91Ó°ÊÓ

With\({\bar x_1},{\bar x_2}, \ldots ,{\bar x_k}\)denoting the\(k\)calculated sample means, the usual estimate of\(\mu \)is simply the average of these me

\(\hat \mu = \bar \bar x = \frac{1}{k}\sum\limits_{i = 1}^k {{{\bar x}_i}} \)

When \({\bar X_1},{\bar X_2}, \ldots ,{\bar X_m}\)is a random sample from a normal distribution, it can be shown that

\(E(S) = {a_n} \cdot \sigma \)

were

\({a_n} = \frac{{\sqrt 2 \Gamma (n/2)}}{{\sqrt {n - 1} \Gamma ((n - 1)/2)}}\)

and\(\Gamma ( \cdot )\)denoted the gamma function. There is a table of\({a_n}\)for\(n = 3,4, \ldots ,8\). Let

\(\bar S = \frac{1}{k}\sum\limits_{i = 1}^k {{S_i}} \)

where\({S_1},{S_2}, \ldots ,{S_k}\)are the sample standard deviations for the\(k\)samples. The

\(\hat \sigma = \frac{{\bar S}}{{{a_n}}}\)

is an unbiased estimator of\(\sigma \).

Control Limits Based on the Sample Standard Deviations

\(\begin{aligned}{l}LCL = \bar \bar x - 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }}\\UCL = \bar \bar x + 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }}\end{aligned}\)

where

\(\begin{aligned}{l}\bar s = \frac{1}{k}\sum\limits_{i = 1}^k {{s_i}} \\\bar \bar x = \frac{1}{k}\sum\limits_{i = 1}^k {{{\bar x}_i}} \end{aligned}\)

Using this, we calculate (we have all values given in the table in exercise, we use the formula above to calculate it)

\(\begin{aligned}{l}\bar \bar x = \frac{1}{{22}}(12.72 + 12.80 + \ldots + 12.94) = 12.95\\\bar s = \frac{1}{{22}}(0.536 + 0.339 + \ldots + 0.541) = 0.526\\{a_5} = 0.940{\rm{ from the table given in the book page }}684.\end{aligned}\)

The Control Limits Based on the Sample Standard Deviations\(\begin{aligned}{l}LCL = \bar \bar x - 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }} = 12.95 - 3 \cdot \frac{{0.526}}{{0.940\sqrt 5 }} = 12.95 - 0.75 = 12.2\\UCL = \bar \bar x + 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }} = 12.95 + 3 \cdot \frac{{0.526}}{{0.940\sqrt 5 }} = 12.95 + 0.75 = 13.7\end{aligned}\)