Chapter 19, Problem 22CE

A new machine has just been installed to produce printed circuit boards. One of the critical measurements is the thickness of the resistance layer. The quality control inspector randomly selects five boards each half-hour, measures the thickness, and records the results. The measurements (in millimeters) for the period 8:00 a.m. to 10:30 a.m. follow.

1. a. Determine the control limits for the mean and the range.
2. b. Plot the control limits for the mean outside diameter and the range.
3. c. Are there any points on the mean or the range chart that are out of control? Comment on the chart.

To determine

Find the control limits for the mean.

Find the control limits for the range.

Answer to Problem 22CE

The control limits for the mean is 86.61, and 88.11.

The control limits for the range is 0, and 2.75.

Explanation of Solution

Calculation:

The mean of the sample means is,

$\overline{\overline{x}}=\frac{{\displaystyle \sum \overline{x}}}{k}$

Here, ${\displaystyle \sum \overline{x}}$ denotes the sum of sample means, and k denotes the number of sample means.

The formula for the control limits is,

$LCL,UCL=\overline{\overline{x}}\pm A_2\overline{R}$

Here, $\overline{\overline{x}}$ denotes the mean of the sample means, $A_2$ is the constant value and $\overline{R}$ denotes the mean of the ranges.

The mean of the sample ranges is,

$\overline{R}=\frac{{\displaystyle \sum R}}{k}$

Here, R denotes the sum of sample ranges, and k denotes the number of sample means.

The formula for the control limits is,

$LCL,UCL=D_3\overline{R},D_4\overline{R}$

Here, $\overline{R}$ denotes the mean of the sample ranges, and $D_3,D_4$ are the constant values.

The sample means are ranges are obtained below:

Time	Thickness (millimeters)
	1	2	3	4	5	Total	Average	Range
8:00	87.1	87.3	87.9	87.0	87.0	436.3	87.26$\left(=\frac{436.3}{5}\right)$	0.9$\left(=87.9-87.0\right)$
8:30	86.9	88.5	87.6	87.5	87.4	437.9	87.58$\left(=\frac{437.9}{5}\right)$	1.6$\left(=88.5-86.9\right)$
9:00	87.5	88.4	86.9	87.6	88.2	438.6	87.72$\left(=\frac{438.6}{5}\right)$	1.5$\left(=88.4-86.9\right)$
9:30	86.0	88.0	87.2	87.6	87.1	435.9	87.18$\left(=\frac{435.9}{5}\right)$	2$\left(=88.0-86.0\right)$
10:00	87.1	87.1	87.1	87.1	87.1	435.5	87.1$\left(=\frac{435.5}{5}\right)$	0$\left(=87.1-87.1\right)$
10:30	88.0	86.2	87.4	87.3	87.8	436.7	87.34$\left(=\frac{436.7}{5}\right)$	1.8$\left(=88.0-86.2\right)$
	Totals						524.18	7.8

The mean of the sample means is,

$\begin{array}{c}\overline{\overline{x}}=\frac{524.18}{6}\\ =87.36\end{array}$

Hence, the mean of the sample means is 87.36.

The mean of the sample ranges is,

$\begin{array}{c}\overline{R}=\frac{7.8}{6}\\ =1.3\end{array}$

Hence, the mean of the sample ranges it as 1.3.

Compute the value of $A_2$ by using factors for control chart.

From the Appendix B: Tables B.10 Factors for Control Charts:

• Locate the value $A_2$ in chart for Averages row.
• Locate the value 5 in n column.
• The intersecting value is 0.577.

The control limits for the mean is,

$\begin{array}{c}LCL,UCL=87.36\pm \left(0.577\times 1.3\right)\\ =87.36\pm 0.75\\ =\left(87.36-0.75,87.36+0.75\right)\\ =86.61,88.11\end{array}$

Hence, the control limits for the mean is 86.61, and 88.11.

Compute the values of $D_3$ and $D_4$ by using factors for control chart.

From the Appendix B: Tables B.10 Factors for Control Charts:

• Locate the value $D_3$ in chart for Ranges row.
• Locate the value 5 in n column.
• The intersecting value is 0.
• Locate the value $D_4$ in chart for Ranges row.
• Locate the value 5 in n column.
• The intersecting value is 2.115.

The control limits for the range is,

$\begin{array}{c}LCL,UCL=0\left(1.3\right),\left(2.115\right)\left(1.3\right)\\ =0,2.75\end{array}$

Hence, the control limits for the range is 0, and 2.75.

To determine

Construct the control limits for the mean.

Construct the control limits for the range.

Answer to Problem 22CE

Output using MINITAB software is given below:

Explanation of Solution

Calculation:

Step by step procedure to obtain $\overline{x}$ chart using MINITAB software is given as,

• Choose Stat > Control Charts > Variables charts for subgroups > Xbar Chart.
• Select option Observations for a subgroup are in one row of columns.
• Enter the column Thickness 1, Thickness 2, Thickness 3, Thickness 4, and Thickness 5.
• Choose Xbar Options.
• In Estimate, select Rbar under Method for estimating standard deviation.
• Click OK.

Step by step procedure to obtain R chart using MINITAB software is given as,

• Choose Stat > Control Charts > Variables charts for subgroups > R Chart.
• Select option Observations for a subgroup are in one row of columns.
• Enter the column Thickness 1, Thickness 2, Thickness 3, Thickness 4, and Thickness 5.
• Choose R Options.
• In Estimate, select Rbar under Method for estimating standard deviation.
• Click OK.

To determine

Check whether there any points on the mean or the range chart that are out of control.

Answer to Problem 22CE

The points are within the control.

Explanation of Solution

Calculation:

For $\overline{x}$ chart:

The limits are 86.61 to 88.11 any value beyond this limits is considered as outlier.

In the chart no values are outside the limits this indicates that the process is in the control which implies that none of values are out of control.

For R chart:

The limits are 0 to 2.75 any value beyond this limits is considered as outlier.

In the chart no values are outside the limits this indicates that the process is in the control which implies that none of values are out of control.