Dr. Hackney STA Solutions pg 173

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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) . ...
View Full Document

Ask a homework question - tutors are online