I have via the ISO standard 16269 found the solution to a problem that I've been working on. Based on a couple of independent samples from a normally distributed population, I would like to determine a limit above which all the remaining samples in the population will fall with a given confidence level.

The standard does not, however, provide sufficient underlying theory for me to get a satisfactory understanding of the statistics.

From the standard follows:

A random sample of n observations has been drawn from a normally distributed population with unknown mean $\mu$ and unknown standard deviation $\sigma$. The sample mean is $\bar{x}$ and the sample standard deviation is $s$.

For given values of $n$, $m$ and $\alpha$ the smallest factor $k$ is required such that one may have at least $100(1-\alpha)%$ confidence that none of $m$ further observations will exceed $\bar{x}+ks$.

$k$ is given by $$\int_0^\infty{g\left(s\right)\int_{-\infty}^\infty{\Phi^m\left(\bar{x}+ks\right)f\left(\bar{x}\right)d\bar{x}ds}}\geq1-\alpha $$

where $$f(\bar{x})=\sqrt{\frac{n}{2\pi}}\exp\left(-\frac{n}{2}\bar{x}^2\right), -\infty\lt\bar{x}\lt\infty $$ $$ g(s)= \frac{\nu^{\nu/2}s^{\nu-1}}{2^{(\nu/2)-1}\Gamma(\frac{\nu}{2})}\exp\left(-\nu s^2/2\right),s\geq0 $$ $$ \Phi(t)=\int_{-\infty}^t{\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}u^2\right)du} $$ and $$ \Gamma\left(\frac{\nu}{2}\right)= \int_0^\infty{x^{\frac{\nu}{2}-1}\exp\left(-x\right)dx}$$ $$ \nu=n-1 $$

I identify the resemblance with the $\chi^2$, originated from the uncertainty of the variance, as well as the normally distributed contribution from the uncertainty of the mean. But I can't get it together.

I am looking for directions/guidance to help me get a better understanding for the expression which defines $k$?

If no help is to be found here perhaps somebody can recommend a good reference on the topic? I have searched for the references in the standard, unfortunately none are available via my university.


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