On 10-Oct-07 10:59:43, elyakhlifi mustapha wrote: > Hello, > 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?
What that tells *me* is that 'don' is a quite long vector: you should find that length(don) = 63252 = 63251+1 as I verified using pchisq(): pchisq(62667.11,63251) [1] 0.04999935 pchisq(63837.16,63251) [1] 0.9499995 So 62667.11 is the value such that, with 63251 degrees of freedom, Prob(chisq <= 62667.11) = 0.05 Prob(chisq <= 63837.16) = 0.95 (and similarly for the other values in your list). The function qchisq() is simply the inverse of pchisq(): where pchisq(x,df) tells you Prob(chisq <= x) with df degrees of freedom, for given x, qchisq(p,df) tells you what value of x will give Prob(chisq <= x) = p for a given value of p. Best wishes, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 10-Oct-07 Time: 13:05:19 ------------------------------ XFMail ------------------------------ ______________________________________________ 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.
