55010Solution9

# 55010Solution9 - STAT550 Homework 9 Solution 6.6-1 Set E (X...

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STAT550 Homework 9 Solution 6.6-1 Set E ( X ) = 1 n n X i =1 x i ˆ μ = ¯ x Since E ( X 2 ) = σ 2 + μ 2 , Set E ( X 2 ) = 1 n n X i =1 x 2 i ˆ σ 2 + ˆ μ 2 = 1 n n X i =1 x 2 i ˆ σ 2 = 1 n n X i =1 ( x i - ¯ x ) 2 = S 2 = V ar ( X ) 6.6-3 (a) E ( X ) = 1 λ = 1 n n X i =1 x i = ¯ x The moment estimator for λ is ˆ λ = 1 ¯ x (b) ˆ λ = 1 ¯ x = 1 1 n n i =1 x i = 1 1 20 (0 . 107 + ... 0 . 049) = 2 . 22 ˆ σ 2 = 1 λ 2 = 1 2 . 22 2 = 0 . 203 8.2-3 (a) Let X be the proﬁt in one day: ¯ X N ( μ = 300 , σ 2 n = 50 2 30 ) E ( n ¯ X ) = nE ( ¯ X ) = = 30 × 300 = 9000 STD ( n ¯ X ) = q n 2 V ar ( ¯ X ) = s 30 2 · 50 2 30 = 273 . 86 P ( n ¯ X > 9500) = 1 - P ( Z 9500 - 9000 273 . 86 ) = 1 - F (1 . 8258) = 0 . 0339 (b) P ( ¯ X < 290) = P ( Z 290 - 300 q 50 2 30 ) = F ( - 1 . 0954) = 0 . 1367 8.3-1 Let X be the assembly time under the new procedures: ¯ X N ( μ = 0 . 9 , σ 2 n = 0 . 2 2 35 ) The 95% CI of ¯ X : ¯ X ± Z 1 - α 2 · s σ 2 n 0 . 9 ± 1 . 96 · s 0 . 2 2 35 [0 . 83 , 0 . 97] 1

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8.3-3 Let X be the grams of fat a packaged food has: ¯ X N ˆ μ = 1 n n X i =1 x i = 12 . 6 , σ 2 n = 1 n n i =1 ( x i - ¯ x ) 2 n = 0 . 412 5 ! The 90% CI of ¯ X : " 12 . 6 ± 1 . 65 · r 0 . 412 5 # [12 . 13 , 13 . 07] The 95% CI of ¯ X : " 12 . 6 ± 1 . 96 · r 0 . 412 5 # [12 . 04 , 13 . 16]
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## This note was uploaded on 09/25/2010 for the course STAT 550 taught by Professor Staff during the Spring '08 term at San Diego State.

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55010Solution9 - STAT550 Homework 9 Solution 6.6-1 Set E (X...

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