es the: a. t -distribution b. F- distribution c. chi-square distribution d. normal distribution 3. The t-distribution for developing a confidence interval for a mean has degrees of freedom. a. n 2 b. n 1 c. n d. n – 1 e. n - 2 4. As the sample size increases, the t-distribution becomes more similar to the ________ distribution. a. normal b. exponential c. multinominal d. chi-square e. binomial 5. A parameter such as is sometimes referred to as a ________ parameter, because many times we need its value even though it is not the parameter of primary interest. a. special b. random c. nuisance d. independent e. dependent Copyright Cengage Learning. Powered by Cognero. Page 1 Name: Class: Date: Chapter 08 6. Two independent samples of sizes 50 and 50 are randomly selected from two populations to test the difference between the population means, . The sampling distribution of the sample mean difference is: a. normally distributed b. approximately normal c. t - distributed with 98 degrees of freedom d. chi-squared distributed with 99 degrees of freedom 7. The general form of a confidence interval is: a. Point Estimate Multiple × Standard Error b. Point Estimate Multiple Standard Error c. Point Estimate ± Multiple × Standard Error d. Point Estimate Multiple ± Standard Error 8. The approximate standard error of the point estimate of the population total is: a. b. c. d. 9. A confidence interval is an interval estimate for which there is a specified degree of certainty that the actual true value of the population parameter will fall within the interval. a. True b. False 10. When samples of size n are drawn from a population, then the sampling distribution of the sample mean is approximately normal, provided that n is reasonably large. a. True b. False 11. The t-distribution and the standard normal distribution are practically indistinguishable as the degrees of freedom increase. a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 2 Name: Class: Date: Chapter 08 THE FOLLOWING ITEMS REQUIRE THE USE OF EXCEL: 12. (A) Compute has a t-distribution with 15 degrees of freedom. (B) Compute has a t-distribution with 150 degrees of freedom. (C) How do you explain the difference between the results obtained in (A) and (B)? (D) Compute where Z is a standard normal random variable. (E) Compare the results of (D) to the results obtained in (A) and (B). How do you explain the difference in these probabilities? : (A) By using the Excel function TDIST (2, 15, 1) we get 0.03197. (B) By using the Excel function TDIST (2, 150, 1) we get 0.02365. (C) The smaller the degrees of freedom, the larger the variance of t, and so the larger the tail probabilities are. (D) By using the Excel function 1 – NORMDIST (2) we get 0.02275. (E) First, the variance of t with a small degree of freedom is larger than a t with a large degree of freedom, which is larger than for a Z. This explains why the probabilities in (A), (B) and (D) increases. Second, when the sample size is large, the degrees of freedom of t are large; and that the t-distribution and the standard normal distribution are practically indistinguishable. This explains why the probabilities in (B) and (D) are close. The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a tdistribution with 15 degrees of freedom. These values represent the probability to the right of the given positive values. Value t -probability 1.00 0.1666 1.20 0.1244 1.40 0.0909 These values represent the positive t- value for a given probability in both tails (sum of both tails). Probability t -value 0.20 1.3406 0.10 1.7531 0.05 2.1315 13. What is the probability of a t-value smaller than 1.00? : 0.8334 Copyright Cengage Learning. Powered by Cognero. Page 3 Name: Class: Date: Chapter 08 14. What is the probability of a t-value larger than 1.20? : 0.1244 15. What is the probability of a t-value between –1.40 and 1.40? : 0.8182 16. What would be the t-value where 0.05 of the values are in the upper tail? : 1.7531 17. What would be the t-values where 0.10 of the values are in both tails (sum of both tails)? : - 1.7531 and 1.7531 18. What would be the t-values where 0.95 of the values would fall within this interval? : - 2.1315 and 2.1315 19. If you increase the confidence level, the confidence interval: a. decreases b. increases c. stays the same d. may increase or decrease, depending on the sample data 20. If you are constructing a confidence interval for a single mean, the confidence interval will with an increase in the sample size. a. decrease b. increase c. stay the same d. increase or decrease, depending on the sample data 21. The interval estimate 18.5 2.5 was developed for a population mean when the sample standard deviation s was 7.5. Had s equaled 15, the interval estimate would be 37 5.0. a. True b. False 22. The 95% confidence interval for the population mean , given that the sample size n 49 and the population standard deviation a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 4 Name: Class: Date: Chapter 08 23. In order to construct a confidence interval estimate of the population mean , the value of must be given. a. True b. False 24. In general, increasing the confidence level will narrow the confidence interval, and decreasing the confidence level widens the interval. a. True b. False 25. If a sample has 20 observations and a 95% confidence estimate for is needed, the appropriate value of t-multiple is 2.093 a. True b. False A sample of 40 country CD recordings of Willie Nelson has been examined. The average playing time of these recordings is 51.3 minutes, and the standard deviation is 5.8 minutes. 26. (A) Construct a 95% confidence interval for the mean playing time of all Willie Nelson recordings. (B) Interpret the confidence interval you constructed in (A). : (A) n 10, 51.3, s 5.8, (B) We are 95% confident that the mean playing time of all Willie Nelson recordings is approximately between. 49.4 and 53.2 minutes. The average annual household income levels of citizens of selected U.S. cities are shown below. City Household City Household City Household Index Income Index Income Index Income 1 $54,300 21 $53,500 41 $61,500 2 $61,800 22 $45,600 42 $53,000 3 $61,400 23 $70,100 43 $51,000 4 $50,800 24 $108,700 44 $55,600 5 $56,200 25 $46,400 45 $51,600 6 $48,300 26 $56,700 46 $57,200 7 $61,600 27 $59,100 47 $54,300 8 $63,200 28 $46,300 48 $51,500 9 $55,200 29 $52,900 49 $53,500 10 $58,000 30 $56,300 50 $61,800 Copyright Cengage Learning. Powered by Cognero. Page 5 Name: Class: Date: Chapter 08 11 $77,600 31 $67,300 51 $44,800 12 $47,600 32 $63,800 52 $57,400 13 $62,700 33 $70,600 53 $48,100 14 $46,200 34 $49,800 54 $52,700 15 $64,300 35 $51,300 55 $57,400 16 $56,000 36 $56,600 56 $65,500 17 $53,400 37 $49,600 57 $59,600 18 $56,800 38 $67,400 58 $62,000 19 $51,200 39 $53,700 59 $49,700 20 $59,000 40 $48,700 60 $54,400 27. (A) Use Excel to obtain a simple random sample of size 10 from this frame. (B) Using the sample generated in (A), construct a 95% confidence interval for the mean average annual household income level of citizens in the selected U.S. cities. Assume that the population consists of all average annual household income levels in the given frame. (C) Interpret the 95% confidence interval constructed in (B). (D) Does the 95% confidence interval contain the actual population mean? If not, explain why not. What proportion of many similarly constructed confidence intervals should include the true population mean value? : (A) The StatTools Random Sample tool under Data Utilities is used to generate a sample of size 10, then the VLOOKUP function is used to get the corresponding incomes. The following sample is obtained: City Household City Household City Household Index Income Index Income Index Income 1 $54,300 21 $53,500 41 $61,500 2 $61,800 22 $45,600 42 $53,000 3 $61,400 23 $70,100 43 $51,000 4 $50,800 24 $108,700 44 $55,600 5 $56,200 25 $46,400 45 $51,600 6 $48,300 26 $56,700 46 $57,200 7 $61,600 27 $59,100 47 $54,300 8 $63,200 28 $46,300 48 $51,500 9 $55,200 29 $52,900 49 $53,500 10 $58,000 30 $56,300 50 $61,800 11 $77,600 31 $67,300 51 $44,800 12 $47,600 32 $63,800 52 $57,400 13 $62,700 33 $70,600 53 $48,100 14 $46,200 34 $49,800 54 $52,700 15 $64,300 35 $51,300 55 $57,400 16 $56,000 36 $56,600 56 $65,500 17 $53,400 37 $49,600 57 $59,600 18 $56,800 38 $67,400 58 $62,000 19 $51,200 39 $53,700 59 $49,700 20 $59,000 40 $48,700 60 $54,400 Copyright Cengage Learning. Powered by Cognero. Page 6 Name: Class: Date: Chapter 08 (B) (C) We are 95% confident that the average annual household income level of all citizens is approximately between $52,737 and $66,243. (D) This confidence interval easily captures the true population mean of $57,043. Approximately 95% of the confidence intervals constructed in this way should contain the true population mean. The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees in 2004 reveals the following family dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. Use StatTools for your calculations. 28. (A) Construct a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation. (B) What assumption about the population distribution must be made to (A)? (C) Interpret the 90% confidence interval constructed in (A). (D) Suppose you used a 95% confidence interval in (A). What would be your ? (E) Suppose the fourth value were 593 instead of 93. What would be your to (A)? What effect does this change have on the confidence interval? (F) Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation. (G) Interpret the 90% confidence interval constructed in (E). : (A) (B) The population of dental expenses must be approximately normally distributed. Copyright Cengage Learning. Powered by Cognero. Page 7 Name: Class: Date: Chapter 08 (C) We are 90% confident that the mean family dental expenses for all employees of this corporation is between $199.26 and $333.74. (D) (E) The additional $500 in dental expenses, divided across the sample of 12, raises the mean by $41.67 and increases the standard deviation by nearly $18.20. The interval width increases over $23 in the process. (F) Copyright Cengage Learning. Powered by Cognero. Page 8 Name: Class: Date: Chapter 08 The widths of 100 elevator rails have been measured. The sample mean and standard deviation of the elevator rails are 2.05 inches and 0.01 inch. 29. (A) Construct a 95% confidence interval for the average width of an elevator rail. Do we need to assume that the width of elevator rails follows a normal distribution? (B) How large a sample of elevator rails would we have to measure to ensure that we could estimate, with 95% confidence, the average diameter of an elevator rail within 0.01 inch? : (A) n 100, The assumption of normality is not crucial here because the sample size is fairly large. (B) 0.01, B 0.01, z – multiple 1.96 A large regional department store is evaluating the effectiveness of its credit card program, which costs it approximately $1m per year to administer. The store believes that for the credit card program to be worthwhile, the administrative costs should be no more than 10% of the total of the average annual account balances. Rather than reviewing each of the 15,000 individual accounts, the store’s analysts randomly selected a sample of 500 average annual balances from the frame. The sample mean and sample standard deviation were $215.75 and $55.90, respectively. 30. (A) Construct a 95% confidence interval for the mean of the average annual credit account balances. (B) Interpret the 95% confidence interval constructed in (A). (C) Use the confidence interval constructed for (A) to help the store evaluate its criteria for whether or not the credit card program is worthwhile. : (A) n 500, 215.75, s $55.90, t–multiple 1.965, (B) The store can be 95% confident that the typical average annual balance of all 15,000 credit card accounts will be somewhere between $210.84 and $220.66. (C) The point estimate of the total of the annual credit account balances is 15,000 ($215.75) $3,236,250. Similarly, the upper and lower confidence interval limits for the total are 15,000($210.84) $3,162,575 and 15,000($220.66) $3,309,925, respectively. The store’s current administrative costs ($1,000,000) are over 30% of the best possible case for the total of the average annual account balances (the upper limit of the interval, $3,309,925), therefore the credit card program is not worthwhile. Copyright Cengage Learning. Powered by Cognero. Page 9 Name: Class: Date: Chapter 08 31. If you decrease the confidence level, the confidence interval . a. decreases b. increases c. stays the same d. may increase or decrease, depending on the sample data 32. If you are constructing a confidence interval for a single mean, the confidence interval will with a decrease in the sample size. a. decrease b. increase c. stay the same d. increase or decrease, depending on the sample data 33. Suppose there are 500 accounts in a population. You sample 50 of them and find a sample mean of $500. What would be your estimate for the population total? a. $5,000 b. $50,000 c. $250,000 d. $500,000 e. None of these choices 34. We can form a confidence interval for the population total T by finding a confidence interval for the population mean in the usual way, and then multiplying the lower and upper limits the confidence interval by the population size N. a. True b. False 35. A 90% confidence interval estimate for a population mean is determined to be 72.8 to 79.6. If the confidence level is reduced to 80%, the confidence interval for becomes narrower. a. True b. False 36. The approximate standard error of the point estimate of the population total is . a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 10 Name: Class: Date: Chapter 08 Auditors of Independent Bank are interested in comparing the reported value of all 1775 customer savings account balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each savings account at the bank, the auditors randomly selected a sample of 100 savings account balances from the frame. The sample mean and sample standard deviations were $505.75 and 360.95, respectively. 37. (A) Construct a 90% confidence interval for the total value of all savings account balances within this bank. Assume that the population consists of all savings account balances in the frame. (B) Interpret the 90% confidence interval constructed in (A). : (A) N 1775, n 100, 505.75, s 360.95, t – multiple 1.6604, ($719,326.70, $1,004,085.8) (B) We are 90% confident that the total balance of all 1775 savings account balances within the bank is approximately between $791,327 and $1,004,086. 38. The standard error of the sampling distribution of the sample proportion , when the sample size n 50 and the population proportion p 0.25, is 0.00375. a. True b. False 39. As a general rule, the normal distribution is used to approximate the sampling distribution of the sample proportion only if the sample size n is greater than 30. a. True b. False 40. If the standard error of the sampling distribution of the sample proportion is 0.0324 for samples of size 200, then the population proportion must be 0.30. a. True b. False 41. The lower limit of the 95% confidence interval for the population proportion p, given that n 300; and 0.10 is 0.1339. a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 11 Name: Class: Date: Chapter 08 42. If a random sample of size 250 is taken from a population, where it is known that the population proportion p 0.4, then the mean of the sampling distribution of the sample proportion is 0.60. a. True b. False 43. The upper limit of the 90% confidence interval for the population proportion p, given that n 100; and 0.20 is 0.2658. a. True b. False 44. The mean of the sampling distribution of the sample proportion , when the sample size n 100 and the population proportion p 0.15, is 15.0. a. True b. False 45. You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of these people had levels over the “safe” count of 200. Construct a 95% confidence interval for the population proportion of people in Iowa with cholesterol levels over 200. : (0.3216, 0.4784). Copyright Cengage Learning. Powered by Cognero. Page 12 Name: Class: Date: Chapter 08 A marketing research consultant hired by Coca-Cola is interested in determining the proportion of customers who favor Coke over other soft drinks. A random sample of 400 consumers was selected from the market under investigation and showed that 53% favored Coca-Cola over other brands. 46. (A) Compute a 95% confidence interval for the true proportion of people who favor Coke. Do the results of this poll convince you that a majority of people favors Coke? (B) Suppose 2,000 (not 400) people were polled and 53% favored Coke. Would you now be convinced that a majority of people favor Coke? Why might your be different than in (A)? (C) How many people would have to be surveyed to be 95% confident that you can estimate the fraction of people who favor Coca-Cola within 1%? : (A) n 400, 0.53 0.53 ± 0.0489 (0.4811, 0.5789). Since confidence interval ranges from about 48% to 57.9%, it is difficult to conclude that a majority of people favors Coke. It could be below 50%. (B) n 2,000, 0.53 0.53 ± 0.0219 (0.5081, .5519). In this case the 95% confidence interval is entirely above 50%, the data is now more convincing than it was previously. (C) 0.53, z – multiple 1.96, B .01. The sample size for proportion is given by: . Copyright Cengage Learning. Powered by Cognero. Page 13 Name: Class: Date: Chapter 08 An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased six years ago. A random sample of 200 customers selected from the automobile dealer’s records indicates that 88 still own cars that were purchased six years earlier. 47. (A) Construct a 95% confidence interval estimate of the population proportion of all customers who still own the cars they purchased six years ago (B) How can the result in (A) be used by the automobile dealer to study satisfaction with cars purchased at the dealership? : (A) 88/200 0.44, z – multiple 1.96 (B) The dealer can infer that the proportion of all customers who still own the cars they purchased at the dealership 6 years earlier is somewhere between 03712 and 0.5088 with a 95% level of confidence. 48. The shape of a chi-square distribution a. is symmetric b. is skewed to the left c. is skewed to the right d. depends on the sample data 49. In developing a confidence interval for the population standard deviation , we make use of the fact that the sampling distribution of the sample standard deviation s is not the normal distribution or the t-distribution, but rather a rightskewed distribution called the chi-square distribution, which (for this procedure) has n – 1 degrees of freedom. a. True b. False 50. The degrees of freedom for the t and chi-square distributions is a numerical parameter of the distribution that defines the precise shape of the distribution. a. True b. False 51. The confidence interval for the population standard deviation s is centered at the point estimate, the sample standard deviation s. a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 14 Name: Class: Date: Chapter 08 The employee benefits manager of a medium size business would like to estimate the proportion of full-time employees who prefer adopting plan A of three available health care plans in the coming annual enrollment period. A reliable frame of the company’s employees and their tentative health care preferences are available. Using Excel, the manager chose a random sample of size 50 from the frame. There were 17 employees in the sample who preferred plan A. 52. (A) Construct a 99% confidence interval for the proportion of company employees who prefer plan A. Assume that the population consists of the preferences of all employees in the frame. (B) Interpret the 99% confidence interval constructed in (A). : (A) (0.1675), 0.5125) (B) We are 99% confident that the true proportion of all employees who prefer plan A is within this confidence interval; that is, between 0.1675 and 0.5125. Senior management of a consulting services firm is concerned about a growing decline in the firm’s weekly number of billable hours. The firm expects each professional employee to spend at least 40 hours per week on work. In an effort to understand this problem better, management would like to estimate the standard deviation of the number of hours their employees spend on work-related activities in a typical week. Rather than reviewing the records of all the firm’s full-time employees, the management randomly selected a sample of size 50 from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, respectively. 53. (A) Construct a 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week. (B) Interpret the 99% confidence interval constructed in (A). (C) Given the target range of 40 to 60 hours of work per week, should senior management be concerned about the number of hours their employees are currently devoting to work? Explain why or why not. : (A) . The confidence interval formulas for (not presented in the book) are: Lower limit 5.936 Upper limit (B) We are 99% confident that the population standard deviation is between 5.936 and 10.057. (C) The best guess for the population mean is 48.5 hours per week, and about 95% of all employees are within 2 standard deviations of this, where we are almost sure (99% sure) that this standard deviation is between 5.9 and 10.1. But even if the standard deviation is only 5.9, then 48.5 2 standard deviations will produce the range 36.7 to 60.3. Maybe management should be concerned. Copyright Cengage Learning. Powered by Cognero. Page 15 Name: Class: Date: Chapter 08 54. When the samples we want to compare are paired in some natural way, such as pretest/posttest for each person or husband/wife pairs, a more appropriate form of analysis is to not compare two separate variables, but their . a. difference b. sum c. ratio d. total e. product 55. Two independent samples of sizes 20 and 25 are randomly selected from two normal populations with equal variances. In order to test the difference between the population means, the test statistic is: a. a standard normal random variable b. approximately standard normal random variable c. t-distributed with 45 degrees of freedom d. t-distributed with 43 degrees of freedom 56. The number of degrees of freedom needed to construct 90% confidence interval for the difference between means when the data are gathered from paired samples, with 15 observations in each sample, is: a. 30 b. 15 c. 28 d. 14 57. An example of a problem where the sample data would be paired is: a. Difference between the means of appraised and sales house prices b. Difference between the proportion of defective items from two suppliers c. Difference in the mean life of two major brands of batteries d. Difference in the mean salaries for graduates in two different academic fields at a university e. None of these options 58. In general, the paired-sample procedure is appropriate when the samples are naturally paired in some way and there is a reasonably large positive correlation between the pairs. In this case, the paired-sample procedure makes more efficient use of the data and generally results in narrower confidence intervals. a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 16 Name: Class: Date: Chapter 08 59. If two random samples of size 40 each are selected independently from two populations whose variances are 35 and 45, then the standard error of the sampling distribution of the sample mean difference, , equals 1.4142. a. True b. False 60. If two random samples of sizes 30 and 35 are selected independently from two populations whose means are 85 and 90, then the mean of the sampling distribution of the sample mean difference, , equals 5. a. True b. False 61. In developing confidence interval for the difference between two population means using two independent samples, we use the pooled estimate in estimating the standard error of the sampling distribution of the sample mean difference if the populations are normal with equal variances. a. True b. False 62. Samples of exam scores for employees before and after a training class would be examples of paired data a. True b. False 63. If two samples contain the same number of observations, then the data must be paired. a. True b. False 64. If we cannot make the strong assumption that the variances of two samples are equal, then we must use the pooled standard deviation in calculating the standard error of a difference between the means. a. True b. False Copyright Cengage Learning. Powered by Cognero. Page 17 Name: Class: Date: Chapter 08 Q-Mart is interested in comparing its male and female customers. Q-Mart would like to know if its female charge customers spend more money, on average, than its male charge customers. They have collected random samples of 25 female customers and 22 male customers. On average, women charge customers spend $102.23 and men charge customers spend $86.46. Some information is shown below. Summary statistics for two samples Female Male Sample sizes 25 22 Sample means 102.23 86.46 Sample standard deviations 93.393 59.695 Confidence interval for difference between means Sample mean difference 15.77 Pooled standard deviation 79.466 Std error of difference 23.23 65. (A) Use a t - value of 2.014 to calculate a 95% confidence interval for the difference between the average female purchase and the average male purchase. Would you conclude that there is a significant difference between females and males in this case? Explain. (B) What are the degrees of freedom for the t-multiple in this calculation? Explain how you would calculate the degrees of freedom in this case. (C) What is the assumption in this case that allows you to use the pooled standard deviation for this confidence interval? Name: Class: Date: Chapter 08 A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The average thickness and the standard deviation of thickness for shift 1, based on a random sample of 40 gaskets, are 10.85 mm and 0.16 mm, respectively. The similar figures for shift 2, based on a random sample of 30 gaskets, are 10.90 mm and 0.19 mm. Let be the difference in thickness between shifts 1 and 2, and assume that the population variances are equal. 66. (A) Construct a 95% confidence interval for . (B) Based on your to (A), are you convinced that the gaskets from shift 2 are, on average, wider than those from shift 1? Why or why not? Copyright Cengage Learning. Powered by Cognero. Page 19 Name: Class: Date: Chapter 08 A market research consultant hired by Coke Classic Company is interested in estimating the difference between the proportions of female and male customers who favor Coke Classic over Pepsi Cola in Chicago. A random sample of 200 consumers from the market under investigation showed the following frequency distribution. Male Female Coke 72 38 110 Pepsi 58 32 90 67. (A) Construct a 95% confidence interval for the difference between the proportions of male and female customers who prefer Coke Classic over Pepsi Cola. (B) Interpret the constructed confidence interval. Copyright Cengage Learning. Powered by Cognero. Page 20 Name: Class: Date: Chapter 08 A real estate agent has collected a random sample of 40 houses that were recently sold in Grand Rapids, Michigan. She is interested in comparing the appraised value and recent selling price (in thousands of dollars) of the houses in this particular market. The values of these two variables for each of the 40 randomly selected houses are shown below. House Value Price House Value Price 1 140.93 140.24 21 136.57 135.35 2 132.42 129.89 22 130.44 121.54 3 118.30 121.14 23 118.13 132.98 4 122.14 111.23 24 130.98 147.53 5 149.82 145.14 25 131.33 128.49 6 128.91 139.01 26 141.10 141.93 7 134.61 129.34 27 117.87 123.55 8 121.99 113.61 28 160.58 162.03 9 150.50 141.05 29 151.10 157.39 10 142.87 152.90 30 120.15 114.55 11 155.55 157.79 31 133.17 139.54 12 128.50 135.57 32 140.16 149.92 13 143.36 151.99 33 124.56 122.08 14 119.65 120.53 34 127.97 136.51 15 122.57 118.64 35 101.93 109.41 16 145.27 149.51 36 131.47 127.29 17 149.73 146.86 37 121.27 120.45 18 147.70 143.88 38 143.55 151.96 19 117.53 118.52 39 136.89 132.54 20 140.13 146.07 40 106.11 114.33 68. (A) Use the sample data to generate a 95% confidence interval for the mean difference between the appraised values and selling prices of the houses sold in Grand Rapids. (B) Interpret the constructed confidence interval fin (A) for the real estate agent. Copyright Cengage Learning. Powered by Cognero. Page 21 Name: Class: Date: Chapter 08 Q-Mart is interested in comparing customer who used its own charge card with those who use other types of credit cards. Q-Mart would like to know if customers who use the Q-Mart card spend more money per visit, on average, than customers who use some other type of credit card. They have collected information on a random sample of 38 charge customers as shown below. On average, the person using a Q-Mart card spends $192.81 per visit and customers using another type of card spend $104.47 per visit. Summary statistics for two samples Q-Mart Other Charges Sample sizes 13 25 Sample means 192.81 104.47 Sample standard deviations 115.243 71.139 Confidence interval for difference between means Sample mean difference 88.34 Pooled standard deviation 88.323 Std error of difference 30.201 69. (A) Using a t - value of 2.0281, calculate a 95% confidence interval for the difference between the average Q-Mart charge and the average charge on another type of credit card. (B) What are the degrees of freedom for the t - multiple in this calculation? Explain how you would calculate the degrees of freedom in this case. (C) What is the assumption in this case that allows you to use the pooled standard deviation for this confidence interval? (D) Would you conclude that there is a significant difference between the two types of customers in this case? Explain. : (A) (B) (C) In order to use the pooled standard deviation for this confidence interval, we must assume that the two population standard deviations are equal ( ). (D) Since the confidence interval in (B) does not include 0, there appears to be a significant difference between the means of the two groups. In this case, it appears as though the Q-Mart charge card holders spend more money than those who use other types of charge cards. 70. In constructing confidence interval estimate for the difference between the means of two populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are as follows: Construct 90% confidence interval for . : t-multiple 1.6646, 20 1.6646(6.776) 20 11.28 (8.72, 31.28) Copyright Cengage Learning. Powered by Cognero. Page 22 Name: Class: Date: Chapter 08 A defensive driving training company is interested in evaluating the relative effectiveness of its two main modes of training; online and traditional classroom. The company has collected a random sample of 300 customers in a particular area, with the following results: Online Classroom Pass 143 128 Fail 17 12 Total 160 140 71. (A) Construct a 95% confidence interval for the difference between the proportions of online and classroom customers who pass the final exam. (B) Interpret the confidence interval obtained in (A). : (A , , , and 0.0244 0.0237 0.048 (B) We are 95% confident that the population difference between these proportions is between –0.1147 and 0.0736. Since the interval includes both negative differences and positive differences, the company really cannot conclude whether one form of training is better than the other or vice-versa. 72. When you calculate the sample size for a proportion, you use an estimate for the population proportion; namely . A conservative value for n can be obtained by using a. 0.01 b. 0.05 c. 0.10 d. 0.50 e. 1.00 73. Confidence intervals are a function of which of the following three things? a. The population, the sample, and the standard deviation b. The sample, the variable of interest, and the degrees of freedom c. The data in the sample, the confidence level, and the sample size d. The sampling distribution, the confidence level, and the degrees of freedom e. The mean, median, and mode Copyright Cengage Learning. Powered by Cognero. Page 23 Name: Class: Date: Chapter 08 74. After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the maximum allowable error (B) must be reduced to just 0.025. If the original calculation led to a sample size of 1000, the sample size will now have to be: a. 2000 b. 4000 c. 1000 d. 8000 75. For a given confidence level, the procedure for controlling interval length usually begins with the specification of a. the point estimate b. the population standard deviation, s c. the sample standard deviation, s d. the interval half-length, B 76. In determining the sample size n for estimating the population proportion p, a conservative value of n can be obtained by using 0.50 as an estimate of p. a. True b. False 77. You are trying to estimate the average amount a family spends on food during a year. In the past, the standard deviation of the amount a family has spent on food during a year has been approximately $1200. If you want to be 99% sure that you have estimated average family food expenditures within $60, how many families do you need to survey? . 78. You have been assigned to determine whether more people prefer Coke to Pepsi. Assume that roughly half the population prefers Coke and half prefers Pepsi. How large a sample would you need to take to ensure that you could estimate, with 95% confidence, the proportion of people preferring Coke within 3% of the actual value? 79. (A) You can be 95% confident that the mean salary for all production managers with at least 15 years of experience is between what two numbers (the t-multiple with 8 degrees of freedom is 2.306)? What assumption are you making about the distribution of salaries? (B) What sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years of experience and have only 5 chances in 100 of being off by more than $4200? 80. (A) Find a 95% confidence interval for the mean account balance on this store’s credit card (the t-multiple with 39 degrees of freedom is 2.0227). (B) What sample size would be needed to ensure that we could estimate the true mean account balance and have only 5 chances in 100 of being off by more than $100? : Name: Class: Date: Chapter 08 The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n 160 units. Suppose that today’s sample contains 14 defectives. 81. (A) Determine a 95% confidence interval for the proportion defective for the process today. (B) Based on your to (A), is it still reasonable to think the overall proportion defective produced by today’s process is actually the targeted 4%? Explain your reasoning. (C) The confidence interval in (A) is based on the assumption of a large sample size. Is this sample size sufficiently large in this example? Explain how you arrived at your . (D) How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today’s sample)? 82. You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the amount a family has spent on food during a year has been approximately $800. If you want to be 95% sure that you estimated average family food expenditures within $50, how many families do you need to survey? : Chapter 08 83. In past years, approximately 25% of all U.S. families purchased potato chips at least once a month. We are interested in determining the fraction of all U.S. families that currently purchase potato chips at least once a month. How many families must we survey if we want to be 99% sure that our estimate of the fraction of U.S. families currently purchasing potato chips at least once a month is accurate within 2%? : 84. There are, generally speaking, two types of statistical inference. They are: a. sample estimation and population estimation b. confidence interval estimation and hypothesis testing c. interval estimation for a mean and point estimation for a proportion d. independent sample estimation and dependent sample estimation e. None of these choices 85. From a sample of 500 items, 30 were found to be defective. The point estimate of the population proportion defective will be: a. 0.06 b. 30.0 c. 16.667 d. None of the above 86. The chi-square distribution for developing a confidence interval for a standard deviation has degrees of freedom. a. n 2 b. n 1 c. n d. n – 1 e. n - 2 Copyright Cengage Learning. Powered by Cognero. Page 27 Name: Class: Date: Chapter 08