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415mid2Solution08 - 2 x n σ 2 y m ∼ N(0 1 Z 025 = 1 96...

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Stat/math 415 mid2 1. Test about one proportion (a) H 0 : P = 0 . 761 H 1 : P < 0 . 761 (b) Test statistic Z = ˆ P - P 0 P 0 (1 - P 0 ) /n N (0 , 1) Critical region { Z : Z < - 1 . 645 } Z 0 . 05 = 1 . 645 Since Z = - 1 . 357 > - 1 . 645 , fail to reject H 0 . Therefore we can not say the U.S.LFP rate for men has fallen under fixed type I error. Next we need to consider the power of our test, which is related with TYPE II error because small power, large type II error, can always drive to fail to reject H 0 . Check type II error before conclusion, and if it is too large to obtain reliability of our test, then we can say there is not enough evidence to reject H 0 . Increasing sample size is one of solution to decrease type II error given fixed type I error. (c) P-value: P = P ( Z ≤ - 1 . 357) = 0 . 087 Because 0 . 05 < P < 0 . 1 , reject H 0 when α = 0 . 1 , but fail to reject H 0 when α = 0 . 05 . 2. Test about two mean (a) H 0 : μ x = μ y μ x 6 = μ y (b) ¯ X - ¯ Y N ( μ x - μ y , σ 2 x n + σ 2 y m ) X and Y are independent. Critical region : { Z : | Z | > 1 . 96 } where K = ¯ X - ¯ Y - (
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Unformatted text preview: 2 x n + σ 2 y m ∼ N (0 , 1) , Z . 025 = 1 . 96 Since Z= 24 . 3-27 . 5 √ 16 20 + 25 30 =-3 . 2 1 . 278 =-2 . 50 <-1 . 96 (c) 24 . 3-27 . 5 ± 1 . 96 q 16 20 + 25 30 ⇔ [-5 . 7 ,-. 7] Because The confidence interval does not include 0, reject H . Therefore the mean height of the red pine in West Virginia is different from that of the main population. 3. Let X ∼ N ( μ, 100) , n = 25 . H : μ = 80 , H 1 : μ < 80 , RR= { ¯ x ≤ 77 . 5 } (a) Power function is K ( μ ) = P { ¯ x ≤ 77 . 5 } . (a) α = P (¯ x ≤ 77 . 5 | μ = 80) = P ( Z ≤ 77 . 5-80 √ 100 25 ) = 0 . 1056 (b) P (¯ x > 77 . 5 | μ = 75) = P ( Z > 77 . 5-75 √ 100 25 ) = 0 . 1056 (b) (c) n = (1 . 96+1 . 96) 2 100 (80-75) 2 = 61 . 4656 , so n=62....
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