I'm told that the $Var(\overline{x}) \approx Var(X)/N$, where $\overline{x}$ is the mean of $N$ elements from $X$. However, I have a data set where these variances seem to differ by a factor of about 30.
A series of Bernoulli trials is run. The input data consists of numbers of successes and attempts made in an hour. I have some $400000$ trials unevenly distributed in around $1000$ hours. I take $M$ random samples of hours, and calculate their means as sum of successes divided by sum of attempts. I expect that if $E$ is the mean of these $M$ means, then their variance will be $E(1-E)/\text{sample size}$, however the real variance is found to be much bigger.
I suspect this effect in large part due to the uneven distribution of trials in hours and the way I sample them. Is that true? If so, can I correct for it? I tried aggregating hours with few events, but it didn't seem to help. Could something else be causing it? Should I expect the real variance to be correct if I use all the data rather than random samples?