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| Suppose we have each of K people (or groups) which we wish to pool each with variances, $$V_text{k}$$ and mean $$m_text{k}$$ based on a sample of size $$n_text{k}$$. | 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 $$\mbox{n}_text{k}$$. |
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| Pooled Variance V = \frac{\sum_text{k}(n_text{k}-1)V_text{k}}{\sum_text{k}(n_text{k}-1)} | \mbox{Pooled Variance V = } \frac{\sum_text{k}(n_text{k}-1)V_text{k}}{\sum_text{k}(n_text{k}-1)} |
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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 $$\mbox{n}_text{k}$$.
$$ \mbox{Pooled Variance V = } \frac{\sum_text{k}(n_text{k}-1)V_text{k}}{\sum_text{k}(n_text{k}-1)} $$
$$ Pooled Mean Standard Error = \sqrt{V / \sum_text{k} n_text{k}} $$