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.

Math


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 .

Math


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.

Math


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.

Math


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.

Math


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.

Math


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 ).

Math