Apologies for a naive question, Can anyone please explain why the formula for pooled variance is the weighted sum and not just the sum of individual samples?
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variance = 1/m x sum samples. so if you want to pool variance of m1 and m2 samples, you first multiply each variance by its corresponding count so you now have sums and then divide by the total counts (m1+m2)... this is the same as doing a weighted sum with m1/(m1+m2), m2(m1+m2) weights – seanv507 Jan 26 '22 at 12:34
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More answers are available by searching our site for [pooled variance](https://stats.stackexchange.com/search?q=pool*+variance). – whuber Jan 26 '22 at 15:31