\documentstyle[12pt]{article}
\begin{document}
\begin{tabbing}
Subject: \= \kill
MEMO \\
To: \> Bob Tichy \\
From: \> Steve Verrill \\
Date: \> March 29, 1999 \\
Subject: \> The plastic lumber standard \\
\end{tabbing}
Hi Bob. I have tentatively answered your five questions.
This memo is in rough draft
form. The computer programs that illustrate it are also in rough
draft form. I encourage you to try them out, but I don't guarantee
that they are bug-free. You can access them from
http://www1.fpl.fs.fed.us/astmplast.html.
I need to turn my attention to other matters for a week. Next Monday
I will check this material one more time. If you have any questions
about this memo, please e-mail me at steve@ws13.fpl.fs.fed.us or call
me at 608-231-9375.
\section*{6.1.3 --- Mean}
\subsection*{One possible approach}
Perhaps they want something like
``The lower one-sided 75\% confidence bound on the secant modulus must
exceed 50,000 psi.'' This confidence bound is calculated as
\begin{equation}
\bar{x} - t_{n-1}(.25) \times s/\sqrt{n}
\end{equation}
where $n$ is the number of specimens tested,
$\bar{x}$ is the average modulus of these specimens,
$s$ is the sample standard deviation, and
$t_{n-1}(.25)$ is taken from a $t$ table. Some
$t_k(.25)$ and $t_k(.125)$ (used below) values are presented in Table 1.
\newpage
\begin{center}
\begin{tabular}{|c|r|r|} \hline
k & $t_k(.25)$ & $t_k(.125)$ \\ \hline \hline
1 & 1.000 & 2.414 \\ \hline
2 & .816 & 1.604 \\ \hline
3 & .765 & 1.423 \\ \hline
4 & .741 & 1.344 \\ \hline
5 & .727 & 1.301 \\ \hline
6 & .718 & 1.273 \\ \hline
7 & .711 & 1.254 \\ \hline
8 & .706 & 1.240 \\ \hline
9 & .703 & 1.230 \\ \hline
10 & .700 & 1.221 \\ \hline
15 & .691 & 1.197 \\ \hline
20 & .687 & 1.185 \\ \hline
25 & .684 & 1.178 \\ \hline
30 & .683 & 1.173 \\ \hline
50 & .679 & 1.164 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 1: $t_k(.25)$ and $t_k(.125)$ values
\end{center}
The probability of failing this test can be calculated as follows:
\[
\mbox{Prob(failure)} = \mbox{Prob} (\bar{x} - t_{n-1}(.25) \times
s/\sqrt{n} \leq 50,000)
\]
\[
= \mbox{Prob} ( \bar{x} - \mu + (\mu - 50000) \leq
t_{n-1}(.25) \times s/\sqrt{n} )
\]
\[
= \mbox{Prob} \left ( (\bar{x} - \mu + (\mu - 50000)) / (s/\sqrt{n}) \leq
t_{n-1}(.25) \right ) .
\]
This is the probability that a noncentral $t$ distribution with $n -
1$ degrees of freedom and noncentrality parameter
\[
\gamma = (\mu - 50000)\sqrt{n}/(COV \times \mu)
\]
is less than or equal to $t_{n-1}(.25)$. (Note that this is always
.75 for $\mu = 50000$ regardless of the values of $COV$ and $n$.)
There are routines that calculate the noncentral $t$ distribution. I
have incorporated them into a WWW program
(see http://www1.fpl.fs.fed.us/astmplast1.html)
so that your committee can
investigate the effects of using this criterion (the one-sided lower 75\%
confidence bound on $\mu$ must exceed 50000) on producers' and
consumers' risks. Committee members can see how the probability of
failure varies as a function of $n$,
$COV$, and $\mu$. An example of a table produced by the program is
provided as Table 2.
\vspace{.25 in}
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|} \hline
& \multicolumn{5}{c|}{Secant modulus in psi} \\ \cline{2-6}
$COV$ & 46000 & 48000 & 50000 & 52000 & 54000 \\ \hline \hline
5 & 1.000 & 0.999 & 0.750 & 0.042 & 0.000 \\ \hline
10 & 1.000 & 0.976 & 0.750 & 0.299 & 0.051 \\ \hline
15 & 0.994 & 0.938 & 0.750 & 0.450 & 0.193 \\ \hline
20 & 0.979 & 0.907 & 0.750 & 0.530 & 0.315 \\ \hline
25 & 0.961 & 0.884 & 0.750 & 0.577 & 0.401 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 2: Probability of failing the test, $n = 10$
\end{center}
\subsection*{Their approach?}
Their Table 2 suggests that they are taking a different
approach. Presumably given a coefficient of variation, they find a
sample size such that the halfwidth of a two-sided 75\% confidence interval on
the mean is 5\% of the mean. Then if the average calculated with this
sample size lies above 50,000 psi, the material is accepted.
The half width of a two-sided 75\% confidence interval on the mean is
\begin{equation}
t_{n-1}(.125) \times s/\sqrt{n} \approx t_{n-1}(.125) \times \mu \times COV /\sqrt{n}
\end{equation}
For this to be 5\% of the mean we need
\begin{equation}
t_{n-1}(.125) \times COV /\sqrt{n} = .05
\end{equation}
In Table 3 below, given a COV value I provide the $n$ value that is
needed to ensure that equation 3 holds. This table generally
agrees with Table 2 of the standard. (If one takes this approach
then the true mean modulus is within 5\% of the sample average for
75\% of the experiments that are conducted. In 12.5\% of the
experiments the true mean will lie more than 5\% below the sample
average. In 12.5\% of the experiments the true mean will lie more
than 5\% above the sample average. In 87.5\% of the experiments
conducted the true mean will exceed
$.95 \times \bar{x}$.)
\vspace{.25 in}
\begin{center}
\begin{tabular}{|c||r|r||r|r|} \hline
COV & n & halfwidth & n & halfwidth \\ \hline \hline
.05 & 3 & .0463 & 2 & .0854 \\ \hline
.10 & 7 & .0481 & 6 & .0531 \\ \hline
.15 & 14 & .0483 & 13 & .0503 \\ \hline
.20 & 23 & .0493 & 22 & .0505 \\ \hline
.25 & 35 & .0495 & 34 & .0502 \\ \hline
.30 & 49 & .0499 & 48 & .0504 \\ \hline
.35 & 67 & .0496 & 66 & .0500 \\ \hline
.40 & 86 & .0499 & 85 & .0503 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 3: Halfwidths of 75\% confidence intervals on the mean (as
fractions of the mean)
\end{center}
What effect does their criterion have on the probability of failure?
The probability of failing their test can be calculated as follows:
\[
\mbox{Prob(failure)} = \mbox{Prob} (\bar{x} \leq 50,000)
\]
\[
= \mbox{Prob} ( \bar{x} - \mu + (\mu - 50000) \leq 0)
\]
\[
= \mbox{Prob} \left ( (\bar{x} - \mu + (\mu - 50000)) / (s/\sqrt{n})
\leq 0 \right ) .
\]
This is the probability that a noncentral $t$ distribution with $n -
1$ degrees of freedom and noncentrality parameter
\[
\gamma = (\mu - 50000)\sqrt{n}/(COV \times \mu)
\]
is less than or equal to $0$. (Note that this is always .5 for $\mu =
50000$ regardless of the values of $COV$ and $n$.)
There are routines that calculate the noncentral $t$ distribution. I
have incorporated them into a WWW program
(see http://www1.fpl.fs.fed.us/astmplast2.html)
so that your committee can
investigate the effects of using this criterion ($n$ must be
determined in accord with Table 3 and then the resulting average
secant modulus must exceed 50000) on producers' and
consumers' risks. Committee members can see how the probability of
failure varies as a function of
$COV$ and $\mu$. An example of a table produced by the program is
provided as Table 4.
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|r|} \hline
& & \multicolumn{5}{c|}{Secant modulus in psi} \\ \cline{3-7}
$COV$ & $n$ & 46000 & 48000 & 50000 & 52000 & 54000 \\ \hline \hline
5 & 3 & 0.999 & 0.926 & 0.500 & 0.091 & 0.005 \\ \hline
10 & 7 & 0.989 & 0.865 & 0.500 & 0.154 & 0.025 \\ \hline
15 & 14 & 0.985 & 0.851 & 0.500 & 0.169 & 0.032 \\ \hline
20 & 23 & 0.981 & 0.841 & 0.500 & 0.178 & 0.038 \\ \hline
25 & 35 & 0.980 & 0.838 & 0.500 & 0.181 & 0.040 \\ \hline
30 & 49 & 0.979 & 0.835 & 0.500 & 0.185 & 0.042 \\ \hline
35 & 67 & 0.979 & 0.835 & 0.500 & 0.184 & 0.042 \\ \hline
40 & 86 & 0.978 & 0.833 & 0.500 & 0.186 & 0.043 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 4: Probability of failing the test
\end{center}
\section*{6.1.3 --- Fifth percentile}
The language of the standard is incorrect in regard to the fifth
percentile. Table 3 of ASTM D2915 enables one to calculate a
one-sided lower 75\% confidence bound on the fifth percentile (not
``the lower 5th percentile value at 75\% confidence'').
This
bound will lie below the true fifth percentile in 75\% of experiments
conducted.
Now let us consider the acceptance sampling problem. Perhaps they
would want a one-sided lower 75\% confidence bound on the fifth
percentile of flexural stress to exceed 1000 psi.
The one-sided lower confidence bound on the fifth percentile is
\begin{equation}
\bar{x} - k_n \times s
\end{equation}
where $\bar{x}$ is the sample mean, $s$ is the sample standard
deviation, and $k_n$ is the appropriate $k$ factor from Table 3 of
D2915. (Since they would be working at a 75\% confidence level and
would be interested in the fifth percentile, the appropriate $k$ value
would be taken from the fourth column [first element equal to 3.152]
of Table 3 of D2915.)
The probability of failing can be calculated as follows:
\[
\mbox{Prob(failure)} = \mbox{Prob} (\bar{x} - k_n \times
s \leq 1000)
\]
\[
= \mbox{Prob} (\bar{x} - \mu + (\mu - 1000) \leq k_n \times
s)
\]
\[
= \mbox{Prob} \left ( (\bar{x} - \mu + (\mu - 1000))/(s/\sqrt{n}) \leq k_n \times
\sqrt{n} \right ) .
\]
This is the probability that a noncentral $t$ distribution with $n -
1$ degrees of freedom and noncentrality parameter
\[
\gamma = (\mu - 1000)\sqrt{n}/(COV \times \mu)
\]
is less than or equal to $k_n \times \sqrt{n}$.
There are routines that calculate $k_n$ and the noncentral $t$ distribution. I
have incorporated them into a WWW program
(see http://www1.fpl.fs.fed.us/astmplast3.html)
so that your committee can
investigate the effects of using this criterion (the parametric one-sided lower 75\%
confidence bound on the fifth percentile must exceed 1000) on producers' and
consumers' risks. Committee members can see how the probability of
failure varies as a function of $n$,
$COV$, and the true fifth percentile value.
An example of a table produced by the program is
provided as Table 5.
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|} \hline
& \multicolumn{5}{c|}{Fifth percentile} \\ \cline{2-6}
$COV$ & 800 & 900 & 1000 & 1100 & 1200 \\ \hline \hline
5 & 1.000 & 1.000 & 0.750 & 0.018 & 0.000 \\ \hline
10 & 1.000 & 0.992 & 0.750 & 0.265 & 0.049 \\ \hline
15 & 0.999 & 0.955 & 0.750 & 0.452 & 0.224 \\ \hline
20 & 0.987 & 0.912 & 0.750 & 0.556 & 0.385 \\ \hline
25 & 0.960 & 0.874 & 0.750 & 0.617 & 0.496 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 5: Probability of failing the test, $n = 10$
\end{center}
\section*{6.3.1.4}
\subsection*{Original mean ``known''}
Let $\mu_1$ denote the known original mean and
$\mu_2$ denote the unknown degraded mean. Perhaps they would want
a one-sided lower 75\% confidence bound on $\mu_2$ to be greater than
$.9 \times \mu_1$.
The one-sided lower confidence bound on $\mu_2$ is
\begin{equation}
\bar{x}_2 - t_{n-1}(.25) \times s_2/\sqrt{n} .
\end{equation}
There is no required sample size here. However
\[
t_{n-1}(.25) \times s_2/\sqrt{n}
\]
is smaller for larger $n$ so a ``good'' producer will increase their
chances of passing by testing larger sample sizes.
The probability of failing can be calculated as follows:
\[
\mbox{Prob(failure)} = \mbox{Prob} (\bar{x_2} - t_{n-1}(.25) \times
s_2/\sqrt{n} \leq .9 \times \mu_1)
\]
\[
= \mbox{Prob} ( \bar{x_2} - \mu_2 + (\mu_2 - .9 \times \mu_1) \leq
t_{n-1}(.25) \times s_2/\sqrt{n} )
\]
\[
= \mbox{Prob} \left ( (\bar{x_2} - \mu_2 + (\mu_2 - .9 \times \mu_1))
/ (s_2/\sqrt{n}) \leq t_{n-1}(.25) \right ) .
\]
This is the probability that a noncentral $t$ distribution with $n -
1$ degrees of freedom and noncentrality parameter
\[
\gamma = (\mu_2 - .9 \times \mu_1)\sqrt{n}/(COV \times \mu_2)
\]
is less than or equal to $t_{n-1}(.25)$. (Note that this is always
.75 for $\mu_2 = .9 \times \mu_1$ regardless of the values of $COV$ and $n$.)
There are routines that calculate the noncentral $t$ distribution. I
have incorporated them into a WWW program
(see http://www1.fpl.fs.fed.us/astmplast4.html)
so that your committee can
investigate the effects of using this criterion (the one-sided lower 75\%
confidence bound on $\mu_2$ must exceed $.9 \times \mu_1$) on producers' and
consumers' risks. Committee members can see how the probability of
failure varies as a function of $n$,
$COV$, and the reduction factor $r = \mu_2/\mu_1$.
An example of a table produced by the program is
provided as Table 6.
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|} \hline
& \multicolumn{5}{c|}{Reduction factor} \\ \cline{2-6}
$COV$ & .86 & .88 & .90 & .92 & .94 \\ \hline \hline
5 & 1.000 & 0.982 & 0.750 & 0.247 & 0.024 \\ \hline
10 & 0.983 & 0.917 & 0.750 & 0.498 & 0.257 \\ \hline
15 & 0.950 & 0.874 & 0.750 & 0.588 & 0.416 \\ \hline
20 & 0.919 & 0.848 & 0.750 & 0.631 & 0.504 \\ \hline
25 & 0.895 & 0.831 & 0.750 & 0.657 & 0.557 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 6: Probability of failing the test, $n = 10$
\end{center}
\subsection*{Original mean estimated}
In this case one would want the one-sided lower 75\% confidence bound on
$\mu_2 - .9 \times \mu_1$ to exceed 0.
The approximate (approximate because we are pooling the sums of
squares and if the $COV$ is constant the standard deviations won't be
constant) one-sided lower 75\%
confidence bound on $\mu_2 - .9 \times \mu_1$ is
\begin{equation}
\bar{x}_2 - .9 \bar{x}_1 - t_{n_1 + n_2 - 2}(.25) \times s_{pooled} \times
\sqrt{1/n_2 + .9^2/n_1}
\end{equation}
where $n_1$ is the sample size for the undegraded sample, $\bar{x}_1$
and $s_1$ are the corresponding sample mean and standard deviation,
$n_2$ is the sample size for the degraded sample, $\bar{x}_2$
and $s_2$ are the corresponding sample mean and standard deviation,
\[
s_{pooled}^2 = \left ( (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 \right )/(n_1
+ n_2 - 2),
\]
and $t_{n_1 + n_2 - 2}(.25)$ is the appropriate value from a $t$ table.
Again there is no required sample size here. And again
\[
t_{n_1 + n_2 - 2}(.25) \times s_{pooled} \times
\sqrt{1/n_2 + .9^2/n_1}
\]
is smaller for larger $n$ so a ``good'' producer will increase their
chances of passing by testing larger sample sizes.
The probability of failing can be approximated as follows:
\[
\mbox{Prob(failure)} = \mbox{Prob}
( \bar{x}_2 - .9 \bar{x}_1 - t_{n_1 + n_2 - 2}(.25) \times s_{pooled} \times
\sqrt{1/n_2 + .9^2/n_1} \leq 0 )
\]
\[
= \mbox{Prob}
\left ( ( \bar{x}_2 - .9 \bar{x}_1 - (\mu_2 - .9\mu_1) + (\mu_2 -
.9\mu_1) ) / (s_{pooled} \times
\sqrt{1/n_2 + .9^2/n_1} ) \leq t_{n_1 + n_2 - 2}(.25) \right ) .
\]
This is the probability that a noncentral $t$ distribution with $n_1 + n_2 -
2$ degrees of freedom and noncentrality parameter
\[
\gamma = (\mu_2 - .9\mu_1)/ \left ( COV \times \mu \times
\sqrt{1/n_2 + .9^2/n_1} \right )
\]
is less than or equal to $t_{n_1 + n_2 - 2}(.25)$ where $\mu = (\mu_1
+ \mu_2)/2$. (Note that this is always
.75 for $\mu_2 = .9\mu_1 $ regardless of the values of $COV$, $n_1$,
and $n_2$.)
There are routines that calculate the noncentral $t$ distribution. I
have incorporated them into a WWW program
(see http://www1.fpl.fs.fed.us/astmplast5.html)
so that your committee can
investigate the effects of using this criterion (the one-sided lower 75\%
confidence bound on $\mu_2 - .9\mu_1$ must exceed 0) on producers' and
consumers' risks. Committee members can see how the probability of
failure varies as a function of $n_1$, $n_2$,
$COV$, and the reduction factor, $r = \mu_2/\mu_1$.
An example of a table produced by the program is
provided as Table 7.
\begin{center}
\begin{tabular}{|r|r|r|r|r|r|} \hline
& \multicolumn{5}{c|}{Reduction factor} \\ \cline{2-6}
$COV$ & .86 & .88 & .90 & .92 & .94 \\ \hline \hline
5 & 0.996 & 0.952 & 0.750 & 0.383 & 0.105 \\ \hline
10 & 0.953 & 0.879 & 0.750 & 0.575 & 0.386 \\ \hline
15 & 0.911 & 0.843 & 0.750 & 0.637 & 0.513 \\ \hline
20 & 0.880 & 0.822 & 0.750 & 0.667 & 0.576 \\ \hline
25 & 0.859 & 0.809 & 0.750 & 0.684 & 0.614 \\ \hline
\end{tabular}
\end{center}
\begin{center}
Table 7: Probability of failing the test, $n_1 = n_2 = 10$
\end{center}
\end{document}