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← Revision 19 as of 2014-07-25 11:10:50 ⇥
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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 to K (n(k)-1) V(k)] / [sum_k to K (n(k) -1)] |
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| {{{ $$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}^text{K} n_text{k}} $$ }}} |
Pooled Mean Standard Error = Sqrt{ [V] / [sum_k to 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 to K (n(k)-1) V(k)] / [sum_k to 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_k to K n(k) ] }