91Ó°ÊÓ

In the exercise is suggested an alternative chart for controlling process variation involves plotting the sample variances and using the control limits given in the exercise. The data is given in Exercise\(11\), which is\(k = 24\),\(n = 6\)and\({s_i},i = 1,2, \ldots ,24\). We need to find square of all\({s_i}\)and find the average value.

\(\begin{aligned}{*{20}{c}}{{\rm{ Day }}}&{s_i^2}&{{\rm{ Day }}}&{s_i^2}\\1&{1.69}&{13}&{1.64}\\2&{0.77}&{14}&{1.30}\\3&{2.04}&{15}&{1.88}\\4&{2.53}&{16}&{1.96}\\5&{2.31}&{17}&{2.50}\\6&{1.61}&{18}&{0.96}\\7&{1.35}&{19}&{1.46}\\8&{0.62}&{20}&{0.56}\\9&{2.19}&{21}&{1.80}\\{10}&{0.64}&{22}&{2.56}\\{11}&{2.02}&{23}&{1.49}\\{12}&{2.72}&{24}&{1.39}\\{}&{}&{}&{}\end{aligned}\)

Therefore, we can calculate

\(\overline {{s^2}} = \frac{1}{k}\sum\limits_{i = 1}^k {s_i^2} = \frac{1}{{24}}\sum\limits_{i = 1}^{24} {s_i^2} = \frac{1}{{24}} \cdot 39.9944 = 1.6664\)

Everything else we need is given in the exercise, therefore

\(\begin{aligned}{l}LCL = \overline {{s^2}} \cdot \chi _{0.999,n - 1}^2 \cdot \frac{1}{{n - 1}} = 1.6664 \cdot 0.210 \cdot \frac{1}{{6 - 1}} = 0.07;\\UCL = \overline {{s^2}} \cdot \chi _{0.001,n - 1}^2 \cdot \frac{1}{{n - 1}} = 1.6664 \cdot 20.515 \cdot \frac{1}{{6 - 1}} = 6.837;\\CL = \frac{{0.07 + 6.387}}{2} = 3.228.\end{aligned}\)

Now we have everything to construct the corresponding chart for the data of Exercise \(11\) . As we can see from the control chart, every point between the control limits\((LCL\) and \(UCL)\)therefore the process seems to be in control.