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Unformatted text preview: Stat 511 Homework 12 Solution Spring 2011 Due: 5pm, Tuesday Apr 26 1. If y Poisson ( ), the pmf of y is f ( y ) = y e y ! Equation (1) on slide 2 of section 26 gives the exponentionalscale family form of a distribution. Write the pmf of y in exponentialscale form and identify , ( ), T ( y ), a ( ), and b ( ). f ( y ) = y e y ! = exp ( y log   log y !) Namely,we have = , ( ) = log( ) , T ( y ) = y, a ( ) = 1 , b ( ) = 2. The data in birth2.txt are from a sociological study in Baltimore. The investigators believed that there is an association between loss of a child during pregnancy and the behaviour in school of a subsequent liveborn child. The prospective research design (looking forward) would be to enroll some mothers who have lost a child and some who havent, then look at school behaviour (problem or not). This is very inefficient because the proportion of school problems is low (thankfully!). Instead, the investigators used a retrospective design, which is common in epidemiology and efficient at examining associations with rare events. The investigators identified 255 problem children and 110 control (nonproblem) children. The mothers were then asked about the birth order (2nd child, 3rd child, ...) and whether they had lost the previous child. Birth order is 2 (2nd child), 3.5 (3rd or 4th child) and 5 (5th or higher child). The response is whether the previous child was lost or not. (The loss rates seem high to me, but thats what they were recorded as). The investigators want to know whether the odds of losing a child are associated with birth order and/or whether or not the subsequent child was a problem in school. (a) Fit an appropriate generalized linear model treating problem as a factor and birth order as a continuous linear covariate. Do not worry about overdispersion for now. Report the odds ratio for being a problem child and the odds ratio for a 1 child increase in birth order. Does being a problem child increase or decrease the probability that your mom lost the previous child? d < read.table(D:/STAT511/data/birth2.txt,as.is=T,header=T) > d$problem <as.factor(d$problem) > d$f < cbind(d$loss,d$noloss) > d.glm < glm(f~problem+birth,data=d,family=binomial) > summary(d.glm) Estimate Std. Error z value Pr(>z) 1 Stat 511 Homework 12 Solution Spring 2011 (Intercept) 2.63005 0.394596.665 2.64e11 *** problemY 0.36724 0.24394 1.505 0.132 birth 0.48683 0.09764 4.986 6.17e07 *** The odds ratio for being a problem child is exp(0.36724)=1.4437 and for a 1 child increase in birth order is exp(0.48683)=1.6272. Being a problem child increases the probability that the Mom lost the previous child by about 44%....
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This note was uploaded on 02/11/2012 for the course STAT 511 taught by Professor Staff during the Spring '08 term at Iowa State.
 Spring '08
 Staff

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