Combined variance estimate for samples of varying sizes I'm working on my master thesis, and something's come up where I don't know if I'm "allowed" to do this and call it good science, I've scoured the internet to no avail of finding my answer, so I figured I'd ask you.
In short, I have a certain unknown distribution generating data, I'm trying to determine this distribution's $\sigma$ (I actually don't care about $\mu$). So what I'm attempting to do is to sample the data, and use the CLT to my advantage for this task. The thing is, these data points aren't sampled in equal size. (and sadly I can't explain why on here, NDA stuff with my master thesis...)  Sometimes I'll have 5 points in a sample, sometimes I'll have 10, sometimes 50 (more often than not, it's <15 sample sizes)
My question is this :

Am I allowed to use the CLT with variable sample sizes to determine the unknown distribution's $\sigma$?

Everywhere I look, I see that $\frac{\sigma}{\sqrt{n}}$ shows up as the sampling distribution's standard deviation, but since my $n$ is not constant, I don't know what to do.
I've actually done a Monte Carlo experiment where I sample a known distribution with random sample sizes (sample size as a RV following an exponential distribution) and the sample distribution's $\sigma$ gets close to the known distribution's $\sigma$ but not quite...
Any help is appreciated. Thanks in advance!
 A: Assuming that all samples are from populations with the same variance $\sigma^2,$ you can estimate the variance as
$$\hat\sigma^2 = \frac{\sum_{i=1}^n (r_i - 1)S_i^2}{\sum_{i=1}^n r_i\; - n},$$
where you have $n$ samples with $r_i\ge 2, i=1,\dots, n$ replications in each,
and $S_i^2$ is the sample variance of the $r_i$ replications in sample $i.$
That is $S_i^2 = \frac{1}{r_i - 1}\sum_{j=1}^{r_i}(X_{ij} - \bar X_i)^2$
and $\bar X_i = \frac{1}{r_i}\sum_{j=1}^{r_i}X_{ij}.$
If data are normal, then the relationship
$\frac{(\sum_{i=1}^n r_i\; - n)\hat\sigma}{\sigma^2}
\sim \mathsf{Chisq}(\nu),$ with $\nu = \sum_{i=1}^n r_i\; - n,$ can be used to
make a confidence interval for $\sigma^2.$
The displayed formula is essentially the formula for the denominator mean square in
an unbalanced one-way ANOVA with $n$ levels of the factor (groups) and $r_i$ replications per factor (group), where all groups are from populations with the same variance $\sigma^2.$ [It is a weighted average of the individual sample variances, where weights are degrees of freedom $r_i-1,$ not numbers $r_i$ of replications.]
