Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 9 Introduction to Hypothesis Testing 1) Hypothesis testing and confidence interval estimation are essentially two totally different statistical procedures and share little in common with each other. Answer: FALSE 2) In hypothesis testing, the null hypothesis should contain the equality sign. Answer: TRUE 3) When the decision maker has control over the null and alternative hypotheses, the alternative hypotheses should be the ʺresearchʺ hypothesis. Answer: TRUE 4) In testing a hypothesis, statements for the null and alternative hypotheses as well as the selection of the level of significance should precede the collection and examination of the data. Answer: TRUE 5) The null and alternate hypotheses must be opposites of each other. Answer: TRUE 6) A sample is used to obtain a 95 percent confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05. Answer: TRUE 7) A conclusion to ʺnot rejectʺ the null hypothesis is the same as the decision to ʺaccept the null hypothesisʺ. Answer: FALSE 8) A one‐tailed hypothesis for a population mean with a significance level equal to .05 will have a critical value equal to z = .45. Answer: FALSE 9) Whenever possible, in establishing the null and alternative hypotheses, the research hypothesis should be made the alternative hypothesis. Answer: TRUE 10) If a hypothesis test is conducted for a population mean, a null and alternative hypothesis of the form: H0 : μ = 100 Ha: μ ≠ 100 will result in a one‐tailed hypothesis test since the sample result can fall in only one tail. Answer: FALSE 11) If the sample data lead the decision maker to reject the null hypothesis, the alpha level is the maximum probability of committing a Type II error. Answer: FALSE 12) The following is an appropriate statement of the null and alternate hypotheses for a test of a population mean: H0 : μ < 50 Ha : μ > 50 Answer: FALSE 13) In a one‐tailed hypothesis test, the larger the significance level, the greater the critical value will be. Answer: FALSE 14) In a two‐tailed hypothesis test the area in each tail of the rejection region is equal to α. Answer: FALSE 15) A local medical center has advertised that the mean wait for services will be less than 15 minutes. Given this claim, the hypothesis test for the population mean should be a one‐tailed test with the rejection region in the lower (left‐hand) tail of the sampling distribution. Answer: TRUE 16) A local medical center has advertised that the mean wait for services will be less than 15 minutes. In an effort to test whether this claim can be substantiated, a random sample of 100 customers was selected and their wait times were recorded. The mean wait time was 17.0 minutes. Based on this sample result, there is sufficient evidence to reject the medical centerʹs claim. Answer: FALSE 9‐1 Copyright © 2014 Pearson Education, Inc. 17) When using the t‐distribution in a hypothesis test, the population does not need to be assumed normally distributed. Answer: FALSE 18) In a hypothesis test, the p‐value measures the probability that the alternative hypothesis is true. Answer: FALSE 19) The Adams Shoe Company believes that the mean size for menʹs shoes is now more than 10 inches. To test this, it has selected a random sample of n = 100 men. Assuming that the test is to be conducted using a .05 level of significance, a p‐value of .07 would lead the company to conclude that its belief is correct. Answer: FALSE 20) In conducting a hypothesis test where the conclusion is to reject the null hypothesis, then either a correct decision has been made or else a Type I error. Answer: TRUE 21) A large tire manufacturing company has claimed that its top line tire will average more than 80,000 miles. If a consumer group wished to test this claim, they would formulate the following null and alternative hypotheses: H0 : μ ≥ 80,000 Hα : μ ≠ 80,000 Answer: FALSE 22) A large tire manufacturing company has claimed that its top line tire will average more than 80,000 miles. If a consumer group wished to test this claim, the research hypothesis would be: Ha : μ > 80,000 miles. Answer: TRUE 23) A report recently published in a major business periodical stated that the average salary for female managers is less than $50,000. If we were interested in testing this, the following null and alternative hypotheses would be established: H0 : μ ≥ 50,000 Hα : μ < 50,000 Answer: TRUE 24) When someone has been accused of a crime the null hypothesis is: H0 : innocent. In this case, a Type I error would be convicting an innocent person. Answer: TRUE 25) Of the two types of statistical errors, the one that decision makers have most control over is Type I error. Answer: TRUE 26) If a hypothesis test leads to incorrectly rejecting the null hypothesis, a Type II statistical error has been made. Answer: FALSE 27) The police chief in a local city claims that the average speed for cars and trucks on a stretch of road near a school is at least 45 mph. If this claim is to be tested, the null and alternative hypotheses are: H0 : μ < 45 mph Ha : μ ≥ 45 mph Answer: FALSE 28) When a battery company claims that their batteries last longer than 100 hours and a consumer group wants to test this claim, the hypotheses should be: H0 : μ ≤ 100 Ha : μ > 100 Answer: TRUE 29) A report recently submitted to the managing partner for a market research company stated ʺthe hypothesis test may have resulted in either a Type I or a Type II error. We wonʹt know which one occurred until later.ʺ This statement is one that we might correctly make for any hypothesis that we have conducted. Answer: FALSE 30) The significance level in a hypothesis test corresponds to the maximum probability that a Type I error will be committed. Answer: TRUE 31) If the probability of a Type I error is set at 0.05, then the probability of a Type II error will be 0.95. Answer: FALSE 32) Type II error is failing to reject the null hypothesis when the null is actually false. Answer: TRUE 9‐2 Copyright © 2014 Pearson Education, Inc. 33) The critical value in a null hypothesis test is called alpha. Answer: FALSE 34) The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the value of that corresponds to the critical value is approximately $14,993.50. Answer: TRUE 35) The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the null and alternative hypotheses to be tested are: H0 : μ ≤ $14,500 HA : μ > $14,500 Answer: TRUE 36) The director of the city Park and Recreation Department claims that the mean distance people travel to the cityʹs greenbelt is more than 5.0 miles. Assuming that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed, the critical value is approximately 4.75 miles. Answer: FALSE 37) The director of the city Park and Recreation Department claims that the mean distance people travel to the cityʹs greenbelt is more than 5.0 miles. Assume that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed. Given this information, if the sample mean is 15.90 miles, the null hypothesis should be rejected. Answer: TRUE 38) A two‐tailed hypothesis test with α = 0.05 is similar to a 95 percent confidence interval. Answer: TRUE 39) The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Given this, the appropriate null and alternative hypotheses are H0 : ≤ $1,700 HA : > $1,700 Answer: FALSE 40) The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Based on this, the null hypothesis should be rejected if > $1,854.66 approximately. Answer: TRUE 41) When using the p‐value method, the null hypothesis is rejected when the calculated p‐value > α. Answer: FALSE 42) Generally, it is possible to appropriately test a null and alternative hypotheses using the test statistic approach and reach a different conclusion than would be reached if the p‐value approach were used. Answer: FALSE 43) A two‐tailed hypothesis test is used when the null hypothesis looks like the following: H0 : = 100. Answer: FALSE 44) When using the p‐value method for a two‐tailed hypothesis, the p‐value is found by finding the area in the tail beyond the test statistic, then doubling it. Answer: TRUE 45) Lube‐Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on a significance level of 0.10, the critical value for the test is approximately z = 1.28. Answer: TRUE 9‐3 Copyright © 2014 Pearson Education, Inc. 46) Lube‐Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the sample standard deviation was 780 miles. Based on this information, the test statistic is approximately t = 1.000. Answer: TRUE 47) Lube‐Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on this information, the p‐value for the hypothesis test is less than 0.10. Answer: FALSE 48) For testing a research hypothesis, the burden of proof that a new product is no better than the original is placed on the new product, and the research hypothesis is formulated as the null hypothesis. Answer: FALSE 49) When deciding the null and alternative hypotheses, the rule of thumb is that if the claim contains the equality (e.g., at least, at most, no different from, etc.), the claim becomes the null hypothesis. If the claim does not contain the equality (e.g., less than, more than, different from), the claim is the alternative hypothesis. Answer: TRUE 50) The executive director of the United Way believes that more than 24 percent of the employees in the high‐tech industry have made voluntary contributions to the United Way. In order to test this statistically, the appropriate null and alternative hypotheses are: H0 : ≤ .24 HA : > .24 Answer: FALSE 51) A company that makes and markets a device that is aimed at helping people quit smoking claims that at least 70 percent of the people who have used the product have quit smoking. To test this, a random sample of n = 100 product users was selected. Of these, 65 people were found to have quit smoking. Given these results, the test statistic value is z = ‐1.0911. Answer: TRUE 52) A company that makes and markets a device that is aimed at helping people quit smoking claims that at least 70 percent of the people who have used the product have quit smoking. To test this, a random sample of n = 100 product users was selected. The critical value for the hypothesis test using a significance level of 0.05 would be approximately ‐1.645. Answer: TRUE 53) A cell phone company believes that 90 percent of its customers are satisfied with their service. They survey n = 30 customers. Based on this, it is acceptable to assume the sample distribution is normally distributed. Answer: FALSE 54) Aceco has a contract with a supplier to ship parts that contain no more than three percent defects. When a large shipment of parts comes in, Aceco samples n = 150. Based on the results of the sample, they either accept the shipment or reject it. If Aceco wants no more than a 0.10 chance of rejecting a good shipment, the cut‐off between accepting and rejecting should be 0.0478 or 4.78 percent of the sample. Answer: TRUE 55) When testing a hypothesis involving population proportions, an increase in sample size will result in a smaller chance of making a Type I statistical error. Answer: FALSE 56) When the hypothesized proportion is close to 0.50, the spread in the sampling distribution of is greater than when the hypothesized proportion is close to 0.0 or 1.0. Answer: TRUE 57) A major package delivery company claims that at least 95 percent of the packages it delivers reach the destination on time. As part of the evidence in a lawsuit against the package company, a random sample of n = 200 packages was selected. A total 9‐4 Copyright © 2014 Pearson Education, Inc. of 188 of these packages were delivered on time. Using a significance level of 0.05, the critical value for this hypothesis test is approximately 0.90. Answer: FALSE 58) A major package delivery company claims that at least 95 percent of the packages it delivers reach the destination on time. As part of the evidence in a lawsuit against the package company, a random sample of n = 200 packages was selected. A total of 188 of these packages were delivered on time. Using a significance level of .05, the test statistic for this test is 59) A cell phone company believes that 90 percent of their customers are satisfied. They survey a sample of n = 100 customers and find that 82 say they are satisfied. In calculating the standard error of the sampling distribution (σp) the proportion to use is 0.82. Answer: FALSE 60) One claim states the IRS conducts audits for not more than 5 percent of total tax returns each year. In order to test this claim statistically, the appropriate null and alternative hypotheses are: H0 : μ ≤ 0.05 Ha : μ > 0.05 Answer: FALSE 61) In a hypothesis test, increasing the sample size will generally result in a smaller chance of making a Type I error since sampling error is likely to be reduced. Answer: TRUE 62) An article in an operations management journal recently stated that a formal hypothesis test rejected the hypothesis that mean employee productivity was less than $45.70 per hour in the wood processing industry. Given this conclusion, it is possible that a Type I statistical error was committed. Answer: TRUE 63) The chance of making a Type II statistical error increases if the ʺtrueʺ population mean is closer to the hypothesized population mean, all other factors held constant. Answer: TRUE 64) Choosing an alpha of 0.01 will cause beta to equal 0.99. Answer: FALSE 65) If a decision maker is concerned that the chance of making a Type II error is too large, one option that will help reduce the risk is to reduce the significance level. Answer: FALSE 66) Type II errors are typically greater for two‐tailed hypothesis tests than for one‐tailed tests. Answer: FALSE 67) If a decision maker wishes to reduce the chance of making a Type II error, one option is to increase the sample size. Answer: TRUE 68) To calculate beta requires making a ʺwhat ifʺ assumption about the true population parameter, where the ʺwhat‐ifʺ value is one that would cause the null hypothesis to be false. Answer: TRUE 69) The probability of a Type II error decreases as the ʺtrueʺ population value gets farther from the hypothesized population value, given that everything else is held constant. Answer: TRUE 70) A city newspaper has stated that the average time required to sell a used car advertised in the paper is less than 5 days. Assuming that the population standard deviation is 2.1 days, if the ʺtrueʺ population mean is 4.1 days and a sample size of n = 49 is used with an alpha equal to 0.05, the probability that the hypothesis test will lead to a Type II error is approximately .0869. Answer: TRUE 71) The director of the city Park and Recreation Department claims that the mean distance people travel to the cityʹs greenbelt is more than 5.0 miles. Assuming that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed, the probability of a Type II error is approximately .4545 when the ʺtrueʺ population mean is 5.5 miles. Answer: FALSE 9‐5 Copyright © 2014 Pearson Education, Inc. 72) A major airline has stated in an industry report that its mean onground time between domestic flights is less than 18 minutes. To test this, the company plans to sample 36 randomly selected flights and use a significance level of 0.10. Assuming that the population standard deviation is known to be 4.0 minutes, the probability that the null hypothesis will be ʺacceptedʺ if the true population mean is 16 minutes is approximately 0.955. Answer: TRUE 73) A major airline has stated in an industry report that its mean onground time between domestic flights is less than 18 minutes. To test this, the company plans to sample 36 randomly selected flights and use a significance level of .10. Assuming that the population standard deviation is known to be 4.0 minutes, if the true population mean is 16 minutes, the decision maker could end up making either a Type I or a Type II error depending on the sample result. Answer: FALSE 74) When someone is on trial for suspicion of committing a crime, the hypotheses are: H0 : innocent HA : guilty Which of the following is correct? A) Type I error is acquitting a guilty person. B) Type I error is convicting an innocent person. C) Type II error is acquitting an innocent person. D) Type II error is convicting an innocent person. 75) Which of the following statements is true? A) The decision maker controls the probability of making a Type I statistical error. B) Alpha represents the probability of making a Type II error. C) Alpha and beta are directly related such that when one is increased the other will increase also. D) The alternative hypothesis should contain the equality. 76) In a hypothesis test involving a population mean, which of the following would be an acceptable formulation? A) H0 : ≤ $1,700 Ha : > $1,700 B) H0 : > $1,700 Ha : ≥ $1,700 C) H0 : μ ≤ $1,700 Ha : μ > $1,700 D) None of the above is a correct formulation. 77) Which of the following would be an appropriate null hypothesis? A) The mean of a population is equal to 55. B) The mean of a sample is equal to 55. C) The mean of a population is greater than 55. D) The mean of a sample is greater than 55. 78) If we are performing a two‐tailed test of whether μ = 100, the probability of detecting a shift of the mean to 105 will be ________ the probability of detecting a shift of the mean to 110. A) less than B) greater than C) equal to D) not comparable to 79) If an economist wishes to determine whether there is evidence that average family income in a community exceeds $25,000. The best null hypothesis is: A) μ = 25,000. B) μ > 25,000. C) μ ≤ 25,000. D) μ ≥ 25,000. 80) If the p value is less than α in a two‐tailed test, A) the null hypothesis should not be rejected. B) the null hypothesis should be rejected. C) a one‐tailed test should be used. D) More information is needed to reach a conclusion about the null hypothesis. 81) A hypothesis test is to be conducted using an alpha = .05 level. This means: A) there is a 5 percent chance that the null hypothesis is true. B) there is a 5 percent chance that the alternative hypothesis is true. C) there is a maximum 5 percent chance that a true null hypothesis will be rejected. D) there is a 5 percent chance that a Type II error has been committed. 82) In a two‐tailed hypothesis test for a population mean, an increase in the sample size will: A) have no effect on whether the null hypothesis is true or false. B) have no effect on the significance level for the test. C) result in a sampling distribution that has less variability. D) All of the above are true. 9‐6 Copyright © 2014 Pearson Education, Inc. 83) The reason for using the t‐distribution in a hypothesis test about the population mean is: A) the population standard deviation is unknown. B) it results in a lower probability of a Type I error occurring. C) it provides a smaller critical value than the standard normal distribution for a given sample size. D) the population is not normally distributed. 84) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, which of the following would be the upper tail critical value? A) 1.28 B) 1.645 C) 1.96 D) 2.575 85) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, which of the following would be the correct formulation of the null and alternative hypotheses? A) H0 : = 16 HA : = 16 B) H0 : μ = 16 HA : μ ≠ 16 C) H0 : μ ≥ 16 HA : μ < 16 D) H0 : ≥ 16 HA : < 16 86) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, how large could the sample mean be before they would reject the null hypothesis? A) 16.2 ounces B) 16.049 ounces C) 15.8 ounces D) 16.041 ounces 87) The cost of a college education has increased at a much faster rate than costs in general over the past twenty years. In order to compensate for this, many students work part‐ or full‐time in addition to attending classes. At one university, it is believed that the average hours students work per week exceeds 20. To test this at a significance level of 0.05, a random sample of n = 20 students was selected and the following values were observed: 26 15 10 40 10 20 30 36 40 0 5 10 20 32 16 12 40 36 10 0 Based on these sample data, which of the following statements is true? A) The standard error of the sampling distribution is approximately 3.04. B) The test statistic is approximately t = 0.13. C) The research hypothesis that the mean hours worked exceeds 20 is not supported by these sample data. D) All of the above are true. 88) Based on these sample data, the critical value expressed in hours: A) is approximately equal to 25.26 hours. B) is approximately equal to 25.0 hours. C) cannot be determined without knowing the population standard deviation. D) is approximately 22 hours. 89) The R.D. Wilson Company makes a soft drink dispensing machine that allows customers to get soft drinks from the machine in a cup with ice. When the machine is running properly, the average number of fluid ounces in the cup should be 14. Periodically the machines need to be tested to make sure that they have not gone out of adjustment. To do this, six cups are filled by the machine and a technician carefully measures the volume in each cup. In one such test, the following data were observed: 14.25 13.7 14.02 14.13 13.99 14.04 Which of the following would be the correct null hypothesis if the company wishes to test the machine? A) H0 : = 14 ounces B) H0 : μ = 14 ounces C) H0 : μ ≠ 14 ounces D) H0 : ≠ 14 ounces 9‐7 Copyright © 2014 Pearson Education, Inc. 90) Based on these sample data, which of the following is true if the significance level is .05? A) No conclusion can be reached about the status of the machine based on a sample size of only six cups. B) The null hypothesis cannot be rejected since the test statistic is approximately t = 0.20, which is not in the rejection region. C) The null hypothesis can be rejected since the sample mean is greater than 14. D) The null can be rejected because the majority of the sample values exceed 14. 91) A concern of Major League Baseball is that games last too long. Some executives in the leagueʹs headquarters believe that the mean length of games this past year exceeded 3 hours (180 minutes). To test this, the league selected a random sample of 80 games and found the following results: = 193 minutes and s = 16 minutes. Based on these results, if the null hypothesis is tested using an alpha level equal to 0.10, which of the following is true? A) The null hypothesis should be rejected if > 182.31. B) The test statistic is t = 1.2924. C) Based on the sample data, the null hypothesis cannot be rejected. D) It is possible that when the hypothesis test is completed, a Type II statistical error has been made. 92) When testing a two‐tailed hypothesis using a significance level of 0.05, a sample size of n = 16, and with the population standard deviation unknown, which of the following is true? A) The null hypothesis can be rejected if the sample mean gets too large or too small compared with the hypothesized mean. B) The alpha probability must be split in half and a rejection region must be formed on both sides of the sampling distribution. C) The test statistic will be a t‐value. D) All of the above are true. 93) A major airline is concerned that the waiting time for customers at its ticket counter may be exceeding its target average of 190 seconds. To test this, the company has selected a random sample of 100 customers and times them from when the customer first arrives at the checkout line until he or she is at the counter being served by the ticket agent. The mean time for this sample was 202 seconds with a standard deviation of 28 seconds. Given this information and the desire to conduct the test using an alpha level of 0.02, which of the following statements is true? A) The chance of a Type II error is 1 ‐ 0.02 = 0.98. B) The test to be conducted will be structured as a two‐tailed test. C) The test statistic will be approximately t = 4.286, so the null hypothesis should be rejected. D) The sample data indicate that the difference between the sample mean and the hypothesized population mean should be attributed only to sampling error. 94) A house cleaning service claims that it can clean a four bedroom house in less than 2 hours. A sample of n = 16 houses is taken and the sample mean is found to be 1.97 hours and the sample standard deviation is found to be 0.1 hours. Using a 0.05 level of significance the correct conclusion is: A) reject the null because the test statistic (‐1.2) is < the critical value (1.7531). B) do not reject the null because the test statistic (1.2) is > the critical value (‐1.7531). C) reject the null because the test statistic (‐1.7531) is < the critical value (‐1.2). D) do not reject the null because the test statistic (‐1.2) is > the critical value (‐1.7531). 95) The manager of an online shop wants to determine whether the mean length of calling time of its customers is significantly more than 3 minutes. A random sample of 100 customers was taken. The average length of calling time in the sample was 3.1 minutes with a standard deviation of 0.5 minutes. At a 0.05 level of significance, it can be concluded that the mean of the population is: A) significantly greater than 3. B) not significantly greater than 3. C) significantly less than 3. D) not significantly different from 3.10. 96) Woof Chow Dog Food Company believes that it has a market share of 25 percent. It surveys n = 100 dog owners and ask whether or not Woof Chow is their regular brand of dog food. The appropriate null and alternate hypotheses are: A) H0 : ρ = .25 Ha : ρ ≠ .25 B) H0 : p = .25 Ha : p ≠ .25 9‐8 Copyright © 2014 Pearson Education, Inc. C) H0 : μ = .25 Ha : μ ≠ .25 D) H0 : p ≤ .25 Ha : p > .25 97) Woof Chow Dog Food Company believes that it has a market share of 25 percent. It surveys n = 100 dog owners and ask whether or not Woof Chow is their regular brand of dog food, and 23 people say yes. Based upon this information, what is the critical value if the hypothesis is to be tested at the 0.05 level of significance? A) 1.28 B) 1.645 C) 1.96 D) 2.575 98) Based upon this information, what is the value of the test statistic? A) ‐0.462 B) ‐0.475 C) 0.462 D) 0.475 99) After completing sales training for a large company, it is expected that the salesperson will generate a sale on at least 15 percent of the calls he or she makes. To make sure that the sales training process is working, a random sample of n = 400 sales calls made by sales representatives who have completed the training have been selected and the null hypothesis is to be tested at 0.05 alpha level. Suppose that a sale is made on 36 of the calls. Based on this information, what is the test statistic for this test? A) Approximately 0.1417 B) About z = ‐3.35 C) z = ‐1.645 D) t = ‐4.567 100) After completing sales training for a large company, it is expected that the salesperson will generate a sale on at least 15 percent of the calls he or she makes. To make sure that the sales training process is working, a random sample of n = 400 sales calls made by sales representatives who have completed the training have been selected and the null hypothesis is to be tested at 0.05 alpha level. Suppose that a sale is made on 36 of the calls. Based on these sample data, which of the following is true? A) The null hypothesis should be rejected since the test statistic falls in the lower tail rejection region. B) The null hypothesis is supported since the sample results do not fall in the rejection region. C) There is insufficient evidence to reject the null hypothesis and the sample proportion is different from the hypothesized proportion due to sampling error. D) It is possible that a Type II statistical error has been committed. 101) Mike runs for the president of the student government and is interested to know whether the proportion of the student body in favor of him is significantly more than 50 percent. A random sample of 100 students was taken. Fifty‐five of them favored Mike. At a 0.05 level of significance, it can be concluded that the proportion of the students in favor of Mike A) is significantly greater than 50 percent because 55 percent of the sample favored him. B) is not significantly greater than 50 percent. C) is significantly greater than 55 percent. D) is not significantly different from 55 percent. 102) Which of the following is not a required step in finding beta? A) Assuming a true value of the population parameter where the null is false B) Finding the critical value based on the null hypothesis C) Converting the critical value from the standard normal distribution to the units of the data D) Finding the power of the test 103) If the Type I error (α) for a given test is to be decreased, then for a fixed sample size n: A) the Type II error (β) will also decrease. B) the Type II error (β) will increase. C) the power of the test will increase. D) a one‐tailed test must be utilized. 104) For a given sample size n, if the level of significance (α) is decreased, the power of the test: A) will increase. B) will decrease. C) will remain the same. D) cannot be determined. 105) The power of a test is measured by its capability of: A) rejecting a null hypothesis that is true. B) not rejecting a null hypothesis that is true. C) rejecting a null hypothesis that is false. D) not rejecting a null hypothesis that is false. 9‐9 Copyright © 2014 Pearson Education, Inc. 106) Which of the following will be helpful if the decision maker wishes to reduce the chance of making a Type II error? A) Increase the level of significance at which the hypothesis test is conducted. B) Increase the sample size. C) Both A and B will work. D) Neither A nor B will be effective. 107) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, if the ʺtrueʺ population mean is 32.0 mpg, what is the probability that the test will lead the consumer group to ʺacceptʺ the claimed mileage for this car? A) About 0.45 B) Approximately 0.0455 C) About 0.9545 D) None of the above 108) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, what is the critical value in terms of miles per gallon that would be needed prior to finding beta? A) 32.5065 B) 33.4935 C) 33.588 D) 32.412 109) Suppose we want to test H0 : μ ≥ 30 versus H1 : μ < 30. Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0 in favor of H1? A) = 28, s = 6 B) = 27, s = 4 C) = 32, s = 2 D) = 26, s = 9 110) A contract calls for the mean diameter of a cylinder to be 1.50 inches. As a quality check, each day a random sample of n = 36 cylinders is selected and the diameters are measured. Assuming that the population standard deviation is thought to be 0.10 inch and that the test will be conducted using an alpha equal to 0.025, what would the probability of a Type II error be? A) Approximately 0.1267 B) About 0.6789 C) 0.975 D) Canʹt be determined without knowing the ʺtrueʺ population mean. 111) A company that sells an online course aimed at helping high‐school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the companyʹs course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, what is the probability that the counseling organization will incorrectly ʺacceptʺ the null hypothesis when, in fact, the true mean increase is actually 95 points? A) Approximately 0.508 B) About 0.492 C) Approximately 0.008 D) Canʹt be determined without knowing the sample results. 112) A company that sells an online course aimed at helping high‐school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the companyʹs course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, find the critical value in terms of improvement in SAT points, which would be needed prior to finding a beta. A) Reject the null if SAT improvement is > 95 points. B) Reject the null if SAT improvement is < 85.065 points. C) Reject the null if SAT improvement is > 95.88 points. D) Reject the null if SAT improvement is > 94.935 points. 113) A recent report in which a major pharmaceutical company released the results of testing that had been done on the cholesterol reduction that people could expect if they use the companyʹs new drug indicated that the Type II error probability for a given ʺtrueʺ mean was 0.1250 based on the sample size of n = 64 subjects. Given this, what was the power of the test under these same conditions? The alpha level used in the test was 0.05. A) 0.95 B) 0.875 C) Essentially zero D) Power would be undefined in this case since the hypothesis would be rejected. 9‐10 Copyright © 2014 Pearson Education, Inc. 114) If the hypothesis test you are conducting is a two‐tailed test, which of the following is a possible step that you could take to increase the power of the test? A) Reduce the sample size B) Increase alpha C) Increase beta D) Use the t‐distribution 115) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, if the ʺtrueʺ population mean is 32.0 mpg, what is the probability that the test will lead the consumer group to reject the claimed mileage for this car? A) About 0.075 B) Approximately 0.95 C) 0.05 D) None of the above 116) For the following z‐test statistic, compute the p‐value assuming that the hypothesis test is a one‐tailed test: z = 1.34. A) 0.0606 B) 0.0815 C) 0.0124 D) 0.0901 117) For the following z‐test statistic, compute the p‐value assuming that the hypothesis test is a one‐tailed test: z = 2.09. A) 0.0172 B) 0.0183 C) 0.0415 D) 0.0611 118) For the following z‐test statistic, compute the p‐value assuming that the hypothesis test is a one‐tailed test: z = ‐1.55. A) 0.0606 B) 0.1512 C) 0.0901 D) 0.0172 119) For the following hypothesis test: With n = 80, σ = 9, and = 47.1, state the decision rule in terms of the critical value of the test statistic. A) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not reject. B) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not reject. C) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not accept. D) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not accept. 120) For the following hypothesis test: With n = 80, σ = 9, and = 47.1, state the calculated value of the test statistic z. A) 3.151 B) ‐2.141 C) 2.087 D) ‐3.121 121) State the appropriate p‐value. A) 0.0183 B) 0.0314 C) 0.0512 D) 0.0218 122) State the conclusion. A) Because the computed value of z = 3.012 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p‐value is less than 0.02 B) Because the computed value of z = 3.012 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p‐value is less than 0.02 C) Because the computed value of z = 2.087 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p‐value is less than 0.02 D) Because the computed value of z = 2.087 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p‐value is less than 0.02 9‐11 Copyright © 2014 Pearson Education, Inc. 123) For the following hypothesis test: With n = 15, s = 7.5, and = 62.2, state the decision rule in terms of the critical value of the test statistic A) This is a two‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than ‐2.1448 or greater than 2.1448. Otherwise, do not reject. B) This is a two‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is less than ‐2.1448 or greater than 2.1448. Otherwise, do not accept. C) This is a two‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than ‐1.1828 or greater than 1.1828. Otherwise, do not reject. D) This is a two‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than ‐1.1828 or greater than 1.1828. Otherwise, do not accept. 124) State the calculated value of the test statistic t. A) 1.014 B) 0.012 C) 0.878 D) 1.312 125)S ate the conclusion. A) Because the computed value of t = 0.878 is not less than ‐2.1448 and not greater than 2.1448, do not reject the null hypothesis. B) Because the computed value of t = 1.312 is not less than ‐2.1448 and not greater than 2.1448, do not reject the null hypothesis. C) Because the computed value of t = 0.878 is not less than ‐2.1448 and not greater than 2.1448, reject the null hypothesis D) Because the computed value of t = 1.312 is not less than ‐2.1448 and not greater than 2.1448, reject the null hypothesis 126) For the following hypothesis: With n = 20, = 71.2, s = 6.9, and α = 0.1, state the decision rule in terms of the critical value of the test statistic. A) This is a one‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, reject. B) This is a one‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, reject. C) This is a one‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, do not reject. D) This is a one‐tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, do not reject. 127) State the calculated value of the test statistic t. A) 1.58 B) 0.78 C) 1.14 D) 0.41 128) State the conclusion. A) Because the computed value of t = 0.78 is not greater than 2.1727, reject the null hypothesis. B) Because the computed value of t = 0.78 is not greater than 2.1727, do not reject the null hypothesis. C) Because the computed value of t = 0.78 is not greater than 1.3277, reject the null hypothesis. D) Because the computed value of t = 0.78 is not greater than 1.3277, do not reject the null hypothesis. 129) The National Club Association does periodic studies on issues important to its membership. The 2012 Executive Summary of the Club Managers Association of America reported that the average country club initiation fee was $31,912. Suppose a random sample taken in 2009 of 12 country clubs produced the following initiation fees: $29,121 $31,472 $28,054 $31,005 $36,295 $32,771 $26,205 $33,299 $25,602 $33,726 $39,731 $27,816 9‐12 Copyright © 2014 Pearson Education, Inc. Based on the sample information, can you conclude at the α = 0.05 level of significance that the average 2009 country club level of significance. initiation fees are lower than the 2008 average? Conduct your test at the A) Because t = ‐0.4324 is not less than t critical = ‐1.4512, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. B) Because t = ‐0.4324 is not less than t critical = ‐1.4512, reject Ho. The 2009 average country club initiation fee is less than the 2008 average. C) Because t = ‐0.5394 is not less than t critical = ‐1.7959, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. D) Because t = ‐0.5394 is not less than t critical = ‐1.7959, reject Ho. The 2009 average country club initiation fee is less than the 2008 average. 130) The director of a state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ 30,000 HA : μ < 30,000 B) H0 : μ ≥ 29,750 HA : μ < 29,750 C) H0 : μ ≤ 30,000 HA : μ > 30,000 D) H0 : μ ≤ 29,750 HA : μ > 29,750 131) Assuming the population standard deviation is known to be $2,500 and the significance level for the test is to be 0.05, what is the critical value (stated in dollars)? A) For alpha = .05 and a one tailed, lower tail test, the critical value is z = ‐1.645. Solving for the critical x‐bar: ‐1.645 = (x‐bar ‐ 30,000)/250, x‐bar = $29,588.75 B) For alpha = .05 and a one tailed, lower tail test, the critical value is z = ‐1.96. Solving for the critical x‐bar: ‐1.96 = (x‐bar ‐ 30,000)/250, x‐bar = $34,211.14 C) For alpha = .05 and a one tailed, lower tail test, the critical value is z = ‐1.645. Solving for the critical x‐bar: ‐1.645 = (x‐bar ‐ 30,000)/250, x‐bar = $34,211.14 D) For alpha = .05 and a one tailed, lower tail test, the critical value is z = ‐1.96. Solving for the critical x‐bar: ‐1.96 = (x‐bar ‐ 30,000)/250, x‐bar = $30,411.25 132) A mail‐order business prides itself in its ability to fill customersʹ orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customersʹ orders is six or less unless the data suggest strongly otherwise. Establish the appropriate null and alternative hypotheses. A) H0 : μ ≥ 6 days Ha : μ < 6 days B) H0 : μ ≤ 6 days Ha : μ > 6 days C) H0 : μ > 6 days Ha : μ ≤ 6 days D) H0 : μ < 6 days Ha : μ ≥ 6 days 133) On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail‐order business is achieving its goal? Use a significance level of 0.025 to answer this question. A) Since 2.2216 < 2.4511, reject H0 and conclude that the mail‐order business is not achieving its goal B) Since 2.7406 > 2.023, reject H0 and conclude that the mail‐order business is not achieving its goal. C) Since 2.4421 > 2.023, reject H0 and conclude that the mail‐order business is not achieving its goal. D) Since 2.2346 < 2.5113, reject H0 and conclude that the mail‐order business is not achieving its goal. 134) On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail‐order business is achieving its goal? Use a significance level of 0.025 to answer this question. Conduct the test using this p‐value. A) Since 0.024 > 0.0041, reject the null hypothesis. B) Since 0.046 > 0.0025, reject the null hypothesis. C) Since 0.0046 < 0.025, reject the null hypothesis. D) Since 0.0024 < 0.041, reject the null hypothesis. 9‐13 Copyright © 2014 Pearson Education, Inc. 135) The makers of Mini‐Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. Use α = 0.05. Establish the appropriate null and alternative hypotheses to be tested for boxes that are supposed to have an average of 24 ounces. A) H0 : μ = 32 ounces Ha : μ ≠ 32 ounces B) H0 : μ = 16 ounces Ha : μ ≠ 16 ounces C) H0 : μ = 22 ounces Ha : μ ≠ 22 ounces D) H0 : μ = 24 ounces Ha : μ ≠ 24 ounces 136) Use α= 0.05. At the end of a particular shift during which the machine was filling 24‐ounce boxes of Mini‐Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. A) Since ‐1.2445 < 1.013 < 1.2445, do not reject H0 and conclude that the filling machine remains all right to operate. B) Since ‐1.2445 < 1.013 < 1.2445, reject H0 and conclude that the filling machine needs to be moderated. C) Since ‐2.1315 < 1.83 < 2.1315, do not reject H0 and conclude that the filling machine remains all right to operate. D) Since ‐2.1315 < 1.83 < 2.1315, reject H0 and conclude that the filling machine needs to be moderated. 137) Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. Conduct the test using a p‐value. A) p‐value = 0.0872 > 0.025; therefore do not reject H0 B) p‐value = 0.0422 > 0.005; therefore do not reject H0 C) p‐value = 0.0314 < 0.105; therefore reject H0 D) p‐value = 0.0121< 0. 0805; therefore reject H0 138) At a recent meeting, the manager of a national call center for a major Internet bank made the statement that the average past‐due amount for customers who have been called previously about their bills is now no larger than $20.00. Other bank managers at the meeting suggested that this statement may be in error and that it might be worthwhile to conduct a test to see if there is statistical support for the call center managerʹs statement. The file called Bank Call Center contains data for a random sample of 67 customers from the call center population. Assuming that the population standard deviation for past due amounts is known to be $60.00, what should be concluded based on the sample data? Test using α = 0.10. A) Because p‐value = 0.4121 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center managerʹs statement that the mean past due amount is $20.00 or less. B) Because p‐value = 0.4121 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center managerʹs statement that the mean past due amount is $20.00 or less. C) Because p‐value = 0.2546 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center managerʹs statement that the mean past due amount is $20.00 or less. D) Because p‐value = 0.2546 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center managerʹs statement that the mean past due amount is $20.00 or less. 139) The U.S. Bureau of Labor Statistics (www.bls.gov) released its Consumer Expenditures report in October 2008. Among its findings is that average annual household spending on food at home was $3,624. Suppose a random sample of 137 households in Detroit was taken to determine whether the average annual expenditure on food at home was less for consumer units in Detroit than in the nation as a whole. The sample results are in the file Detroit Eats. Based on the sample results, can it be concluded at the α = 0.02 level of significance that average consumer‐unit spending for food at home in Detroit is less than the national average? A) Because t = ‐13.2314 is less than the critical t value of ‐1.4126, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average B) Because t = ‐13.2314 is less than the critical t value of ‐1.4126, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average 9‐14 Copyright © 2014 Pearson Education, Inc. C) Because t = ‐15.7648 is less than the critical t value of ‐2.0736, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average. D) Because t = ‐15.7648 is less than the critical t value of ‐2.0736, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average. 140) The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out‐of‐pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out‐of‐pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondentsʹ 2012 out‐of‐pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out‐of‐pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test. A) Because t = ‐2.69 is less than ‐2.3456, do not reject H0 Conclude that 2012 average out‐of‐pocket prescription drug expenses are not lower than the 2002 average. B) Because t = ‐2.69 is less than ‐2.3456, reject H0 Conclude that 2012 average out‐of‐pocket prescription drug expenses are lower than the 2002 average. C) Because t = ‐1.69 is less than ‐0.8712, reject H0 Conclude that 2012 average out‐of‐pocket prescription drug expenses are lower than the 2002 average. D) Because t = ‐1.69 is less than ‐0.8712, do not reject H0 Conclude that 2012 average out‐of‐pocket prescription drug expenses are not lower than the 2002 average. 141) Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. Determine if the process is not producing the tees to specification. Use a significance level of 0.01. A) Since t = 2.1953 < 2.8073 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. B) Since t = 2.1953 < 2.8073 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. C) Since t = 1.2814 < 1.9211 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. D) Since t = 1.2814 < 1.9211 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. 142) If the hypothesis test determines the specification is not being met, the production process will be shut down while causes and remedies are determined. At times this occurs even though the process is functioning to specification. What type of statistical error would this be? A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error. B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error. C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error. D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error. 9‐15 Copyright © 2014 Pearson Education, Inc. 143) Given the following null and alternative Test the hypothesis using α = 0.01 assuming that a sample of n = 200 yielded x = 105 items with the desired attribute. A) Since ‐2.17 > ‐2.33, the null hypothesis is not rejected. B) Since ‐1.86 > ‐1.02, the null hypothesis is not rejected. C) Since ‐2.17 > ‐2.33, the null hypothesis is rejected. D) Since ‐1.86 > ‐1.02, the null hypothesis is rejected. 144) For the following hypothesis test: With n= 64 and p= 0.42, state the decision rule in terms of the critical value of the test statistic A) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.013 or less than ‐2.013. Otherwise, do not reject. B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.013 or greater than ‐2.013. Otherwise, do not reject. C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.575 or less than ‐2.575. Otherwise, do not reject. D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.575 or greater than ‐2.575. Otherwise, do not reject. 145)State the calculated value of the test statistic A) t = 0.4122 B) t = 1.7291 C) z = 0.3266 D) z = 1.2412 146) State the conclusion A) Because the calculated value of the test statistic, t=0.4122, is neither greater than 2.013 nor less than ‐2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than ‐2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than ‐2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. D) Because the calculated value of the test statistic, z = 0.3266, is neither greater than 2.575 nor less than ‐2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. 147) For the following hypothesis test With n = 100 and p = 0.66, state the decision rule in terms of the critical value of the test statistic A) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than the critical value of the test statistic z = ‐1.96. Otherwise, do not reject. B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than the critical value of the test statistic z = ‐1.645. Otherwise, do not reject. C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic z = 1.96. Otherwise, do not reject. D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic z = 1.645. Otherwise, do not reject. 148)State the calculated value of the test statistic. A) 2.7299 B) ‐2.0785 C) 1.4919 D) ‐0.3421 149) State the conclusion. A) Because the computed value of z = ‐2.0785 is less than the critical value of z = ‐1.96, reject the null hypothesis and conclude that the population proportion is less than 0.75. B) Because the computed value of z = ‐0.3412 is less than the critical value of z = ‐1.645, reject the null hypothesis and conclude 9‐16 Copyright © 2014 Pearson Education, Inc. that the population proportion is less than 0.75. C) Because the computed value of z = 1.4919 is greater than the critical value of z = ‐1.96, accept the null hypothesis and conclude that the population proportion is greater than 0.75. D) Because the computed value of z = ‐0.3412 is greater than the critical value of z = ‐1.645, accept the null hypothesis and conclude that the population proportion is greater than 0.75. 150) Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 18% of them had borrowed against it in the last year. A random sample of 100 employees from a local company who have a 401(k) retirement plan found that 14 had borrowed from their plan. Based on the sample results, is it possible to conclude, at the α = 0.025 level of significance, that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 18% reported nationwide? A) The z‐critical value for this lower tailed test is z = ‐1.96. Because ‐1.5430 is greater than the z‐critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. B) The z‐critical value for this lower tailed test is z = ‐1.96. Because ‐1.0412 is greater than the z‐critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. C) The z‐critical value for this lower tailed test is z = 1.96. Because 1.5430 is less than the z‐critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. D) The z‐critical value for this lower tailed test is z = 1.96. Because 1.0412 is less than the z‐critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. 151) An issue that faces individuals investing for retirement is allocating assets among different investment choices. Suppose a study conducted 10 years ago showed that 65% of investors preferred stocks to real estate as an investment. In a recent random sample of 900 investors, 540 preferred real estate to stocks. Is this new data sufficient to allow you to conclude that the proportion of investors preferring stocks to real estate has declined from 10 years ago? Conduct your analysis at the α = 0.02 level of significance. A) Because z = ‐1.915 is not less than ‐2.055, do not reject H0. A higher proportion of investors prefer stocks today than 10 years ago. B) Because z = ‐1.915 is not less than ‐2.055, do not reject H0. A lower proportion of investors prefer stocks today than 10 years ago. C) Because z = ‐3.145 is less than ‐2.055, reject H0. A lower proportion of investors prefer stocks today than 10 years ago. D) Because z = ‐3.145 is less than ‐2.055, reject H0. A higher proportion of investors prefer stocks today than 10 years ago. 152) A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the stateʹs registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling. State the appropriate null and alternative hypotheses. A) H0 : p = 0.58 Ha : p ≠ 0.58 B) H0 : p ≤ 0.55 Ha : p > 0.55 C) H0 : p = 0.55 Ha : p ≠ 0.55 D) H0 : p ≤ 0.58 Ha : p > 0.58 153) A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the stateʹs registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling. Assuming that a significance level of 0.05 is used, what conclusion should the governor reach based on these sample data? A) Since z = 1.1594 < 1.645, do not reject the null hypothesis. 9‐17 Copyright © 2014 Pearson Education, Inc. The sample data do not provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. B) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. C) Since z = 1.1594 < 1.645, do not reject the null hypothesis. The sample data do not provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling. D) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling. 154) A recent article in The Wall Street Journal entitled ʺAs Identity Theft Moves Online, Crime Rings Mimic Big Businessʺ states that 39% of the consumer scam complaints by American consumers are about identity theft. Suppose a random sample of 90 complaints is obtained. Of these complaints, 40 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in The Wall Street Journal? (Test using α= 0.10.) A) Since z = 1.947 > 1.645, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. B) Since z = 2.033 > 1.96, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. C) Since z = 1.341 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. D) Since z = 0.97 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. 155) Because of the complex nature of the U.S. income tax system, many people have questions for the Internal Revenue Service (IRS). Yet, an article published by the Detroit Free Press entitled ʺAssistance: IRS Help Centers Give the Wrong Informationʺ discusses the propensity of IRS staff employees to give incorrect tax information to tax‐payers who call with questions. Then IRS Inspector General Pamela Gardiner told a Senate subcommittee that ʺthe IRS employees at 400 taxpayer assistance centers nationwide encountered 8.5 million taxpayers face‐to‐face last year. The problem: When inspector general auditors posing as taxpayers asked them to answer tax questions, the answers were right 69% of the time.ʺ Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed. State the appropriate null and alternative hypotheses. A) H0 : p = 0.69 Ha : p ≠ 0.69 B) H0 : p = 0.58 Ha : p ≠ 0.58 C) H0 : p > 0.69 Ha : p ≤ 0.69 D) H0 : p > 0.58 Ha : p ≤ 0.58 156)Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed. Using an α= 0.05 level, based on the sample data, what conclusion should be reached about the IRS rate of correct tax returns? and z = ‐1.96. Since A) The z‐critical values from the standard normal table for a two‐tailed test with alpha = 0.05 are z = ‐0.96 > ‐1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.69 rate quoted in the Detroit Free Press article and z = ‐1.96. B) The z‐critical values from the standard normal table for a two‐tailed test with alpha = 0.05 are Since z = ‐0.96 > ‐1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.58 rate quoted in the Detroit Free Press article C) The z‐critical values from the standard normal table for a two‐tailed test with alpha = 0.05 are and z = ‐1.96. Since z= ‐3.19 < ‐1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.69 rate quoted in the Detroit Free Press article. D) The z‐critical values from the standard normal table for a two‐tailed test with alpha = 0.05 are and z = ‐1.96. Since z = ‐3.19 < ‐1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is 9‐18 Copyright © 2014 Pearson Education, Inc. 157) You are given the following null and alternative hypotheses: If the true population mean is 1.25, determine the value of beta. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.40 B) 0.04 C) 0.51 D) 0.80 158) Calculate the power of the test. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.49 B) 0.20 C) 0.96 D) 0.60 159) You are given the following null and alternative hypotheses: If the true population mean is 4,345, determine the value of beta. Assume the population standard deviation is known to be 200 and the sample size is 100. A) 0.9192 B) 0.8233 C) 0.6124 D) 0.0314 160)Calculate the power of the test. Assume the population standard deviation is known to be 200 and the sample size is 100. A) 0.1766 B) 0.3876 C) 0.0808 D) 0.9686 161) You are given the following null and alternative hypotheses: Calculate the probability of committing a Type II error when the population mean is 505, the sample size is 64, and the population standard deviation is known to be 36 A) 0.1562 B) 0.5997 C) 0.3426 D) 0.8888 162) According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people. State the appropriate null and alternative hypotheses. A) H0 : μ > 243 Ha : μ ≤ 243 B) H0 : μ < 243 Ha : μ ≥ 243 C) H0 : μ ≤ 243 Ha : μ > 243 D) H0 : μ ≥ 243 Ha : μ < 243 163) Calculate the probability of committing a Type II error if the true population mean is 230 gallons. Assume that the population standard deviation is known to be 40 gallons. A) 0.0331 B) 0.0712 C) 0.0537 D) 0.1412 164) Swift is the holding company for Swift Transportation Co., Inc., a truckload carrier headquartered in Phoenix, Arizona. Swift operates the largest truckload fleet in the United States. Before Swift switched to its current computer‐based billing system, the average payment time from customers was approximately 40 days. Suppose before purchasing the present billing system, it performed a test by examining a random sample of 24 invoices to see if the system would reduce the average billing time. The sample indicates that the average payment time is 38.7 days. The company that created the billing system indicates that the system would reduce the average billing time to less than 40 days. Conduct a hypothesis test to determine if the new computer‐based billing system would reduce the average billing time to less than 40 days. Assume the standard deviation is known to be 6 days. Use a significance level of 0.025. A) Since z = ‐0.423 > ‐1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer‐based billing system would reduce the average billing time to less than 40 days. 9‐19 Copyright © 2014 Pearson Education, Inc. B) Since z = ‐1.0614 > ‐1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer‐based billing system would reduce the average billing time to less than 40 days. C) z = ‐1.0231 > ‐1.96, we will reject H0, there is sufficient evidence to conclude that the new computer‐based billing system would reduce the average billing time to less than 40 days. D) z = 0.341 > ‐1.96, we will reject H0, there is sufficient evidence to conclude that the new computer‐based billing system would reduce the average billing time to less than 40 days. 165) Waiters at Finegoldʹs Restaurant and Lounge earn most of their income from tips. Each waiter is required to ʺtip‐outʺ a portion of tips to the table bussers and hostesses. The manager has based the ʺtip‐outʺ rate on the assumption that the mean tip is at least 15% of the customer bill. To make sure that this is the correct assumption, he has decided to conduct a test by randomly sampling 60 bills and recording the actual tips. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ 15 Ha : μ < 15 B) H0 : μ ≤ 15 Ha : μ > 15 C) H0 : μ ≥ 9 Ha : μ < 9 D) H0 : μ ≤ 9 Ha : μ > 9 166) Calculate the probability of a Type II error if the true mean is 14%. Assume that the population standard deviation is known to be 2% and that a significance level equal to 0.01 will be used to conduct the hypothesis test. A) 0.0041 B) 0.1251 C) 0.0606 D) 0.4123 167) Nationwide Mutual Insurance, based in Columbus, Ohio, is one of the largest diversified insurance and financial services organizations in the world, with more than $157 billion in assets. Nationwide ranked 108th on the Fortune 100 list in 2008. The company provides a full range of insurance and financial services. In a recent news release Nationwide reported the results of a new survey of 1,097 identity theft victims. The survey shows victims spend an average of 81 hours trying to resolve their cases. If the true average time spent was 81 hours, determine the probability that a test of hypothesis designed to test that the average was less than 85 hours would reject the research hypothesis. Use α= 0.05 and a standard deviation of 50. A) 0.0123 B) 0.5182 C) 0.1241 D) 0.1587 168) According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ $47,413 HA : μ < $47,413 B) H0 : μ < $47,413 HA : μ ≥ $47,413 C) H0 : μ ≤ $47,413 HA : μ > $47,413 D) H0 : μ > $47,413 HA : μ ≤ $47,413 169) Compute the power of the hypothesis test to reject the null hypothesis if the true average starting salary is only $47,000. Assume that the population standard deviation is known to be $4,600 and the test is to be conducted using an alpha level equal to 0.01. A) 0.0872 B) 0.1323 C) 0.8554 D) 0.9812 170) What is meant by the terms Type I and Type II statistical error? Answer: When a null hypothesis is tested using sample information, it is expected that sampling error will exist. It is possible that the sampling error will lead to an error in the conclusion that is reached with respect to the null hypothesis. For example, if the null hypothesis is true, extreme sampling error will push the test statistic into the rejection region. Rejecting a true null hypothesis is called a Type I error. Also, if the null hypothesis is false, the sample data may be such that we do not reject the null hypothesis. ʺAcceptingʺ a false null hypothesis is referred to as a Type II error. 178) Explain why an increase in sample size will reduce the probability of a Type II error but will not impact the probability of a Type I error. Answer: The probability of a Type I error is set by the decision maker and is based on his/her willingness to reject a true null hypothesis. Thus, alpha is independent of the size of the sample. However, the Type II error probability, beta, is affected by the size of the sample. The reason for this is that the standard error for the sampling distribution is found using . Thus, if n is increased, the standard error is reduced. Then, if the standard error is reduced, the ʺtrueʺ population parameter is relatively farther from the hypothesized population value making it easier for the sample data to distinguish between the hypothesized value and the ʺtrueʺ value. Thus, an increase in sample size will reduce beta. 9‐20 Copyright © 2014 Pearson Education, Inc. 179) A small city is considering breaking away from the county school system and starting its own city school system. City leaders believe that more than 60 percent of residents support the idea. A poll of n = 215 residents is taken and 134 people say they support starting a city school district. Using a 0.10 level of significance, conduct a hypothesis test to determine whether this poll supports the belief of city leaders. Answer: Since ʺmore than 60 percentʺ is a strict inequality, this must go in the alternate hypothesis, so the hypotheses are: H0 : p ≤ 0.6 HA : p > 0.6 Since this a proportion, the normal distribution is used, and the test statistic for α = 0.10 is z = 1.28 and the decision rule is: Reject the null if Z > 1.28. The test statistic formula is And the sample proportion is p = 134/215 = 0.6233. z = So the test statistic is z = = = 0.6976 This does not fall in the rejection region since 0.6976 < 1.28, so we do not reject the null. This indicates the proportion of residents who support a city school district is not significantly greater than 0.6. 180) The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. What is the critical value in ounces? Answer: The company will take action if it determines that the fill process is putting too little or too much in the cans on average. Therefore, the research hypothesis is that the population mean is not equal to 12 ounces. Given this, the null and alternative hypotheses are: H0 : μ = 12 ounces Ha : μ ≠ 12 ounces. We note that the hypothesis test is two‐tailed since we can reject it if the sample mean gets too large or too small. Thus, there are actually two critical values, one on each side of the distribution. We find the critical values as follows: μ ± z Because this is a two‐tailed test, we must split the alpha into two parts of 0.025 each. The z‐value from the standard normal distribution for 0.5000 ‐ 0.025 = 0.475 is 1.96. Given this, the critical values are determined as: 12.0 ± 1.96 or 12.0 ± .0653 Thus, the decision rule is: If 11.9347 ≤ ≤ 12.0653 do not reject the null hypothesis, otherwise reject. 181) The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. Using the test statistic approach, what conclusion should the company reach if the sample mean is 12.02 ounces? What type of statistical error may have been committed? Answer: The company will take action if it determines that the fill process is putting too little or too much in the cans on average. Therefore, the research hypothesis is that the population mean is not equal to 12 ounces. Given this, the null and alternative hypotheses are: H0 : μ = 12 ounces Ha : μ ≠ 12 ounces. We note that the hypothesis test is two‐tailed since we can reject if the sample mean gets too large or too small. The critical z‐value from the standard normal table for a two‐tailed test with alpha = 0.05 is 1.96, the decision rule is reject the 9‐21 Copyright © 2014 Pearson Education, Inc. null if z < ‐1.96 or z > 1.96. The test statistic is computed as: z = = = .60 Since 0.60 < 1.96, we do not reject the null hypothesis based on these sample data. So we can conclude that the filling process does not need adjusting. Thus, the statistical error that could have been committed is a Type II error since the null hypothesis was not rejected. If a Type II error was committed this would mean that the filling machine actually needs adjusting, but we failed to detect it. 182) A national car rental chain believes that more than 80 percent of its customers are satisfied with the check‐in process that the company is using. To test this, a random sample of n = 200 customers are surveyed. These sample results show 168 that say they were satisfied. If the test is to be conducted using a .05 level of significance, what is the critical value? Answer: This is a hypothesis about a single population proportion. The appropriate null and alternative hypotheses are: H0 : p ≤ .80 Ha : p > .80 The critical value for this one‐tailed test is found as: p + z The z‐value from the standard normal table for a 0.05 level of significance and one‐tailed test is 1.645. Then the critical value is: .80 + 1.645 = = .8465. Thus, if the sample proportion exceeds 0.8465, we reject the null hypothesis, since = = .84 Since 0.84 is just slightly less than 0.8465, we do not reject the null hypothesis. 183) A contract with a parts supplier calls for no more than .04 defects in the large shipment of parts. To test whether the shipment meets the contract, the receiving company has selected a random sample of n = 100 parts and found 6 defects. If the hypothesis test is to be conducted using a significance level equal to 0.05, what is the test statistic and what conclusion should the company reach based on the sample data? Answer: This is a hypothesis about a single population proportion. The appropriate null and alternative hypotheses are: H0 : p ≤ .04 Ha : p > .04 Note that the research hypothesis is that the shipment contains more than the suggested proportion of defects. The test statistic when testing about proportions is a z‐value computed as: z = . For this situation, the calculated z‐value is z = = 1.02. We compare this value to the z‐value from the standard normal distribution table for 0.45 which is z = 1.645. The decision rule is: If z calculated > 1.645, reject the null hypothesis, otherwise do not reject. Since 1.02 < 1.645, we cannot reject the null hypothesis based on the sample data. Thus, there is insufficient evidence to conclude that the proportion of defects in the shipment exceeds 0.04. 9‐22 Copyright © 2014 Pearson Education, Inc.