l used the following code to generate a sample and then calculated then did a log rank test.can l get the normal version of the logrank eg sqrt of the chisqr(1) will give you the N~(0,1). from my sample can i use the above expression to get the normal dist from the result of the log rank test. thank s=1 while(s!=0){ n=100 m<-matrix(nrow=n,ncol=4) colnames(m)=c("treatmentgrp","strata","censoringTime","survivalTime") for(i in 1:n) m[i,]<-c(sample(c(1,2),1,replace=TRUE),sample(c(1,2),1,replace=TRUE),rexp(1,0.07),rexp(1,0.02)) m<-cbind(m,0) m[m[,3]>m[,4],5]<-1 colnames(m)[5]<-"censoring" act.surv.time<-pmin(m[,"censoringTime"],m[,"survivalTime"]) m<-cbind(m,act.surv.time) print(m) write.table(m,âclipboardâ,sep=â\tâ,col.names=NA) print(â.................â) print(âthe logrank testâ) b=data.frame(m) c= survdiff(Surv(act.surv.time,censoring)~treatmentgrp,data=b) print(c) s=s-1 print(â...............................................................â) }
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