ismch09 - 132 Instructors Solutions Manual for Statistics...

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132 Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition Chapter 9: Hypothesis Testing 9.1 a. European perspective: 0 : H Genetically modified food stuffs are not safe 1 : H They are safe b. U.S. farmer perspective: 0 : H Genetically modified food stuffs are safe 1 : H They are not safe 9-2 0 : B G H T T No difference in the total number of votes between Bush and Gore 1 : B G H T T Bush with more votes 9-3 0 : H No change interest rates in interest rates is warranted 1 : H Reduce interest rates to stimulate the economy 9-4 0 : A B H π π : There is no difference in the percentage of underfilled cereal packages 1 : A B H π π < : Lower percentage after the change 9-5 0 : 16 H μ ; 1 : 16 H μ < ; reject 0 H if Z .10 < -1.28. 15.84 16 .4 16 Z - = = -1.6, therefore, Reject 0 H at the 10% level. 9-6 0 : 50 H μ ; 1 : 50 H μ < ; reject 0 H if Z .10 < -1.28 48.2 50 3 9 Z - = = -1.8, therefore, Reject 0 H at the 10% level. 9-7 a. 0 : 3 H μ = ; 1 : 3 H μ ; reject 0 H if Z .05 > 1.645 3.07 3 .4 64 Z - = = 1.4, therefore, Do Not Reject 0 H at the 5% level. b. p-value = 1 – F Z (1.4) = 1 - .9192 = .0808 c. the p-value would be higher – the graph should show that the p-value now corresponds to the area in both of the tails of the distribution whereas before it was the area in one of the tails. d. A one-sided alternative is more appropriate since we are not interested in detecting possible low levels of impurity, only high levels of impurity.
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133 Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition 9-8 0 1 : 3; : 3; H H μ μ < 2.4 3 1.8 100 Z - = = -3.33, p-value = 1 – F Z (3.33) = 1 - .9996 = .0004, therefore, reject 0 H at levels in excess of .04% 9-9 0 1 : 4; : 4; H H μ μ = reject 0 H if –2.575 > Z .005 > -2.575 4.27 4 1.32 1562 Z - = = 8.08, p-value = 1 – F Z (8.08) = 1 – 1.0000 = .0000, therefore, reject 0 H at any common level of alpha. 9-10 0 1 : 0; : 0; H H μ μ = .078 0 .201 76 Z - = = 3.38, p-value = 2[1 – F Z (3.38)] = 2[1 - .9996] = .0008, therefore, reject 0 H at levels in excess of .08% 9-11 0 1 : 3; : 3; H H μ μ reject 0 H if Z .01 > 2.33 3.31 3 .7 172 Z - = = 5.81, p-value = 1 – F Z (5.81) = 1 – 1.0000 = .0000, therefore, reject 0 H at any common level of alpha. 9-12 0 1 : 0; : 0; H H μ μ = < 2.91 0 11.33 170 Z - - = = -3.35, p-value = 1 – F Z (3.35) = 1 – .9996 = .0004, therefore, reject 0 H at any common level of alpha. 9-13 0 1 : 125.32; : 125.32; H H μ μ = reject 0 H if |t 15, .05/2 | > 2.131 131.78 125.32 25.4 16 t - = = 1.017, therefore, do not reject 0 H at the .05 level. 9.14a. No, the 95% confidence level provides for 2.5% of the area in either tail. This does not correspond to a one-tailed hypothesis test with an alpha of 5% which has 5% of the area in one of the tails. b. Yes.
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Chapter 9: Hypothesis Testing 134 9-15 0 1 : 10; : 10; H H μ μ = < 8.82 10 2.4013 10 t - = = -1.554, p-value = .077, therefore, reject 0 H at levels of alpha greater than 7.7% 9-16 0 1 : 20; : 20; H H μ μ = reject 0 H if |t 8, .05/2 | > 2.306 20.3556 20 .6126 9 t - = = 1.741, therefore, do not reject 0 H at the 5% level 9-17 0 1 : 78.5; : 78.5; H H μ μ = reject 0 H if |t 7, .10/2 | > 1.895 74.5 78.5 6.2335 8 t - = = -1.815, therefore, do not reject 0 H at the 10% level 9.18The population values must be assumed to be normally distributed. 0 1 : 50; : 50; H H μ μ < reject 0 H if t 19, .05 < -1.729 41.3 50 12.2 20 t - = = -3.189, therefore, reject 0 H at the 5% level 9-19 a. 0
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