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Chapter 16: Q27E (page 700)

In some situations, the sizes of sampled specimens vary, and larger specimens are expected to have more defects than smaller ones. For example, sizes of fabric samples inspected for flaws might vary over time. Alternatively, the number of items inspected might change with time. Let

\(\begin{aligned}{l}{u_i} = \frac{{ the number of defects observed at time i}}{{ size of entity inspected at time i}}\\ = \frac{{{x_i}}}{{{g_i}}}\end{aligned}\)

where "size" might refer to area, length, volume, or simply the number of items inspected. Then a \(u\) chart plots \({u_1},{u_2}, \ldots \), has center line \(\bar u\), and the control limits for the \(i\) th observations are \(\bar u \pm 3\sqrt {\bar u/{g_i}} \).

Painted panels were examined in time sequence, and for each one, the number of blemishes in a specified sampling region was determined. The surface area \(\left( {f{t^2}} \right)\) of the region examined varied from panel to panel. Results are given below. Construct a \(u\) chart.

\(\begin{aligned}{*{20}{c}}{ Panel }&{\begin{aligned}{*{20}{c}}{ Area }\\{ Examined }\end{aligned}}&{\begin{aligned}{*{20}{c}}{ No. of }\\{ Blemishes }\end{aligned}}\\1&{.8}&3\\2&{.6}&2\\3&{.8}&3\\4&{.8}&2\\5&{1.0}&5\\6&{1.0}&5\\7&{.8}&{10}\\8&{1.0}&{12}\\9&{.6}&4\\{10}&{.6}&2\\{11}&{.6}&1\\{12}&{.8}&3\\{13}&{.8}&5\\{14}&{1.0}&4\\{15}&{1.0}&6\\{16}&{1.0}&{12}\\{17}&{.8}&3\\{18}&{.6}&3\\{19}&{.6}&5\\{20}&{.6}&1\end{aligned}\)

Short Answer

Expert verified

The process is in-control.

Step by step solution

01

Step 1:To Find the LCL and UCL

The \(u\)chart limits are given in the exercise. Start with formula to obtain \(u_i^\prime s\)

\({u_i} = \frac{{{x_i}}}{{{g_i}}}\)

The data obtained using this formula is

\(3.75,3.33,3.75,2.50,5.00,5.00,12.50,12.00,6.67,3.33,1.67,3.75,6.25,4.00,6.00,12.00,3.75,5.00,8.33,1.67.\)

The center line is

\(\bar u = 5.5125.\)

Depending on\({g_i}\), the\(LCL\)and\(UCL\)are different. For\({g_i} = 0.6\)the\(LCL\)is

\(\begin{aligned}{l}LCL = \bar u - 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 - 3 \cdot 9.0933\\ = 0,\end{aligned}\)

set it to zero because it is negative.

The\(UCL\)is

\(\begin{aligned}{l}UCL = \bar u + 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 + 3 \cdot 9.0933\\ = 14.6.\end{aligned}\)

For\({g_i} = 0.8\)the\(LCL\)is

\(\begin{aligned}{l}LCL = \bar u - 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 - 3 \cdot 7.857\\ = 0\end{aligned}\)

set it to zero because it is negative.

The\(UCL\)is

\(\begin{aligned}{l}UCL = \bar u + 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 + 3 \cdot 7.857\\ = 13.4.\end{aligned}\)

For\({g_i} = 1\)the\(LCL\)is

\(\begin{aligned}{l}LCL = \bar u - 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 - 3 \cdot 7.044\\ = 0,\end{aligned}\)

set it to zero because it is negative.

The\(UCL\)is

\(\begin{aligned}{l}UCL = \bar u + 3\sqrt {\frac{{\bar u}}{{{g_i}}}} = 5.5125 + 3 \cdot 7.044\\ = 12.6.\end{aligned}\)

The following \(u\) charts suggest that the process is in control - no data exceeds \(LCL\) and \(UCL\).

02

Step 2:Final proof

The process is in-control.

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Most popular questions from this chapter

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:

\(\begin{aligned}{*{20}{c}}{Subgroup}&{x\_(1)}&{x\_(2)}&{x\_(3)}&{x\_(4)}\\1&{13.02}&{12.95}&{12.92}&{12.99}\\2&{13.02}&{13.10}&{12.96}&{12.96}\\3&{13.04}&{13.08}&{13.05}&{13.10}\\4&{13.04}&{12.96}&{12.96}&{12.97}\\5&{12.96}&{12.97}&{12.90}&{13.05}\\6&{12.90}&{12.88}&{13.00}&{13.05}\\7&{12.97}&{12.96}&{12.96}&{12.99}\\8&{13.04}&{13.02}&{13.05}&{12.97}\\9&{13.05}&{13.10}&{12.98}&{12.96}\\{10}&{12.96}&{13.00}&{12.99}&{12.99}\\{11}&{12.90}&{13.05}&{12.98}&{12.88}\\{12}&{12.96}&{12.98}&{12.97}&{13.02}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{ }&{}&{}&{}&{}\\{13}&{13.00}&{12.96}&{12.99}&{12.90}\\{14}&{12.88}&{12.94}&{13.05}&{13.00}\\{15}&{12.96}&{12.96}&{13.04}&{12.98}\\{16}&{12.99}&{12.94}&{13.00}&{13.05}\\{17}&{13.05}&{13.02}&{12.88}&{12.96}\\{18}&{13.08}&{13.06}&{13.10}&{13.05}\\{19}&{13.02}&{13.05}&{13.04}&{12.97}\\{20}&{12.96}&{12.90}&{12.97}&{13.05}\\{21}&{12.98}&{12.99}&{12.96}&{13.00}\\{22}&{12.97}&{13.02}&{12.96}&{12.99}\\{23}&{13.04}&{13.00}&{12.98}&{13.10}\\{24}&{13.02}&{12.90}&{13.05}&{12.97}\\{25}&{12.93}&{12.88}&{12.91}&{12.90}\end{aligned}\)

In the case of known 饾泹 and 饾洈, what control limits are necessary for the probability of a single point being outside the limits for an in-control process to be .005?

For what\(\bar x\)values will the LCL in a\(c\)chart be negative?

An alternative to the \(p\) chart for the fraction defective is the np chart for number defective. This chart has \(UCL = n\bar p + 3\sqrt {n\bar p(1 - \bar p)} ,LCL = n\bar p - 3\sqrt {n\bar p(1 - \bar p)} \),

and the number of defectives from each sample is plotted on the chart. Construct such a chart for the data of Example 16.6. Will the use of an \(n p\) chart always give the same message as the use of a \(p\)chart (i.e., are the two charts equivalent)?

Let \(\alpha \) be a number between 0 and 1 , and define a sequence \({W_1},{W_2},{W_3}, \ldots \) by \({W_0} = \mu \)and \({W_t} = \alpha {\bar X_t} + (1 - \alpha ){W_{t - 1}}\) for \(t = 1,2, \ldots .\)Substituting for \({W_{t - 1}}\) its representation in terms of \({\bar X_{t - 1}}\) and \({W_{t - 2}}\), then substituting for \({W_{t - 2}}\), and so on, results in

\(\begin{array}{l}{W_t} = \alpha {{\bar X}_t} + \alpha (1 - \alpha ){{\bar X}_{t - 1}} + \ldots \\ + \alpha {(1 - \alpha )^{t - 1}}{{\bar X}_1} + {(1 - \alpha )^t}\mu \end{array}\)

The fact that \({W_t}\) depends not only on \({\bar X_t}\) but also on averages for past time points, albeit with (exponentially) decreasing weights, suggests that changes in the process mean will be more quickly reflected in the \({W_t}\) 's than in the individual \({\bar X_t}\) 's.

a. Show that \(E\left( {{W_t}} \right) = \mu \).

b. Let \(\sigma _t^2 = V\left( {{W_t}} \right)\), and show that

\(\sigma _t^2 = \frac{{\alpha \left( {1 - {{(1 - \alpha )}^{2\eta }}} \right)}}{{2 - \alpha }} \times \frac{{{\sigma ^2}}}{n}\)

c. An exponentially weighted moving-average control chart plots the \({W_t}\)'s and uses control limits \({\mu _0} \pm 3{\sigma _t}\) (or \(\bar \bar x\) in place of \({\mu _0}\) ). Construct such a chart for the data of Example 16.9, using \({\mu _0} = 40\).
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