hw7sol - Engineering 0020 Probability and Statistics 1...

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Unformatted text preview: Engineering 0020 Probability and Statistics 1 Solutions - Homework #7 Spring, 2007 6. a /2=0.1/2=0.05 Z /2 = 1.645 8439 1.645(100/ 25) [8406.1; 8471.9] b /2=0.08/2=0.04 Z /2 = 1.75 8439 1.75(100/ 25) [8404.0; 8474.0] 7. a Since L=2Z /2 ( / n ), it follows that if the new interval L' has half this length, all else being the same, the new sample size must be 4 n so that L'= 2Z /2 ( / 4 n ) =L/2. b If the sample size becomes 25 n , then all else being the same, the new length is L'= 2Z /2 ( / 25 n ) = L/ 25, i.e., decreased by a factor of 5. 18. a For a 90% lower confidence bound we want to find LCL such that P(LCL< ]=0.9. Now, P[( X - (1.28) S/ n)< ] = P[ X - < (1.28) S/ n)]= P[( X - )/(S/ n) < (1.28)) P[Z<1.28] =0.90 (from Normal tables), since (X bar- )/(S/ n) Z when n is large. Since ( X - (1.28) S/ n) = 4.25 - (1.28)1.30/ 78= 4.06 we thus have P[4.06< ] = 0.90 , i.e., a 90% lower confidence bound for is 4.06 NOTE: This is referred to as a one-sided interval for and represented as [4.06;[4....
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hw7sol - Engineering 0020 Probability and Statistics 1...

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