Re: Chi squared -> p-value - Any formula??

On Jul 29, 2:10 am, "Dave (from the UK)" <see-my-signat...@see-
below.com> wrote:
Ray Koopman wrote:
My favorite simple approximation for the standard normal cdf is

1/(1 + Exp[-.496937z*Sqrt[z^2 + 10.28]]),

for which the absolute error is < 1.58*10^-4.

The corresponding approximation for the complementary cdf (i.e.,

the upper tail area) of a chi-square variable with 1 d.f. is

2/(1 + Exp[.496937*Sqrt[x(x + 10.28)]).

Thanks for that Ray - it was most helpful.

The claimed accuracy of the approximation for the standard normal cdf (<
1.58*10^-4) is somewhat better than that of the approximation Ken gave,
and is also simpler to type, so I decided to try your approximation.

I did some checks of your formula in Mathematica and found largest
*absolute* error of the p-value is 0.0003147. (I tested that for
chi-squared values in the range 0 to 40 in steps of 0.00001)

For chi-squared in the range 0 to 6.63489 (p-values 1.0 down to 0.01),
the maximum fractional error in the p-values was 2.91%.

Extending its range somewhat more, with chi-squared to 10.827566
(p-values of 1.0 down to 0.001) gave an error in the p-values of <= 9.18
%, which is good enough for me.

Just for completeness, I tested it to p-values down to 0.0001
(chi-squared up to 15.136705) at which point the error in p-values is
quite large (16.9%), but realistically for me, I'm only really
interested in p-values down to 0.005 or so, where the error is 4.56% or
less.

So far I've only tested a Mathematica approximation, but assuming Tcl's
maths functions are not broken, it should give similar results in Tcl.

BTW, the reason for using Tcl is that I have a program that was
originally written in that. Its not an ideal language for maths stuff,
but the main program is a chess database, not a maths package.

--
Dave (from the UK)

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It is always of the form: month-y...@xxxxxxxxxxxx
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The normal cdf gives a one-tailed p-value, but the chi-square ccdf is
equivalent to a two-tailed normal p-value, so the error will also be
twice as large. To see the error curves in Mathematica:

Plot[1/(1 + Exp[-.496937z*Sqrt[z^2 + 10.28]]) - .5 Erfc[-z/Sqrt[2]],
{z,-4.5,4.5}];

Plot[2/(1 + Exp[.496937*Sqrt[x(x + 10.28)]]) - Erfc[Sqrt[x/2]],
{x,0,20}];

.