Ovarian cancer example: h(t rx, age) = h(t) exp(β1 rx + β2 age) > cph1 = coxph(surv(futime, fustat)~rx+age, ovarian) > summary(cph1)

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1 Ovarian cancer example: h(t rx, age) = h(t) exp(β1 rx + β2 age) > cph1 = coxph(surv(futime, fustat)~rx+age, ovarian) > summary(cph1) coxph(formula = Surv(futime, fustat) ~ rx + age, data = ovarian) rx age rx age Rsquare= (max possible= ) Likelihood ratio test= 15.9 on 2 df, p= Wald test = 13.5 on 2 df, p= Score (logrank) test = 18.6 on 2 df, p=9.34e-05 The coxph() function gives you the hazard ratio for a one unit change in the predictor as well as the 95% confidence interval. Also given is the Wald statistic for each parameter as well as overall likelihood ratio, wald and score tests. What if we wanted to estimate hr (rx = 1, age = 50 : rx = 2, age = 60)? The point estimate is easily obtained as, exp{ 0.804(1 2) (50 60)} = We can get the variance-covariance matrix for the parameter estimates from the fitted object. > cph1$var [,1] [,2] [1,] [2,] > sqrt(diag(cph1$var)) [1]

2 The 95% CI for hr (rx = 1, age = 50 : rx = 2, age = 60) is, = exp ( ± 1.96 (0.706)) = (0.099, 2.67) Partial likelihood ratio tests What if we want to do a likelihood ratio test for H0 : β2 = 0 in the previous model? Reduced model: >cph1r = coxph(surv(futime, fustat)~rx, ovarian) > summary(cph1r) coxph(formula = Surv(futime, fustat) ~ rx, data = ovarian) rx rx Rsquare= 0.04 (max possible= ) Likelihood ratio test= 1.05 on 1 df, p=0.305 Wald test = 1.03 on 1 df, p=0.310 Score (logrank) test = 1.06 on 1 df, p=0.303

3 > cph1$loglik [1] > cph1r$loglik [1] log L (full model) = log L (reduced model) = X 2 = 2 log L (reduced model) ( 2 log L (full model)) = = > χ1 2 = 3.84 Stratification example Suppose in the previous ovarian cancer example, we stratify by age instead of including it as a predictor in the model. We will create two strata, women over the age of 60 (n=7) and women younger than 60 (n=19). There are 6 events in each group. We fit the following regression model, h(t rx, agegrp) = h agegrp(t) exp(rx β), agegrp = 1, 2. > cph2=coxph(surv(futime, fustat)~rx+strata(age>60), ovarian) > summary(cph2) coxph(formula = Surv(futime, fustat) ~ rx + strata(age > 60), data = ovarian) rx rx Rsquare= 0.02 (max possible= ) Likelihood ratio test= 0.52 on 1 df, p=0.471 Wald test = 0.51 on 1 df, p=0.474 Score (logrank) test = 0.52 on 1 df, p=0.471

4 How does this compare to an additive model with agegrp as a predictor? > cph3=coxph(surv(futime, fustat)~rx+(age>60), ovarian) > summary(cph3) coxph(formula = Surv(futime, fustat) ~ rx + (age > 60), data = ovarian) rx age > 60TRUE rx age > 60TRUE Rsquare= (max possible= ) Likelihood ratio test= 11.8 on 2 df, p= Wald test = 12.2 on 2 df, p= Score (logrank) test = 17.3 on 2 df, p= What about an interaction? > cph4 = coxph(surv(futime, fustat)~rx*(age>60), ovarian) > summary(cph4) coxph(formula = Surv(futime, fustat) ~ rx * (age > 60), data = ovarian) rx age > 60TRUE rx:age > 60TRUE rx age > 60TRUE rx:age > 60TRUE Rsquare= (max possible= ) Likelihood ratio test= 11.8 on 3 df, p= Wald test = 12.1 on 3 df, p= Score (logrank) test = 17.5 on 3 df, p=

5 8.4 Three different treatments were administered to rats who had F98 glioma cells implanted into their brains (Data from Chapter 7 (Exercise 7.7)). The data for the three groups of rats lists the death times (in days) in that exercise. Create two dummy variables, Z1 = 1 if animal is in the radiation only group, 0 otherwise; Z2 = 1 if animal is in the radiation plus BPA group, 0 otherwise. Use the Breslow method of handling ties in the problems below. (a) Estimate β1 and β2 and their respective standard errors. Find a 95% confidence interval for the relative risk of death of an animal radiated only compared to an untreated animal. (b) Test the global hypothesis of no effect of either radiation or radiation plus BPA on survival. Perform the test using all the three tests(wald, likelihood ratio, and score test). (c) Test the hypothesis that the effect a radiated only animal has on survival is the same as the effect of radiation plus BPA (i.e., Test H0: β1 = β2). (d) Find an estimate and a 95% confidence interval for the relative risk of death for a radiation plus BPA animal as compared to a radiated only animal. (e) Test the hypothesis that any radiation given as a treatment (either radiation alone or with BPA) has a different effect on survival than no radiation. Use the likelihood ratio test. (f) Repeat part (e) using a Wald test.

6 Solution: > Ratdata <- read.table(file="ratdata.txt",col.names=c("trtyp","time","dth")) Here, 1 = untreated, 2 = rad, 3 = rad + BPA, actually same data as BNCT >ratcox<coxph(surv(time,dth)~i(trtyp==2)+i(trtyp==3),ratdata,method="breslow") > summary(ratcox) coxph(formula = Surv(time, Dth) ~ I(TrTyp == 2) + I(TrTyp == 3), data = Ratdata, method = "breslow") n= 30 I(TrTyp == 2)TRUE e-03 I(TrTyp == 3)TRUE e-06 I(TrTyp == 2)TRUE I(TrTyp == 3)TRUE Rsquare= (max possible= ) Likelihood ratio test= 27.4 on 2 df, p=1.14e-06 Wald test = 22.4 on 2 df, p=1.34e-05 Score (logrank) test = 31.7 on 2 df, p=1.28e-07 Part (a) > round(c(ratcox$coef, se=sqrt(diag(ratcox$var))),4) I(TrTyp == 2)TRUE I(TrTyp == 3)TRUE se1 se > round(exp(2*(ratcox$coef[1]+c(-1,1)*1.96*sqrt(ratcox$var[1,1]))),4) [1] Part (b) Here, 2df χ 2 test of β1 = β2 = 0 From summary(ratcox): Likelihood ratio test= 27.4 on 2 df, p=1.14e-06 Wald test = 22.4 on 2 df, p=1.34e-05 Score (logrank) test = 31.7 on 2 df, p=1.28e-07

7 Part (c) (d) We want standard error for (β1 hat - β2 hat)*2 = (2,-2) %*% β hat. >c((ratcox$coef %*%c(-1,1) )/sqrt(c(-1,1) %*% ratcox$var %*% c(-1,1))) [1] # highly significant difference: pval > 2*(1- pnorm(2.7588)) [1] > c(estrr = exp(ratcox$coef %*% c(-1,1)), CI=exp(2*(ratcox$coef %*% c(-1,1) + c(-1,1)*1.96*sqrt(c(-1,1) %*% ratcox$var %*% c(-1,1))))) EstRR CI1 CI Part (e) - (f) If we fit using Cox model combining both radiation groups, then >ratcox1<-coxph(surv(time,dth)~ I(TrTyp>1), Ratdata, method="breslow") > summary(ratcox1) coxph(formula = Surv(time, Dth) ~ I(TrTyp > 1), data = Ratdata, method = "breslow") n= 30 I(TrTyp > 1)TRUE e-05 I(TrTyp > 1)TRUE Rsquare= (max possible= ) Likelihood ratio test= 18.9 on 1 df, p=1.36e-05 Wald test = 18.9 on 1 df, p=1.41e-05 Score (logrank) test = 26.2 on 1 df, p=3.12e-07 > ratcoxalt <- coxph(surv(time,dth) ~ I(TrTyp>1), Ratdata, robust=t, method="breslow")

8 > summary(ratcoxalt) coxph(formula = Surv(time, Dth) ~ I(TrTyp > 1), data = Ratdata, method = "breslow", robust = T) n= 30 coef exp(coef) se(coef) robust se z p I(TrTyp > 1)TRUE e-06 I(TrTyp > 1)TRUE Rsquare= (max possible= ) Likelihood ratio test= 18.9 on 1 df, p=1.36e-05 Wald test = 22.6 on 1 df, p=1.97e-06 Score (logrank) test = 26.2 on 1 df, p=3.12e-07, Robust = 14.9 p= (Note: the likelihood ratio and score tests assume independence of observations within a cluster, the Wald and robust score tests do not). > sqrt(c(naive.var=ratcoxalt$naive, Robust.var=ratcoxalt$var)) Naive.var Robust.var Part (e) It seems to ask for test of β1 = 0 OR β2 = 0. We can do it using a Cox-model fit with the combined single covariate = Z1 + Z2.

9 > plot(survfit(ratcox)) > plot(survfit(ratcox1)) > coxhaz <- basehaz(ratcox, FALSE) > coxhaz hazard time Fig 1.1(ratcox) Fig 1.2(ratcox1) Model fitting syntax above is for non time-dependent covariates in Cox model. Timedependent covariate Cox model is also available, but with slightly different data structure: data-frame or list no longer has time and status columns, but instead start, stop, event. This is a counting-process data-structure introduced by Andersen and Gill (1982). (a,b] defines as start = a and stop = b. Syntax of model-fitting becomes: > coxph(surv(start,stop,event) ~ x + strata(reg),data=dfram)