I have set of random values with the same distribution $y_1, \ldots, y_N$

$$T = \frac{1}{N}\sum_{j = 1}^{N} y_j$$

I want to find the confidence interval for the mathematical expectation $E(Y)$

I can use an approximation:

$T∼N(μ,\frac{\sigma^2}{N})$, where $\sigma=\frac{\sum^N_{i=1}(y_i−T)^2}{N−1}$.

This approximation is valid asymptotically. Then my confidence interval is $T±1.96 \frac{\sigma}{\sqrt{N}}$.

Of course, this method has a mistake, because I only have an approximate Normal distribution and must use the sample variance.

How can I deal with this inaccuracy and make a final confidence interval?

  • 1
    $\begingroup$ Replacing $\sigma^2$ by the sample variance is a good approximation if n is large. The convergence rate relies on additional assumptions over moments of higher order. You can also obtain a nonparametric bootstrap interval for the mean by resampling with replacement and obtaining the corresponding means and compare it with the CLT approximation. $\endgroup$
    – user10525
    May 27, 2012 at 8:49

1 Answer 1


Procrastinator is right. There is no special way to make an approximate CI exact. to get an exact interval you would need to use the exact distribution for the sample mean based on an assumed distribution for the population. But there is a theorem about the order of accuracy of the normal approximation. It is called the Berry-Esseen Theorem. You can find it in any book on advanced probability.

  • 2
    $\begingroup$ The good ol' Berry-Esseen Theorem - Now on Wikipedia $\endgroup$
    – Dason
    May 27, 2012 at 17:44

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