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) ] }