Chapter 11 solution.docx - Chapter 11 11.7 Rejection region z < z x n 980 1000 200 100 z.005 2.575 or z > z.005 = 2.575 1.00 p-value = 2P(Z < 1.00 =

Chapter 11 solution.docx - Chapter 11 11.7 Rejection region...

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Chapter 11 11.7 Rejection region: z < 575 . 2 005 . z or z > 005 . z = 2.575 00 . 1 100 / 200 1000 980 n / x z p-value = 2P(Z < –1.00) = 2(.1587) = .3174 There is not enough evidence to infer that 1000. 11.11 Rejection region: z > 01 . z = 2.33 00 . 5 100 / 20 70 80 n / x z p-value = p(z > 5.00) = 0 There is enough evidence to infer that > 70.
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11.15 a. 00 . 1 25 / 5 20 21 n / x z p-value = 2P(Z > 1.00) = 2(1 – .8413) = .3174 b. 00 . 2 25 / 5 20 22 n / x z p-value = 2P(Z > 2.00) = 2(1 – .9772) = .0456 c. 00 . 3 25 / 5 20 23 n / x z p-value = 2P(Z > 3.00) = 2(1 – .9987) = .0026 d. The value of the test statistic increases and the p-value decreases. 11.17 a. 00 . 4 100 / 25 1000 990 n / x z p-value = P(Z < –4.00) = 0 b. 00 . 2 100 / 50 1000 990 n / x z p-value = P(Z < –2.00) = .0228
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c. 00 . 1 100 / 100 1000 990 n / x z p-value = P(Z < –1.00) = .1587 d. d. The value of the test statistic increases and the p-value increases. 11.29 : H 0 = 50 : H 1 > 50 89 . 3 18 / 10 50 17 . 59 n / x z p-value = P(Z > 3.89) = 0
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