Chapter 6.S, Problem 7P

a)

Summary Introduction

To determine: The value of $\overline{\overline{x}}$.

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 7P

Hence, the value of $\overline{\overline{x}}$ is 155.16mm.

Explanation of Solution

Given information:

The following information is given:

Day	Mean (MM)	Range (MM)
1	156.9	4.2
2	153.2	4.6
3	153.6	4.1
4	155.5	5
5	156.6	4.5

Determine the value of $\overline{\overline{x}}$:

Day	Mean (MM)
1	156.9
2	153.2
3	153.6
4	155.5
5	156.6
Total	775.8

Calculate the $\overline{\overline{x}}$:

It is calculated by dividing the sum of mean values and the number of days.

$\begin{array}{c}\overline{\overline{x}}=\frac{\text{Sum of mean}}{\text{Number of days}}\\ =\frac{156.9+153.2+153.6+155.5+156.6}{5}\\ ==\frac{775.8}{5}\\ =155.16\end{array}$

Hence, the value of $\overline{\overline{x}}$ is 155.16mm.

b)

Summary Introduction

To determine: The value of $\overline{R}$.

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 7P

Hence, the value of $\overline{R}$ is 4.48mm.

Explanation of Solution

Given information:

The following information is given:

Day	Mean (MM)	Range (MM)
1	156.9	4.2
2	153.2	4.6
3	153.6	4.1
4	155.5	5
5	156.6	4.5

Determine the value of $\overline{R}$:

Day	Range (MM)
1	4.2
2	4.6
3	4.1
4	5
5	4.5
Total	22.4

Calculate the $\overline{R}$:

It is calculated by dividing the sum of mean values and the number of days.

$\begin{array}{c}\overline{R}=\frac{\text{Sum of range}}{\text{Number of days}}\\ =\frac{4.2+4.6+4.1+5.0+4.5}{5}\\ =\frac{22.4}{5}\\ =4.48\end{array}$

Hence, the value of $\overline{R}$ is 4.48mm.

c)

Summary Introduction

To plot: The UCL and LCL for $\overline{x}$.

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 7P

Hence, the value of UCL is 156.54mm and LCL is 153.78mm.

Explanation of Solution

Given information:

The following information is given:

Day	Mean (MM)	Range (MM)
1	156.9	4.2
2	153.2	4.6
3	153.6	4.1
4	155.5	5
5	156.6	4.5

Determine the UCL and LCL of $\overline{x}$:

Formulae to determine Upper Control Limit and Lower Control Limit are given as follows:

${\text{UCL}}_{\overline{x}}=\overline{\overline{x}}+{\text{A}}_2\times \overline{R}$

${\text{LCL}}_{\overline{x}}=\overline{\overline{x}}-{\text{A}}_2\times \overline{R}$

Here, the overall mean is $\overline{\overline{x}}=155.16\text{mm}$, the average range is $\overline{R}=4.48\text{mm}$ and A₂ is the Mean factor derived from standard tables showing factors for computing control charts.

Given the sample size of 10, the Mean factor ${\text{A}}_2=0.308$ for $\sigma =3$ (refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by adding the multiple of average range and mean factor with the overall mean.

$\begin{array}{c}{\text{UCL}}_{\overline{x}}=\overline{\overline{x}}+{\text{A}}_2\times \overline{R}\\ =155.16+0.308\times 4.48\\ =156.54\text{mm}\end{array}$

Lower control limit can be calculated by subtracting the multiple of average range and mean factor from the overall mean.

$\begin{array}{c}{\text{LCL}}_{\overline{x}}=\overline{\overline{x}}-{\text{A}}_2\times \overline{R}\\ =155.16-0.308\times 4.48\\ =153.78\text{mm}\end{array}$

Hence, upper control limit is 156.54mm and 153.78mm.

Plot the values:

d)

Summary Introduction

To plot: The UCL and LCL for $\overline{R}$.

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 7P

Hence, the value of UCL is 7.96mm and LCL is 0.999mm.

Explanation of Solution

Given information:

The following information is given:

Day	Mean (MM)	Range (MM)
1	156.9	4.2
2	153.2	4.6
3	153.6	4.1
4	155.5	5
5	156.6	4.5

Determine the UCL and LCL of $\overline{R}$:

Formulae to determine Upper Control Limit and Lower Control Limit are given as follows:

${\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}$

${\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}$

Here, the average range is $\overline{R}=4.48\text{mm}$ and D₃ is the lower range factor and D₄ is the upper range factor.

Given the sample size of 10 for $\sigma =3$, the Lower range factor ${\text{D}}_{\text{3}}=0.223$ and the Upper range factor ${\text{D}}_{\text{4}}=1.777$ (refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by multiplying upper range factor and average range.

$\begin{array}{c}{\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}\\ =1.777\times 4.48\text{mm}\\ =\text{7}\text{.96}\text{mm}\end{array}$

Upper control limit can be calculated by multiplying lower range factor and average range.

$\begin{array}{c}{\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}\\ =0.223\times 4.48\text{mm}\\ =\text{0}\text{.999}\text{mm}\end{array}$

Hence, upper control limit is 7.96mm and 0.999mm.

Plot the values:

e)

Summary Introduction

To determine: The UCL and LCL for $\overline{x}$ if the true diameter mean is 155mm.

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 7P

Hence, the value of UCL is 156.38mm and LCL is 153.62mm.

Explanation of Solution

Given information:

The following information is given:

Day	Mean (MM)	Range (MM)
1	156.9	4.2
2	153.2	4.6
3	153.6	4.1
4	155.5	5
5	156.6	4.5

Given the target value of the diameter $\mu =155\text{mm}$, calculate the upper control limits ${\text{UCL}}_{\overline{x}}$ and ${\text{LCL}}_{\overline{x}}$ using the formula

${\text{UCL}}_{\overline{x}}=\mu +{\text{A}}_2\times \overline{R}$

${\text{LCL}}_{\overline{x}}=\mu -{\text{A}}_2\times \overline{R}$

Here,

The target value of the diameter $\mu =155\text{mm}$, the average range $\overline{R}=4.48\text{mm}$ and A₂ is the Mean factor derived from standard tables showing factors for computing control charts.

Given the sample size of 10, the Mean factor ${\text{A}}_2=0.308$ for $\sigma =3$(refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by adding the multiple of average range and mean factor with the overall mean.

$\begin{array}{c}{\text{UCL}}_{\overline{x}}=\mu +{\text{A}}_2\times \overline{R}\\ =155+0.308\times 4.48\\ =156.38\text{mm}\end{array}$

Lower control limit can be calculated by subtracting the multiple of average range and mean factor from the overall mean.

$\begin{array}{c}{\text{LCL}}_{\overline{x}}=\mu -{\text{A}}_2\times \overline{R}\\ =155-0.308\times 4.48\\ =153.62\text{mm}\end{array}$

Hence, for the $\overline{x}$ chart, the upper control limit is ${\text{UCL}}_{\overline{x}}=156.38\text{mm}$ and the lower control limit is ${\text{LCL}}_{\overline{x}}=153.62\text{mm}$.