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 .

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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.

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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.

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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

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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.

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