Dr. Hackney STA Solutions pg 173

# Dr. Hackney STA Solutions pg 173 - Chapter 11 Analysis of...

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Unformatted text preview: Chapter 11 Analysis of Variance and Regression 11.1 a. The first order Taylor's series approximation is Var[g(Y )] [g ()]2 VarY = [g ()]2 v(). b. If we choose g(y) = g (y) = y a 1 v(x) dx, then a dg () d = d d 1 v(x) dx = 1 v() , by the Fundamental Theorem of Calculus. Then, for any , Var[g (Y )] dg () d 1 v() 2 v() = 1. dg () d 2 11.2 a. v() = , g (y) = y, = 1 , 2 Varg (Y ) v() = 1/4, independent of . b. To use the Taylor's series approximation, we need to express everything in terms of = EY = np. Then v() = (1 - /n) and 2 2 dg () 1 1 1 1 = = . d n 4n(1 - /n) 1- 2 n n Therefore Var[g (Y )] dg () d 2 2 v() = 1 , 4n independent of , that is, independent of p. () 1 1 c. v() = K2 , dgd = and Var[g (Y )] K2 = K, independent of . 11.3 a. g (y) is clearly continuous with the possible exception of = 0. For that value use l'H^pital's rule to get o y - 1 (log y)y lim = lim = log y. 0 0 1 b. From Exercise 11.1, we want to find v() that satisfies y -1 = Taking derivatives d dy y -1 = y -1 = d dy y a y a 1 v(x) dx. 1 v(x) dx = 1 v(y) . ...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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