Chapter 10, Problem 20P

a)

Summary Introduction

To determine: The mean of each sample.

Answer to Problem 20P

The mean of each sample is shown in Table 1.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Calculation of mean of each sample:

Sample
Sl. No.	1	2	3	4
1	4.5	4.6	4.5	4.7
2	4.2	4.5	4.6	4.6
3	4.2	4.4	4.4	4.8
4	4.3	4.7	4.4	4.5
5	4.3	4.3	4.6	4.9
Mean	4.3	4.5	4.5	4.7

Table 1

Excel Worksheet:

Sample 1:

$\begin{array}{l}=\frac{4.5+4.2+4.2+4.3+4.3}{5}\\ =4.3\end{array}$

The mean is calculated by adding each sample points. Adding the points 4.5, 4.2, 4.2, 4.3 and 4.3 and dividing by 5 gives mean of 4.3. The same process is followed for finding mean for other samples.

Hence, the mean of each sample is shown in Table 1

b)

Summary Introduction

To determine: The mean and standard deviation when the process parameters are unknown.

Answer to Problem 20P

The mean and standard deviation when the process parameters are unknown are 4.5 and 0.192.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Calculation of mean and standard deviation:

Table 1 provides the mean for each sample points.

$\begin{array}{c}\overline{\overline{\text{X}}}=\frac{{\displaystyle \sum _{\text{n=1}}^{\text{5}}\overline{\text{X}}}}{\text{n}}\\ =\frac{4.3+4.5+4.5+4.7}{4}\\ =4.5\end{array}$ (1)

The mean is calculated by adding each mean of the samples. Adding the points 4.3, 4.5, 4.5 and 4.7 and dividing by 4 gives mean of 4.5.

$\begin{array}{c}\text{S}=\sqrt{\frac{1}{\text{N}-1}{{\displaystyle \sum _{\text{i}=\text{1}}^{\text{N}}\left({\text{x}}_{\text{i}}-\overline{\text{x}}\right)}}^2}\\ =0.192\end{array}$

The standard deviation is calculated using the above formula and substituting the values of mean in the above formula and the resultant of 0.192 is obtained.

Hence, the mean and standard deviation when the process parameters are unknown are 4.5 and 0.192.

c)

Summary Introduction

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

Answer to Problem 20P

The mean and standard deviation of the sampling distribution is4.5 and 0.086 respectively.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Calculation of mean and standard deviation of the sampling distribution:

From calculation of mean of each samples, the mean for sampling distribution can be computed, the mean for sampling distribution is 4.5 (refer equation (1)).

$\begin{array}{c}{\sigma }_{\overline{\text{x}}}=\frac{\sigma }{\sqrt{\text{n}}}\\ =\frac{0.192}{\sqrt{5}}\\ =0.086\end{array}$ (2)

The standard deviation of the sampling distribution is calculated by dividing 0.192 with the square root of 5 which gives the resultant as 0.086.

Hence, the mean and standard deviation of the sampling distribution is 4.5 and 0.086 respectively.

d)

Summary Introduction

To determine: The three-sigma control limit for the process and alpha risk provided by them.

Answer to Problem 20P

The three-sigma control limits for the process are 4.758 and 4.242.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Calculation of three-sigma control limit for the process:

$\begin{array}{l}=4.5\pm 3.00\left(0.086\right)\\ =4.5\pm 0.258\end{array}$

$\begin{array}{c}\text{UCL}=4.5+0.258\\ =4.758\\ \text{LCL}=4.5-0.258\\ =4.242\end{array}$ (3)

The three-sigma control limits for the process is calculated by multiplying 3.00 with 0.086 (refer equation (2)) and the resultant is added with 4.5 to get an upper control limit which is 4.758 and subtracted to get lower control limit which is 4.242. Using z-factor table z = +3.00 corresponds to 0.4987.

$\begin{array}{c}=0.5–0.4987\\ =0.0013\text{ in the tail}.\\ =2\times 0.0013\\ =0.0026\end{array}$

The alpha risk is calculated to be as 0.0026.

Hence, the three-sigma control limits for the process are 4.758 and 4.242.

e)

Summary Introduction

To determine: The alpha risk for control limits of 4.14 and 4.86.

Answer to Problem 20P

The alpha risk for control limits of 4.14 and 4.86 is +4.1.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Formula:

$\text{z}=\frac{\text{x}-\text{μ}}{\text{σ}}$

Calculation alpha risk for control limits of 4.14 and 4.86:

$\begin{array}{c}\text{z}=\frac{4.86-4.5}{0.086}\\ =\frac{0.36}{0.086}\\ =+4.19\end{array}$

The alpha risk is calculated by dividing the difference of 4.86 and 4.5 with 0.086 which gives +4.19 which is the risk is close to zero.         

Hence, the alpha risk for control limits of 4.14 and 4.86 is +4.1

f)

Summary Introduction

To determine: Whether any of the sample means are beyond the control limits.

Answer to Problem 20P

There are no sample means which lies beyond the control limits.

Explanation of Solution

Given information:

Sample
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Determination of whether any of the sample means are beyond the control limits:

Table 1 provides the sample means for each sample. From observation, it can be found that each sample mean are within the control limit of 4.14 and 4.86. Therefore, each sample means lies within the control limits of 4.14 and 4.86.

Hence, there are no sample means which lies beyond the control limits.

g)

Summary Introduction

To determine: Whether any of the samples are beyond the control limits.

Answer to Problem 20P

All points are within control limits.

Explanation of Solution

Given information:

SAMPLE
1	2	3	4
4.5	4.6	4.5	4.7
4.2	4.5	4.6	4.6
4.2	4.4	4.4	4.8
4.3	4.7	4.4	4.5
4.3	4.3	4.6	4.9

Formula:

Mean Chart:

$\begin{array}{l}\text{UCL}=\overline{\overline{\text{X}}}+{\text{A}}_{\text{2}}\overline{\text{R}}\\ \text{LCL}=\overline{\overline{\text{X}}}-{\text{A}}_{\text{2}}\overline{\text{R}}\end{array}$

Range Chart:

$\begin{array}{c}\text{UCL}={\text{D}}_4\overline{\text{R}}\\ \text{LCL}={\text{D}}_3\overline{\text{R}}\end{array}$

Calculation of upper and lower control limits:

	SAMPLE
	1	2	3	4
	4.5	4.6	4.5	4.7
	4.2	4.5	4.6	4.6
	4.2	4.4	4.4	4.8
	4.3	4.7	4.4	4.5
	4.3	4.3	4.6	4.9
Mean	4.3	4.5	4.5	4.7
Range	.3	.4	.2	.4

From factors of three-sigma chart, A₂ = 0.58; D₃ = 0; D₄ = 2.11.

Mean control chart:

$\begin{array}{c}\overline{\overline{\text{X}}}=\frac{{\displaystyle \sum _{\text{i}=\text{1}}^4\overline{\text{X}}}}{\text{n}}\\ =\frac{4.3+4.5+4.5+4.7}{4}\\ =4.5\end{array}$

Upper control limit:

$\begin{array}{c}\text{UCL}=\overline{\overline{\text{X}}}+{\text{A}}_{\text{2}}\overline{\text{R}}\\ =4.5+\left(0.58\times 0.325\right)\\ =4.689\end{array}$

The Upper control limit is calculated by adding the product of 0.58 and 0.325 with 4.5 which yields 4.689.

Lower control limit:

$\begin{array}{c}\text{LCL}=\overline{\overline{\text{X}}}-{\text{A}}_{\text{2}}\overline{\text{R}}\\ =4.5-\left(0.58\times 0.325\right)\\ =4.311\end{array}$

The Lower control limit is calculated by subtracting the product of 0.58 and 0.325 with 4.5 which yields 4.311.

The UCL and LCL for mean charts are 4.686 and 4.311. (4)

A graph is plotted using the UCL and LCL and mean values which shows the points are within the control limits.

Range control chart:

$\begin{array}{c}\overline{\text{R}}=\frac{{\displaystyle \sum _{i=1}^4\text{R}}}{4}\\ =\frac{0.3+0.4+0.2+0.4}{4}\\ =0.325\end{array}$

Upper control limit:

$\begin{array}{c}\text{UCL}={\text{D}}_4\overline{\text{R}}\\ =2.11\times 0.325\\ =0.686\end{array}$

The Upper control limit is calculated by multiplying 2.11 with 0.325 which yields 0.686.

Lower control limit:

$\begin{array}{c}\text{LCL}={\text{D}}_3\overline{\text{R}}\\ =0\times 0.325\\ =0.0\end{array}$

The lower control limit is calculated by multiplying 0 with 0.325 which yields 0.0.

A graph is plotted using the UCL, LCL and Range values which shows that the points are within the control region.

Hence, all points are within control limits.

h)

Summary Introduction

To explain: The reason for variations in control limits.

Answer to Problem 20P

Use of different measures for dispersion to the measure the standard deviation and range.

Explanation of Solution

Reason for variations in control limits:

The control limits vary because in equation (3) and (4) because of the use of different measure for dispersion to measure the standard deviation and range.

Hence, the difference arises due to the use of different measures for dispersion to the measure the standard deviation and range.

i)

Summary Introduction

To determine: The control limits for the process and whether the process will be in control.

Answer to Problem 20P

The process is out of control with UCL=4.641 and LCL=4.159.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Mean}=4.4\\ \text{Standard}\text{deviation}=0.18\end{array}$

Determination of control limits of the process:

Sample mean is given in Table 1.

$\begin{array}{c}=4.4\pm 3\left(\frac{0.18}{\sqrt{5}}\right)\\ =4.4\pm 0.241\\ \text{UCL}=4.641\\ \text{LCL}=4.159\end{array}$

To calculate the control limits 0.18 is divided by root of 5 and is multiplied by 3 and the resultant is added to 4.4 to give UCL which is 4.641 and subtracted from 4.4 to get the LCL which is 4.159.

The graph shows that the some of the points are above the control limits which make the process to be out of control.

Hence, the process is out of control with UCL=4.641 and LCL=4.159.