finalrevsol3

finalrevsol3 - Solutions to Final Review Problems- Part 2...

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Solutions to Final Review Problems- Part 2 15. ˆ p = 34 / 100 = 0 . 34 (a) Hypotheses: H 0 : p = 0 . 29 H a : p 6 = 0 . 29 Test Statistic: Z = ˆ p - p 0 q p 0 (1 - p 0 ) n = 0 . 34 - 0 . 29 q (0 . 29)(0 . 71) 100 1 . 10 Since this is a two-tailed Z -test, the P-value is the area under the Z curve above and below the test statistic and it’s negative. P-value = 2 · P ( Z > 1 . 10) = 2(0 . 1357) = 0 . 2714 (b) P-value > 0 . 10 implies fail to reject H 0 at 0 . 10 level of signiFcance. (c) Anything above the P-value. ±or example α = 0 . 28. 16. (a) ( n - 1) S 2 χ 2 α/ 2 ,n - 1 , ( n - 1) S 2 χ 2 1 - α/ 2 ,n - 1 (10 - 1)24 2 23 . 587 , (10 - 1)24 2 1 . 735 ! (219 . 7821 , 2987 . 896) (b) Hypotheses: H 0 : σ 2 = 25 2 H a : σ 2 < 25 2 Test Statistic: W = ( n - 1) S 2 σ 2 0 = (10 - 1)24 2 25 2 = 8 . 2944 Rejection Region: The rejection region is the lower tail (with area 0
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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