Re: significance in contingency table
From: Aleks Jakulin (a_jakulin_at_@hotmail.com)
Date: 10/14/04
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Date: Thu, 14 Oct 2004 23:57:47 +0200
François Charton
> I understand the need of finding the law of the statistics under the
> null hypothesis (independence).
>
> The first approach seems the most natural, for the law of the
> the formula hints at a Chi-square with one degree of freedom: the
> data can be supposed to be binomial, for large samples, it converges
> normal, and the statistics seems to be homogeneous with its square
> (once you center it and normalise it).
Not sure exactly what you did here.
If you don't want to give up asymptotics, a possible solution is to
use standardized Pearson's residuals. If n_ij is the true count in
cell {ij}, and e_ij is the expected count under the null, the
standardized Pearson's residual is (n_ij - e_ij) / Sqrt[e_ij (1-p_i*)
(1-p_*j) ]. It has asymptotically a standard normal distribution. But
as I wrote in the previous post, with an increasing number of rows
and columns, you have the multiple testing problem: it is increasingly
likely that these residuals will be large just by chance.
If your null is independence, and with all the computer power today,
why not do permutation testing? Or, if the null is dependence, use
nonparametric bootstrap.
Aleks
-- mag. Aleks Jakulin http://www.ailab.si/aleks/ Artificial Intelligence Laboratory, Faculty of Computer and Information Science, University of Ljubljana, Slovenia.
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