商务统计学(第四版)课后习题答案第九章

308 Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests CHAPTER 9 9.1 H0 is used to denote the null hypothesis. 9.2 H1 is used to denote the alternative hypothesis. 9.3 is used to denote the significance level, or the chance of committing a Type I error. 9.4 is used to denote the risk or the chance of committing a Type II error. 9.5 1 represents the power of a statistical test that is, the probability of correctly rejecting the null hypothesis when in reality the null hypothesis is false and should be rejected. 9.6 is the probability of making a Type I error that is, the probability of incorrectly rejecting the null hypothesis when in reality the null hypothesis is true and should not be rejected. 9.7 is the probability of making a Type II error that is, the probability of incorrectly failing to reject the null hypothesis when it is false. 9.8 The power of a test is the complement, which is (1 - ), of the probability of making a Type II error. 9.9 It is possible to incorrectly reject a true null hypothesis because the mean of a single sample can fall in the rejection region even though the hypothesized population mean is true. 9.10 It is possible to not reject a false null hypothesis because the mean of a single sample can fall in the non-rejection region even though the hypothesized population mean is false. 9.11 will increase. 9.12 Other things being equal, the closer the hypothesized mean is to the actual mean, the larger the risk of committing a Type II error will be. 9.13 is the probability of incorrectly convicting the defendant when he is innocent. is the probability of incorrectly failing to convict the defendant when he is guilty. 9.14 Under the French judicial system, unlike ours in the United States, the null hypothesis assumes the defendant is guilty, the alternative hypothesis assumes the defendant is innocent. A Type I error would be not convicting a guilty person and a Type II error would be convicting an innocent person. Solutions to End-of-Section and Chapter Review Problems 309 9.15 (a) A Type I error is the mistake of approving an unsafe drug. A Type II error is not approving a safe drug. (b) The consumer groups are trying to avoid a Type I error. To them, the risk of mistakenly approving an unsafe drug is most important and, therefore they want to try to avoid a Type I error. (c) The industry lobbyists are trying to avoid a Type II error. To them, the risk of not approving a safe drug is most important and, therefore, they want to try to avoid a Type II error. (d) To lower both Type I and Type II errors, the FDA can require more information and evidence in the form of more rigorous testing. This can easily translate into longer time to approve a new drug. 9.16 H0: = 20 minutes. 20 minutes is adequate travel time between classes. H1: 20 minutes. 20 minutes is not adequate travel time between classes. 9.17 H0: = 70 pounds. The cloth has a mean breaking strength of 70 pounds. H1: 70 pounds. The cloth has a mean breaking strength that differs from 70 pounds. 9.18 H0: = 1.00. The mean amount of paint per can is one gallon. H1: 1.00. The mean amount of paint per can differs from one gallon. 9.19 H0: = 375 hours. The mean life of the manufacturers light bulbs is equal to 375 hours. H1: 375 hours. The mean life of the manufacturers light bulbs differs from 375 hours. 9.20 Decision rule: Reject H0 if Z + 1.96. Decision: Since Zcalc = + 2.21 is greater than Zcrit = + 1.96, reject H0. 9.21 Decision rule: Reject H0 if Z + 1.645. 9.22 Decision rule: Reject H0 if Z + 2.58. 9.23 Decision: Since Zcalc = 2.61 is less than Zcrit = 2.58, reject H0. 9.24 p-value = 2(1 .9772) = 0.0456 9.25 Since the p-value of 0.0456 is less than the 0.10 level of significance, the statistical decision is to reject the null hypothesis. 9.26 p-value = 0.1676 9.27 At 0.01 level of significance, the statistical decision is to not reject the null hypothesis. 310 Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests 9.28 (a) H0: = 70 pounds. The cloth has a mean breaking strength of 70 pounds. H1: 70 pounds. The cloth has a mean breaking strength that differs from 70 pounds. Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 69.170 3.5 49 = 1.80 Decision: Since Zcalc = 1.80 is between the critical bounds of 1.96, do not reject H0. There is not enough evidence to conclude that the cloth has a mean breaking strength that differs from 70 pounds. (b) p-value = 2(0.0359) = 0.0718 Interpretation: The probability of getting a sample of 49 pieces that yield a mean strength that is farther away from the hypothesized population mean than this sample is 0.0718 or 7.18%. (c) Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 69.170 1.75 49 = 3.60 Decision: Since Zcalc = 3.60 is less than the lower critical bound of 1.96, reject H0. There is enough evidence to conclude that the cloth has a mean breaking strength that differs from 70 pounds. (d) Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 69 70 3.5 49 = 2.00 Decision: Since Zcalc = 2.00 is less than the lower critical bound of 1.96, reject H0. There is enough evidence to conclude that the cloth has a mean breaking strength that differs from 70 pounds. 9.29 (a) H0: = 1.00. The mean amount of paint per can is one gallon. H1: 1.00. The mean amount of paint per can differs from one gallon. Decision rule: Reject H0 if Z + 2.58. Test statistic: Z = X n = 0.9951.00 0.02 50 = 1.77 Decision: Since Zcalc = 1.77 is between the critical bounds of 2.58, do not reject H0. There is not enough evidence to conclude that the mean amount of paint per one- gallon can differs from one gallon. (b) p-value = 2(0.0384) = 0.0768 Interpretation: The probability of getting a sample of 50 cans that will yield a mean amount that is farther away from the hypothesized population mean than this sample is 0.0768. (c) X Z n = 0.995 2.58 0.02 50 0.98771.0023 (d) Same decision. The confidence interval includes the hypothesized value of 1.00. Solutions to End-of-Section and Chapter Review Problems 311 9.30 (a) H0: = 375 hours. The mean life of the manufacturers light bulbs is equal to 375 hours. H1: 375 hours. The mean life of the manufacturers light bulbs differs from 375 hours. Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 350 375 100 64 = 2.00 Decision: Since Zcalc = 2.00 is below the critical bound of 1.96, reject H0. There is enough evidence to conclude that the mean life of the manufacturers light bulbs differs from 375 hours. (b) p-value = 2(0.0228) = 0.0456. Interpretation: The probability of getting a sample of 64 light bulbs that will yield a mean life that is farther away from the hypothesized population mean than this sample is 0.0456. (c) 100 350 1.96 64 XZ n = 325.50374.50 (d) The results are the same. The confidence interval formed does not include the hypothesized value of 375 hours. 9.31 (a) H0: = 2.00 liters. The mean amount of soft drink placed in 2-liter bottles at the local bottling plant is equal to 2 liters. H1: 2.00 liters. The mean amount of soft drink placed in 2-liter bottles at the local bottling plant differs from 2 liters. Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 1.99 2.00 0.05 100 = 2.00 Decision: Since Zcalc = 2.00 is below the lower critical bounds of 1.96, reject H0. There is enough evidence to conclude that the amount of soft drink placed in 2-liter bottles at the local bottling plant differs from 2 liters. (b) p-value = 2(0.0228) = 0.0456. Interpretation: The probability of getting a sample of 100 bottles that will yield a mean amount that is farther away from the hypothesized population mean than this sample is 0.0456. (c) X Z n =1.99 1.96 0.05 100 1.98021.9998 (d) The results here are the same. The confidence interval formed does not include 2.00. 312 Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests 9.32 (a) 01 :8 :8HH= Decision rule: Reject 0 Hif Z 1.96 Test statistic: 7.983 8 0.8 .15/50 Z = Decision: Do not reject 0 H. There is insufficient evidence to conclude that the amount of salad dressing placed in 8 oz bottle is significantly different from 8 oz. (b) p-value = 0.4229. The probability of observing a Z test statistic more extreme than 0.8 is 0.4229 if the population mean is indeed 8 oz. (c) Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 7.983 8 0.05 50 = 2.40 Decision: Since Zcalc = 2.40 is less than the lower critical bound of 1.96, reject H0. There is enough evidence to conclude that the machine is filling bottles improperly. (d) Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = 7.952 8 0.15 50 = 2.26 Decision: Since Zcalc = 2.26 is less than the lower critical bound of 1.96, reject H0. There is enough evidence to conclude that the machine is filling bottles improperly. Solutions to End-of-Section and Chapter Review Problems 313 9.33 (a) H0: = $160. The mean amount of cash withdrawn per customer from the ATM machine is equal to $160. H1: $160. The mean amount of cash withdrawn per customer from the ATM machine is not equal to $160. Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = $172 $160 $30 36 = 2.40 Decision: Since Zcalc = 2.40 is greater than the upper critical bound of 1.96, reject H0. There is enough evidence to conclude that the mean amount of cash withdrawn per customer from the ATM machine is not equal to $160. (b) p-value = 0.0164. The probability of observing a Z test statistic more extreme than 2.40 is 0.0164 if the population mean is indeed $160. (c) Decision rule: Reject H0 if Z 2.58. Test statistic: Z = X n = $172 $160 $30 36 = 2.40 Decision: Since Zcalc = 2.40 is between the critical bounds of Z = 2.58, do not reject H0. There is not enough evidence to conclude that the mean amount of cash withdrawn per customer from the ATM machine is not equal to $160. (d) Decision rule: Reject H0 if Z + 1.96. Test statistic: Z = X n = $172 $160 $24 36 = 3.00 Decision: Since Zcalc = 3.00 is greater than the upper critical bound of 1.96, reject H0. There is enough evidence to conclude that the mean amount of cash withdrawn per customer from the ATM machine is not equal to $160. 9.34 Z = 2.33 9.35 Since Zcalc = 2.39 is greater than Zcrit = 2.33, reject H0. 9.36 Z = 2.33 9.37 Since Zcalc = 1.15 is greater than Zcrit = 2.33, do not reject H0. 9.38 p-value = 1 0.9772 = 0.0228 9.39 Since the p-value = 0.0228 is less than = 0.05, reject H0. 9.40 p-value = 0.0838 9.41 Since the p-value = 0.0838 is greater than = 0.01, do not reject H0. 9.42 p-value = ()1.380.9162P Z 0.01, do not reject the null hypothesis. 314 Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests 9.44 H0: 2.8 feet. The mean length of steel bars produced is at least 2.8 feet and the production equipment does not need immediate adjustment. H1: 2.8 feet. The mean length of steel bars produced is less than 2.8 feet and the production equipment does need immediate adjustment. (a) Decision rule: If Z 1.645, reject H0. Test statistic: Z = X n = 2.732.8 0.2 25 = 1.75 Decision: Since Zcalc = 1.75 is less than Zcrit = 1.645, reject H0. There is enough evidence to conclude the production equipment needs adjustment. (b) Decision rule: If p-value 0.05, reject H0. Test statistic: Z = X n = 2.732.8 0.2 25 = 1.75 p-value = 0.0401 Decision: Since p value = 0.0401 is less than = 0.05, reject H0. There is enough evidence to conclude the production equipment needs adjustment. (c) The probability of obtaining a sample whose mean is 2.73 feet or less when the null hypothesis is true is 0.0401. (d) The conclusions are the same. 9.45 H0: 25 min. The mean delivery time is not less than 25 minutes. H1: 25 min. The mean delivery time is less than 25 minutes. (a) Decision rule: If Z 2.3263 or when the p-value $160. The mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is greater than $160. Decision rule: If Z 1.645, reject H0. Test statistic: Z = X n = $172 $160 $30 36 = 2.40 Decision: Since Zcalc = 2.40 is greater than 1.645, reject H0. There is enough evidence to conclude that the mean amount of money withdrawn from the ATM machines per customer transaction over a weekend is greater than $160. (b) Decision rule: If p-value 1.2902, reject H0. Test statistic: t = X S n = $315.40 $300.00 $43.20 100 = 3.5648 Decision: Since tcalc = 3.5648 is above the critical bound of t = 1.2902, reject H0. There is enough evidence to conclude that the mean cost of textbooks per semester at a large university is more than $300. (b) H0: $300. The mean cost of textbooks per semester at a large university is no more than $300. H1: $300. The mean cost of textbooks per semester at a large university is more than $300. Decision rule: d.f. = 99. If t 1.6604, reject H0. Test statistic: t = X S n = $315.40 $300.00 $75.00 100 = 2.0533 Decision: Since tcalc = 2.0533 is above the critical bound of t = 1.6604, reject H0. There is enough evidence to conclude that the mean cost of textbooks per semester at a large university is more than $300. Solutions to End-of-Section and Chapter Review Problems 317 9.54 (c) H0: $300. The mean cost of textbooks per semester at a large university is no cont. more than $300. H1: $300. The mean cost of textbooks per semester at a large university is more than $300. Decision rule: d.f. = 99. If t 1.2902, reject H0. Test statistic: t = X S n = $305.11 $300.00 $43.20 100 = 1.1829 Decision: Since tcalc = 1.1829 is below the critical bound of t = 1.2902, do not reject H0. There is not enough evidence to conclude that the mean cost of textbooks per semester at a large university is more than $300. 9.55 01 :22 :22HH= Decision rule: Reject 0 H if |t| 1.6766 d.f. = 49 Test statistic: 25.3622 3.2816 7.24 50 X t S n = Decision: Since |t| 1.6766, reject 0 H. There is enough evidence to conclude that the mean shopping time at the local supermarket is different from the claimed value of 22.2 minutes. 9.56 H0: 3.7 The mean waiting time is no less than 3.7 minutes. H1: 3.7 The mean waiting time is less than 3.7 minutes. Decision rule: d.f. = 63. If t 2.0096 or p-value 4 The mean life of the batteries is more than 400 hours. Decision rule: Reject 0 H if t 1.7823 Test statistic: 473.46-400 =1.2567 /210.77/ 13 X t sn = Decision: Since t 4 The mean life of the batteries is more than 400 hours. Decision rule: Reject 0 H if t 1.7823 Test statistic: 550.38-400 =1.7195 /315.33/ 13 X t sn = Decision: Since t 2.0555 d.f. = 26 Test statistic: 43.888945 0.2284 25.2835