Fitting distributions
Hola! I think the original posters problem was to fit a weibull. this could be done with: library(gnlm)
x <- rweibull(100, 2, 3) y <- rep(1,100)
fit.dist(x,y,"Weibull")
Weibull distribution, n = 100
mean variance alpha.hat mu.hat
2.438586 1.840581 1.890002 6.782797
-log likelihood AIC
-296.4451 -294.4451
yi ni pi.hat pi.tilde like.comp resid
1 1.8744688 1 0.01 0.300583890 -3.403142 -5.300156
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Kjetil Halvorsen
Prof Brian Ripley wrote:
On Thu, 6 Sep 2001, kjetil halvorsen wrote:
Hola! I think some of this is avaliable as fit.dist in Jim Lindsays package gnlm.
Not according to his help page:
\name{fit.dist}
\title{Fit Probability Distributions to Frequency Data}
Not the same problem.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272860 (secr)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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