I am trying to compare two different methods of estimating confidence intervals: a parametric approach that uses the assumption that the sample is t-distributed (i.e. the formulas that are given here: Wikipedia: Normal Distribution), and bootstrapping.

The procedure is rather simple: For every sample size $5 < N <200$, I generate a sample with $N$ normally distributed random numbers. I then calculate the confidence interval using the parametric formulas for every such sample. Then, I do the same with bootstrap: For each sample, I draw 1000 sub-samples with replacement, calculate their mean and standard deviation, sort these values, and cut off the top and bottom 2.5%, which should give me the 95%-confidence interval ($\alpha=5\%$).

For each sample size, I then plot the width of the confidence interval, both for the one that I got from the parametric approach and the bootstrap. I also calculate their difference as: [(Width of Parametric CI) - (Width of Bootstrap CI)] / (Width of Parametric CI) * 100, to get a percentage.

I would expect that for increasing sample size, the difference between the two methods vanishes. As shown in the graphic below, this is indeed the case for the mean of the sample. However for the standard deviation, the parametric model and bootstrap do not really seem to converge against the same value, or at least significantly slower than the mean.

I wonder: Is there a reason why it works for the mean, but not for the standard deviation? Did I just simply make a mistake when implementing all this (if desired, I can post the Python source code that was used to generate the graphic)?

enter image description here

Thanks in advance for all your ideas and suggestions :)


1 Answer 1


I would say that since sample mean is $\overline{X}=\frac{\sum{X_i}}{n}$, its convergence rate is 1/n, which is also the convergence rate of sample variance. But the convergence rate of sample standard deviation $ 1/\sqrt{n} $.

  • $\begingroup$ I ran a couple of more tests and it seems like this is indeed the correct answer, so thank you! :) $\endgroup$
    – der_herr_g
    Commented Oct 10, 2014 at 3:41

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