I recently came across Analysis of Variance (ANOVA), and it seems that the total sum of squares of the residuals (or "within groups / treatments") divided by the degrees of freedom is Chi-squared distributed. I've also been told that this is the case for "between groups / treatments" under the null hypothesis.

I was wondering why this is the case, and roughly how we might go about showing this (I'm not after a rigorous proof, though).

One thing I'm wondering behind this question is are they both genuinely Chi-squared distributed, or is it an asymptotic result? (i.e. it holds in the limit as our sample size tends to infinity).



1 Answer 1


After several hours of trying to figure this out, I think I've got it to a decent extent. I don't have the time or LaTeX skills (or whatever is used to type equations etc. here) to provide a full answer. Here's what is hopefully a useful sketch of an answer to anyone else asking the same question, though:

Firstly, you don't quite get Chi-squared distributions as described in the question. The sums of squares before they're divided by the degrees of freedom are proportional to a random variable with Chi-squared distribution, where the constant of proportionality is the variance of the observations (assumed to be constant for all treatments / subgroups). This constant will cancel when dividing the M.S, still giving an F-distribution.

Secondly, the distributions are exact (assuming that our observations are perfectly normally distributed to begin with, which may be a stretch), so they're not just (proportional to) Chi-squared in a limit or anything.

Lastly, I was able to figure out why they're (proportional to) Chi-squared for the case of equal replication (slightly non-rigorously, but good enough for me), and this was a super-useful result / link. en.wikipedia.org/wiki/Cochran%27s_theorem#Sample_mean_and_sample_variance


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