Chapter 10, Problem 5P

a)

Summary Introduction

To determine: The fraction defective in each sample.

Introduction: Quality is a measure of excellence or a state of being free from deficiencies, defects and important variations. It is obtained by consistent and strict commitment to certain standards to attain uniformity of a product to satisfy consumers’ requirement.

Answer to Problem 5P

The fraction defective is shown in Table 1.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Calculation of fraction defective in each sample:

n	200
Sample	1	2	3	4
Number with errors	4	2	5	9
Prop defective	0.02	0.01	0.025	0.045

Excel Worksheet:

The proportion defective is calculated by dividing the number of errors with the number of samples. For sample 1, the number of errors 4 is divided by 200 which give 0.02 as prop defective.

Hence, the fraction defective is shown in Table 1.

b)

Summary Introduction

To determine: The estimation for fraction defective when true fraction defective for the process is unknown.

Introduction: Quality is a measure of excellence or a state of being free from deficiencies, defects and important variations. It is obtained by consistent and strict commitment to certain standards to attain uniformity of a product to satisfy consumers’ requirement.

Answer to Problem 5P

The fraction defective is 0.025.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Calculation of fraction defective:

The fraction defective is calculated when true fraction defective is unknown.

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{defective}=4+2+5+9\\ =20\end{array}$

Total number of defective is calculated by adding the number of errors, (4+2+5+9) which accounts to 20

$\overline{\text{p}}=\frac{\text{Total}\text{number}\text{of}\text{defective}}{\text{Total}\text{number}\text{of}\text{observation}}$

$\begin{array}{c}\overline{\text{p}}=\frac{\text{20}}{4\times 200}\\ =0.025\end{array}$ (1)

The fraction defective is calculated by dividing total number of defective with total number of observation which is 20 is divided with the product of 4 and 200 which is 0.025.

Hence, the fraction defective is 0.025.

c)

Summary Introduction

To determine: The estimate of mean and standard deviation of the sampling distribution of fraction defective for samples for the size.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

The estimate of mean and standard deviation of the sampling distribution is 0.025 and 0.011.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Estimate of mean and standard deviation of the sampling distribution:

Mean = 0.025 (from equation (1))

$\begin{array}{c}\text{Std dev}\text{.}=\sqrt{\frac{\overline{\text{p}}\left(1-\overline{\text{p}}\right)}{\text{n}}}\\ =\sqrt{\frac{0.025\left(1-0.025\right)}{200}}\\ =0.011\end{array}$

The estimate for mean is shown in equation (1) and standard deviation is calculated by substituting the value which yields 0.011.

Hence, estimate of mean and standard deviation of the sampling distribution is 0.025 and 0.011.

d)

Summary Introduction

To determine: The control limits that would give an alpha risk of 0.03 for the process.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

The control limits that would give an alpha risk of 0.03 for the process are 0.0489 and 0.0011.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Control limits that would give an alpha risk of 0.03 for the process:

$\begin{array}{c}\text{Alpha}\text{risk}=0.03\\ =\frac{0.03}{2}\\ =0.015\end{array}$

0.015 is in each tail and using z-factor table, value that corresponds to 0.5000 – 0.0150 is 0.4850 which is z = 2.17.

$\begin{array}{c}\text{UCL}=\text{0}\text{.025}+\left(\text{2}\text{.17}\times \text{0}\text{.011}\right)\\ =\text{0}\text{.0489}\\ \text{LCL}=\text{0}\text{.025}-\left(\text{2}\text{.17}\times \text{0}\text{.011}\right)\\ =\text{0}\text{.0011}\end{array}$ (2)

The UCL is calculated by adding 0.025 with the product of 2.17 and 0.011 which gives 0.0489 and LCL is calculated by subtracting 0.025 with the product of 2.17 and 0.011 which yields 0.0011.

Hence, the control limits that would give an alpha risk of 0.03 for the process are 0.0489 and 0.0011.

e)

Summary Introduction

To determine: The alpha risks that control limits 0.47 and 0.003 will provide.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

The alpha risks that control limits 0.47 and 0.003 will provide is 0.0456.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Alpha risks that control limits 0.47 and 0.003 will provide:

The following equation z value can be calculated,

$\begin{array}{c}\text{z}\left(0.011\right)=0.047-0.025\\ \text{z}=\frac{0.022}{0.011}\\ \text{z}=2.00\end{array}$

From z factor table, the probability value which corresponds to z = 2.00 is 0.4772, on each tail,

$\begin{array}{c}=0.5000-0.4772\\ =0.0228\\ =2\times 0.0228\\ =0.0456\end{array}$

0.0228 is observed on each tail and doubling the value gives 0.0456 which is the alpha risk.

Hence, alpha risks that control limits 0.47 and 0.003 will provide is 0.0456

f)

Summary Introduction

To determine: Whether the process is in control when using 0.047 and 0.003.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has an upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

The alpha risks that control limits 0.047 and 0.003 will provide is 0.0456.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Calculation of fraction defective in each sample:

n	200
Sample	1	2	3	4
Number with errors	4	2	5	9
Prop defective	0.02	0.01	0.025	0.045

UCL = 0.047 & LCL = 0.003

Graph:

A graph is plotted using UCL, LCL and prop defective values which show that all the sample points are well within the control limits which makes the process to be in control.

Hence, the process is within control for the limits 0.047 & 0.003.

g)

Summary Introduction

To determine: The mean and standard deviation of the sampling distribution.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

The mean and standard deviation of the sampling distribution is 0.02&0.0099.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Long run fraction defective of the process is 0.02

Calculation of mean and standard deviation of the sampling distribution:

Fraction defective in each sample:

n	200
Sample	1	2	3	4
Number with errors	4	2	5	9
Prop defective	0.02	0.01	0.025	0.045

The mean is calculated by taking average for the proportion defective,

$\begin{array}{l}=\frac{0.02+0.01+0.025+0.045}{4}\\ =0.02\end{array}$

The values of the proportion defective are added and divided by 4 which give 0.02.

The standard deviation is calculated using the formula,

$$ $\begin{array}{c}\text{Std dev}\text{.}=\sqrt{\frac{\overline{\text{p}}\left(1-\overline{\text{p}}\right)}{\text{n}}}\\ =\sqrt{\frac{0.02\left(1-0.02\right)}{200}}\\ =0.099\end{array}$

The standard deviation is calculated by substituting the values in the above formula and taking square root for the resultant value which yields 0.099.

Hence, mean and standard deviation of the sampling distribution is 0.02&0.0099.

h)

Summary Introduction

To construct: A control chart using two sigma control limits and check whether the process is in control.

Introduction:

Control chart:

It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.

Answer to Problem 5P

Control chart is constructed using two-sigma control limits and the chart shows that the process is not in control.

Explanation of Solution

Given information:

Sample	1	2	3	4
Number with errors	4	2	5	9

Fraction defective in each sample:

n	200
Sample	1	2	3	4
Number with errors	4	2	5	9
Prop defective	0.02	0.01	0.025	0.045

Calculation of control limits:

$\begin{array}{c}\text{Control limits}=\overline{\text{p}}\pm 2\sqrt{\frac{\overline{\text{p}}\left(1-\overline{\text{p}}\right)}{\text{n}}}\\ =0.02\pm 2\sqrt{\frac{0.02\left(1-0.02\right)}{200}}\\ =0.02\pm 0.0198\end{array}$

$\begin{array}{c}\text{UCL}=0.02+0.0198\\ =0.0398\\ \text{LCL}=0.02-0.0198\\ =0.0002\end{array}$

The control limits are calculated using the above formula and substituting the values and taking square root gives the control limits of the UCL and LCL which are 0.0398 and 0.0002 respectively.

Graph:

A graph is plotted using the fraction defective, UCL and LCL values which shows that one sample points is beyond the control region which makes the process to be out of control.

Hence, control chart is constructed using two-sigma control limits and the chart shows that the process is not in control.