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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