Following a recent down vote I have been trying to check my understanding of the Pearson Chi Squared test. I usually use the chi squared statistic (or reduced chi squared statistic) for fitting or checking the resulting fit. In this case the variance is not usually the expected number of counts in a table or histogram but some experimentally determined variance. Either way, I was always under the impression that the test still used the asymptotic normality of the multinomial PDF (i.e. my test statistic is

Q=(n−Nm)⊤V−1(n−Nm)

and (n−Nm) is asymptotically multinormal where V is is covariance matrix). Therefore Q has a chi-squared distribution given large n so using the expected number of counts as the denominator in the statistic becomes valid for large n. Its possible that this is only true for histograms, I haven’t analysed a small table of data in years.

Is there a more subtle argument that I am missing? I would be interested in a reference, or even better a short explanation. (Although its possible I just got down voted for omitting the word asymptotic, which I concede is rather important.)

**Answer**

A Chi-square test is designed to analyze categorical data. That means that the data has been counted and divided into categories. It will not work with parametric or continuous data. So it doesn’t work to determine resulting fit in every instance.

**Attribution***Source : Link , Question Author : Bowler , Answer Author : BradHanks*