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HW 5 Solutions - HW 5 Solutions 4.146 We require P(|Y |...

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HW 5 Solutions 4.146 We require P (| Y – μ| ≤ k σ) ≥ .90 = 1 – 1/ k 2 . Solving for k , we see that k = 3.1622. Thus, the necessary interval is | Y – 25,000| ≤ (3.1622)(4000) or 12,351 ≤ 37,649. 4.154 a. μ = 7, σ 2 = 2(7) = 14. b. Note that σ = 14 = 3.742. The value 23 is (23 – 7)/3.742 = 4.276 standard deviations above the mean, which is unlikely. 9.17 . Thus, by Theorem 9.2, X Y is a consistent estimator of μ 1 – μ 2 . 9.20 Since E ( Y ) = np and V ( Y ) = npq , we have that E ( Y / n ) = p and V ( Y / n ) = pq / n . Thus, Y / n is consistent since it is unbiased and its variance goes to 0 with n . 9.26 a. We have that ) ( ) ( ) ( ) ( ) ( ) ( ε - θ - ε + θ = ε + θ ε - θ n n n F F Y P . If ε > θ, 1 ) ( ) ( = ε + θ n F and 0 ) ( ) ( = ε - θ n F . Thus, ) ( ) ( ε + θ ε - θ n Y P = 1. If ε < θ, 1 ) ( ) ( = ε + θ n F , ( 29 n n F θ ε - θ = ε - θ ) ( ) ( . So, ( 29 n n Y P θ ε - θ - = ε + θ ε - θ 1 ) ( ) ( . b. The result follows from ( 29 [ ] 1 1 lim ) ( lim ) ( = - = ε + θ ε - θ θ ε - θ n n n n Y P . 9.39 Thus is sufficient by Factorization Theorem. 9.44 With β known, the likelihood is ( 29 ) 1 ( 1 ) ( + α - = α β α = α n i i n n y L . By Theorem 9.4, U = = n i i Y 1 is sufficient for α with ( 29 ) 1 ( ) , ( + α - α β α = α u u g n n and h ( y ) = 1. 9.51 Again, using the indicator notation, the density is
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