# Re: [R] Weibull survival modeling with covariate

From: Watalu, Y. (aka Wataru) <wataru_at_crayon.se.uec.ac.jp>
Date: Thu 09 Jun 2005 - 12:19:49 EST

Hi,

survreg() uses a different parametrization, say

F(x, Wshape, Wscale) = 1-exp(-Wscale*(x^Wshape))), and fits a parametric model with these formulas.

Wshape = 1/"Scale" (calculated by survreg())    log(Wscale) = model with covariates

Is it correct?

Thanks a lot.

Watalu

> I was wondering if someone familiar
> with survival analysis can help me with
> the following.
> I would like to fit a Weibull curve,
> that may be dependent on a covariate,
> my dataframe "labdata" that has the
> fields "cov", "time", and "censor". Do
> I do the following?
> wieb<-survreg(Surv(labdata\$time,
> dist="weibull")
>
> This returns:
>> weib
> Call:
> survreg(formula = Surv(labdata\$time,
> labdata\$censor) ~ labdata\$cov,
> dist = "weibull")
>
> Coefficients:
> (Intercept) labdata\$cov
> 8.091955112 0.001552897
>
> Scale= 0.7532474
>
> Loglik(model)= -12633.6
> Loglik(intercept only)= -12734.8
> Chisq= 202.41 on 1 degrees of
> freedom, p= 0
> n= 5496
>
>
> I am not quite sure how to use the
> output. I see that it gives the Scale
> parameter. How do I find the Shape
> paramater as a function of the
> covariate?
>
> Thank you,
> Steven
>
> ---------------------------------------
> -------------------------
> Steven Shechter
> PhD Candidate in Industrial Engineering
> University of Pittsburgh
> www.pitt.edu/~sms13
>
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