Q&A Hero

Q: A credit card company operates two customer service centers: one in Boise and one in Richmond. Callers to the service centers dial a single number, and a computer program routs callers to the center having the fewest calls waiting. As part of a customer service review program, the credit card center would like to determine whether the average length of a call (not including hold time) is different between the two centers. The managers of the customer service centers are willing to assume that the populations of interest are normally distributed with equal variances. Suppose a random sample of phone calls to the two centers is selected and the following results are reported: BoiseRichmondSample Size120135Sample Mean(seconds)195216Sample St. Dev. (seconds)35.1037.80Using the sample results, develop a 90% confidence interval estimate for the difference between the two population means.A) -29.3124 (1 - 2) -18.6876B) -24.2412 (1 - 2) -17.7588C) -26.2941 (1 - 2) -11.8059D) -28.5709 (1 - 2) -13.4291

Q: Construct a 95% confidence interval estimate for the difference between two population means based on the following information: Population 1Population 21 = 3552 = 3201 = 502 = 40n1 = 50n2 = 80A) 25.41 (1 - 2) 44.59B) 35.41 (1 - 2) 40.59C) 15.741 (1 - 2) 54.16D) 22.13 (1 - 2) 47.87

Q: The following information is based on independent random samples taken from two normally distributed populations having equal variances: n1 = 24 n2 = 28 1 = 130 2 = 125 s1 = 19 s2 = 17.5 Based on the sample information, determine the 95% confidence interval estimate for the difference between the two population means. A) -6.23 < (1 - 2) < 14.23 B) -4.81 < (1 - 2) < 16.81 C) -5.17 < (1 - 2) < 15.17 D) -3.25 < (1 - 2) < 17.25

Q: The following information is based on independent random samples taken from two normally distributed populations having equal variances: n1 = 15n2 = 131 = 502 = 53s1 = 5s2 = 6Based on the sample information, determine the 90% confidence interval estimate for the difference between the two population means.A) -6.54 (1 - 2) 0.54B) -4.25 (1 - 2) 1.25C) -5.98 (1 - 2) 1.88D) -2.25 (1 - 2) 5.25

Q: A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on a 95 percent confidence level, what is the upper limit of the confidence interval estimate? A) 0.2340 B) 0.1034 C) -0.031 D) -0.018

Q: A study was recently conducted at a major university to determine whether there is a difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on these sample data, and testing at the 0.10 level of significance, what is the value of the test statistic? A) Approximately z = 1.645 B) About z = -3.04 C) Approximately z = 3.45 D) About z = 1.96

Q: Suppose that two population proportions are being compared to test whether there is any difference between them. Assume that the test statistic has been calculated to be z = 2.21. Find the p-value for this situation. A) p-value = 0.0136 B) p-value = 0.4864 C) p-value = 0.0272 D) p-value = 0.9728

Q: There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, which of the following is true? A) Based the sample data, the null hypothesis of equal population proportions is rejected since the test statistic exceeds the critical value. B) Based on the sample data, the null hypothesis should not be rejected since the test statistic of z = 2.04 exceeds the critical value of z = 1.96. C) According to the test results, the hypothesis should be rejected since the test statistic value, z = 1.54, falls in the rejection region. D) Based on the sample data, there is not sufficient evidence to conclude that a difference exists between the proportion of search hits since the test statistic, z = 1.54, does not fall in the rejection region.

Q: There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, what is the critical value for the hypothesis test? A) z = 196 B) z = 1.645 C) t = 2.0456 D) z = 2.575

Q: There are a number of highly touted search engines for finding things of interest on the Internet. Recently a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, what is the null hypothesis to be tested?A) H0 : 1 = 2B) H0 : p1 p2C) H0 : 1 = 2D) H0 : p1 = p2

Q: Suppose a survey is taken of two groups of people where each person is asked a yes/no question and the proportion of people who answer yes is calculated for each group. Which of the following is true about a hypothesis test of the difference in the two proportions? A) Normality can be assumed if the sample size for each population is at least 30. B) The t-distribution should be used if the standard deviations are unknown. C) The standard deviation must be assumed equal. D) Normality can be assumed if, in each group, at least 5 people say yes, and at 5 people say no.

Q: An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable response to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. The samples resulted in 57 males that liked the ad and 47 females that liked the ad. Based on this information, what is the value of the test statistic? A) z = 1.645 B) z = 1.42 C) t = 2.234 D) z = 1.024

Q: An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable response to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. Given this information, what is the critical value? A) z = 1.645 B) t = 1.96 C) z = 1.96 D) Can't be determined without knowing the results of the sample.

Q: An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable responses to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. Given this, what is the appropriate null hypothesis?A) H0 : 1 = 2B) H0 : p1 p2C) H0 : 1 = 2D) H0 : p1 = p2

Q: To increase productivity, workers went through a training program. The management wanted to know the effectiveness of the program. A sample of seven workers was taken and their daily production rates before and after the training are shown below. Worker Before After 1 18 22 2 23 25 3 25 27 4 22 25 5 20 24 6 21 19 7 19 20 Based on the data, the training program is: A) effective. B) ineffective. C) neither effective nor ineffective. D) None of the above

Q: Assume that 10 people join a weight loss program for 3 months. Each person's weight both before and after the program is recorded and the number of pounds each person lost is found. The following summarizes the results for the 10 people: Mean weight lost = 9 pounds Sample standard deviation of weight lost = 4.6 pounds Assume that the hypothesis test will be conducted to determine whether or not the weight loss program is effective using a 0.05 level of significance. What is the value of the test statistic? A) t = 6.19 B) t = 1.96 C) z = 1.96 D) z = 6.19

Q: Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Mike" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are: Player New Club Leading Club 1 236.4 237.2 2 202.5 200.4 3 245.6 240.8 4 257.4 259.3 5 223.5 218.9 6 205.3 200.6 7 266.7 258.9 8 240 236.5 9 278.9 280.5 10 211.4 206.5 What is the critical value for the appropriate hypothesis test if the test is conducted using a 0.05 level of significance? A) z = 1.645 B) t = 1.7341 C) t = 1.8331 D) t = 2.2622

Q: Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are: PlayerNew ClubLeading Club1236.4237.22202.5200.43245.6240.84257.4259.35223.5218.96205.3200.67266.7258.98240236.59278.9280.510211.4206.5What is an appropriate null hypothesis to be tested?A) H0 : 1 = 2B) H0 : 1 2C) H0 : d 0D) H0 : D = 0

Q: In testing for differences between the means of two paired populations, an appropriate null hypothesis would be:A) H0 : D = 2B) H0 : D = 0C) H0 : D < 0D) H0 : D > 0

Q: If we are testing for the difference between the means of two paired populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to: A) 39 B) 38 C) 19 D) 18

Q: The t-test for the mean difference between 2 related populations assumes that the respective: A) sample sizes are equal. B) sample variances are equal. C) populations are approximately normal or sample sizes are large. D) All of the above

Q: Suppose that a group of 10 people join a weight loss program for 3 months. Each person's weight is recorded at the beginning and at the end of the 3-month program. To test whether the weight loss program is effective, the data should be treated as: A) independent samples using the normal distribution. B) paired samples using the t-distribution. C) independent samples using the t-distribution. D) independent proportions.

Q: Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players is: Player New Club Leading Club 1 236.4 237.2 2 202.5 200.4 3 245.6 240.8 4 257.4 259.3 5 223.5 218.9 6 205.3 200.6 7 266.7 258.9 8 240 236.5 9 278.9 280.5 10 211.4 206.5 Based on these sample data, what is the point estimate for the difference between the mean distance for the new driver versus the leading driver? A) 2.81 B) 1.55 C) -3.45 D) 233.4

Q: Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see how it compares with the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. Given this description of the planned test, which of the following statements is true? A) The test won't be meaningful if only five balls are hit by each player with each club. B) The samples in this case are called paired samples since the same players are hitting both golf clubs. C) The test will be invalid unless different players are used to hit each club so that the samples will be independent. D) The samples are independent because each player is independent of the other players.

Q: The U.S. Golf Association provides a number of services for its members. One of these is the evaluation of golf equipment to make sure that the equipment satisfies the rules of golf. For example, they regularly test the golf balls made by the various companies that sell balls in the United States. Recently they undertook a study of two brands of golf balls with the objective to see whether there is a difference in the mean distance that the two golf ball brands will fly off the tee. To conduct the test, the U.S.G.A. uses a robot named "Iron Byron," which swings the club at the same speed and with the same swing pattern each time it is used. The following data reflect sample data for a random sample of balls of each brand. Brand A: 234 236 230 227 234 233 228 229 230 238 Brand B: 240 236 241 236 239 243 230 239 243 240 Given this information, what is the test statistic for testing whether the two population means are equal? A) t = 1.115 B) t = 1.96 C) t = -4.04 D) t = -2.58

Q: Assume that you are testing the difference in the means of two independent populations at the 0.05 level of significance. The null hypothesis is: H0 : A - ² 0 and you have found the test statistic is What should you conclude?A) The mean of pop. A is greater than the mean of pop. B because p-value < .B) The mean of pop. A is greater than the mean of pop. B because p-value > .C) There is no significant difference in the two means because p-value > .D) The mean of pop. B is greater than the mean of pop. A because p-value < .

Q: There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available: North End Other Sample Size 20 10 Sample Mean Increase $4,010 $3,845 Sample St. Deviations $1,800 $1,750 Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the critical value? A) z = 1.578 B) t = 1.7011 C) t = 0.2388 D) t = 2.0484

Q: There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available: North End Other Sample Size 20 10 Sample Mean Increase $4,010 $3,845 Sample St. Deviations $1,800 $1,750 Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the value of the test statistic? A) z = 1.578 B) t = 1.7011 C) t = 0.2388 D) t = 0.3944

Q: There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compared with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. Based on the information provided, the research (or alternate) hypothesis is:A) 1 = 2B) 1 2C) 1 > 2D) 1 < 2

Q: A hypothesis test for the difference between two means is considered a two-tailed test when: A) the population variances are equal. B) the null hypothesis states that the population means are equal. C) the alpha level is 0.10 or higher. D) the standard deviations are unknown.

Q: Under what conditions can the t-distribution be correctly employed to test the difference between two population means? A) When the samples from the two populations are small and the population variances are unknown B) When the two populations of interest are assumed to be normally distributed C) When the population variances are assumed to be equal D) All of the above

Q: In conducting a hypothesis test for the difference between two population means where the standard deviations are known and the null hypothesis is:H0 : A - β 0What is the p-value assuming that the test statistic has been found to be z = 2.52?A) 0.0059B) 0.9882C) 0.0118D) 0.4941

Q: A commuter has two different routes available to drive to work. She wants to test whether route A is faster than route B. The best hypotheses are:A) H0 : A - β 0HA : A- β < 0B) H0 : A - β 0HA : A- β > 0C) H0 : A - β = 0HA : A- β 0D) H0 : A - β < 0HA : A- Îβ 0

Q: A recent study posed the question about whether Japanese managers are more motivated than American managers. A randomly selected sample of each was administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below. AmericanJapaneseSample Size211100Mean SSATL Score65.7579.83Population Std. Dev11.076.41Which of the following is the correct the null and alternative hypotheses to determine if the average SSATL score of Japanese managers differs from the average SSATL score of American managers?A) H0 : A - J 0 versus H1 : A - J < 0B) H0 : A - J 0 versus H1 : A - J > 0C) H0 : A - J = 0 versus H1 : A - J 0D) H0 : A - J = 0 versus H1 : A - J 0

Q: A recent study posed the question about whether Japanese managers are more motivated than American managers. A randomly selected sample of each was administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below. American Japanese Sample Size 211 100 Mean SSATL Score 65.75 79.83 Population Std. Dev 11.07 6.41 Judging from the way the data were collected, which test would likely be most appropriate? A) Related samples t-test for mean difference B) Pooled-variance t-test for the difference in means C) Independent samples Z-test for the difference in means D) Related samples Z-test for mean difference

Q: Given the following information, calculate the degrees of freedom that should be used in the pooled-standard deviation t-test. A) df = 41 B) df = 39 C) df = 16 D) df = 25

Q: If we are testing for the difference between the means of two independent populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to: A) 39 B) 38 C) 19 D) 18

Q: A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling: In-House Credit Card National Credit Card Sample Size: 86 113 Mean Monthly Purchases: $45.67 $39.87 Standard Deviation: $10.90 $12.47 Suppose that the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level of .05, what is the value of the test statistic assuming the standard deviations are known? A) t = 3.49 B) z = 11.91 C) z = 2.86 D) z = 3.49

Q: The management of a department store is interested to estimate the difference in the amount of money spent by female and male shoppers. You are given the following information. Female Shoppers Male Shoppers Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 A 95 percent confidence interval estimate for the difference between the average purchases of the customers using the two different credit cards is: A) 49 to 64 B) 11.68 to 18.32 C) 125 to 140 D) 8 to 10

Q: A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling: In-House Credit Card National Credit Card Sample Size: 86 113 Mean Monthly Purchases: $45.67 $39.87 Standard Deviation: $10.90 $12.47 Given this information, which of the following statements is true? A) If either of the sample sizes is increased, the resulting confidence interval will have a smaller margin of error. B) If the confidence level were changed from 95 percent to 90 percent, the margin of error in the estimate would be reduced. C) Neither A nor B are true. D) Both A and B are true.

Q: A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling: In-House Credit Card National Credit Card Sample Size: 86 113 Mean Monthly Purchases: $45.67 $39.87 Standard Deviation: $10.90 $12.47 Based on these sample data, what is the lower limit for the 95 percent confidence interval estimate for the difference between population means? A) About $5.28 B) Approximately $4.85 C) Approximately $2.54 D) Approximately $3.41

Q: When estimating a confidence interval for the difference between 2 means using the method where sample variances are pooled, which of the following assumptions is not needed? A) The populations are normally distributed. B) The populations have equal variances. C) The samples are independent. D) The sample sizes are equal.

Q: A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for random independent samples of cars from each population. The variances are assumed equal and the populations normally distributed. With Device Without Device 22.6 26.9 23.4 24.4 28.4 20.8 29.0 20.8 29.3 20.2 20.0 26.0 28.1 25.6 Given this data, what is the upper limit for a 95 percent confidence interval estimate for the difference in mean mpg? A) Approximately 3.88 mpg B) About 5.44 mpg C) Just under 25.0 D) None of the above

Q: A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for independent random samples of cars from each population. The variances are assumed equal and the populations normally distributed. With Device Without Device 22.6 26.9 23.4 24.4 28.4 20.8 29.0 20.8 29.3 20.2 20.0 26.0 28.1 25.6 Given this data, what is the critical value if the owners wish to have a 90 percent confidence interval estimate? A) t = 2.015 B) t = 1.7823 C) z = 1.645 D) z = 1.96

Q: If the population variances are assumed to be known in an application where a manager wishes to estimate the difference between two population means, the 95 percent confidence interval estimate can be developed using which of the following critical values? A) z = 1.645 B) z = 1.96 C) t value that depends on the sample sizes from the two populations D) z = 2.575

Q: If a manager wishes to develop a confidence interval estimate for estimating the difference between two population means, an increase in the size of the samples used will result in: A) an increase in the size of the critical value. B) a wider confidence interval. C) a more precise confidence interval. D) a less precise confidence interval

Q: A direct retailer that sells clothing on the Internet has two distribution centers and wants to determine if there is a difference between the proportion of customer order shipments that contain errors (wrong color, wrong size, etc.). It calculates a 95 percent confidence interval on the difference in the sample proportions to be -0.012 to 0.037. Based on this, it can conclude that the distribution centers differ significantly for the proportion of orders with errors.

Q: An Internet service provider is interested in estimating the proportion of homes in a particular community that have computers but do not already have Internet access. To do this, the company has selected a random sample of n = 200 homes and made calls. A total of 188 homes responded to the survey question with 38 saying that they had a computer with no Internet access. The 95 percent confidence interval estimate for the true population proportion is approximately 0.1447 - 0.2595.

Q: A major manufacturer of home electronics is interested in determining whether customers have a preference between two new speaker designs for their home entertainment centers. To test this, the design department manager has selected a random sample of customers and shown them the first design. A second sample of customers is shown design 2. The manager then asks each customer whether they prefer the new design they were shown over the one they currently own. The following results were observed: Design 1 Design 2 Sample size n1 = 150 n2 = 80 Number preferring new x1 = 65 x2 = 58 Based on these data and a significance level equal to 0.05, the test statistic is approximately -4.22 and thus the null hypothesis should be rejected.

Q: A major manufacturer of home electronics is interested in determining whether customers have a preference between two new speaker designs for their home entertainment centers. To test this, the design department manager has selected a random sample of customers and shown them the first design. A second sample of customers is shown design 2. The manager then asks each customer whether they prefer the new design they were shown over the one they currently own. The following results were observed: Design 1Design 2Sample sizen1 = 150n2 = 80Number preferring newx1 = 65x2 = 58Based on these data and a significance level equal to 0.05, the appropriate null and alternative hypotheses are:H0 : 1 2Ha : 1 < 2

Q: A direct retailer that sells clothing on the Internet has two distribution centers and wants to determine if there is a difference between the proportion of customer order shipments that contain errors (wrong color, wrong size, etc.). It takes a sample of orders from each distribution center and obtain the following results: Distribution Center 1 Distribution Center 2 Number of orders 120 145 Number of errors 4 6 Based on these data it can proceed with assuming the normal distribution for each of the proportion sampling distributions.

Q: An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results. Division 1 Division 2 Sample Size n1 = 100 n2 = 100 Errors found x1 = 13 x2 = 8 Based on this information and using a significance level equal to 0.05, the test statistic for the hypothesis test is approximately 1.153 and, therefore, the null hypothesis is not rejected.

Q: An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results. Division 1 Division 2 Sample Size n1 = 100 n2 = 100 Errors found x1 = 13 x2 = 8 Based on this information, and using a significance level equal to 0.05, the pooled estimator for the overall proportion is = .1050.

Q: An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results. Division 1 Division 2 Sample Size n1 = 100 n2 = 100 Errors found x1 = 13 x2 = 8 Based on this information and using a significance level equal to 0.05, the critical value from the standard normal table is z = 1.645.

Q: An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. The null and alternative hypotheses that will be tested are:H0 : 1 2Ha : 1 < 2

Q: The test statistic that is used when testing hypotheses about the difference between two population proportions is the t-value from the t-distribution.

Q: To test whether Model A and Model B cars have the same MPG, the first step is to select two independent random samples of drivers and assign one of them to drive Model A and the other Model B.

Q: Two placement exams are available that students can take to determine which math class they should begin with in their freshman year. It is believed that there is no difference in the mean scores that would be received for the two tests. To test this using a 0.05 level of significance, a randomly selected group of students took both tests and had their scores recorded. The following data were obtained: Student Test A Test B 1 78 82 2 86 74 3 74 79 4 72 93 5 75 80 6 68 82 7 77 99 Based on these data, the test statistic is approximately t = -1.892.

Q: When conducting a hypothesis test to determine whether or not two groups differ, using paired samples rather than independent samples has the advantage of controlling for sources of variation that might distort the conclusions of the study.

Q: If the point estimate in a paired difference estimation example does not fall in the resulting confidence interval, the decision maker can conclude that the two populations likely have different means.

Q: In comparing two populations using paired differences, after the difference is found for each pair, the method for testing whether the mean difference is equal to 0 becomes the same as was used for a one-sample hypothesis test with unknown standard deviation.

Q: If you are interested in estimating the difference between the means of two samples that have been paired, the point estimate for this difference is the mean value of the paired differences.

Q: If the sample data lead us to suspect that the variances of the two populations are not equal, the t-test statistic and the degrees of freedom must be adjusted accordingly.

Q: In order to test the difference in populations means, samples were collected for two independent populations where the variances are assumed equal and the population normally distributed. The following data resulted: Population 1 Population 2 = 112 = 107 s = 14 s = 17 n = 25 n = 28 The value of the pooled standard deviation is 15.66.

Q: The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The following data were reported recently where the values represent the percent moisture in the wood: Pine Fir 13.1 19.8 10.2 13.5 14.7 20.3 17.8 20.5 17.6 21.4 19.2 23.4 7.4 14.6 13.3 18.8 17.3 10.7 Based on these data, the critical t value from the t-distribution will be 1.7459 if the significance level is set at 0.05 and variances are presumed equal.

Q: The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The hypothesis test that will be conducted using an alpha = 0.05 level will be a two-tailed test.

Q: The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The following data were reported recently where the values represent the percent of moisture in the wood: PineFir13.119.810.213.514.720.317.820.517.621.419.223.47.414.613.318.817.3 10.7 The null and alternative hypotheses to be tested areH0 : p fHa : p < f

Q: To find the pooled standard deviation involves taking a weighted average of the two sample variances, then finding its square root.

Q: There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts Joint Accounts n = 20 n = 30 s = $256 s = $300 = $1,123 = $1,245 Based upon these data, the critical value from the t-distribution for testing the difference between the two population means using a significance level of 0.05 is t = 1.6772.

Q: There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts Joint Accounts n = 20 n = 30 s = $256 s = $300 = $1,123 = $1,245 Based upon these data, assuming that the populations are normally distributed with equal variances, the test statistic for testing whether the two populations have equal means is approximately -1.49.

Q: When performing a hypothesis test for the difference between the means of two independent populations where the standard deviations are known, it is necessary to use the pooled standard deviation in calculating the test statistic.

Q: The t-distribution can be used to test hypotheses about the difference between two population means given the following two assumptions: - each population is normally distributed, and - the two populations have equal variances.

Q: Two samples are said to be independent if they are collected at different points in time.

Q: When performing a hypothesis test for the difference between the means of two independent populations, where the standard deviations are known, the variances must be assumed equal.

Q: In order to make the test for the difference between two population means valid, the sample size in each independent sample must be the same.

Q: The t-distribution is still applicable even when there are small violations of the assumptions for the case when the variances for two populations are unknown. This is particularly true when the sample sizes are approximately equal.

Q: All other things held constant, increasing the level of confidence for a confidence interval estimate for the difference between two population means will result in a wider confidence interval estimate.

Q: Increasing the size of the samples in a study to estimate the difference between two population means will increase the level of confidence that a decision maker can have regarding the interval estimate.

Q: Recently the managers for a large retail department store stated that a study has revealed that female shoppers spend on average 23.5 minutes longer in the store per visit than do male shoppers. Based on this information, the managers can be confident that female shoppers, as a population, do spend longer times in the store than do males shoppers, as a population.

Q: In estimating a confidence interval for the difference between two means, when the samples are independent and the standard deviations are unknown, it can be acceptable for there to be small violations of the assumptions of normality and equal variances, especially when the sample sizes are equal.