On Fri, 7 Sep 2001, kjetil halvorsen wrote:
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
From the help page again:
breaks: If TRUE, `y' contains breaks between categories instead of
mid-points.
so according the help page it is assuming grouped observations
specified by midpoints. That is not the same problem, as I did say.
Brian is correct.
Who knows if the help page is right?
Yes it is.
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.
Actually, it is rather gnlr in the same library that will do much of the proposed fitdist. Jim
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)
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-- 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 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
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