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| $$\mbox{Pooled Variance V = } \frac{\sum_text{k} (n_text{k}-1) \mbox{V}_text{k}}{\sum_text{k}(n_text{k} -1)} $$ | Then pooling the K variances we have |
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| $$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}n_text{k}} $$ | $$\mbox{Pooled Variance V = } \frac{\sum_text{k}^text{K} (n_text{k}-1) \mbox{V}_text{k}}{\sum_text{k}^text{K} (n_text{k} -1)} $$ 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
$$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}^text{K} n_text{k}} $$