91Ó°ÊÓ

Assume\(k\)independently selected samples, where every sample consisting of\(n\)observations of a normally distributed variable. Sample standard deviations are\({s_1},{s_2}, \ldots ,{s_k}\), and

\(\bar s = \frac{{\sum {{s_i}} }}{k}\)

where we plot values\({s_1},{s_2}, \ldots ,{s_k}\)on the\({\rm{S}}\)Chart. The center line is on height\(\bar s\).

Upper Control Limit and Lower Control Limit for\({\rm{S}}\)control chart are

\(\begin{aligned}{l}LCL = \bar s - 3\bar s\frac{{\sqrt {1 - a_n^2} }}{{{a_n}}}\\UCL = \bar s + 3\bar s\frac{{\sqrt {1 - a_n^2} }}{{{a_n}}}\end{aligned}\)

Where LCL will be negative for\(n \le 5\). When this happens it is customary to take\({\rm{LCL}} = 0\)

Values for\({a_n}\)are given in the table on the page\(684\)for\(n = 3,4, \ldots ,8\).

\(\begin{aligned}{*{20}{c}}{\rm{n}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}&{\rm{7}}&{\rm{8}}\\{{\rm{a\_(n)}}}&{{\rm{.886}}}&{{\rm{.921}}}&{{\rm{.940}}}&{{\rm{.952}}}&{{\rm{.959}}}&{{\rm{.965}}}\end{aligned}\)

From the Exercise\({\rm{11}}\), we have that\(k = 24\)and

\(\sum\limits_{i = 1}^{24} {{s_i}} = 30.34\)

which implies that

\(\bar s = \frac{1}{{24}}\sum\limits_{i = 1}^{24} {{s_i}} = \frac{1}{{24}} \cdot 30.35 = 1.2642\)

Also, from the Exercise\({\rm{11}}\)each sample standard deviation is based on\(n = 6\)observations of the refractive index of fiber-optic cable. Therefore, we look the table and see that\({a_6} = 0.952\).

\(\begin{aligned}{*{20}{c}}{\rm{n}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}&{\rm{7}}&{\rm{8}}\\{{\rm{a\_(n)}}}&{{\rm{.886}}}&{{\rm{.921}}}&{{\rm{.940}}}&{{\rm{.952}}}&{{\rm{.959}}}&{{\rm{.965}}}\end{aligned}\)

Now we can calculate the limits

\(\begin{aligned}{l}LCL = \bar s - 3\bar s\frac{{\sqrt {1 - a_n^2} }}{{{a_n}}} = 1.2642 - 3 \cdot 1.2642 \cdot \frac{{\sqrt {1 - {{0.952}^2}} }}{{0.952}} = 0.45\\UCL = \bar s + 3\bar s\frac{{\sqrt {1 - a_n^2} }}{{{a_n}}} = 1.2642 + 3 \cdot 1.2642 \cdot \frac{{\sqrt {1 - {{0.952}^2}} }}{{0.952}} = 2.484\\CL = \bar s = 30.34.\end{aligned}\)

Since the biggest value is\({s_{12}} = 1.65\)and the smallest is\({s_{20}} = 0.75\)and the following is true

\(LCL = 0.45 < 0.75 = {s_{20}} < {s_{12}} = 1.65 < UCL = 2.484\)

all values are between the specification limits so the process seems to be in control in respect to variability.