TITLE: Calculating uncertainty in standard deviation
QUESTION [9 upvotes]: I have a distribution with literally an infinite number of potential data points.  I need the standard deviation.  I generate about a hundred points and take the standard deviation of the points.  This gives a hopefully good approximation of the true standard deviation, but it won't, of course, be exact.  How do I estimate the uncertainty in the standard deviation?  This seems like a very basic question, but web searching hasn't provided any solution.  If I missed it somehow, my apologies.

REPLY [2 votes]: This is pretty standard and can be answered by searching "Confidence interval of a standard deviation." Here are the steps:
Step 1) Pick a confidence level. The confidence level is the probability of your interval estimate containing the actual population standard deviation. Common choices for confidence levels are 90%, 95%, 99%. I'll work through the steps for a 90% confidence interval.
Step 2) Use a chi-squared distribution to find the left and right critical values $\chi^2_L, \chi^2_R$ for your chosen confidence level. The degrees of freedom are the sample size minus one, in this case, $99$. For your example, the critical values for 90% confidence would be approximately $\chi^2_L = 77.93$, $\chi^2_R = 124.32$
Step 3) Use your sample standard deviation $s$ and sample size $n$ to find the left and right endpoints of the confidence interval for the population standard deviation $\sigma$ via the formula: $$s\sqrt{ \frac{n-1}{\chi^2_R}} < \sigma < s\sqrt{ \frac{n-1}{\chi^2_L}}.$$ In your example, whatever your value for $s$ was, you can be 90% confident that the true value of $\sigma$ is between $s \sqrt{ \frac{99}{124.32}} = 0.892s$ on the low end, and $s \sqrt{ \frac{99}{77.93}} = 1.127s$ on the high end.