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# solutionset12 - Problem Set 12 Solutions STAT 3022, section...

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Problem Set 12 Solutions STAT 3022, section 001 12/13/2011 20.9. Donner Party. For females: log (odds) = 3 . 2 - (0 . 078 × age ) For males: log (odds) = 1 . 6 - (0 . 078 × age ) (a) The estimated probabilities of survival are: For 25 year-old men: log ± ˆ π 1 - ˆ π ² = 1 . 6 - (0 . 078)(25) = - 0 . 35 ˆ π = exp( - 0 . 35) 1 + exp( - 0 . 35) = 0 . 4134 For 50 year-old men: log ± ˆ π 1 - ˆ π ² = 1 . 6 - (0 . 078)(50) = - 2 . 3 ˆ π = exp( - 2 . 3) 1 + exp( - 2 . 3) = 0 . 0911 For 25 year-old women: log ± ˆ π 1 - ˆ π ² = 3 . 2 - (0 . 078)(25) = 1 . 25 ˆ π = exp(1 . 25) 1 + exp(1 . 25) = 0 . 7773 For 50 year-old women: log ± ˆ π 1 - ˆ π ² = 3 . 2 - (0 . 078)(50) = - 0 . 7 ˆ π = exp( - 0 . 7) 1 + exp( - 0 . 7) = 0 . 3318 (b) When the estimated probability of survival is ˆ π = 0 . 5, then log ± ˆ π 1 - ˆ π ² = log ± 0 . 5 0 . 5 ² = log(1) = 0 (i) For men, the age corresponding to ˆ π = 0 . 5 is 0 = 1 . 6 - (0 . 078 × age ) age = 20 . 51 (ii) For women, the age corresponding to ˆ π = 0 . 5 is 0 = 3 . 2 - (0 . 078 × age ) age = 41 . 03

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20.10. Odds Ratio. Let ω A = the odds at A , and ω B = the odds at B . Then log ( ω A ) - log ( ω B ) = ( β 0 + β 1 A + β 2 x 2 + ... ) - ( β 0 + β 1 B + β 2 x 2 + ... ) = β 1 A - β 1 B = β 1 ( A - B ) Note, however, that the left-hand side can be written log ( ω A ) - log ( ω B ) = log ± ω A ω B ² and so taking the anti-logarithm of both sides yields ω A ω B = exp ( β 1 ( A - B )) 20.11. Space Shuttle. > shuttle <- read.csv("http://users.stat.umn.edu/~graalum/data/ch20/ex2011.csv",header=T) (a) R output given below: > mshut <- glm(FAILURE~TEMP,family=binomial,data=shuttle) > summary(mshut) Call: glm(formula = FAILURE ~ TEMP, family = binomial, data = shuttle) Deviance Residuals: Min 1Q Median 3Q Max -1.2125 -0.8253 -0.4705 0.5907 2.0512 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 10.87535 5.70291 1.907 0.0565 . TEMP -0.17132 0.08344 -2.053 0.0400 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 28.975 on 23 degrees of freedom Residual deviance: 23.030 on 22 degrees of freedom AIC: 27.030 Number of Fisher Scoring iterations: 4 The ﬁtted logistic regression model is log ± ˆ π 1 - ˆ π ² = 10 . 875 - 0 . 171( temp )
(b) Wish to test H 0 : β 1 = 0 vs. H a : β 1 < 0. The test statistic is z = ˆ β 1 SE( ˆ β 1 ) = - 0 . 171 0 . 083 = - 2 . 053 The one-sided p -value is P ( Z < - 2 . 053) = 0 . 020 We reject H 0 and conclude that the odds of O-ring failure decrease with temperature. > pnorm(-2.053)

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## This note was uploaded on 01/22/2012 for the course STAT 3022 taught by Professor Staff during the Fall '08 term at Minnesota.

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solutionset12 - Problem Set 12 Solutions STAT 3022, section...

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