# HW7 - Stat 415 HW 7 10.36 R ={Z-0 ^ ^ 0 > z 0 ^ > z so we...

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Stat 415 HW 7 10.36 R = { Z : θ - θ 0 σ ˆ θ > z α } , so we reject ˆ θ - θ 0 > z α σ ˆ θ θ 0 < ˆ θ - z α σ ˆ θ where the left-hand side is the 100% lower conﬁdence bound for θ . 10.37 ˆ p - z 0 . 01 q ˆ p (1 - ˆ p ) n = 0 . 57 By 10.36, p 0 [0 , 0 . 57) which is rejection region, so we reject H 0 . This does not conﬂict with the answer to exercise 10.15. 10.41 (i) H 0 : μ 1 - μ 2 = 0 , H a : μ 1 - μ 2 6 = 0 Under H 0 , Y 1 - Y 2 N ± 0 , σ 2 1 n 1 + σ 2 2 n 2 ² , s 2 1 p σ 2 1 ,s 2 2 p σ 2 2 by the law of large sample. (ii) Z = ¯ Y 1 - ¯ Y 2 r s 2 1 n 1 + s 2 2 n 2 = 1 . 58 (iii) R = { Z ; | Z | > 1 . 96 } , P-value= [ | Z | > 1 . 58 | H 0 ] = 0 . 1142 > 0 . 05 (iv) Not reject H 0 in a test at level α = 0 . 1 because p-value > 0 . 05 and also 1.58 is not in the R. 10.44 1. (i) H 0 : p 0 = 0 . 85 , H a : p 0 > 0 . 85 (ii) Z = ˆ p - p 0 q p 0 (1 - p 0 ) n = 5 . 34 (iii) R = { Z > 2 . 33 } (iv) Reject H 0 because of 5.34>2.33 2. P-value = [ Z > 5 . 34 | H 0 ] = 0 . 000 . Hence H 0 can be rejected for any value α > 0 . 001 10.53 1. (i) H 0 : μ = 45 H a : μ < 45 Small sample from normal and unknown σ 2 (ii) Z = ¯ Y - μ 0 s/ n t 17 (iii) R = { Z < - 1 . 739 (= t α, 17 ) } P = [ Z < - 3 . 24 | H 0 ] = 0 . 002407253 (iv) The p-value is less than 0.05, or -3.24<-1.739, so reject

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## This note was uploaded on 05/17/2008 for the course MATH 415 taught by Professor Yuzhang during the Spring '08 term at Penn State.

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HW7 - Stat 415 HW 7 10.36 R ={Z-0 ^ ^ 0 > z 0 ^ > z so we...

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