How to estimate sample mean and variance from derived data Hi,
(I hope this is not too basic)
I basically have a set of data from the same underlying distribution (which I would like to estimate), but I only have available the mean and N from partitions of the data.  I can estimate the mean using a weighted average, and, I thought, the sample variance using weighted variance (e.g:  https://stat.ethz.ch/pipermail/r-help/2008-July/168762.html)
However, I can't make much sense of the results. And I suspect I'm misunderstanding how/what weighted sample average should work.
Edit: From a set of (unobserved) data points all drawn from some distribution with some $\mu$ and $\sigma$, I can observe only the average and cardinality for partitions of the data.  I.e. my observed values $X_i$ are each the average of some partition of size $N_i$, and thus have, $\mu_i=\mu$ and a $\sigma_i=\sigma/\sqrt{N_i}$.
Question: how can I estimate $\mu$ and $\sigma$ from the observed $X_i$s?  Note that I can't observe the variance (or any other parameters) for the partitions.
The goal is to identify partitions that do not fit the expected distribution, i.e. identify partitions where $X_i$ is too many $\sigma_i$s from $\mu$.  
 A: Note: New answer.
If you don't know the variances $Var X_1,...,VarX_m$, you have to make parametric assumptions about the distribution of the data.  For example, under the assumption of a Poisson distribution, the the maximum likelihood estimator of the variance would simply be the aggregated mean $EX$.  Under the assumption of the normal distribution, the intuitive answer
$$VarX = (1/m)\sum_i N_i (EX_i-\mu)^2$$
is the maximum likelihood estimator.
Assuming your distirbution is normal $(\mu, \sigma^2)$, then the maximum likelihood estimate of $\mu$ is the aggregated mean $(1/N)\sum_i N_i EX_i$.  The profile likelihood as a function of $\sigma^2$ is then
$$L(\sigma^2) = \prod_i \frac{1}{\sqrt{2\pi\sigma^2/N_i}} exp\left(\frac{-(EX_i - \mu)^2}{2\sigma^2/N_i}\right).$$
The maximum likelihood estimate of $\sigma^2$ is the maximizing argument of the above.  For computational convenience we take the logarithm of the likelihood as it is easier to maximize.
$$\log L(\sigma^2) = \sum_i -\frac{1}{2}\log(2\pi/N_i) - \frac{1}{2}\log(\sigma^2) - \frac{(EX_i - \mu)^2}{2\sigma^2/N_i}$$
Setting the derivative to zero,
$$0=\frac{\partial L(\sigma^2)}{\partial \sigma^2} = \sum_i -\frac{1}{2\sigma^2} + \frac{(EX_i-\mu)^2}{2/N_i}\frac{1}{\sigma^4}$$
multiplying by $\sigma^4$,
$$ 0 = \sum_i -\frac{1}{2}\sigma^2 + \frac{N_i(EX_i-\mu)^2}{2}$$
which yields the maximum likelihood estimator
$$\sigma^2 = (1/m)\sum_i N_i (EX_i-\mu)^2.$$
For general parametric $f(\theta)$ you need the densities of the convolutions $f * \cdots * f(\theta)$, which makes the problem much more difficult outside of a few special parametric families.  However if all of your $N_i$ are large, you can use the normal distribution as an approximation by the Central Limit Theorem.
A: @charles: You're misusing the word "sample" in the usual way; you should say two samples of sizes $N_1$ and $N_2$.
However (and this is the occasion for making this an "answer" rather than a comment), I think a conceptually more efficient way to think about this is to recall that $\text{var}(X) = E(\text{var}(X\mid Y)) + \text{var}(E(X \mid Y))$.  I.e. the total variance is the explained variance plus the unexplained variance.  See Law of total variance.
