Estimator of the variance of a population from different groups Suppose a population can be divided into 4 different groups (A, B, C, D) and you take the sample variance for a parameter of each group.
A proposed estimator for the population variance is
$$s^2_A/n_A + s^2_B/n_B + s^2_C/n_C + s^2_D/n_D,$$
where $n_A, n_B, n_C$ and $n_D$ are the number of observations in each group.
In other words, the proposed estimator for the population variance is the average of the estimators of the conditional variance calculated before.
Is this population variance estimator correct?

If you do not have the sample means, you cannot back out the variance that you would have calculated if you did so when you had all of the observations and not just the summary statistics. Consider the following scenarios.
Scenario 1: $A=\{1,2,3\}$, $B=\{1,2,3\}$
Scenario 2: $A=\{1,2,3\}$, $B=\{7,8,9\}$
In both scenarios, the groups have the same variance. However, the pooling of all six numbers results in different variances. If all you have are $n_A$, $s^2_A$, $n_B$, and $s^2_B$, you have no way to distinguish between these two scenarios, even though $\{1,2,3,1,2,3\}$ and $\{1,2,3,7,8,9\}$ have different variances.
This is not to be confused with finding the common sample variance as is done in ANOVA. There, you do not need the sample means, but it also addresses a completely different question.
