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The new director of a local YMCA has been told by his predecessors that the average member has belonged for 8.7 years. Examining a random sample of 15 membership files, he finds the mean length of of 2.5 years. Assuming the population is approximately normally distributed, and using the 0.05 level, does this result suggest that the actual mean length of membership to be 7.2 years, with a standard deviation membership may be some value other than 8.7 years?A scrap metal dealer claims that the mean of his cash sales is "no more than $80," but an Inter nal Revenue Service agent believes the dealer is untruthful. Observing a sample of 20 cash customers, the agent finds the mean purchase to be $91, with a standard deviation of $21. Assuming the population is approximately normally distributed, and using the 0.05 level of significance, is the agent's suspicion confirmed?Taxco, a firm specializing in the preparation of income tax returns, claims the mean refund for customers who received refunds last year was $150. For a random sample of 12 customers who received refunds last year, the mean amount was found to be $125, with a standard deviation of $43. Assuming that the population is approximately normally distributed, and using the 0.10 level in a two-tail test, do these results suggest that Taxco's assertion may be accurate?During 2008, college work-study students earned a mean of $1478. Assume that a sample consisting of 45 of the work-study students at a large university was found to have earned a mean of $1503 during that year, with a standard deviation of $210. Would a one-tail test at the 0.05 level suggest the average earnings of this university's work-study students were significantly higher than the national mean?According to the Federal Reserve Board, the mean net worth of U.S. households headed by persons 75 years or older is $840,000. Suppose a simple random sample of 50 households in this age group is obtained from a certain region of the United States and is found to have a mean net worth of $815,000, with a standard deviation of $120,000. From these sample results, and using the 0.05 level of significance in a two-tail test, comment on whether the mean net worth for all the region's households in this age category might not be the same as the mean value reported for their counterparts across the nation.Using the sample results in Exercise 10.40, construct and interpret the 95% confidence interval for the population mean. Is the hypothesized population mean ($840,000) within the interval? Given the presence or absence of the $640.000 value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 10.49? Reference: Exercise 10.49 According to the Federal Reserve Board, the mean net worth of U.S. households headed by persons 75 years or older is $840,000. Suppose a simple random sample of 50 households in this age group is obtained from a certain region of the United States and is found to have a mean net worth of $615,000, with a standard deviation of $120,000. From these sample results, and using the 0.05 level of significance in a two-tail test, comment on whether the mean net worth for all the region's households in this age category might not be the same as the mean value reported for their counterparts across the nation.It has been reported that the average life for halogen lightbulbs is 4000 hours. Learning of this figure, a plant manager would like to find out whether the vibration and temperature conditions that the facility's bulbs encounter might be having an adverse effect on the service life of bulbs in her plant. In a test involving 15 halogen bulbs installed in various locations around the plant, she finds the average life for bulbs in the sample is 3882 hours, with a standard deviation of 200 hours. Assuming the population of halogen bulb lifetimes to be approximately normally distributed, and using the 0.025 level of significance, do the test results tend to support the manager's suspicion that adverse conditions might be detrimental to the operating lifespan of halogen lightbulbs used in her plant?In response to an inquiry from its national office, the manager of a local bank has stated that her bank's average service time for a drive-through customer is 93 seconds. A student intern working at the bank happens to be taking a statistics course and is curious as to whether the true average might be some value other than 93 seconds. The intern observes a simple random sample of 50 drive-through customers whose average service time is 89.5 seconds, with a standard deviation of 11.3 seconds. From these sample results, and using the 0.05 level of significance, what conclusion would the student reach with regard to the bank manager's claim?Using the sample results in Exercise 10.52, construct and interpret the 95% confidence interval for the population mean. Is the hypothesized population mean (93 seconds) within the interval? Given the presence or absence of the 93 seconds value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 10.52? Reference: Exercise 10.52 In response to an inquiry from its national office, the manager of a local bank has stated that her bank's average service time for a drive-through customer is 93 seconds. A student intern working at the bank happens to be taking a statistics course and is curious as to whether the true average might be some value other than 93 seconds. The intern observes a simple random sample of 50 drive-through customers whose average service time is 89.5 seconds, with a standard deviation of 11.3 seconds. From these sample results, and using the 0.05 level of significance, what conclusion would the student reach with regard to the bank manager's claim?The U.S. Census Bureau says the 52-question "long form" received by 1 in 3 households during the 2000 census takes a mean of 38 minutes to complete. Suppose a simple random sample of 35 persons is given the form, and their mean time to complete it is 36.8 minutes, with a standard deviation of 4.0 minutes. From these sample results, and using the 0.10 level of significance, would it seem that the actual population mean time for completion might be some value other than 38 minutes?Using the sample results in Exercise 10.54, construct and interpret the 90% confidence interval for the population mean. Is the hypothesized population mean (38 minutes) within the interval? Given the presence or absence of the 38 minutes value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 10.54? Reference: Exercise 10.54 The U.S. Census Bureau says the 52-question "long form" received by 1 in 8 households during the 2000 census takes a mean of 38 minutes to complete. Suppose a simple random sample of 35 persons is given the form, and their mean time to complete it is 36.8 minutes, with a standard deviation of 4.0 minutes. From these sample results, and using the 0.10 level of significance, would it seem that the actual population mean time for completion might be some value other than 38 minutes?( DATA SET ) Note: Exercises require a computer and statistical software. The International Council of Shopping Centers reports that the average teenager spends $57 during a shopping trip to the mall. The promotions director of a local mall has used a variety of strategies to attract area teens to his mall, including live bands and "teenappreciation days" that feature special bargains for this age group. He believes teen shoppers at his mall respond to his promotional efforts by shopping there more often and spending more when they do. Mall management decides to evaluate the promotions director's success by surveying a simple random sample of 45 local teens and finding out how much they spent on their most recent shopping visit to the mall. The results are listed in data file XR10056. Use a suitable hypothesis test in examining whether the mean mall shopping expenditure for teens in this area might be higher than for U.S. teens as a whole. Identify and interpret the p-value for the test. Using the 0.025 level of significance, what conclusion do you reach?( DATA SET ) Note: Exercises require a computer and statistical software. Using the sample data in Exercise 10.57, construct and interpret the 95%% confidence interval for the population mean. Is the hypothesized population mean ($817) within the interval? Given the presence or absence of the $817 value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 10.57? Reference: Exercise 10.57 According to the Insurance Information Institute, the mean annual expenditure for automobile insurance for U.S. motorists is $817. Suppose that a government official in North Carolina has surveyed a simple random sample of 80 residents of her state, and that their auto insurance expenditures for the most recent year are in data file XR10057. Based on these data, examine whether the mean annual auto insurance expenditure for motorists in North might be different from the $817 for the country as a whole. Identify and interpret the p-value for the test. Using the 0.05 level of significance, what conclusion do you reach?When carrying out a hypothesis test for a population proportion, under what conditions is it appropriate to use the normal distribution as an approximation to the (theoretically correct) binomial distribution?For a simple random sample, n = 200 and p = 0.34. At the 0.01 level, test H: w = 0.40 versus H. : T # 0.40.A simple random sample of 300 items is selected from a large shipment, and testing reveals that 4% of the sampled items are defective. The supplier claims that no more than 2% of the items in the shipment are defective. Carry out an appropriate hypothesis test and comment on the credibility of the supplier's claim.According to the human resources director of a plant, no more than 5%% of employees hired in the past year have violated their preemployment agreement not to use any of five illegal drugs. The agreement specified that random urine checks could be carried out to ascertain compliance In a random sample of 400 employees, screening detected at least one of these drugs in the systems of 8% of those tested. At the 0.025 level, is the human resources director's claim credible? Determine and interpret the p-value for the test.In the past, 44% of those taking a public accounting qualifying exam have passed the exam on their first try. Lately, the availability of exam preparation books and tutoring sessions may have improved the likelihood of an individual's passing on his or her first try. In a sample of 250 recent applicants, 130 passed on their first attempt. At the 0.05 level of significance, can we conclude that the proportion passing on the first try has increased? Determine and interpret the p-value for the test.According to the National Association of Home Builders, 55%% of new single-family homes built during 2005 had a fireplace. Suppose a nation wide homebuilder has claimed that its homes are a cross section of America," but a simple random sample of 800 of its single-family homes built during that year included only 50.0%% that had a fireplace. Using the 0.05 level of significance in a two-tail test, examine whether the percentage of sample homes having a fireplace could have differed from 55% simply by chance. Determine and interpret the p-value for the test.Based on the sample results in Exercise 10.89, construct and interpret the 95% confidence interval for the population proportion. Is the hypothesized proportion (0.55) within the interval? Given the presence or absence of the 0.55 value within the interval, is this consistent with the findings of the hypothesis test conducted in Exercise 10.89? Reference: Exercise 10.69 According to the National Association of Home Builders, 55%% of new single-family homes built during 2005 had a fireplace. Suppose a nation wide homebuilder has claimed that its homes are "a cross section of America," but a simple random sample of 800 of its single-family homes built during that year included only 50.0%% that had a fireplace. Using the 0.05 level of significance in a two-tail test, examine whether the percentage of sample homes having a fireplace could have differed from 55% simply by chance. Determine and interpret the p-value for the test
