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Chapter 16: Q45 SE (page 714)

Resistance observations (ohms) for subgroups of a certain type of register gave the following summary quantities:

\(\begin{array}{*{20}{c}}i&{{n_i}}&{{{\overline x }_i}}&{{s_i}}&i&{{n_i}}&{{{\overline x }_i}}&{{s_i}}\\1&4&{430.0}&{22.5}&{11}&4&{445.2}&{27.3}\\2&4&{418.2}&{20.6}&{12}&4&{430.1}&{22.2}\\3&3&{435.5}&{25.1}&{13}&4&{427.2}&{24.0}\\4&4&{427.6}&{22.3}&{14}&4&{439.6}&{23.3}\\5&4&{444.0}&{21.5}&{15}&3&{415.9}&{31.2}\\6&3&{431.4}&{28.9}&{16}&4&{419.8}&{27.5}\\7&4&{420.8}&{25.4}&{17}&3&{447.0}&{19.8}\\8&4&{431.4}&{24.0}&{18}&4&{434.4}&{23.7}\\9&4&{428.7}&{21.2}&{19}&4&{422.2}&{25.1}\\{10}&4&{440.1}&{25.8}&{20}&4&{425.7}&{24.4}\\{}&{}&{}&{}&{}&{}&{}&{}\end{array}\)

Short Answer

Expert verified

Use different charts and methods.

Step by step solution

01

Step 1:To Find the determine required values

Use hint to determine required values. The sum of\(n_i^\prime s\)are

\(\sum {{n_i}} = 4 + 4 + \ldots + 4 + 4 = 76;\)

the sum of product of\(n_i^\prime s\)with corresponding\(\bar x_i^\prime {\rm{s}}\)are

\(\begin{array}{l}\sum {{n_i}} {{\bar x}_i} = 4 \times 430 + 4 \times 418.2 + \ldots + 4 \times 425.7\\ = 32,729.4\end{array}\)

The average is

\(\bar \bar x = 430.65\)

From the hint,\({s^2}\)is

\({s^2} = \frac{{\sum {\left( {{n_i} - 1} \right)} s_i^2}}{{\sum {\left( {{n_i} - 1} \right)} }} = \frac{{27,380.16 - 5661.4}}{{76 - 20}} = 590.0279,\)

which results in

\(s = 24.2905.\)

\(S\)chart, for variation and\(n = 3\), the limits are

\(\begin{array}{l}LCL = 24.2905 - \frac{{3 \times 24.2905 \times \sqrt {1 - 0.88{6^2}} }}{{0.886}}\mathop = \limits^{(1)} 0\\UCL = 24.2905 + \frac{{3 \times 24.2905 \times \sqrt {1 - 0.88{6^2}} }}{{0.886}} = 62.43.\end{array}\)

(1) : the value is less than zero, so set it to zero.

For variation and\(n = 4\), the limits are

\(\begin{array}{l}LCL = 24.2905 - \frac{{3 \times 24.2905 \times \sqrt {1 - 0.92{1^2}} }}{{0.921}}\mathop = \limits^{(1)} 0\\UCL = 24.2905 + \frac{{3 \times 24.2905 \times \sqrt {1 - 0.92{1^2}} }}{{0.921}} = 55.11.\end{array}\)

For location and\(n = 3\), the limits are

\(\begin{array}{l}LCL = 430.65 - 47.49 = 383.16\\UCL = 430.65 + 47.49 = 478.14\end{array}\)

For location and\(n = 4\), the limits are

\(\begin{array}{l}LCL = 430.65 - 39.56 = 391.09\\UCL = 430.65 + 39.56 = 470.21\end{array}\)

02

Step 2:Final proof

Use different charts and methods.

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Most popular questions from this chapter

Some sources advocate a somewhat more restrictive type of doubling-sampling plan in which \({r_1} = {c_2} + 1\); that is, the lot is rejected if at either stage the (total) number of defectives is at least \({r_1}\) (see the book by Montgomery). Consider this type of sampling plan with \({n_1} = 50,{n_2} = 100,{c_1} = 1\), and \({r_1} = 4\). Calculate the probability of lot acceptance when \(p = .02,.05\), and .10.

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.

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

The following numbers are observations on tensile strength of synthetic fabric specimens selected from a production process at equally spaced time intervals. Construct appropriate control charts, and comment (assume an assignable cause is identifiable for any out of-control observations).

\(\begin{array}{*{20}{r}}{ 1. }&{51.3}&{51.7}&{49.5}&{12.}&{49.6}&{48.4}&{50.0}\\{ 2. }&{51.0}&{50.0}&{49.3}&{13.}&{49.8}&{51.2}&{49.7}\\{ 3. }&{50.8}&{51.1}&{49.0}&{14.}&{50.4}&{49.9}&{50.7}\\{ 4. }&{50.6}&{51.1}&{49.0}&{15.}&{49.4}&{49.5}&{49.0}\\{ 5. }&{49.6}&{50.5}&{50.9}&{16.}&{50.7}&{49.0}&{50.0}\\{ 6. }&{51.3}&{52.0}&{50.3}&{17.}&{50.8}&{49.5}&{50.9}\\{ 7. }&{49.7}&{50.5}&{50.3}&{18.}&{48.5}&{50.3}&{49.3}\\{ 8. }&{51.8}&{50.3}&{50.0}&{19.}&{49.6}&{50.6}&{49.4}\\{9.}&{48.6}&{50.5}&{50.7}&{20.}&{50.9}&{49.4}&{49.7}\\{ 10. }&{49.6}&{49.8}&{50.5}&{21.}&{54.1}&{49.8}&{48.5}\\{ 11. }&{49.9}&{50.7}&{49.8}&{22.}&{50.2}&{49.6}&{51.5}\end{array}\)

Refer to Example \(16.11\), in which a single-sample plan with \(n = 50\) and \(c = 2\) was employed.

a. Calculate \(AOQ\) for \(p = .01,.02, \ldots ,.10\). What does this suggest about the value of \(p\) for which \(AOQ\) is a maximum and the corresponding \(AOQL\) ?

b. Determine the value of \(p\) for which \(AOQ\) is a maximum and the corresponding value of \(AOQL\). (Hint: Use calculus.)

c. For \(N = 2000\), calculate ATI for the values of \(p\) given in part (a).

The target value for the diameter of a certain type of driveshaft is \(.75\) in. The size of the shift in the average diameter considered important to detect is \(.002\) in. Sample average diameters for successive groups of \(n = 4\)shafts are as follows: \(.7507,.7504,.7492,.7501,.7503\), \(.7510,.7490,.7497,.7488,.7504,.7516,.7472,.7489\), \(.7483,.7471,.7498,.7460,.7482,.7470,.7493,.7462\), \(.7481\). Use the computational form of the CUSUM procedure with \(h = .003\) to see whether the process mean remained on target throughout the time of observation.
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