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I have two sets of independent samples from the same distribution. For each, I have calculated

  1. sample weighted mean (u1, u2)
  2. sample weighted std deviation (d1, d2)
  3. sample weighted std error (e1, e2)

The question below gives a way to calculate the pooled statistics in the 'unweighted' case

How to calculate pooled variance of two groups given known group variances, means, and sample sizes?

So, how can I calculate this for the weighted case?

Edit 1: To clarify, EACH sample has a 'weight' associated with it which represents a 'confidence' associated with it 0 < weight < 1

Edit 2: To clarify further, each sample represents return on money invested in a particular investment. The weight of a sample is the fraction of the total money invested in a given interval. The two groups of samples are from two different time intervals. The total money invested in the two intervals is not the same

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user1827356
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    A general solution is explained in my post at http://stats.stackexchange.com/a/16609. There are, however, multiple different meanings of "weights" in statistics. (For this reason some software such as Stata supports up to *four* different kinds of weight calculations!) In light of this could you clarify what your weights are intended to mean or explain how they arise? – whuber May 13 '14 at 20:01
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    Thanks for the quick turnaround. I've editied to clarify what weights mean in my case – user1827356 May 13 '14 at 20:08
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    Well done! That's the best way to respond. But now I must apologize, because I need to follow up with yet another question: how were these "confidences" computed and what are they intended to mean? It could also help us if you were to explain the intention or purpose behind computing pooled statistics in the first place. – whuber May 13 '14 at 20:11
  • Added more info in Question – user1827356 May 13 '14 at 20:23
  • I have weighted data from 4 waves of surveys with 50% overlap. I now want to combine the data to have an annual set and compute totals. Should I divide the weights by 4? because right now the totals seems funny and too large. –  Mar 05 '15 at 12:19

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