Chapter 6.S, Problem 20P

a)

Summary Introduction

To establish: Upper and lower control limits for the control chart and graph the data.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

Answer to Problem 20P

The upper control limit is 0.05812 and lower control limit is 0.

Explanation of Solution

Given information:

The following information is given:

Sample	1	2	3	4	5	6	7	8	9	10
Defectives	5	7	4	4	6	3	5	6	2	8

Compute the defective rate $\left(\overline{p}\right)$:

Defective rate is calculated by dividing the sum of defectives and multiple of number of samples and sample size. Hence, the defective rate is 0.025.

$\begin{array}{c}\overline{p}=\frac{\text{Sum of defectives}}{\text{Number of samples}\times \text{Sample size}}\\ =\frac{5+7+4+4+6+3+5+6+2+8}{10\times 200}\\ =\frac{50}{2,000}\\ =0.025\end{array}$

Determine $\left({\sigma }_{\overline{p}}\right)$ standard deviation of the sampling distribution:

Sample size is given as 200 and the defective rate is calculated as 0.025. Substitute the values in above equation. Hence, the standard deviation is 0.01104.

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}\\ =\sqrt{\frac{0.025\times \left(1-0.025\right)}{200}}\\ =\sqrt{\frac{0.025\times 0.975}{200}}\\ =0.01104\end{array}$

Determine UCL:

It can be calculated by adding the defective rate with the value attained by multiplying the number of standard deviation for setting the limit and standard deviation of the sampling distribution.

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.025+3\times 0.01104\\ =0.05812\end{array}$

Hence, the UCL value is 0.05812.

Determine LCL:

It can be calculated by subtracting the defective rate with the value attained by multiplying the number of standard deviation for setting the limit and standard deviation of the sampling distribution.

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.025-3\times 0.01104\\ =-0.00812\\ =0\end{array}$

Hence, the LCL value is 0.

Draw a p-chart for the attained values:

b)

Summary Introduction

To determine: Whether the process been in control.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

Answer to Problem 20P

The process been in control.

Explanation of Solution

Given information:

The following information is given:

Sample	1	2	3	4	5	6	7	8	9	10
Defectives	5	7	4	4	6	3	5	6	2	8

Determine whether the process in control:

Since all the values are within the upper and lower control limits, the process seems to be in control.

The last two samples, sample number #9 and sample number #10 are showing a wider dispersion from the center line. Further samples need to be drawn and verified whether the defective rates increase in dispersion.

c)

Summary Introduction

To establish: Upper and lower control limits for the control chart and graph the data, if the sample size were 100.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

Answer to Problem 20P

The upper control limit is 0.11537 and lower control limit is 0.

Explanation of Solution

Given information:

The following information is given:

Sample	1	2	3	4	5	6	7	8	9	10
Defectives	5	7	4	4	6	3	5	6	2	8

Compute the defective rate $\left(\overline{p}\right)$:

Defective rate is calculated by dividing the sum of defectives and multiple of number of samples and sample size. Hence, the defective rate is 0.05.

$\begin{array}{c}\overline{p}=\frac{\text{Sum of defectives}}{\text{Number of samples}\times \text{Sample size}}\\ =\frac{5+7+4+4+6+3+5+6+2+8}{10\times 100}\\ =\frac{50}{1,000}\\ =0.05\end{array}$

Determine $\left({\sigma }_{\overline{p}}\right)$ standard deviation of the sampling distribution:

Sample size is given as 100 and the defective rate is calculated as 0.05. Substitute the values in above equation. Hence, the standard deviation is 0.02179.

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}\\ =\sqrt{\frac{0.05\times \left(1-0.05\right)}{100}}\\ =\sqrt{\frac{0.05\times 0.95}{100}}\\ =0.02179\end{array}$

Determine UCL:

It can be calculated by adding the defective rate with the value attained by multiplying the number of standard deviation for setting the limit and standard deviation of the sampling distribution.

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.05+3\times 0.02179\\ =0.11537\end{array}$

Hence, the UCL value is 0.11537.

Determine LCL:

It can be calculated by subtracting the defective rate with the value attained by multiplying the number of standard deviation for setting the limit and standard deviation of the sampling distribution.

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.05-3\times 0.02179\\ =-0.01537\\ =0\end{array}$

Hence, the LCL value is 0.

Draw a p-chart for the attained values:

Since all the values are within the upper and lower control limits, the process seems to be in control. This is expected, as the observations were within the limits even when the limits were narrower.