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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_${k}$. | 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 $n_text{k}$. |
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| Pooled Variance V = \frac{\Sum_text{k}(n_{k}-1)V_text{k}}{\Sum_text{k}(n_text{k}-1)} | Pooled Variance V = \frac{\Sum_text{k}(n_text{k}-1)V_text{k}}{\Sum_text{k}(n_text{k}-1)} |
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, $V_text{k}$ and mean $m_text{k}$ based on a sample of size $n_text{k}$.
$$ Pooled Variance V = \frac{\Sum_text{k}(n_text{k}-1)V_text{k}}{\Sum_text{k}(n_text{k}-1)} $$
$$ Pooled Mean Standard Error = \Sqrt{V / \sum_text{k}n_text{k}}