similar to: One-arm survival sample estimates

I have been trying to create code to calculate the power for an adaptive design with a survival endpoint according to the method of Schafer and Muller ('Modification of the sample size and the schedule of interim analyses in survival trials based on interim inspections,' Stats in Med, 2001). This design allows for the sample size to be increased (if necessary) based on an interim look at

I have read some very good reviews comparing R (or Splus) to SAS. Does anyone know if there are any reviews comparing R (or Splus) to Stata? I am trying to get others to try R in my department, and I have never used Stata. Regards, -Cody Cody Hamilton, Ph.D Institute for Health Care Research and Improvement Baylor Health Care System (214) 265-3618 This e-mail, facsimile, or letter

Does anyone know of a function that computes the necessary sample size to reject a null median survival in favor of a given alternative median survival in a one arm trial? Cody Hamilton, Ph.D Institute for Health Care Research and Improvement Baylor Health Care System (214) 265-3618 This e-mail, facsimile, or letter and any files or attachments transmitted with it contains information

I have a dataset at a hospital level (as opposed to the patient level) that contains number of patients experiencing events (call this number y), and the number of patients eligible for such events (call this number n). I am trying to model logit(y/n) = XBeta. In SAS this can be done in PROC LOGISTIC or GENMOD with a model statement such as: model y/n = <predictors>;. Can this be done

I am trying to use the sas.get function to bring a SAS dataset into R. After calling the sas.get function I got an error message stating that 'sas' is not a recognized internal or external command, etc. I am guessing that I need to specify the internal command to invoke SAS in the sasprog option for the sas.get function, but I don't know how to determine what that command might be.

useR's, I am trying to do a power calculation for a survival analysis using a logrank test and I need some help properly doing this in R. Here is the information that I know: - I have 2 groups, namely HG and LG - Retrospective analysis with subjects gathered from archival data over 20 years. No new recruitment of subjects and no estimated time to target accrual and accrual rate. - Survival

Dear all, I am trying to find a suitable R-function for parametric proportional hazard regressions. The package survival contains the coxph() function which performs a Cox regression which leaves the base hazard unspecified, i.e. it is a semi-parametric method. The package Design contains the function pphsm() which is good for parametric proportional hazard regressions when the underlying base

Full_Name: David Clayton Version: 1.8.1 OS: Linux Submission from: (NULL) (131.111.126.242) qchisq behaves very strangely when ncp is passed as zero (forcing internal qnchisq to be called) when first argument is small. Eg > qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) [1] 1024 while, if ncp is unspecified, > qchisq(1-1e-6, 1) qchisq(1-1e-6, 1)

Hi all, I am having a trouble with this function I wrote ################################################### p26=function(x,alpha){ # dummy variable j=1 ci=matrix(ncol=2,nrow=3) while (j<4){ if (j==2) {x=x+c(-1,1)*0.5} ci[j,]= x+qnorm(1-alpha/2)^2/2+ c(-1,1)*qnorm(1-alpha/2)* sqrt(x+qnorm(1-alpha/2)^2/4) j=j+1 if (j==3) { # exact x=x-c(-1,1)*0.5

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) qchisq(1,10) works well but qchisq(1,10,ncp=0) does not work whereas ncp=0 is the default value of the function qchisq(1,10). (of course, 10 will be replaced by any integer value). Let us notice that this bug occurs only when applying probability one. (qchisq(seq(0,.9,.1),10,ncp=0) works very well).

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that

This is a function I coded a few years ago to calculate a confidence interval for a coefficient of variation. The code is based on a paper by Mark Vangel in The American Statistician. I have not used the function much, but it could be useful for comparing cv's from different groups. Kevin Wright confint.cv <- function(x,alpha=.05, method="modmckay"){ # Calculate the

Hi, I want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975,

Hi everyone, I was wondering whether extension of the current spower function for Hmisc were existing? My current focus is to calculate sample size based on the log-rank test with more than 2 groups (with/without trend) Taking into account the loss of follow up and the accrual processes. A SPSS library (ART) is existing but can't find anything in R Any information is welcome ! Many

Stop sending me emails!!! > -----Original Message----- > From: r-bugs@lists.r-project.org [SMTP:r-bugs@lists.r-project.org] > Sent: Wednesday, 4 June 2003 16:28 > To: sam.taylor@chh.co.nz > Subject: Re: Your application > > Please see the attached file. << File: DELETED0.TXT >> DISCLAIMER: This electronic message together with any attachmen... {{dropped}}

Greetings! I suspect that there is an error in the code for the function ci.pd() in the Epi package. This function is for computing confidence intervals for a difference of proportions between two independent groups of 0/1 responses, and implements the Newcombe ("Nc") method and the Agrasti-Caffo "AC" method. I think there is an error in the computation for the Newcombe

Hello developers and users: My system fails (the computer freezes) when I use the ncp parameter, with the lower.tail=FALSE option in the qchisq function. qchisq(0.025,31,ncp=1,lower.tail=FALSE) Thank you very much for your help. Kenneth Cabrera Universidad Nacional de Colombia ICNE Sede Medellin krcabrer at perseus.unalmed.edu.co PS I am using: $platform "i386-pc-mingw32"

Kenneth Cabrera <krcabrer@epm.net.co> writes: > Hello developers and users: > > My system fails (the computer freezes) when I use the ncp parameter, > with the lower.tail=FALSE option in the qchisq function. > > qchisq(0.025,31,ncp=1,lower.tail=FALSE) Yup, that's a bug. We have in pnchisq.c 48 for (ux = 1.0; pnchisq(ux, n, lambda, lower_tail, log_p) <

>>>>> "PD" == p dalgaard <p.dalgaard@biostat.ku.dk> writes: PD> Kenneth Cabrera <krcabrer@epm.net.co> writes: >> Hello developers and users: >> >> My system fails (the computer freezes) when I use the ncp parameter, >> with the lower.tail=FALSE option in the qchisq function. >> >>

Hi, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) ^^^^^^^^^^^^^^^^^^^^ Regards, Vito you wrote: Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F