1. Lecture 7 GLMs II Binomial Family Olivier MISSA, om502@york.ac.ukom502@york.ac.uk Advanced Research Skills

2. 2 Outline Continue our Introduction to Generalized Linear Models. In this lecture: Illustrate the use of GLMs for proportion and binary data.

3. 3 Binary & Proportion data tend to follow the Binomial distribution The Canonical link of this glm family is the logit function: The variance reaches a maximum for intermediate values of p and a minimum at either 0% or 100%. Reminder

4. 4 In R, binary/proportion data can be entered into a model as a response in three different ways: as a numeric vector (holding the number or proportion of successes) as a logical vector or a factor (TRUE or the first factor level will be considered successes). as a two-column matrix (the first column holding the number of successes and the second column the number of failures). Three ways to work with binary data

5. 5 Toxicity to tobacco budworm (moth) of different doses of trans-cypermethrin. Batches of 20 moths (of each sex) were put in contact for three days with increasing doses of the pyrethroid. 1 st Example Dose (micrograms) Sex 1 2 4 8 16 32 Male 1 4 9 13 18 20 Female 0 2 6 10 12 16 Number of dead moths out of 20 tested > (dose <- rep(2^(0:5), 2)) [1] 1 2 4 8 16 32 1 2 4 8 16 32 > numdead <- c(1,4,9,13,18,20,0,2,6,10,12,16) > (sex <- factor( rep( c("M","F"), c(6,6) ) )) [1] M M M M M M F F F F F F Levels: F M > SF <- cbind(numdead, numalive=20-numdead)

6. 6 1 st Example > modb <- glm(SF ~ sex*dose, family=binomial) > summary(modb) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.71578 0.32233 -5.323 1.02e-07 *** sexM -0.21194 0.51523 -0.411 0.68082 dose 0.11568 0.02379 4.863 1.16e-06 *** sexM:dose 0.18156 0.06692 2.713 0.00666 ** --- (Dispersion parameter for binomial family taken to be 1) Null deviance: 124.876 on 11 degrees of freedom Residual deviance: 18.164 on 8 degrees of freedom AIC: 56.275 What is modelled is the proportion of successes

7. 7 1 st Example > ldose <- log2(dose) > modb2 <- glm(SF ~ sex*ldose, family=binomial) > summary(modb2) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.9935 0.5527 -5.416 6.09e-08 *** sexM 0.1750 0.7783 0.225 0.822 ldose 0.9060 0.1671 5.422 5.89e-08 *** sexM:ldose 0.3529 0.2700 1.307 0.191 --- (Dispersion parameter for binomial family taken to be 1) Null deviance: 124.8756 on 11 degrees of freedom Residual deviance: 4.9937 on 8 degrees of freedom AIC: 43.104

8. 8 1 st Example > drop1(modb2, test="Chisq") Single term deletions Model: SF ~ sex * ldose Df Deviance AIC LRT Pr(Chi) 4.994 43.104 sex:ldose 1 6.757 42.867 1.763 0.1842 > modb3 <- update(modb2, ~. – sex:ldose) > summary(modb3) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.4732 0.4685 -7.413 1.23e-13 *** sexM 1.1007 0.3558 3.093 0.00198 ** ldose 1.0642 0.1311 8.119 4.70e-16 *** --- (Dispersion parameter for binomial family taken to be 1) Null deviance: 124.876 on 11 degrees of freedom Residual deviance: 6.757 on 9 degrees of freedom AIC: 42.867

9. 9 1 st Example > drop1(modb3, test="Chisq") Single term deletions Model: SF ~ sex + ldose Df Deviance AIC LRT Pr(Chi) 6.757 42.867 sex 1 16.984 51.094 10.227 0.001384 ** ldose 1 118.799 152.909 112.042 < 2.2e-16 *** > shapiro.test(residuals(modb3), type="deviance") Shapiro-Wilk normality test data: residuals(modb3, type = "deviance") W = 0.9666, p-value = 0.8725

10. 10 1 st Example > par(mfrow=c(2,2)) > plot(modb3)

11. 11 1 st Example > plot( c(0,1) ~ c(1,32), type="n", log="x", xlab="dose", ylab="Probability") > text(dose, numdead/20, labels=as.character(sex) ) > ld <- seq(0,32,0.5) > lines (ld, predict(modb3, data.frame(ldose=log2(ld), sex=factor(rep("M", length(ld)), levels=levels(sex))), type="response") ) > lines (ld, predict(modb3, data.frame(ldose=log2(ld), sex=factor(rep("F", length(ld)), levels=levels(sex))), type="response"), lty=2, col="red" )

12. 12 1 st Example > modbp <- glm(SF ~ sex*ldose, family=binomial(link="probit")) > AIC(modbp) [1] 41.87836 > modbc <- glm(SF ~ sex*ldose, family=binomial(link="cloglog")) > AIC(modbc) [1] 43.8663 > AIC(modb3) [1] 42.86747

13. 13 > summary(modb3) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.4732 0.4685 -7.413 1.23e-13 *** sexM 1.1007 0.3558 3.093 0.00198 ** ldose 1.0642 0.1311 8.119 4.70e-16 *** --- > exp(modb3$coeff) ## careful it may be misleading (Intercept) sexM ldose 0.031019 3.006400 2.898560 ## odds ration: p / (1-p) > exp(modb3$coeff[1]+modb3$coeff[2]) ## odds for males (Intercept) 0.09325553 1 st Example logit scale Every doubling of the dose will lead to an increase in the odds of dying over surviving by a factor of 2.899

14. 14 Erythrocyte Sedimentation Rate in a group of patients. Two groups : 20 (ill) mm/hour Q: Is it related to globulin & fibrinogen level in the blood ? 2 nd Example > data("plasma", package="HSAUR") > str(plasma) 'data.frame': 32 obs. of 3 variables: $ fibrinogen: num 2.52 2.56 2.19 2.18 3.41 2.46 3.22 2.21... $ globulin : int 38 31 33 31 37 36 38 37 39 41... $ ESR : Factor w/ 2 levels "ESR 20": 1 1... > summary(plasma) fibrinogen globulin ESR Min. :2.090 Min. :28.00 ESR < 20:26 1st Qu.:2.290 1st Qu.:31.75 ESR > 20: 6 Median :2.600 Median :36.00 Mean :2.789 Mean :35.66 3rd Qu.:3.167 3rd Qu.:38.00 Max. :5.060 Max. :46.00

15. 15 2 nd Example > stripchart(globulin ~ ESR, vertical=T, data=plasma, xlab="Erythrocyte Sedimentation Rate (mm/hr)", ylab="Globulin blood level", method="jitter" ) > stripchart(fibrinogen ~ ESR, vertical=T, data=plasma, xlab="Erythrocyte Sedimentation Rate (mm/hr)", ylab="Fibrinogen blood level", method="jitter" )

16. 16 2 nd Example > mod1 <- glm(ESR~fibrinogen, data=plasma, family=binomial) > summary(mod1) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -6.8451 2.7703 -2.471 0.0135 * fibrinogen 1.8271 0.9009 2.028 0.0425 * --- (Dispersion parameter for binomial family taken to be 1) Null deviance: 30.885 on 31 degrees of freedom Residual deviance: 24.840 on 30 degrees of freedom AIC: 28.840 > mod2 <- glm(ESR~fibrinogen+globulin, data=plasma, family=binomial) > AIC(mod2) [1] 28.97111 factor

17. 17 2 nd Example > anova(mod1, mod2, test="Chisq") Analysis of Deviance Table Model 1: ESR ~ fibrinogen Model 2: ESR ~ fibrinogen + globulin Resid. Df Resid. Dev Df Deviance P(>|Chi|) 1 30 24.8404 2 29 22.9711 1 1.8692 0.1716 > summary(mod2) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -12.7921 5.7963 -2.207 0.0273 * fibrinogen 1.9104 0.9710 1.967 0.0491 * globulin 0.1558 0.1195 1.303 0.1925 --- (Dispersion parameter for binomial family taken to be 1) Null deviance: 30.885 on 31 degrees of freedom Residual deviance: 22.971 on 29 degrees of freedom AIC: 28.971 The difference in terms of Deviance between these models is not significant, which leads us to select the least complex model

18. 18 2 nd Example > shapiro.test(residuals(mod1, type="deviance")) Shapiro-Wilk normality test data: residuals(mod1, type = "deviance") W = 0.6863, p-value = 5.465e-07 > par(mfrow=c(2,2)) > plot(mod1)