Re: New Pearson's chi-square test?

From: Richard Ulrich (Rich.Ulrich_at_comcast.net)
Date: 01/18/05 

Date: Mon, 17 Jan 2005 23:13:07 -0500

On 17 Jan 2005 18:44:35 -0800, glenbarnett@geocities.com (Glen) wrote:

> Richard Ulrich (Rich.Ulrich@comcast.net) said:
> > A random, standard-normal variable can be thought of
> > as a "chi" so that the squared version is a chi-squared.
> > And then, the sum of k independent chi-squared
> > variables is a X^2 with k degrees of freedom.
>
> A "chi" random variable has in the past been the square root
> of a chi-square random variable. Standard deviations from normal
> samples have a multiple of a chi-distribution.
>
> See for example
> http://mathworld.wolfram.com/ChiDistribution.html which also
> gives a couple of book references.
>

Thanks for the correction and the reference.

That was ignorant of me, rather than just careless.
I don't actually recall any discussions or uses of
the 'chi'; and I've mainly mentioned it in exactly
this context, in explanations of chi-squared.

> A google search turns up several more references.
>
> So a chi(1) is the absolute value of a standard normal. Hence
> adding up the squares of independent chi(1) distributions will as
> Rich says, give a chi-squared(k).
>
> The chi has to some extent fallen out of common use.

- Okay. 

-- 
Rich Ulrich, wpilib@pitt.edu
http://www.pitt.edu/~wpilib/index.html