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Magazine Chapter 16: Q10E (page 689)
When installing a bath faucet, it is important to properly fasten the threaded end of the faucet stem to the water supply line. The threaded stem dimensions must meet product specifications, otherwise malfunction and leakage may occur. Authors of "Improving the Process Capability of a Boring Operation by the Application of Statistical Techniques" (Intl. J. Sci. Engr. Research, Vol. 3, Issue\(5\), May\(2012\)) investigated the production process of a particular bath faucet manufactured in India. The article reported the threaded stem diameter (target value being\(13\;mm\)) of each faucet in\(25\)samples of size\(4\)as shown here:
\(\begin{aligned}{*{20}{c}}{Subgroup}&{x\_(1)}&{x\_(2)}&{x\_(3)}&{x\_(4)}\\1&{13.02}&{12.95}&{12.92}&{12.99}\\2&{13.02}&{13.10}&{12.96}&{12.96}\\3&{13.04}&{13.08}&{13.05}&{13.10}\\4&{13.04}&{12.96}&{12.96}&{12.97}\\5&{12.96}&{12.97}&{12.90}&{13.05}\\6&{12.90}&{12.88}&{13.00}&{13.05}\\7&{12.97}&{12.96}&{12.96}&{12.99}\\8&{13.04}&{13.02}&{13.05}&{12.97}\\9&{13.05}&{13.10}&{12.98}&{12.96}\\{10}&{12.96}&{13.00}&{12.99}&{12.99}\\{11}&{12.90}&{13.05}&{12.98}&{12.88}\\{12}&{12.96}&{12.98}&{12.97}&{13.02}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{ }&{}&{}&{}&{}\\{13}&{13.00}&{12.96}&{12.99}&{12.90}\\{14}&{12.88}&{12.94}&{13.05}&{13.00}\\{15}&{12.96}&{12.96}&{13.04}&{12.98}\\{16}&{12.99}&{12.94}&{13.00}&{13.05}\\{17}&{13.05}&{13.02}&{12.88}&{12.96}\\{18}&{13.08}&{13.06}&{13.10}&{13.05}\\{19}&{13.02}&{13.05}&{13.04}&{12.97}\\{20}&{12.96}&{12.90}&{12.97}&{13.05}\\{21}&{12.98}&{12.99}&{12.96}&{13.00}\\{22}&{12.97}&{13.02}&{12.96}&{12.99}\\{23}&{13.04}&{13.00}&{12.98}&{13.10}\\{24}&{13.02}&{12.90}&{13.05}&{12.97}\\{25}&{12.93}&{12.88}&{12.91}&{12.90}\end{aligned}\)
Short Answer
\(L{\rm{ }}C{\rm{ }}L = 12.914{\rm{ }};{\rm{ }}U{\rm{ }}C{\rm{ }}L = 13.063{\rm{ }};\) The process is in control.
Step by step solution
Given
To obtain an estimate of based on the sample range, note that if\({X_1},{X_2}, \ldots ,{X_n}\)form a random sample from a normal distribution, then
\(\sigma R = \sigma \cdot \left\{ {\max \left( {{Z_1}, \ldots ,{Z_n}} \right) - \min \left( {{Z_1}, \ldots ,{Z_n}} \right)} \right\}\)
where\({Z_1},{Z_2}, \ldots ,{Z_n}\)are independent standard random variables.
Now denote the ranges for the\(k\)samples in the quality control data set by\({r_1},{r_2}, \ldots ,{r_k}\). We have
\(\hat \sigma = \frac{{\frac{1}{k}\sum\limits_{i = 1}^k {{r_i}} }}{{{b_n}}} = \frac{{\bar r}}{{{b_n}}}\)
where\({b_n} = E\)(range of a standard normal sample). There is a table for values of\({b_n}\).
Control Limits Based on the Sample Ranges
\(\begin{aligned}{l}LCL = \bar \bar x - 3 \cdot \frac{{\bar r}}{{{b_n}\sqrt n }}\\UCL = \bar \bar x + 3 \cdot \frac{{\bar r}}{{{b_n}\sqrt n }}\end{aligned}\)
Where
\(\begin{aligned}{l}\bar r = \frac{1}{k}\sum\limits_{i = 1}^k {{r_i}} \\\bar \bar x = \frac{1}{k}\sum\limits_{i = 1}^k {{{\bar x}_i}} \end{aligned}\)
Find the control limits on the sample based
We calculate the following (as the average of each row)
\(\begin{aligned}{l}{{\bar x}_1} = 12.97\\{{\bar x}_2} = 13.01\\{{\bar x}_3} = 13.0675\\{{\bar x}_4} = 12.9825,\\{{\bar x}_{25}} = 12.905\end{aligned}\)
therefore,
\(\bar \bar x = \frac{1}{{25}}\sum\limits_{i = 1}^{25} {{{\bar x}_i}} = \frac{1}{{25}}(12.97 + 13.01 + \ldots + 12.905) = 12.9888\)
From the table we have
\(\,{b_4} = 2.05\)
Now we find\({r_i}\)for\(i = 1,2, \ldots ,25\)as maximum values of all\({x_1},{x_2},{x_3},{x_4}\)in one row minus the minimum values of all\({x_1},{x_2},{x_3},{x_4}\)in one row. For example
\({r_1} = {x_1} - {x_3} = 0.1\)
Find now\(\bar r\)as
\(\bar r = \frac{1}{{25}}\sum\limits_{i = 1}^{25} {{r_i}} = 0.102\)
Finally, the Control Limits Based on the Sample Ranges
\(\begin{aligned}{l}LCL = \bar \bar x - 3 \cdot \frac{{\bar r}}{{{b_{n\sqrt n }}}} = 12.9888 - 3 \cdot \frac{{0.102}}{{2.058\sqrt 4 }} = 12.9888 - 0.748 = 12.914\\UCL = \bar \bar x + 3 \cdot \frac{{\bar r}}{{{b_n}\sqrt n }} = 12.9888 + 3 \cdot \frac{{0.102}}{{2.058\sqrt 4 }} = 12.9888 + 0.748 = 13.063\end{aligned}\)
All values are between the control limits, therefore, the process is in control.
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