# hw7sol - Engineering 0020 Probability and Statistics 1...

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Engineering 0020 Probability and Statistics 1 Solutions - Homework #7 Spring, 2007 6. a α /2=0.1/2=0.05 Z α /2 = 1.645 8439 ± 1.645(100/ 25) [8406.1; 8471.9] b α /2=0.08/2=0.04 Z α /2 = 1.75 8439 ± 1.75(100/ 25) [8404.0; 8474.0] 7. a Since L=2Z α /2 ( σ / 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 α /2 ( σ / 4 n ) =L/2. b If the sample size becomes 25 n , then all else being the same, the new length is L'= 2Z α /2 ( σ / 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< μ ]=0.9. Now, P[( X - (1.28) S/ n)< μ ] = P[ X - μ < (1.28) S/ n)]= P[( X - μ )/(S/ n) < (1.28)) P[Z<1.28] =0.90 (from Normal tables), since (X bar - μ )/(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< μ ] = 0.90 , i.e., a 90% lower confidence bound for μ is 4.06 NOTE: This is referred to as a one-sided interval for μ and represented as [4.06; ) b Since n =78 is a large sample, we can use ( x ± Z α /2 s/ n) With α =0.1, Z α /2 =1.645 4.25 ± 1.645 (1.3/8.83) [4.008; 4.492] 32. The interval for a small sample from a Normal population with unknown S.D.

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