Is the Berry-Esseen theorem useful for justifying normality?

The Kolmogorov-Smirnov (KS) test tells one how confident they can be that a sample comes from a hypothesized distribution. It is my understanding that this test can be used to justify whether or not the underlying distribution of a sample is normal. If with high confidence one can say that the underlying distribution of a data source is normal then the z- or t-interval can be applied to compute confidence intervals for the mean of the data.

My question has to do with the inverse of this problem. Say we have data with a known (non-normal) distribution and want to justify under what parameters the distribution is “normal enough” to use z- and t-intervals for the mean. One way of comparing a distribution to its normal approximation that I see a lot is to compute the Kolmogorov metric between two distributions, i.e. for a random variable X and it’s normal approximation XN(E[X],var[X]) the Kolmogorov metric is
Note that the above is not referring to the KS statistic but rather the maximum discrepancy between two theoretical cdf’s. One famous result using this approach is the Berry-Esseen theorem which puts an upper bound on the Kolmogorov metric for an arbitrary distribution and it’s normal approximation.

My question is this: Is there an accepted value for the Kolmogorov metric that has been used to justify sufficient normality for the purposes of confidence intervals?

Perhaps a equivalent question would be: Does the Berry-Esseen theorem have any practical application (in an analogous way to the KS-test) for justifying normality?


No. Different confidence interval procedures have difference sensitivities to normality, and various requirements of the work can influence how much deviation from the theoretical coverage is acceptable.^{\dagger}

For instance, while confidence intervals for the mean that are based on the t-test are known for their robustness to deviations from normality, confidence intervals for variance that are based on the F-test are known for their lack of robustness to such deviations.

^{\dagger}Is 94.5\% coverage good enough for a 95\% confidence interval? Under most circumstances, I would say so, but perhaps you have an application where that is inadequate.

Source : Link , Question Author : Aaron Hendrickson , Answer Author : Dave

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