Prediction vs. tolerance interval I want to predict a factor and also get an upper bound for my prediction. It seems that what I want is prediction interval. but prediction interval only specifies the bound for one next prediction not all. Recently I found tolerance interval definition. Can I use it like prediction interval for future samples?
 A: Prediction intervals can indeed provide a interval for multiple forthcoming observations. The most common usage of the prediction interval is however when the number of forthcoming observations is one. 
For the normal distribution this is just a special case of the more general version with multiple forthcoming observations. As is seen in the standard ISO standard 16269.
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 
  $$ \Gamma\left(\frac{\nu}{2}\right)= \int_0^\infty{x^{\frac{\nu}{2}-1}\exp\left(-x\right)dx}$$ 
  $$ \nu=n-1 $$

