecca. Rebecca is very tall compared to other 14 year old girls. Rebecca is taller than Matt. Matt is shorter than the average 14 year old boy. 74. Construct a histogram of the IPO data (see ). Use the following information to answer the next three exercises: Ninety homeowners were asked the number of estimates they obtained before having their homes fumigated. Let X = the number of estimates. x Relative Frequency Cumulative Relative Frequency 1 0.3 2 0.2 4 0.4 5 0.1 75. Complete the cumulative frequency column. 76. Calculate the sample mean (a), the sample standard deviation (b) and the percent of the estimates that fall at or below four (c). 77. Calculate the median, M , the first quartile, Q 1 , the third quartile, Q 3 . Then construct a box plot of the data. 78. The middle 50% of the data are between _____ and _____. Use the following information to answer the next three exercises: Seventy 5 th and 6 th graders were asked their favorite dinner. Pizza Hamburgers Spaghetti Fried shrimp 5th grader 15 6 9 0 6th grader 15 7 10 8 79. Find the probability that one randomly chosen child is in the 6th grade and prefers fried shrimp. 32 70 8 32 8 8 8 70 80. Find the probability that a child does not prefer pizza. 30 70 30 40 40 70 1 81. Find the probability a child is in the 5 th grade given that the child prefers spaghetti. 9 19 9 70 9 30 19 70 82. A sample of convenience is a random sample. true false 83. A statistic is a number that is a property of the population. true false 84. You should always throw out any data that are outliers. true false 85. Lee bakes pies for a small restaurant in Felton, CA. She generally bakes 20 pies in a day, on average. Of interest is the number of pies she bakes each day. Define the random variable X . State the distribution for X . Find the probability that Lee bakes more than 25 pies in any given day. 86. Six different brands of Italian salad dressing were randomly selected at a supermarket. The grams of fat per serving are 7, 7, 9, 6, 8, 5. Assume that the underlying distribution is normal. Calculate a 95% confidence interval for the population mean grams of fat per serving of Italian salad dressing sold in supermarkets. 87. Given: uniform, exponential, normal distributions. Match each to a statement below. mean = median ≠ mode mean > median > mode mean = median = mode Chapter 10 Use the following information to answer the next three exercises: In a survey at Kirkwood Ski Resort the following information was recorded: 0–10 11–20 21–40 40+ Ski 10 12 30 8 Snowboard 6 17 12 5 Suppose that one person from was randomly selected. 88. Find the probability that the person was a skier or was age 11–20. 89. Find the probability that the person was a snowboarder given they were age 21–40. 90. Explain which of the following are true and which are false. Sport and age are independent events. Ski and age 11–20 are mutually exclusive events. P (Ski AND age 21–40) < P (Ski|age 21–40) P (Snowboard OR age 0–10) < P (Snowboard|age 0–10) 91. The average length of time a person with a broken leg wears a cast is approximately six weeks. The standard deviation is about three weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the thirty people. 92. The distribution for X is uniform. What can we say for certain about the distribution for X ¯ when n = 1? The distribution for X ¯ is still uniform with the same mean and standard deviation as the distribution for X . The distribution for X ¯ is normal with the different mean and a different standard deviation as the distribution for X . The distribution for X ¯ is normal with the same mean but a larger standard deviation than the distribution for X . The distribution for X ¯ is normal with the same mean but a smaller standard deviation than the distribution for X . 93. The distribution for X is uniform. What can we say for certain about the distribution for ∑ X when n = 50? distribution for ∑ X is still uniform with the same mean and standard deviation as the distribution for X . The distribution for ∑ X is normal with the same mean but a larger standard deviation as the distribution for X . The distribution for ∑ X is normal with a larger mean and a larger standard deviation than the distribution for X . The distribution for ∑ X is normal with the same mean but a smaller standard deviation than the distribution for X . Use the following information to answer the next three exercises: A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots. 94. State the approximate distribution for X ¯ , the distribution for the average lengths of baby carrots in 16 five-pound bags. X ¯ ~ ______ 95. Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches. 96. Find the probability that x ¯ is between two and 2.25 inches. Use the following information to answer the next three exercises: At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of five minutes and a standard deviation of two minutes. 97. Find the 90 th percentile of waiting time in minutes. 98. Find the median waiting time for one student. 99. Find the probability that the average waiting time for 40 students is at least 4.5 minutes. Chapter 11 Use the following information to answer the next four exercises: Suppose that the time that owners keep their cars (purchased new) is normally distributed with a mean of seven years and a standard deviation of two years. We are interested in how long an individual keeps his car (purchased new). Our population is people who buy their cars new. 100. Sixty percent of individuals keep their cars at most how many years? 101. Suppose that we randomly survey one person. Find the probability that person keeps their car less than 2.5 years. 102. If we are to pick individuals ten at a time, find the distribution for the mean car length ownership. 103. If we are to pick ten individuals, find the probability that the sum of their ownership time is more than 55 years. 104. For which distribution is the median not equal to the mean? Uniform Exponential Normal Student t 105. Compare the standard normal distribution to the Student’s t -distribution, centered at zero. Explain which of the following are true and which are false. As the number surveyed increases, the area to the left of –1 for the Student’s t -distribution approaches the area for the standard normal distribution. As the degrees of freedom decrease, the graph of the Student’s t -distribution looks more like the graph of the standard normal distribution. If the number surveyed is 15, the normal distribution should never be used. Use the following information to answer the next five exercises: We are interested in the checking account balance of twenty-year-old college students. We randomly survey 16 twenty-year-old college students. We obtain a sample mean of $640 and a sample standard deviation of $150. Let X = checking account balance of an individual twenty year old college student. 106. Explain why we cannot determine the distribution of X . 107. If you were to create a confidence interval or perform a hypothesis test for the population mean checking account balance of twenty-year-old college students, what distribution would you use? 108. Find the 95% confidence interval for the true mean checking account balance of a twenty-year-old college student. 109. What type of data is the balance of the checking account considered to be? 110. What type of data is the number of twenty-year-olds considered to be? 111. On average, a busy emergency room gets a patient with a shotgun wound about once per week. We are interested in the number of patients with a shotgun wound the emergency room gets per 28 days. Define the random variable X . State the distribution for X . Find the probability that the emergency room gets no patients with shotgun wounds in the next 28 days. Use the following information to answer the next two exercises: The probability that a certain slot machine will pay back money when a quarter is inserted is 0.30. Assume that each play of the slot machine is independent from each other. A person puts in 15 quarters for 15 plays. 112. Is the expected number of plays of the slot machine that will pay back money greater than, less than or the same as the median? Explain your answer. 113. Is it likely that exactly eight of the 15 plays would pay back money? Justify your answer numerically. 114. A game is played with the following rules: it costs $10 to enter. a fair coin is tossed four times. if you do not get four heads or four tails, you lose your $10. if you get four heads or four tails, you get back your $10, plus $30 more. Over the long run of playing this game, what are your expected earnings? 115. The mean grade on a math exam in Rachel’s class was 74, with a standard deviation of five. Rachel earned an 80. The mean grade on a math exam in Becca’s class was 47, with a standard deviation of two. Becca earned a 51. The mean grade on a math exam in Matt’s class was 70, with a standard deviation of eight. Matt earned an 83. Find whose score was the best, compared to their own class. Justify your answer numerically. Use the following information to answer the next two exercises: A random sample of 70 compulsive gamblers were asked the number of days they go to casinos per week. The results are given in the following graph: 116. Find the number of responses that were five. 117. Find the mean, standard deviation, the median, the first quartile, the third quartile and the IQR . 118. Based upon research at De Anza College, it is believed that about 19% of the student population speaks a language other than English at home. Suppose that a study was done this year to see if that percent has decreased. Ninety-eight students were randomly surveyed with the following results. Fourteen said that they speak a language other than English at home. State an appropriate null hypothesis. State an appropriate alternative hypothesis. Define the random variable, P ′. Calculate the test statistic. Calculate the p -value. At the 5% level of decision, what is your decision about the null hypothesis? What is the Type I error? What is the Type II error? 119. Assume that you are an emergency paramedic called in to rescue victims of an accident. You need to help a patient who is bleeding profusely. The patient is also considered to be a high risk for contracting hepatitis. Assume that the null hypothesis is that the patient does not have hepatitis. What is a Type I error? 120. It is often said that Californians are more casual than the rest of Americans. Suppose that a survey was done to see if the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals. Fifty of each was surveyed with the following results. Fifteen Californians wear jeans to work and six non-Californians wear jeans to work. Let C = Californian professional; NC = non-Californian professional State appropriate null and alternate hypotheses. Define the random variable. Calculate the test statistic and p -value. At the 5% significance level, what is your decision? What is the Type I error? What is the Type II error? Use the following information to answer the next two exercises: A group of Statistics students have developed a technique that they feel will lower their anxiety level on statistics exams. They measured their anxiety level at the start of the quarter and again at the end of the quarter. Recorded is the paired data in that order: (1000, 900); (1200, 1050); (600, 700); (1300, 1100); (1000, 900); (900, 900). 121. This is a test of (pick the best answer): large samples, independent means small samples, independent means dependent means 122. State the distribution to use for the test. Chapter 12 Use the following information to answer the next two exercises: A recent survey of U.S. teenage pregnancy was answered by 720 girls, age 12–19. Six percent of the girls surveyed said they have been pregnant. We are interested in the true proportion of U.S. girls, age 12–19, who have been pregnant. 123. Find the 95% confidence interval for the true proportion of U.S. girls, age 12–19, who have been pregnant. 124. The report also stated that the results of the survey are accurate to within ±3.7% at the 95% confidence level. Suppose that a new study is to be done. It is desired to be accurate to within 2% of the 95% confidence level. What is the minimum number that should be surveyed? 125. Given: X ~ Exp ( 1 3 ) . Sketch the graph that depicts: P ( x > 1). Use the following information to answer the next three exercises: The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the mean amount of money a customer spends in one trip to the supermarket is $72. 126. Find the probability that one customer spends less than $72 in one trip to the supermarket? 127. Suppose five customers pool their money. How much money altogether would you expect the five customers to spend in one trip to the supermarket (in dollars)? 128. State the distribution to use if you want to find the probability that the mean amount spent by five customers in one trip to the supermarket is less than $60. Chapter 13 Use the following information to answer the next two exercises: Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the probability of water rationing is 10%. However, in any year, the probability of water rationing is 5%. 129. What is the probability of both a drought and water rationing occurring? 130. Out of the years with water rationing, find the probability that there is a drought. Use the following information to answer the next three exercises: Apple Pumpkin Pecan Female 40 10 30 Male 20 30 10 131. Suppose that one individual is randomly chosen. Find the probability that the person’s favorite pie is apple or the person is male. 132. Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan. 133. Conduct a hypothesis test to determine if favorite pie type and gender are independent. Use the following information to answer the next two exercises: Let’s say that the probability that an adult watches the news at least once per week is 0.60. 134. We randomly survey 14 people. On average, how many people do we expect to watch the news at least once per week? 135. We randomly survey 14 people. Of interest is the number that watch the news at least once per week. State the distribution of X . X ~ _____ 136. The following histogram is most likely to be a result of sampling from which distribution? Chi-Square Geometric Uniform Binomial 137. The ages of De Anza evening students is known to be normally distributed with a population mean of 40 and a population standard deviation of six. A sample of six De Anza evening students reported their ages (in years) as: 28; 35; 47; 45; 30; 50. Find the probability that the mean of six ages of randomly chosen students is less than 35 years. Hint: Find the sample mean. 138. A math exam was given to all the fifth grade children attending Country School. Two random samples of scores were taken. The null hypothesis is that the mean math scores for boys and girls in fifth grade are the same. Conduct a hypothesis test. n x ¯ s 2 Boys 55 82 29 Girls 60 86 46 139. In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. Conduct a hypothesis test. 140. Which of the following is preferable when designing a hypothesis test? Maximize α and minimize β Minimize α and maximize β Maximize α and β Minimize α and β Use the following information to answer the next three exercises: 120 people were surveyed as to their favorite beverage (non-alcoholic). The results are below. Beverage/Age 0–9 10–19 20–29 30+ Totals Milk 14 10 6 0 30 Soda 3 8 26 15 52 Juice 7 12 12 7 38 Totals 24 330 44 22 120 141. Are the events of milk and 30+: independent events? Justify your answer. mutually exclusive events? Justify your answer. 142. Suppose that one person is randomly chosen. Find the probability that person is 10–19 given that they prefer juice. 143. Are “Preferred Beverage” and “Age” independent events? Conduct a hypothesis test. 144. Given the following histogram, which distribution is the data most likely to come from? uniform exponential normal chi-square Solutions Chapter 3 1. c. parameter 2. a. population 3. b. statistic 4. d. sample 5. e. variable 6. quantitative continuous 7. 2.27 3.04 –1, 4, 4 8. Answers will vary. Chapter 4 9. c. (0.80)(0.30) 10. b. No, and they are not mutually exclusive either. 11. a. all employed adult women 12. 0.5773 13. 0.0522 14. b. The middle fifty percent of the members lost from 2 to 8.5 lbs. 15. c. All of the data have the same value. 16. c. The lowest data value is the median. 17. 0.279 18. b. No, I expect to come out behind in money. 19. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, …25 20. B (25, 0.04) 21. 0.0165 22. 1 23. c. quantitative discrete 24. all words used by Tom Clancy in his novels Chapter 5 25. 24% 27% 26. qualitative 27. 0.36 28. 0.7636 29. No No 30. B (10, 0.76) 31. 0.9330 32. X = the number of questions posted to the statistics listserv per day. X = 0, 1, 2,… X ~ P (2) 0 33. $150 34. Matt 35. false true false false 36. 16 37. first quartile: 2 second quartile: 2 third quartile: 3 38. 0.5 39. 7 15 40. 2 15 Chapter 6 41. true true False – the median and the mean are the same for this symmetric distribution. true 42. 8 8 P ( x < k ) = 0.65 = ( k – 3) ( 1 10 ) . k = 9.5 43. False – 3 4 of the data are at most five. True – each quartile has 25% of the data. False – that is unknown. False – 50% of the data are four or less. 44. d. G and H are independent events. 45. False – J and K are independent so they are not mutually exclusive which would imply dependency (meaning P ( J AND K ) is not 0). False – see answer c. True – P ( J OR K ) = P ( J ) + P ( K ) – P ( J AND K ) = P ( J ) + P ( K ) – P ( J ) P ( K ) = 0.3 + 0.6 – (0.3)(0.6) = 0.72. Note the P ( J AND K ) = P ( J ) P ( K ) because J and K are independent. False – J and K are independent so P ( J ) = P ( J | K ) 46. a. P (5) Chapter 7 47. a. U (0, 4) 48. b. 2 hour 49. a. 1 4 50. 0.7165 4.16 0 51. c. 5 years 52. c. exponential 53. 0.63 54. B (14, 0.20) 55. B (14, 0.20) Chapter 8 56. c. the mean amount of weight lost by 15 people on the special weight loss diet. 57. 0.9951 58. 12.99 59. c. 1 2 60. b. 0.60 61. c. N (60, 5.477) 62. 0.9990 63. a. eight days 64. c. 0.7500 65. a. 80% 66. b. 35% 67. b. no 68. b. quantitative continuous 69. c. 150 70. d. 0.06 71. c. 0.44 72. b. 0 Chapter 9 73. d. Matt is shorter than the average 14 year old boy. 74. Answers will vary. 75. x Relative Frequency Cumulative Relative Frequency 1 0.3 0.3 2 0.2 0.2 4 0.4 0.4 5 0.1 0.1 76. 2.8 1.48 90% 77. M = 3; Q 1 = 1; Q 3 = 4 78. 1 and 4 79. d. 8 70 80. c. 40 70 81. a. 9 19 82. b. false 83. b. false 84. b. false 85. X = the number of pies Lee bakes every day. P (20) 0.1122 86. CI: (5.25, 8.48) 87. uniform exponential normal Chapter 10 88. 77 100 89. 12 42 90. false false true false 91. N (180, 16.43) 92. a. The distribution for X ¯ is still uniform with the same mean and standard deviation as the distribution for X . 93. c. The distribution for ∑ X is normal with a larger mean and a larger standard deviation than the distribution for X . 94. N ( 2 , 0.25 16 ) 95. Answers will vary. 96. 0.5000 97. 7.6 98. 5 99. 0.9431 Chapter 11 100. 7.5 101. 0.0122 102. N (7, 0.63) 103. 0.9911 104. b. Exponential 105. true false false 106. Answers will vary. 107. Student’s t with df = 15 108. (560.07, 719.93) 109. quantitative continuous data 110. quantitative discrete data 111. X = the number of patients with a shotgun wound the emergency room gets per 28 days P (4) 0.0183 112. greater than 113. No; P ( x = 8) = 0.0348 114. You will lose $5. 115. Becca 116. 14 117. Sample mean = 3.2 Sample standard deviation = 1.85 Median = 3 Q 1 = 2 Q 3 = 5 IQR = 3 118. d. z = –1.19 e. 0.1171 f. Do not reject the null hypothesis. 119. We conclude that the patient does have the HIV virus when, in fact, the patient does not. 120. c. z = 2.21; p = 0.0136 d. Reject the null hypothesis. e. We conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is not greater. f. We cannot conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is greater. 121. c. dependent means 122. t 5 Chapter 12 123. (0.0424, 0.0770) 124. 2,401 125. Check student's solution. 126. 0.6321 127. $360 128. N ( 72 , 72 5 ) Chapter 13 129. 0.02 130. 0.40 131. 100 140 132. 10 60 133. p -value = 0; Reject the null hypothesis; conclude that they are dependent events 134. 8.4 135. B (14, 0.60) 136. d. Binomial 137. 0.3669 138. p -value = 0.0006; reject the null hypothesis; conclude that the averages are not equal 139. p -value = 0; reject the null hypothesis; conclude that the proportion of males is higher 140. Minimize α and β 141. No Yes, P ( M AND 30+) = 0 142. 12 38 143. No; p -value = 0 144. a. uniform References Data from the San Jose Mercury News . Baran, Daya. “20 Percent of Americans Have Never Used Email.” Webguild.org, 2010. Available online at: http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed October 17, 2013). Data from Parade Magazine .", "section": "Review Exercises (Ch 3-13)", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Practice Tests (1-4) and Final Exams Practice Test 1 1.1: Definitions of Statistics, Probability, and Key Terms Use the following information to answer the next three exercises. A grocery store is interested in how much money, on average, their customers spend each visit in the produce department. Using their store records, they draw a sample of 1,000 visits and calculate each customer’s average spending on produce. 1 . Identify the population, sample, parameter, statistic, variable, and data for this example. population sample parameter statistic variable data 2 . What kind of data is “amount of money spent on produce per visit”? qualitative quantitative-continuous quantitative-discrete 3 . The study finds that the mean amount spent on produce per visit by the customers in the sample is $12.84. This is an example of a: population sample parameter statistic variable 1.2: Data, Sampling, and Variation in Data and Sampling Use the following information to answer the next two exercises. A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a short survey including information about how many times they have visited the club in the past week. 4 . What kind of a sampling design is this? cluster stratified simple random systematic 5 . “Number of visits per week” is what kind of data? qualitative quantitative-continuous quantitative-discrete 6 . Describe a situation in which you would calculate a parameter, rather than a statistic. 7 . The U.S. federal government conducts a survey of high school seniors concerning their plans for future education and employment. One question asks whether they are planning to attend a four-year college or university in the following year. Fifty percent answer yes to this question; that fifty percent is a: parameter statistic variable data 8 . Imagine that the U.S. federal government had the means to survey all high school seniors in the U.S. concerning their plans for future education and employment, and found that 50 percent were planning to attend a 4-year college or university in the following year. This 50 percent is an example of a: parameter statistic variable data Use the following information to answer the next three exercises. A survey of a random sample of 100 nurses working at a large hospital asked how many years they had been working in the profession. Their answers are summarized in the following (incomplete) table. 9 . Fill in the blanks in the table and round your answers to two decimal places for the Relative Frequency and Cumulative Relative Frequency cells. # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 25 5–10 30 > 10 empty 10 . What proportion of nurses have five or more years of experience? 11 . What proportion of nurses have ten or fewer years of experience? 12 . Describe how you might draw a random sample of 30 students from a lecture class of 200 students. 13 . Describe how you might draw a stratified sample of students from a college, where the strata are the students’ class standing (first through final years). 14 . A manager wants to draw a sample, without replacement, of 30 employees from a workforce of 150. Describe how the chance of being selected will change over the course of drawing the sample. 15 . The manager of a department store decides to measure employee satisfaction by selecting four departments at random, and conducting interviews with all the employees in those four departments. What type of survey design is this? cluster stratified simple random systematic 16 . A popular American television sports program conducts a poll of viewers to see which team they believe will win the NFL (National Football League) championship this year. Viewers vote by calling a number displayed on the television screen and telling the operator which team they think will win. Do you think that those who participate in this poll are representative of all football fans in America? 17 . Two researchers studying vaccination rates independently draw samples of 50 children, ages 3–18 months, from a large urban area, and determine if they are up to date on their vaccinations. One researcher finds that 84 percent of the children in her sample are up to date, and the other finds that 86 percent in his sample are up to date. Assuming both followed proper sampling procedures and did their calculations correctly, what is a likely explanation for this discrepancy? 18 . A high school increased the length of the school day from 6.5 to 7.5 hours. Students who wished to attend this high school were required to sign contracts pledging to put forth their best effort on their school work and to obey the school rules; if they did not wish to do so, they could attend another high school in the district. At the end of one year, student performance on statewide tests had increased by ten percentage points over the previous year. Does this improvement prove that a longer school day improves student achievement? 19 . You read a newspaper article reporting that eating almonds leads to increased life satisfaction. The study was conducted by the Almond Growers Association, and was based on a randomized survey asking people about their consumption of various foods, including almonds, and also about their satisfaction with different aspects of their life. Does anything about this poll lead you to question its conclusion? 20 . Why is non-response a problem in surveys? 1.3: Frequency, Frequency Tables, and Levels of Measurement 21 . Compute the mean of the following numbers, and report your answer using one more decimal place than is present in the original data: 14, 5, 18, 23, 6 1.4: Experimental Design and Ethics 22 . A psychologist is interested in whether the size of tableware (bowls, plates, etc.) influences how much college students eat. He randomly assigns 100 college students to one of two groups: the first is served a meal using normal-sized tableware, while the second is served the same meal, but using tableware that it 20 percent smaller than normal. He records how much food is consumed by each group. Identify the following components of this study. population sample experimental units explanatory variable treatment response variable 23 . A researcher analyzes the results of the SAT (Scholastic Aptitude Test) over a five-year period and finds that male students on average score higher on the math section, and female students on average score higher on the verbal section. She concludes that these observed differences in test performance are due to genetic factors. Explain how lurking variables could offer an alternative explanation for the observed differences in test scores. 24 . Explain why it would not be possible to use random assignment to study the health effects of smoking. 25 . A professor conducts a telephone survey of a city’s population by drawing a sample of numbers from the phone book and having her student assistants call each of the selected numbers once to administer the survey. What are some sources of bias with this survey? 26 . A professor offers extra credit to students who take part in her research studies. What is an ethical problem with this method of recruiting subjects? 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs Use the following information to answer the next four exercises. The midterm grades on a chemistry exam, graded on a scale of 0 to 100, were: 62, 64, 65, 65, 68, 70, 72, 72, 74, 75, 75, 75, 76,78, 78, 81, 83, 83, 84, 85, 87, 88, 92, 95, 98, 98, 100, 100, 740 27 . Do you see any outliers in this data? If so, how would you address the situation? 28 . Construct a stem plot for this data, using only the values in the range 0–100. 29 . Describe the distribution of exam scores. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30 . In a class of 35 students, seven students received scores in the 70–79 range. What is the relative frequency of scores in this range? Use the following information to answer the next three exercises. You conduct a poll of 30 students to see how many classes they are taking this term. Your results are: 1; 1; 1; 1 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4; 4; 4; 4 5; 5; 5; 5 31 . You decide to construct a histogram of this data. What will be the range of your first bar, and what will be the central point? 32 . What will be the widths and central points of the other bars? 33 . Which bar in this histogram will be the tallest, and what will be its height? 34 . You get data from the U.S. Census Bureau on the median household income for your city, and decide to display it graphically. Which is the better choice for this data, a bar graph or a histogram? 35 . You collect data on the color of cars driven by students in your statistics class, and want to display this information graphically. Which is the better choice for this data, a bar graph or a histogram? 2.3: Measures of the Location of the Data 36 . Your daughter brings home test scores showing that she scored in the 80 th percentile in math and the 76 th percentile in reading for her grade. Interpret these scores. 37 . You have to wait 90 minutes in the emergency room of a hospital before you can see a doctor. You learn that your wait time was in the 82 nd percentile of all wait times. Explain what this means, and whether you think it is good or bad. 2.4: Box Plots Use the following information to answer the next three exercises. 1; 1; 2; 3; 4; 4; 5; 5; 6; 7; 7; 8; 9 38 . What is the median for this data? 39 . What is the first quartile for this data? 40 . What is the third quartile for this data? Use the following information to answer the next four exercises. This box plot represents scores on the final exam for a physics class. 41 . What is the median for this data, and how do you know? 42 . What are the first and third quartiles for this data, and how do you know? 43 . What is the interquartile range for this data? 44 . What is the range for this data? 2.5: Measures of the Center of the Data 45 . In a marathon, the median finishing time was 3:35:04 (three hours, 35 minutes, and four seconds). You finished in 3:34:10. Interpret the meaning of the median time, and discuss your time in relation to it. Use the following information to answer the next three exercises. The value, in thousands of dollars, for houses on a block, are: 45; 47; 47.5; 51; 53.5; 125. 46 . Calculate the mean for this data. 47 . Calculate the median for this data. 48 . Which do you think better reflects the average value of the homes on this block? 2.6: Skewness and the Mean, Median, and Mode 49 . In a left-skewed distribution, which is greater? the mean the media the mode 50 . In a right-skewed distribution, which is greater? the mean the median the mode 51 . In a symmetrical distribution what will be the relationship among the mean, median, and mode? 2.7: Measures of the Spread of the Data Use the following information to answer the next four exercises. 10; 11; 15; 15; 17; 22 52 . Compute the mean and standard deviation for this data; use the sample formula for the standard deviation. 53 . What number is two standard deviations above the mean of this data? 54 . Express the number 13.7 in terms of the mean and standard deviation of this data. 55 . In a biology class, the scores on the final exam were normally distributed, with a mean of 85, and a standard deviation of five. Susan got a final exam score of 95. Express her exam result as a z -score, and interpret its meaning. 3.1: Terminology Use the following information to answer the next two exercises. You have a jar full of marbles: 50 are red, 25 are blue, and 15 are yellow. Assume you draw one marble at random for each trial, and replace it before the next trial. Let P ( R ) = the probability of drawing a red marble. Let P ( B ) = the probability of drawing a blue marble. Let P ( Y ) = the probability of drawing a yellow marble. 56 . Find P ( B ). 57 . Which is more likely, drawing a red marble or a yellow marble? Justify your answer numerically. Use the following information to answer the next two exercises. The following are probabilities describing a group of college students. Let P ( M ) = the probability that the student is male Let P ( F ) = the probability that the student is female Let P ( E ) = the probability the student is majoring in education Let P ( S ) = the probability the student is majoring in science 58 . Write the symbols for the probability that a student, selected at random, is both female and a science major. 59 . Write the symbols for the probability that the student is an education major, given that the student is male. 3.2: Independent and Mutually Exclusive Events 60 . Events A and B are independent. If P ( A ) = 0.3 and P ( B ) = 0.5, find P ( A AND B ). 61 . C and D are mutually exclusive events. If P ( C ) = 0.18 and P ( D ) = 0.03, find P ( C OR D ). 3.3: Two Basic Rules of Probability 62 . In a high school graduating class of 300, 200 students are going to college, 40 are planning to work full-time, and 80 are taking a gap year. Are these events mutually exclusive? Use the following information to answer the next two exercises. An archer hits the center of the target (the bullseye) 70 percent of the time. However, she is a streak shooter, and if she hits the center on one shot, her probability of hitting it on the shot immediately following is 0.85. Written in probability notation: P ( A ) = P ( B ) = P (hitting the center on one shot) = 0.70 P ( B | A ) = P(hitting the center on a second shot, given that she hit it on the first) = 0.85 63 . Calculate the probability that she will hit the center of the target on two consecutive shots. 64 . Are P ( A ) and P ( B ) independent in this example? 3.4: Contingency Tables Use the following information to answer the next three exercises. The following contingency table displays the number of students who report studying at least 15 hours per week, and how many made the honor roll in the past semester. Honor roll No honor roll Total Study at least 15 hours/week 200 Study less than 15 hours/week 125 193 Total 1,000 65 . Complete the table. 66 . Find P (honor roll|study at least 15 hours per week). 67 . What is the probability a student studies less than 15 hours per week? 68 . Are the events “study at least 15 hours per week” and “makes the honor roll” independent? Justify your answer numerically. 3.5: Tree and Venn Diagrams 69 . At a high school, some students play on the tennis team, some play on the soccer team, but neither plays both tennis and soccer. Draw a Venn diagram illustrating this. 70 . At a high school, some students play tennis, some play soccer, and some play both. Draw a Venn diagram illustrating this. Practice Test 1 Solutions 1.1: Definitions of Statistics, Probability, and Key Terms 1 . population: all the shopping visits by all the store’s customers sample: the 1,000 visits drawn for the study parameter: the average expenditure on produce per visit by all the store’s customers statistic: the average expenditure on produce per visit by the sample of 1,000 variable: the expenditure on produce for each visit data: the dollar amounts spent on produce; for instance, $15.40, $11.53, etc 2 . c 3 . d 1.2: Data, Sampling, and Variation in Data and Sampling 4 . d 5 . c 6 . Answers will vary. Sample Answer: Any solution in which you use data from the entire population is acceptable. For instance, a professor might calculate the average exam score for her class: because the scores of all members of the class were used in the calculation, the average is a parameter. 7 . b 8 . a 9 . # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 25 0.25 0.25 5–10 30 0.30 0.55 > 10 45 0.45 1.00 10 . 0.75 11 . 0.55 12 . Answers will vary. Sample Answer: One possibility is to obtain the class roster and assign each student a number from 1 to 200. Then use a random number generator or table of random number to generate 30 numbers between 1 and 200, and select the students matching the random numbers. It would also be acceptable to write each student’s name on a card, shuffle them in a box, and draw 30 names at random. 13 . One possibility would be to obtain a roster of students enrolled in the college, including the class standing for each student. Then you would draw a proportionate random sample from within each class (for instance, if 30 percent of the students in the college are first-year students, then 30 percent of your sample would be drawn from the first-year class). 14 . For the first person picked, the chance of any individual being selected is one in 150. For the second person, it is one in 149, for the third it is one in 148, and so on. For the 30th person selected, the chance of selection is one in 121. 15 . a 16 . No. There are at least two chances for bias. First, the viewers of this particular program may not be representative of American football fans as a whole. Second, the sample will be self-selected, because people have to make a phone call in order to take part, and those people are probably not representative of the American football fan population as a whole. 17 . These results (84 percent in one sample, 86 percent in the other) are probably due to sampling variability. Each researcher drew a different sample of children, and you would not expect them to get exactly the same result, although you would expect the results to be similar, as they are in this case. 18 . No. The improvement could also be due to self-selection: only motivated students were willing to sign the contract, and they would have done well even in a school with 6.5 hour days. Because both changes were implemented at the same time, it is not possible to separate out their influence. 19 . At least two aspects of this poll are troublesome. The first is that it was conducted by a group who would benefit by the result—almond sales are likely to increase if people believe that eating almonds will make them happier. The second is that this poll found that almond consumption and life satisfaction are correlated, but does not establish that eating almonds causes satisfaction. It is equally possible, for instance, that people with higher incomes are more likely to eat almonds, and are also more satisfied with their lives. 20 . You want the sample of people who take part in a survey to be representative of the population from which they are drawn. People who refuse to take part in a survey often have different views than those who do participate, and so even a random sample may produce biased results if a large percentage of those selected refuse to participate in a survey. 1.3: Frequency, Frequency Tables, and Levels of Measurement 21 . 13.2 1.4: Experimental Design and Ethics 22 . population: all college students sample: the 100 college students in the study experimental units: each individual college student who participated explanatory variable: the size of the tableware treatment: tableware that is 20 percent smaller than normal response variable: the amount of food eaten 23 . There are many lurking variables that could influence the observed differences in test scores. Perhaps the boys, on average, have taken more math courses than the girls, and the girls have taken more English classes than the boys. Perhaps the boys have been encouraged by their families and teachers to prepare for a career in math and science, and thus have put more effort into studying math, while the girls have been encouraged to prepare for fields like communication and psychology that are more focused on language use. A study design would have to control for these and other potential lurking variables (anything that could explain the observed difference in test scores, other than the genetic explanation) in order to draw a scientifically sound conclusion about genetic differences. 24 . To use random assignment, you would have to be able to assign people to either smoke or not smoke. Because smoking has many harmful effects, this would not be an ethical experiment. Instead, we study people who have chosen to smoke, and compare them to others who have chosen not to smoke, and try to control for the other ways those two groups may differ (lurking variables). 25 . Sources of bias include the fact that not everyone has a telephone, that cell phone numbers are often not listed in published directories, and that an individual might not be at home at the time of the phone call; all these factors make it likely that the respondents to the survey will not be representative of the population as a whole. 26 . Research subjects should not be coerced into participation, and offering extra credit in exchange for participation could be construed as coercion. In addition, this method will result in a volunteer sample, which cannot be assumed to be representative of the population as a whole. 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 27 . The value 740 is an outlier, because the exams were graded on a scale of 0 to 100, and 740 is far outside that range. It may be a data entry error, with the actual score being 74, so the professor should check that exam again to see what the actual score was. 28 . Stem Leaf 6 2 4 5 5 8 7 0 2 2 4 5 5 5 6 8 8 8 1 3 3 4 5 7 8 9 2 5 8 8 10 0 0 29 . Most scores on this exam were in the range of 70–89, with a few scoring in the 60–69 range, and a few in the 90–100 range. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30 . R F = 7 35 = 0.2 31 . The range will be 0.5–1.5, and the central point will be 1. 32 . Range 1.5–2.5, central point 2; range 2.5–3.5, central point 3; range 3.5–4.5, central point 4; range 4.5–5.5., central point 5. 33 . The bar from 3.5 to 4.5, with a central point of 4, will be tallest; its height will be nine, because there are nine students taking four courses. 34 . The histogram is a better choice, because income is a continuous variable. 35 . A bar graph is the better choice, because this data is categorical rather than continuous. 2.3: Measures of the Location of the Data 36 . Your daughter scored better than 80 percent of the students in her grade on math and better than 76 percent of the students in reading. Both scores are very good, and place her in the upper quartile, but her math score is slightly better in relation to her peers than her reading score. 37 . You had an unusually long wait time, which is bad: 82 percent of patients had a shorter wait time than you, and only 18 percent had a longer wait time. 2.4: Box Plots 38 . 5 39 . 3 40 . 7 41 . The median is 86, as represented by the vertical line in the box. 42 . The first quartile is 80, and the third quartile is 92, as represented by the left and right boundaries of the box. 43 . IQR = 92 – 80 = 12 44 . Range = 100 – 75 = 25 2.5: Measures of the Center of the Data 45 . Half the runners who finished the marathon ran a time faster than 3:35:04, and half ran a time slower than 3:35:04. Your time is faster than the median time, so you did better than more than half of the runners in this race. 46 . 61.5, or $61,500 47 . 49.25 or $49,250 48 . The median, because the mean is distorted by the high value of one house. 2.6: Skewness and the Mean, Median, and Mode 49 . c 50 . a 51 . They will all be fairly close to each other. 2.7: Measures of the Spread of the Data 52 . Mean: 15 Standard deviation: 4.3 μ = 10 + 11 + 15 + 15 + 17 + 22 6 = 15 s = ∑ ( x − x ¯ ) 2 n − 1 = 94 5 = 4.3 53 . 15 + (2)(4.3) = 23.6 54 . 13.7 is one standard deviation below the mean of this data, because 15 – 4.3 = 10.7 55 . z = 95 − 85 5 = 2.0 Susan’s z -score was 2.0, meaning she scored two standard deviations above the class mean for the final exam. 3.1: Terminology 56 . P ( B ) = 25 90 = 0.28 57 . Drawing a red marble is more likely. P ( R ) = 50 80 = 0.62 P ( Y ) = 15 80 = 0.19 58 . P ( F AND S ) 59 . P ( E | M ) 3.2: Independent and Mutually Exclusive Events 60 . P ( A AND B ) = (0.3)(0.5) = 0.15 61 . P ( C OR D ) = 0.18 + 0.03 = 0.21 3.3: Two Basic Rules of Probability 62 . No, they cannot be mutually exclusive, because they add up to more than 300. Therefore, some students must fit into two or more categories (e.g., both going to college and working full time). 63 . P ( A and B ) = ( P ( B | A ))( P ( A )) = (0.85)(0.70) = 0.595 64 . No. If they were independent, P ( B ) would be the same as P ( B | A ). We know this is not the case, because P ( B ) = 0.70 and P ( B | A ) = 0.85. 3.4: Contingency Tables 65 . Honor roll No honor roll Total Study at least 15 hours/week 482 200 682 Study less than 15 hours/week 125 193 318 Total 607 393 1,000 66 . P (honor roll|study at least 15 hours per week) = 482 682 , or about 0.707 67 . P ( study less than 15 hours per week) = 318 1000 , or 0.318 68 . Let P ( S ) = study at least 15 hours per week Let P ( H ) = makes the honor roll From the table, P ( S ) = 0.682, P ( H ) = 0.607, and P ( S AND H ) =0.482. If P ( S ) and P ( H ) were independent, then P ( S AND H ) would equal ( P ( S ))( P ( H )). However, ( P ( S ))( P ( H )) = (0.682)(0.607) = 0.414, while P ( S AND H ) = 0.482. Therefore, P ( S ) and P ( H ) are not independent. 3.5: Tree and Venn Diagrams 69 . 70 . Practice Test 2 4.1: Probability Distribution Function (PDF) for a Discrete Random Variable Use the following information to answer the next five exercises. You conduct a survey among a random sample of students at a particular university. The data collected includes their major, the number of classes they took the previous semester, and amount of money they spent on books purchased for classes in the previous semester. 1. If X = student’s major, then what is the domain of X ? 2. If Y = the number of classes taken in the previous semester, what is the domain of Y ? 3 . If Z = the amount of money spent on books in the previous semester, what is the domain of Z ? 4 . Why are X , Y , and Z in the previous example random variables? 5 . After collecting data, you find that for one case, z = –7. Is this a possible value for Z ? 6 . What are the two essential characteristics of a discrete probability distribution? Use this discrete probability distribution represented in this table to answer the following six questions. The university library records the number of books checked out by each patron over the course of one day, with the following result: x P ( x ) 0 0.20 1 0.45 2 0.20 3 0.10 4 0.05 7 . Define the random variable X for this example. 8 . What is P ( x > 2)? 9 . What is the probability that a patron will check out at least one book? 10 . What is the probability a patron will take out no more than three books? 11 . If the table listed P (4) as 0.15, how would you know that there was a mistake? 12 . What is the average number of books taken out by a patron? 4.2: Mean or Expected Value and Standard Deviation Use the following information to answer the next five exercises. Three jobs are open in a company: one in the accounting department, one in the human resources department, and one in the sales department. The accounting job receives 30 applicants, the human resources department receives 40 applicants, and the sales department receives 60 applicants. 13 . If X = the number of applications for a job, use this information to fill in . x P ( x ) x P ( x ) 14 . What is the mean number of applicants per department? 15 . What is the PDF for X ? 16 . Add a fourth column to the table, for ( x – μ ) 2 P ( x ). 17 . What is the standard deviation of X ? 4.3: Binomial Distribution 18 . In a binomial experiment, if p = 0.65, what does q equal? 19 . What are the required characteristics of a binomial experiment? 20 . Joe conducts an experiment to see how many times he has to flip a coin before he gets four heads in a row. Does this qualify as a binomial experiment? Use the following information to answer the next three exercises. In a particularly community, 65 percent of households include at least one person who has graduated from college. You randomly sample 100 households in this community. Let X = the number of households including at least one college graduate. 21 . Describe the probability distribution of X . 22 . What is the mean of X ? 23 . What is the standard deviation of X ? Use the following information to answer the next four exercises. Joe is the star of his school’s baseball team. His batting average is 0.400, meaning that for every ten times he comes to bat (an at-bat), four of those times he gets a hit. You decide to track his batting performance his next 20 at-bats. 24 . Define the random variable X in this experiment. 25 . Assuming Joe’s probability of getting a hit is independent and identical across all 20 at-bats, describe the distribution of X . 26 . Given this information, what number of hits do you predict Joe will get? 27 . What is the standard deviation of X ? 4.4: Geometric Distribution 28 . What are the three major characteristics of a geometric experiment? 29 . You decide to conduct a geometric experiment by flipping a coin until it comes up heads. This takes five trials. Represent the outcomes of this trial, using H for heads and T for tails. 30 . You are conducting a geometric experiment by drawing cards from a normal 52-card pack, with replacement, until you draw the Queen of Hearts. What is the domain of X for this experiment? 31 . You are conducting a geometric experiment by drawing cards from a normal 52-card deck, without replacement, until you draw a red card. What is the domain of X for this experiment? Use the following information to answer the next three exercises. In a particular university, 27 percent of students are engineering majors. You decide to select students at random until you choose one that is an engineering major. Let X = the number of students you select until you find one that is an engineering major. 32 . What is the probability distribution of X ? 33 . What is the mean of X ? 34 . What is the standard deviation of X ? 4.5: Hypergeometric Distribution 35 . You draw a random sample of ten students to participate in a survey, from a group of 30, consisting of 16 boys and 14 girls. You are interested in the probability that seven of the students chosen will be boys. Does this qualify as a hypergeometric experiment? List the conditions and whether or not they are met. 36 . You draw five cards, without replacement, from a normal 52-card deck of playing cards, and are interested in the probability that two of the cards are spades. What are the group of interest, size of the group of interest, and sample size for this example? 4.6: Poisson Distribution 37 . What are the key characteristics of the Poisson distribution? Use the following information to answer the next three exercises. The number of drivers to arrive at a toll booth in an hour can be modeled by the Poisson distribution. 38 . If X = the number of drivers, and the average numbers of drivers per hour is four, how would you express this distribution? 39 . What is the domain of X ? 40 . What are the mean and standard deviation of X ? 5.1: Continuous Probability Functions 41 . You conduct a survey of students to see how many books they purchased the previous semester, the total amount they paid for those books, the number they sold after the semester was over, and the amount of money they received for the books they sold. Which variables in this survey are discrete, and which are continuous? 42 . With continuous random variables, we never calculate the probability that X has a particular value, but always speak in terms of the probability that X has a value within a particular range. Why is this? 43 . For a continuous random variable, why are P ( x < c ) and P ( x ≤ c ) equivalent statements? 44 . For a continuous probability function, P ( x < 5) = 0.35. What is P ( x > 5), and how do you know? 45 . Describe how you would draw the continuous probability distribution described by the function f ( x ) = 1 10 for 0 ≤ x ≤ 10 . What type of a distribution is this? 46 . For the continuous probability distribution described by the function f ( x ) = 1 10 for 0 ≤ x ≤ 10 , what is the P (0 < x < 4)? 5.2: The Uniform Distribution 47 . For the continuous probability distribution described by the function f ( x ) = 1 10 for 0 ≤ x ≤ 10 , what is the P (2 < x < 5)? Use the following information to answer the next four exercises. The number of minutes that a patient waits at a medical clinic to see a doctor is represented by a uniform distribution between zero and 30 minutes, inclusive. 48 . If X equals the number of minutes a person waits, what is the distribution of X ? 49 . Write the probability density function for this distribution. 50 . What is the mean and standard deviation for waiting time? 51 . What is the probability that a patient waits less than ten minutes? 5.3: The Exponential Distribution 52 . The distribution of the variable X , representing the average time to failure for an automobile battery, can be written as: X ~ Exp ( m ). Describe this distribution in words. 53 . If the value of m for an exponential distribution is ten, what are the mean and standard deviation for the distribution? 54 . Write the probability density function for a variable distributed as: X ~ Exp (0.2). 6.1: The Standard Normal Distribution 55 . Translate this statement about the distribution of a random variable X into words: X ~ (100, 15). 56 . If the variable X has the standard normal distribution, express this symbolically. Use the following information for the next six exercises. According to the World Health Organization, distribution of height in centimeters for girls aged five years and no months has the distribution: X ~ N (109, 4.5). 57 . What is the z -score for a height of 112 inches? 58 . What is the z -score for a height of 100 centimeters? 59 . Find the z -score for a height of 105 centimeters and explain what that means In the context of the population. 60 . What height corresponds to a z -score of 1.5 in this population? 61 . Using the empirical rule, we expect about 68 percent of the values in a normal distribution to lie within one standard deviation above or below the mean. What does this mean, in terms of a specific range of values, for this distribution? 62 . Using the empirical rule, about what percent of heights in this distribution do you expect to be between 95.5 cm and 122.5 cm? 6.2: Using the Normal Distribution Use the following information to answer the next four exercises. The distributor of lotto tickets claims that 20 percent of the tickets are winners. You draw a sample of 500 tickets to test this proposition. 63 . Can you use the normal approximation to the binomial for your calculations? Why or why not. 64 . What are the expected mean and standard deviation for your sample, assuming the distributor’s claim is true? 65 . What is the probability that your sample will have a mean greater than 100? 66 . If the z -score for your sample result is –2.00, explain what this means, using the empirical rule. 7.1: The Central Limit Theorem for Sample Means (Averages) 67 . What does the central limit theorem state with regard to the distribution of sample means? 68 . The distribution of results from flipping a fair coin is uniform: heads and tails are equally likely on any flip, and over a large number of trials, you expect about the same number of heads and tails. Yet if you conduct a study by flipping 30 coins and recording the number of heads, and repeat this 100 times, the distribution of the mean number of heads will be approximately normal. How is this possible? 69 . The mean of a normally-distributed population is 50, and the standard deviation is four. If you draw 100 samples of size 40 from this population, describe what you would expect to see in terms of the sampling distribution of the sample mean. 70 . X is a random variable with a mean of 25 and a standard deviation of two. Write the distribution for the sample mean of samples of size 100 drawn from this population. 71 . Your friend is doing an experiment drawing samples of size 50 from a population with a mean of 117 and a standard deviation of 16. This sample size is large enough to allow use of the central limit theorem, so he says the standard deviation of the sampling distribution of sample means will also be 16. Explain why this is wrong, and calculate the correct value. 72 . You are reading a research article that refers to “the standard error of the mean.” What does this mean, and how is it calculated? Use the following information to answer the next six exercises. You repeatedly draw samples of n = 100 from a population with a mean of 75 and a standard deviation of 4.5. 73 . What is the expected distribution of the sample means? 74 . One of your friends tries to convince you that the standard error of the mean should be 4.5. Explain what error your friend made. 75 . What is the z -score for a sample mean of 76? 76 . What is the z -score for a sample mean of 74.7? 77 . What sample mean corresponds to a z -score of 1.5? 78 . If you decrease the sample size to 50, will the standard error of the mean be smaller or larger? What would be its value? Use the following information to answer the next two questions. We use the empirical rule to analyze data for samples of size 60 drawn from a population with a mean of 70 and a standard deviation of 9. 79 . What range of values would you expect to include 68 percent of the sample means? 80 . If you increased the sample size to 100, what range would you expect to contain 68 percent of the sample means, applying the empirical rule? 7.2: The Central Limit Theorem for Sums 81 . How does the central limit theorem apply to sums of random variables? 82 . Explain how the rules applying the central limit theorem to sample means, and to sums of a random variable, are similar. 83 . If you repeatedly draw samples of size 50 from a population with a mean of 80 and a standard deviation of four, and calculate the sum of each sample, what is the expected distribution of these sums? Use the following information to answer the next four exercises. You draw one sample of size 40 from a population with a mean of 125 and a standard deviation of seven. 84 . Compute the sum. What is the probability that the sum for your sample will be less than 5,000? 85 . If you drew samples of this size repeatedly, computing the sum each time, what range of values would you expect to contain 95 percent of the sample sums? 86 . What value is one standard deviation below the mean? 87 . What value corresponds to a z -score of 2.2? 7.3: Using the Central Limit Theorem 88 . What does the law of large numbers say about the relationship between the sample mean and the population mean? 89 . Applying the law of large numbers, which sample mean would expect to be closer to the population mean, a sample of size ten or a sample of size 100? Use this information for the next three questions. A manufacturer makes screws with a mean diameter of 0.15 cm (centimeters) and a range of 0.10 cm to 0.20 cm; within that range, the distribution is uniform. 90 . If X = the diameter of one screw, what is the distribution of X ? 91 . Suppose you repeatedly draw samples of size 100 and calculate their mean. Applying the central limit theorem, what is the distribution of these sample means? 92 . Suppose you repeatedly draw samples of 60 and calculate their sum. Applying the central limit theorem, what is the distribution of these sample sums? Practice Test 2 Solutions Probability Distribution Function (PDF) for a Discrete Random Variable 1 . The domain of X = {English, Mathematics,….], i.e., a list of all the majors offered at the university, plus “undeclared.” 2 . The domain of Y = {0, 1, 2, …}, i.e., the integers from 0 to the upper limit of classes allowed by the university. 3 . The domain of Z = any amount of money from 0 upwards. 4 . Because they can take any value within their domain, and their value for any particular case is not known until the survey is completed. 5 . No, because the domain of Z includes only positive numbers (you can’t spend a negative amount of money). Possibly the value –7 is a data entry error, or a special code to indicated that the student did not answer the question. 6 . The probabilities must sum to 1.0, and the probabilities of each event must be between 0 and 1, inclusive. 7 . Let X = the number of books checked out by a patron. 8 . P ( x > 2) = 0.10 + 0.05 = 0.15 9 . P ( x ≥ 0) = 1 – 0.20 = 0.80 10 . P ( x ≤ 3) = 1 – 0.05 = 0.95 11 . The probabilities would sum to 1.10, and the total probability in a distribution must always equal 1.0. 12 . x ¯ = 0(0.20) + 1(0.45) + 2(0.20) + 3(0.10) + 4(0.05) = 1.35 Mean or Expected Value and Standard Deviation 13 . x P ( x ) x P ( x ) 30 30/130 6.92 40 40/130 12.31 60 60/130 27.69 14 . x ¯ = 6.92 + 12.31 + 27.69 = 46.92 15 . P ( x = 30) = 0.23 P ( x = 40) = 0.31 P ( x = 60) = 0.46 16 . x P ( x ) xP ( x ) ( x – μ ) 2 P ( x ) 30 0.23 6.92 (30 – 46.92) 2 (0.23) = 66.09 40 0.31 12.31 (40 – 46.92) 2 (0.31) = 14.75 60 0.46 46.92 (60 – 46.92) 2 (0.46) = 78.93 ∑ x = ( 66.09 + 14.75 + 78.93 ) = 12.64 Binomial Distribution 18 . q = 1 – 0.65 = 0.35 19 . There are a fixed number of trials. There are only two possible outcomes, and they add up to 1. The trials are independent and conducted under identical conditions. 20 . No, because there are not a fixed number of trials 21 . X ~ B (100, 0.65) 22 . μ = np = 100(0.65) = 65 23 . σ x = n p q = 100 ( 0.65 ) ( 0.35 ) = 4.77 24 . X = Joe gets a hit in one at-bat (in one occasion of his coming to bat) 25 . X ~ B (20, 0.4) 26 . μ = np = 20(0.4) = 8 27 . σ x = n p q = 20 ( 0.40 ) ( 0.60 ) = 2.19 4.4: Geometric Distribution 28 . A series of Bernoulli trials are conducted until one is a success, and then the experiment stops. At least one trial is conducted, but there is no upper limit to the number of trials. The probability of success or failure is the same for each trial. 29 . T T T T H 30 . The domain of X = {1, 2, 3, 4, 5, ….n}. Because you are drawing with replacement, there is no upper bound to the number of draws that may be necessary. 31 . The domain of X = {1, 2, 3, 4, 5, 6, 7, 8., 9, 10, 11, 12…27}. Because you are drawing without replacement, and 26 of the 52 cards are red, you have to draw a red card within the first 27 draws. 32 . X ~ G (0.24) 33 . μ = 1 p = 1 0.27 = 3.70 34 . σ = 1 − p p 2 = 1 − 0.27 0.27 2 = 3.16 4.5: Hypergeometric Distribution 35 . Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials). 36 . The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five. 4.6: Poisson Distribution 37 . A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known. 38 . X ~ P (4) 39 . The domain of X = {0, 1, 2, 3, …..) i.e., any integer from 0 upwards. 40 . μ = 4 σ = 4 = 2 5.1: Continuous Probability Functions 41 . The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold. 42 . Because for a continuous random variable, P ( x = c ) = 0, where c is any single value. Instead, we calculate P ( c < x < d ), i.e., the probability that the value of x is between the values c and d . 43 . Because P ( x = c ) = 0 for any continuous random variable. 44 . P ( x > 5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1. 45 . This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at 1 10 and 0. 46 . P ( 0 < x < 4 ) = ( 4 − 0 ) ( 1 10 ) = 0.4 5.2: The Uniform Distribution 47 . P ( 2 < x < 5 ) = ( 5 − 2 ) ( 1 10 ) = 0.3 48 . X ~ U (0, 15) 49 . f ( x ) = 1 b − a for ( a ≤ x ≤ b ) so f ( x ) = 1 30 for ( 0 ≤ x ≤ 30 ) 50 . μ = a + b 2 = 0 + 30 5 = 15.0 σ = ( b − a ) 2 12 = ( 30 − 0 ) 2 12 = 8.66 51 . P ( x < 10 ) = ( 10 ) ( 1 30 ) = 0.33 5.3: The Exponential Distribution 52 . X has an exponential distribution with decay parameter m and mean and standard deviation 1 m . In this distribution, there will be a relatively large numbers of small values, with values becoming less common as they become larger. 53 . μ = σ = 1 m = 1 10 = 0.1 54 . f ( x ) = 0.2 e –0.2 x where x ≥ 0. 6.1: The Standard Normal Distribution 55 . The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15. 56 . X ~ N (0,1) 57 . z = x − μ σ so z = 112 − 109 4.5 = 0.67 58 . z = x − μ σ so z = 100 − 109 4.5 = − 2.00 59 . z = 105 − 109 4.5 = −0.89 This girl is shorter than average for her age, by 0.89 standard deviations. 60 . 109 + (1.5)(4.5) = 115.75 cm 61 . We expect about 68 percent of the heights of girls of age five years and zero months to be between 104.5 cm and 113.5 cm. 62 . We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean. 6.2: Using the Normal Distribution 63 . Yes, because both np and nq are greater than five. np = (500)(0.20) = 100 and nq = 500(0.80) = 400 64 . μ = n p = ( 500 ) ( 0.20 ) = 100 σ = n p q = 500 ( 0.20 ) ( 0.80 ) = 8.94 65 . Fifty percent, because in a normal distribution, half the values lie above the mean. 66 . The results of our sample were two standard deviations below the mean, suggesting it is unlikely that 20 percent of the lotto tickets are winners, as claimed by the distributor, and that the true percent of winners is lower. Applying the Empirical Rule, If that claim were true, we would expect to see a result this far below the mean only about 2.5 percent of the time. 7.1: The Central Limit Theorem for Sample Means (Averages) 67 . The central limit theorem states that if samples of sufficient size drawn from a population, the distribution of sample means will be normal, even if the distribution of the population is not normal. 68 . The sample size of 30 is sufficiently large in this example to apply the central limit theorem. This theorem ] states that for samples of sufficient size drawn from a population, the sampling distribution of the sample mean will approach normality, regardless of the distribution of the population from which the samples were drawn. 69 . You would not expect each sample to have a mean of 50, because of sampling variability. However, you would expect the sampling distribution of the sample means to cluster around 50, with an approximately normal distribution, so that values close to 50 are more common than values further removed from 50. 70 . X ¯ ∼ N ( 25 , 0.2 ) because X ¯ ∼ N ( μ x , σ x n ) 71 . The standard deviation of the sampling distribution of the sample means can be calculated using the formula ( σ x n ) , which in this case is ( 16 50 ) . The correct value for the standard deviation of the sampling distribution of the sample means is therefore 2.26. 72 . The standard error of the mean is another name for the standard deviation of the sampling distribution of the sample mean. Given samples of size n drawn from a population with standard deviation σ x , the standard error of the mean is ( σ x n ) . 73 . X ~ N (75, 0.45) 74 . Your friend forgot to divide the standard deviation by the square root of n . 75 . z = x ¯ − μ x σ x = 76 − 75 4.5 = 2.2 76 . z = x ¯ − μ x σ x = 74.7 − 75 4.5 = −0.67 77 . 75 + (1.5)(0.45) = 75.675 78 . The standard error of the mean will be larger, because you will be dividing by a smaller number. The standard error of the mean for samples of size n = 50 is: ( σ x n ) = 4.5 50 = 0.64 79 . You would expect this range to include values up to one standard deviation above or below the mean of the sample means. In this case: 70 + 9 60 = 71.16 and 70 − 9 60 = 68.84 so you would expect 68 percent of the sample means to be between 68.84 and 71.16. 80 . 70 + 9 100 = 70.9 and 70 − 9 100 = 69.1 so you would expect 68 percent of the sample means to be between 69.1 and 70.9. Note that this is a narrower interval due to the increased sample size. 7.2: The Central Limit Theorem for Sums 81 . For a random variable X , the random variable ΣX will tend to become normally distributed as the size n of the samples used to compute the sum increases. 82 . Both rules state that the distribution of a quantity (the mean or the sum) calculated on samples drawn from a population will tend to have a normal distribution, as the sample size increases, regardless of the distribution of population from which the samples are drawn. 83 . Σ X ∼ N ( n μ x , ( n ) ( σ x ) ) so Σ X ∼ N ( 4000 , 28.3 ) 84 .The probability is 0.50, because 5,000 is the mean of the sampling distribution of sums of size 40 from this population. Sums of random variables computed from a sample of sufficient size are normally distributed, and in a normal distribution, half the values lie below the mean. 85 . Using the empirical rule, you would expect 95 percent of the values to be within two standard deviations of the mean. Using the formula for the standard deviation is for a sample sum: ( n ) ( σ x ) = ( 40 ) ( 7 ) = 44.3 so you would expect 95 percent of the values to be between 5,000 + (2)(44.3) and 5,000 – (2)(44.3), or between 4,911.4 and 588.6. 86 . μ − ( n ) ( σ x ) = 5000 − ( 40 ) ( 7 ) = 4955.7 87 . 5000 + ( 2.2 ) ( 40 ) ( 7 ) = 5097.4 7.3: Using the Central Limit Theorem 88 . The law of large numbers says that as sample size increases, the sample mean tends to get nearer and nearer to the population mean. 89 . You would expect the mean from a sample of size 100 to be nearer to the population mean, because the law of large numbers says that as sample size increases, the sample mean tends to approach the population mea. 90 . X ~ N (0.10, 0.20) 91 . X ¯ ∼ N ( μ x , σ x n ) and the standard deviation of a uniform distribution is b − a 12 . In this example, the standard deviation of the distribution is b − a 12 = 0.10 12 = 0.03 so X ¯ ∼ N ( 0.15 , 0.003 ) 92 . Σ X ∼ N ( ( n ) ( μ x ) , ( n ) ( σ x ) ) so Σ X ∼ N ( 9.0 , 0.23 ) Practice Test 3 8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal Use the following information to answer the next seven exercises. You draw a sample of size 30 from a normally distributed population with a standard deviation of four. 1 . What is the standard error of the sample mean in this scenario, rounded to two decimal places? 2 . What is the distribution of the sample mean? 3 . If you want to construct a two-sided 95% confidence interval, how much probability will be in each tail of the distribution? 4 . What is the appropriate z -score and error bound or margin of error ( EBM ) for a 95% confidence interval for this data? 5 . Rounding to two decimal places, what is the 95% confidence interval if the sample mean is 41? 6 . What is the 90% confidence interval if the sample mean is 41? Round to two decimal places 7 . Suppose the sample size in this study had been 50, rather than 30. What would the 95% confidence interval be if the sample mean is 41? Round your answer to two decimal places. 8 . For any given data set and sampling situation, which would you expect to be wider: a 95% confidence interval or a 99% confidence interval? 8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t 9 . Comparing graphs of the standard normal distribution ( z -distribution) and a t -distribution with 15 degrees of freedom ( df ), how do they differ? 10 . Comparing graphs of the standard normal distribution ( z -distribution) and a t -distribution with 15 degrees of freedom ( df ), how are they similar? Use the following information to answer the next five exercises. Body temperature is known to be distributed normally among healthy adults. Because you do not know the population standard deviation, you use the t-distribution to study body temperature. You collect data from a random sample of 20 healthy adults and find that your sample temperatures have a mean of 98.4 and a sample standard deviation of 0.3 (both in degrees Fahrenheit). 11 . What is the degrees of freedom ( df ) for this study? 12 . For a two-tailed 95% confidence interval, what is the appropriate t -value to use in the formula? 13 . What is the 95% confidence interval? 14 . What is the 99% confidence interval? Round to two decimal places. 15 . Suppose your sample size had been 30 rather than 20. What would the 95% confidence interval be then? Round to two decimal places 8.3: Confidence Interval for a Population Proportion Use this information to answer the next four exercises. You conduct a poll of 500 randomly selected city residents, asking them if they own an automobile. 280 say they do own an automobile, and 220 say they do not. 16 . Find the sample proportion and sample standard deviation for this data. 17 . What is the 95% two-sided confidence interval? Round to four decimal places. 18 . Calculate the 90% confidence interval. Round to four decimal places. 19 . Calculate the 99% confidence interval. Round to four decimal places. Use the following information to answer the next three exercises. You are planning to conduct a poll of community members age 65 and older, to determine how many own mobile phones. You want to produce an estimate whose 95% confidence interval will be within four percentage points (plus or minus) the true population proportion. Use an estimated population proportion of 0.5. 20 . What sample size do you need? 21 . Suppose you knew from prior research that the population proportion was 0.6. What sample size would you need? 22 . Suppose you wanted a 95% confidence interval within three percentage points of the population. Assume the population proportion is 0.5. What sample size do you need? 9.1: Null and Alternate Hypotheses 23 . In your state, 58 percent of registered voters in a community are registered as Republicans. You want to conduct a study to see if this also holds up in your community. State the null and alternative hypotheses to test this. 24 . You believe that at least 58 percent of registered voters in a community are registered as Republicans. State the null and alternative hypotheses to test this. 25 . The mean household value in a city is $268,000. You believe that the mean household value in a particular neighborhood is lower than the city average. Write the null and alternative hypotheses to test this. 26 . State the appropriate alternative hypothesis to this null hypothesis: H 0 : μ = 107 27 . State the appropriate alternative hypothesis to this null hypothesis: H 0 : p < 0.25 9.2: Outcomes and the Type I and Type II Errors 28 . If you reject H 0 when H 0 is correct, what type of error is this? 29 . If you fail to reject H 0 when H 0 is false, what type of error is this? 30 . What is the relationship between the Type II error and the power of a test? 31 . A new blood test is being developed to screen patients for cancer. Positive results are followed up by a more accurate (and expensive) test. It is assumed that the patient does not have cancer. Describe the null hypothesis, the Type I and Type II errors for this situation, and explain which type of error is more serious. 32 . Explain in words what it means that a screening test for TB has an α level of 0.10. The null hypothesis is that the patient does not have TB. 33 . Explain in words what it means that a screening test for TB has a β level of 0.20. The null hypothesis is that the patient does not have TB. 34 . Explain in words what it means that a screening test for TB has a power of 0.80. 9.3: Distribution Needed for Hypothesis Testing 35 . If you are conducting a hypothesis test of a single population mean, and you do not know the population variance, what test will you use if the sample size is 10 and the population is normal? 36 . If you are conducting a hypothesis test of a single population mean, and you know the population variance, what test will you use? 37 . If you are conducting a hypothesis test of a single population proportion, with np and nq greater than or equal to five, what test will you use, and with what parameters? 38 . Published information indicates that, on average, college students spend less than 20 hours studying per week. You draw a sample of 25 students from your college, and find the sample mean to be 18.5 hours, with a standard deviation of 1.5 hours. What distribution will you use to test whether study habits at your college are the same as the national average, and why? 39 . A published study says that 95 percent of American children are vaccinated against measles, with a standard deviation of 1.5 percent. You draw a sample of 100 children from your community and check their vaccination records, to see if the vaccination rate in your community is the same as the national average. What distribution will you use for this test, and why? 9.4: Rare Events, the Sample, Decision, and Conclusion 40 . You are conducting a study with an α level of 0.05. If you get a result with a p-value of 0.07, what will be your decision? 41 . You are conducting a study with α = 0.01. If you get a result with a p -value of 0.006, what will be your decision? Use the following information to answer the next five exercises. According to the World Health Organization, the average height of a one-year-old child is 29”. You believe children with a particular disease are smaller than average, so you draw a sample of 20 children with this disease and find a mean height of 27.5” and a sample standard deviation of 1.5”. 42 . What are the null and alternative hypotheses for this study? 43 . What distribution will you use to test your hypothesis, and why? 44 . What is the test statistic and the p -value? 45 . Based on your sample results, what is your decision? 46 . Suppose the mean for your sample was 25.0. Redo the calculations and describe what your decision would be. 9.5: Additional Information and Full Hypothesis Test Examples 47 . You conduct a study using α = 0.05. What is the level of significance for this study? 48 . You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H 0 : μ = 35.5 H a : μ ≠ 35.5 Will you conduct a one-tailed or two-tailed test? 49 . You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H 0 : μ ≥ 35.5 H a : μ < 35.5 Will you conduct a one-tailed or two-tailed test? Use the following information to answer the next three exercises. Nationally, 80 percent of adults own an automobile. You are interested in whether the same proportion in your community own cars. You draw a sample of 100 and find that 75 percent own cars. 50 . What are the null and alternative hypotheses for this study? 51 . What test will you use, and why? 10.1: Comparing Two Independent Population Means with Unknown Population Standard Deviations 52 . You conduct a poll of political opinions, interviewing both members of 50 married couples. Are the groups in this study independent or matched? 53 . You are testing a new drug to treat insomnia. You randomly assign 80 volunteer subjects to either the experimental (new drug) or control (standard treatment) conditions. Are the groups in this study independent or matched? 54 . You are investigating the effectiveness of a new math textbook for high school students. You administer a pretest to a group of students at the beginning of the semester, and a posttest at the end of a year’s instruction using this textbook, and compare the results. Are the groups in this study independent or matched? Use the following information to answer the next two exercises. You are conducting a study of the difference in time at two colleges for undergraduate degree completion. At College A, students take an average of 4.8 years to complete an undergraduate degree, while at College B, they take an average of 4.2 years. The pooled standard deviation for this data is 1.6 years 55 . Calculate Cohen’s d and interpret it. 56 . Suppose the mean time to earn an undergraduate degree at College A was 5.2 years. Calculate the effect size and interpret it. 57 . You conduct an independent-samples t-test with sample size ten in each of two groups. If you are conducting a two-tailed hypothesis test with α = 0.01, what p-values will cause you to reject the null hypothesis? 58 . You conduct an independent samples t -test with sample size 15 in each group, with the following hypotheses: H 0 : μ ≥ 110 H a : μ < 110 If α = 0.05, what t -values will cause you to reject the null hypothesis? 10.2: Comparing Two Independent Population Means with Known Population Standard Deviations Use the following information to answer the next six exercises. College students in the sciences often complain that they must spend more on textbooks each semester than students in the humanities. To test this, you draw random samples of 50 science and 50 humanities students from your college, and record how much each spent last semester on textbooks. Consider the science students to be group one, and the humanities students to be group two. 59 . What is the random variable for this study? 60 . What are the null and alternative hypotheses for this study? 61 . If the 50 science students spent an average of $530 with a sample standard deviation of $20 and the 50 humanities students spent an average of $380 with a sample standard deviation of $15, would you not reject or reject the null hypothesis? Use an alpha level of 0.05. What is your conclusion? 62 . What would be your decision, if you were using α = 0.01? 10.3: Comparing Two Independent Population Proportions Use the information to answer the next six exercises. You want to know if proportion of homes with cable television service differs between Community A and Community B. To test this, you draw a random sample of 100 for each and record whether they have cable service. 63 . What are the null and alternative hypotheses for this study 64 . If 65 households in Community A have cable service, and 78 households in community B, what is the pooled proportion? 65 . At α = 0.05, will you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed. 66 . Using an alpha value of 0.01, would you reject the null hypothesis? What is your conclusion? 65 households in Community A have cable service, and 78 households in community B. 100 households in each community were surveyed. 10.4: Matched or Paired Samples Use the following information to answer the next five exercises. You are interested in whether a particular exercise program helps people lose weight. You conduct a study in which you weigh the participants at the start of the study, and again at the conclusion, after they have participated in the exercise program for six months. You compare the results using a matched-pairs t-test, in which the data is {weight at conclusion – weight at start}. You believe that, on average, the participants will have lost weight after six months on the exercise program. 67 . What are the null and alternative hypotheses for this study? 68 . Calculate the test statistic, assuming that x ¯ d = –5, s d = 6, and n = 30 (pairs). 69 . What are the degrees of freedom for this statistic? 70 . Using α = 0.05, what is your decision regarding the effectiveness of this program in causing weight loss? What is the conclusion? 71 . What would it mean if the t -statistic had been 4.56, and what would have been your decision in that case? 11.1: Facts About the Chi-Square Distribution 72 . What is the mean and standard deviation for a chi-square distribution with 20 degrees of freedom? 11.2: Goodness-of-Fit Test Use the following information to answer the next four exercises. Nationally, about 66 percent of high school graduates enroll in higher education. You perform a chi-square goodness of fit test to see if this same proportion applies to your high school’s most recent graduating class of 200. Your null hypothesis is that the national distribution also applies to your high school. 73 . What are the expected numbers of students from your high school graduating class enrolled and not enrolled in higher education? 74 . Fill out the rest of this table. Observed ( O ) Expected ( E ) O – E ( O – E )2 ( O − E ) 2 z Enrolled 145 Not enrolled 55 75 . What are the degrees of freedom for this chi-square test? 76 . What is the chi-square test statistic and the p -value. At the 5% significance level, what do you conclude? 77 . For a chi-square distribution with 92 degrees of freedom, the curve _____________. 78 . For a chi-square distribution with five degrees of freedom, the curve is ______________. 11.3: Test of Independence Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for first-year students and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing. 79 . Compute the expected values for the cells. Cell = Yes Cell = No First-year 100 150 Senior 200 50 80 . Compute ( O − E ) 2 z for each cell, where O = observed and E = expected. 81 . What is the chi-square statistic and degrees of freedom for this study? 82 . At the α = 0.5 significance level, what is your decision regarding the null hypothesis? 11.4: Test of Homogeneity 83 . You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test? 11.5: Comparison Summary of the Chi-Square Tests: Goodness-of-Fit, Independence and Homogeneity 84 . A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level. Ethnic Group \\ Response Type Favor Oppose No Opinion Row Total White / Non-Hispanic 234 433 43 710 Latino 147 106 19 272 African American 24 41 6 71 Asian American 54 48 16 118 Column Total 459 628 84 1171 85 . In a test of homogeneity, what must be true about the expected value of each cell? 86 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence? 87 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity? 11.6: Test of a Single Variance 88 . A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this? Practice Test 3 Solutions 8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal 1 . σ n = 4 30 = 0.73 2 . normal 3 . 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution. 4 . z -score = 1.96; E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.73 ) = 1.4308 5 . 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding. 6 . The z -value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085. The 90% confidence interval is 41 ± 1.20 = (39.80, 42.20). The calculator function Zinterval answer is (40.78, 41.23). Answers differ due to rounding. 7 . The standard error of measurement is: σ n = 4 50 = 0.57 E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.57 ) = 1.12 The 95% confidence interval is 41 ± 1.12 = (39.88, 42.12). The calculator function Zinterval answer is (40.84, 41.16). Answers differ due to rounding. 8 . The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution. 8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t 9 . The t -distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”). 10 . Both distributions are symmetrical and centered at zero. 11 . df = n – 1 = 20 – 1 = 19 12 . You can get the t -value from a probability table or a calculator. In this case, for a t -distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e., t α 2 = 2.093 . The calculator function is invT(0.975, 19). 13 . E B M = t α 2 ( s n ) = ( 2.093 ) ( 0.3 20 ) = 0.140 98.4 ± 0.14 = (98.26, 98.54). The calculator function Tinterval answer is (98.26, 98.54). 14 . t α 2 = 2.861. The calculator function is invT(0.995, 19). E B M = t α 2 ( s n ) = ( 2.861 ) ( 0.3 20 ) = 0.192 98.4 ± 0.19 = (98.21, 98.59). The calculator function Tinterval answer is (98.21, 98.59). 15 . df = n – 1 = 30 – 1 = 29. t α 2 = 2.045 E B M = z t ( s n ) = ( 2.045 ) ( 0.3 30 ) = 0.112 98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51). 8.3: Confidence Interval for a Population Proportion 16 . p ′ = 280 500 = 0.56 q ′ = 1 − p ′ = 1 − 0.56 = 0.44 s = p q n = 0.56 ( 0.44 ) 500 = 0.0222 17 . Because you are using the normal approximation to the binomial, z α 2 = 1.96 . Calculate the error bound for the population ( EBP ): E B P = z a 2 p q n = 1.96 ( 0.222 ) = 0.0435 Calculate the 95% confidence interval: 0.56 ± 0.0435 = (0.5165, 0.6035). The calculator function 1-PropZint answer is (0.5165, 0.6035). 18 . z α 2 = 1.64 E B P = z a 2 p q n = 1.64 ( 0.0222 ) = 0.0364 0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965) 19 . z α 2 = 2.58 E B P = z a 2 p q n = 2.58 ( 0.0222 ) = 0.0573 0.56 ± 0.05 = (0.5127, 0.6173). The calculator function 1-PropZint answer is (0.5028, 0.6172). 20 . EBP = 0.04 (because 4% = 0.04) z α 2 = 1.96 for a 95% confidence interval n = z 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.04 2 = 0.9604 0.0016 = 600.25 You need 601 subjects (rounding upward from 600.25). 21 . n = n 2 p q E B P 2 = 1.96 2 ( 0.6 ) ( 0.4 ) 0.04 2 = 0.9220 0.0016 = 576.24 You need 577 subjects (rounding upward from 576.24). 22 . n = n 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.03 2 = 0.9604 0.0009 = 1067.11 You need 1,068 subjects (rounding upward from 1,067.11). 9.1: Null and Alternate Hypotheses 23 . H 0 : p = 0.58 H a : p ≠ 0.58 24 . H 0 : p ≥ 0.58 H a : p < 0.58 25 . H 0 : μ ≥ $268,000 H a : μ < $268,000 26 . H a : μ ≠ 107 27 . H a : p ≥ 0.25 9.2: Outcomes and the Type I and Type II Errors 28 . a Type I error 29 . a Type II error 30 . Power = 1 – β = 1 – P (Type II error). 31 . The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment. 32 . The screening test has a ten percent probability of a Type I error, meaning that ten percent of the time, it will detect TB when it is not present. 33 . The screening test has a 20 percent probability of a Type II error, meaning that 20 percent of the time, it will fail to detect TB when it is in fact present. 34 . Eighty percent of the time, the screening test will detect TB when it is actually present. 9.3: Distribution Needed for Hypothesis Testing 35 . The Student’s t -test. 36 . The normal distribution or z -test. 37 . The normal distribution with μ = p and σ = p q n 38 . t 24 . You use the t -distribution because you don’t know the population standard deviation, and the degrees of freedom are 24 because df = n – 1. 39 . X ¯ ~ N ( 0.95 , 0.051 100 ) Because you know the population standard deviation, and have a large sample, you can use the normal distribution. 9.4: Rare Events, the Sample, Decision, and Conclusion 40 . Fail to reject the null hypothesis, because α ≤ p 41 . Reject the null hypothesis, because α ≥ p . 42 . H 0 : μ ≥ 29.0” H a : μ < 29.0” 43 . t 19 . Because you do not know the population standard deviation, use the t -distribution. The degrees of freedom are 19, because df = n – 1. 44 . The test statistic is –4.4721 and the p -value is 0.00013 using the calculator function TTEST. 45 . With α = 0.05, reject the null hypothesis. 46 . With α = 0.05, the p -value is almost zero using the calculator function TTEST so reject the null hypothesis. 9.5: Additional Information and Full Hypothesis Test Examples 47 . The level of significance is five percent. 48 . two-tailed 49 . one-tailed 50 . H 0 : p = 0.8 H a : p ≠ 0.8 51 . You will use the normal test for a single population proportion because np and nq are both greater than five. 10.1: Comparing Two Independent Population Means with Unknown Population Standard Deviations 52 . They are matched (paired), because you interviewed married couples. 53 . They are independent, because participants were assigned at random to the groups. 54 . They are matched (paired), because you collected data twice from each individual. 55 . d = x ¯ 1 − x ¯ 2 s p o o l e d = 4.8 − 4.2 1.6 = 0.375 This is a small effect size, because 0.375 falls between Cohen’s small (0.2) and medium (0.5) effect sizes. 56 . d = x ¯ 1 − x ¯ 2 s p o o l e d = 5.2 − 4.2 1.6 = 0.625 The effect size is 0.625. By Cohen’s standard, this is a medium effect size, because it falls between the medium (0.5) and large (0.8) effect sizes. 57 . p -value < 0.01. 58 . You will only reject the null hypothesis if you get a value significantly below the hypothesized mean of 110. 10.2: Comparing Two Independent Population Means with Known Population Standard Deviations 59 . X ¯ 1 − X ¯ 2 , i.e., the mean difference in amount spent on textbooks for the two groups. 60 . H 0 : X ¯ 1 − X ¯ 2 ≤ 0 H a : X ¯ 1 − X ¯ 2 > 0 This could also be written as: H 0 : X ¯ 1 ≤ X ¯ 2 H a : X ¯ 1 > X ¯ 2 61 . Using the calculator function 2-SampTtest, reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students. 62 . Using the calculator function 2-SampTtest, reject the null hypothesis. At the 1% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students. 10.3: Comparing Two Independent Population Proportions 63 . H 0 : p A = p B H a : p A ≠ p B 64 . p c = x A + x A n A + n A = 65 + 78 100 + 100 = 0.715 65 . Using the calculator function 2-PropZTest, the p-value = 0.0417. Reject the null hypothesis. At the 5% significance level, here is sufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service. 66 . Using the calculator function 2-PropZTest, the p -value = 0.0417. Do not reject the null hypothesis. At the 1% significance level, there is insufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service. 10.4: Matched or Paired Samples 67 . H 0 : x ¯ d ≥ 0 H a : x ¯ d < 0 68 . t = – 4.5644 69 . df = 30 – 1 = 29. 70 . Using the calculator function TTEST, the p -value = 0.00004 so reject the null hypothesis. At the 5% level, there is sufficient evidence to conclude that the participants lost weight, on average. 71 . A positive t -statistic would mean that participants, on average, gained weight over the six months. 11.1: Facts About the Chi-Square Distribution 72 . μ = df = 20 σ = 2 ( d f ) = 40 = 6.32 11.2: Goodness-of-Fit Test 73 . Enrolled = 200(0.66) = 132. Not enrolled = 200(0.34) = 68 74 . Observed (O) Expected (E) O – E (O – E)2 ( O − E ) 2 z Enrolled 145 132 145 – 132 = 13 169 169 132 = 1.280 Not enrolled 55 68 55 – 68 = –13 169 169 68 = 2.485 75 . df = n – 1 = 2 – 1 = 1. 76 . Using the calculator function Chi-square GOF – Test (in STAT TESTS), the test statistic is 3.7656 and the p-value is 0.0523. Do not reject the null hypothesis. At the 5% significance level, there is insufficient evidence to conclude that high school most recent graduating class distribution of enrolled and not enrolled does not fit that of the national distribution. 77 . approximates the normal 78 . skewed right 11.3: Test of Independence 79 . Cell = Yes Cell = No Total First-year 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250 Senior 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250 Total 300 200 500 80 . ( 100 − 150 ) 2 150 = 16.67 ( 150 − 100 ) 2 100 = 25 ( 200 − 100 ) 2 150 = 16.67 ( 50 − 100 ) 2 100 = 25 81 . Chi-square = 16.67 + 25 + 16.67 + 25 = 83.34. df = ( r – 1)( c – 1) = 1 82 . p -value = P (Chi-square, 83.34) = 0 Reject the null hypothesis. You could also use the calculator function STAT TESTS Chi-Square – Test. 11.4: Test of Homogeneity 83 . The table has five rows and two columns. df = ( r – 1)( c – 1) = (4)(1) = 4. 11.5: Comparison Summary of the Chi-Square Tests: Goodness-of-Fit, Independence and Homogeneity 84 . Using the calculator function (STAT TESTS) Chi-square Test, the p -value = 0. Reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the poll responses independent of the participants’ ethnic group. 85 . The expected value of each cell must be at least five. 86 . H 0 : The variables are independent. H a : The variables are not independent. 87 . H 0 : The populations have the same distribution. H a : The populations do not have the same distribution. 11.6: Test of a Single Variance 88 . H 0 : σ 2 ≤ 5 H a : σ 2 > 5 Practice Test 4 12.1 Linear Equations 1 . Which of the following equations is/are linear? y = –3 x y = 0.2 + 0.74 x y = –9.4 – 2 x A and B A, B, and C 2 . To complete a painting job requires four hours setup time plus one hour per 1,000 square feet. How would you express this information in a linear equation? 3 . A statistics instructor is paid a per-class fee of $2,000 plus $100 for each student in the class. How would you express this information in a linear equation? 4 . A tutoring school requires students to pay a one-time enrollment fee of $500 plus tuition of $3,000 per year. Express this information in an equation. 12.2: Slope and Y -intercept of a Linear Equation Use the following information to answer the next four exercises. For the labor costs of doing repairs, an auto mechanic charges a flat fee of $75 per car, plus an hourly rate of $55. 5 . What are the independent and dependent variables for this situation? 6 . Write the equation and identify the slope and intercept. 7 . What is the labor charge for a job that takes 3.5 hours to complete? 8 . One job takes 2.4 hours to complete, while another takes 6.3 hours. What is the difference in labor costs for these two jobs? 12.3: Scatter Plots 9 . Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression. 10 . Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression. 11 . Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression. 12 . Describe the pattern in this scatter plot, and decide whether the X and Y variables would be good candidates for linear regression. 12.4: The Regression Equation Use the following information to answer the next four exercises. Height (in inches) and weight (In pounds) in a sample of college students have a linear relationship with the following summary statistics: x ¯ = 68.4 y ¯ =141.6 s x = 4.0 s y = 9.6 r = 0.73 Let Y = weight and X = height, and write the regression equation in the form: y ^ = a + b x 13 . What is the value of the slope? 14 . What is the value of the y intercept? 15 . Write the regression equation predicting weight from height in this data set, and calculate the predicted weight for someone 68 inches tall. 12.5: Correlation Coefficient and Coefficient of Determination 16 . The correlation between body weight and fuel efficiency (measured as miles per gallon) for a sample of 2,012 model cars is –0.56. Calculate the coefficient of determination for this data and explain what it means. 17 . The correlation between high school GPA and first-year college GPA for a sample of 200 university students is 0.32. How much variation in first-year college GPA is not explained by high school GPA? 18 . Rounded to two decimal places what correlation between two variables is necessary to have a coefficient of determination of at least 0.50? 12.6: Testing the Significance of the Correlation Coefficient 19 . Write the null and alternative hypotheses for a study to determine if two variables are significantly correlated. 20 . In a sample of 30 cases, two variables have a correlation of 0.33. Do a t -test to see if this result is significant at the α = 0.05 level. Use the formula: t = r n − 2 1 − r 2 21 . In a sample of 25 cases, two variables have a correlation of 0.45. Do a t -test to see if this result is significant at the α = 0.05 level. Use the formula: t = r n − 2 1 − r 2 12.7: Prediction Use the following information to answer the next two exercises. A study relating the grams of potassium ( Y ) to the grams of fiber ( X ) per serving in enriched flour products (bread, rolls, etc.) produced the equation: y ^ = 25 + 16 x 22 . For a product with five grams of fiber per serving, what are the expected grams of potassium per serving? 23 . Comparing two products, one with three grams of fiber per serving and one with six grams of fiber per serving, what is the expected difference in grams of potassium per serving? 12.8: Outliers 24 . In the context of regression analysis, what is the definition of an outlier, and what is a rule of thumb to evaluate if a given value in a data set is an outlier? 25 . In the context of regression analysis, what is the definition of an influential point, and how does an influential point differ from an outlier? 26 . The least squares regression line for a data set is y ^ = 5 + 0.3 x and the standard deviation of the residuals is 0.4. Does a case with the values x = 2, y = 6.2 qualify as an outlier? 27 . The least squares regression line for a data set is y ^ = 2.3 − 0.1 x and the standard deviation of the residuals is 0.13. Does a case with the values x = 4.1, y = 2.34 qualify as an outlier? 13.1: One-Way ANOVA 28 . What are the five basic assumptions to be met if you want to do a one-way ANOVA? 29 . You are conducting a one-way ANOVA comparing the effectiveness of four drugs in lowering blood pressure in hypertensive patients. What are the null and alternative hypotheses for this study? 30 . What is the primary difference between the independent samples t -test and one-way ANOVA? 31 . You are comparing the results of three methods of teaching geometry to high school students. The final exam scores X 1 , X 2 , X 3 , for the samples taught by the different methods have the following distributions: X 1 ~ N (85, 3.6) X 1 ~ N (82, 4.8) X 1 ~ N (79, 2.9) Each sample includes 100 students, and the final exam scores have a range of 0–100. Assuming the samples are independent and randomly selected, have the requirements for conducting a one-way ANOVA been met? Explain why or why not for each assumption. 32 . You conduct a study comparing the effectiveness of four types of fertilizer to increase crop yield on wheat farms. When examining the sample results, you find that two of the samples have an approximately normal distribution, and two have an approximately uniform distribution. Is this a violation of the assumptions for conducting a one-way ANOVA? 13.2: The F Distribution Use the following information to answer the next seven exercises. You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain among calves whose feed includes one of the supplements. You have four groups of 30 calves (one is a control group receiving the usual feed, but no supplement). You will conduct a one-way ANOVA after one year to see if there are difference in the mean weight for the four groups. 33 . What is SS within in this experiment, and what does it mean? 34 . What is SS between in this experiment, and what does it mean? 35 . What are k and i for this experiment? 36 . If SS within = 374.5 and SS total = 621.4 for this data, what is SS between ? 37 . What are MS between , and MS within , for this experiment? 38 . What is the F Statistic for this data? 39 . If there had been 35 calves in each group, instead of 30, with the sums of squares remaining the same, would the F Statistic be larger or smaller? 13.3: Facts About the F Distribution 40 . Which of the following numbers are possible F Statistics? 2.47 5.95 –3.61 7.28 0.97 41 . Histograms F 1 and F 2 below display the distribution of cases from samples from two populations, one distributed F 3,15 and one distributed F 5,500 . Which sample came from which population? 42 . The F Statistic from an experiment with k = 3 and n = 50 is 3.67. At α = 0.05, will you reject the null hypothesis? 43 . The F Statistic from an experiment with k = 4 and n = 100 is 4.72. At α = 0.01, will you reject the null hypothesis? 13.4: Test of Two Variances 44 . What assumptions must be met to perform the F test of two variances? 45 . You believe there is greater variance in grades given by the math department at your university than in the English department. You collect all the grades for undergraduate classes in the two departments for a semester, and compute the variance of each, and conduct an F test of two variances. What are the null and alternative hypotheses for this study? Practice Test 4 Solutions 12.1 Linear Equations 1 . e. A, B, and C. All three are linear equations of the form y = mx + b . 2 . Let y = the total number of hours required, and x the square footage, measured in units of 1,000. The equation is: y = x + 4 3 . Let y = the total payment, and x the number of students in a class. The equation is: y = 100( x ) + 2,000 4 . Let y = the total cost of attendance, and x the number of years enrolled. The equation is: y = 3,000( x ) + 500 12.2: Slope and Y-intercept of a Linear Equation 5 . The independent variable is the hours worked on a car. The dependent variable is the total labor charges to fix a car. 6 . Let y = the total charge, and x the number of hours required. The equation is: y = 55 x + 75 The slope is 55 and the intercept is 75. 7 . y = 55(3.5) + 75 = 267.50 8 . Because the intercept is included in both equations, while you are only interested in the difference in costs, you do not need to include the intercept in the solution. The difference in number of hours required is: 6.3 – 2.4 = 3.9. Multiply this difference by the cost per hour: 55(3.9) = 214.5. The difference in cost between the two jobs is $214.50. 12.3: Scatter Plots 9 . The X and Y variables have a strong linear relationship. These variables would be good candidates for analysis with linear regression. 10 . The X and Y variables have a strong negative linear relationship. These variables would be good candidates for analysis with linear regression. 11 . There is no clear linear relationship between the X and Y variables, so they are not good candidates for linear regression. 12 . The X and Y variables have a strong positive relationship, but it is curvilinear rather than linear. These variables are not good candidates for linear regression. 12.4: The Regression Equation 13 . r ( s y s x ) = 0.73 ( 9.6 4.0 ) = 1.752 ≈ 1.75 14 . a = y ¯ − b x ¯ = 141.6 − 1.752 ( 68.4 ) = 21.7632 ≈ 21.76 15 . y ^ = 21.76 + 1.75 ( 68 ) = 140.76 12.5: Correlation Coefficient and Coefficient of Determination 16 . The coefficient of determination is the square of the correlation, or r 2 . For this data, r 2 = (–0.56)2 = 0.3136 ≈ 0.31 or 31%. This means that 31 percent of the variation in fuel efficiency can be explained by the bodyweight of the automobile. 17 . The coefficient of determination = 0.32 2 = 0.1024. This is the amount of variation in first-year college GPA that can be explained by high school GPA. The amount that cannot be explained is 1 – 0.1024 = 0.8976 ≈ 0.90. So about 90 percent of variance in first-year college GPA in this data is not explained by high school GPA. 18 . r = r 2 0.5 = 0.707106781 ≈ 0.71 You need a correlation of 0.71 or higher to have a coefficient of determination of at least 0.5. 12.6: Testing the Significance of the Correlation Coefficient 19 . H 0 : ρ = 0 H a : ρ ≠ 0 20 . t = r n − 2 1 − r 2 = 0.33 30 − 2 1 − 0.33 2 = 1.85 The critical value for α = 0.05 for a two-tailed test using the t 29 distribution is 2.045. Your value is less than this, so you fail to reject the null hypothesis and conclude that the study produced no evidence that the variables are significantly correlated. Using the calculator function tcdf, the p -value is 2tcdf(1.85, 10^99, 29) = 0.0373. Do not reject the null hypothesis and conclude that the study produced no evidence that the variables are significantly correlated. 21 . t = r n − 2 1 − r 2 = 0.45 25 − 2 1 − 0.45 2 = 2.417 The critical value for α = 0.05 for a two-tailed test using the t 24 distribution is 2.064. Your value is greater than this, so you reject the null hypothesis and conclude that the study produced evidence that the variables are significantly correlated. Using the calculator function tcdf, the p-value is 2tcdf(2.417, 10^99, 24) = 0.0118. Reject the null hypothesis and conclude that the study produced evidence that the variables are significantly correlated. 12.7: Prediction 22 . y ^ = 25 + 16 ( 5 ) = 105 23 . Because the intercept appears in both predicted values, you can ignore it in calculating a predicted difference score. The difference in grams of fiber per serving is 6 – 3 = 3 and the predicted difference in grams of potassium per serving is (16)(3) = 48. 12.8: Outliers 24 . An outlier is an observed value that is far from the least squares regression line. A rule of thumb is that a point more than two standard deviations of the residuals from its predicted value on the least squares regression line is an outlier. 25 . An influential point is an observed value in a data set that is far from other points in the data set, in a horizontal direction. Unlike an outlier, an influential point is determined by its relationship with other values in the data set, not by its relationship to the regression line. 26 . The predicted value for y is: y ^ = 5 + 0.3 x = 5.6 . The value of 6.2 is less than two standard deviations from the predicted value, so it does not qualify as an outlier. Residual for (2, 6.2): 6.2 – 5.6 = 0.6 (0.6 < 2(0.4)) 27 . The predicted value for y is: y ^ = 2.3 – 0.1(4.1) = 1.89. The value of 2.32 is more than two standard deviations from the predicted value, so it qualifies as an outlier. Residual for (4.1, 2.34): 2.32 – 1.89 = 0.43 (0.43 > 2(0.13)) 13.1: One-Way ANOVA 28 . Each sample is drawn from a normally distributed population All samples are independent and randomly selected. The populations from which the samples are draw have equal standard deviations. The factor is a categorical variable. The response is a numerical variable. 29 . H 0 : μ 1 = μ 2 = μ 3 = μ 4 H a : At least two of the group means μ 1, μ 2, μ 3, μ 4 are not equal. 30 . The independent samples t -test can only compare means from two groups, while one-way ANOVA can compare means of more than two groups. 31 . Each sample appears to have been drawn from a normally distributed populations, the factor is a categorical variable (method), the outcome is a numerical variable (test score), and you were told the samples were independent and randomly selected, so those requirements are met. However, each sample has a different standard deviation, and this suggests that the populations from which they were drawn also have different standard deviations, which is a violation of an assumption for one-way ANOVA. Further statistical testing will be necessary to test the assumption of equal variance before proceeding with the analysis. 32 . One of the assumptions for a one-way ANOVA is that the samples are drawn from normally distributed populations. Since two of your samples have an approximately uniform distribution, this casts doubt on whether this assumption has been met. Further statistical testing will be necessary to determine if you can proceed with the analysis. 13.2: The F Distribution 33 . SS within is the sum of squares within groups, representing the variation in outcome that cannot be attributed to the different feed supplements, but due to individual or chance factors among the calves in each group. 34 . SS between is the sum of squares between groups, representing the variation in outcome that can be attributed to the different feed supplements. 35 . k = the number of groups = 4 n 1 = the number of cases in group 1 = 30 n = the total number of cases = 4(30) = 120 36 . SS total = SS within + SS between so SS between = SS total – SS within 621.4 – 374.5 = 246.9 37 . The mean squares in an ANOVA are found by dividing each sum of squares by its respective degrees of freedom ( df ). For SS total , df = n – 1 = 120 – 1 = 119. For SS between , df = k – 1 = 4 – 1 = 3. For SS within , df = 120 – 4 = 116. MS between = 246.9 3 = 82.3 MS within = 374.5 116 = 3.23 38 . F = M S b e t w e e n M S w i t h i n = 82.3 3.23 = 25.48 39 . It would be larger, because you would be dividing by a smaller number. The value of MS between would not change with a change of sample size, but the value of MS within would be smaller, because you would be dividing by a larger number ( df within would be 136, not 116). Dividing a constant by a smaller number produces a larger result. 13.3: Facts About the F Distribution 40 . All but choice c, –3.61. F Statistics are always greater than or equal to 0. 41 . As the degrees of freedom increase in an F distribution, the distribution becomes more nearly normal. Histogram F 2 is closer to a normal distribution than histogram F 1, so the sample displayed in histogram F 1 was drawn from the F 3,15 population, and the sample displayed in histogram F 2 was drawn from the F 5,500 population. 42 . Using the calculator function Fcdf, p -value = Fcdf(3.67, 1E, 3,50) = 0.0182. Reject the null hypothesis. 43 . Using the calculator function Fcdf, p -value = Fcdf(4.72, 1E, 4, 100) = 0.0016 Reject the null hypothesis. 13.4: Test of Two Variances 44 . The samples must be drawn from populations that are normally distributed, and must be drawn from independent populations. 45 . Let σ M 2 = variance in math grades, and σ E 2 = variance in English grades. H 0 : σ M 2 ≤ σ E 2 H a : σ M 2 > σ E 2 Practice Final Exam 1 Use the following information to answer the next two exercises: An experiment consists of tossing two, 12-sided dice (the numbers 1–12 are printed on the sides of each die). Let Event A = both dice show an even number. Let Event B = both dice show a number more than eight 1 . Events A and B are: mutually exclusive. independent. mutually exclusive and independent. neither mutually exclusive nor independent. 2 . Find P ( A | B ). 2 4 16 144 4 16 2 144 3 . Which of the following are TRUE when we perform a hypothesis test on matched or paired samples? Sample sizes are almost never small. Two measurements are drawn from the same pair of individuals or objects. Two sample means are compared to each other. Answer choices b and c are both true. Use the following information to answer the next two exercises: One hundred eighteen students were asked what type of color their bedrooms were painted: light colors, dark colors, or vibrant colors. The results were tabulated according to gender. Light colors Dark colors Vibrant colors Female 20 22 28 Male 10 30 8 4 . Find the probability that a randomly chosen student is male or has a bedroom painted with light colors. 10 118 68 118 48 118 10 48 5 . Find the probability that a randomly chosen student is male given the student’s bedroom is painted with dark colors. 30 118 30 48 22 118 30 52 Use the following information to answer the next two exercises: We are interested in the number of times a teenager must be reminded to do their chores each week. A survey of 40 parents and guardians was conducted. shows the results of the survey. x P ( x ) 0 2 40 1 5 40 2 3 14 40 4 7 40 5 4 40 6 . Find the probability that a teenager is reminded two times. 8 8 40 6 40 2 7 . Find the expected number of times a teenager is reminded to do their chores. 15 2.78 1.0 3.13 Use the following information to answer the next two exercises: On any given day, approximately 37.5% of the cars parked in the De Anza parking garage are parked crookedly. We randomly survey 22 cars. We are interested in the number of cars that are parked crookedly. 8 . For every 22 cars, how many would you expect to be parked crookedly, on average? 8.25 11 18 7.5 9 . What is the probability that at least ten of the 22 cars are parked crookedly. 0.1263 0.1607 0.2870 0.8393 10 . Using a sample of 15 Stanford-Binet IQ scores, we wish to conduct a hypothesis test. Our claim is that the mean IQ score on the Stanford-Binet IQ test is more than 100. It is known that the standard deviation of all Stanford-Binet IQ scores is 15 points. The correct distribution to use for the hypothesis test is: Binomial Student's t Normal Uniform Use the following information to answer the next three exercises: De Anza College keeps statistics on the pass rate of students who enroll in math classes. In a sample of 1,795 students enrolled in Math 1A (1st quarter calculus), 1,428 passed the course. In a sample of 856 students enrolled in Math 1B (2nd quarter calculus), 662 passed. In general, are the pass rates of Math 1A and Math 1B statistically the same? Let A = the subscript for Math 1A and B = the subscript for Math 1B. 11 . If you were to conduct an appropriate hypothesis test, the alternate hypothesis would be: H a : p A = p B H a : p A > p B H o : p A = p B H a : p A ≠ p B 12 . The Type I error is to: conclude that the pass rate for Math 1A is the same as the pass rate for Math 1B when, in fact, the pass rates are different. conclude that the pass rate for Math 1A is different than the pass rate for Math 1B when, in fact, the pass rates are the same. conclude that the pass rate for Math 1A is greater than the pass rate for Math 1B when, in fact, the pass rate for Math 1A is less than the pass rate for Math 1B. conclude that the pass rate for Math 1A is the same as the pass rate for Math 1B when, in fact, they are the same. 13 . The correct decision is to: reject H 0 not reject H 0 There is not enough information given to conduct the hypothesis test Kia, Alejandra, and Iris are runners on the track teams at three different schools. Their running times, in minutes, and the statistics for the track teams at their respective schools, for a one mile run, are given in the table below: Running Time School Average Running Time School Standard Deviation Kia 4.9 5.2 0.15 Alejandra 4.2 4.6 0.25 Iris 4.5 4.9 0.12 14 . Which student is the BEST when compared to the other runners at her school? Kia Alejandra Iris Impossible to determine Use the following information to answer the next two exercises: The following adult ski sweater prices are from the Gorsuch Ltd. Winter catalog: $212, $292, $278, $199, $280, $236 Assume the underlying sweater price population is approximately normal. The null hypothesis is that the mean price of adult ski sweaters from Gorsuch Ltd. is at least $275. 15 . The correct distribution to use for the hypothesis test is: Normal Binomial Student's t Exponential 16 . The hypothesis test: is two-tailed. is left-tailed. is right-tailed. has no tails. 17 . Sara, a statistics student, wanted to determine the mean number of books that college professors have in their office. She randomly selected two buildings on campus and asked each professor in the selected buildings how many books are in their office. Sara surveyed 25 professors. The type of sampling selected is simple random sampling. systematic sampling. cluster sampling. stratified sampling. 18 . A clothing store would use which measure of the center of data when placing orders for the typical \"middle\" customer? mean median mode IQR 19 . In a hypothesis test, the p -value is the probability that an outcome of the data will happen purely by chance when the null hypothesis is true. called the preconceived alpha. compared to beta to decide whether to reject or not reject the null hypothesis. Answer choices A and B are both true. Use the following information to answer the next three exercises: A community college offers classes 6 days a week: Monday through Saturday. Maria conducted a study of the students in her classes to determine how many days per week the students who are in her classes come to campus for classes. In each of her 5 classes she randomly selected 10 students and asked them how many days they come to campus for classes. Each of her classes are the same size. The results of her survey are summarized in . Number of Days on Campus Frequency Relative Frequency Cumulative Relative Frequency 1 2 2 12 .24 3 10 .20 4 .98 5 0 6 1 .02 1.00 20 . Combined with convenience sampling, what other sampling technique did Maria use? simple random systematic cluster stratified 21 . How many students come to campus for classes four days a week? 49 25 30 13 22 . What is the 60 th percentile for the this data? 2 3 4 5 Use the following information to answer the next two exercises: The following data are the results of a random survey of 110 Reservists called to active duty to increase security at California airports. Number of Dependents Frequency 0 11 1 27 2 33 3 20 4 19 23 . Construct a 95% confidence interval for the true population mean number of dependents of Reservists called to active duty to increase security at California airports. (1.85, 2.32) (1.80, 2.36) (1.97, 2.46) (1.92, 2.50) 24 . The 95% confidence interval above means: Five percent of confidence intervals constructed this way will not contain the true population aveage number of dependents. We are 95% confident the true population mean number of dependents falls in the interval. Both of the above answer choices are correct. None of the above. 25 . X ~ U (4, 10). Find the 30 th percentile. 0.3000 3 5.8 6.1 26 . If X ~ Ex p(0.8), then P ( x < μ ) = __________ 0.3679 0.4727 0.6321 cannot be determined 27 . The lifetime of a computer circuit board is normally distributed with a mean of 2,500 hours and a standard deviation of 60 hours. What is the probability that a randomly chosen board will last at most 2,560 hours? 0.8413 0.1587 0.3461 0.6539 28 . A survey of 123 reservists called to active duty as a result of the September 11, 2001, attacks was conducted to determine the proportion that were married. Eighty-six reported being married. Construct a 98% confidence interval for the true population proportion of reservists called to active duty that are married. (0.6030, 0.7954) (0.6181, 0.7802) (0.5927, 0.8057) (0.6312, 0.7672) 29 . Winning times in 26 mile marathons run by world class runners average 145 minutes with a standard deviation of 14 minutes. A sample of the last ten marathon winning times is collected. Let x = mean winning times for ten marathons. The distribution for x is: N ( 145 , 14 10 ) N ( 145 , 14 ) t 9 t 10 30 . Suppose that Phi Beta Kappa honors the top one percent of college and university seniors. Assume that grade point means (GPA) at a certain college are normally distributed with a 2.5 mean and a standard deviation of 0.5. What would be the minimum GPA needed to become a member of Phi Beta Kappa at that college? 3.99 1.34 3.00 3.66 The number of people living on American farms has declined steadily during the 20 th century. Here are data on the farm population (in millions of persons) from 1935 to 1980. Year 1935 1940 1945 1950 1955 1960 1965 1970 1975 1980 Population 32.1 30.5 24.4 23.0 19.1 15.6 12.4 9.7 8.9 7.2 31 . The linear regression equation is y ^ = 1166.93 – 0.5868 x . What was the expected farm population (in millions of persons) for 1980? 7.2 5.1 6.0 8.0 32 . In linear regression, which is the best possible SSE ? 13.46 18.22 24.05 16.33 33 . In regression analysis, if the correlation coefficient is close to one what can be said about the best fit line? It is a horizontal line. Therefore, we can not use it. There is a strong linear pattern. Therefore, it is most likely a good model to be used. The coefficient correlation is close to the limit. Therefore, it is hard to make a decision. We do not have the equation. Therefore, we cannot say anything about it. Use the following information to answer the next three exercises: A study of the career plans of young women and men sent questionnaires to all 722 members of the senior class in the College of Business Administration at the University of Illinois. One question asked which major within the business program the student had chosen. Here are the data from the students who responded. Does the data suggest that there is a relationship between the gender of students and their choice of major? Female Male Accounting 68 56 Administration 91 40 Economics 5 6 Finance 61 59 34 . The distribution for the test is: Chi 2 8 . Chi 2 3 . t 721 . N ( 0 , 1 ) . 35 . The expected number of female who choose finance is: 37. 61. 60. 70. 36 . The p -value is 0.0127 and the level of significance is 0.05. The conclusion to the test is: there is insufficient evidence to conclude that the choice of major and the gender of the student are not independent of each other. there is sufficient evidence to conclude that the choice of major and the gender of the student are not independent of each other. there is sufficient evidence to conclude that students find economics very hard. there is in sufficient evidence to conclude that more females prefer administration than males. 37 . An agency reported that the work force nationwide is composed of 10% professional, 10% clerical, 30% skilled, 15% service, and 35% semiskilled laborers. A random sample of 100 San Jose residents indicated 15 professional, 15 clerical, 40 skilled, 10 service, and 20 semiskilled laborers. At α = 0.10 does the work force in San Jose appear to be consistent with the agency report for the nation? Which kind of test is it? Chi 2 goodness of fit Chi 2 test of independence Independent groups proportions Unable to determine Practice Final Exam 1 Solutions Solutions 1 . b. independent 2 . c. 4 16 3 . b. Two measurements are drawn from the same pair of individuals or objects. 4 . b. 68 118 5 . d. 30 52 6 . b. 8 40 7 . b. 2.78 8 . a. 8.25 9 . c. 0.2870 10 . c. Normal 11 . d. H a : p A ≠ p B 12 . b. conclude that the pass rate for Math 1A is different than the pass rate for Math 1B when, in fact, the pass rates are the same. 13 . b. not reject H 0 14 . c. Iris 15 . c. Student's t 16 . b. is left-tailed. 17 . c. cluster sampling 18 . b. median 19 . a. the probability that an outcome of the data will happen purely by chance when the null hypothesis is true. 20 . d. stratified 21 . b. 25 22 . c. 4 23 . a. (1.85, 2.32) 24 . c. Both above are correct. 25 . c. 5.8 26 . c. 0.6321 27 . a. 0.8413 28 . a. (0.6030, 0.7954) 29 . a. N 145 14 10 30 . d. 3.66 31 . b. 5.1 32 . a. 13.46 33 . b. There is a strong linear pattern. Therefore, it is most likely a good model to be used. 34 . b. Chi 2 3 . 35 . d. 70 36 . b. There is sufficient evidence to conclude that the choice of major and the gender of the student are not independent of each other. 37 . a. Chi 2 goodness-of-fit Practice Final Exam 2 1 . A study was done to determine the proportion of teenagers that own a car. The population proportion of teenagers that own a car is the: statistic. parameter. population. variable. Use the following information to answer the next two exercises: value frequency 0 1 1 4 2 7 3 9 6 4 2 . The box plot for the data is: 3 . If six were added to each value of the data in the table, the 15 th percentile of the new list of values is: six one seven eight Use the following information to answer the next two exercises: Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the probability of water rationing is ten percent. However, in any year, the probability of water rationing is five percent. 4 . What is the probability of both a drought and water rationing occurring? 0.05 0.01 0.02 0.30 5 . Which of the following is true? Drought and water rationing are independent events. Drought and water rationing are mutually exclusive events. None of the above Use the following information to answer the next two exercises: Suppose that a survey yielded the following data: Favorite Pie gender apple pumpkin pecan female 40 10 30 male 20 30 10 6 . Suppose that one individual is randomly chosen. The probability that the person’s favorite pie is apple or the person is male is _____. 40 60 60 140 120 140 100 140 7 . Suppose H 0 is: Favorite pie and gender are independent. The p -value is ______. ≈ 0 1 0.05 cannot be determined Use the following information to answer the next two exercises: Let’s say that the probability that an adult watches the news at least once per week is 0.60. We randomly survey 14 people. Of interest is the number of people who watch the news at least once per week. 8 . Which of the following statements is FALSE? X ~ B (14 0.60) The values for x are: {1 ,2 ,3 ,... ,14}. μ = 8.4 P ( X = 5) = 0.0408 9 . Find the probability that at least six adults watch the news at least once per week. 6 14 0.8499 0.9417 0.6429 10 . The following histogram is most likely to be a result of sampling from which distribution? chi-square with df = 6 exponential uniform binomial 11 . The ages of campus day and evening students is known to be normally distributed. A sample of six campus day and evening students reported their ages (in years) as: {18, 35, 27, 45, 20, 20}. What is the error bound for the 90% confidence interval of the true average age? 11.2 22.3 17.5 8.7 12 . If a normally distributed random variable has µ = 0 and σ = 1, then 97.5% of the population values lie above: –1.96. 1.96. 1. –1. Use the following information to answer the next three exercises. The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the average amount of money a customer spends in one trip to the supermarket is $72. 13 . What is the probability that one customer spends less than $72 in one trip to the supermarket? 0.6321 0.5000 0.3714 1 14 . How much money altogether would you expect the next five customers to spend in one trip to the supermarket (in dollars)? 72 72 2 5 5184 360 15 . If you want to find the probability that the mean amount of money 50 customers spend in one trip to the supermarket is less than $60, the distribution to use is: N (72, 72) N ( 72 , 72 50 ) Exp (72) E x p ( 1 72 ) Use the following information to answer the next three exercises: The amount of time it takes a fourth grader to carry out the trash is uniformly distributed in the interval from one to ten minutes. 16 . What is the probability that a randomly chosen fourth grader takes more than seven minutes to take out the trash? 3 9 7 9 3 10 7 10 17 . Which graph best shows the probability that a randomly chosen fourth grader takes more than six minutes to take out the trash given that they have already taken more than three minutes? 18 . We should expect a fourth grader to take how many minutes to take out the trash? 4.5 5.5 5 10 Use the following information to answer the next three exercises: At the beginning of the quarter, the amount of time a student waits in line at the campus cafeteria is normally distributed with a mean of five minutes and a standard deviation of 1.5 minutes. 19 . What is the 90 th percentile of waiting times (in minutes)? 1.28 90 7.47 6.92 20 . The median waiting time (in minutes) for one student is: 5. 50. 2.5. 1.5. 21 . Find the probability that the average wait time for ten students is at most 5.5 minutes. 0.6301 0.8541 0.3694 0.1459 22 . A sample of 80 software engineers in Silicon Valley is taken and it is found that 20% of them earn approximately $50,000 per year. A point estimate for the true proportion of engineers in Silicon Valley who earn $50,000 per year is: 16. 0.2. 1. 0.95. 23 . If P ( Z < z α ) = 0.1587 where Z ~ N (0, 1), then α is equal to: –1. 0.1587. 0.8413. 1. 24 . A professor tested 35 students to determine their entering skills. At the end of the term, after completing the course, the same test was administered to the same 35 students to study their improvement. This would be a test of: independent groups. two proportions. matched pairs, dependent groups. exclusive groups. A math exam was given to all the third grade children attending ABC School. Two random samples of scores were taken. n x ¯ s Boys 55 82 5 Girls 60 86 7 25 . Which of the following correctly describes the results of a hypothesis test of the claim, “There is a difference between the mean scores obtained by third grade girls and boys at the 5% level of significance”? Do not reject H 0 . There is insufficient evidence to conclude that there is a difference in the mean scores. Do not reject H 0 . There is sufficient evidence to conclude that there is a difference in the mean scores. Reject H 0 . There is insufficient evidence to conclude that there is no difference in the mean scores. Reject H 0 . There is sufficient evidence to conclude that there is a difference in the mean scores. 26 . In a survey of 80 males, 45 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. The correct conclusion is that: there is insufficient information to conclude that the proportion for males is the same as the proportion for females. there is insufficient information to conclude that the proportion for males is not the same as the proportion for females. there is sufficient evidence to conclude that the proportion for males is higher than the proportion for females. not enough information to make a conclusion. 27 . From past experience, a statistics teacher has found that the average score on a midterm is 81 with a standard deviation of 5.2. This term, a class of 49 students had a standard deviation of 5 on the midterm. Do the data indicate that we should reject the teacher’s claim that the standard deviation is 5.2? Use α = 0.05. Yes No Not enough information given to solve the problem 28 . Three loading machines are being compared. Ten samples were taken for each machine. Machine I took an average of 31 minutes to load packages with a standard deviation of two minutes. Machine II took an average of 28 minutes to load packages with a standard deviation of 1.5 minutes. Machine III took an average of 29 minutes to load packages with a standard deviation of one minute. Find the p -value when testing that the average loading times are the same. p -value is close to zero p -value is close to one not enough information given to solve the problem Use the following information to answer the next three exercises: A corporation has offices in different parts of the country. It has gathered the following information concerning the number of bathrooms and the number of employees at seven sites: Number of employees x 650 730 810 900 102 107 1150 Number of bathrooms y 40 50 54 61 82 110 121 29 . Is the correlation between the number of employees and the number of bathrooms significant? Yes No Not enough information to answer question 30 . The linear regression equation is: ŷ = 0.0094 − 79.96 x ŷ = 79.96 + 0.0094 x ŷ = 79.96 − 0.0094 x ŷ = − 0.0094 + 79.96 x 31 . If a site has 1,150 employees, approximately how many bathrooms should it have? 69 91 91,954 We should not be estimating here. 32 . Suppose that a sample of size ten was collected, with x ¯ = 4.4 and s = 1.4. H 0 : σ 2 = 1.6 vs. H a : σ 2 ≠ 1.6. Which graph best describes the results of the test? Sixty-four backpackers were asked the number of days since their latest backpacking trip. The number of days is given in : # of days 1 2 3 4 5 6 7 8 Frequency 5 9 6 12 7 10 5 10 33 . Conduct an appropriate test to determine if the distribution is uniform. The p -value is > 0.10. There is insufficient information to conclude that the distribution is not uniform. The p -value is < 0.01. There is sufficient information to conclude the distribution is not uniform. The p -value is between 0.01 and 0.10, but without alpha ( α ) there is not enough information There is no such test that can be conducted. 34 . Which of the following statements is true when using one-way ANOVA? The populations from which the samples are selected have different distributions. The sample sizes are large. The test is to determine if the different groups have the same means. There is a correlation between the factors of the experiment. Practice Final Exam 2 Solutions Solutions 1 . b. parameter. 2 . a. 3 . c. seven 4 . c. 0.02 5 . c. none of the above 6 . d. 100 140 7 . a. ≈ 0 8 . b. The values for x are: {1, 2, 3,..., 14} 9 . c. 0.9417. 10 . d. binomial 11 . d. 8.7 12 . a. –1.96 13 . a. 0.6321 14 . d. 360 15 . b. N ( 72 , 72 50 ) 16 . a. 3 9 17 . d. 18 . b. 5.5 19 . d. 6.92 20 . a. 5 21 . b. 0.8541 22 . b. 0.2 23 . a. –1. 24 . c. matched pairs, dependent groups. 25 . d. Reject H 0 . There is sufficient evidence to conclude that there is a difference in the mean scores. 26 . c. there is sufficient evidence to conclude that the proportion for males is higher than the proportion for females. 27 . b. no 28 . b. p -value is close to 1. 29 . b. No 30 . c. y ^ = 79.96 x – 0.0094 31 . d. We should not be estimating here. 32 . a. 33 . a. The p -value is > 0.10. There is insufficient information to conclude that the distribution is not uniform. 34 . c. The test is to determine if the different groups have the same means.", "section": "Practice Tests (1-4) and Final Exams", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Data Sets Lap Times The following tables provide lap times from Terri Vogel's log book. Times are recorded in seconds for 2.5-mile laps completed in a series of races and practice runs. Race Lap Times (in seconds) Lap 1 Lap 2 Lap 3 Lap 4 Lap 5 Lap 6 Lap 7 Race 1 135 130 131 132 130 131 133 Race 2 134 131 131 129 128 128 129 Race 3 129 128 127 127 130 127 129 Race 4 125 125 126 125 124 125 125 Race 5 133 132 132 132 131 130 132 Race 6 130 130 130 129 129 130 129 Race 7 132 131 133 131 134 134 131 Race 8 127 128 127 130 128 126 128 Race 9 132 130 127 128 126 127 124 Race 10 135 131 131 132 130 131 130 Race 11 132 131 132 131 130 129 129 Race 12 134 130 130 130 131 130 130 Race 13 128 127 128 128 128 129 128 Race 14 132 131 131 131 132 130 130 Race 15 136 129 129 129 129 129 129 Race 16 129 129 129 128 128 129 129 Race 17 134 131 132 131 132 132 132 Race 18 129 129 130 130 133 133 127 Race 19 130 129 129 129 129 129 128 Race 20 131 128 130 128 129 130 130 Practice Lap Times (in seconds) Lap 1 Lap 2 Lap 3 Lap 4 Lap 5 Lap 6 Lap 7 Practice 1 142 143 180 137 134 134 172 Practice 2 140 135 134 133 128 128 131 Practice 3 130 133 130 128 135 133 133 Practice 4 141 136 137 136 136 136 145 Practice 5 140 138 136 137 135 134 134 Practice 6 142 142 139 138 129 129 127 Practice 7 139 137 135 135 137 134 135 Practice 8 143 136 134 133 134 133 132 Practice 9 135 134 133 133 132 132 133 Practice 10 131 130 128 129 127 128 127 Practice 11 143 139 139 138 138 137 138 Practice 12 132 133 131 129 128 127 126 Practice 13 149 144 144 139 138 138 137 Practice 14 133 132 137 133 134 130 131 Practice 15 138 136 133 133 132 131 131 Stock Prices The following table lists initial public offering (IPO) stock prices for all 1999 stocks that at least doubled in value during the first day of trading. IPO Offer Prices $17.00 $23.00 $14.00 $16.00 $12.00 $26.00 $20.00 $22.00 $14.00 $15.00 $22.00 $18.00 $18.00 $21.00 $21.00 $19.00 $15.00 $21.00 $18.00 $17.00 $15.00 $25.00 $14.00 $30.00 $16.00 $10.00 $20.00 $12.00 $16.00 $17.44 $16.00 $14.00 $15.00 $20.00 $20.00 $16.00 $17.00 $16.00 $15.00 $15.00 $19.00 $48.00 $16.00 $18.00 $9.00 $18.00 $18.00 $20.00 $8.00 $20.00 $17.00 $14.00 $11.00 $16.00 $19.00 $15.00 $21.00 $12.00 $8.00 $16.00 $13.00 $14.00 $15.00 $14.00 $13.41 $28.00 $21.00 $17.00 $28.00 $17.00 $19.00 $16.00 $17.00 $19.00 $18.00 $17.00 $15.00 $14.00 $21.00 $12.00 $18.00 $24.00 $15.00 $23.00 $14.00 $16.00 $12.00 $24.00 $20.00 $14.00 $14.00 $15.00 $14.00 $19.00 $16.00 $38.00 $20.00 $24.00 $16.00 $8.00 $18.00 $17.00 $16.00 $15.00 $7.00 $19.00 $12.00 $8.00 $23.00 $12.00 $18.00 $20.00 $21.00 $34.00 $16.00 $26.00 $14.00 References Data compiled by Jay R. Ritter of University of Florida using data from Securities Data Co. and Bloomberg .", "section": "Data Sets", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Group and Partner Projects Univariate Data Student Learning Objectives The student will design and carry out a survey. The student will analyze and graphically display the results of the survey. Instructions As you complete each task below, check it off. Answer all questions in your summary. ____ Decide what data you are going to study. Here are two examples, but you may NOT use them: number of M&M's per bag, number of pencils students have in their backpacks. ____ Are your data discrete or continuous? How do you know? ____ Decide how you are going to collect the data (for instance, buy 30 bags of M&M's; collect data from the World Wide Web). ____ Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random (using a random number generator) sampling. Do not use convenience sampling. Which method did you use? Why did you pick that method? ____ Conduct your survey. Your data size must be at least 30. ____ Summarize your data in a chart with columns showing data value, frequency, relative frequency and cumulative relative frequency. Answer the following (rounded to two decimal places): x ¯ = _____ s = _____ First quartile = _____ Median = _____ 70 th percentile = _____ ____ What value is two standard deviations above the mean? ____ What value is 1.5 standard deviations below the mean? ____ Construct a histogram displaying your data. ____ In complete sentences, describe the shape of your graph. ____ Do you notice any potential outliers? If so, what values are they? Show your work in how you used the potential outlier formula to determine whether or not the values might be outliers. ____ Construct a box plot displaying your data. ____ Does the middle 50% of the data appear to be concentrated together or spread apart? Explain how you determined this. ____ Looking at both the histogram and the box plot, discuss the distribution of your data. Assignment Checklist You need to turn in the following typed and stapled packet, with pages in the following order: ____ Cover sheet : name, class time, and name of your study ____ Summary page : This should contain paragraphs written with complete sentences. It should include answers to all the questions above. It should also include statements describing the population under study, the sample, a parameter or parameters being studied, and the statistic or statistics produced. ____ URL for data, if your data are from the World Wide Web ____ Chart of data, frequency, relative frequency, and cumulative relative frequency ____ Page(s) of graphs: histogram and box plot Continuous Distributions and Central Limit Theorem Student Learning Objectives The student will collect a sample of continuous data. The student will attempt to fit the data sample to various distribution models. The student will validate the central limit theorem. Instructions As you complete each task below, check it off. Answer all questions in your summary. Part I: Sampling ____ Decide what continuous data you are going to study. (Here are two examples, but you may NOT use them: the amount of money a student spent on college supplies this term, or the length of time distance telephone call lasts.) ____ Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random (using a random number generator) sampling. Do not use convenience sampling. What method did you use? Why did you pick that method? ____ Conduct your survey. Gather at least 150 pieces of continuous, quantitative data . ____ Define (in words) the random variable for your data. X = _______ ____ Create two lists of your data: (1) unordered data, (2) in order of smallest to largest. ____ Find the sample mean and the sample standard deviation (rounded to two decimal places). x ¯ = ______ s = ______ ____ Construct a histogram of your data containing five to ten intervals of equal width. The histogram should be a representative display of your data. Label and scale it. Part II: Possible Distributions ____ Suppose that X followed the following theoretical distributions. Set up each distribution using the appropriate information from your data. ____ Uniform: X ~ U ____________ Use the lowest and highest values as a and b . ____ Normal: X ~ N ____________ Use x ¯ to estimate for μ and s to estimate for σ . ____ Must your data fit one of the above distributions? Explain why or why not. ____ Could the data fit two or three of the previous distributions (at the same time)? Explain. ____ Calculate the value k (an X value) that is 1.75 standard deviations above the sample mean. k = _________ (rounded to two decimal places) Note: k = x ¯ + (1.75) s ____ Determine the relative frequencies ( RF ) rounded to four decimal places. NOTE R F = frequency total number surveyed RF ( X < k ) = ______ RF ( X > k ) = ______ RF ( X = k ) = ______ NOTE You should have one page for the uniform distribution, one page for the exponential distribution, and one page for the normal distribution. ____ State the distribution: X ~ _________ ____ Draw a graph for each of the three theoretical distributions. Label the axes and mark them appropriately. ____ Find the following theoretical probabilities (rounded to four decimal places). P ( X < k ) = ______ P ( X > k ) = ______ P ( X = k ) = ______ ____ Compare the relative frequencies to the corresponding probabilities. Are the values close? ____ Does it appear that the data fit the distribution well? Justify your answer by comparing the probabilities to the relative frequencies, and the histograms to the theoretical graphs. Part III: CLT Experiments ______ From your original data (before ordering), use a random number generator to pick 40 samples of size five. For each sample, calculate the average. ______ On a separate page, attached to the summary, include the 40 samples of size five, along with the 40 sample averages. ______ List the 40 averages in order from smallest to largest. ______ Define the random variable, X ¯ , in words. X ¯ = _______________ ______ State the approximate theoretical distribution of X ¯ . X ¯ ~ ______________ ______ Base this on the mean and standard deviation from your original data. ______ Construct a histogram displaying your data. Use five to six intervals of equal width. Label and scale it. Calculate the value k ¯ (an X ¯ value) that is 1.75 standard deviations above the sample mean. k ¯ = _____ (rounded to two decimal places) Determine the relative frequencies ( RF ) rounded to four decimal places. RF ( X ¯ < k ¯ ) = _______ RF ( X ¯ > k ¯ ) = _______ RF ( X ¯ = k ¯ ) = _______ Find the following theoretical probabilities (rounded to four decimal places). P ( X ¯ < k ¯ ) = _______ P ( X ¯ > k ¯ ) = _______ P ( X ¯ = k ¯ ) = _______ ______ Draw the graph of the theoretical distribution of X . ______ Compare the relative frequencies to the probabilities. Are the values close? ______ Does it appear that the data of averages fit the distribution of X ¯ well? Justify your answer by comparing the probabilities to the relative frequencies, and the histogram to the theoretical graph. In three to five complete sentences for each, answer the following questions. Give thoughtful explanations. ______ In summary, do your original data seem to fit the uniform, exponential, or normal distributions? Answer why or why not for each distribution. If the data do not fit any of those distributions, explain why. ______ What happened to the shape and distribution when you averaged your data? In theory, what should have happened? In theory, would “it” always happen? Why or why not? ______ Were the relative frequencies compared to the theoretical probabilities closer when comparing the X or X ¯ distributions? Explain your answer. Assignment Checklist You need to turn in the following typed and stapled packet, with pages in the following order: ____ Cover sheet : name, class time, and name of your study ____ Summary pages : These should contain several paragraphs written with complete sentences that describe the experiment, including what you studied and your sampling technique, as well as answers to all of the questions previously asked questions ____ URL for data, if your data are from the World Wide Web ____ Pages, one for each theoretical distribution , with the distribution stated, the graph, and the probability questions answered ____ Pages of the data requested ____ All graphs required Hypothesis Testing-Article Student Learning Objectives The student will identify a hypothesis testing problem in print. The student will conduct a survey to verify or dispute the results of the hypothesis test. The student will summarize the article, analysis, and conclusions in a report. Instructions As you complete each task, check it off. Answer all questions in your summary. ____ Find an article in a newspaper, magazine, or on the internet which makes a claim about ONE population mean or ONE population proportion. The claim may be based upon a survey that the article was reporting on. Decide whether this claim is the null or alternate hypothesis. ____ Copy or print out the article and include a copy in your project, along with the source. ____ State how you will collect your data. (Convenience sampling is not acceptable.) ____ Conduct your survey. You must have more than 50 responses in your sample. When you hand in your final project, attach the tally sheet or the packet of questionnaires that you used to collect data. Your data must be real. ____ State the statistics that are a result of your data collection: sample size, sample mean, and sample standard deviation, OR sample size and number of successes. ____ Make two copies of the appropriate solution sheet. ____ Record the hypothesis test on the solution sheet, based on your experiment. Do a DRAFT solution first on one of the solution sheets and check it over carefully. Have a classmate check your solution to see if it is done correctly. Make your decision using a 5% level of significance. Include the 95% confidence interval on the solution sheet. ____ Create a graph that illustrates your data. This may be a pie or bar graph or may be a histogram or box plot, depending on the nature of your data. Produce a graph that makes sense for your data and gives useful visual information about your data. You may need to look at several types of graphs before you decide which is the most appropriate for the type of data in your project. ____ Write your summary (in complete sentences and paragraphs, with proper grammar and correct spelling) that describes the project. The summary MUST include: Brief discussion of the article, including the source Statement of the claim made in the article (one of the hypotheses). Detailed description of how, where, and when you collected the data, including the sampling technique; did you use cluster, stratified, systematic, or simple random sampling (using a random number generator)? As previously mentioned, convenience sampling is not acceptable. Conclusion about the article claim in light of your hypothesis test; this is the conclusion of your hypothesis test, stated in words, in the context of the situation in your project in sentence form, as if you were writing this conclusion for a non-statistician. Sentence interpreting your confidence interval in the context of the situation in your project Assignment Checklist Turn in the following typed (12 point) and stapled packet for your final project: ____ Cover sheet containing your name(s), class time, and the name of your study ____ Summary , which includes all items listed on summary checklist ____ Solution sheet neatly and completely filled out. The solution sheet does not need to be typed. ____ Graphic representation of your data , created following the guidelines previously discussed; include only graphs which are appropriate and useful. ____ Raw data collected AND a table summarizing the sample data ( n , x ¯ and s ; or x , n , and p ’, as appropriate for your hypotheses); the raw data does not need to be typed, but the summary does. Hand in the data as you collected it. (Either attach your tally sheet or an envelope containing your questionnaires.) Bivariate Data, Linear Regression, and Univariate Data Student Learning Objectives The students will collect a bivariate data sample through the use of appropriate sampling techniques. The student will attempt to fit the data to a linear model. The student will determine the appropriateness of linear fit of the model. The student will analyze and graph univariate data. Instructions As you complete each task below, check it off. Answer all questions in your introduction or summary. Check your course calendar for intermediate and final due dates. Graphs may be constructed by hand or by computer, unless your instructor informs you otherwise. All graphs must be neat and accurate. All other responses must be done on the computer. Neatness and quality of explanations are used to determine your final grade. Part I: Bivariate Data Introduction ____State the bivariate data your group is going to study. Here are two examples, but you may NOT use them: height vs. weight and age vs. running distance. ____Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random sampling (using a random number generator) sampling. Convenience sampling is NOT acceptable. ____Conduct your survey. Your number of pairs must be at least 30. ____Print out a copy of your data. Analysis ____On a separate sheet of paper construct a scatter plot of the data. Label and scale both axes. ____State the least squares line and the correlation coefficient. ____On your scatter plot, in a different color, construct the least squares line. ____Is the correlation coefficient significant? Explain and show how you determined this. ____Interpret the slope of the linear regression line in the context of the data in your project. Relate the explanation to your data, and quantify what the slope tells you. ____Does the regression line seem to fit the data? Why or why not? If the data does not seem to be linear, explain if any other model seems to fit the data better. ____Are there any outliers? If so, what are they? Show your work in how you used the potential outlier formula in the Linear Regression and Correlation chapter (since you have bivariate data) to determine whether or not any pairs might be outliers. Part II: Univariate Data In this section, you will use the data for ONE variable only. Pick the variable that is more interesting to analyze. For example: if your independent variable is sequential data such as year with 30 years and one piece of data per year, your x -values might be 1971, 1972, 1973, 1974, …, 2000. This would not be interesting to analyze. In that case, choose to use the dependent variable to analyze for this part of the project. _____Summarize your data in a chart with columns showing data value, frequency, relative frequency, and cumulative relative frequency. _____Answer the following question, rounded to two decimal places: Sample mean = ______ Sample standard deviation = ______ First quartile = ______ Third quartile = ______ Median = ______ 70th percentile = ______ Value that is 2 standard deviations above the mean = ______ Value that is 1.5 standard deviations below the mean = ______ _____Construct a histogram displaying your data. Group your data into six to ten intervals of equal width. Pick regularly spaced intervals that make sense in relation to your data. For example, do NOT group data by age as 20-26,27-33,34-40,41-47,48-54,55-61 . . . Instead, maybe use age groups 19.5-24.5, 24.5-29.5, . . . or 19.5-29.5, 29.5-39.5, 39.5-49.5, . . . _____In complete sentences, describe the shape of your histogram. _____Are there any potential outliers? Which values are they? Show your work and calculations as to how you used the potential outlier formula in Descriptive Statistics (since you are now using univariate data) to determine which values might be outliers. _____Construct a box plot of your data. _____Does the middle 50% of your data appear to be concentrated together or spread out? Explain how you determined this. _____Looking at both the histogram AND the box plot, discuss the distribution of your data. For example: how does the spread of the middle 50% of your data compare to the spread of the rest of the data represented in the box plot; how does this correspond to your description of the shape of the histogram; how does the graphical display show any outliers you may have found; does the histogram show any gaps in the data that are not visible in the box plot; are there any interesting features of your data that you should point out. Due Dates Part I, Intro: __________ (keep a copy for your records) Part I, Analysis: __________ (keep a copy for your records) Entire Project, typed and stapled: __________ ____ Cover sheet: names, class time, and name of your study ____ Part I: label the sections “Intro” and “Analysis.” ____ Part II: ____ Summary page containing several paragraphs written in complete sentences describing the experiment, including what you studied and how you collected your data. The summary page should also include answers to ALL the questions asked above. ____ All graphs requested in the project ____ All calculations requested to support questions in data ____ Description: what you learned by doing this project, what challenges you had, how you overcame the challenges NOTE Include answers to ALL questions asked, even if not explicitly repeated in the items above.", "section": "Group and Partner Projects", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Solution Sheets Hypothesis Testing with One Sample Class Time: __________________________ Name: _____________________________________ H 0 : _______ H a : _______ In words, CLEARLY state what your random variable X ¯ or P ′ represents. State the distribution to use for the test. What is the test statistic? What is the p -value? In one or two complete sentences, explain what the p -value means for this problem. Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p -value. Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences . Alpha: _______ Decision: _______ Reason for decision: _______ Conclusion: _______ Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. Hypothesis Testing with Two Samples Class Time: __________________________ Name: _____________________________________ H 0 : _______ H a : _______ In words, clearly state what your random variable X ¯ 1 − X ¯ 2 , P ′ 1 − P ′ 2 or X ¯ d represents. State the distribution to use for the test. What is the test statistic? What is the p -value? In one to two complete sentences, explain what the p-value means for this problem. Use the previous information to sketch a picture of this situation. CLEARLY label and scale the horizontal axis and shade the region(s) corresponding to the p -value. Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences . Alpha: _______ Decision: _______ Reason for decision: _______ Conclusion: _______ In complete sentences, explain how you determined which distribution to use. The Chi-Square Distribution Class Time: __________________________ Name: ____________________________________ H 0 : _______ H a : _______ What are the degrees of freedom? State the distribution to use for the test. What is the test statistic? What is the p -value? In one to two complete sentences, explain what the p -value means for this problem. Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p -value. Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and write appropriate conclusions, using complete sentences. Alpha: _______ Decision: _______ Reason for decision: _______ Conclusion: _______ F Distribution and One-Way ANOVA Class Time: __________________________ Name: ____________________________________ H 0 : _______ H a : _______ df ( n ) = ______ df ( d ) = _______ State the distribution to use for the test. What is the test statistic? What is the p -value? Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p -value. Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and write appropriate conclusions, using complete sentences . Alpha: _______ Decision: _______ Reason for decision: _______ Conclusion: _______", "section": "Solution Sheets", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Mathematical Phrases, Symbols, and Formulas English Phrases Written Mathematically When the English says: Interpret this as: X is at least 4. X ≥ 4 The minimum of X is 4. X ≥ 4 X is no less than 4. X ≥ 4 X is greater than or equal to 4. X ≥ 4 X is at most 4. X ≤ 4 The maximum of X is 4. X ≤ 4 X is no more than 4. X ≤ 4 X is less than or equal to 4. X ≤ 4 X does not exceed 4. X ≤ 4 X is greater than 4. X > 4 X is more than 4. X > 4 X exceeds 4. X > 4 X is less than 4. X < 4 There are fewer X than 4. X < 4 X is 4. X = 4 X is equal to 4. X = 4 X is the same as 4. X = 4 X is not 4. X ≠ 4 X is not equal to 4. X ≠ 4 X is not the same as 4. X ≠ 4 X is different than 4. X ≠ 4 Formulas Formula 1: Factorial n ! = n ( n − 1 ) ( n − 2 ) . . . ( 1 ) 0 ! = 1 Formula 2: Combinations ( n r ) = n ! ( n − r ) ! r ! Formula 3: Binomial Distribution X ~ B ( n , p ) P ( X = x ) = ( n x ) p x q n − x , for x = 0 , 1 , 2 , . . . , n Formula 4: Geometric Distribution X ~ G ( p ) P ( X = x ) = q x − 1 p , for x = 1 , 2 , 3 , . . . Formula 5: Hypergeometric Distribution X ~ H ( r , b , n ) P ( X = x ) = ( ( r x ) ( b n − x ) ( r + b n ) ) Formula 6: Poisson Distribution X ~ P ( μ ) P ( X = x ) = μ x e − μ x ! Formula 7: Uniform Distribution X ~ U ( a , b ) f ( X ) = 1 b − a , a < x < b Formula 8: Exponential Distribution X ~ E x p ( m ) f ( x ) = m e − m x m > 0 , x ≥ 0 Formula 9: Normal Distribution X ~ N ( μ , σ 2 ) f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 , – ∞ < x < ∞ Formula 10: Student's t -distribution X ~ t d f f ( x ) = ( 1 + x 2 n ) − ( n + 1 ) 2 Γ ( n + 1 2 ) nπ Γ ( n 2 ) X = Z Y n Z ~ N ( 0 , 1 ), Y ~ Χ d f 2 , n = degrees of freedom Formula 11: Chi-Square Distribution X ~ Χ d f 2 f ( x ) = x n − 2 2 e − x 2 2 n 2 Γ ( n 2 ) , x > 0 , n = positive integer and degrees of freedom Formula 12: F Distribution X ~ F d f ( n ) , d f ( d ) d f ( n ) = degrees of freedom for the numerator d f ( d ) = degrees of freedom for the denominator f ( x ) = Γ ( u + v 2 ) Γ ( u 2 ) Γ ( v 2 ) ( u v ) u 2 x ( u 2 − 1 ) [ 1 + ( u v ) x − 0.5 ( u + v ) ] X = Y u W v , Y , W are chi-square Symbols and Their Meanings Symbols and their Meanings Chapter (1st used) Symbol Spoken Meaning Sampling and Data The square root of same Sampling and Data π Pi 3.14159… (a specific number) Descriptive Statistics Q 1 Quartile one the first quartile Descriptive Statistics Q 2 Quartile two the second quartile Descriptive Statistics Q 3 Quartile three the third quartile Descriptive Statistics IQR interquartile range Q 3 – Q 1 = IQR Descriptive Statistics x ¯ x-bar sample mean Descriptive Statistics μ mu population mean Descriptive Statistics s s x sx s sample standard deviation Descriptive Statistics s 2 s x 2 s squared sample variance Descriptive Statistics σ σ x σx sigma population standard deviation Descriptive Statistics σ 2 σ x 2 sigma squared population variance Descriptive Statistics Σ capital sigma sum Probability Topics { } brackets set notation Probability Topics S S sample space Probability Topics A Event A event A Probability Topics P ( A ) probability of A probability of A occurring Probability Topics P ( A | B ) probability of A given B prob. of A occurring given B has occurred Probability Topics P ( A OR B ) prob. of A or B prob. of A or B or both occurring Probability Topics P ( A AND B ) prob. of A and B prob. of both A and B occurring (same time) Probability Topics A ′ A-prime, complement of A complement of A, not A Probability Topics P ( A ') prob. of complement of A same Probability Topics G 1 green on first pick same Probability Topics P ( G 1 ) prob. of green on first pick same Discrete Random Variables PDF prob. distribution function same Discrete Random Variables X X the random variable X Discrete Random Variables X ~ the distribution of X same Discrete Random Variables B binomial distribution same Discrete Random Variables G geometric distribution same Discrete Random Variables H hypergeometric dist. same Discrete Random Variables P Poisson dist. same Discrete Random Variables λ Lambda average of Poisson distribution Discrete Random Variables ≥ greater than or equal to same Discrete Random Variables ≤ less than or equal to same Discrete Random Variables = equal to same Discrete Random Variables ≠ not equal to same Continuous Random Variables f ( x ) f of x function of x Continuous Random Variables pdf prob. density function same Continuous Random Variables U uniform distribution same Continuous Random Variables Exp exponential distribution same Continuous Random Variables k k critical value Continuous Random Variables f ( x ) = f of x equals same Continuous Random Variables m m decay rate (for exp. dist.) The Normal Distribution N normal distribution same The Normal Distribution z z -score same The Normal Distribution Z standard normal dist. same The Central Limit Theorem CLT Central Limit Theorem same The Central Limit Theorem X ¯ X -bar the random variable X -bar The Central Limit Theorem μ x mean of X the average of X The Central Limit Theorem μ x ¯ mean of X -bar the average of X -bar The Central Limit Theorem σ x standard deviation of X same The Central Limit Theorem σ x ¯ standard deviation of X -bar same The Central Limit Theorem Σ X sum of X same The Central Limit Theorem Σ x sum of x same Confidence Intervals CL confidence level same Confidence Intervals CI confidence interval same Confidence Intervals EBM error bound for a mean same Confidence Intervals EBP error bound for a proportion same Confidence Intervals t Student's t -distribution same Confidence Intervals df degrees of freedom same Confidence Intervals t α 2 student t with a /2 area in right tail same Confidence Intervals p ′ ; p ^ p -prime; p -hat sample proportion of success Confidence Intervals q ′ ; q ^ q -prime; q -hat sample proportion of failure Hypothesis Testing H 0 H -naught, H -sub 0 null hypothesis Hypothesis Testing H a H-a , H -sub a alternate hypothesis Hypothesis Testing H 1 H -1, H -sub 1 alternate hypothesis Hypothesis Testing α alpha probability of Type I error Hypothesis Testing β beta probability of Type II error Hypothesis Testing X 1 ¯ − X 2 ¯ X 1-bar minus X 2-bar difference in sample means Hypothesis Testing μ 1 − μ 2 mu -1 minus mu -2 difference in population means Hypothesis Testing P ′ 1 − P ′ 2 P 1-prime minus P 2-prime difference in sample proportions Hypothesis Testing p 1 − p 2 p 1 minus p 2 difference in population proportions Chi-Square Distribution Χ 2 Ky -square Chi-square Chi-Square Distribution O Observed Observed frequency Chi-Square Distribution E Expected Expected frequency Linear Regression and Correlation y = a + bx y equals a plus b-x equation of a line Linear Regression and Correlation y ^ y -hat estimated value of y Linear Regression and Correlation r \"r\" sample correlation coefficient Linear Regression and Correlation ρ rho (\"row\") population correlation coefficient Linear Regression and Correlation ε error same Linear Regression and Correlation SSE Sum of Squared Errors same Linear Regression and Correlation 1.9 s 1.9 times s cut-off value for outliers F -Distribution and ANOVA F F -ratio F -ratio", "section": "Mathematical Phrases, Symbols, and Formulas", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "NOTEs for the TI-83, 83+, 84, 84+ Calculators Quick Tips Legend represents a button press [ ] represents yellow command or green letter behind a key < > represents items on the screen To adjust the contrast Press , then hold to increase the contrast or to decrease the contrast. To capitalize letters and words Press to get one capital letter, or press , then to set all button presses to capital letters. You can return to the top-level button values by pressing again. To correct a mistake If you hit a wrong button, just hit and start again. To write in scientific notation Numbers in scientific notation are expressed on the TI-83, 83+, 84, and 84+ using E notation, such that... 4.321 E 4 = 4 .321 × 10 4 4.321 E –4 = 4 .321 × 10 –4 To transfer programs or equations from one calculator to another: Both calculators: Insert your respective end of the link cable cable and press , then [LINK] . Calculator receiving information: Use the arrows to navigate to and select <RECEIVE> Press . Calculator sending information: Press appropriate number or letter. Use up and down arrows to access the appropriate item. Press to select item to transfer. Press right arrow to navigate to and select <TRANSMIT> . Press . NOTE ERROR 35 LINK generally means that the cables have not been inserted far enough. Both calculators: Insert your respective end of the link cable cable Both calculators: press , then [QUIT] to exit when done. Manipulating One-Variable Statistics NOTE These directions are for entering data with the built-in statistical program. Sample Data We are manipulating one-variable statistics. Data Frequency –2 10 –1 3 0 4 1 5 3 8 To begin: Turn on the calculator. Access statistics mode. Select <4:ClrList> to clear data from lists, if desired. , Enter list [L1] to be cleared. , [L1] , Display last instruction. , [ENTRY] Continue clearing remaining lists in the same fashion, if desired. , , [L2] , Access statistics mode. Select <1:Edit . . .> Enter data. Data values go into [L1] . (You may need to arrow over to [L1] ). Type in a data value and enter it. (For negative numbers, use the negate (-) key at the bottom of the keypad). , , Continue in the same manner until all data values are entered. In [L2] , enter the frequencies for each data value in [L1] . Type in a frequency and enter it. (If a data value appears only once, the frequency is \"1\"). , Continue in the same manner until all data values are entered. Access statistics mode. Navigate to <CALC> . Access <1:1-var Stats> . Indicate that the data is in [L1] ... , [L1] , ...and indicate that the frequencies are in [L2] . , [L2] , The statistics should be displayed. You may arrow down to get remaining statistics. Repeat as necessary. Drawing Histograms NOTE We will assume that the data is already entered. We will construct two histograms with the built-in STATPLOT application. The first way will use the default ZOOM. The second way will involve customizing a new graph. Access graphing mode. , [STAT PLOT] Select <1:plot 1> to access plotting - first graph. Use the arrows navigate go to <ON> to turn on Plot 1. <ON> , Use the arrows to go to the histogram picture and select the histogram. Use the arrows to navigate to <Xlist> . If \"L1\" is not selected, select it. , [L1] , Use the arrows to navigate to <Freq> . Assign the frequencies to [L2] . , [L2] , Go back to access other graphs. , [STAT PLOT] Use the arrows to turn off the remaining plots. Be sure to deselect or clear all equations before graphing. To deselect equations: Access the list of equations. Select each equal sign (=). Continue, until all equations are deselected. To clear equations: Access the list of equations. Use the arrow keys to navigate to the right of each equal sign (=) and clear them. Repeat until all equations are deleted. To draw default histogram: Access the ZOOM menu. Select <9:ZoomStat> . The histogram will show with a window automatically set. To draw custom histogram: Access window mode to set the graph parameters. X min = –2.5 X max = 3.5 X s c l = 1 (width of bars) Y min = 0 Y max = 10 Y s c l = 1 (spacing of tick marks on y -axis) X r e s = 1 Access graphing mode to see the histogram. To draw box plots: Access graphing mode. , [STAT PLOT] Select <1:Plot 1> to access the first graph. Use the arrows to select <ON> and turn on Plot 1. Use the arrows to select the box plot picture and enable it. Use the arrows to navigate to <Xlist> . If \"L1\" is not selected, select it. , [L1] , Use the arrows to navigate to <Freq> . Indicate that the frequencies are in [L2] . , [L2] , Go back to access other graphs. , [STAT PLOT] Be sure to deselect or clear all equations before graphing using the method mentioned above. View the box plot. , [STAT PLOT] Linear Regression Sample Data The following data is real. The percent of declared ethnic minority students at De Anza College for selected years from 1970–1995 was: Year Student Ethnic Minority Percentage 1970 14.13 1973 12.27 1976 14.08 1979 18.16 1982 27.64 1983 28.72 1986 31.86 1989 33.14 1992 45.37 1995 53.1 The independent variable is \"Year,\" while the independent variable is \"Student Ethnic Minority Percent.\" Student Ethnic Minority Percentage By hand, verify the scatterplot above. NOTE The TI-83 has a built-in linear regression feature, which allows the data to be edited.The x -values will be in [L1] ; the y -values in [L2] . To enter data and do linear regression: ON Turns calculator on. Before accessing this program, be sure to turn off all plots. Access graphing mode. , [STAT PLOT] Turn off all plots. , Round to three decimal places. To do so: Access the mode menu. , [STAT PLOT] Navigate to <Float> and then to the right to <3> . All numbers will be rounded to three decimal places until changed. Enter statistics mode and clear lists [L1] and [L2] , as describe previously. , Enter editing mode to insert values for x and y . , Enter each value. Press to continue. To display the correlation coefficient: Access the catalog. , [CATALOG] Arrow down and select <DiagnosticOn> ... , , r and r 2 will be displayed during regression calculations. Access linear regression. Select the form of y = a + bx . , The display will show: LinReg y = a + bx a = –3176.909 b = 1.617 r = 2 0.924 r = 0.961 This means the Line of Best Fit (Least Squares Line) is: y = –3176.909 + 1.617 x Percent = –3176.909 + 1.617 (year #) The correlation coefficient r = 0.961 To see the scatter plot: Access graphing mode. , [STAT PLOT] Select <1:plot 1> To access plotting - first graph. Navigate and select <ON> to turn on Plot 1. <ON> Navigate to the first picture. Select the scatter plot. Navigate to <Xlist> . If [L1] is not selected, press , [L1] to select it. Confirm that the data values are in [L1] . <ON> Navigate to <Ylist> . Select that the frequencies are in [L2] . , [L2] , Go back to access other graphs. , [STAT PLOT] Use the arrows to turn off the remaining plots. Access window mode to set the graph parameters. X min = 1970 X max = 2000 X s c l = 10 (spacing of tick marks on x -axis) Y min = − 0.05 Y max = 60 Y s c l = 10 (spacing of tick marks on y -axis) X r e s = 1 Be sure to deselect or clear all equations before graphing, using the instructions above. Press the graph button to see the scatter plot. To see the regression graph: Access the equation menu. The regression equation will be put into Y1. Access the vars menu and navigate to <5: Statistics> . , Navigate to <EQ> . <1: RegEQ> contains the regression equation which will be entered in Y1. Press the graphing mode button. The regression line will be superimposed over the scatter plot. To see the residuals and use them to calculate the critical point for an outlier: Access the list. RESID will be an item on the menu. Navigate to it. , [LIST] , <RESID> Confirm twice to view the list of residuals. Use the arrows to select them. , The critical point for an outlier is: 1.9 V SSE n - 2 where: n = number of pairs of data SSE = sum of the squared errors Σ residual 2 Store the residuals in [L3] . , , [L3] , Calculate the Σresidual 2 n - 2 . Note that n - 2 = 8 , [L3] , , , Store this value in [L4] . , , [L4] , Calculate the critical value using the equation above. , , , , , [V] , , [LIST] , , , , [L4] , , , Verify that the calculator displays: 7.642669563. This is the critical value. Compare the absolute value of each residual value in [L3] to 7.64. If the absolute value is greater than 7.64, then the (x, y) corresponding point is an outlier. In this case, none of the points is an outlier. To obtain estimates of y for various x -values: There are various ways to determine estimates for \" y. \" One way is to substitute values for \" x \" in the equation. Another way is to use the on the graph of the regression line. TI-83, 83+, 84, 84+ instructions for distributions and tests Distributions Access DISTR (for \"Distributions\"). For technical assistance, visit the Texas Instruments website at http://www.ti.com and enter your calculator model into the \"search\" box. Binomial Distribution binompdf( n , p , x ) corresponds to P ( X = x ) binomcdf( n , p , x ) corresponds to P (X ≤ x) To see a list of all probabilities for x : 0, 1, . . . , n , leave off the \" x \" parameter. Poisson Distribution poissonpdf(λ, x ) corresponds to P ( X = x ) poissoncdf(λ, x ) corresponds to P ( X ≤ x ) Continuous Distributions (general) − ∞ uses the value –1EE99 for left bound ∞ uses the value 1EE99 for right bound Normal Distribution normalpdf( x , μ , σ ) yields a probability density function value (only useful to plot the normal curve, in which case \" x \" is the variable) normalcdf(left bound, right bound, μ , σ ) corresponds to P (left bound < X < right bound) normalcdf(left bound, right bound) corresponds to P (left bound < Z < right bound) – standard normal invNorm( p , μ , σ ) yields the critical value, k : P ( X < k ) = p invNorm( p ) yields the critical value, k : P ( Z < k ) = p for the standard normal Student's t -Distribution tpdf( x , df ) yields the probability density function value (only useful to plot the student- t curve, in which case \" x \" is the variable) tcdf(left bound, right bound, df ) corresponds to P (left bound < t < right bound) Chi-square Distribution Χ 2 pdf( x , df ) yields the probability density function value (only useful to plot the chi 2 curve, in which case \" x \" is the variable) Χ 2 cdf(left bound, right bound, df ) corresponds to P (left bound < Χ 2 < right bound) F Distribution Fpdf( x , dfnum , dfdenom ) yields the probability density function value (only useful to plot the F curve, in which case \" x \" is the variable) Fcdf(left bound,right bound, dfnum , dfdenom ) corresponds to P (left bound < F < right bound) Tests and Confidence Intervals Access STAT and TESTS . For the confidence intervals and hypothesis tests, you may enter the data into the appropriate lists and press DATA to have the calculator find the sample means and standard deviations. Or, you may enter the sample means and sample standard deviations directly by pressing STAT once in the appropriate tests. Confidence Intervals ZInterval is the confidence interval for mean when σ is known. TInterval is the confidence interval for mean when σ is unknown; s estimates σ. 1-PropZInt is the confidence interval for proportion. NOTE The confidence levels should be given as percents (ex. enter \" 95 \" or \" .95 \" for a 95% confidence level). Hypothesis Tests Z-Test is the hypothesis test for single mean when σ is known. T-Test is the hypothesis test for single mean when σ is unknown; s estimates σ. 2-SampZTest is the hypothesis test for two independent means when both σ's are known. 2-SampTTest is the hypothesis test for two independent means when both σ's are unknown. 1-PropZTest is the hypothesis test for single proportion. 2-PropZTest is the hypothesis test for two proportions. Χ 2 -Test is the hypothesis test for independence. Χ 2 GOF-Test is the hypothesis test for goodness-of-fit (TI-84+ only). LinRegTTEST is the hypothesis test for Linear Regression (TI-84+ only). NOTE Input the null hypothesis value in the row below \" Inpt .\" For a test of a single mean, \" μ∅ \" represents the null hypothesis. For a test of a single proportion, \" p∅ \" represents the null hypothesis. Enter the alternate hypothesis on the bottom row.", "section": "NOTEs for the TI-83, 83+, 84, 84+ Calculators", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}
{"text": "Tables This module contains links to government site tables used in statistics. Tables (NIST/SEMATECH e-Handbook of Statistical Methods, http://www.itl.nist.gov/div898/handbook/, January 3, 2009) Critical Values of the Student's t Distribution Cumulative Distribution Function of the Standard Normal Distribution Critical Values of the Chi-Square Distribution Upper Critical Values of the F Distribution Tables for Probability Distributions", "section": "Tables", "book": "Introductory Statistics 2e", "subject": "Math", "source": "https://openstax.org/details/books/introductory-statistics-2e"}