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Sorry, my stats is rusty and I'm failing to find a quick answer to what is likely to be a very basic question.

If I take n measurements of a property that has an inherent mean mu and standard deviation s, I would estimate mu from the sample mean, and quantify the uncertainty in this estimate with the standard error. But what about the standard deviation? (I guess this is sort of a population s.d., though of a potentially infinite population) Presumably I'd estimate it from the sample standard deviation, but what is the right way to quantify its uncertainty? Clearly a value based upon 5 values is less reliable than one using 500.

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  • $\begingroup$ @DeltaIV Thanks, that may well be what I'm after though I'm not familiar with the terms given. Clearly I need to hit the textbooks more. $\endgroup$ – beldaz Aug 14 '16 at 11:36
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Its common to assume that the distribution-variance of the sample and that of the entire population are similar, so no surprise you didn't find an answer quickly.

None the less you can use this formula to calculate the CI of the standard deviation(as a factor of your sample size)

CI:=[SDSQRT((n-1)/CHIINV((alpha/2), n-1)), SDSQRT((n-1)/CHIINV(1-alpha/2), n-1))]

Where alpha is your confidence target and CHIINV(CHI Inverse Function) returns the one-tailed chi-statistic given a target probability.

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  • $\begingroup$ Thanks. I haven't come across this approach to qualifying things (I'm a lapsed physicist, and stats was a bit cursory). I'd assumed there was an equivalent to the standard error, but as you suggest this probably isn't normally required since sample variance probably converges rapidly to population's for any reasonable sample size. $\endgroup$ – beldaz Aug 14 '16 at 11:43
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    $\begingroup$ @DaFanat, in the OP of stats.stackexchange.com/questions/156518/…, its mentioned that SE_s=(SE_s^2)^0.5, where SE_s^2=(2sigma^4/(N-1))^0.5. I am wondering why there is a difference between the two formulas? Or do they lead to the same values? $\endgroup$ – alpha_989 Sep 16 '17 at 22:36

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