Say we have a normal distribution with unknown mean and variance and we take $n$ samples from it. Then, the usual confidence interval for the mean uses the $t$-distribution. But if we know that the variance is less than, say, 100, can we use that knowledge to make a smaller confidence interval for the mean?

  • $\begingroup$ You may be able to get around the problem by adopting a Bayesian approach. You could assign (say) $\sigma^2 \sim U(0,100)$ and then assign a flat prior to $\mu$. However, I suspect the impact of any approach will depend on how close the true $\sigma^2$ is to the upper limit $\endgroup$
    – jcken
    Jul 27, 2021 at 10:27
  • $\begingroup$ What sample size? $\endgroup$
    – BruceET
    Jul 27, 2021 at 14:25
  • $\begingroup$ Related: stats.stackexchange.com/questions/517043/… ; math.stackexchange.com/questions/4158937/… $\endgroup$
    – Peter O.
    Jul 30, 2021 at 6:15

1 Answer 1


It should be possible, but the benefit may not be worth the effort. A Wald confidence interval does not account for the sampling variability of the estimated standard error, essentially treating it as known. Inverting this test produces a confidence interval very similar to inverting the t-interval. Any interval you construct that accounts for the sampling variability of the standard error, while knowing that the unknown true population variance is less than 100, will be wider than a Wald interval and shorter than a t-interval. Let me know if I have made any mistakes.


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