Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Solutions: Assignment 7 November 1, 2005 N.D. 1 a. X bin ( n, p ) E [ X ] = np V ( X ) = np (1 - p ) E p ] = E [ X n ] = 1 n E [ X ] = 1 n [ np ] = p Bias = ( E p ] - p ) = p - p = 0 b. V p ) = V ( X n ) = 1 n 2 V ( X ) = 1 n 2 ( np (1 - p )) = p (1 - p ) n c. S.E p ) = p V p ) = r p (1 - p ) n 1
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d. MSE ( X ) = E [(ˆ p - p ) 2 ] = ( bias p )) 2 + V p ) = 0 + V p ) = p (1 - p ) n e. S.E. p ) 0 . 10 S.E p ) = p V p ) = r p (1 - p ) n r p (1 - p ) n 0 . 10 p (1 - p ) n 0 . 10 2 p (1 - p ) 0 . 1 2 n largest left side can be is when p = . 5 . 5 2 . 1 2 n n 25 N.D. 2 a. For large n, X is approx. N ( np, np (1 - p )) ˆ p = X n So ˆ p is just X multiplied by a constant, 1 n Thus ˆ p also has a normal distribution. From above, we showed E p ] = p , SE p ) = q p (1 - p ) n 2
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So because ˆ p is normal, we can standardize it using the familiar methods, to get Z = ˆ θ - θ S.E.θ = ˆ p - p p (1 - p ) /n distributed N (0 , 1) b. Z = ˆ p - p p p (1 - p ) /n So P ( - z α/ 2 Z z α/ 2 ) = 1 - α P ( - z α/ 2 ˆ p - p p p (1 - p ) /n z α/ 2 ) = 1 - α P(-z α/ 2 p p (1 - p ) /n ˆ p - p z α/ 2 p p (1 - p ) /n ) = 1 - α P(-ˆ p - z α/ 2 p p (1 - p ) /n ≤ - p ≤ - ˆ p + z α/ 2 p p (1 - p ) /n ) = 1 - α P(ˆ p - z α/ 2 p p (1 - p
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hw7 solutions - Solutions: Assignment 7 November 1, 2005...

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