Faculty of Science FINAL EXAMINATION Mathematics MATH 523 Generalized Linear Models

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1 Faculty of Science FINAL EXAMINATION Mathematics MATH 523 Generalized Linear Models Examiner: Professor K.J. Worsley Associate Examiner: Professor R. Steele Date: Thursday, April 17, 2008 Time: 14:00-17:00 hours INSTRUCTIONS: Answer all questions. Any books, notes or calculators may be brought into the exam. Computer printout and tables are provided at the end of the exam. Each part of each question is worth approximately equal marks. This exam comprises this cover, 5 pages of questions, 10 pages of computer printout, 3 pages of figures, and 4 pages of tables (23 pages in all). 1

2 MATH 523 FINAL EXAM, April 17, We wish to test for the effect of X on Y. What are the consequences, if any, on inference about the effect of X if (a) We ignore an effect of Z (i.e. we base our inference on Y X) when in fact Y is related to Z. (b) We allow for an effect of Z (i.e. we base our inference on Y Z + X) when in fact Y is not related to Z. (c) We allow for an interaction between Z and X (i.e. Y Z + X + Z : X). we base our inference on 2. The Great Plague of London ( ) was an outbreak of bubonic plague that struck London and was particularly violent during the hot months of August and September of The following table gives the number of deaths during nine consecutive weeks starting with the week of September 4th, The variables were: week: consecutive weeks starting with the week of September 4th, 1665 all: number of deaths due to all diseases plague: number of plague deaths week all plague We shall assume that the number of plague deaths (plague) is a binomial random variable out of the number of deaths due to all diseases (all). A logistic regression was fitted to a polynomial in the number of weeks (week = 1,..., 9). (a) Estimate the degree of the polynomial by carrying out appropriate tests. (b) Is there any evidence for a polynomial effect beyond degree 3?

3 MATH 523 FINAL EXAM, April 17, (c) For the quadratic model, draw a rough sketch of the fitted proportion of plague deaths against time for say 20 weeks before and after the nine weeks considered here. (d) Using the quadratic model, estimate the time i. when the proportion of plague deaths was a maximum, ii. when the proportion of plague deaths would equal 1%. 3. The data set hillrace.dat gives data on record times in Scottish hill races. The variables are: dist: distance in miles height: total height gained during the route, in feet time: record time in minutes dist height time Greenmantle Carnethy Craig Dunain Ben Rha Ben Lomond Goatfell Bens of Jura Cairnpapple Scolty Traprain Lairig Ghru Dollar Lomonds Cairn Table Eildon Two Cairngorm Seven Hills Knock Hill dist height time Black Hill Creag Beag Kildcon Hill Meall Ant-Suidhe Half Ben Nevis Cow Hill N Berwick Law Creag Dubh Burnswark Largo Law Criffel Acmony Ben Nevis Knockfarrel Two Breweries Cockleroi Moffat Chase

4 MATH 523 FINAL EXAM, April 17, We shall assume that time (time) has a normal distribution with an identity link to a linear model in distance (dist) and height (height). (a) From Figure 2.3 and the printout of the residuals it appears that observation 18 (Knock Hill race) is an outlier. Two models were fitted: with all the data, and all the data except observation 18. i. Explain how you can test that observation 18 is an outlier using the summaries of these two models. ii. Carry out the test, making a Bonferroni correction by using the small listing of critical values that follow. (b) After omitting observation 18, test that the effect of distance on running time depends on the height. Does increased height make the distance effect larger or smaller? (c) Check the final model for i. equal variance; ii. outliers; iii. normality. 4. The following table gives remission times of 43 leukemia patients, from Gehan, E.A. (1965) A generalized Wilcoxon test for comparing arbitrarily single-censored samples. Biometrika 52: Some were treated with the drug 6-mercaptopurine (6-MP), known as Purinethol, and the rest are controls. The trial was designed as matched pairs, both withdrawn from the trial when either came out of remission. The variables were: pair: label for pair censoring: censoring, 0/1 time: remission time in weeks

5 MATH 523 FINAL EXAM, April 17, Control 6-MP pair time censoring time censoring We shall assume that remission time (time) has an exponential distribution with a log link to a linear model in treatment (treatment; 1=control, 2=6-MP). (a) Test that the drug 6-MP has an effect on the remission time, i. ignoring the pairs, ii. allowing for the pairs. (b) Several authors have ignored the pairs. Were the pairs really different? (c) Using the first model, estimate the remission time for a leukemia patient who is given the drug 6-MP, and the probability that the remission time will be longer than 26 weeks. 5. Self-reported data on ear infections in 287 swimmers between 15 and 29 years old was collected in New South Wales (Australia). The variables are: frq: Frequent Ocean Swimmer: 1 = occasional swimmer, 2 = frequent swimmer loc: location: 1 = beach swimmer, 4 = non-beach swimmer Age: 2 (aged 15-19), 3 (aged 20-25), or 4 (aged 25-29) sex 1 (male) or 2 (female) inf = number of ear infections

6 MATH 523 FINAL EXAM, April 17, We shall assume that the number of ear infections (inf) is a Poisson random variable with a log link to a linear model. (a) Is there any evidence for an effect of Frequent Ocean Swimmer (frq; 1=no, 2=yes), location (loc), age as a factor (Age) or sex (sex; 1=male, 2=female) on number of ear infections (inf)? (b) Is the goodness of fit test valid for the model in (a)? If so, do it; if not, say why not. (c) It appears that the model in (a) does not fit, and that there is extra-poisson variation. To account for this, we shall assume that Var(inf) = φµ, where E(inf) = µ and φ is unknown. Estimate φ, and using this estimate, repeat (a). (d) From now on we shall assume that there is extra-poisson variation and φ is unknown. Is the effect of age linear or non-linear? Test this. (e) Is there a linear effect of age? Test this. (f) Does the effect of location depend on whether the person is a frequent ocean swimmer or not? Test this. (g) Another way of allowing for extra-poisson variation is to assume that the number of ear infections (inf) is a Poisson random variable with a mean that is itself an exponential random variable. It can then be shown that the marginal distribution of inf1 = inf + 1 is geometric. The likelihood of inf1 can be calculated by assuming that 1 has a binomial distribution with inf1 trials. Show that if there is a log link from θ = E(inf) to a linear model, then there is a logistic link from the binomial probability to the negative of the linear model. (h) All the above models were re-fitted assuming the geometric distribution. Explain why all the parameter estimates are roughly the same, but their standard deviations are different. (i) Explain why the parameter estimates for the model frq+loc+frq:loc are exactly the same. (j) Choose an appropriate model to estimate the expected number of ear infections for a 21-year old female who is a frequent ocean swimmer, and who usually swims in a non-beach location (such as a coral reef).

7 MATH 523 FINAL EXAM, April 17, R version ( ) Copyright (C) 2007 The R Foundation for Statistical Computing > ################ Question 2 ################## > > m<-matrix(scan("c:/keith/teaching/smalldatasets/plague.dat"),10,2,t) Read 18 items > all<-m[:,1] > plague<-m[:,2] > p<-plague/all > week<-1:9 > postscript("c:/keith/teaching/523/2008/fig1.eps",horizontal=false) > par(mfrow=c(2,2)) > plot(week,p,main="fig 2.1 Plague mortality") > glm1<-glm(p~week,family=binomial,weight=all) > summary(glm1) glm(formula = p ~ week, family = binomial, weights = all) (Intercept) <2e-16 *** week <2e-16 *** (Dispersion parameter for binomial family taken to be 1) Null deviance: on 8 degrees of freedom Residual deviance: on 7 degrees of freedom AIC: Number of Fisher Scoring iterations: 4 > week2<-week^2 > glm2<-glm(p~week+week2,family=binomial,weight=all) > summary(glm2) glm(formula = p ~ week + week2, family = binomial, weights = all) (Intercept) <2e-16 *** week <2e-16 *** week <2e-16 *** (Dispersion parameter for binomial family taken to be 1)

8 MATH 523 FINAL EXAM, April 17, Null deviance: on 8 degrees of freedom Residual deviance: on 6 degrees of freedom AIC: Number of Fisher Scoring iterations: 3 > week3<-week^3 > glm3<-glm(p~week+week2+week3,family=binomial,weight=all) > summary(glm3) glm(formula = p ~ week + week2 + week3, family = binomial, weights = all) (Intercept) < 2e-16 *** week e-12 *** week ** week (Dispersion parameter for binomial family taken to be 1) Null deviance: on 8 degrees of freedom Residual deviance: on 5 degrees of freedom AIC: 103 Number of Fisher Scoring iterations: 3 > lines(week,fitted(glm2)) > > ################ Question 3 ################## > > m<-matrix(scan("c:/keith/teaching/smalldatasets/hillrace.dat"),35,3,t) Read 105 items > dist<-m[,1] > height<-m[,2] > time<-m[,3] > plot(dist,time,main="fig 3.1: Time vs dist") > plot(height,time,main="fig 3.2: Time vs height") > glm1<-glm(time~dist+height) > summary(glm1) glm(formula = time ~ dist + height)

9 MATH 523 FINAL EXAM, April 17, Estimate Std. Error t value Pr(> t ) (Intercept) * dist e-12 *** height e-06 *** (Dispersion parameter for gaussian family taken to be ) Null deviance: on 34 degrees of freedom Residual deviance: on 32 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 > plot(fitted(glm1),resid(glm1),main="fig 3.3: Residuals") > abline(0,0) > dev.off() windows 2 > cbind(time,fitted(glm1),resid(glm1)) time > time0<-time[-18] > dist0<-dist[-18]

10 MATH 523 FINAL EXAM, April 17, > height0<-height[-18] > glm2<-glm(time0~dist0+height0) > summary(glm2) glm(formula = time0 ~ dist0 + height0) Estimate Std. Error t value Pr(> t ) (Intercept) e-05 *** dist < 2e-16 *** height e-11 *** (Dispersion parameter for gaussian family taken to be ) Null deviance: on 33 degrees of freedom Residual deviance: on 31 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 > postscript("c:/keith/teaching/523/2008/fig2.eps",horizontal=false) > par(mfrow=c(2,2)) > plot(fitted(glm2),resid(glm2),main="fig 3.4: Omitting obs 18") > abline(0,0) > glm3<-glm(time0~dist0+height0+height0:dist0) > summary(glm3) glm(formula = time0 ~ dist0 + height0 + height0:dist0) Estimate Std. Error t value Pr(> t ) (Intercept) dist e-11 *** height dist0:height ** (Dispersion parameter for gaussian family taken to be ) Null deviance: on 33 degrees of freedom Residual deviance: on 30 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 > plot(fitted(glm3),resid(glm3),main="fig 3.5: With interaction")

11 MATH 523 FINAL EXAM, April 17, > abline(0,0) > > vl<-predict.lm(glm3,se.fit=t)$se.fit^2 > sc<-summary(glm3)$dispersion > z<-resid(glm3)/sqrt(sc-vl) > z > hist(z,main="figure 3.6 Normalized residuals, z") > u<-pnorm(z) > hist(u,main="figure 3.7 Uniforms, u") > dev.off() windows 2 > uhat<-((1:34)-0.5)/34 > u<-sort(u) > postscript("c:/keith/teaching/523/2008/fig3.eps",horizontal=false) > par(mfrow=c(2,2)) > plot(uhat,u,xaxs="i",xlim=c(0,1),yaxs="i", + ylim=c(0,1),main="figure 3.8 P-P plot") > abline(0,1) > zhat<-qnorm(uhat) > z<-sort(z) > plot(zhat,z,main="figure 3.9 Q-Q plot") > abline(0,1) > u-uhat > > ################ Question 4 ################## > > m<-matrix(scan("c:/keith/teaching/smalldatasets/remiss.dat"),42,3,t) Read 126 items > pair<-factor(m[,1]) > time<-m[,2] > censoring<-m[,3] > treatment<-factor(c(rep(c(1,2,2,1),10),c(1,2))) > rate<-censoring/time > summary(glm(rate~treatment,family=poisson,weight=time))

12 MATH 523 FINAL EXAM, April 17, glm(formula = rate ~ treatment, family = poisson, weights = time) (Intercept) < 2e-16 *** treatment *** (Dispersion parameter for poisson family taken to be 1) Null deviance: on 41 degrees of freedom Residual deviance: on 40 degrees of freedom AIC: Inf Number of Fisher Scoring iterations: 6 > summary(glm(rate~treatment+pair,family=poisson,weight=time)) glm(formula = rate ~ treatment + pair, family = poisson, weights = time) (Intercept) treatment e-05 *** pair * pair pair pair pair pair pair * pair pair * pair pair pair pair pair pair * pair pair pair pair pair (Dispersion parameter for poisson family taken to be 1)

13 MATH 523 FINAL EXAM, April 17, Null deviance: on 41 degrees of freedom Residual deviance: on 20 degrees of freedom AIC: Inf Number of Fisher Scoring iterations: 6 > > ################ Question 5 ################## > > m<-matrix(scan("c:/keith/teaching/smalldatasets/ear.dat"),287,5,t) Read 1435 items > frq<-factor(m[,1]) > loc<-factor(m[,2]) > Age<-factor(m[,3]) > sex<-factor(m[,4]) > inf<-m[,5] > summary(glm(inf~frq+loc+age+sex,family=poisson)) glm(formula = inf ~ frq + loc + Age + sex, family = poisson) (Intercept) < 2e-16 *** frq e-09 *** loc e-07 *** Age ** Age sex (Dispersion parameter for poisson family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 281 degrees of freedom AIC: Number of Fisher Scoring iterations: 6 > hist(fitted(glm(inf~frq+loc+age+sex,family=poisson)),main="figure 5.1 Fitted values") > dev.off() windows 2 > age<-m[,3] > summary(glm(inf~frq+loc+age+sex,family=poisson)) glm(formula = inf ~ frq + loc + age + sex, family = poisson)

14 MATH 523 FINAL EXAM, April 17, (Intercept) e-10 *** frq e-09 *** loc e-06 *** age * sex (Dispersion parameter for poisson family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 282 degrees of freedom AIC: Number of Fisher Scoring iterations: 6 > summary(glm(inf~frq+loc+frq:loc,family=poisson)) glm(formula = inf ~ frq + loc + frq:loc, family = poisson) (Intercept) < 2e-16 *** frq e-07 *** loc e-06 *** frq2:loc (Dispersion parameter for poisson family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 283 degrees of freedom AIC: Number of Fisher Scoring iterations: 6 > summary(glm(inf~frq+loc,family=poisson)) glm(formula = inf ~ frq + loc, family = poisson) (Intercept) < 2e-16 *** frq e-09 *** loc e-07 ***

15 MATH 523 FINAL EXAM, April 17, (Dispersion parameter for poisson family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 284 degrees of freedom AIC: Number of Fisher Scoring iterations: 6 > inf1<-inf+1 > y<-1/inf1 > summary(glm(y~frq+loc+age+sex,family=binomial,weight=inf1)) glm(formula = y ~ frq + loc + Age + sex, family = binomial, weights = inf1) (Intercept) e-10 *** frq *** loc ** Age * Age sex (Dispersion parameter for binomial family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 281 degrees of freedom AIC: Number of Fisher Scoring iterations: 4 > summary(glm(y~frq+loc+age+sex,family=binomial,weight=inf1)) glm(formula = y ~ frq + loc + age + sex, family = binomial, weights = inf1) (Intercept) e-05 *** frq *** loc ** age sex (Dispersion parameter for binomial family taken to be 1)

16 MATH 523 FINAL EXAM, April 17, Null deviance: on 286 degrees of freedom Residual deviance: on 282 degrees of freedom AIC: Number of Fisher Scoring iterations: 4 > summary(glm(y~frq+loc+frq:loc,family=binomial,weight=inf1)) glm(formula = y ~ frq + loc + frq:loc, family = binomial, weights = inf1) (Intercept) e-09 *** frq *** loc ** frq2:loc (Dispersion parameter for binomial family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 283 degrees of freedom AIC: Number of Fisher Scoring iterations: 4 > summary(glm(y~frq+loc,family=binomial,weight=inf1)) glm(formula = y ~ frq + loc, family = binomial, weights = inf1) (Intercept) e-10 *** frq *** loc ** (Dispersion parameter for binomial family taken to be 1) Null deviance: on 286 degrees of freedom Residual deviance: on 284 degrees of freedom AIC: Number of Fisher Scoring iterations: 4

17 MATH 523 FINAL EXAM, April 17, Fig 2.1 Plague mortality Fig 3.1: Time vs dist p time week dist Fig 3.2: Time vs height Fig 3.3: Residuals time resid(glm1) height fitted(glm1)

18 MATH 523 FINAL EXAM, April 17, Fig 3.4: Omitting obs 18 Fig 3.5: With interaction resid(glm2) resid(glm3) fitted(glm2) fitted(glm3) Figure 3.6 Normalized residuals, z Figure 3.7 Uniforms, u Frequency Frequency z u

19 MATH 523 FINAL EXAM, April 17, Figure 3.8 P P plot Figure 3.9 Q Q plot u z uhat zhat Figure 5.1 Fitted values Frequency fitted(glm(inf ~ frq + loc + Age + sex, family = poisson))

20 MATH 523 FINAL EXAM, April 17, Upper tail probabilities of the standard Normal distribution z

21 MATH 523 FINAL EXAM, April 17, Quantiles of the t distribution Quantiles of the χ 2 distribution Degrees of Upper tail probability Upper tail probability freedom

22 MATH 523 FINAL EXAM, April 17, Quantiles of the F distribution, P = 0.05 Denominator degrees of Numerator degrees of freedom freedom

23 MATH 523 FINAL EXAM, April 17, Quantiles of the Bonferroni t distribution, P (T > t) = 0.025/n Degrees of freedom n Quantiles of the Bonferroni t distribution, P (T > t) = 0.005/n Degrees of freedom n End of exam