= Pooled sample variance
{title2=$S_p^2=(S_{XX}+S_{YY})/(m+n-2)$}
For two independent normal samples of sizes $m,n$ with a common unknown variance, pooling their within-sample sums of squares gives the <unbiased estimator> $S_p^2=(S_{XX}+S_{YY})/(m+n-2)$. The numerator divided by the common population variance has a <chi-squared distribution> with $m+n-2$ degrees of freedom and is independent of both sample means. Therefore $(\bar X-\bar Y-(\mu_X-\mu_Y))/(S_p\sqrt{1/m+1/n})$ has a <Student t-distribution>. This exact construction requires a positive residual number of degrees of freedom and the common-variance assumption.
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