A few months ago, I posted a query regarding code for a sample size estimate for a one arm survival trial. Below is some code I created to calculate such an estimate - perhaps it may be of some use. #cox.pow computes sample size for a one arm survival trial. med.0 is the null median #survival, med.a is the alternative median survival, a.time is the accrual time, and #f.time is the follow up (assumes constant accrual and two sided test). Approach taken #from Lawless 1982, p. 108 cox.pow<-function(med.0,med.a,a.time,f.time,alpha,beta) { #initialize number of events to 0 r<-0 #find necessary number of events while (qchisq((1-alpha/2),(2*r+1))*(med.0/med.a) >qchisq(beta,(2*r+1)) ) {r<-r+1} #find proportion of patients that will experience event for given median survivals, #accrual and follow up times (formula from Cox 1983) p <- 1 - (exp(-0.69*f.time/med.a)*(1 - exp(-0.69*a.time/med.a))/(0.69*a.time/med.a)) #sample size is r/p n<-round(r/p,0) return(n) } #compute necessary sample size for alternative median 9.3 vs. null median 6.3 with 20 #months accrual and 4 months follow up cox.pow(6.3,9.3,20,4,.05,.2) Regards, -Cody Cody Hamilton, Ph.D Institute for Health Care Research and Improvement Baylor Health Care System This e-mail, facsimile, or letter and any files or attachmen...{{dropped}}