On Fri, Jun 24, 2005 at 11:27:28AM -0400, sms13+ at pitt.edu
wrote:> Hi,
> I was wondering if someone can help me
> interpret the results of running
> weibreg.
>
> I run the following and get the
> following R output.
> > weibreg(Surv(time, censor)~covar)
> fit$fail = 0
> Call:
> weibreg(formula = Surv(time,
> censor)~covar)
>
> Covariate Mean Coef
> Rel.Risk L-R p Wald p
> covar 319.880 -0.002 0.998
> 0.000
>
> log(scale) 0.000 8.239
> 3786.326 0.000
> log(shape) 0.000 0.265
> 1.304 0.000
>
> Events 172
> Total time at risk 845891
> Max. log. likelihood -1609.4
> LR test statistic 34.4
> Degrees of freedom 3
> Overall p-value 1.65026e-07
>
>
> I would just like to find the estimated
> mean survival time as a function of the
> covariate in the model, but am not sure
> how to use this output to find that.
> Any help would be greatly appreciated.
The fitted model is a distribution with hazard function
h(t; a, b, z) = (a/b)(t/b)^(a-1)exp(beta*z),
where a = "baseline shape" and b = "baseline scale". z is
your "covar" and
beta is the estimated regression coefficient. It is an easy exercise to
show that this is the hazard function of a Weibull distribution with shape
a and scale b*exp(-beta*z/a). Thus the mean is
E(T) = b*exp(-beta*z/a)*gamma(1+1/a) ## See ?Weibull
Here gamma is the usual gamma function, see ?gamma. (I notice in the R
documentation of the Weibull distribution that "E(X) = b
Gamma(1+1/a)",
which is an error; the G should be g (lowercase).)
In your case, a = 1.304, b = 3786.326, beta = -0.002, so
E(T) = 3495 * exp(0.00153 * z)
(given my calculations are correct).
Hth,
G??ran
>
> Thank you,
> Steven
>
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--
G??ran Brostr??m tel: +46 90 786 5223
Professor and Head
Department of Statistics fax: +46 90 786 6614
Ume?? University http://www.stat.umu.se/~goran.brostrom/
SE-90187 Ume??, Sweden e-mail: gb at stat.umu.se