Chapter 15, Problem 91SE

a.

To determine

Construct the $\overline{X}\text{ and }R$ chart for the given process.

Revise the control limits if needed.

Answer to Problem 91SE

The $\overline{X}\text{ and }R$ chart for the given process is shown below:

The revised control limits for $\overline{X}$ chart are CL is 64.00028, LCL is 63.98191 and UCL is 64.01865.

The trial control limits for $R$ chart are CL is 0.01796, LCL is 0 and UCL is 0.04623.

Explanation of Solution

Calculation:

The data shows the diameter of fuse pins used in an aircraft engine taken for 25 samples and the sample subgroup size is 3.

Construction of $\overline{X}\text{ and }R$ chart is as follows:

Software procedure:

Step-by-step procedure to construct $\overline{X}\text{ and }R$ using MINITAB is given below:

• Choose Stat > Control Charts > Variables Charts for Subgroups >Xbar-R.
• Choose Observations for a subgroup are in one row of columns and then enter columns of X1, X2 and X3.
• Click OK.

Interpretation:

The $\overline{X}\text{ and }R$ control chart is constructed for the given data and from the $\overline{X}$ chart it can be observed that the data point 10 is falling outside the upper control limit.

Hence, the process is out of statistical control.

Revised control limits after eliminating the out of control data point:

Software procedure:

Step-by-step procedure to construct $\overline{X}\text{ and }R$ using MINITAB is given below:

• Choose Stat > Control Charts > Variables Charts for Subgroups >Xbar-R.
• Choose Observations for a subgroup are in one row of columns and then enter columns of X1, X2 and X3.
• Click OK.

Output obtained from MINITAB is given below:

Interpretation:

The $\overline{X}\text{ and }R$ control chart is constructed for the given data and from the $\overline{X}$ chart it can be observed that all the data point lies within the upper and lower control limits.

Hence, the process is said to be in statistical control.

b.

To determine

Estimate the process mean and standard deviation.

Answer to Problem 91SE

The estimate of the process mean is 64.

The estimate of the process standard deviation is 0.0106.

Explanation of Solution

Calculation:

From part (a) it can be observed that the $\overline{\overline{x}}$ is 64 and the $\overline{r}$ is 0.01796.

Process mean:

$\begin{array}{c}\widehat{\mu }=\overline{\overline{x}}\\ =64\end{array}$

Thus, the estimate of the process mean is 64.

Process standard deviation:

$\widehat{\sigma }=\frac{\overline{r}}{d_2}$

Where,

$\overline{r}$ represent the mean of the sample range.

$d_2$ represent the constant value obtained from the table XI.

The value of $d_2$ is calculated as follows:

• Locate the value of 3 under the column of $n^{*}$.
• Under the same row, locate the value for $A_2$.

Thus, the value of $d_2$ is 1.693.

Substitute 0.01796 as $\overline{r}$ and 1.693 as $d_2$.

$\begin{array}{c}\widehat{\sigma }=\frac{0.01796}{1.693}\\ =0.0106\end{array}$

Thus, the estimate of the process standard deviation is 0.0106.

c.

To determine

Calculate the estimate of PCR.

Identify whether the process meets the minimum capability level of $PCR\ge 1.33$.

Answer to Problem 91SE

The estimate of PCR is 0.63.

No, the process does not meet the minimum capability level of $PCR\ge 1.33$.

Explanation of Solution

Calculation:

The given information is that the process specifications are stated at $64\pm 0.02$.

The upper specification limit and lower specification limit are calculated as follows:

$\begin{array}{c}USL=64+0.02\\ =64.02\end{array}$

$\begin{array}{c}LSL=64-0.02\\ =63.98\end{array}$

PCR:

$PCR=\frac{USL-LSL}{6\widehat{\sigma }}$

Substitute USL as 64.02, LSL as 63.98 and $\widehat{\sigma }$ as 0.0106,

$\begin{array}{c}PCR=\frac{64.02-63.98}{6\left(0.0106\right)}\\ =\frac{0.04}{0.0636}\\ =0.628\\ \approx 0.63\end{array}$

Thus, the value of PCR is 0.63.

Conclusion:

The process does not meet the minimum capability level of 1.33 since the estimate of PCR is 0.63 which is less than 1.33.

d.

To determine

Calculate the estimate of $PC{R}_k$.

Conclude about the process capability using the estimate of $PC{R}_k$.

Answer to Problem 91SE

The estimate of $PC{R}_k$ is 0.63.

Explanation of Solution

Calculation:

$PC{R}_k$:

$PC{R}_k=\text{min}\left[\frac{USL-\overline{\overline{x}}}{3\widehat{\sigma }},\frac{\overline{\overline{x}}-LSL}{3\widehat{\sigma }}\right]$

Substitute USL as 64.02, LSL as 63.98,$\widehat{\sigma }$ as 0.0106 and $\overline{\overline{x}}$ as 64.

$\begin{array}{c}PC{R}_k=\text{min}\left[\frac{64.02-64}{3\left(0.0106\right)},\frac{64-63.98}{3\left(0.0106\right)},\right]\\ =\min \left[\frac{0.02}{0.0318},\frac{0.02}{0.0318}\right]\\ =\frac{0.02}{0.0318}\\ \approx 0.63\end{array}$

Thus, the value of $PC{R}_k$ is 0.63.

Conclusion:

The process does not meet the minimum capability level of 1.33 since the estimate of $PC{R}_k$ is 0.63 which is less than 1.33.

Hence, the process is not capable of operating within specification limits which is due to the values of PCR and $PC{R}_k$ are less than unity.

e.

To determine

Find the new value for the variance.

Answer to Problem 91SE

The new value for the variance is 0.00001089.

Explanation of Solution

Calculation:

The given information is that to change this process as a 6 sigma process, the process variance should be decreased in such a way that the $PC{R}_k$ is 2.0.

$PC{R}_k$:

$PC{R}_k=\frac{\overline{\overline{x}}-LSL}{3\widehat{\sigma }}$

Substitute LSL as 63.98, $\overline{\overline{x}}$ as 64 and $PC{R}_k$ as 2.0,

$\begin{array}{c}2.0=\frac{64-63.98}{3\left(\sigma \right)}\\ 6.0\sigma =0.02\\ \sigma =\frac{0.02}{6.0}\\ \sigma =0.0033\end{array}$

Thus, the value of variance should be $0.00001089\left(={0.0033}^2\right)$.

f.

To determine

Find the probability that the shift is detected in the next sample.

Find the ARL value after the shift.

Answer to Problem 91SE

The probability that the shift is detected in the next sample is 0.079.

The ARL value after the shift is 12.7.

Explanation of Solution

Calculation:

Assume that the process mean has shifted to 64.01

The probability of shift detected on the next sample is calculated by using the formula.

$P\left(x\le X\right)=P\left(\frac{x-\mu }{\sigma }\right)$

Where,

x represents the observed value of x.

$\mu$ represent the process mean.

$\sigma$ represent the process standard deviation.

The probability of shift detected on the next sample is calculated as follows:

$P\left(LCL<\overline{X}<UCL\right)=P\left(\frac{LCL-\overline{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}<Z<\frac{UCL-\overline{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\right)$

Substitute 63.982 as LCL, 64.01865 as UCL, 64.01, 0.0106 as $\sigma$ and 3 as n,

$\begin{array}{c}P\left(63.98<\overline{X}<64.02\right)=P\left(\frac{63.982-64.01}{\frac{0.0106}{\sqrt{3}}}<Z<\frac{64.01865-64.01}{\frac{0.0106}{\sqrt{3}}}\right)\\ =P\left(\frac{-0.0281}{0.006119}<Z<\frac{0.00865}{0.006119}\right)\\ =P\left(-4.59<Z<1.41\right)\end{array}$

Software procedure:

Step-by-step procedure to find the probability value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X value and Middle Tail for the region of the curve to shade.
• Enter the X1 value as –4.59 and X2 value as 1.41.
• Click OK.

Output obtained from MINITAB is given below:

Thus, the probability of detecting a shift in the next sample is given below:

$\begin{array}{c}P\left(\text{Detecting the shift}\right)=1-0.921\\ =0.079\end{array}$

Hence, the probability of detecting a shift in the next sample is 0.079.

ARL:

It is the average run length of the control chart. ARL is the average number of data points which are plotted to indicate the out-of-control condition.

$\text{ARL}=\frac{1}{p}$

Where,

p represents the probability that a data points will exceed the control limits.

From part (b) it can be observed that the value of p is 0.079.

$\begin{array}{c}\text{ARL}=\frac{1}{0.079}\\ \approx 12.7\end{array}$

Thus, the ARL value is 12.7.

Conclusion:

Thus, the average run length after the shift is 12.7.