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1 Chapter 9 Problems and Solutions - Hypothesis Tests 1. The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is $600 or less. A member of the hotel’s accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager’s claim. a. Which form of the hypotheses should be used to test the manager’s claim? Explain. H0: μ ≥ 600 H0: μ ≤ 600 H0: μ = 600 Ha: μ < 600 Ha: μ > 600 Ha: μ ≠ 600 b. c. What conclusion is appropriate when H0 cannot be rejected? What conclusion is appropriate when H0 can be rejected? 3. A production line operation is designed to fill cartons with laundry detergent to a mean weight of 32 ounces. A sample of cartons is periodically selected and weighed to deter- mine whether underfilling or overfilling is occurring. If the sample data lead to a conclu- sion of underfilling or overfilling, the production line will be shut down and adjusted to obtain proper filling. a. Formulate the null and alternative hypotheses that will help in deciding whether to shut down and adjust the production line. b. Comment on the conclusion and the decision when H0 cannot be rejected. c. Comment on the conclusion and the decision when H0 can be rejected. 4. Because of high production-changeover time and costs, a director of manufacturing must convince management that a proposed manufacturing method reduces costs before the new method can be implemented. The current production method operates with a mean cost of $220 per hour. A research study will measure the cost of the new method over a sample production period. a. Develop the null and alternative hypotheses most appropriate for this study. b. Comment on the conclusion when H0 cannot be rejected. c. Comment on the conclusion when H0 can be rejected. 5. Nielsen reported that young men in the United States watch 56.2 minutes of prime-time TV daily (The Wall Street Journal Europe, November 18, 2003). A researcher believes that young men in Germany spend more time watching prime-time TV. A sample of German young men will be selected by the researcher and the time they spend watching TV in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. H0: μ ≤ 56.2 Ha: μ > 56.2 a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error? 2 6. The label on a 3-quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error? 10. Consider the following hypothesis test: H0: μ ≤ 25 Ha: μ > 25 A sample of 40 provided a sample mean of 26.4. The population standard deviation is 6. a. b. c. d. 11. Compute the value of the test statistic. What is the p-value? At α = .01, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? Consider the following hypothesis test: H0: μ = 15 Ha: μ ≠ 15 A sample of 50 provided a sample mean of 14.15. The population standard deviation is 3. a. b. c. d. 13. Compute the value of the test statistic. What is the p-value? At α = .05, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? Consider the following hypothesis test: H0: μ ≤ 50 Ha: μ > 50 A sample of 60 is used and the population standard deviation is 8. Use the critical value approach to state your conclusion for each of the following sample results. Use α = .05. a. b. c. x¯ = 52.5 x¯ = 51 x¯ = 51.8 3 17. Wall Street securities firms paid out record year-end bonuses of $125,500 per employee for 2005 (Fortune, February 6, 2006). Suppose we would like to take a sample of employ- ees at the Jones & Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of $125,500 for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones & Ryan were different from the population mean. b. Suppose a sample of 40 Jones & Ryan employees showed a sample mean year-end bonus of $118,000. Assume a population standard deviation of u = $30,000 and com- pute the p-value. c. With α = .05 as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach. 20. For the United States, the mean monthly Internet bill is $32.79 per household (CNBC, January 18, 2006). A sample of 50 households in a southern state showed a sample mean of $30.63. Use a population standard deviation of a = $5.60. a. Formulate hypotheses for a test to determine whether the sample data support the con- clusion that the mean monthly Internet bill in the southern state is less than the national mean of $32.79. b. What is the value of the test statistic? c. What is the p-value? d. At α = .01, what is your conclusion? 26. Consider the following hypothesis test: H0: μ = 100 Ha: μ ≠ 100 A sample of 65 is used. Identify the p-value and state your conclusion for each of the following sample results. Use α = .05. a. b. c. x¯ = 103 and s = 11.5 x¯ = 96.5 and s = 11.0 x¯ = 102 and s = 10.5 28. A shareholders’ group, in lodging a protest, claimed that the mean tenure for a chief executive officer (CEO) was at least nine years. A survey of companies reported in The Wall Street Journal found a sample mean tenure of x̄ = 7.27 years for CEOs with a standard deviation of s = 6.38 years (The Wall Street Journal, January 2, 2007). a. Formulate hypotheses that can be used to challenge the validity of the claim made by the shareholders’ group. b. Assume 85 companies were included in the sample. What is the p-value for your hypothesis test? c. At α = .01, what is your conclusion? 4 30. Time Warner Inc.’s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10, 2003). Assume that for a sample of 40 days during the first half of 2003, the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the p-value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application? Chapter 9 Solutions 1. a. H0: 600 Manager’s claim. Ha: > 600 b. We are not able to conclude that the manager’s claim is wrong. c. The manager’s claim can be rejected. We can conclude that > 600. 3. a. H0: = 32 Ha: 32 Specified filling weight Overfilling or underfilling exists b. There is no evidence that the production line is not operating properly. Allow the production process to continue. c. Conclude 32 and that overfilling or underfilling exists. Shut down and adjust the production line. 4. a. H0: 220 Ha: < 220 Research hypothesis to see if mean cost is less than $220. b. We are unable to conclude that the new method reduces costs. c. Conclude < 220. Consider implementing the new method based on the conclusion that it lowers the mean cost per hour. 5 5. a. The Type I error is rejecting H0 when it is true. This error occurs if the researcher concludes that young men in Germany spend more than 56.2 minutes per day watching prime-time TV when the national average for Germans is not greater than 56.2 minutes. b. The Type II error is accepting H0 when it is false. This error occurs if the researcher concludes that the national average for German young men is 56.2 minutes when in fact it is greater than 56.2 minutes. 6. a. H0: 1 The label claim or assumption. Ha: > 1 b. Claiming > 1 when it is not. This is the error of rejecting the product’s claim when the claim is true. c. Concluding 1 when it is not. In this case, we miss the fact that the product is not meeting its label specification. 10. a. z x 0 26.4 25 1.48 / n 6 / 40 b. Upper tail p-value is the area to the right of the test statistic Using normal table with z = 1.48: p-value = 1.0000 - .9306 = .0694 Using Excel: p-value = 1 - NORMSDIST(1.48) = .0694 c. p-value > .01, do not reject H0 d. Reject H0 if z 2.33 1.48 < 2.33, do not reject H0 11. a. z x 0 / n 14.15 15 3/ 50 2.00 b. Because z < 0, p-value is two times the lower tail area Using normal table with z = -2.00: p-value = 2(.0228) = .0456 Using Excel: p-value = 2*NORMSDIST(-2.00) = .0456 c. p-value .05, reject H0 d. Reject H0 if z -1.96 or z 1.96 -2.00 -1.96, reject H0 6 Reject H0 if z 1.645 13. a. b. z z z c. x 0 / n x 0 / n x 0 / n 52.5 50 8 / 60 51 50 8 / 60 b. .97 < 1.645, do not reject H0 1.74 17. a. H0: 125,500 z 2.42 1.645, reject H0 .97 51.8 50 8 / 60 2.42 1.74 1.645, reject H0 Ha: 125,500 x 0 118, 000 125,500 1.58 / n 30, 000 / 40 Because z < 0, p-value is two times the lower tail area Using normal table with z = -1.58: p-value = 2(.0571) = .1142 Using Excel: p-value = 2*NORMSDIST(-1.58) = .1141 c. p-value > .05, do not reject H0. We cannot conclude that the year-end bonuses paid by Jones & Ryan differ significantly from the population mean of $125,500. d. Reject H0 if z -1.96 or z 1.96 z = -1.58; cannot reject H0 20. a. H0: 32.79 z b. Ha: < 32.79 x 0 30.63 32.79 2.73 n 5.6 50 c. Lower tail p-value is area to left of the test statistic. Using normal table with z = -2.73: p-value = .0032. Using Excel: p-value = NORMSDIST(-2.73) = .0032 H d. p-value .01; reject 0 . Conclude that the mean monthly internet bill is less in the southern state. 7 26. a. t x 0 s/ n 103 100 11.5 / 65 2.10 Degrees of freedom = n – 1 = 64 Because t > 0, p-value is two times the upper tail area Using t table; area in upper tail is between .01 and .025; therefore, p-value is between .02 and .05. Using Excel: p-value = TDIST(2.10,64,2) = .0397 p-value .05, reject H0 b. t x 0 s/ n 96.5 100 11/ 65 2.57 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between .005 and .01; therefore, p-value is between .01 and .02. Using Excel: p-value = TDIST(2.57,64,2) = .0125 p-value .05, reject H0 c. t x 0 s/ n 102 100 10.5 / 65 1.54 Because t > 0, p-value is two times the upper tail area Using t table: area in upper tail is between .05 and .10; therefore, p-value is between .10 and .20. Using Excel: p-value = TDIST(1.54,64,2) = .1285 p-value > .05, do not reject H0 28. a. H0: 9 b. t x 0 s/ n Ha: 7.27 9 6.38 / 85 2.50 < 9 Degrees of freedom = n – 1 = 84 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between .005 and .01 Using Excel: p-value = TDIST(2.50,84,1) = .0072 8 c. p-value .01; reject H0. The mean tenure of a CEO is significantly lower than 9 years. The claim of the shareholders group is not valid. 30. a. H0: = 600 b. t x 0 s/ n Ha: 612 600 65 / 40 1.17 600 df = n - 1 = 39 Because t > 0, p-value is two times the upper tail area Using t table: area in upper tail is between .10 and .20; therefore, p-value is between .20 and .40. Using Excel: p-value = TDIST(1.17,39,2) = .2491 c. With = .10 or less, we cannot reject H0. We are unable to conclude there has been a change in the mean CNN viewing audience. d. The sample mean of 612 thousand viewers is encouraging but not conclusive for the sample of 40 days. Recommend additional viewer audience data. A larger sample should help clarify the situation for CNN..