Elyakhlifi Mustapha wrote: > > I want to use the quantile function so I read the doc but I don't > understand with this > > > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) > [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 > 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 > 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 > > Can you help me please? > Notice that, if you want adequate help, you must provide all information. You didn't give don, so we can't reproduce the output.
However, since qchisq(0.5, df) for large df is very close to df, it's possible to infer that length(don)-1 = 63251. If you plot dchisq(x, df=63251) (for example: x <- ((-1000):1000) + 63251 plot(x, dchisq(x, df=63251), type="l") or y <- rchisq(10000, df=63251) hist(y) or any other way to view the probability density function), it's easy so see that chi2 for large n is normal-like, with mean close to df, and standard deviation close to sqrt(df). Alberto Monteiro ______________________________________________ R-help@r-project.org 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.
