91Ó°ÊÓ

Let\(p\)denote the probability that a single sample yields an\(\bar x\)value outside the control limits; that is,

\(p = P\left( {\bar X < {\mu _0} - 3 \cdot \sigma /\sqrt n {\rm{ or }}\bar X > {\mu _0} + 3 \cdot \sigma /\sqrt n } \right).\)

Define random variable\(Y\)by

\(Y\)the first\(i\)for which\({\bar X_i}\)falls outside the control limits

If we think of each sample number as a trial and an out-of-control sample as a success, then\(Y\)is the number of (independent) trials necessary to observe a success. This\(Y\)has a geometric distribution, and it can be shown that

\(E(Y) = \frac{1}{p}.\)

The acronym ARL (for average run length) is often used in place of\({\rm{E}}(Y)\). Because\(p = \alpha \)a for an in-control process, we have

\({\rm{ARL}} = E(Y) = \frac{1}{p} = \frac{1}{\alpha }\)

Therefore, using the control limits

\(\begin{aligned}{l}{\mu _0} + 2.81\frac{\sigma }{{\sqrt n }}\\{\mu _0} - 2.81\frac{\sigma }{{\sqrt n }}\end{aligned}\)

our\(p\)from above is

\(p = P\left( {\bar X < {\mu _0} - 2.81 \cdot \sigma /\sqrt n {\rm{ or }}\bar X > {\mu _0} + 2.81 \cdot \sigma /\sqrt n } \right)\)

From which, when we standardize it, we get

\(\begin{aligned}{l}p = P\left( {\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt n }} < - 2.81{\rm{ or }}\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt n }} > 2.81} \right)\\ = 2 \cdot (0.5 - \Phi (2.81)) = = 2 \cdot (0.5 - 0.4975) = 0.005\end{aligned}\)

The acronym\({\rm{ARL}}\)(for average run length) is

\({\rm{ARL}} = \frac{1}{p} = \frac{1}{{0.005}} = 200\)

\(\alpha = P\left( {} \right.\)a single sample gives a point outside the control limits when\(\left. {\mu = {\mu _1}} \right) = \)\( = P\left( {\bar X > {\mu _1} + 2.81 \cdot \frac{\sigma }{{\sqrt n }}} \right.or\bar X < {\mu _1} - 2.81 \cdot \frac{\sigma }{{\sqrt n }}when\mu = P\left( {\frac{{\bar X - {\mu _1}}}{{\sigma /\sqrt n }} > 2.81} \right.or\frac{{\bar X - {\mu _1}}}{{\sigma /\sqrt n }} < - 2.81\) \(when\left. {\mu = {\mu _1}} \right)\)

Using\({\mu _1} = {\mu _0} + \sigma \)and\(n = 4\)we have

\(P\left( {\frac{{\bar X - {\mu _0} - \sigma }}{{\sigma /\sqrt 4 }} > 2.81{\rm{ or }}\frac{{\bar X - {\mu _0} - \sigma }}{{\sigma /\sqrt 4 }} < - 2.81} \right)\)

We now standardize by adding\(\sqrt 4 \)from each term inside the parentheses\(\begin{aligned}{l}P\left( {\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 4 }} > 2.81 + \sqrt 4 {\rm{ or }}\frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 4 }} < - 2.81 + \sqrt 4 } \right) = \\ = P(Z > 2.81 + \sqrt 4 {\rm{ or }}Z < - 2.81 + \sqrt 4 ) = \\ = P(Z > 4.81{\rm{ or }}Z < - 0.81) = \\ = (0.5 - \Phi (0.81)) + (0.5 - \Phi (4.81)) = \\ = 0.5 - 0.2910 + 0.5 - 0.5 = 0.209\end{aligned}\)

The acronym\({\rm{ARL}}\)(for average run length) is

\({\rm{ARL}} = \frac{1}{\alpha } = \frac{1}{{0.209}} = 4.78\)

For the\(3\)-sigma chart we have

\(\begin{aligned}{l}LCL = {\rm{ lower control limit }} = \mu - 3 \cdot \frac{\sigma }{{\sqrt n }}\\UCL = {\rm{ upper control limit }} = \mu + 3 \cdot \frac{\sigma }{{\sqrt n }}\end{aligned}\)

It follows that

\(P\left( {\mu - 3{\sigma _{\bar X}} \le \bar X \le \mu + 3{\sigma _{\bar X}}} \right) = P( - 3 \le Z \le 3) = 0.9974\)

where\(Z\)is a standard normal random variable. Or what we need

\(p = 1 - 0.9974 = 0.0026\)

From this

\({\rm{ARL}} = \frac{1}{{0.0026}} = 385\)

for an in control process

Suppose that, at a particular time point, the process mean shifts to\(\mu = {\mu _0} + \Delta \sigma \). If we define

\(Y = \)the first\(i\)subsequent to the shift for which a sample generates an out-of-control signal,

it is again true that

\({\rm{ARL}} = E(Y) = \frac{1}{p}\)

but now

\(p = 1 - \beta \)

There is table that gives selected ARLs for a 3-sigma chart when\(n = 4\). However, we can calculate the probability of an out of control point

\(\begin{aligned}{l}p = 1 - P\left( { - 3 - \sqrt 4 < \frac{{\bar X - {\mu _0}}}{{\sigma /\sqrt 4 }} < - 3 + \sqrt 4 } \right) = \\ = 1 - P( - 5 < Z < 1) = 1 - (\Phi (1) + \Phi (5)) = \\ = 0.1587\end{aligned}\)

Therefore, the ARL is

\({\rm{ARL}} = \frac{1}{{0.1587}} = 6.3\)