Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
12th Edition
ISBN: 9781259580093
Author: William J Stevenson
Publisher: McGraw-Hill Education
bartleby

Concept explainers

Question
Book Icon
Chapter 10, Problem 20P

a)

Summary Introduction

To determine: The mean of each sample.

a)

Expert Solution
Check Mark

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:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 20P , additional homework tip  1

Sample 1:

=4.5+4.2+4.2+4.3+4.35=4.3

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.

b)

Expert Solution
Check Mark

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.

X¯¯=n=15X¯n=4.3+4.5+4.5+4.74=4.5 (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.

S=1N1i=1N(xi-x¯)2=0.192

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.

c)

Expert Solution
Check Mark

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)).

σx¯=σn=0.1925=0.086 (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.

d)

Expert Solution
Check Mark

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:

=4.5±3.00(0.086)=4.5±0.258

UCL=4.5+0.258=4.758LCL=4.50.258=4.242 (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.

=0.5  0.4987 = 0.0013 in the tail.=2×0.0013=0.0026

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.

e)

Expert Solution
Check Mark

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:

z=x-μσ

Calculation alpha risk for control limits of 4.14 and 4.86:

z=4.86-4.50.086=0.360.086=+4.19

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.

f)

Expert Solution
Check Mark

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.

g)

Expert Solution
Check Mark

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:

UCL=X¯¯+A2R¯LCL=X¯¯-A2R¯

Range Chart:

UCL=D4R¯LCL=D3R¯

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:

X¯¯=i=14X¯n=4.3+4.5+4.5+4.74=4.5

Upper control limit:

UCL=X¯¯+A2R¯=4.5+(0.58×0.325)=4.689

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:

LCL=X¯¯-A2R¯=4.5(0.58×0.325)=4.311

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)

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 20P , additional homework tip  2

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

Range control chart:

R¯=i=14R4=0.3+0.4+0.2+0.44=0.325

Upper control limit:

UCL=D4R¯=2.11×0.325=0.686

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

Lower control limit:

LCL=D3R¯=0×0.325=0.0

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

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 20P , additional homework tip  3

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.

h)

Expert Solution
Check Mark

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.

i)

Expert Solution
Check Mark

Answer to Problem 20P

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

Explanation of Solution

Given information:

Mean=4.4Standarddeviation=0.18

Determination of control limits of the process:

Sample mean is given in Table 1.

=4.4±3(0.185)=4.4±0.241UCL=4.641LCL=4.159

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.

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 20P , additional homework tip  4

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
An automatic filling machine is used to fill 1-liter bottles of cola.  The machine’s output is approximately normal with a mean of 1.0 liter and standard deviation of .01 liter. Output is monitored using means of samples of 25 observations.  Determine upper and lower control limits that will include roughly 97% of the sample means when the process is in control. Using Appendix B, Table A to find the value of Z corresponding to the mean control limits.
A Quality Analyst wants to construct a control chart for determining whether three machines, all producing the same product, are under control with regard to a particular quality variable. Accordingly, he sampled four units of output from each machine, with the following results: Machine Measurements #1 17 15 15 17 #2 16 25 18 25 # 3 23 24 23 22 What is the estimate of the process mean for whenever it is under control? What is the sample average range based upon this limited sample? What are the x-bar chart upper and lower control limits?
At Gleditsia Triacanthos Company, a certain manufactured part is deemed acceptable if its length is between 12.45 to 12.55 inches. The process is normally distributed with an average of 12.49 inches and a standard deviation of 0.014 inches. a) is the process capable of meeting specifications? b) Does the process meet specifications?
Knowledge Booster
Background pattern image
Operations Management
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, operations-management and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Practical Management Science
Operations Management
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:Cengage,
Text book image
Operations Management
Operations Management
ISBN:9781259667473
Author:William J Stevenson
Publisher:McGraw-Hill Education
Text book image
Operations and Supply Chain Management (Mcgraw-hi...
Operations Management
ISBN:9781259666100
Author:F. Robert Jacobs, Richard B Chase
Publisher:McGraw-Hill Education
Text book image
Business in Action
Operations Management
ISBN:9780135198100
Author:BOVEE
Publisher:PEARSON CO
Text book image
Purchasing and Supply Chain Management
Operations Management
ISBN:9781285869681
Author:Robert M. Monczka, Robert B. Handfield, Larry C. Giunipero, James L. Patterson
Publisher:Cengage Learning
Text book image
Production and Operations Analysis, Seventh Editi...
Operations Management
ISBN:9781478623069
Author:Steven Nahmias, Tava Lennon Olsen
Publisher:Waveland Press, Inc.