I have a number of samples $X_1,\ldots,X_n $ from Gaussian distributions centered at $\mu_1,\ldots,\mu_n $ respectively.

I need to construct a confidence interval for the biggest difference between means $\max_{1\leq i<j\leq n}|\mu_i-\mu_j|.$

How can this be done? Probably that interval should be wider than the ordinary t-test confidence interval for the difference between two means (due to selection bias), but tighter than, for example, confidence interval based on Tukey's range test (since only two of all the means are needed to be compared).

This seems like a textbook question, would you please suggest where should I look for the answer? Thanks!

Update. The task which I think might be solved with this confidence interval is the following.

In a framework of equivalence testing the ordinary hypotheses $H_0\colon\mu_1=\mu_2 $ and the alternative $H_1\colon\mu_1\neq\mu_2 $ are usually swapped, resulting in null like $H_0\colon|\mu_1-\mu_2|\geq \varepsilon $ against the alternative $H_1\colon|\mu_1-\mu_2|<\varepsilon $ (see, for example, "Testing Statistical Hypotheses of Equivalence"). The problem with this approach is that $\varepsilon $ must be specified beforehand, while sometimes it would be more convenient to report something like the smallest $\varepsilon $ at which the null hypotheses of nonequivalence is rejected.

For two-sample problem confidence interval for the difference $|\mu_1-\mu_2| $ provides a reliable alternative to equivalence testing: as far as I understood, $\varepsilon $ might be chosen as the smallest meaningful $\varepsilon $ such as $[-\varepsilon, \varepsilon] $ contains that interval.

For multiple samples the situation is not clear to me. Sometimes it is suggested to construct CIs for all pairwise differences, but if we want all of them to be simultaneously valid, they must be enlarged with, say for simplicity, Bonferroni correction. But we actually don't need all of CIs to hold, because to claim the equivalence of all the means only $\max_{1\leq i<j\leq n}|\mu_i-\mu_j|<\varepsilon $ is needed.

And, of course, the assumption of normality is not really desired, so it would be perfect to get rid of it in the end, but it might be kept for a start.


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