practice3soln

# practice3soln - Key to Practice Problem 3 1. a) = exp(- 3 ....

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Unformatted text preview: Key to Practice Problem 3 1. a) = exp(- 3 . 7771 + 0 . 1449 x ) 1 + exp(- 3 . 7771 + 0 . 1449 x ) where x = LI . b) 1- = exp(- 3 . 7771 + 0 . 1449 15) = 0 . 2011 , so the odds for remission vs. not remission is 0.2011. c) = odds 1 + odds = . 2011 1 + 0 . 2011 = . 01674 Probability of remission is less than probability of no remission for LI = 15. c) exp(0 . 1449) = 1 . 1559. d) odds at x = 15 is 0.2011 odds at x = 20 is 0 . 2011 1 . 1559 5 = 0 . 415 . e) CI for is 0 . 1449 1 . 96(0 . 0593) = (0 . 0287 , . 2611), CI for exp( ) = (1 . 0291 , 1 . 2983) . f) Likelihood ratio chi-square (- 2log L ) nothing- (- 2log L ) full = 34 . 372- 26 . 073 = 8 . 299, Wald chi-square z = 0 . 1449 / . 0593 = 2 . 4435. so 2 = z 2 = 5 . 97 . g) = 0 . 08879 and CI=(0.01809,0.3401). 2. a) logit ( ) =- 3 . 596- . 8678 d +2 . 4044 v, where d = 0 , 1 for black and white, respectively; v = 0 , 1 for black and white, respectively....
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## This note was uploaded on 07/08/2011 for the course STA 6127 taught by Professor Mukherjee during the Fall '08 term at University of Florida.

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practice3soln - Key to Practice Problem 3 1. a) = exp(- 3 ....

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