Here is a form that implements the theory described below.

For a complete discussion of tolerance regions see
Irwin Guttman's *Statistical Tolerance Regions: Classical and
Bayesian*, 1970, Hafner Publishing Company, Darien, Connecticut.
Here we work through the calculations needed to obtain a one-sided,
lower, 95% content, 75% confidence tolerance region for a normal
distribution.

For a one-sided, lower, 75% confidence bound on the fifth percentile of a normal
distribution (which corresponds to a one-sided, lower, 95% content, 75% confidence
tolerance region) we have
$$.75=\mathrm{Prob}(\backslash bar\{x\}-ks\mu -1.645\sigma )$$
$$=\mathrm{Prob}(\backslash bar\{x\}-\mu +1.645\sigma ks)$$
$$=\mathrm{Prob}\left(\right(\backslash bar\{x\}-\mu +1.645\sigma )/sk)$$
$$=\mathrm{Prob}\left(\right(\backslash bar\{x\}-\mu +1.645\sigma )/(s/\sqrt{n}\left)k\sqrt{n}\right)$$
$$=\mathrm{Prob}\left(\right[(\backslash bar\{x\}-\mu +1.645\sigma )/(\sigma /\sqrt{n})]/(s/\sigma \left)k\sqrt{n}\right)$$
$$=\mathrm{Prob}\left(\backslash mbox\{N\}\right(1.645\sqrt{n},1)/\sqrt{\chi \_n-1^2/(n-1)}k\sqrt{n})$$
$$=\mathrm{Prob}\left(\mathrm{NCT}\right(1.645\sqrt{n},n-1\left)k\sqrt{n}\right),$$
or
$$\mathrm{NCT}^-1(1.645\sqrt{n},n-1)\left(.75\right)=k\sqrt{n},$$
$$\mathrm{NCT}^-1(1.645\sqrt{n},n-1)\left(.75\right)/\sqrt{n}=k.$$
This last equation can be used to calculate the exact value of
$k$
provided that one has access to a noncentral T inverse routine.
(Here
$\mathrm{NCT}(1.645\sqrt{n},n-1)$
denotes a noncentral T
distribution with noncentrality parameter
$1.645\sqrt{n}$
and
$n-1$
degrees of freedom.)

FORTRAN and C code to calculate the noncentral
T inverse can be found in the DCDFLIB library of probability
distribution functions. DCDFLIB is a
public domain library of ``routines for cumulative distribution
functions, their inverses, and their parameters.'' It was produced
by Barry Brown, James Lovato, and
Kathy Russell of the Department of Biomathematics,
M.D. Anderson Cancer Center, The University of Texas.
DCDFLIB can be found
at
http://odin.mdacc.tmc.edu/anonftp/ .

For a one-sided, lower, 75\% confidence bound on the fifth percentile of a normal distribution (which corresponds to a one-sided, lower, 95\% content, 75\% confidence tolerance region) we have \[ .75 = \mbox{Prob}(\bar{x} - ks \leq \mu - 1.645 \sigma) \] \[ = \mbox{Prob}(\bar{x} - \mu + 1.645 \sigma \leq ks) \] \[ = \mbox{Prob}((\bar{x} - \mu + 1.645 \sigma)/s \leq k) \] \[ = \mbox{Prob}((\bar{x} - \mu + 1.645 \sigma)/(s/\sqrt{n}) \leq k\sqrt{n}) \] \[ = \mbox{Prob}([(\bar{x} - \mu + 1.645 \sigma)/(\sigma/\sqrt{n})]/(s/\sigma) \leq k\sqrt{n}) \] \[ = \mbox{Prob}(\mbox{N}(1.645\sqrt{n},1)/\sqrt{\chi_{n-1}^2/(n-1)} \leq k\sqrt{n}) \] \[ = \mbox{Prob}(\mbox{NCT}(1.645\sqrt{n},n-1) \leq k\sqrt{n}) , \] or \[ \mbox{NCT}^{-1}(1.645 \sqrt{n},n-1)(.75) = k \sqrt{n} , \] \[ \mbox{NCT}^{-1}(1.645 \sqrt{n},n-1)(.75)/\sqrt{n} = k . \] This last equation can be used to calculate the exact value of $k$ provided that one has access to a noncentral T inverse routine. (Here $\mbox{NCT}(1.645\sqrt{n},n-1)$ denotes a noncentral T distribution with noncentrality parameter $1.645\sqrt{n}$ and $n-1$ degrees of freedom.)FORTRAN and C code to calculate the noncentral T inverse can be found in the DCDFLIB library of probability distribution functions. DCDFLIB is a public domain library of ``routines for cumulative distribution functions, their inverses, and their parameters.'' It was produced by Barry Brown, James Lovato, and Kathy Russell of the Department of Biomathematics, M.D. Anderson Cancer Center, The University of Texas. DCDFLIB can be found at http://odin.mdacc.tmc.edu/anonftp/ .

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Last modified on 4/1/2003.

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