On Wed, Jan 14, 2004 at 10:45:56PM +0100, Göran Broström wrote:
On Wed, Jan 14, 2004 at 09:10:51PM +0100, Fredrik Lundgren wrote:
Dear R experts,
How should lambda and gamma (with std.errors) be calculated for a weibull model
with age as an independent predictor? I have assumed that this can be done with
survreg with e. g. (summary(survreg(Surv(time, status) ~ age, dist = 'weibull')) )
and predict.survreg with e.g. (predict(model, se.fit = T, newdata =
data.frame(age = seq(50, 80, 5)) but unfortunately I'm uncapable to sort out how
to get the lambda and gamma values (with std.errors). I haven't found any example
of this in the help pages and would really appreciate any help!
In my package 'eha', function 'weibreg', you will find short discussion of the
different parametrizations of the Weibull distribution. Weibull (in base) and
weibreg (eha) use the same parametrization, different from the one in
survreg. See the help page for weibreg. Oops, I can spot an error in that page;
the reference to 'dgamma' should really be to 'dweibull'.
Göran
To elaborate further, you should maybe be satisfied with the standard errors
you get on the log scale. Calculate confidence intervals (or whatever you
want the se's for) on that scale, and transform these intervals to any scale
you like. Usually much better than doing it in reverse order, ie, calculating
se's (via the delta method) for 'lambda' and 'gamma', and then the confidence
intervals.
Göran
With best wishes and thanks in advance for any help
Fredrik Lundgren
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Göran Broströmtel: +46 90 786 5223
Department of Statistics fax: +46 90 786 6614
Umeå University http://www.stat.umu.se/egna/gb/
SE-90187 Umeå, Sweden e-mail: [EMAIL PROTECTED]
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