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# 409Hw09ans - STAT 409 Fall 2009 Homework#9(due Friday...

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STAT 409 Homework #9 Fall 2009 (due Friday, October 30, by 4:00 p.m.) From the textbook: 8.2-16 H 0 : σ 2 = 30 vs. H 1 : σ 2 = 80. Right – tailed. Recall: If X 1 , X 2 , … , X n are i.i.d. N ( μ , σ 2 ) , then ( 2 2 σ S 1 n - is χ 2 ( n – 1 ) . a) Test Statistic: ( 30 18 1 2 2 0 2 2 s σ s χ = = - n . Reject H 0 if 2 2 α χ χ ( n – 1 ) = 2 0.05 χ ( 18 ) = 28.87 . 30 18 2 s > 28.87 s 2 > 48.116667 . b) P ( Type II Error ) = P ( Do NOT Reject H 0 | H 0 is not true ) = P ( S 2 < 48.116667 | σ 2 = 80 ) = P ( ( 2 2 σ S 1 n - < 80 116667 . 48 18 | σ 2 = 80 ) = P ( χ 2 ( 18 ) < 10.82625 ) 0.10 , since 2 0.90 χ ( 18 ) = 10.86 . EXCEL: =1-CHIDIST(CHIINV(0.05,18)*30/80,18) 0.0984 .

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9.1-4 H 0 : μ = 80 vs. H 1 : μ > 80 Right – tailed. σ 2 = 100, σ = 10, n = 25. The critical (rejection) region is Reject H 0 if x 83. a) K( μ ) = P( Reject H 0 | μ ) = P( x 83 | μ ) = - 25 10 83 Z P μ = 1 – - Φ 2 83 μ . b) α = P( Reject H 0 | μ = 80 ) = K(80) = 1 – Φ (1.5) = 1 – 0.9332 = 0.0668 . c) K(80) = 1 – Φ (1.5) = 1 – 0.9332 = 0.0668 . K(83) = 1 – Φ (0) = 1 – 0.5000 = 0.5000 . K(86) = 1 – Φ (– 1.5) = 1 – 0.0668 = 0.9332 . d) e) x = 83.41. z = 25 10 80 41 . 83 - = 1.705 . P-value = right tail = P( Z 1.705 ) = 0.0441 .
9.1-6 H 0 : μ = 715 vs. H 1 : μ < 715 Left – tailed. σ = 140, n = 25. The critical (rejection) region is Reject H 0 if x 668.94. a) K( μ ) = P( Reject H 0 | μ ) = P( x 668.94 | μ ) = - 25 140 94 . 668 Z P μ = - Φ 28 94 . 668 μ . b) α = P( Reject H 0 | μ = 715 ) = K(715) = Φ (– 1.645) = 0.05 . c) K(668.94) = Φ (0) = 0.50 . K(622.88) = Φ (1.645) = 0.95 . d) e) x = 16698 / 25 = 667.92 668.94. Reject H 0 . f) x = 667.92. z = 25 140 715 92 . 667 - = 1.68 . P-value = left tail = P( Z 1.68 ) = 0.0465 .

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9.1-7 H 0 : μ = 715 vs. H 1 : μ < 715 Left – tailed. σ = 140, n = 25. K( μ ) = P( Reject H 0 | μ ) = P( x c | μ ) = - n c 140 Z P μ = - Φ n c 140 μ . Want K(715) = 0.05, K(650) = 0.90. n c 140 715 - = – 1.645, n c 140 650 - = 1.282. n c n c 140 715 140 650 - - - = n 140 650 715 - = 140 65 n = 1.282 + 1.645 = 2.927. n = 65 927 . 2 140 6.3043. n 39.7443. Round up n = 40 . c = 715 – 40 140 645 . 1 678.58 . OR c = 650 + 40 140 282 . 1 678.38 .
9.1-10 H 0 : λ = 0.5 vs. H 1 : λ > 0.5 Right – tailed. Reject H 0 if = 8 1 i i x 8 . a) significance level α = P ( Reject H 0 | H 0 is true ) = P ( = 8 1 X i i 8 | λ = 0.5 ) . = 8 1 X i i is distributed Poisson ( 0.5 × 8 = 4 ) . P ( Poisson ( 4 ) 8 ) = 1 – P ( Poisson ( 4 ) 7 ) = 1 – 0.949 = 0.051 . b) Power = P ( Reject H 0 | H 0 is false ) = P ( = 8 1 X i i 8 | λ ) = ( 29 ( = - 8 8 !

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