[R] Weibull survival modeling with covariate

[Watalu, Y. (aka Wataru)]

Hi,

I'm also wondering which expression the survreg() uses
for Weibull regression.  Referring to help(survreg) and
help(survreg.distributions), I guess survreg() fits the
following model.

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$time,
> labadata$censor)~labdata$cov,
> dist="weibull")
>
> This returns:
>> weib
> Call:
> survreg(formula = Surv(labdata$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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