This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 onesided interval for μ and represented as [4.06;[4....
View
Full
Document
This homework help was uploaded on 04/10/2008 for the course ENGR 0020 taught by Professor Rajgopal during the Spring '08 term at Pittsburgh.
 Spring '08
 Rajgopal

Click to edit the document details