|
Size: 384
Comment:
|
Size: 604
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 3: | Line 3: |
| 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 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 $$\mbox{n}_text{k}$$. |
| Line 5: | Line 5: |
| $$ Pooled Variance V = \frac{\sum_text{k}(n_text{k}-1)V_text{k}}{\sum_text{k}(n_text{k}-1)} $$ |
Then pooling the K variances we have |
| Line 9: | Line 7: |
| $$ Pooled Mean Standard Error = \sqrt{V / \sum_text{k} n_text{k}} |
$$\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)} $$ and we can use this pooled variance to obtain the standard error of the mean since $$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}^text{K} n_text{k}} $$ |
How do I obtain a pooled mean standard error?
Suppose we wish to pool over each of K people (or groups) with the k-th individual having variance, $$\mbox{V}_text{k}$$ and mean $$\mbox{m}_text{k}$$ based on a sample of size $$\mbox{n}_text{k}$$.
Then pooling the K variances we have
$$\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)} $$
and we can use this pooled variance to obtain the standard error of the mean since
$$\mbox{Pooled Mean Standard Error = } \frac{\mbox{V}}{\sum_text{k}^text{K} n_text{k}} $$