On Thu, 21 Sep 2006, Anaid Diaz wrote: > Hi > I am using R to fit a survival function to my data > (with a weibull distribution). > > Data: Survival of individuals in relation to 4 > treatments ('a','b','c','g') > > syntax: > ---- > survreg(Surv(date2)~males2, dist='weibull') > > But I have some problems interpreting the outcome and > getting the parameters for each curve. > > --------- Value Std. Error z > p > --------- (Intercept) 2.788 0.147 19.022 > 1.13e-80 > --------- males2b -0.107 0.207 -0.519 > 6.04e-01 > --------- males2c -0.486 0.586 -0.831 > 4.06e-01 > --------- males2g 0.580 0.207 2.798 > 5.15e-03 > --------- Log(scale) -1.116 0.139 -8.007 > 1.18e-15 > --------- > --------- Scale= 0.328 > > > I know from Venables & Ripley (2002) that the > parameters of this function should be two:
"this function" being which function? As help(survreg.distributions) says The Weibull distribution is not parameterised the same way as in 'rweibull'. In your model the survival time is a variable whose logarithm has distribution 2.788-0.107males2b-0.486males2c+0.580males2g+epsilon*0.328 where epsilon has cdf F=1-e^{-e^t}. As in the example on help(survreg.distributions) shows, a survreg Weibull model with linear predictor M and scale S corresponds to R's weibull distribution with scale exp(M) and shape 1/S. -thomas ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.