Chapter 31, Problem 31.53E

a.

To determine

To find: The center line and control limits should be drawn on the s chart and $\overline{x}$ chart.

Answer to Problem 31.53E

For the s chart, the center line is 0.0141, the lower control limit is 0, and the upper control limit is 0.0295.

For $\overline{x}$ chart, the center line is 74, the lower control limit is 73.9799, and the upper control limit is 74.0201.

Explanation of Solution

Given info:

The target diameter for ring is $\mu =74\text{millimeters}$ and $\sigma =0.015\text{millimeter}$.

Calculation:

s chart:

Center line for an s chart:

$\begin{array}{c}CL=c_4\sigma \\ =0.9400\left(0.015\right)\\ =\text{0}\text{.0141}\end{array}$

Upper control limits:

From Table 31.3: Control chart constants, the value of $B_6$ corresponding with the sample size $n=5$ is 1.964.

$\begin{array}{c}UCL=B_6\sigma \\ =1.964\left(0.015\right)\\ =\text{0}\text{.0295}\end{array}$

Lower control limits:

From Table 31.3: Control chart constants, the value of $B_5$ corresponding with the sample size $n=5$ is 0.

$\begin{array}{c}LCL=B_5\widehat{\sigma }\\ =0\left(0.015\right)\\ =0\end{array}$

Thus, the center line is 0.0141, the lower control limit is 0, and the upper control limit is 0.0295.

$\overline{x}$chart:

Center line for a $\overline{x}$ chart:

$\begin{array}{c}CL=\mu \\ =74\end{array}$

Upper control limits:

$\begin{array}{c}UCL=\mu +3\frac{\sigma }{\sqrt{n}}\\ =74+3\frac{0.015}{\sqrt{5}}\\ =74+0.0201\\ =\text{74}\text{.0201}\end{array}$

Lower control limits:

$\begin{array}{c}LCL=\mu -3\frac{\sigma }{\sqrt{n}}\\ =74-3\frac{0.015}{\sqrt{5}}\\ =74-0.0201\\ =\text{73}\text{.9799}\end{array}$

Thus, the center line is 74, the lower control limit is 73.9799, and the upper control limit is 74.0201.

b.

To determine

To find: The center line and control limits should be drawn on the s chart and $\overline{x}$ chart when $\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$.

To find: The center line and control limits should be drawn on the s chart and $\overline{x}$ chart when $\mu =74\text{mm}$ and $\sigma =0.005\text{mm}$.

Answer to Problem 31.53E

The control limits are tabulated below:

Mean and SD	s chart			$\overline{x}$ chart
	LCL	CL	UCL	LCL	CL	UCL
$\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$	0	0.0138	0.0313	73.9225	73.945	73.9675
$\mu =74\text{mm}$ and $\sigma =0.005\text{mm}$	0	0.0046	0.0104	73.9225	74	74.0075

The issues that might arise for the manufacturer of the engine when the two parts from the different manufacturers are assembled is the statistical control process does not provide required information about the data.

Explanation of Solution

Given info:

The target diameter for ring is $\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$.

Calculation:

When $\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$, the control limits for s chart is as follows:

Center line for an s chart:

From Table 31.3: Control chart constants, the value of $c_4$ corresponding with the sample size $n=4$ is 0.9213.

$\begin{array}{c}CL=c_4\sigma \\ =0.9213\left(0.015\right)\\ =\text{0}\text{.0138}\end{array}$

Upper control limits:

From Table 31.3: Control chart constants, the value of $B_6$ corresponding with the sample size $n=4$ is 2.088.

$\begin{array}{c}UCL=B_6\sigma \\ =2.088\left(0.015\right)\\ =\text{0}\text{.0313}\end{array}$

Lower control limits:

From Table 31.3: Control chart constants, the value of $B_5$ corresponding with the sample size $n=4$ is 0.

$\begin{array}{c}LCL=B_5\widehat{\sigma }\\ =0\left(0.015\right)\\ =0\end{array}$

Thus, the center line is 0.0138, the lower control limit is 0, and the upper control limit is 0.0313.

When $\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$, the control limits for $\overline{x}$ chart is as follows:

Center line for a $\overline{x}$ chart:

$\begin{array}{c}CL=\mu \\ =73.945\end{array}$

Upper control limits:

$\begin{array}{c}UCL=\mu +3\frac{\sigma }{\sqrt{n}}\\ =73.945+3\frac{0.015}{\sqrt{4}}\\ =73.945+\text{0}\text{.0225}\\ =\text{73}\text{.9675}\end{array}$

Lower control limits:

$\begin{array}{c}LCL=\mu -3\frac{\sigma }{\sqrt{n}}\\ =73.945-3\frac{0.015}{\sqrt{4}}\\ =73.945-\text{0}\text{.0225}\\ =\text{73}\text{.9225}\end{array}$

Thus, the center line is 73.945, the lower control limit is 73.9225, and the upper control limit is 73.9675.

When $\mu =74\text{mm}$ and $\sigma =0.005\text{mm}$, the control limits for s chart is as follows:

Center line for an s chart:

From Table 31.3: Control chart constants, the value of $c_4$ corresponding with the sample size $n=4$ is 0.9213.

$\begin{array}{c}CL=c_4\sigma \\ =0.9213\left(0.005\right)\\ =\text{0}\text{.0046}\end{array}$

Upper control limits:

From Table 31.3: Control chart constants, the value of $B_6$ corresponding with the sample size $n=4$ is 2.088.

$\begin{array}{c}UCL=B_6\sigma \\ =2.088\left(0.005\right)\\ =\text{0}\text{.0104}\end{array}$

Lower control limits:

From Table 31.3: Control chart constants, the value of $B_5$ corresponding with the sample size $n=4$ is 0.

$\begin{array}{c}LCL=B_5\widehat{\sigma }\\ =0\left(0.005\right)\\ =0\end{array}$

Thus, the center line is 0.0046, the lower control limit is 0, and the upper control limit is 0.0104.

When $\mu =74\text{mm}$ and $\sigma =0.005\text{mm}$, the control limits for $\overline{x}$ chart is as follows:

Center line for a $\overline{x}$ chart:

$\begin{array}{c}CL=\mu \\ =74\end{array}$

Upper control limits:

$\begin{array}{c}UCL=\mu +3\frac{\sigma }{\sqrt{n}}\\ =74+3\frac{0.005}{\sqrt{4}}\\ =74+0.0075\\ =\text{74}\text{.0075}\end{array}$

Lower control limits:

$\begin{array}{c}LCL=\mu -3\frac{\sigma }{\sqrt{n}}\\ =74-3\frac{0.005}{\sqrt{4}}\\ =74-0.0075\\ =\text{73}\text{.9925}\end{array}$

Thus, the center line is 74, the lower control limit is 73.9225, and the upper control limit is 74.0075.

Justification:

For s chart, the control limits are wider when using $\mu =73.945\text{mm}$ and $\sigma =0.015\text{mm}$.

For $\overline{x}$ chart, the control limits are wider when using $\mu =74\text{mm}$ and $\sigma =0.005\text{mm}$.

When changing the target value and sigma, the result of the process variation should be changed. Moreover, it is not necessarily capable of meeting the requirement of piston diameter and difficult to identifying the defective items. Thus, the statistical control process does not provide information about the data when the two parts from the different manufacturers are assembled.