Chapter 6.S, Problem 8P

a)

Summary Introduction

To determine: The value of control limits.

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 8P

The value of control limits is 16.08 and 15.92.

Explanation of Solution

Given information:

The following information is given:

The mean of the sample means $\overline{\overline{x}}=16$

Calculate $\left({\sigma }_{\overline{x}}\right)$ the standard deviation of the means:

It is given that the population mean $\overline{\overline{x}}=16$ and the population standard deviation $\sigma =0.12$ using the below formula:

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

Here, the sample size $n=9$

Substitute the values of $\sigma =0.12$ and $n=9$ in the above formula to obtain the value of ${\sigma }_{\overline{x}}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.12}{\sqrt{9}}\\ =\frac{0.12}{3}\\ =0.04\end{array}$

Hence, the standard deviation of the means ${\sigma }_{\overline{x}}=0.04$

Derive the values of upper control limits (UCL) and lower control limits (LCL):

Calculate UCL and LCL using the below formulae:

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

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

Here,

The value of $z=2$, population mean $\overline{\overline{x}}=16$ and the standard deviation of the means ${\sigma }_{\overline{x}}=0.04$.

Compute the value of UCL by substituting in UCL formula, the values of $\overline{\overline{x}}=16$, ${\sigma }_{\overline{x}}=0.04$ and $z=2$

$\begin{array}{c}\text{UCL}=\overline{\overline{x}}+z\times {\sigma }_{\overline{x}}\\ =1\text{6}+2\times \left(0.04\right)\\ =16+0.08\\ =16.08\end{array}$

Hence, the upper control limit is $\text{UCL}=16.08\text{lb}$

Compute the value of LCL by substituting in LCL formula, the values of $\overline{\overline{x}}=16$, ${\sigma }_{\overline{x}}=0.04$ and $z=2$

$\begin{array}{c}\text{LCL}=\overline{\overline{x}}-z\times {\sigma }_{\overline{x}}\\ =16-2\times \left(0.04\right)\\ =16-0.08\\ =15.92\end{array}$

Hence, the lower control limit is $\text{LCL}=15.92\text{lb}$

b)

Summary Introduction

To determine: The value of control limits.

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 8P

The value of control limits is 16.12 and 15.88.

Explanation of Solution

Given information:

The following information is given:

The mean of the sample means $\overline{\overline{x}}=16$

Calculate $\left({\sigma }_{\overline{x}}\right)$ the standard deviation of the means:

It is given that the population mean $\overline{\overline{x}}=16$ and the population standard deviation $\sigma =0.12$ using the below formula:

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

Here, the sample size $n=9$

Substitute the values of $\sigma =0.12$ and $n=9$ in the above formula to obtain the value of ${\sigma }_{\overline{x}}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.12}{\sqrt{9}}\\ =\frac{0.12}{3}\\ =0.04\end{array}$

Hence, the standard deviation of the means ${\sigma }_{\overline{x}}=0.04$

Derive the values of upper control limits (UCL) and lower control limits (LCL):

Calculate UCL and LCL using the below formulae:

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

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

Here,

The value of $z=3$, population mean $\overline{\overline{x}}=16$ and the standard deviation of the means ${\sigma }_{\overline{x}}=0.04$.

Compute the value of UCL by substituting in UCL formula, the values of $\overline{\overline{x}}=16$, ${\sigma }_{\overline{x}}=0.04$ and $z=3$

$\begin{array}{c}\text{UCL}=\overline{\overline{x}}+z\times {\sigma }_{\overline{x}}\\ =1\text{6}+3\times \left(0.04\right)\\ =16+0.12\\ =16.12\end{array}$

Hence, the upper control limit is $\text{UCL}=16.12\text{lb}$

Compute the value of LCL by substituting in LCL formula, the values of $\overline{\overline{x}}=16$, ${\sigma }_{\overline{x}}=0.04$ and $z=3$

$\begin{array}{c}\text{LCL}=\overline{\overline{x}}-z\times {\sigma }_{\overline{x}}\\ =16-2\times \left(0.04\right)\\ =16-0.12\\ =15.88\end{array}$

Hence, the lower control limit is $\text{LCL}=15.88\text{lb}$

The control limits become wider when three standard deviations are used instead of two standard deviations. The process is now allowed wider latitude in terms of natural variations.