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Suppose that we make measurements of an effect, and we know that the values that we obtain follow a normal distribution. But we don't know the mean nor the variance. The hypothesis is that 95% of the measurements are below a threshold $\theta \in \mathbb{R}$. So far, I have tested this with a binomial test. If I have $n$ measurements $x_1,\dots,x_n$ the binomial test that I have used so fare does not use the actual values of the $x_i$ but just the fact that they are greater or smaller than $\theta$. I was wondering if there is a "better" way which uses the actual values of the $x_i$ and maybe the fact we have a normal distribution.

Adrian
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    Welcome to CV Adrian. Your question is formulated in terms of an undefined "$x.$" If by that you mean the random variable "$X,$" then your question makes no sense because there is no test involved in the event $X\le\theta$ and the chance of that event cannot be computed since the distribution of $X$ is unknown. Perhaps you could more clearly get your question across by telling us what your actual statistical problem is, rather than attempting to state it abstractly and mathematically. – whuber Nov 15 '23 at 12:18
  • Is the question if there is a more powerful test than testing the proportion of times your observations exceed $\theta?$ – Dave Nov 15 '23 at 12:19
  • @Dave Exactly I was wondering if the proper values of the $x_i$ could give any benefit for the power – Adrian Nov 15 '23 at 14:43
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    Yes, the Normality assumption permits tests that are more powerful than the nonparametric Binomial test. The general approach is described at https://stats.stackexchange.com/questions/63366. Analytical formulas for the Normal tolerance limits are available in places like Hahn & Meeker, Statistical Intervals. – whuber Nov 15 '23 at 22:12
  • @whuber Thank you very much that was exactly what I was looking for. – Adrian Nov 16 '23 at 08:23

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