192014-07-25 11:10:50PeterWatson182013-10-29 14:34:28PeterWatson172013-10-29 14:33:53PeterWatson162013-03-08 10:17:15localhostconverted to 1.6 markup152007-04-26 16:27:37PeterWatson142007-04-26 16:23:55PeterWatson132007-04-26 16:23:10PeterWatson122007-04-26 16:22:15PeterWatson112007-04-26 16:20:12PeterWatson102007-04-26 16:18:32PeterWatson92007-04-26 16:16:53PeterWatson82007-04-26 16:12:48PeterWatson72007-04-26 16:09:53PeterWatson62007-04-26 09:35:25PeterWatson52007-04-26 09:34:13PeterWatson42007-04-26 09:32:50PeterWatson32007-04-26 09:32:24PeterWatson22007-04-26 09:30:36PeterWatson12007-04-26 09:29:17PeterWatson

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) ] }