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 means

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

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

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

where

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

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

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

From the exercise, we need to subtract the\({22^{{\rm{nd }}}}\)values from the sum which are\(98.34\)for the sample mean and\(1.6\)for the sample standard deviation. Since we deleted one value, we need to divide with one less\(k\)(\(k = 24 - 1 = 23\)now). Therefore, we have

\(\begin{aligned}{l}\bar \bar x = \frac{1}{{23}} \cdot (2317.07 - 98.34) = 96.47\\\bar s = \frac{1}{{23}} \cdot (30.34 - 1.6) = 1.264.\end{aligned}\)

From the table given in the book, we find that

\({a_6} = 0.952\)

Finally, the Control Limits Based on the Sample Standard Deviations

\(\begin{aligned}{l}LCL = \bar \bar x - 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }} = 96.47 - 3 \cdot \frac{{1.264}}{{0.952\sqrt 6 }} = 96.47 - 1.61 = 94.86\\UCL = \bar \bar x + 3 \cdot \frac{{\bar s}}{{{a_n}\sqrt n }} = 96.47 - 3 \cdot \frac{{1.264}}{{0.952\sqrt 6 }} = 96.47 + 1.61 = 98.08\end{aligned}\)

All values are between the control limits, therefore the process is in control now.