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Unformatted text preview: (1p ) on the interval [0 , . 4], the derivative f ( p ) is never zero, so the only critical points are the endpoints p = 0 and p = 0 . 4, so the largest that p (1p ) can be is 0 . 4(1. 4). Hence, n ≥ z 2 . 005 p (1p ) (0 . 05) 2 = 2 . 58 2 (0 . 40)(0 . 60) (0 . 05) 2 = 639 . 01 , so we take n = 640 . 1 Section 5.4 Problem 10 G . E ( Y 2 ) = Z θ y 2 1 θ dy = θ 2 3 , so since we want E ( ˆ θ ) = θ 2 , and E ( Y 2 ) = θ 2 / 3, we can simply let ˆ θ = 3 Y 2 to get an unbiased ˆ θ . Problem 12. For each i , E [( Y iμ ) 2 ] = σ 2 by the deﬁnition of σ 2 . Therefore, E " 1 n n X i =1 ( Y iμ ) 2 # = 1 n n X i =1 E ± ( Y iμ ) 2 ² = 1 n nσ 2 = σ 2 Problem 15. From Example 3.9.9, we know that Var( W ) = σ 2 n , so: E ( W 2 ) = Var( W ) + E ( W ) 2 = σ 2 n + μ 2 , so lim n →∞ E ( W 2 ) = lim n →∞ ³ σ 2 n + μ 2 ´ = μ 2 . 2...
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 Spring '07
 Xu,Lily
 Statistics, Harshad number, Super Bowl, Super Bowl XXIX

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