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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 have each of K people (or groups) which we wish to pool each with variances, $$\mbox{V}_text{k}$$ and mean $$\mbox{m}_text{k}$$ based on a sample of size $$\mbox{n}_text{k}$$. $$\mbox{Pooled Variance V = } \frac{\sum_text{k} (n_text{k}-1) \mbox{V}_text{k}}{\sum_text{k}(n_text{k} -1)} $$ $$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}n_text{k}} $$ |
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| Pooled Variance V = \frac{\Sum_{k}(n_k}-1)V_{k}}{\Sum_{k}(n_{k}-1)} | Pooled Mean Standard Error = \sqrt{V / \sum_text{k} n_text{k}} |
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$$ Pooled Mean Standard Error = \Sqrt{V / \sum_{k}n_{k}} |
How do I obtain a pooled mean standard error?
Suppose we have each of K people (or groups) which we wish to pool each with variances, $$\mbox{V}_text{k}$$ and mean $$\mbox{m}_text{k}$$ based on a sample of size $$\mbox{n}_text{k}$$.
$$\mbox{Pooled Variance V = } \frac{\sum_text{k} (n_text{k}-1) \mbox{V}_text{k}}{\sum_text{k}(n_text{k} -1)} $$
$$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}n_text{k}} $$
$$ Pooled Mean Standard Error = \sqrt{V / \sum_text{k} n_text{k}} $$