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| Suppose we have each of K people (or groups) which we wish to pool each with variances, V_$k$ and mean m$_k$ based on a sample of size n_$k}$. | Suppose we wish to pool over each of K people (or groups) with the k-th individual having variance V(k) and mean m(k) based on a sample of size n(k). |
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| $$ Pooled Variance V = \frac{\Sum_{k}(n_k}-1)V_{k}}{\Sum_{k}(n_{k}-1)} $$ |
Then pooling the K variances we have |
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| $$ Pooled Mean Standard Error = \Sqrt{V / \sum_{k}n_{k}} |
Pooled Variance V = [sum k to K (n(k)-1) V(k)] / [sum_k to K (n(k) -1)] and we can use this pooled variance to obtain the standard error of the mean since Pooled Mean Standard Error = \sqrt{ [V] / [sum_k to K n(k) ] $$ |
How do I obtain a pooled mean standard error?
Suppose we wish to pool over each of K people (or groups) with the k-th individual having variance V(k) and mean m(k) based on a sample of size n(k).
Then pooling the K variances we have
Pooled Variance V = [sum k to K (n(k)-1) V(k)] / [sum_k to K (n(k) -1)]
and we can use this pooled variance to obtain the standard error of the mean since
Pooled Mean Standard Error = \sqrt{ [V] / [sum_k to K n(k) ] $$