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Consider the pooled variance as described in this wiki article https://en.wikipedia.org/wiki/Pooled_variance :

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My observation is that the weighting terms

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simply cancel out the Bessel's correction weighting factors. So then the result is simply

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But then we no longer have count-based weightings on each of the variance terms! I must be missing something here - any insights/corrections?

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You are correct that this cancellation occurs, so for each sub-sample you get

$$(n_k-1) s_k^2 = \sum_{i=1}^{n_k} (y_{k,i} - \bar{y}_k)^2.$$

Each of these terms is a sum of $n_k$ squared deviations of the individual data points from the sub-sample mean. There is no need for any further weighting on the terms, since the summation already gives a quantity that is proportionate to the count of terms.

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  • $\begingroup$ oh ya - the summation is over the individual sample sizes! wrote it down but still "missed" it. thx. $\endgroup$
    – WestCoastProjects
    Commented Oct 3, 2019 at 13:32

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