Applied Statistics: Lent Term

Transcription

1 Applied Statistics: Lent Term Example: Survival data analysis Right Censored Survival Data The data are from a clinical trial investigating the effect of the drug 6-mercaptopurine (6-MP) on the remission times (in weeks) of leukaemia patients. There are two groups of patients in this study: those who were randomised to receive 6-MP and those who were randomised to receive the placebo. These data can be found in the file Leukaem, which is a comma separated value file. Also recorded in this data-set is information on the logarithm (base 10) of the white blood cell count (log(wbc)) and the gender of the 42 patients. In addition, the data-set contains a censoring indicator which identifies those patients who came out of remission (CENSOR=1) and those who were censored (CENSOR=0). Read into R the data-set and call it Leukaem. Examine the data-set. Do you observe any initial differences between the two therapy groups? Kaplan-Meier curves Attach the survival package: library(survival) To create a survival object that describes the right censored survival data to R, we issue the following command: Surv(time=Leukaem$RMISTIME,event=Leukaem$CENSOR) where the two arguments correspond to the observed survival time (time) and the event status indicator (event). Type?Surv to find out more about Surv. The overall survival curve with confidence band can be obtained as follows: plot(survfit(surv(rmistime,censor)~1,data=leukaem),conf.int=t,col=2) The tabular form of this curve can also be obtained: summary(survfit(surv(rmistime,censor)~1,data=leukaem)) The survival curves for the two therapy groups can be displayed as follows:

2 plot(survfit(surv(rmistime,censor)~drug,data=leukaem),col=2:3,lty=1:2,xlab="remission times in weeks",ylab="proportion in remission") legend(25,0.8,c("placebo","6-mp"), lty=1:2,col=2:3) title("kaplan-meier curves for the Leukaemia Patients") Plot now the survival curves by sex and then by log(wbc). Note that for the log(wbc) variable you need to decide how to categorise it. Comparing survival curves From looking at the survival curves for the two drug groups there appears to be a difference in the remission experiences of the 6-MP and Placebo groups, with patients in the 6-MP group remaining, overall, in remission longer than patients in the Placebo group. To confirm this we perform the log-rank test. survdiff(surv(rmistime,censor)~drug,data=leukaem) Perform now the log-rank test on the gender and the categorised log(wbc) variables. What are the results? Was it appropriate to use the log-rank test for these variables? Cox proportional hazards model Below are the commands for fitting a particular Cox proportional hazards model. Use these commands to find the "best" fitting model. leuk.ph <- coxph(surv(rmistime,censor)~drug,data=leukaem) summary(leuk.ph) Do you understand the output? Is there a drug*log(wbc) interaction? Interpret the results. Can you figure out how to plot the predicted survival curve for a person who is on 6-MP and has a log(wbc) of 4.00? (Hint:?survfit) Assessing the validity of the proportional hazards assumption We now assess whether the proportional hazards assumption is valid for the drug variable. First graphically, plot(survfit(surv(rmistime,censor)~drug,data=leukaem),col=2:3,lty=1:2,fun="cloglog", xlim=c(1,40), ylim= c(-3,3), xlab="time of remission (weeks)",ylab="log(-log(s(t)))") Then analytically, leuk.ph <- coxph(surv(rmistime,censor)~drug,data=leukaem) cox.zph(leuk.ph) # using the default K-M transformation of time Test whether the proportional hazards assumption is being satisfied for the sex and log(wbc) variables (and any other variable you create or use). If the PH assumption is not satisfied, re-fit the Cox model but now stratify by the variable not satisfying the

3 PH assumption. This is done by using the function strata("variable name") as an argument on the right hand side of the formula in coxph(.). That is, coxph(surv(rmistime,censor) ~ strata(.) +...,data=leukaem). Now find the most appropriate model and interpret your results. Other R commands which may be useful are step and basehaz. Below is the Leukaem data used for the above practical. RMISTIME DRUG CENSOR LOGWBC SEX

4 Male =1; Female =0; 6-MP =1 Placebo =0 and CENSOR=1 indicates not in remission anymore. Remission is a disease symptom-free state.

5 Right Censored and Left Truncated Survival Data The data-set used in this part of the practical is taken from Klein and Moeschberger (2003). It consists of the ages (in months) of 462 elderly individuals (97 males and 365 females) who were resident at the retirement centre, Channing House, during the period January 1964 to July 1975 when they died or left the centre, as well as the ages (in months) at which the individuals entered the retirement community. Interest lies in the life times of the individuals. However, in addition to right censoring, the life times in this data-set are subject to left truncation as an individual must survive to a sufficient age to enter the retirement community. Individuals who die at an early age are excluded from the study. Also, included in the data-set is the gender of the individuals (coded: Male=1 and Female=2). The data channing can be found in library(kmsurv). To get access to the data once the KMsurv package is loaded type data(channing) Familiarise yourself with the data-set. Do you notice anything peculiar? Modified Kaplan-Meier curves Before estimating the Kaplan-Meier estimator we need to tell R that this is a survival data-set (i.e. a survival object) with right censoring and left truncation. We do this through use of the function Surv(), once again. srvobj <- Surv(time = channing$ageentry, time2 = channing$age, event = channing$death, type="counting")) Can you tell why this does not work? To overcome this problem we create a new data-set called channing1. channing1 <- channing[(channing$time!=0),] Now repeat the earlier step using channing1 instead of channing to create the survival object. Below we have used the function with srvobj <- with(channing1, Surv(time = ageentry, time2 = age, event = death, type="counting")) srvobj The overall estimated K-M curve can now be calculated. summary(survfit(srvobj~1)) plot(survfit(srvobj~1), xlab ="Age at death (in months)", ylab = "Estimated Conditional Proportion Surviving") How do you interpret this estimator?

6 The estimated survival curves by gender can also be calculated. plot(survfit(srvobj~gender,data=channing1), lty=1:2, col=1:2, xlab = "Age at death (in months)", ylab = "Estimated Conditional Proportion Surviving") legend(200,0.9,c("male","female"),lty=1:2,col=1:2) Do you notice anything peculiar? Let us now look at the males in a bit more detail. summary(survfit(srvobj~1,data=channing1,subset=(gender==1))) Do you understand what has happened here? Does this estimator have any real meaning? Let us now estimate the conditional survival curves for males and females separately, given that they have survived up to age 68 years old (i.e. 816 months). channing2 <- channing1[(channing1$ageentry>=816),] srvobj2 <- Surv(time = channing2$ageentry, time2 = channing2$age, event = channing2$death, type="counting") srvobj2 plot(survfit(srvobj2~gender,data=channing2), lty=1:2, col=1:2, xlim = c(800,1250), xlab = "Age at death (in months)", ylab = "Estimated Conditional Proportion Surviving") legend(1100,0.9,c("male","female"),lty=1:2,col=1:2) Is there any evidence (by eye-balling) of a difference between the survival experiences of males and females? Can you figure out how to estimate the conditional survival probabilities for males and females separately, given that they have survived up to age 75 years old? Cox proportional hazards model Let us now investigate formally whether there is an effect of gender on the survival experiences of the individuals in Channing House. channing.cox <- coxph(srvobj~factor(gender),data=channing1) summary(channing.cox) Can you interpret the results obtained? What is the hazards ratio for males versus females? The R function relevel is a useful command when one would like to change the reference level of a factor variable. See if you can figure out how to apply it above. Let us end by checking the proportional hazards assumption analytically. cox.zph(channing.cox) Is there any evidence for non-proportionality of hazards?