Talk:Yates's correction for continuity

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"Using the chi-squared distribution to interpret Pearson's chi-squared statistic requires one to assume that the discrete probability of observed binomial frequencies in the table can be approximated by the continuous chi-squared distribution. This assumption is not quite correct, and introduces some error." This is false. The chi-square distribution is an approximate distribution for the statistic, not the data (frequencies). For the statistic to be approx. *chi-square*, Poisson counts within cells should have a distribution well aproximated by a *gaussian* distribution. — Preceding unsigned comment added by Gklimmt (talk • contribs) 09:10, 24 September 2013 (UTC)[reply]

A sentence should be added that with contingency tables having two rows and two columns Fischer's test should be used as it provides exact results without need of correction (chi2 does not) - unless numbers are very large since chi2 results get more correct with increasing numbers; moreover chi2 is faster when using slow computers. Fischer's test does not work with tables having more than two rows or more than two columns. ___

I thought that Fischers Exact test should be used when there are expected frequencies less than 5? [Anonymous]

Wouldn't Yates's continuity correction actually be different from Yates's chi-squared test? One could say the former is used in the latter, no? - dcljr 20:53, 23 Sep 2004 (UTC)

Article currently says subtract 0.5 from each table entry. I thought you should move each table entry by 0.5 towards the "expected" value. Anyone no for sure?

I also believe that the Fisher Exact test should be used for testing low frequency contingency table. I am wondering if the original author can look into this and edit the text to reflect what is commonly accepted. Hong 20:24, 14 March 2006 (UTC)[reply]

I've tagged this article for cleanup - there seems to be a broken math formula under where it mentions an example 2x2 table. Deathanatos 00:45, 28 September 2006 (UTC)[reply]