Hw11ans - STAT 410 Homework #11 (due Friday, April 11, by...

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STAT 410 Homework #11 Spring 2008 (due Friday, April 11, by 3:00 p.m.) 1. Let X 1 , X 2 , … , X 25 be a random sample from a N ( μ , σ 2 = 100 ) population, and suppose the null hypothesis H 0 : μ = 100 is to be tested. a) If the alternative hypothesis is H 1 : μ 100, compute the power of the appropriate test at μ 1 = 101. Use α = 0.05. b) If the alternative hypothesis is H 1 : μ > 100, compute the power of the appropriate test at μ 1 = 101. Use α = 0.05. c) For the test in (a) compute the p-value associated with X = 103.5. d) For the test in (b) compute the p-value associated with X = 103.5. σ 2 = 100. σ = 10. n = 25. α = 0.05. σ is known. Test Statistic: 25 10 100 X X Z ± 0 - = - = n . a) H 0 : μ = 100 vs. H 1 : μ 100. Two - tailed. Rejection Region: Reject H 0 if Z < 2 z α - = – 1.960 or Z > 2 z = 1.960. 25 10 960 . 1 100 X - < = 96.08 or 25 10 960 . 1 100 X + > = 103.92. Power = P ( Reject H 0 | H 0 is NOT true ) = P ( X < 96.08 | μ = 101 ) + P ( X > 103.92 | μ = 101 ) = ± ² ³ ³ ³ ´ µ - > + ± ² ³ ³ ³ ´ µ - < 25 10 101 103.92 Z P 25 10 101 96.08 Z P = P ( Z < – 2.46 ) + P ( Z > 1.46 ) = 0.0069 + 0.0721 = 0.0790 .
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b) H 0 : μ = 100 vs. H 1 : μ > 100. Right - tailed. Rejection Region: Reject H 0 if Z > z α = 1.645. . 25 10 645 . 1 100 X + > = 103.29. Power = P ( Reject H 0 | H 0 is NOT true ) = P ( X > 103.29 | μ = 101 ) = ± ² ³ ³ ³ ´ µ - > 25 10 101 103.29 Z P = P ( Z > 1.145 ) 2 1251 . 0 1271 . 0 + = 0.1261 .
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X = 103.5. σ 2 = 100. σ = 10. n = 25. σ
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Hw11ans - STAT 410 Homework #11 (due Friday, April 11, by...

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