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Chapter 16: Q17E (page 695)

Subgroups of power supply units are selected once each hour from an assembly line, and the high-voltage output of each unit is determined.

a. Suppose the sum of the resulting sample ranges for\(30\)subgroups, each consisting of four units, is\(85.2\). Calculate control limits for an\(R\)chart.

b. Repeat part (a) if each subgroup consists of eight units and the sum is\(106.2\).

Short Answer

Expert verified

a. \(LCL = 0,UCL = 6.48,CL = 2.84;\)

b. \(LCL = 0.48,UCL = 6.60,CL = 3,54\)

Step by step solution

01

Determine the LCL and UCL

Let\({r_1},{r_2}, \ldots ,{r_k}\)denote the\(k\)sample ranges and

\(\bar r = \frac{{\sum {{r_i}} }}{k}.\)

We plot values\({r_1},{r_2}, \ldots ,{r_k}\)on the\(R\)control chart.

Upper Control Limit and Lower Control Limit for\(R\)control chart are

\(\begin{aligned}{l}LCL = \bar r - 3\bar r\frac{{{c_n}}}{{{b_n}}}\\UCL = \bar r + 3\bar r\frac{{{c_n}}}{{{b_n}}}\end{aligned}\)

Where LCL will be negative for\(n \le 6\). When this happens it is customary to take\({\rm{LCL}} = 0\).

Values for\({c_n}\)and\({b_n}\)are given in the tables on pages\(693\)and\(685\), respectively, for\(n = 3,4, \ldots ,8\).

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}n&3&4&5&6&7&8\\{{b_n}}&{1.693}&{2.058}&{2.325}&{2.536}&{2.706}&{2.844}\end{aligned}\\\begin{aligned}{*{20}{c}}n&3&4&5&6&7&8\\{{c_n}}&{.888}&{.880}&{.864}&{.848}&{.833}&{.820}\end{aligned}\end{aligned}\)

02

A)Step 1: Find the control limits for R chart

From the exercise, we have that\(k = 30,n = 4\)and

\(\sum\limits_{i = 1}^{30} {{r_i}} = 85.2.\)

Therefore, we have that

\(\bar r = \frac{1}{{30}}85.2 = 2.84.\)

From the tables, for\(n = 4\)we have

\(\begin{aligned}{l}{b_4} = 2.058\\{c_4} = 0.880.\end{aligned}\)

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}n&3&4&5&6&7&8\\{{b_n}}&{1.693}&{2.058}&{2.325}&{2.536}&{2.706}&{2.844}\end{aligned}\\\begin{aligned}{*{20}{c}}n&3&4&5&6&7&8\\{{c_n}}&{.888}&{.880}&{.864}&{.848}&{.833}&{.820}\end{aligned}\end{aligned}\)

The\({\rm{3}}\)-sigma control limits for an\({\rm{R}}\)control chart are

\(\begin{aligned}{l}LCL = \bar r - 3\bar r\frac{{{c_n}}}{{{b_n}}} = 2.84 - 3 \cdot 2.84 \cdot \frac{{0.880}}{{2.058}} = - 0.80\\UCL = \bar r + 3\bar r\frac{{{c_n}}}{{{b_n}}} = 2.84 + 3 \cdot 2.84 \cdot \frac{{0.880}}{{2.058}} = 6.48\\CL = \bar r = 2.84\end{aligned}\)

But as we mentioned, for\(n \le 6\) the LCL is negative and it is customary to use \({\rm{LCL}} = 0\) !

03

B)Step 3: Repeat the value and substitute the values

From the exercise, we have that\(k = 30,n = 8\)and

\(\sum\limits_{i = 1}^{30} {{r_i}} = 106.2.\)

Therefore, we have that

\(\bar r = \frac{1}{{30}}106.2 = 3.54.\)

From the tables, for\(n = 4\)we have

\(\begin{aligned}{l}{b_4} = 2.844;\\{c_4} = 0.820.\end{aligned}\)

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{\rm{n}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}&{\rm{7}}&{\rm{8}}\\{{\rm{b\_(n)}}}&{{\rm{1}}{\rm{.693}}}&{{\rm{2}}{\rm{.058}}}&{{\rm{2}}{\rm{.325}}}&{{\rm{2}}{\rm{.536}}}&{{\rm{2}}{\rm{.706}}}&{{\rm{2}}{\rm{.844}}}\end{aligned}\\\begin{aligned}{*{20}{c}}{\rm{n}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}&{\rm{7}}&{\rm{8}}\\{{\rm{c\_(n)}}}&{{\rm{.888}}}&{{\rm{.880}}}&{{\rm{.864}}}&{{\rm{.848}}}&{{\rm{.833}}}&{{\rm{.820}}}\end{aligned}\end{aligned}\)

The 3 -sigma control limits for an\({\rm{R}}\)control chart are

\(\begin{aligned}{l}LCL = \bar r - 3\bar r\frac{{{c_n}}}{{{b_n}}} = 3.54 - 3 \cdot 3.54 \cdot \frac{{0.820}}{{2.844}} = 0.48\\UCL = \bar r + 3\bar r\frac{{{c_n}}}{{{b_n}}} = 3.54 + 3 \cdot 3.54 \cdot \frac{{0.820}}{{2.844}} = 6.60\\CL = \bar r = 3.54\end{aligned}\)

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Most popular questions from this chapter

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).

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

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 \({\mu _0} = 40\).

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?

A sample of 200 ROM computer chips was selected on each of 30 consecutive days, and the number of nonconforming chips on each day was as follows: \(10,18,24,17,37,19,7,25,11,24,29,15,16,21,18,17,15,22,12\$ ,\$ 20,17,18,12,24,30,16,11,20,14,28\)Construct a\(p\)chart and examine it for any out-of-control points.

Containers of a certain treatment for septic tanks are supposed to contain \(16oz\) of liquid. A sample of five containers is selected from the production line once each hour, and the sample average content is determined. Consider the following results: \(15.992,16.051,16.066,15.912\), \(16.030,16.060,15.982,15.899,16.038,16.074,16.029\), \(15.935,16.032,15.960,16.055\). Using \(\Delta = .10\) and\(h = .20\), employ the computational form of the CUSUM procedure to investigate the behaviour of this process.
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