Calculate standard deviation with only the averages from a number of samples I am trying to estimate the standard deviation of a population based on 100 samples each with 1,000,000 data points. I have been told it is acceptable to do this using only the average from each of the 100 samples (rather than using all 100,000,000 data points), but it seems to me that this would be quite inaccurate.
My question is, is this a reasonable way to estimate the population standard deviation and, if so, what is the correct formula to use? Are there any adjustments that need to be made?
Thanks in advance!
 A: To proceed in general, suppose we have $n$ samples that each contain $m$ values, generated as IID variables from an underlying superpopulation with mean parameter $\mu$ and standard deviation parameter $\sigma$.  Now, suppose we define the quantities:
$$\bar{X}_k \equiv \frac{1}{m} \sum_{j=1}^m X_{k,j}
\quad \quad \quad 
\bar{X}_* \equiv \frac{1}{n} \sum_{i=1}^n \bar{X}_k
\quad \quad \quad 
S_{*}^2 \equiv \frac{1}{n-1} \sum_{i=1}^n (\bar{X}_k - \bar{X}_*)^2.$$
The values $\bar{X}_k$ are the sample means of the individual samples $k=1,...,n$ and the values $\bar{X}_*$ and $S_*^2$ are the sample mean and sample variance of the set of sample means.  Using standard moment rules, it can easily be shown that:
$$\mathbb{E}(\bar{X}_k) = \mu
\quad \quad \quad \quad \quad 
\mathbb{V}(\bar{X}_k) = \frac{\sigma^2}{m}
\quad \quad \quad \quad \quad 
\mathbb{E}(S_*^2) = \frac{\sigma^2}{m}.$$
Consequently, you can obtain an estimator for the standard deviation parameter as $\hat{\sigma} = \sqrt{m} \cdot S_*$.  This estimator is biased, but it can be bias-corrected if you want (see here), based on the normality of the sample means (by appeal to the CLT).  The bias-corrected estimator is:
$$\hat{\sigma}_\text{UB} = \sqrt{\frac{(n-1)m}{2}} \cdot \frac{\Gamma(\tfrac{n-1}{2})}{\Gamma(\tfrac{n}{2})} \cdot S_*.$$
In your problem you have $n=100$ and $m=1000000$ which gives the unbiased estimator:
$$\hat{\sigma}_\text{UB} = \sqrt{49500000} \cdot \frac{\Gamma(49.5)}{\Gamma(50)} \cdot S_* = 1002.528 \cdot S_*.$$
