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# HW4 - 8.59 B = 2 = 10 8.108 p = 0.67 n = 415 ^ 95 CI for p...

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8.59 B = 2 = 10 , n = 4 σ 2 B 2 = 100 8.108 ˆ p = 0 . 67 , n = 415 95% CI for p is ˆ p ± 1 . 96 q ˆ p (1 - ˆ p ) n = 0 . 832 ± 0 . 0153 = (0 . 817 , 0 . 847) 9.2 a E μ 1 ) = 1 2 2 μ = μ, E μ 2 ) = 1 4 μ + ( n - 2) μ 2( n - 2) + 1 4 μ, E μ 3 ) = n = μ b V μ 1 ) = 1 4 (2 σ 2 ) = σ 2 2 , V μ 2 ) = 2 16 ( σ 2 ) + ( n - 2) σ 2 4( n - 2) 2 = σ 2 8 + σ 2 4( n - 2) , V μ 3 ) = σ 2 n Hence the eﬃcient of ˆ μ 3 relative to ˆ μ 1 is V μ 1 ) V μ 3 ) = n 2 and the eﬃciency of ˆ μ 3 relative to ˆ μ 2 is V μ 2 ) V μ 3 ) = n 8 + n 4( n - 2) = n 2 8( n - 2) 9.6 V ( ˆ λ 1 ) = V h 1 2( Y 1 + Y 2 ) i = λ 2 , V ( ˆ λ 2 ) = λ n eﬀ ( ˆ λ 1 , ˆ λ 2 ) = 2 n , so using all of the data is much more eﬃcient than using only two points. 9.9 By Theorem 9.1, E ( ˆ θ 1 ) = E ( ˆ θ 2 ) = θ, lim n →∞ V ( ˆ θ
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