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