hw3_sol-1

# hw3_sol-1 - 10-40 given x = 1014 hours = 25 hours n = 20(a...

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10-40 given: 1014 x = hours, σ = 25 hours, n = 20 (a) 95% 2-sided interval 1 95 % 0.05 / 2 0.025 a a = -= = from Table II in the Appendix / 2 0.05/2 1.96 zz a == / 2 /2 101 4 1.96(2 5 20 ) 101 4 1.96(2 5 20) x zn x aa s ms m - ≤≤+ - ≤+ [1003.04,1024.96] m ⇒∈ (b) 95% lower C.I. 1 0.9 5 0.05 a = from Table IV in the Appendix: 0.05 1.645 a 101 4 1.645(2 5 1004.8 x a sm m m -≤ ⇒≤ 10-42 2 1.96 25 5 96.04 97 z n E z E n n a a s s  =   = = = ⇒=

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11-2.a) Claim: the yield is less than 90% 0 1 : 90 : 90 H H claim m m < 2 5 , 90.48 , 5 , 0.05 nx sa = = == 90.4 8 90 0.48 55 obs z - using the critical point 0.05 1.645 zz a Since (0.4 8 1.645) obs a - - , 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 m m = 0.05 , 20 , 25 , 1014 as = === 0 101 4 1000 2.5 2 5 20 obs x z n m s - - = using the p-value p-valu
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## This note was uploaded on 04/02/2008 for the course ISE 225 taught by Professor Palmer during the Spring '08 term at USC.

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hw3_sol-1 - 10-40 given x = 1014 hours = 25 hours n = 20(a...

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