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