[R] chi2
(Ted Harding)
Ted.Harding at manchester.ac.uk
Wed Oct 10 14:05:22 CEST 2007
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.
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Date: 10-Oct-07 Time: 13:05:19
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