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ISM_Chapter_11

ISM_Chapter_11 - Chapter 11 11.1 H 0 The drug is not safe...

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Chapter 11 11.1 0 H : The drug is not safe and effective 1 H : The drug is safe and effective 11.2 0 H : I will complete the Ph.D. 1 H : I will not be able to complete the Ph.D. 11.3 0 H : The batter will hit one deep 1 H : The batter will not hit one deep 11.4 0 H : Risky investment is more successful 1 H : Risky investment is not more successful 11.5 1 H : The plane is on fire 1 H : The plane is not on fire 11.6 The defendant in both cases was O. J. Simpson. The verdicts were logical because in the criminal trial the amount of evidence to convict is greater than the amount of evidence required in a civil trial. The two juries concluded that there was enough (preponderance of) evidence in the civil trial, but not enough evidence (beyond a reasonable doubt) in the criminal trial. All p-values and probabilities of Type II errors were calculated manually using Table 3 in Appendix B. 11.7 Rejection region: z < 575 . 2 005 . - = - z or z > 005 . z = 2.575 00 . 1 100 / 200 1000 980 n / x z - = - = σ μ - = p-value = 2P(Z < –1.00) = 2(.5 – .3413) = 2(.1587) = .3174 There is not enough evidence to infer that μ 1000. 215

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11.8 Rejection region: z > 03 . z = 1.88 60 . 9 / 5 50 51 n / x z = - = σ μ - = p-value = P(Z > .60) = .5 – .2257 = .2743 There is not enough evidence to infer that μ > 50. 11.9 Rejection region: z < 28 . 1 10 . - = - z 75 . 1 25 / 2 15 3 . 14 n / x z - = - = σ μ - = p-value = P(Z < –1.75) = .5 – .4599 = .0401 There is enough evidence to infer that μ < 15. 11.10 Rejection region: z < 96 . 1 025 . - = - z or z > 025 . z = 1.96 0 100 / 10 100 100 n / x z = - = σ μ - = p-value = 2P(Z > 0) = 2(.5) = 1.00 There is not enough evidence to infer that μ 100. 216
11.11 Rejection region: z > 01 . z = 2.33 00 . 5 100 / 20 70 80 n / x z = - = σ μ - = p-value = p(z > 5.00) = 0 There is enough evidence to infer that μ > 70. 11.12 Rejection region: z < 645 . 1 05 . - = - z 33 . 1 100 / 15 50 48 n / x z - = - = σ μ - = p-value = P(Z < –1.33) = .5 – .4082= .0918 There is not enough evidence to infer that μ < 50. 11.13a. 20 . 1 9 / 5 50 52 n / x z = - = σ μ - = p-value = P(Z > 1.20) = .5 – .3849 = .1151 b. 00 . 2 25 / 5 50 52 n / x z = - = σ μ - = 217

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p-value = P(Z > 2.00) = .5 – .4772 = .0228. c. 00 . 4 100 / 5 50 52 n / x z = - = σ μ - = p-value = P(Z > 4.00) = 0. d. The value of the test statistic increases and the p-value decreases. 11.14a. 60 . 9 / 50 200 190 n / x z - = - = σ μ - = p-value = P(Z < –.60) = .5 – .2257 = .2743 b. 00 . 1 9 / 30 200 190 n / x z - = - = σ μ - = p-value = P(Z < –1.00) = .5 – .3413 = .1587 c 00 . 3 9 / 10 200 190 n / x z - = - = σ μ - = p-value = P(Z < –3.00) = .5 – .4987 = .0013 d. The value of the test statistic decreases and the p-value decreases. 11.15 a. 00 . 1 25 / 5 20 21 n / x z = - = σ μ - = p-value = 2P(Z > 1.00) = 2(.5 – .3413) = .3174 b. 00 . 2 25 / 5 20 22 n / x z = - = σ μ - = p-value = 2P(Z > 2.00) = 2(.5 – .4772) = .0456 c. 00 . 3 25 / 5 20 23 n / x z = - = σ μ - = p-value = 2P(Z > 3.00) = 2(.5 – .4987) = .0026 d. The value of the test statistic increases and the p-value decreases. 11.16 a. 25 . 1 100 / 8 100 99 n / x z - = - = σ μ - = p-value = 2P(Z < –1.25) = 2(.5 – .3944) = .2112 b. 88 . 50 / 8 100 99 n / x z - = - = σ μ - = p-value = 2P(Z < –.88) = 2(.5 – .3106) = .3788 c. 56 . 20 / 8 100 99 n / x z - = - = σ μ - = p-value = 2P(Z < –.56) = 2(.5 – .2123) = .5754 218
d. The value of the test statistic increases and the p-value increases. 11.17 a. 00 . 4 100 / 25 1000 990 n / x z - = - = σ μ - = p-value = P(Z < –4.00) = 0 b. 00 . 2 100 / 50 1000 990 n / x z - = - = σ μ - = p-value = P(Z < –2.00) = .5 – .4772 = .0228 c. 00 . 1 100 / 100 1000 990 n / x z - = - = σ μ - = p-value = P(Z < –1.00) = .5 – .3413 = .1587 d. d. The value of the test statistic increases and the p-value increases.

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