I would like to estimate the standard deviation of a population. The data I have is the size and mean of each of $k$ independent uniform subsamples: $(n_1, \mu_1), (n_2, \mu_2), \ldots, (n_k, \mu_k)$. Here, the $i$'th subsample has $n_i$ elements, and those $n_i$ elements have a mean of $\mu_i$.
What is the best way to utilize all $2k$ values to produce the best estimate of the population standard deviation $\sigma$?
Let $\mu$ be the population mean and $\sigma$ be the population SD. You have observations $x_i$ of $k$ independent variables $X_i$ having a common mean $\mu$ and variances $\sigma^2/n_i,$ $i=1,2,\ldots, k.$
You are noncommittal about the sense of "best." To make this specific, let us find the Least Squares estimator of $\sigma.$ This is a Weighted Least Squares problem with weights $1/(1/n_i)=n_i,$ whence
$$\hat\mu = \frac{1}{n_1+n_2+\cdots+n_k}\, \left(n_1x_1 + n_2x_2 + \cdots + n_k x_k\right)$$
with residuals
$$r_i = x_i - \hat\mu$$
which entails
$$\hat\sigma^2 = \frac{1}{k-1}\, \left(n_1 r_1^2 + n_2 r_2^2 + \cdots + n_k r_k^2\right).$$
We may therefore take
$$\hat \sigma = \sqrt{\hat\sigma^2}$$
as a "best" estimate of $\sigma.$
As a check, straightforward algebra will verify that $E[\hat\mu] = \mu$ and $E[\hat\sigma^2]=\sigma^2,$ standard properties of Least Squares estimates. As another check, these calculations reproduce exactly the output of the lm function in R using the $n_i$ in its weights argument.
