hw3_sol

# hw3_sol - Homework 3 Solution 10-40 given x = 1014 hours =...

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Homework 3 Solution 10-40 given: 1014 x = hours, σ = 25 hours, n = 20 (a) 95% 2-sided interval 1 95% 0.05 / 2 0.025 α = - = = from Table II in the Appendix / 2 0.05 / 2 1.96 z z = = / 2 / 2 1014 1.96(25 20) 1014 1.96(25 20) x z n x z n σ μ - + - + [1003.04,1024.96] (b) 95% lower C.I. 1 0.95 0.05 = - = from Table IV in the Appendix: 0.05 1.645 z z = = 1014 1.645(25 20) 1004.8 x z n - - 10-42 2 / 2 / 2 1.96 25 5 96.04 97 z n E z E n n = = = = = =

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11-2.a) Claim: the yield is less than 90% 0 1 : 90 : 90 H H claim μ < 2 5, 90.48, 5, 0.05 n x σ α = = = = 90.48 90 0.48 5 5 obs z - = = using the critical point 0.05 1.645 z z = = Since (0.48 1.645) obs z z > - > - , we should fail to reject the null hypothesis (H 0 ). The yield is not less than 90%. 11-5 Claim: the mean life the light bulbs is 1000 hours. 0 1 : 1000 : 1000 H claim H = 0.05, 20, 25, 1014 n x = = = = 0 1014 1000 2.5 25 20 obs x z n - - = = = using the p-value
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## This note was uploaded on 01/04/2011 for the course ISE 31602D taught by Professor Kurtpalmer during the Fall '10 term at USC.

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hw3_sol - Homework 3 Solution 10-40 given x = 1014 hours =...

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