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I have a series of simulation results and I would like to demonstrate how the confidence interval for the mean and standard deviation decrease as the number of results increases.

I have done this for the mean using the inbuilt MeanCI function as shown below (I've replaced my simulation results with random data).

Needs["HypothesisTesting`"]
data = Table[RandomReal[{-1, 1}], {i, 1, 100}];
movingData = Table[data[[1 ;; i]], {i, 2, Length[data]}];
meanCI = MeanCI[#, ConfidenceLevel -> .80] & /@ movingData;
ListPlot[Transpose[meanCI]]

enter image description here

What I would like to do is create a similar chart showing how the standard deviation confidence interval decreases. There isn't a standard function for StandardDeviationCI but is there a way to use the tools inside the HypothesisTesting package to get a similar result?

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The standard deviation confidence interval may be obtained via Sqrt@VarianceCI[sample]. If $X_1, X_2,\dots,X_n$ are iid observations from a normal distribution, then $${(n-1)\,s^2\over\sigma^2} \approx \chi^2_{n-1}\,.$$ The confidence interval for $\sigma$ or $\sigma^2$ can be computed from the sample and the $\chi^2$ distribution. This is what VarianceCI does. The sample means of a normal distribution are distributed according to a Student $t$ distribution if $\sigma$ is unknown. The confidence interval can be computed from the $t$ distribution, and this is what MeanCI does.

The OP's example data is from a uniform distribution, not a normal distribution. When the sample size is over 50, the error appears to be practically negligible (compared to the appropriate UniformSumDistribution). I don't know how to compute the exact distribution of the standard deviation, but since the sampling distribution approaches a normal distribution, I will assume that VarianceCI will be adequately approximate for larger sample sizes.

The plot below shows the sample standard deviation $s$ bracketed by the confidence interval for $\sigma$. One may notice that the confidence interval is not symmetric. That is because the \chi^2 distribution is skewed.

Needs["HypothesisTesting`"];
SeedRandom[1];
data = RandomReal[{-1, 1}, 100];
sdci = Table[
   Sqrt@VarianceCI[data[[1 ;; i]], ConfidenceLevel -> .80], {i, 2, Length[data]}];
ListPlot[Transpose[sdci] ~Append~
  Table[StandardDeviation[data[[1 ;; i]]], {i, 2, Length[data]}]]

Mathematica graphics

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  • $\begingroup$ This is similar to the approach I took the first time I looked at it. It seemed a little strange to me that the confidence interval isn't symmetrical around the standard deviation (the top band is thicker than the bottom). The other function, MeanCI/ VarianceCI produce symmetrical confidence intervals. $\endgroup$
    – Cam
    Sep 23, 2013 at 5:05
  • $\begingroup$ MeanCI produces a symmetric interval $\bar x \pm s \cdot t_{\alpha/2}/\sqrt{n}$, but VarianceCI produces an asymmetrical interval, $(n-1) s^2$ divided by $\chi^2_{\alpha/2,n-1}$ and $\chi^2_{1-\alpha/2,n-1}$. The distribution $\chi^2$ is skewed, hence an asymmetrical CI; the $t$ distribution is symmetric, hence a symmetric CI. $\endgroup$
    – Michael E2
    Sep 23, 2013 at 5:11
  • $\begingroup$ Nice,VarianceCI does indeed produce a asymmetrical interval. I think I was just a bit confused as to why it was different from MeanCI. It would also explain why there isn't a StandardDeviationCI, because it's a trivial extension to VarianceCI. $\endgroup$
    – Cam
    Sep 23, 2013 at 5:20

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