Chapter 6.S, Problem 13P

a)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 100 and 3-sigma 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 13P

Hence, the UCL is 0.05145 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 100 and 3-sigma limits:

${\text{UCL}}_{\text{p}}=\overline{p}\text{+}z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

${\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 3-sigma limit.

${\sigma }_p$ is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}$:

${\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}$

The given values of $\overline{p}=1.5\%$ and the sample size $n=100$ and obtain the value of ${\sigma }_{\stackrel{⌢}{\text{p}}}$

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{0.015\times \left(0.985\right)}{100}}\\ =\sqrt{\frac{0.014775}{100}}\\ =\sqrt{0.00014775}\\ =0.012155\end{array}$

Hence, the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$

The given values of $\overline{p}=1.5\%$ and $z=3$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$ has been determine. Substitute the values in UCL formula to determine the value of UCL:

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015+3\times 0.01215\\ =0.05145\end{array}$

Hence, the upper control chart limit ${\text{UCL}}_{\text{p}}=0.05145$

The given values of $\overline{p}=1.5\%$ and $z=3$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$ has been determine. Substitute the values in LCL formula to determine the value of LCL:

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015-3\times 0.01215\\ =-0.02145\\ =0\end{array}$

Hence, lower control chart limit cannot be negative therefore ${\text{LCL}}_{\text{p}}=0$

b)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 50 and 3-sigma 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 13P

Hence, the UCL is 0.06657 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 50 and 3-sigma limits:

${\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

${\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 3-sigma limit.

${\sigma }_p$ is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}$:

${\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}$

The given values of $\overline{p}=1.5\%$ and the sample size $n=50$ and obtain the value of ${\sigma }_{\stackrel{⌢}{\text{p}}}$

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{0.015\times \left(0.985\right)}{50}}\\ =\sqrt{\frac{0.014775}{50}}\\ =\sqrt{0.0002955}\\ =0.017190\end{array}$

Hence, the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$

The given values of $\overline{p}=1.5\%$ and $z=3$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$ has been determine. Substitute the values in UCL formula to determine the value of UCL:

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015+3\times 0.017190\\ =0.06657\end{array}$

Hence, the upper control chart limit ${\text{UCL}}_{\text{p}}=0.06657$

The given values of $\overline{p}=1.5\%$ and $z=3$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$ has been determine. Substitute the values in LCL formula to determine the value of LCL:

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015-3\times 0.017190\\ =-0.03657\\ =0\end{array}$

Hence, lower control chart limit cannot be negative therefore ${\text{LCL}}_{\text{p}}=0$

c)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 100 and 2-sigma 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 13P

Hence, the UCL is 0.0393 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 100 and 2-sigma limits:

${\text{UCL}}_{\text{p}}=\overline{p}\text{+}z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

${\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 2-sigma limit.

${\sigma }_p$ is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}$:

${\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}$

The given values of $\overline{p}=1.5\%$ and the sample size $n=100$ and obtain the value of ${\sigma }_{\stackrel{⌢}{\text{p}}}$

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{0.015\times \left(0.985\right)}{100}}\\ =\sqrt{\frac{0.014775}{100}}\\ =\sqrt{0.00014775}\\ =0.012155\end{array}$

Hence, the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$

The given values of $\overline{p}=1.5\%$ and $z=2$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$ has been determine. Substitute the values in UCL formula to determine the value of UCL:

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015+2\times 0.01215\\ =0.0393\end{array}$

Hence, the upper control chart limit ${\text{UCL}}_{\text{p}}=0.0393$

The given values of $\overline{p}=1.5\%$ and $z=2$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.012155$ has been determine. Substitute the values in LCL formula to determine the value of LCL:

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015-2\times 0.01215\\ =-0.0093\\ =0\end{array}$

Hence, lower control chart limit cannot be negative therefore ${\text{LCL}}_{\text{p}}=0$

d)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 50 and 2-sigma 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 13P

Hence, the UCL is 0.04938 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 50 and 2-sigma limits:

${\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

${\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}$

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 2-sigma limit.

${\sigma }_p$ is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}$:

${\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{\overline{p}\times \left(1-\overline{p}\right)}{n}}$

The given values of $\overline{p}=1.5\%$ and the sample size $n=50$ and obtain the value of ${\sigma }_{\stackrel{⌢}{\text{p}}}$

$\begin{array}{c}{\sigma }_{\stackrel{⌢}{\text{p}}}=\sqrt{\frac{0.015\times \left(0.985\right)}{50}}\\ =\sqrt{\frac{0.014775}{50}}\\ =\sqrt{0.0002955}\\ =0.017190\end{array}$

Hence, the standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$

The given values of $\overline{p}=1.5\%$ and $z=2$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$ has been determine. Substitute the values in UCL formula to determine the value of UCL:

$\begin{array}{c}{\text{UCL}}_{\text{p}}=\overline{p}+z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015+2\times 0.017190\\ =0.04938\end{array}$

Hence, the upper control chart limit ${\text{UCL}}_{\text{p}}=0.04938$

The given values of $\overline{p}=1.5\%$ and $z=2$ and ${\sigma }_{\stackrel{⌢}{\text{p}}}=0.017190$ has been determine. Substitute the values in LCL formula to determine the value of LCL:

$\begin{array}{c}{\text{LCL}}_{\text{p}}=\overline{p}-z\times {\sigma }_{\stackrel{⌢}{\text{p}}}\\ =0.015-3\times 0.017190\\ =-0.01938\\ =0\end{array}$

Hence, lower control chart limit cannot be negative therefore ${\text{LCL}}_{\text{p}}=0$

e)

Summary Introduction

To determine: What happen to ${\widehat{\sigma }}_p$ if the sample size is larger.

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

Explanation of Solution

Determine what happen to ${\widehat{\sigma }}_p$ if the sample size is larger:

The standard deviation of the sampling distribution ${\sigma }_{\stackrel{⌢}{\text{p}}}$. reduces when the sample size increases. However, in the calculation the sample size is in the denominator.

f)

Summary Introduction

To determine: Why LCL cannot be less than 0.

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

Explanation of Solution

Determine why LCL cannot be less than 0:

The lower control limit for defectives cannot be less than zero because it becomes negative. A negative lower control limit is meaningless.