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 > > With best wishes and thanks in advance for any help > > Fredrik Lundgren > > ______________________________________________ > [EMAIL PROTECTED] mailing list > https://www.stat.math.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Göran Broström tel: +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] ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
