Chapter 6.S, Problem 10P

a)

Summary Introduction

To determine: The value of ${\sigma }_x$.

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 10P

The value of ${\sigma }_x$ is 0.61.

Explanation of Solution

Given information:

The following information is given:

Sample size is given as 5 and process standard deviation is given as 1.36.

Determine ${\sigma }_x$:

The standard deviation of the sample means denoted by ${\sigma }_{\overline{x}}$ is calculated by the following formula

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Here,

σ refers to process standard deviation

n refers to the sample size.

The given values are $\sigma =1.36$ and sample size $n=5$, substitute the values in the above formula:

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{1.36}{\sqrt{5}}\\ =\frac{1.36}{2.23}\\ =0.61\end{array}$

The standard deviation of the sample means ${\sigma }_{\overline{x}}=0.61$

b)

Summary Introduction

To determine: The control limits for the mean chart if the value of z is 3.

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 10P

The UCL value of $\overline{x}$ is 11.83 and the LCL value is 8.17.

Explanation of Solution

Given information:

The following information is given:

Sample size is given as 5 and process standard deviation is given as 1.36.

Determine the control limits for the mean chart if the value of z is 3:

Formulae to calculate control limits:

$\text{UCL}=\overline{\overline{x}}+z\times {\sigma }_{\overline{x}}$ (1)

$\text{LCL}=\overline{\overline{x}}-z\times {\sigma }_{\overline{x}}$ (2)

Here,

The value of $z=3$

$\overline{\overline{x}}$ is the mean of the sample means

the standard deviation of the mean ${\sigma }_{\overline{x}}=0.61$

Calculate the average for each sample:

Working note:

Average for sample #1:

It is calculated by dividing the total of sample #1 and sample size.

$\begin{array}{c}\text{Average}=\frac{\text{Total of Sample \#1}}{\text{Sample size}}\\ =\frac{51}{5}\\ =10\end{array}$

Note: The same continues for all the samples.

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

It is calculated by dividing the sum of average of all the samples and the number of samples. Hence, the value of $\overline{\overline{x}}$ is 10.

$\begin{array}{c}\overline{\overline{x}}=\frac{\text{Average of all samples}}{\text{Number of samples}}\\ =\frac{10+10+10.2+10.2+10.4+9.8+9.6+9.8+9.8+10}{10}\\ =10\end{array}$

Substitute the values in equation (1)to determine the value of UCL as follows:

$\begin{array}{c}\text{UCL}=\overline{\overline{x}}+z\times {\sigma }_{\overline{x}}\\ =\text{10}+3\times \left(0.61\right)\\ =10+1.83\\ =11.83\end{array}$

Hence, the UCL value is 11.83.

Substitute the values in equation (2) to determine the value of LCL as follows:

$\begin{array}{c}\text{LCL}=\overline{\overline{x}}-z\times {\sigma }_{\overline{x}}\\ =\text{10}-3\times \left(0.61\right)\\ =10-1.83\\ =8.17\end{array}$

Hence, the LCL value is 8.17.

c)

Summary Introduction

To determine: The control limits for the range chart.

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 10P

The UCL value of R-chart is 6.9795 and the LCL value is 0.

Explanation of Solution

Given information:

The following information is given:

Sample size is given as 5 and process standard deviation is given as 1.36.

Determine the control limits for the mean chart if the value of z is 3:

Formulae to calculate control limits:

${\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}$ (3)

${\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}$ (4)

Here,

${\text{D}}_{\text{3}}$ refers to the lower range factor

${\text{D}}_{\text{4}}$ refers to the upper range factor

$\overline{R}$ refers to the average range

Substitute the values in equation (3) to determine the value of UCL as follows:

$\begin{array}{c}{\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}\\ =2.115\times 3.3\\ \text{=6}\text{.9795}\end{array}$

Hence, the UCL value is 6.9795.

Substitute the values in equation (4) to determine the value of LCL as follows:

$\begin{array}{c}{\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}\\ =0\end{array}$

Hence, the LCL value is 0.

d)

Summary Introduction

To determine: Whether the process is 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 10P

The process is in statistical control.

Explanation of Solution

Given information:

The following information is given:

Sample size is given as 5 and process standard deviation is given as 1.36.

Plot the sample mean values in the $\overline{x}$ control chart where ${\text{UCL}}_{\overline{x}}=11.83$ and ${\text{LCL}}_{\overline{x}}=8.17$

Plot the sample mean values in theR-control chart where ${\text{UCL}}_{\text{R}}\text{=6}\text{.9795}\text{oz}$ and ${\text{LCL}}_{\text{R}}\text{=0}$

The sample range values lie well within the upper control limit and lower control limits.

The process is in statistical control.