**Assignment 1: Discussion**

You are working with a company selling building material to builders. You predict the quarterly purchases of customers based on their current purchases by

using a linear regression model. These predictions, however, are not very accurate. Discuss at least three reasons why these predictions may not be accurate and offer three ways in which you can increase the likelihood of accurately predicting your customers’ purchases.

Post your response to the Discussion Area by Friday, March 8, 2013.

**Assignment 2: Controls**

As a quality analyst you are also responsible for controlling the weight of a box of cereal. The Operations Manager asks you to identify the ways in which statistical quality control methods can be applied to the weights of the boxes. Provide your recommendations to the Operations Manager in a two-three page report. Using the data provided in the

Doc Sharing

area labeled M4A2Data, create Xbar and R charts.

Your report should indicate the following along with valid justifications of your answers:

- The control limits of the weights of the boxes.
- Nonrandom patterns or trends, if any.
- If the process is in control.
- The appropriate action if the process is not in control.

Submit your handout and summary to the

M4: Assignment 2 Dropbox

by **Monday, March 11, 2013**.

>

Dataand Answers

1.3

0

0

0

6.300 6.260

0

26.320

6.330 6.320

30

6.330

6.360 6.290

0

4 6.300 6.2900

6.340 6.290

0

56.390 6.295

66.330

6.330 6.292

76.323

6.400 6.289

1

86.471 6.286

5

96.498 6.283

5

10 6.2800

6.525 6.280

5

116.390 6.335 6.390 6.277

6.400 6.339 6.400 6.274

As a quality analyst you are also responsible for controlling the weight of a box of cereal. The Operations Manager asks you to identify the ways in which statistical quality control methods can be applied to the weights of the boxes. Provide your recommendations to the Operations Manager in a two-page report. Using the data provided in the Doc Sharing area, create Xbar and R charts. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Your report should indicate the following along with valid justifications of your answers: | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

The control limits of the weights of the boxes. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Nonrandom patterns or trends, if any. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

If the process is in control. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

The appropriate action if the process is not in control. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Sample | Box | 1 | Box 2 | Box | 3 | Sample Means | Maximum | Minimum | Sample | Range | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6 | 0 | 6.2 | 8 | 6. | 26 | 6.280 | 0.0 | 4 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.3 | 20 | 6.330 | 6.3 | 23 | 0.0 | 10 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.2 | 9 | 6.360 | 6.32 | 7 | 0.07 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.340 | 6.31 | 0.0 | 5 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.295 | 6.3 | 15 | 6.390 | 6.333 | 0.095 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.292 | 6.3 | 19 | 6.3 | 14 | 0.038 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.289 | 6.400 | 6.337 | 0. | 11 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.286 | 6.327 | 6.471 | 6.361 | 0. | 18 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.283 | 6.331 | 6.498 | 6.371 | 0. | 21 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.335 | 6.5 | 25 | 6.38 | 0. | 24 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

6.277 | 6.339 | 0.1 | 13 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

12 | 6.274 | 6.343 | 0.126 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

See all the other sheets in this file. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

(a) The control limits are: (a) For Mean weight, LCL = 6.3087 ounces, UCL = 6.3696 ounces and (b) For Range, LCL = 0.0324 ounes, UCL = 0.1964 ounce | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

(b) At the time measurements were taken on the 5th through 9th samples, the mean weights and the range of weights appears to have been increasing in a somewhat linear pattern. It appears that the filling machine has gradually gone off setting during this period. This is the only non-random variation observed at the time of taking the 12 measurements. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

(c ) The process is not in control since we see points lying outside the control limits in both x-bar chart and R chart. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

(d) Corrective action in terms of resetting the machine is required. Trials should be run and again more samples drawn to check if the process has come under control. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

See all the sheets for the calculations and tables. The process is in statistical control. |

Chart

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

10 10 10 10

11 11 11 11

XBar

LCL-X Center-X UCL-X

Sample Mean

Control Chart

LCL

XBar

UCL

424 6.3266666667

6.3087307879

6.3391515152

6.3695722424

6.31

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.38

6.3087307879

6.3391515152

6.3695722424

6.3087307879

6.3391515152

6.3695722424

6.339

6.3087307879

6.3391515152

6.3695722424

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

10 10 10 10

11 11 11 11

Range

LCL-R Center-R UCL-RSample Range

Control Chart CalculationsLCL

RBarUCL

0.01 0.0323649091 0.1143636364 0.19636236360.07

0.0323649091

0.1143636364

0.1963623636

0.0323649091

0.1143636364

0.1963623636

0.095

0.0323649091

0.1143636364

0.1963623636

0.038

0.0323649091

0.1143636364

0.1963623636

0.0323649091

0.1143636364

0.1963623636

0.185

0.0323649091

0.1143636364

0.1963623636

0.215

0.0323649091

0.1143636364

0.1963623636

0.245

0.0323649091

0.1143636364

0.1963623636

0.113

0.0323649091

0.1143636364

0.1963623636

0.126

0.0323649091

0.1143636364

0.1963623636

XBar Range LCL-R Center-R UCL-R LCL-X Center-X UCL-X

2 6.3266666667 0.07 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

3 6.31 0.05 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

4 6.3333333333 0.095 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

5 6.3136666667 0.038 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

6 6.3373333333 0.111 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

7 6.3613333333 0.185 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

8 6.3706666667 0.215 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

9 6.38 0.245 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

10 6.3353333333 0.113 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

11 6.339 0.126 0.0323649091 0.1143636364 0.1963623636 6.3087307879 6.3391515152 6.3695722424

Number |

Calculations

Control Chart CalculationsData

12 2 0

3 04 0

RBar 0.1143636364 5 06 0

7

89

0.0323649091 10

Center 0.1143636364 110.1963623636 12 0.283 1.717

1314

6.3391515152 15

0.223

0.266 16

17

1819

Lower Control Limit 6.3087307879 20 Center 6.3391515152 21 Upper Control Limit 6.3695722424 22 23 24 25 26 Factor value not available. Possible error in sample/subgroup size. Factor value not available. Possible error in sample/subgroup size.

Control Chart Factors Table. | ||||||

Subgroup size | D3 | D4 | A2 | |||

Sample/Subgroup Size | 3.267 | 1.880 | ||||

2.575 | 1.023 | |||||

R Chart Intermediate Calculations | 2.282 | 0.729 | ||||

2.114 | 0.577 | |||||

D3 Factor | 0.283 | 2.004 | 0.483 | |||

D4 Factor | 1.7 | 17 | 0.076 | 1.924 | 0.419 | |

0.136 | 1.864 | 0.373 | ||||

R Chart Control Limits | 0.184 | 1.8 | 16 | 0.337 | ||

Lower Control Limit | 0.223 | 1.777 | 0.308 | |||

0.256 | 1.744 | 0.285 | ||||

Upper Control Limit | 0.266 | |||||

0.307 | 1.693 | 0.249 | ||||

XBar Chart Intemediate Calculations | 0.328 | 1.672 | 0.235 | |||

Average of Subgroup Averages | 0.347 | 1.653 | ||||

A2 Factor | 0.363 | 1.637 | 0.212 | |||

A2 Factor * RBar | 0.0304207273 | 0.378 | 1.622 | 0.203 | ||

0.391 | 1.609 | 0.194 | ||||

XBar Chart Control Limits | 0.404 | 1.596 | 0.187 | |||

0.415 | 1.585 | 0.180 | ||||

0.425 | 1.575 | 0.173 | ||||

0.435 | 1.565 | 0.167 | ||||

0.443 | 1.557 | 0.162 | ||||

0.452 | 1.548 | 0.157 | ||||

0.459 | 1.541 | 0.153 | ||||

Factor value not available. Possible error in sample/subgroup size. | ||||||

Steps in Constructing the Xbar Chart |

1. Find the mean of each subgroup Xbar(1), Xbar(2), Xbar(3)… Xbar(k) and the grand mean of all subgroups using:

2. Find the UCL and LCL using the following equations:

A(3) can be found in the following table:

n A(3) n A(3)

2 2.659 6 1.287

3 1.954 7 1.182

4 1.628 8 1.099

5 1.427 9 1.032

3. Plot the LCL, UCL, centerline, and subgroup means

4. Interpret the data using the following guidelines to determine if the process is in control:

· One point outside the 3 sigma control limits

· Eight successive points on the same side of the centerline

· Six successive points that increase or decrease

· Two out of three points that are on the same side of the centerline, both at a distance exceeding 2 sigmas from the centerline

· Four out of five points that are on the same side of the centerline, four at a distance exceeding 1 sigma from the centerline f. Using an average run length (ARL) for determining process anomalies

Example: The following data consists of 20 sets of three measurements of the diameter of an engine shaft.

n

#1

#2

#3

StdDev

Xbar

1

2.0000

1.9998

2.0002

0.0002

2.0000

2

1.9998

2.0003

2.0002

0.0003

2.0001

3

1.9998

2.0001

2.0005

0.0004

2.0001

4

1.9997

2.0000

2.0004

0.0004

2.0000

5

2.0003

2.0003

2.0002

0.0001

2.0003

6

2.0004

2.0003

2.0000

0.0002

2.0002

7

1.9998

1.9998

1.9998

0.0000

1.9998

8

2.0000

2.0001

2.0001

0.0001

2.0001

9

2.0005

2.0000

1.9999

0.0003

2.0001

10

1.9995

1.9998

2.0001

0.0003

1.9998

11

2.0002

1.9999

2.0001

0.0002

2.0001

12

2.0002

1.9998

2.0005

0.0004

2.0002

13

2.0000

2.0001

1.9998

0.0002

2.0000

14

2.0000

2.0002

2.0004

0.0002

2.0002

15

1.9994

2.0001

1.9996

0.0004

1.9997

16

1.9999

2.0003

1.9993

0.0005

1.9998

17

2.0002

1.9998

2.0004

0.0003

2.0001

18

2.0000

2.0001

2.0001

0.0001

2.0001

19

1.9997

1.9994

1.9998

0.0002

1.9996

20

2.0003

2.0007

1.9999

0.0004

2.0003

Sbar chart limits:

SBAR = 0.0002

UCL = B(4) x SBAR = 2.568 x .0002 = 0.0005136

LCL = B(3) x SBAR = 0 x .0002 = 0.00

Xbar chart limits:

XDBLBAR = 2.0000

UCL = XDBLBAR + A(3) x SBAR = 2.000+1.954 x .0002 = 2.0003908

LCL = XDBLBAR – A(3) x SBAR = 2.000-1.954 x 0002 = 1.9996092

S-Chart:

Xbar Chart:

Xbar and R Charts (1 of 2)

Theoretical Control Limits for Xbar Charts:

Although theoretically possible, since we do not know either the population process mean or standard deviation, these formulas cannot be used directly and both must be estimated from the process itself. First the R chart is constructed. If the R chart validates that the process variation is in statistical control, the XBAR chart is constructed.

1. Find the mean of each subgroup Xbar(1), Xbar(2), Xbar(3)… Xbar(k) and the grand mean of all subgroups using:

2. Find the UCL and LCL using the following equations:

· UCLX-bar = X-double-bar + A2 (R-bar)

· LCLX-bar = X-double-bar – A2 (R-bar)

A(2) can be found in the following table:

n A(2)

n A(2)

2 1.880

6 .483

3 1.023

7 .419

4 .729

8 .373

5 .577

9 .337

3. Plot the LCL, UCL, centerline, and subgroup means

4. Interpret the data using the following guidelines to determine if the process is in control:

· One point outside the 3 sigma control limits

· eight successive points on the same side of the centerline

· Six successive points that increase or decrease

· Two out of three points that are on the same side of the centerline,

both at a distance exceeding 2 sigmas from the centerline

· Four out of five points that are on the same side of the centerline, four at a distance exceeding 1 sigma from the centerline

· Using an average run length (ARL) for determining process anomalies.

Using the navigation on the left, please proceed to the next page.

Xbar and R Charts (2 of 2)

Example: The following data consists of 20 sets of three measurements of the diameter of an engine shaft.

n

1

2

3

Range

Xbar

1

2.0000

1.9998

2.0002

0.0004

2.0000

2

1.9998

2.0003

2.0002

0.0005

2.0001

3

1.9998

2.0001

2.0005

0.0007

2.0001

4

1.9997

2.0000

2.0004

0.0007

2.0000

5

2.0003

2.0003

2.0002

0.0001

2.0003

6

2.0004

2.0003

2.0000

0.0004

2.0002

7

1.9998

1.9998

1.9998

0.0000

1.9998

8

2.0000

2.0001

2.0001

0.0001

2.0001

9

2.0005

2.0000

1.9999

0.0006

2.0001

10

1.9995

1.9998

2.0001

0.0006

1.9998

11

2.0002

1.9999

2.0001

0.0003

2.0001

12

2.0002

1.9998

2.0005

0.0007

2.0002

13

2.0000

2.0001

1.9998

0.0003

2.0000

14

2.0000

2.0002

2.0004

0.0004

2.0002

15

1.9994

2.0001

1.9996

0.0007

1.9997

16

1.9999

2.0003

1.9993

0.0010

1.9998

17

2.0002

1.9998

2.0004

0.0006

2.0001

18

2.0000

2.0001

2.0001

0.0001

2.0001

19

1.9997

1.9994

1.9998

0.0004

1.9996

20

2.0003

2.0007

1.9999

0.0008

2.0003

Rbar chart limits:

RBAR = 0.0005

UCL=D(4)*RBAR = 2.574 * .0005 = 0.001287

LCL=D(3)*RBAR = 0.000 * .0005 = 0.000

Xbar chart limits:

XDBLBAR = 2.0000

UCL = XDBLBAR + A(2) x RBAR = 2.000+1.023 x .0005 = 2.0005115

LCL = XDBLBAR – A(2) x RBAR = 2.000-1.023 x.0005 = 1.9994885

R-Chart:

Xbar-Chart:

Using the navigation on the left, please proceed to the next page.

CHAPTER <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<**b**>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>**1**7: Pro

Chapter Outline

**
17.1
** Quality: Its Meaning

an**d**

a Historical Perspective

**
17.2
** Statistical Process Control and Causes o

**17.3
** Sampling a Process, Rational Subgrouping, and Control Charts

**
17.4
** and R Charts

**
17. 5
** Pattern Analysis

**
17. 6
** Comparison of a Process with Specifications: Capability Studies

**
17.7
** Charts for Fraction Non

**
17.8
** Cause-and-Effect and Defect Concentration Diagrams (Optional)

This chapter explains how to use **
control charts
** to improve business processes. Basically, a control chart is a graphical device that helps us determine when a process is “out of control.” The information provided by a control chart helps us discover the causes of un

We begin this chapter by tracing the history of the U.S. quality movement. Then we study __control charts__ for monitoring the

In order to demonstrate the ideas of this chapter, we employ three case studies:

**The Hole Location Case:** A manufacturer of automobile air conditioner compressors uses control charts to reduce variation in the locations of a hose connection hole that is punched in the outer housing (or shell) of the compressor.

**The Hot Chocolate Temperature Case:** The food service staff at a university dining hall wishes to avoid possible litigation by making sure that it does not serve excessively hot beverages. The staff uses control charts to find and eliminate causes of unusual variations in hot chocolate temperatures.

**The Camshaft Case:** An automobile manufacturer wishes to improve the process it uses to harden a part in a camshaft assembly. The manufacturer uses control charts and process capability studies to reduce the sources of process variation that are responsible for a 12 percent rework rate and a 9 percent scrap rate. After the process variation is reduced, virtually all of the hardened parts meet specifications (note: this case is included in the supplementary exercises).

17.1: Quality: Its Meaning and a Historical Perspective

What is quality?

It is not easy to define quality, and a number of different definitions have been proposed. One definition that makes sense is **fitness for use.** Here the user of a product or service can be an individual, a manufacturer, a retailer, or the like. For instance, an individual who purchases a television set or a videocassette recorder expects the unit to be defect free and to provide years of reliable, high-performance service. If the TV or VCR performs as desired, it is fit for use. Another definition of quality that makes sense says that **quality is the extent to which customers feel that a product or service exceeds their needs and expectations.** For instance, if the VCR purchaser believes the unit exceeds all the needs and expectations he or she had for the VCR when it was purchased, then the customer is satisfied with the unit’s quality.

Three types of quality can be considered: **
quality of design, quality of conformance,** and

The marketing research arm of a company must determine what the customer seeks in each of these dimensions. Consumer research is used to develop a product or service concept—a combination of design characteristics that exceeds the expectations of a large number of consumers. This concept is translated into a design. The design includes specifications that, if met, will satisfy consumer wants and needs. A production process is then developed to meet the design specifications. In order to do this, variables that can control the process must be identified, and the relationships between input variables and final quality characteristics must be understood. The manufacturer expresses quality characteristics as measurable variables that can be tracked and used to monitor and improve the performance of the process. Service call analysis often leads to product or service redesigns in order to improve the product or service concept. It is extremely important that the initial design be a good one so that excessive redesigns and customer dissatisfaction can be avoided.

History of the quality movement

In the 1700s and 1800s, master craftsmen and their apprentices were responsible for designing and building products. Quantities of goods produced were small, and product quality was controlled by expert workmanship. Master craftsmen had a great deal of pride in their work, and quality was not a problem. However, the introduction of mass production in the late 1800s and early 1900s changed things. Production processes became very complex, with many workers (rather than one skilled craftsman) responsible for the final product. Inevitably, product quality characteristics displayed variation. In particular, Henry Ford developed the moving assembly line at Ford Motor Company. As assembly line manufacturing spread, quality became a problem. Production managers were rewarded for meeting production quotas, and quality suffered. To make mass-produced products more **consistent**, inspectors were hired to check product quality. However, 100 percent **inspection** proved to be costly, and people started to look for alternatives.

Much of the early work in quality control was done at Bell Telephone (now known as American Telephone and Telegraph or AT&T). The Bell System and Western Electric, the manufacturing arm of Bell Telephone, formed the Inspection Engineering Department to deal with quality problems. In 1924 Walter Shewhart of Bell Telephone Laboratories introduced the concept of statistical quality control—controlling quality of mass-produced goods. Shewhart believed that variation always exists in manufactured products, and that the variation can be studied, monitored, and controlled using statistics. In particular, Shewhart developed a statistical tool called the **
control chart.** Such a chart is a graph that can tell a company when a process needs to be adjusted and when the process should be left alone. In the late 1920s Harold F. Dodge and Harold G. Romig, also of Bell Telephone Laboratories, introduced

During World War II statistical quality control became widespread. Faced with the task of producing large quantities of high-quality war matériel, industry turned to statistical methods, failure analysis, vendor certification, and early product design. The U.S. War Department required that suppliers of war matériel employ acceptance sampling, and its use became commonplace. Statistical control charts were also used, although not as widely as acceptance sampling.

In 1946 the **American Society for Quality Control (ASQC)** was established to encourage the use of quality improvement methods. The organization sponsors training programs, seminars, and publications dealing with quality issues. In spite of the efforts of the ASQC, however, interest in quality in American industry diminished after the war. American business had little competition in the world market—Europe and Japan were rebuilding their shattered economies. Tremendous emphasis was placed on increased production because firms were often unable to meet the demand for their products. Profits were high, and the concern for quality waned. As a result, postwar American managers did not understand the importance of quality and process improvement, and they were not informed about quality improvement techniques.

However, events in Japan took a different turn. After the war, Japanese industrial capacity was crippled. Productivity was very low, and products were of notoriously bad quality. In those days, products stamped “Made in Japan” were generally considered to be “cheap junk.” The man credited with turning this situation around is W. Edwards Deming. Deming, born in 1900, earned a Ph.D. in mathematical physics from Yale University in 1927. He then went to work in a Department of Agriculture–affiliated laboratory. Deming, who had learned statistics while studying physics, applied statistics to experiments conducted at the laboratory. Through this work, Deming was introduced to Walter Shewhart, who explained his theories about using statistical control charts to improve quality and productivity. During World War II, Deming was largely responsible for teaching 35,000 American engineers and technical people how to use statistics to improve the quality of war matériel. After the war, the Allied command sent a group of these engineers to Japan. Their mission was to improve the Japanese communication system. In doing so, the engineers employed the statistical methods they had learned, and Deming’s work was brought to the attention of the **Union of Japanese Scientists and Engineers (JUSE).** Deming, who had started his own consulting firm in 1946, was asked by the JUSE to help increase Japanese productivity. In July 1950 Deming traveled to Japan and gave a series of lectures titled “Elementary Principles of the Statistical Control of Quality” to a group of 230 Japanese managers. Deming taught the Japanese how to use statistics to determine how well a system can perform, and taught them how to design process improvements to make the system operate better and more efficiently. He also taught the Japanese that the more quality a producer builds into a product, the less it costs. Realizing the serious nature of their economic crisis, the Japanese adopted Deming’s ideas as a philosophy of doing business. Through Deming, the Japanese found that by listening to the wants and needs of consumers and by using statistical methods for process improvement in production, they could export high-quality products to the world market.

Although American business was making only feeble attempts to improve product quality in the 1950s and 1960s, it was able to maintain a dominant competitive position. Many U.S. companies focused more on marketing and financial strategies than on product and production. But the Japanese and other foreign competitors were making inroads. By the 1970s, the quality of many Japanese and European products (for instance, automobiles, television sets, and electronic equipment) became far superior to their American-made counterparts. Also, rising prices made consumers more quality conscious—people expected high quality if they were going to pay high prices. As a result, the market shares of U.S. firms rapidly decreased. Many U.S. firms were severely injured or went out of business.

Meanwhile, Deming continued teaching and preaching quality improvement. While Deming was famous in Japan, he was relatively unknown in the United States until 1980. In June 1980 Deming was featured in an NBC television documentary titled “If Japan Can, Why Can’t We?” This program, written and narrated by then–NBC correspondent Lloyd Dobyns, compared Japanese and American industrial productivity and credited Deming for Japan’s success. Within days, demand for Deming’s consulting services skyrocketed. Deming consulted with many major U.S. firms. Among these firms are The Ford Motor Company, General Motors Corporation, and The Procter & Gamble Company. Ford, for instance, began consulting with Deming in 1981. Donald Petersen, who was Ford’s chairman and chief executive officer at the time, became a Deming disciple. By following the Deming philosophy, Ford, which was losing 2 billion dollars yearly in 1980, attempted to create a quality culture. Quality of Ford products was greatly improved, and the company again became profitable. The 1980s saw many U.S. companies adopt a philosophy of continuous improvement of quality and productivity in all areas of their businesses—manufacturing, accounting, sales, finance, personnel, marketing, customer service, maintenance, and so forth. This overall approach of applying quality principles to all company activities is called **
total quality management (TQM)
** or

The fundamental ideas behind Deming’s approach to quality and productivity improvement are contained in his “**14 points.**” These are a set of managerial principles that, if followed, Deming believed would enable a company to improve quality and productivity, reduce costs, and compete effectively in the world market. We briefly summarize the 14 points in Table 17.1. For more complete discussions of these points, see Bowerman and O’Connell (1996), Deming (1986), Walton (1986), Scherkenbach (1987), or Gitlow, Gitlow, Oppenheim, and Oppenheim (1989). Deming stressed that implementation of the 14 points requires both changes in management philosophy and the use of statistical methods. In addition, Deming believed that it is necessary to follow all of the points, not just some of them.

Table 17.1: W. Edwards Deming’s 14 Points

In 1988 the first **Malcolm Baldrige National Quality Awards** were presented. These awards, presented by the U.S. Commerce Department, are named for the late Malcolm Baldrige, who was Commerce Secretary during the Reagan administration. The awards were established to promote quality awareness, to recognize quality achievements by U.S. companies, and to publicize successful quality strategies. The Malcolm Baldrige National Quality Award Consortium, formed by the ASQC (now known as the ASQ) and the American Productivity and Quality Center, administers the award. The Baldrige award has become one of the most prestigious honors in American business. Annual awards are given in three categories—manufacturing, service, and small business. Winners include companies such as Motorola Inc., Xerox Corporation Business Products and Systems, the Commercial Nuclear Fuel Division of Westinghouse Electric Corporation, Milliken and Company, Cadillac Division, General Motors Corporation, Ritz Carlton Hotels, and AT&T Consumer Communications.

Finally, the 1990s saw the adoption of an international quality standards system called **
ISO 9000.** More than 90 countries around the globe have adopted the ISO 9000 standards for their companies, as have many multinational corporations (including AT&T, 3M, IBM, Motorola, and DuPont). As a brief introduction to ISO 9000, we quote “Is ISO 9000 for You?” published by CEEM Information Systems:

What Is ISO 9000?

ISO 9000 is a series of international standards for quality assurance management systems. It establishes the organizational structure and processes for assuring that the production of goods or services meets a consistent and agreed-upon level of quality for a company’s customers.

The ISO 9000 series is unique in that it applies to a very wide range of organizations and industries encompassing both the manufacturing and service sectors.

Why Is ISO 9000 Important?

ISO 9000 is important for two reasons. First … the discipline imposed by the standard for processes influencing your quality management systems can enhance your company’s quality consistency. Whether or not you decide to register your company to ISO 9000 standards, your implementing such discipline can achieve greater efficiency in your quality control systems.

Second … more and more companies, both here at home and internationally, are requiring their suppliers to be ISO 9000 registered. To achieve your full market potential in such industries, registration is becoming essential. Those companies who become registered have a distinct competitive advantage, and sales growth in today’s demanding market climate requires every advantage you can muster.1

Clearly, quality has finally become a crucially important issue in American business. The quality revolution now affects every area in business. But the Japanese continue to mount new challenges. For years, the Japanese have used **designed statistical experiments** to develop new processes, find and remedy process problems, improve product performance, and improve process efficiency. Much of this work is based on the insights of Genichi Taguchi, a Japanese engineer. His methods of experimental design, the so-called **Taguchi methods,** have been heavily used in Japan since the 1960s. Although Taguchi’s methodology is controversial in statistical circles, the use of experimental design gives the Japanese a considerable advantage over U.S. competitors because it enables them to design a high level of quality into a product before production begins. Some U.S. manufacturers have begun to use experimental design techniques to design quality into their products. It will be necessary for many more U.S. companies to do so in order to remain competitive in the future—a challenge for the 21st century.

** 17.2**: Statistical Process Control and Causes of Process Variation

Statistical process control

**
Statistical process control (SPC)
** is a systematic method for analyzing process data (quality characteristics) in which we monitor and study the process variation. The goal is to stabilize the process and to reduce the amount of process variation. The ultimate goal is

Before the widespread use of SPC, quality control was based on an inspection approach. Here the product is first made, and then the final product is inspected to eliminate defective items. This is called **action on the output** of the process. The emphasis here is on detecting defective product that has already been produced. This is costly and wasteful because, if defective product is produced, the bad items must be (1) **scrapped,** (2) **reworked or reprocessed** (that is, fixed), or (3) **downgraded** (sold off at a lower price). In fact, the cost of bad quality (scrap, rework, and so on) can be tremendously high. It is not unusual for this cost to be as high as 10 to 30 percent or more of a company’s dollar sales.

In contrast to the inspection approach, SPC emphasizes integrating quality improvement into the process. Here the goal is **preventing bad quality by taking appropriate action on the process.** In order to accomplish this goal, we must decide when actions on the process are needed. The focus of much of this chapter is to show how such decisions can be made.

Causes of process variation

In order to understand SPC methodology, we must realize that the variations we observe in quality characteristics are caused by different sources. These sources include factors such as equipment (machines or the like), materials, people, methods and procedures, the environment, and so forth. Here we must distinguish between usual process variation and **unusual process variation.** Usual process variation results from what we call

__common causes of process variation__.

**
Common causes
** are sources of variation that have the potential to influence all process observations. That is, these sources of variation are inherent to the current process design.

Common cause variation can be substantial. For instance, obsolete or poorly maintained equipment, a poorly designed process, and inadequate instructions for workers are examples of common causes that might significantly influence all process output. As an example, suppose that we are filling 16-ounce jars with grape jelly. A 25-year-old, obsolete filler machine might be a common cause of process variation that influences all the jar fills. While (in theory) it might be possible to replace the filler machine with a new model, we might have chosen not to do so, and the obsolete filler causes all the jar fills to exhibit substantial variation.

Common causes also include small influences that would cause slight variation even if all conditions are held as constant as humanly possible. For example, in the jar fill situation, small variations in the speed at which jars move under the filler valves, slight floor vibrations, and small differences between filler valve settings would always influence the jar fills even when conditions are held as constant as possible. Sometimes these small variations are described as being due to “chance.”

Together, the important and unimportant common causes of variation determine the **usual process variability.** That is, these causes determine the amount of variation that exists when the process is operating routinely. We can reduce the amount of **common cause variation** by removing some of the important common causes. **Reducing common cause variation is usually a management responsibility.** For instance, replacing obsolete equipment, redesigning a plant or process, or improving plant maintenance would require management action.

In addition to common cause variation, processes are affected by a different kind of variation called **assignable cause variation** (sometimes also called **special cause** or **specific cause variation).**

**
Assignable causes
** are sources of unusual process variation. These are intermittent or permanent changes in the process that are not common to all process observations and that may cause important process variation. Assignable causes are usually of short duration, but they can be persistent or recurring conditions.

For example, in the jar filling situation, one of the filler valves may become clogged so that some jars are being substantially underfilled (or perhaps are not filled at all). Or a relief operator might incorrectly set the filler so that all jars are being substantially overfilled for a short period of time. As another example, suppose that a bank wishes to study the length of time customers must wait before being served by a teller. If a customer fills out a banking form incorrectly, this might cause a temporary delay that increases the waiting time for other customers. Notice that **assignable causes** such as these can often be remedied by local supervision—for instance, by a production line foreman, a machine operator, a head bank teller, or the like. **One objective of SPC is to detect and eliminate assignable causes of process variation. By doing this, we reduce the amount of process variation. This results in improved quality.**

It is important to point out that an assignable cause could be beneficial—that is, it could be an unusual process variation resulting in unusually good process performance. In such a situation, we wish to discover the root cause of the variation, and then we wish to incorporate this condition into the process if possible. For instance, suppose we find that a process performs unusually well when a raw material purchased from a particular supplier is used. It might be desirable to purchase as much of the raw material as possible from this supplier.

When a process exhibits only common cause variation, it will operate in a stable fashion. That is, in the absence of any unusual process variations, the process will display a constant amount of variation around a constant mean. On the other hand, if assignable causes are affecting the process, then the process will not be stable—unusual variations will cause the **process mean** or variability to change over time. It follows that

1 When a process is influenced only by **common cause variation, the process will be in statistical control.
**

2 When a process is influenced by **one or more assignable causes, the process will not be in statistical control.**

In general, in order to bring a process into **statistical control,** we must find and eliminate undesirable assignable causes of process variation, and we should (if feasible) build desirable assignable causes into the process. When we have done these things, the process is what we call a **stable, common cause system.** This means that the process operates in a consistent fashion and is **predictable.** Since there are no unusual process variations, the process (as currently configured) is doing all it can be expected to do.

When a process is in statistical control, management can evaluate the process capability. That is, it can assess whether the process can produce output meeting customer or producer requirements. If it does not, action by local supervision will not remedy the situation—remember, the assignable causes (the sources of process variation that can be dealt with by local supervision) have already been removed. Rather, some fundamental change will be needed in order to reduce common cause variation. For instance, perhaps a new, more modern filler machine must be purchased and installed. This will require action by management.

Finally, the SPC approach is really a philosophy of doing business. It is an entire firm or organization that is focused on a single goal: continuous quality and productivity improvement. The impetus for this philosophy must come from management. Unless management is supportive and directly involved in the ongoing quality improvement process, the SPC approach will not be successful.

Exercises for Sections 17.1 and 17.2

CONCEPTS

17.1 Write an essay comparing the management philosophy that Dr. Deming advocated in his 14 points to the management styles you have been exposed to in your personal work experiences. Do you think Dr. Deming’s philosophy is preferable to the management styles you have seen in practice? Which of the 14 points do you agree with? Which do you disagree with?

17.2 Write a paragraph explaining how common causes of process variation differ from assignable causes of process variation.

METHODS AND APPLICATIONS

__17.3__ In this exercise we consider several familiar processes. In each case, describe several common causes and several assignable causes that might result in variation in the given quality characteristic.

a Process: getting ready for school or work in the morning.

Quality characteristic: the time it takes to get ready.

b Process: driving, walking, or otherwise commuting from your home or apartment to school or work.

Quality characteristic: the time it takes to commute.

c Process: studying for and taking a statistics exam.

Quality characteristic: the score received on the exam.

d Process: starting your car in the morning.

Quality characteristic: the time it takes to start your car.

__17.4__ Form a group of three or four students in your class. As a group project, select a familiar process and determine a variable that measures the quality of some aspect of the output of this process. Then list some common causes and assignable causes that might result in variation of the variable you have selected for the process. Discuss your lists in class.

17.3: Sampling a Process, Rational Subgrouping, and Control Charts

In order to find and eliminate assignable causes of process variation, we sample output from the process. To do this, we first decide which **process variables**—that is, which process characteristics—will be studied. Several graphical techniques (sometimes called prestatistical tools) are used here. Pareto charts (see Section 2.1) help identify problem areas and opportunities for improvement. Cause-and-effect diagrams (see optional Section 17.8) help uncover sources of process variation and potentially important process variables. The goal is to identify process variables that can be studied in order to decrease the gap between customer expectations and process performance.

Whenever possible and economical, it is best to study a **quantitative,** rather than a **categorical,** process variable. For example, suppose we are filling 16-ounce jars with grape jelly, and suppose specifications state that each jar should contain between 15.95 and 16.05 ounces of jelly. If we record the fill of each sampled jar by simply noting that the jar either “meets specifications” (the fill is between 15.95 and 16.05 ounces) or “does not meet the specifications,” then we are studying a **categorical process variable.** However, if we measure and record the amount of grape jelly contained in the jar (say, to the nearest one-hundredth of an ounce), then we are studying a **quantitative process variable.** Actually measuring the fill is best because this tells us **how close** we are to the specification limits and thus provides more information. As we will soon see, this additional information often allows us to decide whether to take action on a process by using a relatively small number of measurements.

**When we study a quantitative process variable, we say that we are employing measurement data.** To analyze such data, we take a series of samples (usually called

Each subgroup is typically observed over a short period of time—a period of time in which the process operating characteristics do not change much. That is, we employ

__rational subgroups__.

Rational Subgroups

**
Rational subgroups
** are selected so that,

In order to obtain rational subgroups, we must determine the frequency with which subgroups will be selected. For example, we might select a subgroup once every 15 minutes, once an hour, or once a day. In general, we should observe subgroups often enough to detect important process changes. For instance, suppose we wish to study a process, and suppose we feel that workers’ shift changes (that take place every eight hours) may be an important source of process variation. In this case, rational subgroups can be obtained by selecting a subgroup during each eight-hour shift. Here shift changes will occur between subgroups. Therefore, if shift changes are an important source of variation, the rational subgroups will enable us to observe the effects of these changes by comparing plot points for different subgroups (shifts). However, in addition, suppose hourly machine resets are made, and we feel that these resets may also be an important source of process variation. In this case, rational subgroups can be obtained by selecting a subgroup during each hour. Here machine resets will occur between subgroups, and we will be able to observe their effects by comparing plot points for different subgroups (hours). If in this situation we selected one subgroup each eight-hour shift, we would not obtain rational subgroups. This is because hourly machine resets would occur within subgroups, and we would not be able to observe the effects of these resets by comparing plot points for different shifts. In general, it is very important to try to identify important sources of variation (potential assignable causes such as shift changes, resets, and so on) before deciding how subgroups will be selected. As previously stated, constructing a __cause-and-effect diagram__ helps uncover these sources of variation (see optional Section 17.8).

Once we determine the sampling frequency, we need to determine the subgroup size—that is, the number of measurements that will be included in each subgroup—and how we will actually select the measurements in each subgroup. It is recommended that the **subgroup size be held constant.** Denoting this constant subgroup size as n, we typically choose n to be from 2 to 6, with n = 4 or 5 being a frequent choice. To illustrate how we can actually select the subgroup measurements, suppose we select a subgroup of 5 units every hour from the output of a machine that produces 100 units per hour. We can select these units by using a **consecutive, periodic,** or **random** sampling process. If we employ consecutive sampling, we would select 5 consecutive units produced by the machine at the beginning of (or at some time during) each hour. Here **production conditions**—machine operator, machine setting, raw material batch, and so forth—**will be as constant as possible within the subgroup.** Such a subgroup provides a “freeze-frame picture” of the process at a particular point in time. Thus the **chance of variations occurring within the subgroups is minimized.** If we use periodic sampling, we would select 5 units periodically through each hour. For example, since the machine produces 100 units per hour, we could select the 1st, 21st, 41st, 61st, and 81st units produced. If we use random sampling, we would use a random number table to randomly select 5 of the 100 units produced during each hour. If production conditions are really held fairly constant during each hour, then consecutive, periodic, and random sampling will each provide a similar representation of the process. If production conditions vary considerably during each hour, and if we are able to recognize this variation by using a periodic or random sampling procedure, this would tell us that we should be sampling the process more often than once an hour. Of course, if we are using periodic or random sampling every hour, we might not realize that the process operates with considerably less variation during shorter periods (perhaps because we have not used a consecutive sampling procedure). We therefore might not recognize the extent of the hourly variation.

Lastly, it is important to point out that we **must also take subgroups for a period of time that is long enough to give potential sources of variation a chance to show up.** If, for instance, different batches of raw materials are suspected to be a significant source of process variation, and if we receive new batches every few days, we may need to collect subgroups for several weeks in order to assess the effects of the batch-to-batch variation. **A statistical rule of thumb says that we require at least 20 subgroups of size 4 or 5 in order to judge statistical control and in order to obtain reasonable estimates of the process mean and variability.** However, practical considerations may require the collection of much more data.

We now look at two more concrete examples of subgrouped data.

EXAMPLE 17.1: The Hole Location Case2

A manufacturer produces automobile air conditioner compressor shells. The compressor shell is basically the outer metal housing of the compressor. Several holes of various sizes must be punched into the shell to accommodate hose connections that must be made to the compressor. If any one of these holes is punched in the wrong location, the compressor shell becomes a piece of scrap metal (at considerable cost to the manufacturer). Figure 17.1(a) illustrates a compressor shell (note the holes that have been punched in the housing). Experience with the hole-punching process suggests that substantial changes (machine resets, equipment lubrication, and so forth) can occur quite frequently—as often as two or three times an hour. Because we wish to observe the impact of these changes if and when they occur, rational subgroups are obtained by selecting a subgroup every 20 minutes or so. Specifically, about every 20 minutes five compressor shells are consecutively selected from the process output. For each shell selected, a measurement that helps to specify the location of a particular hole in the compressor shell is made. The measurement is taken by measuring from one of the edges of the compressor shell (called the trim edge) to the bottom of the hole [see Figure 17.1(a)]. Obviously, it is not possible to measure to the center of the hole because you cannot tell where it is! The target value for the measured dimension is 3.00inches. Of course, the manufacturer would like as little variation around the target as possible. Figure 17.1(b) gives the measurements obtained for 20 subgroups that were selected between 8 a.m. and 2:20 p.m. on a particular day. Here a subgroup consists of the five measurements labeled 1 through 5 in a single row in the table. Notice that Figure 17.1(b) also gives the mean, , and the range, R, of the measurements in each subgroup. In the next section we will see how to use the subgroup means and ranges to detect when unusual process variations have taken place.

Figure 17.1: The Compressor Shell and the Hole Location Data HoleLoc

EXAMPLE 17.2: The Hot Chocolate Temperature Case3

Since 1994 a number of consumers have filed and won large claims against national fast-food chains as a result of being scalded by excessively hot beverages such as coffee, tea, and hot chocolate. Because of such litigation, the food service staff at a university dining hall wishes to study the temperature of the hot chocolate dispensed by its hot chocolate machine. The dining hall staff believes that there might be substantial variations in hot chocolate temperatures from meal to meal. Therefore, it is decided that at least one subgroup of hot chocolate temperatures will be observed during each meal—breakfast (6:30 a.m. to 10 a.m.), lunch (11 a.m. to 1:30 p.m.), and dinner (5 p.m. to 7:30 p.m.). In addition, since the hot chocolate machine is heavily used during most meals, the dining hall staff also believes that hot chocolate temperatures might vary substantially from the beginning to the end of a single meal. It follows that the staff will obtain rational subgroups by selecting a subgroup a half hour after the beginning of each meal and by selecting another subgroup a half hour prior to the end of each meal. Specifically, each subgroup will be selected by pouring three cups of hot chocolate over a 10-minute time span using periodic sampling (the second cup will be poured 5 minutes after the first, and the third cup will be poured 5 minutes after the second). The temperature of the hot chocolate will be measured by a candy thermometer (to the nearest degree Fahrenheit) immediately after each cup is poured.

Table 17.2 gives the results for 24 subgroups of three hot chocolate temperatures taken at each meal served at the dining hall over a four-day period. Here a subgroup consists of the three temperatures labeled 1 through 3 in a single row in the table. The table also gives the mean, , and the range, R, of the temperatures in each subgroup. In the next section we will use the subgroup means and ranges to detect unusual process variations (that is, to detect assignable causes).

Table 17.2: 24 Subgroups of Three Hot Chocolate Temperatures (Measurements to the Nearest Degree Fahrenheit) HotChoc

Subgrouped data are used to determine when assignable causes of process variation exist. Typically, we analyze subgrouped data by plotting summary statistics for the subgroups versus time. The resulting plots are often called **graphs of process performance.** For example, the subgroup means and the **subgroup ranges** of the hole location measurements in Figure 17.1(b) are plotted in time order on graphs of process performance in the Excel output of Figure 17.2. The subgroup means ( values) and ranges (R values) are plotted on the vertical axis, while the time sequence (in this case, the subgroup number) is plotted on the horizontal axis. The values and R values for corresponding subgroups are lined up vertically. The plot points on each graph are connected by line segments as a visual aid. However, the lines between the plot points do not really say anything about the process performance between the observed subgroups. Notice that the subgroup means and ranges vary over time.

Figure 17.2: Excel Output of Graphs of Performance (Subgroup Means and Ranges) for the Hole Location Data in Figure 17.1(b)

If we consider the plot of subgroup means, very high and very low points are undesirable—they represent large deviations from the target hole location dimension (3 inches). If we consider the plot of subgroup ranges, very high points are undesirable (high variation in the hole location dimensions), while very low points are desirable (little variation in the hole location dimensions).

We now wish to answer a very basic question. **Is the variation that we see on the graphs of performance due to the usual process variation (that is, due to common causes), or is the variation due to one or more assignable causes (unusual variations)?** It is possible that unusual variations have occurred and that action should be taken to reduce the variation in production conditions. It is also possible that the variation in the plot points is caused by common causes and that (given the current configuration of the process) production conditions have been held as constant as possible. For example, do the high points on the plot in Figure 17.2 suggest that one or more assignable causes have increased the hole location dimensions enough to warrant corrective action? As another example, do the high points on the R plot suggest that excess variability in the hole location dimensions exists and that corrective action is needed? Or does the lowest point on the R plot indicate that an improvement in process performance (reduction in variation) has occurred due to an assignable cause?

We can answer these questions by converting the graphs of performance shown in Figure 17.2 into

**control charts. In general, by converting graphs of performance into control charts, we can (with only a small chance of being wrong) determine whether observed process variations are unusual (due to assignable causes). That is, the purpose of a control chart is to monitor a process so we can take corrective action in response to assignable causes when it is needed. This is called statistical process monitoring. The use of “seat of the pants intuition” has not been found to be a particularly effective way to decide whether observed process performance is unusual. By using a control chart, we can reduce our chances of making two possible errors—(1) taking action when none is needed and (2) not taking action when action is needed.**

A control chart employs a **center line** (denoted **CNL**) and two control limits—an **upper control limit** (denoted **UCL**) and a **lower control limit** (denoted **LCL**). The center line represents the average performance of the process when it is in a state of statistical control—that is, when only common cause variation exists. The upper and lower control limits are horizontal lines situated above and below the center line. These control limits are established so that, when the process is in control, almost all plot points will be between the upper and lower limits. In practice, the control limits are used as follows:

1 If all observed plot points are **between the LCL and UCL** (and if no unusual patterns of points exist—this will be explained later), we have no evidence that assignable causes exist and we assume that the process is in statistical control. In this case, **only common causes of process variation exist,** and **no action to remove assignable causes is taken on the process.** If we were to take such action, we would be **unnecessarily tampering** with the process.

2 If we observe one or more plot points **outside the control limits,** then we have evidence that the process is **out of control due to one or more assignable causes.** Here **we must take action on the process** to remove these assignable causes.

In the next section we begin to discuss how to construct control charts. Before doing this, however, we must emphasize the importance of **documenting** a process while the subgroups of data are being collected. The time at which each subgroup is taken is recorded, and the person who collected the data is also recorded. Any process changes (machine resets, adjustments, shift changes, operator changes, and so on) must be documented. Any potential sources of variation that may significantly affect the process output should be noted. If the process is not well documented, it will be very difficult to identify the root causes of unusual variations that may be detected when we analyze the subgroups of data.

17.4: and R Charts

**Chapter 21**

**and
R
charts
** are the most commonly used control charts for measurement data (such charts are often called

Before seeing how to construct **and R charts**, we should mention that it is also possible to monitor the process variability by using a chart for **subgroup standard deviations.** Such a chart is called an **
s chart.** However, the overwhelming majority of practitioners use R charts rather than s charts. This is partly due to historical reasons. When control charts were developed, electronic calculators and computers did not exist. It was, therefore, much easier to compute a subgroup range than it was to compute a subgroup standard deviation. For this reason, the use of R charts has persisted. Some people also feel that it is easier for factory personnel (some of whom may have little mathematical background) to understand and relate to the subgroup range. In addition, while the standard deviation (which is computed using all the measurements in a subgroup) is a better measure of variability than the range (which is computed using only two measurements), the R chart usually suffices. This is because and R charts usually employ small subgroups—as mentioned previously, subgroup sizes are often between 2 and 6. For such subgroup sizes, it can be shown that using subgroup ranges is almost as effective as using subgroup standard deviations.

To construct and R charts, suppose we have observed rational subgroups of n measurements over successive time periods (hours, shifts, days, or the like). We first calculate the mean and range R for each subgroup, and we construct graphs of performance for the values and for the R values (as in Figure 17.2). In order to calculate center lines and control limits, let denote the mean of the subgroup of n measurements that is selected in a particular time period. Furthermore, assume that the population of all process measurements that could be observed in any time period is normally distributed with mean μ and standard deviation σ, and also assume successive process measurements are statistically **independent.**4 Then, if μ and σ stay constant over time, the sampling distribution of subgroup means in any time period is normally distributed with mean μ and standard deviation . It follows that (in any time period) 99.73 percent of all possible values of the subgroup mean are in the interval

This fact is illustrated in Figure 17.3. It follows that we can set a center line and control limits for the chart as

If an observed subgroup mean is inside these control limits, we have no evidence to suggest that the process is out of control. However, if the subgroup mean is outside these limits, we conclude that μ and/or σ have changed, and that the process is out of control. The chart limits are illustrated in Figure 17.3.

Figure 17.3: An Illustration of Chart Control Limits with the Process Mean μ and Process Standard Deviation σ Known

If the process is in control, and thus μ and σ stay constant over time, it follows that μ and σ are the mean and standard deviation of all possible process measurements. For this reason, we call μ the process mean and σ the **process standard deviation.** Since in most real situations we do not know the true values of μ and σ, we must estimate these values. If the process is in control, an appropriate estimate of the process mean μ is

( is **pronounced “x double bar”**). It follows that the center line for the chart is

To obtain control limits for the chart, we compute

It can be shown that an appropriate estimate of the process standard deviation σ is where d2 is a constant that depends on the subgroup size n. Although we do not present a development of d2 here, it intuitively makes sense that, for a given subgroup size, our best estimate of the process standard deviation should be related to the average of the subgroup ranges (). The number d2 relates these quantities. Values of d2 are given in Table 17.3 for subgroup sizes n = 2 through n = 25. At the end of this section we further discuss why we /d2 use to estimate the process standard deviation.

Table 17.3: Control Chart Constants for and R Charts

Substituting the estimate of μ and the estimate /d2 of σ into the limits

we obtain

Finally, we define

and rewrite the control limits as

Here we call A2 a **
control chart constant.** As the formula for A2 implies, this control chart constant depends on the subgroup size n. Values of A2 are given in Table 17.3 for subgroup sizes n = 2 through n = 25.

The center line for the R chart is

Furthermore, assuming normality, it can be shown that there are control chart constants D4 and D3 so that

Here the control chart constants D4 and D3 also depend on the subgroup size n. Values of D4 and D3 are given in Table 17.3 for subgroup sizes n = 2 through n = 25. We summarize the center lines and control limits for and R charts in the following box:

and R Chart Center Lines and Control Limits

EXAMPLE 17.3: The Hole Location Case

Consider the hole location data for air conditioner compressor shells that is given in Figure 17.1 (page 751). In order to calculate and R chart control limits for this data, we compute

Looking at Table 17.3, we see that when the subgroup size is n = 5, the control chart constants needed for and R charts are A2 = .577 and D4 = 2.114. It follows that center lines and control limits are

Since D3 is not listed in Table 17.3 for a subgroup size of n = 5, the R chart does not have a lower control limit. Figure 17.4 presents the MINITAB output of the and R charts for the hole location data. Note that the center lines and control limits that we have just calculated are shown on the and R charts.

Figure 17.4: MINITAB Output of and R Charts for the Hole Location Data

Control limits such as those computed in Example 17.3 are called **trial control limits.** Theoretically, control limits are supposed to be computed using subgroups collected while the process is in statistical control. However, it is impossible to know whether the process is in control until we have constructed the control charts. If, after we have set up the and R charts, we find that the process is in control, we can use the charts to monitor the process.

If the charts show that the process is not in statistical control (for example, there are plot points outside the control limits), we must find and eliminate the assignable causes before we can calculate control limits for monitoring the process. In order to understand how to find and eliminate assignable causes, we must understand how changes in the process mean and the process variation show up on and R charts. To do this, consider Figures ** 17.5** and

Figure 17.5: A Shift of the Process Mean Shows Up on the Chart

Figure 17.6: An Increase in the Process Variation Shows Up on Both the and R Charts

1 The sample ranges plotted on the R chart increase and go out of control.

1 The sample means plotted on the chart become more variable (because, since σ increases, increases) and go out of control.

Since changes in the process mean and in the process variation show up on the chart, we do not begin by analyzing the chart. This is because, if there were out-of-control sample meanson the chart, we would not know whether the process mean or the process variation had changed. Therefore, it might be more difficult to identify the assignable causes of the out-of-control sample means because the assignable causes that would cause the process mean to shift could be very different from the assignable causes that would cause the process variation to increase. For instance, unwarranted frequent resetting of a machine might cause the process level to shift up and down, while improper lubrication of the machine might increase the process variation.

In order to simplify and better organize our analysis procedure, we begin by analyzing the R chart, which reflects only changes in the process variation. Specifically, we first identify and eliminate the assignable causes of the out-of-control sample ranges on the R chart, and then we analyze the chart. The exact procedure is illustrated in the following example.

EXAMPLE 17.4: The Hole Location Case

Consider the and R charts for the hole location data that are given in Figure 17.4. To develop control limits that can be used for ongoing control, we first examine the R chart. We find two points above the UCL on the R chart. This indicates that excess within-subgroup variability exists at these points. We see that the out-of-control points correspond to subgroups 7 and 17. Investigation reveals that, when these subgroups were selected, an inexperienced, newly hired operator ran the operation while the regular operator was on break. We find that the inexperienced operator is not fully closing the clamps that fasten down the compressor shells during the hole punching operation. This is causing excess variability in the hole locations. This assignable cause can be **eliminated** by thoroughly retraining the newly hired operator.

Since we have identified and corrected the assignable cause associated with the points that are out of control on the R chart, we can drop subgroups 7 and 17 from the data set. We recalculate center lines and control limits by using the remaining 18 subgroups. We first recompute (omitting and R values for subgroups 7 and 17)

Notice here that has not changed much (see Figure 17.4), but has been reduced from .085 to .0683. Using the new and values, revised control limits for the chart are

The revised UCL for the R chart is

Since D3 is not listed for subgroups of size 5, the R chart does not have a LCL. Here the reduction in has reduced the UCL on the R chart from .1797 to .1444 and has also narrowed the control limits for the chart. For instance, the UCL for the chart has been reduced from 3.0552 to 3.0457. The MINITAB output of the and R charts employing these revised center lines and control limits is shown in Figure 17.7.

Figure 17.7: MINITAB Output of and R Charts for the Hole Location Data: Subgroups 7 and 17 Omitted

We must now check the revised R chart for statistical control. We find that the chart shows good control: there are no other points outside the control limits or long runs of points on either side of the center line. Since the R chart is in good control, we can analyze the revised chart. We see that two plot points are above the UCL on the chart. Notice that these points were not outside our original trial control limits in Figure 17.4. However, the elimination of the assignable cause and the resulting reduction in has narrowed the chart control limits so that these points are now out of control. Since the R chart is in control, the points on the chart that are out of control suggest that the process level has shifted when subgroups 1 and 12 were taken. Investigation reveals that these subgroups were observed immediately after start-up at the beginning of the day and immediately after start-up following the lunch break. We find that, if we allow a five-minute machine warm-up period, we can eliminate the process level problem.

Since we have again found and eliminated an assignable cause, we must compute newly revised center lines and control limits. Dropping subgroups 1 and 12 from the data set, we recompute

Using the newest and values, we compute newly revised control limits as follows:

Again, the R chart does not have a LCL. We obtain the newly revised and R charts that are shown in the MINITAB output of Figure 17.8. We see that all the points on each chart are inside their respective control limits. This says that the actions taken to remove assignable causes have brought the process into statistical control. However, it is important to point out that, although the process is in statistical control, **this does not necessarily mean that the process is capable of producing products that meet the customer’s needs.** That is, while the control charts tell us that no assignable causes of process variation remain, the charts do not (directly) tell us anything about how much common cause variation exists. If there is too much common cause variability, the process will not meet customer or manufacturer specifications. We talk more about this later.

Figure 17.8: MINITAB Output of and R Charts for the Hole Location Data: Subgroups 1, 7, 12, and 17 Omitted. The Charts Show Good Control.

**When both the** **and R charts are in statistical control, we can use the control limits for ongoing process monitoring.** New and R values for subsequent subgroups are plotted with respect to these limits. Plot points outside the control limits indicate the existence of assignable causes and the need for action on the process. The appropriate corrective action can often be taken by local supervision. Sometimes management intervention may be needed. For example, if the assignable cause is out-of-specification raw materials, management may have to work with a supplier to improve the situation. The ongoing control limits occasionally need to be updated to include newly observed data. However, since employees often seem to be uncomfortable working with limits that are frequently changing, it is probably a good idea to update center lines and control limits only when the new data would substantially change the limits. Of course, if an important process change is implemented, new data must be collected, and we may need to develop new center lines and control limits from scratch.

Sometimes it is not possible to find an assignable cause, or it is not possible to eliminate the assignable cause even when it can be identified. In such a case, it is possible that the original (or partially revised) trial control limits are good enough to use; this will be a subjective decision. Occasionally, it is reasonable to drop one or more subgroups that have been affected by an assignable cause that cannot be eliminated. For example, the assignable cause might be an event that very rarely occurs and is unpreventable. If the subgroup(s) affected by the assignable cause have a detrimental effect on the control limits, we might drop the subgroups and calculate revised limits. Another alternative is to collect new data and use them to calculate control limits.

In the following box we summarize the most important points we have made regarding the analysis of and R charts:

Analyzing and R Charts to Establish Process Control

1 Remember that it is important to use both the chart and the R chart to study the process.

2 Begin by analyzing the R chart for statistical control.

a Find and eliminate assignable causes that are indicated by the R chart.

b Revise both the and R chart control limits, dropping data for subgroups corresponding to assignable causes that have been found and eliminated in 2a.

c Check the revised R chart for control.

d Repeat 2a, b, and c as necessary until the R chart shows statistical control.

3 When the R chart is in statistical control, the chart can be properly analyzed.

a Find and eliminate assignable causes that are indicated by the chart.

b Revise both the and R chart control limits, dropping data for subgroups corresponding to assignable causes that have been found and eliminated in 3a.

c Check the revised chart (and the revised R chart) for control.

d Repeat 3a, b, and c (or, if necessary, 2a, b, and c and 3a, b, and c) as needed until both the and R charts show statistical control.

4 When both the and R charts are in control, use the control limits for process monitoring.

a Plot and R points for newly observed subgroups with respect to the established limits.

b If either the chart or the R chart indicates a lack of control, take corrective action on the process.

5 Periodically update the and R control limits using all relevant data (data that describe the process as it now operates).

6 When a major process change is made, develop new control limits if necessary.

EXAMPLE 17.5: The Hole Location Case

We consider the hole location problem and the revised and R charts shown in Figure 17.8. Since the process has been brought into statistical control, we may use the control limits in Figure 17.8 to monitor the process. This would assume that we have used an appropriate subgrouping scheme and have observed enough subgroups to give potential assignable causes a chance to show up. In reality, we probably want to collect considerably more than 20 subgroups before setting control limits for ongoing control of the process.

We assume for this example that the control limits in Figure 17.8 are reasonable. Table 17.4 gives four subsequently observed subgroups of five hole location dimensions. The subgroup means and ranges for these data are plotted with respect to the ongoing control limits in the MINITAB output of Figure **17.9**. We see that the R chart remains in control, while the mean for subgroup 24 is above the UCL on the chart. This tells us that an assignable cause has increased the process mean. Therefore, action is needed to reduce the process mean.

Table 17.4: Four Subgroups of Five Hole Location Dimensions Observed after Developing Control Limits for Ongoing Process Monitoring

Figure 17.9: MINITAB Output of and R Charts for the Hole Location Data: Ongoing Control

EXAMPLE 17.6: The Hot Chocolate Temperature Case

Consider the hot chocolate data given in Table 17.2 (page 752). In order to set up and R charts for these data, we compute

and

Looking at Table 17.3 (page 757), we see that the and R control chart constants for the subgroup size n = 3 are A2 = 1.023 and D4 = 2.574. It follows that we calculate center lines and control limits as follows:

Since D3 is not given in Table 17.3 for the subgroup size n = 3, the R chart does not have a lower control limit.

The and R charts for the hot chocolate data are given in the MegaStat output of Figure **17.10**. We see that the R chart is in **good statistical control**, while the chart is out of control with three subgroup means above the UCL and with one subgroup mean below the LCL. Looking at the chart, we see that the subgroup means that are above the UCL were observed during lunch (note subgroups 4, 10, and 22). Investigation and process documentation reveal that on these days the hot chocolate machine was not turned off between breakfast and lunch. Discussion among members of the dining hall staff further reveals that, because there is less time between breakfast and lunch than there is between lunch and dinner or dinner and breakfast, the staff often fails to turn off the hot chocolate machine between breakfast and lunch. Apparently, this is the reason behind the higher hot chocolate temperatures observed during lunch. Investigation also shows that the dining hall staff failed to turn on the hot chocolate machine before breakfast on Thursday (see subgroup 19)—in fact, a student had to ask that the machine be turned on. This caused the subgroup mean for subgroup 19 to be far below the chart LCL. The dining hall staff concludes that the hot chocolate machine needs to be turned off after breakfast and then turned back on 15 minutes before lunch (prior experience suggests that it takes the machine 15 minutes to warm up). The staff also concludes that the machine should be turned on 15 minutes before each meal. In order to ensure that these actions are taken, an automatic timer is purchased to turn on the hot chocolate machine at the appropriate times. This brings the process into statistical control. Figure **17.11** shows and R charts with revised control limits calculated using the subgroups that remain after the subgroups for the out-of-control lunches (subgroups 3, 4, 9, 10, 21, and 22) and the out-of-control breakfast (subgroups 19 and 20) are eliminated from the data set. We see that these revised control charts are in statistical control.

Figure 17.10: MegaStat Output of and R Charts for the Hot Chocolate Temperature Data

Figure 17.11: MegaStat Output of Revised and R Charts for the Hot Chocolate Temperature Data. The Process Is Now in Control.

Having seen how to interpret and R charts, we are now better prepared to understand why **we estimate** the process standard deviation σ by /d2. Recall that when μ and σ are known, the chart control limits are . The standard deviation σ in these limits is the process standard deviation **when the process is in control.** When this standard deviation is unknown, we estimate σ **as if the process is in control, even though the process might not be in control.** The quantity /d2 is an appropriate estimate of σ because **is the average of individual ranges computed from rational subgroups—subgroups selected so that the chances that important process changes occur within a subgroup are minimized.** Thus each subgroup range, and therefore /d2, estimates the process variation as if the process were in control. Of course, we could also compute the standard deviation of the measurements in each subgroup, and employ the average of the subgroup standard deviations to estimate σ. The key is not whether we use ranges or standard deviations to measure the variation within the subgroups. Rather, the key is that we must calculate a measure of variation for each subgroup and then must average the separate measures of subgroup variation in order to estimate the process variation as if the process is in control.

Exercises for Sections 17.3 and 17.4

CONCEPTS

17.5 Explain (1) the purpose of an chart, (2) the purpose of an R chart, (3) why both charts are needed.

17.6 Explain why the initial control limits calculated for a set of subgrouped data are called “trial control limits.”

17.7 Explain why a change in process variability shows up on both the and R charts.

17.8 In each of the following situations, what conclusions (if any) can be made about whether the process mean is changing? Explain your logic.

a R chart out of control.

b R chart in control, chart out of control.

c Both and R charts in control.

METHODS AND APPLICATIONS

17.9 Table 17.5 gives five subgroups of measurement data. Use these data to Measure 1

a Find and R for each subgroup.

b Find and .

c Find A2 and D4.

d Compute and R chart center lines and control limits.

Table 17.5: Five Subgroups of Measurement Data Measure 1

17.10 In the book Tools and Methods for the Improvement of Quality, Gitlow, Gitlow, Oppenheim, and Oppenheim discuss a resort hotel’s efforts to improve service by reducing variation in the time it takes to clean and prepare rooms. In order to study the situation, five rooms are selected each day for 25 consecutive days, and the time required to clean and prepare each room is recorded. The data obtained are given in Table 17.6. RoomPrep

a Calculate the subgroup mean and range R for each of the first two subgroups.

b Show that minutes and that = 2.696 minutes.

c Find the control chart constants A2 and D4 for the cleaning and preparation time data. Does D3 exist? What does this say?

d Find the center line and control limits for the chart for this data.

e Find the center line and control limit for the R chart for this data.

f Set up (plot) the and R charts for the cleaning time data.

g Are the and R charts in control? Explain.

Table 17.6: Daily Samples of Five Room Cleaning and Preparation Times RoomPrep

17.11 A pizza restaurant monitors the size (measured by the diameter) of the 10-inch pizzas that it prepares. Pizza crusts are made from doughs that are prepared and prepackaged in boxes of 15 by a supplier. Doughs are thawed and pressed in a pressing machine. The toppings are added, and the pizzas are baked. The wetness of the doughs varies from box to box, and if the dough is too wet or greasy, it is difficult to press, resulting in a crust that is too small. The first shift of workers begins work at 4 p.m., and a new shift takes over at 9 p.m. and works until closing. The pressing machine is readjusted at the beginning of each shift. The restaurant takes five consecutive pizzas prepared at the beginning of each hour from opening to closing on a particular day. The diameter of each baked pizza in the subgroups is measured, and the pizza crust diameters obtained are given in Table 17.7. Use the pizza crust diameter data to do the following: PizzaDiam

Table 17.7: 10 Samples of Pizza Crust Diameters PizzaDiam

a Show that and = .84.

b Find the center lines and control limits for the and R charts for the pizza crust data.

c Set up the and R charts for the pizza crust data.

d Is the R chart for the pizza crust data in statistical control? Explain.

e Is the chart for the pizza crust data in statistical control? If not, use the chart and the information given with the data to try to identify any assignable causes that might exist.

f Suppose that, based on the chart, the manager of the restaurant decides that the employees do not know how to properly adjust the dough pressing machine. Because of this, the manager thoroughly trains the employees in the use of this equipment. Because an assignable cause (incorrect adjustment of the pressing machine) has been found and eliminated, we can remove the subgroups affected by this unusual process variation from the data set. We therefore drop subgroups 1 and 6 from the data. Use the remaining eight subgroups to show that we obtain revised center lines of and = .825.

g Use the revised values of and to compute revised and R chart control limits for the pizza crust diameter data. Set up and R charts using these revised limits. Be sure to omit subgroup means and ranges for subgroups 1 and 6 when setting up these charts.

h Has removing the assignable cause brought the process into statistical control? Explain.

**17.12** A chemical company has collected 15 daily subgroups of measurements of an important chemical property called “acid value” for one of its products. Each subgroup consists of six acid value readings: a single reading was taken every four hours during the day, and the readings for a day are taken as a subgroup. The 15 daily subgroups are given in Table 17.8. AcidVal

Table 17.8: 15 Subgroups of Acid Value Measurements for a Chemical Process AcidVal

a Show that for these data and = 2.06.

b Set up and R charts for the acid value data. Are these charts in statistical control?

c On the basis of these charts, is it possible to draw proper conclusions about whether the mean acid value is changing? Explain why or why not.

d Suppose that investigation reveals that the out-of-control points on the R chart (the ranges for subgroups 1 and 10) were caused by an equipment malfunction that can be remedied by redesigning a mechanical part. Since the assignable cause that is responsible for the large ranges for subgroups 1 and 10 has been found and eliminated, we can remove subgroups 1 and 10 from the data set. Show that using the remaining 13 subgroups gives revised center lines of and = 1.4385.

e Use the revised values of and to compute revised and R chart control limits for the acid value data. Set up the revised and R charts, making sure to omit subgroup means and ranges for subgroups 1 and 10.

f Are the revised and R charts for the remaining 13 subgroups in statistical control? Explain. What does this result tell us to do?

**17.13** The data in Table 17.9 consist of 30 subgroups of measurements that specify the location of a “tube hole” in an air conditioner compressor shell. Each subgroup contains the tube hole dimension measurement for five consecutive compressor shells selected from the production line. The first 15 subgroups were observed on March 21, and the second 15 subgroups were observed on March 22. As indicated in Table 17.9, the die press used in the hole punching operation was changed after subgroup 5 was observed, and a die repair was made after subgroup 25 was observed. TubeHole

Table 17.9: 30 Subgroups of Tube Hole Location Dimensions for Air Conditioner Compressor Shells TubeHole

a Show that for the first 15 subgroups (observed on March 21) we have and = 6.1333.

b Set up and R charts for the 15 subgroups that were observed on March 21 (do not use any of the March 22 data). Do these and R charts show statistical control?

c Using the control limits you computed by using the 15 subgroups observed on March 21, set up and R charts for all 30 subgroups. That is, add the subgroup means and ranges for March 22 to your and R charts, but use the limits you computed from the March 21 data.

d Do the and R charts obtained in part c show statistical control? Explain.

e Does it appear that changing to die press #628 is an assignable cause? Explain. (Note that the die press is the machine that is used to punch the tube hole.)

f Does it appear that making a die repair is an assignable cause? Explain.

**17.14** A company packages a bulk product in bags with a 50-pound label weight. During a typical day’s operation of the fill process, 22 subgroups of five bag fills are observed. Using the observed data, and are calculated to be 52.9364 pounds and 1.6818 pounds, respectively. When the 22 ’s and 22 R’s are plotted with respect to the appropriate control limits, the first 6 subgroups are found to be out of control. This is traced to a mechanical start-up problem, which is remedied. Using the remaining 16 subgroups, and are calculated to be 52.5875 pounds and 1.2937 pounds, respectively.

a Calculate appropriate revised and R chart control limits.

b When the remaining 16 ’s and 16 R’s are plotted with respect to the appropriate revised control limits, they are found to be within these limits. What does this imply?

**17.15** In the book Tools and Methods for the Improvement of Quality, Gitlow, Gitlow, Oppenheim, and Oppenheim discuss an example of using and R charts to study tuning knob diameters. In their problem description the authors say this:

A manufacturer of high-end audio components buys metal tuning knobs to be used in the assembly of its products. The knobs are produced automatically by a subcontractor using a single machine that is supposed to produce them with a constant diameter. Nevertheless, because of persistent final assembly problems with the knobs, management has decided to examine this process output by requesting that the subcontractor keep an x-bar and R chart for knob diameter.

On a particular day the subcontractor selects four knobs every half hour and carefully measures their diameters. Twenty-five subgroups are obtained, and these subgroups (along with their subgroup means and ranges) are given in Table 17.10. KnobDiam

Table 17.10: 25 Subgroups of Tuning Knob Diameters KnobDiam

a For these data show that and = 5.16. Then use these values to calculate control limits and to set up and R charts for the 25 subgroups of tuning knob diameters. Do these and R charts indicate the existence of any assignable causes? Explain.

b An investigation is carried out to find out what caused the large range for subgroup 23. The investigation reveals that a water pipe burst at 7:25 p.m. and that the mishap resulted in water leaking under the machinery used in the tuning knob production process. The resulting disruption is the apparent cause for the out-of-control range for subgroup 23. The water pipe is mended, and since this fix is reasonably permanent, we are justified in removing subgroup 23 from the data set. Using the remaining 24 subgroups, show that revised center lines are and = 4.88.

c Use the revised values of and to set up revised and R charts for the remaining 24 subgroups of diameters. Be sure to omit the mean and range for subgroup 23.

d Looking at the revised R chart, is this chart now in statistical control? What does your answer say about whether we can use the chart to decide if the process mean is changing?

e Looking at the revised chart, is this chart in statistical control? What does your answer tell us about the process mean?

f An investigation is now undertaken to find the cause of the very high values for subgroups 10, 11, 12, and 13. We again quote Gitlow, Gitlow, Oppenheim, and Oppenheim:

The investigation leads to the discovery that … a keyway wedge had cracked and needed to be replaced on the machine. The mechanic who normally makes this repair was out to lunch, so the machine operator made the repair. This individual had not been properly trained for the repair; for this reason, the wedge was not properly aligned in the keyway, and the subsequent points were out of control. Both the operator and the mechanic agree that the need for this repair was not unusual. To correct this problem it is decided to train the machine operator and provide the appropriate tools for making this repair in the mechanic’s absence. Furthermore, the maintenance and engineering staffs agree to search for a replacement part for the wedge that will not be so prone to cracking.

Since the assignable causes responsible for the very high values for subgroups 10, 11, 12, and 13 have been found and eliminated, we remove these subgroups from the data set. Show that removing subgroups 10, 11, 12, and 13 (in addition to the previously removed subgroup 23) results in the revised center lines and = 5.25. Then use these revised values to set up revised and R charts for the remaining 20 subgroups.

g Are all of the subgroup means and ranges for these newly revised and R charts inside their respective control limits?

17.5: Pattern Analysis

Chapter 21

When we observe a plot point outside the control limits on a control chart, we have strong evidence that an assignable cause exists. In addition, several other data patterns indicate the presence of assignable causes. Precise description of these patterns is often made easier by **dividing the control band into zones—designated A, B, and C.** Zone boundaries are set at points that are one and two standard deviations (of the plotted statistic) on either side of the center line. We obtain six zones—each zone being one standard deviation wide—with three zones on each side of the center line. The zones that stretch one standard deviation above and below the center line are designated as **
C zones.** The zones that extend from one to two standard deviations away from the center line are designated as

B zones.

A zones.

Figure 17.12: Zone Boundaries

Finally, Figure 17.12(b) shows (based on a normal distribution of plot points) the percentages of points that we would expect to observe in each zone when the process is in statistical control. For instance, we would expect to observe 34.13 percent of the plot points in the upper portion of zone C.

For an chart, if the distribution of process measurements is reasonably normal, then the distribution of subgroup means will be approximately normal, and the percentages shown in Figure 17.12 apply. That is, the plotted subgroup means for an “in control” chart should look as if they have been randomly selected from a normal distribution. Any distribution of plot points that looks very different from the expected percentages will suggest the existence of an assignable cause.

Various companies (for example, Western Electric [AT&T] and Ford Motor Company) have established sets of rules for identifying assignable causes; use of such rules is called **
pattern analysis.** We now summarize some commonly accepted rules. Note that many of these rules are illustrated in Figures 17.13, 17.14, 17.15, and

Figure 17.13: A Plot Point outside the Control Limits

Figure 17.14: Two Out of Three Consecutive Plot Points in Zone A (or beyond)

Figure 17.15: Four Out of Five Consecutive Plot Points in Zone B (or beyond)

**Source:** H. Gitlow, S. Gitlow, A. Oppenheim, and R. Oppenheim, Tools and Methods for the Improvement of Quality, pp. 191–93, 209–211. Copyright © 1989. Reprinted by permission of McGraw-Hill Companies, Inc.

Figure 17.16: Other Out-of-Control Patterns

Pattern Analysis for and R Charts

If one or more of the following conditions exist, it is reasonable to conclude that one or more assignable causes are present:

1 One plot point beyond zone A (that is, outside the **three standard deviation control limits**)—see Figure 17.13.

2 Two out of three consecutive plot points in zone A (or beyond) on one side of the center line of the control chart. Sometimes a zone boundary that separates zones A and B is called a **two standard deviation warning limit.** It can be shown that, if the process is in control, then the likelihood of observing two out of three plot points beyond this warning limit (even when no points are outside the control limits) is very small. Therefore, such a pattern signals an assignable cause. Figure 17.14 illustrates this pattern. Specifically, note that plot points 5 and 6 are two consecutive plot points in zone A and that plot points 19 and 21 are two out of three consecutive plot points in zone A.

3 Four out of five consecutive plot points in zone B (or beyond) on one side of the center line of the control chart. Figure 17.15 illustrates this pattern. Specifically, note that plot points 2, 3, 4, and 5 are four consecutive plot points in zone B and that plot points 12, 13, 15, and 16 are four out of five consecutive plot points in zone B (or beyond).

4 **A run of at least eight plot points.** Here we define a **
run
** to be

5 A **nonrandom pattern** of plot points. Such a pattern might be an increasing or **decreasing trend, a fanning-out or funneling-in pattern, a cycle, an alternating pattern,** or any other pattern that is very inconsistent with the percentages given in Figure 17.12 (see parts (d) through (h) of Figure 17.16).

**If none of the patterns or conditions in 1 through 5 exists,** then the process shows good statistical control—or is said to be **“in control.” A process that is in control should not be tampered with.** On the other hand, **if one or more of the patterns in 1 through 5 exist, action must be taken to find the cause of the out-of-control pattern(s)** (which should be eliminated if the assignable cause is undesirable).

It is tempting to use many rules to decide when an assignable cause exists. However, if we use too many rules we can end up with an unacceptably high chance of a **false out-of-control signal** (that is, an out-of-control signal when there is no assignable cause present). For most control charts, the use of the rules just described will yield an overall probability of a false signal in the range of 1 to 2 percent.

EXAMPLE 17.7: The Hole Location Case

Figure **17.17** shows ongoing and R charts for the hole location problem. Here the chart includes zone boundaries with zones A, B, and C labeled. Notice that the first out-of-control condition (one plot point beyond zone A) exists. Looking at the last five plot points on the chart, we see that the third out-of-control condition (four out of five consecutive plot points in zone B or beyond) also exists.

Figure 17.17: Ongoing and R Charts for the Hole Location Data—Zones A, B, and C Included

Exercises for Section 17.5

CONCEPTS

17.16 When a process is in statistical control:

a What percentage of the plot points on an chart will be found in the C zones (that is, in the middle 1/3 of the chart’s “control band”)?

b What percentage of the plot points on an chart will be found in either the C zones or the B zones (that is, in the middle 2/3 of the chart’s “control band”)?

c What percentage of the plot points on an chart will be found in the C zones, the B zones, or the A zones (that is, in the chart’s “control band”)?

17.17 Discuss how a sudden increase in the process mean shows up on the chart.

**17.18** Discuss how a sudden decrease in the process mean shows up on the chart.

**17.19** Discuss how a steady increase in the process mean shows up on the chart. Also, discuss how a steady decrease in the process mean shows up on the chart.

**17.20** Explain what we mean by a “false out-of-control signal.”

METHODS AND APPLICATIONS

**17.21** In the June 1991 issue of Quality Progress, Gunter presents several control charts. Four of these charts are reproduced in Figure 17.18. For each chart, find any evidence of a lack of statistical control (that is, for each chart identify any evidence of the existence of one or more assignable causes). In each case, if such evidence exists, clearly explain why the plot points indicate that the process is not in control.

Figure 17.18: Charts for Exercise 17.21

Source: B. Gunter, “Process Capability Studies Part 3: The Tale of the Charts,” Quality Progress (June 1991), pp. 77–82. Copyright © 1991. American Society for Quality. Used with permission.

**17.22** In the book Tools and Methods for the Improvement of Quality, Gitlow, Gitlow, Oppenheim, and Oppenheim present several control charts in a discussion and exercises dealing with pattern analysis. These control charts, which include appropriate A, B, and C zones, are reproduced in Figure 17.19. For each chart, identify any evidence of a lack of statistical control (that is, for each chart identify any evidence suggesting the existence of one or more assignable causes). In each case, if such evidence exists, clearly explain why the plot points indicate that the process is not in control.

Figure 17.19: Charts for Exercise 17.22

Source: H. Gitlow, S. Gitlow, A. Oppenheim, and R. Oppenheim, Tools and Methods for the Improvement of Quality, pp. 191–93, 209–11. Copyright © 1989. Reprinted by permission of McGraw-Hill Companies, Inc.

**17.23** Consider the tuning knob diameter data given in Table 17.10 (page 770). Recalling that the subgroup size is n = 4 and that and = 5.16 for these data, KnobDiam

a Calculate all of the zone boundaries for the chart.

b Calculate all of the R chart zone boundaries that are either 0 or positive.

**17.24** Given what you now know about pattern analysis, examine each of the following and R charts for evidence of lack of statistical control. In each case, explain any evidence indicating the existence of one or more assignable causes.

a The pizza crust diameter and R charts of Exercise 17.11 (page 766). PizzaDiam

b The acid value and R charts of Exercise 17.12 (page 767). AcidVal

c The tube hole location and R charts of Exercise 17.13 (page 768). TubeHole

17.6: Comparison of a Process with Specifications: Capability Studies

If we have a process in statistical control, we have found and eliminated the assignable causes of process variation. Therefore, the individual process measurements fluctuate over time with a **constant standard deviation σ
** around a

As will be shown in Example 17.9, one way to study the capability of a process that is in statistical control is to construct a **histogram** from a set of individual process measurements. The histogram can then be compared with the product specification limits. In addition, we know that if all possible individual process measurements are normally distributed with mean μ and standard deviation σ, then 99.73 percent of these measurements will be in the interval [μ − 3σ, μ + 3σ]. Estimating μ and σ by and, /d2, we obtain the __natural tolerance limits__6 for the process.

Natural Tolerance Limits

The natural tolerance limits for a normally distributed process that is in statistical control are

where d2 is a constant that depends on the subgroup size n. Values of d2 are given in Table 17.3 (page 757) for subgroup sizes n = 2 to n = 25. These limits contain approximately 99.73 percent of the individual process measurements.

If the natural tolerance limits are inside the specification limits, then almost all (99.73 percent) of the individual process measurements are produced within the specification limits. In this case we say that the process is **
capable
** of meeting specifications. Furthermore, if we use and R charts to monitor the process, then as long as the process remains in statistical control, the process will continue to meet the specifications. If the natural tolerance limits are wider than the specification limits, we say that the process is

EXAMPLE 17.8: The Hot Chocolate Temperature Case

Consider the and R chart analysis of the hot chocolate temperature data. Suppose the dining hall staff has determined that all of the hot chocolate it serves should have a temperature between 130°F and 150°F. Recalling that the and R charts of Figure 17.11 (page 765) show that the process has been brought into control with and with = 4.75, we find that is an estimate of the mean hot chocolate temperature, and that /d2 = 4.75/1.693 = 2.81 is an estimate of the standard deviation of all the hot chocolate temperatures. Here d2 = 1.693 is obtained from Table 17.3 (page 757) corresponding to the subgroup size n = 3. Assuming that the temperatures are approximately normally distributed, the natural tolerance limits

tell us that approximately 99.73 percent of the individual hot chocolate temperatures will be between 132.31°F and 149.15°F. Since these natural tolerance limits are inside the specification limits (130°F to 150°F), almost all the temperatures are within the specifications. Therefore, the hot chocolate–making process is capable of meeting the required temperature specifications. Furthermore, if the process remains in control on its and R charts, it will continue to meet specifications.

EXAMPLE 17.9: The Hole Location Case

Again consider the hole punching process for air conditioner compressor shells. Recall that we were able to get this process into a state of statistical control with and = .0675 by removing several assignable causes of process variation.

Figure 17.20 gives a relative frequency histogram of the 80 individual hole location measurements used to construct the and R charts of Figure 17.8 (page 761). This histogram suggests that the population of all individual hole location dimensions is approximately normally distributed.

Figure 17.20: A Relative Frequency Histogram of the Hole Location Data (Based on the Data with Subgroups 1, 7, 12, and 17 Omitted)

Since the process is in statistical control, = 3.0006 is an estimate of the process mean, and /d2 = .0675/2.326 = .0290198 is an estimate of the process standard deviation. Here d2 = 2.326 is obtained from Table 17.3 (page 757) corresponding to the subgroup size n = 5. Furthermore, the natural tolerance limits

tell us that almost all (approximately 99.73 percent) of the individual hole location dimensions produced by the hole punching process are between 2.9135 inches and 3.0877 inches.

Suppose a major customer requires that the hole location dimension must meet specifications of 3.00 ± .05 inches. That is, the customer requires that every individual hole location dimension must be between 2.95 inches and 3.05 inches. The natural tolerance limits, [2.9135, 3.0877], which contain almost all individual hole location dimensions, are wider than the specification limits [2.95, 3.05]. This says that some of the hole location dimensions are outside the specification limits. Therefore, the process is not capable of meeting the specifications. Note that the histogram in Figure 17.20 also shows that some of the hole location dimensions are outside the specification limits.

Figure 17.21 illustrates the situation, assuming that the individual hole location dimensions are normally distributed. The figure shows that the natural tolerance limits are wider than the specification limits. The shaded areas under the normal curve make up the fraction of product that is outside the specification limits. Figure 17.21 also shows the calculation of the estimated fraction of hole location dimensions that are out of specification. We estimate that 8.55 percent of the dimensions do not meet the specifications.

Figure 17.21: Calculating the Fraction out of Specification for the Hole Location Data. Specifications Are 3.00 ± .05.

Since the process is not capable of meeting specifications, it must be improved by removing common cause variation. This is management’s responsibility. Suppose engineering and management conclude that the excessive variation in the hole locations can be reduced by redesigning the machine that punches the holes in the compressor shells. Also suppose that after a research and development program is carried out to do this, the process is run using the new machine and 20 new subgroups of n = 5 hole location measurements are obtained. The resulting and R charts (not given here) indicate that the process is in control with and = .0348. Furthermore, a histogram of the 100 hole location dimensions used to construct the and R charts indicates that all possible hole location measurements are approximately normally distributed. It follows that we estimate that almost all individual hole location dimensions are contained within the new natural tolerance limits

As illustrated in Figure 17.22, these tolerance limits are within the specification limits 3.00 ± .05. Therefore, the new process is now capable of producing almost all hole location dimensions inside the specifications. The new process is capable because the estimated process standard deviation has been substantially reduced (from /d2 = .0675/2.326 = .0290 for the old process to /d2 = .0348/2.326 = .0149613 for the redesigned process).

Figure 17.22: A Capable Process: The Natural Tolerance Limits Are within the Specification Limits

Next, note that (for the improved process) the z value corresponding to the lower specification limit (2.95) is

This says that the lower specification limit is 3.36 estimated process standard deviations below . Since the lower natural tolerance limit is 3 estimated process standard deviations below , there is a **leeway** of .36 estimated process standard deviations between the lower natural tolerance limit and the lower specification limit (see Figure 17.22). Also, note that the z value corresponding to the upper specification limit (3.05) is

This says that the upper specification limit is 3.33 estimated process standard deviations above . Since the upper natural tolerance limit is 3 estimated process standard deviations above , there is a leeway of .33 estimated process standard deviations between the upper natural tolerance limit and the upper specification limit (see Figure 17.22). Because some leeway exists between the natural tolerance limits and the specification limits, the distribution of process measurements (that is, the curve in Figure 17.22) can shift slightly to the right or left (or can become slightly more spread out) without violating the specifications. Obviously, the more leeway, the better.

To understand why process leeway is important, recall that a process must be in statistical control before we can assess the capability of the process. In fact:

In order to demonstrate that a company’s product meets customer requirements, the company must present

1 and R charts that are in statistical control.

2 Natural tolerance limits that are within the specification limits.

However, even if a capable process shows good statistical control, the process mean and/or the process variation will occasionally change (due to new assignable causes or unexpected recurring problems). If the process mean shifts and/or the process variation increases, a process will need some leeway between the natural tolerance limits and the specification limits in order to avoid producing out-of-specification product. We can determine the amount of process leeway (if any exists) by defining what we call the **
sigma level capability
** of the process.

Sigma Level Capability

The sigma level capability of a process is the number of estimated process standard deviations between the estimated process mean, , and the specification limit that is closest to .

For instance, in the previous example the lower specification limit (2.95) is 3.36 estimated standard deviations below the estimated process mean, , and the upper specification limit (3.05) is 3.33 estimated process standard deviations above . It follows that the upper specification limit is closest to the estimated process mean , and because this specification limit is 3.33 estimated process standard deviations from , we say that the hole punching process has 3.33 sigma capability.

If a process has a sigma level capability of three or more, then there are at least three estimated process standard deviations between and the specification limit that is closest to . It follows that, if the distribution of process measurements is normally distributed, then the process is capable of meeting the specifications. For instance, Figure 17.23(a) illustrates a process with three sigma capability. This process is just barely capable—that is, there is no process leeway. Figure 17.23(b) illustrates a process with **six sigma capability.** This process has three standard deviations of leeway. In general, we see that if a process is capable, the sigma level capability expresses the amount of process leeway. The higher the sigma level capability, the more process leeway. More specifically, for a capable process, the sigma level capability minus three gives the number of estimated standard deviations of process leeway. For example, since the hole punching process has 3.33 sigma capability, this process has 3.33 − 3 = .33 estimated standard deviations of leeway.

Figure 17.23: Sigma Level Capability and Process Leeway

The difference between three sigma and six sigma capability is dramatic. To illustrate this, look at Figure 17.23(a), which shows that a normally distributed process with three sigma capability produces 99.73 percent good quality (the area under the distribution curve between the specification limits is .9973). On the other hand, Figure 17.23(b) shows that a normally distributed process with six sigma capability produces 99.9999998 percent good quality. Said another way, if the process mean is centered between the specifications, and if we produce large quantities of product, then a normally distributed process with three sigma capability will produce an average of 2,700 defective products per million, while a normally distributed process with six sigma capability will produce an average of only .002 defective products per million.

In the long run, however, process shifts due to assignable causes are likely to occur. It can be shown that, if we monitor the process by using an chart that employs a typical subgroup size of 4 to 6, the largest sustained shift of the process mean that might remain undetected by the chart is a shift of 1.5 process standard deviations. In this worst case, it can be shown that a normally distributed three sigma capable process will produce an average of 66,800 defective products per million (clearly unacceptable), while a normally distributed six sigma capable process will produce an average of only 3.4 defective products per million. Therefore, if a six sigma capable process is monitored by and R charts, then, when a process shift occurs, we can detect the shift (by using the control charts), and we can take immediate corrective action before a substantial number of defective products are produced.

This is, in fact, how control charts are supposed to be used to prevent the production of defective product. That is, our strategy is

Prevention Using Control Charts

1 Reduce common cause variation in order to create leeway between the natural tolerance limits and the specification limits.

2 Use control charts to establish statistical control and to monitor the process.

3 When the control charts give out-of-control signals, take immediate action on the process to reestablish control before out-of-specification product is produced.

Since 1987, a number of U.S. companies have adopted a **six sigma philosophy.** In fact, these companies refer to themselves as **six sigma companies.** It is the goal of these companies to achieve six sigma capability for all processes in the entire organization. For instance, Motorola, Inc., the first company to adopt a six sigma philosophy, began a five-year quality improvement program in 1987. The goal of Motorola’s companywide defect reduction program is to achieve six sigma capability for all processes—for instance, manufacturing processes, delivery, information systems, order completeness, accuracy of transactions records, and so forth. As a result of its six sigma plan, Motorola claims to have saved more than $1.5 billion. The corporation won the Malcolm Baldrige National Quality Award in 1988, and Motorola’s six sigma plan has become a model for firms that are committed to quality improvement. Other companies that have adopted the six sigma philosophy include IBM, Digital Equipment Corporation, and General Electric.

To conclude this section, we make two comments. First, it has been traditional to measure process capability by using what is called the

C p k

**index.** This index is calculated by dividing the sigma level capability by three. For example, since the hole punching process illustrated in Figure 17.22 has a sigma level capability of 3.33, the C p k for this process is 1.11. In general, if C p k is at least 1, then the sigma level capability of the process is at least 3 and thus the process is capable. Historically, C p k has been used because its value relative to the number 1 describes the process capability. We prefer using sigma level capability to characterize process capability because we believe that it is more intuitive.

Second, when a process is in control, then the estimates /d2 and s of the process standard deviation will be very similar. This implies that we can compute the natural tolerance limits by using the alternative formula For example, since the mean and standard deviation of the 80 observations used to construct the and R charts in Figure 17.8 (page 761) are and s = .028875, we obtain the natural tolerance limits

These limits are very close to those obtained in Example 17.9, [2.9135, 3.0877], which were computed by using the estimate /d2 = .0290198 of the process standard deviation. Use of the alternative formula is particularly appropriate when there are long-run process variations that are not measured by the subgroup ranges (in which case /d2 underestimates the process standard deviation). Since statistical control in any real application of SPC will not be perfect, some people believe that this version of the natural tolerance limits is the most appropriate.

Exercises for Section 17.6

CONCEPTS

**17.25** Write a short paragraph explaining why a process that is in statistical control is not necessarily capable of meeting customer requirements (specifications).

**17.26** Explain the interpretation of the natural tolerance limits for a process. What assumptions must be made in order to properly make this interpretation? How do we check these assumptions?

**17.27** Explain how the natural tolerance limits compare to the specification limits when

a A process is capable of meeting specifications.

b A process is not capable of meeting specifications.

**17.28** For each of the following, explain

a Why it is important to have leeway between the natural tolerance limits and the specification limits.

b What is meant by the sigma level capability for a process.

c Two reasons why it is important to achieve six sigma capability.

METHODS AND APPLICATIONS

**17.29** Consider the room cleaning and preparation time situation in Exercise 17.10 (page 766). We found that and R charts based on subgroups of size 5 for this data are in statistical control with minutes and = 2.696 minutes. RoomPrep

a Assuming that the cleaning and preparation times are approximately normally distributed, calculate a range of values that contains almost all (approximately 99.73 percent) of the individual cleaning and preparation times.

b Find reasonable estimates of the maximum and minimum times needed to clean and prepare an individual room.

c Suppose the resort hotel wishes to specify that every individual room should be cleaned and prepared in 20 minutes or less. Is this upper specification being met? Explain. Note here that there is no lower specification, since we would like cleaning times to be as short as possible (as long as the job is done properly).

d If the upper specification for room cleaning and preparation times is 20 minutes, find the sigma level capability of the process. If the upper specification is 30 minutes, find the sigma level capability.

**17.30** Suppose that and R charts based on subgroups of size 3 are used to monitor the moisture content of a type of paper. The and R charts are found to be in statistical control with percent and = .4 percent. Further, a histogram of the individual moisture content readings suggests that these measurements are approximately normally distributed.

a Compute the natural tolerance limits (limits that contain almost all the individual moisture content readings) for this process.

b If moisture content specifications are 6.0 percent ± .5 percent, is this process capable of meeting the specifications? Why or why not?

c Estimate the fraction of paper that is out of specification.

d Find the sigma level capability of the process.

**17.31** A grocer has a contract with a produce wholesaler that specifies that the wholesaler will supply the grocer with grapefruit that weigh at least .75 pounds each. In order to monitor the grapefruit weights, the grocer randomly selects three grapefruit from each of 25 different crates of grapefruit received from the wholesaler. Each grapefruit’s weight is determined and, therefore, 25 subgroups of three grapefruit weights are obtained. When and R charts based on these subgroups are constructed, we find that these charts are in statistical control with and = .11. Further, a histogram of the individual grapefruit weights indicates that these measurements are approximately normally distributed.

a Calculate a range of values that contains almost all (approximately 99.73 percent) of the individual grapefruit weights.

b Find a reasonable estimate of the maximum weight of a grapefruit that the grocer is likely to sell.

c Suppose that the grocer’s contract with its produce supplier specifies that grapefruits are to weigh a minimum of .75 lb. Is this lower specification being met? Explain. Note here that there is no upper specification, since we would like grapefruits to be as large as possible.

d If the lower specification of .75 lb. is not being met, estimate the fraction of grapefruits that weigh less than .75 lb. Hint: Find an estimate of the standard deviation of the individual grapefruit weights.

**17.32** Consider the pizza crust diameters for 10-inch pizzas given Exercise 17.11 (pages 766–767). We found that, by removing an assignable cause, we were able to bring the process into statistical control with and = .825. PizzaDiam

a Recalling that the subgroup size for the pizza crust and R charts is 5, and assuming that the pizza crust diameters are approximately normally distributed, calculate the natural tolerance limits for the diameters.

b Using the natural tolerance limits, estimate the largest diameter likely to be sold by the restaurant as a 10-inch pizza.

c Using the natural tolerance limits, estimate the smallest diameter likely to be sold by the restaurant as a 10-inch pizza.

d Are all 10-inch pizzas sold by this restaurant really at least 10 inches in diameter? If not, estimate the fraction of pizzas that are not at least 10 inches in diameter.

**17.33** Consider the bag fill situation in Exercise 17.14 (page 769). We found that the elimination of a start-up problem brought the filling process into statistical control with and = 1.2937.

a Recalling that the fill weight and R charts are based on subgroups of size 5, and assuming that the fill weights are approximately normally distributed, calculate the natural tolerance limits for the process.

b Suppose that management wishes to reduce the mean fill weight in order to save money by “giving away” less product. However, since customers expect each bag to contain at least 50 pounds of product, management wishes to leave some process leeway. Therefore, after the mean fill weight is reduced, the lower natural tolerance limit is to be no less than 50.5 lb. Based on the natural tolerance limits, how much can the mean fill weight be reduced? If the product costs $2 per pound, and if 1 million bags are sold per year, what is the yearly cost reduction achieved by lowering the mean fill weight?

**17.34** Suppose that a normally distributed process (centered at target) has three sigma capability. If the process shifts 1.5 sigmas to the right, show that the process will produce defective products at a rate of 66,800 per million.

**17.35** Suppose that a product is assembled using 10 different components, each of which must meet specifications for five different quality characteristics. Therefore, we have 50 different specifications that potentially could be violated. Further suppose that each component possesses three sigma capability (process centered at target) for each quality characteristic. Then, if we assume normality and independence, find the probability that all 50 specifications will be met.

17.7: Charts for Fraction Nonconforming

Sometimes, rather than collecting measurement data, we inspect items and simply decide whether each item conforms to some desired criterion (or set of criteria). For example, a fuel tank does or does not leak, an order is correctly or incorrectly processed, a batch of chemical product is acceptable or must be reprocessed, or plastic wrap appears clear or too hazy. When an inspected unit does not meet the desired criteria, it is said to be **
nonconforming
** (or defective). When an inspected unit meets the desired criteria, it is said to be

conforming

The control chart that we set up for this type of data is called a **
p chart.** To construct this chart, we observe subgroups of n units over time. We inspect or test the n units in each subgroup and determine the number d of these units that are nonconforming. We then calculate for each subgroup

and we plot the values versus time on the p chart. If the process being studied is in statistical control and producing a fraction p of nonconforming units, and if the units inspected are independent, then the number of nonconforming units d in a subgroup of n units inspected can be described by a binomial distribution. If, in addition, n is large enough so that np is greater than 2,7 then both d and the fraction of nonconforming units are approximately described by normal distributions. Furthermore, the population of all possible values has mean = p and standard deviation

Therefore, if p is known we can compute three standard deviation control limits for values of by setting

However, since it is unlikely that p will be known, we usually must estimate p from process data. The estimate of p is

Substituting for p, we obtain the following:

Center Line and Control Limits for a p Chart

The control limits calculated using these formulas are considered to be trial control limits. Plot points above the upper control limit suggest that one or more assignable causes have increased the process fraction nonconforming. Plot points below the lower control limit may suggest that an improvement in the process performance has been observed. However, plot points below the lower control limit may also tell us that an inspection problem exists. Perhaps defective items are still being produced, but for some reason the inspection procedure is not finding them. If the chart shows a lack of control, assignable causes must be found and eliminated and the trial control limits must be revised. Here data for subgroups associated with assignable causes that have been eliminated will be dropped, and data for newly observed subgroups will be added when calculating the revised limits. This procedure is carried out until the process is in statistical control. When control is achieved, the limits can be used to monitor process performance. **The process capability for a process that is in statistical control is expressed using**

**, the estimated process fraction nonconforming.** When the process is in control and is too high to meet internal or customer requirements, common causes of process variation must be removed in order to reduce . This is a management responsibility.

EXAMPLE 17.10

To improve customer service, a corporation wishes to study the fraction of incorrect sales invoices that are sent to its customers. Every week a random sample of 100 sales invoices sent during the week is selected, and the number of sales invoices containing at least one error is determined. The data for the last 30 weeks are given in Table 17.11. To construct a p chart for these data, we plot the fraction of incorrect invoices versus time. Since the true overall fraction p of incorrect invoices is unknown, we estimate p by (see Table 17.11)

Table 17.11: Sales Invoice Data—100 Invoices Sampled Weekly Invoice

Since = 100(.023) = 2.3 is greater than 2, the population of all possible values has an approximate normal distribution if the process is in statistical control. Therefore, we calculate the center line and control limits for the p chart as follows:

Since the LCL calculates negative, there is no lower control limit for this p chart. The MegaStat output of the p chart for these data is shown in Figure 17.24. We note that none of the plot points is outside the control limits, and we fail to see any nonrandom patterns of points. We conclude that the process is in statistical control with a relatively constant process fraction nonconforming of = .023. That is, the process is stable with an average of approximately 2.3 incorrect invoices per each 100 invoices processed. Since no assignable causes are present, there is no reason to believe that any of the plot points have been affected by unusual process variations. That is, it will not be worthwhile to look for unusual circumstances that have changed the average number of incorrect invoices per 100 invoices processed. If an average of 2.3 incorrect invoices per each 100 invoices is not acceptable, then management must act to remove common causes of process variation. For example, perhaps sales personnel need additional training or perhaps the invoice itself needs to be redesigned.

Figure 17.24: MegaStat Output of a p Chart for the Sales Invoice Data

In the previous example, subgroups of 100 invoices were randomly selected each week for 30 weeks. In general, subgroups must be taken often enough to detect possible sources of variation in the process fraction nonconforming. For example, if we believe that shift changes may significantly influence the process performance, then we must observe at least one subgroup per shift in order to study the shift-to-shift variation. Subgroups must also be taken long enough to allow the major sources of process variation to show up. As a general rule, at least 25 subgroups will be needed to estimate the process performance and to test for process control.

We have said that the size n of each subgroup should be large enough so that np (which is usually estimated by ) is greater than 2 (some practitioners prefer np to be greater than 5). Since we often monitor a p that is quite small (.05 or .01 or less), n must often be quite large. Subgroup sizes of 50 to 200 or more are common. Another suggestion is to choose a subgroup size that is large enough to give a positive lower control limit (often, when employing a p chart, smaller subgroup sizes give a calculated lower control limit that is negative). A positive LCL is desirable because it allows us to detect opportunities for process improvement. Such an opportunity exists when we observe a plot point below the LCL. If there is no LCL, it would obviously be impossible to obtain a plot point below the LCL. It can be shown that

A condition that guarantees that the subgroup size is large enough to yield a **positive lower control limit for a p chart is**

where **
p0 is an initial estimate of the fraction nonconforming** produced by the process. This condition is appropriate when three standard deviation control limits are employed.

For instance, suppose experience suggests that a process produces 2 percent nonconforming items. Then, in order to construct a p chart with a positive lower control limit, the subgroup size employed must be greater than

As can be seen from this example, for small values of p0 the above condition may require very large subgroup sizes. For this reason, it is not crucial that the lower control limit be positive.

We have thus far discussed how often—that is, over what specified periods of time (each hour, shift, day, week, or the like)—we should select subgroups. We have also discussed how large each subgroup should be. We next consider how we actually choose the items in a subgroup. One common procedure—which often yields large subgroup sizes—is to include in a subgroup **all (that is, 100 percent) of the units produced in a specified period of time.** For instance, a subgroup might consist of all the units produced during a particular hour. When employing this kind of scheme, we must carefully consider the independence assumption. The binomial distribution assumes that successive units are produced independently. It follows that a p chart would not be appropriate if the likelihood of a unit being defective depends on whether other units produced in close proximity are defective. Another procedure is to **randomly select the units in a subgroup from all the units produced in a specified period of time.** This was the procedure used in Example 17.10 to obtain the subgroups of sales invoices. As long as the subgroup size is small relative to the total number of units produced in the specified period, the units in the randomly selected subgroup should probably be independent. However, if the rate of production is low, it could be difficult to obtain a large enough subgroup when using this method. In fact, even if we inspect 100 percent of the process output over a specified period, and even if the production rate is quite high, it might still be difficult to obtain a large enough subgroup. This is because (as previously discussed) we must select subgroups often enough to detect possible assignable causes of variation. If we must select subgroups fairly often, the production rate may not be high enough to yield the needed subgroup size in the time in which the subgroup must be selected.

In general, the large subgroup sizes that are required can make it difficult to set up useful p charts. For this reason, we sometimes (especially when we are monitoring a very small p) relax the requirement that np be greater than 2. Practice shows that even if np is somewhat smaller than 2, we can still use the three standard deviation p chart control limits. In such a case, we detect assignable causes by looking for points outside the control limits and by looking for runs of points on the same side of the center line. In order for the distribution of all possible values to be sufficiently normal to use the pattern analysis rules we presented for charts, must be greater than 2. In this case we carry out pattern analysis for a p chart as we do for an chart (see Section 17.5), and we use the following zone boundaries:

Here, when the LCL calculates negative, it should not be placed on the control chart. Zone boundaries, however, can still be placed on the control chart as long as they are not negative.

Exercises for Section 17.7

CONCEPTS

**17.36** In your own words, define a nonconforming unit.

**17.37** Describe two situations in your personal life in which you might wish to plot a control chart for fraction nonconforming.

**17.38** Explain why it can sometimes be difficult to obtain rational subgroups when using a control chart for fraction nonconforming.

METHODS AND APPLICATIONS

**17.39** Suppose that = .1 and n = 100. Calculate the upper and lower control limits, UCL and LCL, of the corresponding p chart.

**17.40** Suppose that = .04 and n = 400. Calculate the upper and lower control limits, UCL and LCL, of the corresponding p chart.

**17.41** In the July 1989 issue of Quality Progress, William J. McCabe discusses using a p chart to study a company’s order entry system. The company was experiencing problems meeting the promised 60-day delivery schedule. An investigation found that the order entry system frequently lacked all the information needed to correctly process orders. Figure 17.25 gives a p chart analysis of the percentage of orders having missing information.

a From Figure 17.25 we see that = .527. If the subgroup size for this p chart is n = 250, calculate the upper and lower control limits, UCL and LCL.

b Is the p chart of Figure 17.25 in statistical control? That is, are there any assignable causes affecting the fraction of orders having missing information?

c On the basis of the p chart in Figure 17.25, McCabe says,

The process was stable and one could conclude that the cause of the problem was built into the system. The major cause of missing information was salespeople not paying attention to detail, combined with management not paying attention to this problem. Having sold the product, entering the order into the system was generally left to clerical people while the salespeople continued selling.

Can you suggest possible improvements to the order entry system?

Figure 17.25: A p Chart for the Fraction of Orders with Missing Information

Source: W. J. McCabe, “Examining Processes Improves Operations,” Quality Progress (July 1989), pp. 26–32. Copyright © 1989 American Society for Quality. Used with permission.

**17.42** In the book Tools and Methods for the Improvement of Quality, Gitlow, Gitlow, Oppenheim, and Oppenheim discuss a data entry operation that makes a large number of entries every day. Over a 24-day period, daily samples of 200 data entries are inspected. Table 17.12 gives the number of erroneous entries per 200 that were inspected each day. DataErr

a Use the data in Table 17.12 to compute . Then use this value of to calculate the control limits for a p chart of the data entry operation, and set up the p chart. Include zone boundaries on the chart.

Table 17.12: The Number of Erroneous Entries for 24 Daily Samples of 200 Data Entries DataErr

b Is the data entry process in statistical control, or are assignable causes affecting the process? Explain.

c Investigation of the data entry process is described by Gitlow, Gitlow, Oppenheim, and Oppenheim as follows:

In our example, to bring the process under control, management investigated the observations which were out of control (days 8 and 22) in an effort to discover and remove the **special causes** of variation in the process. In this case, management found that on day 8 a new operator had been added to the workforce without any training. The logical conclusion was that the new environment probably caused the unusually high number of errors. To ensure that this special cause would not recur, the company added a one-day training program in which data entry operators would be acclimated to the work environment.

A team of managers and workers conducted an investigation of the circumstances occurring on day 22. Their work revealed that on the previous night one of the data entry consoles malfunctioned and was replaced with a standby unit. The standby unit was older and slightly different from the ones currently used in the department. The repairs on the regular console were not expected to be completed until the morning of day 23. To correct this special source of variation, the team recommended purchasing a spare console that would match the existing equipment and disposing of the outdated model presently being used as the backup. Management then implemented the suggestion.

Since the assignable causes on days 8 and 22 have been found and eliminated, we can remove the data for these days from the data set. Remove the data and calculate the new value of . Then set up a revised p chart for the remaining 22 subgroups.

d Did the actions taken bring the process into statistical control? Explain.

**17.43** In the July 1989 issue of Quality Progress, William J. McCabe discusses using a p chart to study the percentage of errors made by 21 buyers processing purchase requisitions. The p chart presented by McCabe is shown in Figure 17.26. In his explanation of this chart, McCabe says,

The causes of the errors … could include out-of-date procedures, unreliable office equipment, or the perceived level of management concern with errors. These causes are all associated with the system and are all under management control.

Focusing on the 21 buyers, weekly error rates were calculated for a 30-week period (the data existed, but weren’t being used). A p-chart was set up for the weekly department error rate. It showed a 5.2 percent average rate for the department. In week 31, the manager called the buyers together and made two statements: “I care about errors because they affect our costs and delivery schedules,” and “I am going to start to count errors by individual buyers so I can understand the causes.” The p-chart … shows an almost immediate drop from 5.2 percent to 3.1 percent.

The explanation is that the common cause system (supervision, in this case) had changed; the improvement resulted from eliminating buyer sloppiness in the execution of orders. The p-chart indicates that buyer errors are now stable at 3.1 percent. The error rate will stay there until the common cause system is changed again.

Figure 17.26: p Chart for the Weekly Department Error Rate for 21 Buyers Processing Purchase Requisitions

Source: W. J. McCabe, “Examining Processes Improves Operations,” Quality Progress (July 1989), pp. 26–32. Copyright © 1989 American Society for Quality. Used with permission.

a The p chart in Figure 17.26 shows that = .052 for weeks 1 through 30. Noting that the subgroup size for this chart is n = 400, calculate the control limits UCL and LCL for the p chart during weeks 1 through 30.

b The p chart in Figure 17.26 shows that after week 30 the value of is reduced to .031. Assuming that the process has been permanently changed after week 30, calculate new control limits based on = .031. If we use these new control limits after week 30, is the improved process in statistical control? Explain.

**17.44** The customer service manager of a discount store monitors customer complaints. Each day a random sample of 100 customer transactions is selected. These transactions are monitored, and the number of complaints received concerning these transactions during the next 30 days is recorded. The numbers of complaints received for 20 consecutive daily samples of 100 transactions are, respectively, 2, 5, 10, 1, 5, 6, 9, 4, 1, 7, 1, 5, 7, 4, 5, 4, 6, 3, 10, and 5. Complaints

a Use the data to compute . Then use this value of to calculate the control limits for a p chart of the complaints data. Set up the p chart.

b Are the customer complaints for this 20-day period in statistical control? That is, have any unusual problems caused an excessive number of complaints during this period? Explain why or why not.

c Suppose the discount store receives 13 complaints in the next 30 days for the 100 transactions that have been randomly selected on day 21. Should the situation be investigated? Explain why or why not.

17.8: Cause-and-Effect and Defect Concentration Diagrams (Optional)

We saw in Chapter 2 that Pareto charts are often used to identify quality problems that require attention. When an opportunity for improvement has been identified, it is necessary to examine potential causes of the problem or defect (the undesirable effect). Because many processes are complex, there are often a very large number of possible causes, and it may be difficult to focus on the important ones. In this section we discuss two diagrams that can be employed to help uncover potential causes of process variation that are resulting in the undesirable effect.

The **
cause-and-effect diagram
** was initially developed by Japanese quality expert

The causes identified by the team are organized into a cause-and-effect diagram as follows:

1 After clearly stating the problem, write it in an effect box at the far right of the diagram. Draw a horizontal (center) line connected to the effect box.

2 Identify major potential cause categories. Write them in boxes that are connected to the center line. Various approaches can be employed in setting up these categories. For example, Figure 17.27 is a cause-and-effect diagram for “why tables are not cleared quickly” in a restaurant. This diagram employs the categories

Figure 17.27: A Cause-and-Effect Diagram for “Why Tables Are Not Cleared Quickly” in a Restaurant

Source: M. Gaudard, R. Coates, and L. Freeman, “Accelerating Improvement,” Quality Progress (October 1991), pp. 81–88. Copyright © 1991. American Society for Quality. Used with permission.

3 Identify subcauses and classify these according to the major potential cause categories identified in step 2. Identify new major categories if necessary. Place subcauses on the diagram as branches. See Figure 17.27.

4 Try to decide which causes are most likely causing the problem or defect. Circle the most likely causes. See Figure 17.27.

After the cause-and-effect diagram has been constructed, the most likely causes of the problem or defect need to be studied. It is usually necessary to collect and analyze data in order to find out if there is a relationship between likely causes and the effect. We have studied various statistical methods (for instance, control charts, scatter plots, ANOVA, and regression) that help in this determination.

A **
defect concentration diagram
** is a picture of the product. It depicts all views—for example, front, back, sides, bottom, top, and so on. The various kinds of defects are then illustrated on the diagram. Often, by examining the locations of the defects, we can discern information concerning the causes of the defects. For example, in the October 1990 issue of Quality Progress, The Juran Institute presents a defect concentration diagram that plots the locations of chips in the enamel finish of a kitchen range. This diagram is shown in Figure 17.28. If the manufacturer of this range plans to use protective packaging to prevent chipping, it appears that the protective packaging should be placed on the corners, edges, and burners of the range.

Figure 17.28: Defect Concentration Diagram Showing the Locations of Enamel Chips on Kitchen Ranges

Source: “The Tools of Quality Part V: Check Sheets,” from QI Tools: Data Collection Workbook, p. 11. Copyright © 1989. Juran Institute, Inc. Reprinted with permission from Juran Institute, Inc.

Exercises for Section 17.8

CONCEPTS

**17.45** Explain the purpose behind constructing (a) a cause-and-effect diagram and (b) a defect concentration diagram.

**17.46** Explain how to construct (a) a cause-and-effect diagram and (b) a defect concentration diagram.

METHODS AND APPLICATIONS

**17.47** In the January 1994 issue of Quality Progress, Hoexter and Julien discuss the quality of the services delivered by law firms. One aspect of such service is the quality of attorney–client communication. Hoexter and Julien present a cause-and-effect diagram for “poor client–attorney telephone communications.” This diagram is shown in Figure 17.29.

a Using this diagram, what (in your opinion) are the most important causes of poor client–attorney telephone communications?

b Try to improve the diagram. That is, try to add causes to the diagram.

Figure 17.29: A Cause-and-Effect Diagram for “Poor Client–Attorney Telephone Communications”

Source: R. Hoexter and M. Julien, “Legal Eagles Become Quality Hawks,” Quality Progress (January 1994), pp. 31–33. Copyright © 1994 American Society for Quality. Used with permission.

**17.48** In the October 1990 issue of Quality Progress, The Juran Institute presents an example that deals with the production of integrated circuits. The article describes the situation as follows:

The manufacture of integrated circuits begins with silicon slices that, after a sequence of complex operations, will contain hundreds or thousands of chips on their surfaces. Each chip must be tested to establish whether it functions properly. During slice testing, some chips are found to be defective and are rejected. To reduce the number of rejects, it is necessary to know not only the percentage but also the locations and the types of defects. There are normally two major types of defects: functional and parametric. A functional reject occurs when a chip does not perform one of its functions. A parametric reject occurs when the circuit functions properly, but a parameter of the chip, such as speed or power consumption, is not correct.

Figure 17.30 gives a defect concentration diagram showing the locations of rejected chips within the integrated circuit. Only those chips that had five or more defects during the testing of 1,000 integrated circuits are shaded. Describe where parametric rejects tend to be, and describe where functional rejects tend to be.

Figure 17.30: Defect Concentration Diagram Showing the Locations of Rejected Chips on Integrated Circuits

Source: “The Tools of Quality Part V: Check Sheets,” from QI Tools: Data Collection Workbook, p. 12. Copyright © 1989 Juran Institute, Inc. Used with permission.

**17.49** In the September 1994 issue of Quality Progress, Franklin P. Schargel presents a cause-and-effect diagram for the “lack of quality in schools.” We present this diagram in Figure 17.31.

a Identify and circle the causes that you feel contribute the most to the “lack of quality in schools.”

b Try to improve the diagram. That is, see if you can add causes to the diagram.

Figure 17.31: A Cause-and-Effect Diagram on the “Lack of Quality in Schools”

Source: F. P. Schargel, “Teaching TQM in an Inner City High School,” Quality Progress (September 1994), pp. 87–90. Copyright © 1994 American Society for Quality. Used with permission.

Chapter Summary

In this chapter we studied how to improve business processes by using control charts. We began by considering several meanings of quality, and we discussed the history of the quality movement in the United States. We saw that Walter Shewhart introduced statistical quality control while working at Bell Telephone Laboratories during the 1920s and 30s, and we also saw that W. Edwards Deming taught the Japanese how to use statistical methods to improve product quality following World War II. When the quality of Japanese products surpassed that of American-made goods, and when, as a result, U.S. manufacturers lost substantial shares of their markets, Dr. Deming consulted and lectured extensively in the United States. This sparked an American reemphasis on quality that continues to this day. We also briefly presented **Deming’s 14 Points,** a set of management principles that, if followed, Deming believed would enable a company to improve quality and productivity, reduce costs, and gain competitive advantage.

We next learned that processes are influenced by common cause variation (inherent variation) and by assignable cause variation (unusual variation), and we saw that a control chart signals when assignable causes exist. Then we discussed how to sample a process. In particular, we explained that effective control charting requires **rational subgrouping.** Such subgroups minimize the chances that important process variations will occur within subgroups, and they maximize the chances that such variations will occur between subgroups.

Next we studied and R charts in detail. We saw that charts are used to monitor and stabilize the process mean (level), and that R charts are used to monitor and stabilize the process variability. In particular, we studied how to construct and R charts by using **control chart constants,** how to recognize out-of-control conditions by employing zone boundaries and **
pattern analysis,** and how to use and R charts to get a process into statistical control.

While it is important to bring a process into statistical control, we learned that it is also necessary to meet the customer’s or **manufacturer’s requirements** (or specifications). Since statistical control does not guarantee that the process output meets specifications, we must carry out a **capability study** after the process has been brought into control. We studied how this is done by computing **
natural tolerance limits,** which are limits that contain almost all the individual process measurements. We saw that, if the natural tolerance limits are inside the specification limits, then the process is

capable

of meeting the specifications. We also saw that we can measure how capable a process is by using

sigma level capability,

We continued by studying **
p charts,** which are charts for fraction nonconforming. Such charts are useful when it is not possible (or when it is very expensive) to measure the quality characteristic of interest.

We concluded this chapter with an optional section on how to construct **
cause-and-effect diagrams
** and

It should be noted that two useful types of control charts not discussed in this chapter are **individuals charts** and c charts. These charts are discussed in Appendix L of the CD-ROM included with this book.

Glossary of Terms

**acceptance sampling:**

A statistical sampling technique that enables us to accept or reject a quantity of goods (the lot) without inspecting the entire lot. (page 744)

**assignable causes (of process variation):**

Unusual sources of process variation. Also called special causes or **specific causes** of process variation. (page 748)

**capable process:**

A process that has the ability to produce products or services that meet customer or manufacturer requirements (specifications). (page 776)

**cause-and-effect diagram:**

A diagram that enumerates (lists) the potential causes of an undesirable effect. (page 789)

**common causes (of process variation):**

Sources of process variation that are inherent to the process design—that is, sources of usual process variation. (page 747)

**conforming unit (nondefective):**

An inspected unit that meets a set of desired criteria. (page 783)

**control chart:**

A graph of process performance that includes a center line and two control limits—an upper control limit, UCL, and a lower control limit, LCL. Its purpose is to detect assignable causes. (page 754)

C p k index:

A process’s sigma level capability divided by 3. (page 781)

**defect concentration diagram:**

An illustration of a product that depicts the locations of defects that have been observed. (page 790)

**ISO 9000:**

A series of international standards for quality assurance management systems. (page 746)

**natural tolerance limits:**

Assuming a process is in statistical control and assuming process measurements are normally distributed, limits that contain almost all (approximately 99.73 percent) of the individual process measurements. (page 775)

**nonconforming unit (defective):**

An inspected unit that does not meet a set of desired criteria. (page 783)

**pattern analysis:**

Looking for patterns of plot points on a control chart in order to find evidence of assignable causes. (page 771)

p chart:

A control chart on which the proportion nonconforming (in subgroups of size n) is plotted versus time. (page 783)

**quality of conformance:**

How well a process is able to meet the requirements (specifications) set forth by the process design. (page 743)

**quality of design:**

How well the design of a product or service meets and exceeds the needs and expectations of the customer. (page 743)

**quality of performance:**

How well a product or service performs in the marketplace. (page 743)

**rational subgroups:**

Subgroups of process observations that are selected so that the chances that process changes will occur between subgroups is maximized. (page 750)

R chart:

A control chart on which subgroup ranges are plotted versus time. It is used to monitor the process variability (or spread). (page 754)

**run:**

A sequence of plot points on a control chart that are of the same type—for instance, a sequence of plot points above the center line. (page 772)

**sigma level capability:**

The number of estimated process standard deviations between the estimated process mean, , and the specification limit that is closest to . (page 779)

**statistical process control (SPC):**

A systematic method for analyzing process data in which we monitor and study the process variation. The goal is continuous process improvement. (page 747)

**subgroup:**

A set of process observations that are grouped together for purposes of control charting. (page 750)

**total quality management (TQM):**

Applying quality principles to all company activities. (page 745)

**variables control charts:**

Control charts constructed by using measurement data. (page 754)

chart (x-bar chart):

A control chart on which subgroup means are plotted versus time. It is used to monitor the process mean (or level). (page 754)

Important Formulas

Center line and control limits for an chart: page 757

Center line and control limits for an R chart: page 757

Zone boundaries for an chart: page 771

Zone boundaries for an R chart: page 771

Natural tolerance limits for normally distributed process measurements: page 775

Sigma level capability: page 779

Cpk index: page 781

Center line and control limits for a p chart: page 784

Zone boundaries for a p chart: page 786

Supplementary Exercises

Exercises **17.50** through **17.53** are based on a case study adapted from an example presented in the paper “Managing with Statistical Models” by James C. Seigel (1982). Seigel’s example concerned a problem encountered by Ford Motor Company.

The Camshaft Case Camshaft

An automobile manufacturer produces the parts for its vehicles in many different locations and transports them to assembly plants. In order to keep the assembly operations running efficiently, it is vital that all parts be within specification limits. One important part used in the assembly of V6 engines is the engine camshaft, and one important quality characteristic of this camshaft is the case hardness depth of its eccentrics. A camshaft eccentric is a metal disk positioned on the camshaft so that as the camshaft turns, the eccentric drives a lifter that opens and closes an engine valve. The V6 engine camshaft and its eccentrics are illustrated in Figure 17.32. These eccentrics are hardened by a process that passes the camshaft through an electrical coil that “cooks” or “bakes” the camshaft. Studies indicate that the hardness depth of the eccentric labeled in Figure 17.32 is representative of the hardness depth of all the eccentrics on the camshaft. Therefore, the hardness depth of this representative eccentric is measured at a specific location and is regarded to be the **hardness depth of the camshaft.** The optimal or target hardness depth for a camshaft is 4.5 mm. In addition, specifications state that, in order for the camshaft to wear properly, the hardness depth of a camshaft must be between 3.0 mm and 6.0 mm.

Figure 17.32: A Camshaft and Related Parts

The automobile manufacturer was having serious problems with the process used to harden the camshaft. This problem was resulting in 12 percent rework and 9 percent scrap, or a total of 21 percent out-of-specification camshafts. The hardening process was automated. However, adjustments could be made to the electrical coil employed in the process. To begin study of the process, a problem-solving team selected 30 daily subgroups of n = 5 hardened camshafts and measured the hardness depth of each camshaft. For each subgroup, the team calculated the mean and range R of the n = 5 hardness depth readings. The 30 subgroups are given in Table 17.13. The subgroup means and ranges are plotted in Figure 17.33. These means and ranges seem to exhibit substantial variability, which suggests that the hardening process was not in statistical control; we will compute control limits shortly.

Table 17.13: Hardness Depth Data for Camshafts (Coil #1) Camshaft

Figure 17.33: Graphs of Performance ( and R) for Hardness Depth Data (Using Coil #1)

Although control limits had not yet been established, the problem-solving team took several actions to try to stabilize the process while the 30 subgroups were being collected:

1 At point A, which corresponds to a low average and a high range, the power on the coil was increased from 8.2 to 9.2.

2 At point B the problem-solving team found a bent coil. The coil was straightened, although at point B the subgroup mean and range do not suggest that any problem exists.

3 At point C, which corresponds to a high average and a high range, the power on the coil was decreased to 8.8.

4 At point D, which corresponds to a low average and a high range, the coil shorted out. The coil was straightened, and the team designed a gauge that could be used to check the coil spacing to the camshaft.

5 At point E, which corresponds to a low average, the spacing between the coil and the camshaft was decreased.

6 At point F, which corresponds to a low average and a high range, the first coil (Coil #1) was replaced. Its replacement (Coil #2) was a coil of the same type.

17.50 Using the data in Table 17.13: Camshaft

a Calculate and and then find the center lines and control limits for and R charts for the camshaft hardness depths.

b Set up the and R charts for the camshaft hardness depth data.

c Are the and R charts in statistical control? Explain.

Examining the actions taken at points A through E (in Figure 17.33), the problem-solving team learned that the power on the coil should be roughly 8.8 and that it is important to monitor the spacing between the camshaft and the coil. It also learned that it may be important to check for bent coils. The problem-solving team then (after replacing Coil #1 with Coil #2) attempted to control the hardening process by using this knowledge. Thirty new daily subgroups of n = 5 hardness depths were collected. The and R charts for these subgroups are given in Figure 17.34.

Figure 17.34: and R Charts for Hardness Depth Data Using Coil #2 (Same Type as Coil #1)

**17.51** Using the values of and in Figure 17.34: Camshaft

a Calculate the control limits for the chart in Figure 17.34.

b Calculate the upper control limit for the R chart in Figure 17.34.

c Are the and R charts for the 30 new subgroups using Coil #2 (which we recall was of the same type as Coil #1) in statistical control? Explain.

**17.52** Consider the and R charts in Figure 17.34. Camshaft

a Calculate the natural tolerance limits for the improved process.

b Recalling that specifications state that the hardness depth of each camshaft must be between 3.0 mm. and 6.0 mm., is the improved process capable of meeting these specifications? Explain.

c Use and to estimate the fraction of hardness depths that are out of specification for the improved process.

Since the hardening process shown in Figure 17.34 was not capable, the problem-solving team redesigned the coil to reduce the common cause variability of the process. Thirty new daily subgroups of n = 5 hardness depths were collected using the redesigned coil, and the resulting and R charts are given in Figure 17.35.

Figure 17.35: and R Charts for Hardness Depth Data Using a Redesigned Coil

17.53 Using the values of and given in Figure 17.35: Camshaft

a Calculate the control limits for the and R charts in Figure 17.35.

b Is the process (using the redesigned coil) in statistical control? Explain.

c Calculate the natural tolerance limits for the process (using the redesigned coil).

d Is the process (using the redesigned coil) capable of meeting specifications of 3.0 mm. to 6.0 mm.? Explain. Also find and interpret the sigma level capability.

**17.54** A bank officer wishes to study how many credit cardholders attempt to exceed their established credit limits. To accomplish this, the officer randomly selects a weekly sample of 100 of the cardholders who have been issued credit cards by the bank, and the number of cardholders who have attempted to exceed their credit limit during the week is recorded. The numbers of cardholders who exceeded their credit limit in 20 consecutive weekly samples of 100 cardholders are, respectively, 1, 4, 9, 0, 4, 6, 0, 3, 8, 5, 3, 5, 2, 9, 4, 4, 3, 6, 4, and 0. Construct a control chart for the data and determine if the data are in statistical control. If 12 cardholders in next week’s sample of 100 cardholders attempt to exceed their credit limit, should the bank regard this as unusual variation in the process? CredLim

Appendix 17.1: Control Charts Using MINITAB

The instruction blocks in this section each begin by describing the entry of data into the Minitab Data window. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.1 for further information about entering data, saving data, and printing results when using MINITAB.

**
p control chart
** similar to Figure 17.24 on page 785 (data file: Invoice.MTW):

• In the Data window, enter the 30 weekly error counts from Table 17.11 (page 785) into column C1 with variable name Invoice.

• Select **Stat: Control Charts: Attributes Charts: p.**

• In the p Chart dialog box, enter Invoice into the Variables window.

• Enter 100 in the “Subgroup size” window to indicate that each error count is based on a sample of 100 invoices.

• Click OK in the p Chart dialog box.

• The p control chart will be displayed in a graphics window.

Appendix 17.2: Control Charts Using MegaStat

The instructions in this section begin by describing the entry of data into an Excel worksheet. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.2 for further information about entering data, saving data, and printing results in Excel. Please refer to Appendix 1.3 for more information about using MegaStat.

**X-bar and R charts** in Figure 17.10 on page 764 (data file: HotChoc.xlsx):

• In cells A1, A2, and A3, enter the column labels Temp1, Temp2, and Temp3.

• In columns A, B, and C, enter the hot chocolate temperature data as 24 rows of 3 measurements, as laid out in the columns headed 1, 2, and 3 in Table 17.2 on page 752. When entered in this way, each row is a subgroup (sample) of three temperatures. Calculated means and ranges (as in Table 17.2) need not be entered—only the raw data is needed.

• Select **Add-Ins: MegaStat: Quality Control Process Charts.**

• In the “Quality Control Process Charts” dialog box, click on “Variables (Xbar and R).”

• Use the autoexpand feature to select the range A1.C25 into the Input Range window. Here each row in the selected range is a subgroup (sample) of measurements.

• Click OK in the “Quality Control Process Charts” dialog box.

• The requested control charts are placed in an output file and may be edited using standard Excel editing features. See Appendix 1.3 (page 41) for additional information about editing Excel graphics.

p control chart

in Figure 17.24 on page 785 (data file: Invoice.xlsx):

• Enter the 30 weekly error counts from Table 17.11 (page 785) into Column A with the label Invoice in cell A1.

• Select Add-Ins: MegaStat: Quality Control Process Charts.

• In the “Quality Control Process Charts” dialog box, select “Proportion nonconforming (p).”

• Use the autoexpand feature to enter the range A1.A31 into the Input Range window.

• Enter the subgroup (sample) size (here equal to 100) into the Sample size box.

• Click OK in the “Quality Control Process Charts” dialog box.

**A c chart for nonconformities** (discussed in Appendix X of this book) can be obtained by entering data as for the p chart and by selecting “Number of defects (c).”

1 Source: CEEM Information Services, “Is ISO 9000 for You?” 1993.

2 The data for this case were obtained from a metal fabrication plant located in the Cincinnati, Ohio, area. For confidentiality, we have agreed to withhold the company’s name.

3 The data for this case were collected for a student’s term project with the cooperation of the Food Service at Miami University, Oxford, Ohio.

4 Basically, statistical independence here means that successive process measurements do not display any kind of pattern over time.

5 When D3 is not listed, the theoretical lower control limit for the R chart is negative. In this case, some practitioners prefer to say that the LCLR equals 0. Others prefer to say that the LCLR does not exist because a range R equal to 0 does not indicate that an assignable cause exists and because it is impossible to observe a negative range below LCLR. We prefer the second alternative. In practice, it makes no difference.

6 There are a number of alternative formulas for the natural tolerance limits. Here we give the version that is the most clearly related to using and R charts. At the end of this section we present an alternative formula.

7 Some statisticians believe that this condition should be np > 5. However, for p charts many think np > 2 is sufficient.

(Bowerman 742)

Bowerman, Bruce L. Business Statistics in Practice, 5th Edition. McGraw-Hill Learning Solutions, 022008.

CHAPTER <<<<<<<<<<<<<<<<<<<<<**b**>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>b>**1**** 3**: S

is

C**h**apter Outline

__13.1__

The Simple Linear Regression Mo** d**el

**and**

the Least Squares Point Estimates

**
13.2
** Model Assumptions and the Standard Error

**
13.3
** Testing the Signi

y

-Intercept

**
13.4
** Confidence and Prediction Intervals

**13. 5
** Simple Coefficients of Determination and Correlation (This section may be read anytime after reading Section 13.1)

**13. 6
** Testing the Significance of the Population Correlation Coefficient (Optional)

**
13.7
** An

F

Test for the Model

**
13.8
** Residual Analysis (Optional)

**
13.9
** Some Shortcut Formulas (Optional)

**Managers often ma ke decisions by studying the relationships between variables, and process improvements can often be made by understanding how changes in one or more variables affect the process output. Regression analysis is a statistical technique in which we use observed data to relate a variable of interest, which is called the dependent (or response) variable, to one or more independent (or predictor) variables. The objective is to build a regression model,** or

In the next three chapters we give a thorough presentation of regression analysis. We begin in this chapter by presenting **simple linear regression** analysis. Using this technique is appropriate when we are relating a dependent variable to a single independent variable and when a straight-line model describes the relationship between these two variables. We explain many of the methods of this chapter in the context of two new cases:

**The Fuel Consumption Case:** A management consulting firm uses simple linear regression analysis to predict the weekly amount of fuel (in millions of cubic feet of natural gas) that will be required to heat the homes and businesses in a small city on the basis of the week’s average hourly temperature. A natural gas company uses these predictions to improve its gas ordering process. One of the gas company’s objectives is to reduce the fines imposed by its pipeline transmission system when the company places inaccurate natural gas orders.

**The QHIC Case:** The marketing department at Quality Home Improvement Center (QHIC) uses simple linear regression analysis to predict home upkeep expenditure on the basis of home value. Predictions of home upkeep expenditures are used to help determine which homes should be sent advertising brochures promoting QHIC’s products and services.

13.1: The Simple Linear Regression Model and the Least Squares Point Estimates

__Chapter 14__

The __simple linear regression model__

The simple linear regression model assumes that the relationship between the **
dependent variable, which is denoted y,** and the

independent variable, denoted x,

EXAMPLE 13.1: The Fuel Consumption Case: Reducing Natural Gas Transmission Fines

Part 1: The natural gas transmission problem

When the natural gas industry was deregulated in 1993, natural gas companies became responsible for acquiring the natural gas needed to heat the homes and businesses in the cities they serve. To do this, natural gas companies purchase natural gas from marketers (usually through long-term contracts) and periodically (perhaps daily, weekly, or monthly) place orders for natural gas to be transmitted by pipeline transmission systems to their cities. There are hundreds of pipeline transmission systems in the United States, and many of these systems supply a large number of cities. For instance, the map in Figure 13.1 illustrates the pipelines of and the cities served by the Columbia Gas System.

Figure 13.1: The Columbia Gas System

**Source:** Columbia Gas System 1995 Annual Report.

© Reprinted courtesy of Columbia Gas System.

To place an order (called a nomination) for an amount of natural gas to be transmitted to its city over a period of time (day, week, month), a natural gas company makes its best prediction of the city’s natural gas needs for that period. The natural gas company then instructs its marketer(s) to deliver this amount of gas to its pipeline transmission system. If most of the natural gas companies being supplied by the transmission system can predict their cities’ natural gas needs with reasonable accuracy, then the overnominations of some companies will tend to cancel the undernominations of other companies. As a result, the transmission system will probably have enough natural gas to efficiently meet the needs of the cities it supplies.

In order to encourage natural gas companies to make accurate transmission nominations and to help control costs, pipeline transmission systems charge, in addition to their usual fees, transmission fines. A natural gas company is charged a transmission fine if it substantially undernominates natural gas, which can lead to an excessive number of unplanned transmissions, or if it substantially overnominates natural gas, which can lead to excessive storage of unused gas. Typically, pipeline transmission systems allow a certain percentage nomination error before they impose a fine. For example, some systems do not impose a fine unless the actual amount of natural gas used by a city differs from the nomination by more than 10 percent. Beyond the allowed percentage nomination error, fines are charged on a sliding scale—the larger the nomination error, the larger the transmission fine.

Part 2: The fuel consumption data

Suppose we are analysts in a management consulting firm. The natural gas company serving a small city has hired the consulting firm to develop an accurate way to predict the amount of fuel (in millions of cubic feet—MMcf—of natural gas) that will be required to heat the city. Because the pipeline transmission system supplying the city evaluates nomination errors and assesses fines weekly, the natural gas company wants predictions of future weekly fuel consumptions. Moreover, since the pipeline transmission system allows a 10 percent nomination error before assessing a fine, the natural gas company would like the actual and predicted weekly fuel consumptions to differ by no more than 10 percent. Our experience suggests that weekly fuel consumption substantially depends on the average hourly temperature (in degrees Fahrenheit) measured in the city during the week. Therefore, we will try to predict the **dependent (response) variable** weekly fuel consumption (y) on the basis of the **independent (predictor) variable** average hourly temperature (x) during the week. To this end, we observe values of y and x for eight weeks. The data are given in Table 13.1. In Figure ** 13.2** we give an Excel output of a scatter plot of y versus x. This plot shows (1) a tendency for the fuel consumptions to decrease in a straight-line fashion as the temperatures increase and (2) a scattering of points around the straight line. A regression model describing the relationship between y and x must represent these two characteristics. We now develop such a model.1

Table 13.1: The Fuel Consumption Data FuelCon 1

Figure 13.2: Excel Output of a Scatter Plot of y versus x

Part 3: The simple linear regression model

To begin, suppose that there were an exact straight line relationship between y (weekly fuel consumption) and x (average hourly temperature). Specifically, suppose that we knew that when the average hourly temperature is 0°F, weekly fuel consumption is 15.5 MMcf of natural gas. Also, suppose that we knew that, for each one-degree increase in average hourly temperature, weekly fuel consumption decreases by .1 MMcf of natural gas. In this case, the straight line relating y to x would have y__-intercept__ 15.5 and ** slope** −.1 and would be given by the equation

y = 15.5 − .1x

For example, if we knew that the average hourly temperature in a future week will be 40°F, then it would follow that fuel consumption in the future week will be

y = 15.5 − .1(40) = 11.5 MMcf of natural gas

In reality, Figure 13.2 shows that the relationship between y and x is not exactly a straight line. However, we can use a modified straight-line equation—called the **simple linear regression model—**to relate y to x. The simple linear regression model employs a y-intercept β0, a slope β1, and an ** error term** ε (the meanings of which will soon be discussed) and is expressed as follows:

y = β0 + β1x + ε

This model says that the values of y can be represented by a **mean level**—β0 + β1x—that changes in a straight line fashion as x changes, combined with random fluctuations—described by the error term ε—that cause the values of y to deviate from the mean level. Here:

1 The mean level β0 + β1x which we denote as μy, is the mean of the fuel consumptions (y) that would be observed in all weeks having an average hourly temperature of x. Furthermore, when we say that the mean level changes in a straight line fashion as x changes, we mean that different mean fuel consumptions corresponding to different values of x form a straight line. This line, which we call the **line of means,** is defined by the equation μy = β0 + β1x. The line of means has y-intercept β0 and slope β1, and the values of β0 and β1 determine the values of μy for different values of x. For example, Table 13.1 tells us that the average hourly temperature in week 1 was 28°F and the average hourly temperature in week 5 was 45.9°F. It follows that the mean fuel consumption for all weeks that have an average hourly temperature of 28°F is β0 + β1(28). Similarly, the mean fuel consumption for all weeks that have an average hourly temperature of 45.9°F is β0 + β1(45.9). Because we do not know the true values of β0 and β1, we cannot actually calculate these mean fuel consumptions. However, when we learn (in the next subsection) how to estimate β0 and β1, we will be able to estimate these means. For now, note that these means are explicitly identified in Figure 13.3, which illustrates the line of means. These means—and the mean fuel consumptions that correspond to the other average hourly temperatures in Table 13.1—are depicted as triangles that lie on the line of means.

Figure 13.3: The Simple Linear Regression Model Relating Weekly Fuel Consumption (y) to Average Hourly Temperature (x)

2 The **
y-intercept** β0 of the line of means can be understood by considering Figure 13.3. As illustrated in this figure, the y-intercept β0 is the mean fuel consumption for all weeks that have an average hourly temperature of 0°F. However, since we have not observed any temperatures near 0°F, we do not have any data to tell us whether the line of means describes mean fuel consumption when the average hourly temperature is 0°F. Therefore, although β0 is an important component of the line of means, its interpretation is of dubious practical value. More will be said about this later.

3 The slope β1 of the line of means can also be understood by considering Figure 13.3. As illustrated in this figure, the slope β1 is the change in mean weekly fuel consumption that is associated with a one-degree increase in average hourly temperature.

4 The error term ε of the simple linear regression model describes the effect on weekly fuel consumption of all factors other than the average hourly temperature. Such factors would include the average hourly wind velocity and the amount of cloud cover in the city. For example, Figure 13.3 shows that the error term for the first week is positive. Therefore, the observed fuel consumption y = 12.4 in the first week is above the corresponding mean fuel consumption for all weeks when x = 8. As another example, Figure 13.3 also shows that the error term for the fifth week is negative. Therefore, the observed fuel consumption y = 9.4 in the fifth week is below the corresponding mean fuel consumption for all weeks when x = 45.9. Of course, since we do not know the true values of β0 and β1 the relative positions of the quantities pictured in Figure 13.3 are only hypothetical.

With the fuel consumption example as background, we are ready to define the **
simple linear regression model relating the dependent variable y to the independent variable x.** We suppose that we have gathered n observations—each observation consists of an observed value of x and its corresponding value of y. Then:

The Simple Linear Regression Model

The **simple linear (or straight line) regression model** is: y = β0 + β1x + ε Here

1 μy = β0 + β1x is the **mean value** of the dependent variable y when the value of the independent variable is x.

2 β0 is the

y

–**intercept.** β0 is the mean value of y when x equals zero.2

3 β1 is the **
slope.** β1 is the change (amount of increase or decrease) in the mean value of y associated with a one-unit increase in x. If β1 is positive, the mean value of y increases as x increases. If β1 is negative, the mean value of y decreases as x increases.

4 ε is an **
error term
** that describes the effects on y of all factors other than the value of the independent variable x.

This model is illustrated in Figure ** 13.4** (note that x0 in this figure denotes a specific value of the independent variable x). The y-intercept β0 and the slope β1 are called

Figure 13.4: The Simple Linear Regression Model (Here the Slope β1 Is Positive)

The fuel consumption data in Table 13.1 were observed sequentially over time (in eight consecutive weeks). When data are observed in time sequence, the data are called **
time series data.** Many applications of regression utilize such data. Another frequently used type of data is called

EXAMPLE 13.2: The QHIC Case

Quality Home Improvement Center (QHIC) operates five stores in a large metropolitan area. The marketing department at QHIC wishes to study the relationship between x, home value (in thousands of dollars), and y, yearly expenditure on home upkeep (in dollars). A random sample of 40homeowners is taken and asked to estimate their expenditures during the previous year on the types of home upkeep products and services offered by QHIC. Public records of the county auditor are used to obtain the previous year’s assessed values of the homeowner’s homes. The resulting x and y values are given in Table 13.2. Because the 40 observations are for the same year (for different homes), these data are cross-sectional.

Table 13.2: The QHIC Upkeep Expenditure Data QHIC

The MINITAB output of a scatter plot of y versus x is given in Figure __13.5__. We see that the observed values of y tend to increase in a straight-line (or slightly curved) fashion as x increases. Assuming that the mean value of y given x has a straight-line relationship, it is reasonable to relate y to x by using the simple linear regression model having a positive slope (β1 > 0)

Figure 13.5: MINITAB Plot of Upkeep Expenditure versus Value of Home

y = β0 + β1x + ε

The slope β1 is the change (increase) in mean dollar yearly upkeep expenditure that is associated with each $1,000 increase in home value. In later examples the marketing department at QHIC will use predictions given by this simple linear regression model to help determine which homes should be sent advertising brochures promoting QHIC’s products and services.

We have interpreted the slope β1 of the simple linear regression model to be the change in the mean value of y associated with a one-unit increase in x. We sometimes refer to this change as the effect of the independent variable x on the dependent variable y. However, we cannot prove that a change in an independent variable causes a change in the dependent **variable.** Rather, regression can be used only to establish that the two variables move together and that the independent variable contributes information for predicting the dependent variable. For instance, regression analysis might be used to establish that as liquor sales have increased over the years, college professors’ salaries have also increased. However, this does not prove that increases in liquor sales cause increases in college professors’ salaries. Rather, both variables are influenced by a third variable—long-run growth in the national economy.

The least squares **point estimate**s

Suppose that we have gathered n observations (x1, y1), (x2, y2), …, (xn, yn),

where

each observation consists of a value of an independent variable x and a corresponding value of a dependent variable y. Also, suppose that a scatter plot of the n observatives indicates that the simple linear regression model

relates y to x.

In order to estimate the y-intercept β0 and the slope β1 of the line of means of this model, we could visually draw a line—called an **estimated regression line**—through the scatter plot. Then, we could read the y-intercept and slope off the estimated regression line and use these values as the point estimates of β0 and β1. Unfortunately, if different people visually drew lines through the scatter plot, their lines would probably differ from each other. What we need is the “best line” that can be drawn through the scatter plot. Although there are various definitions of what this best line is, one of the most useful best lines is the **least squares line**.

To understand the least squares line, we let

ŷ = b0 + b1x

denote the general equation of an estimated regression line drawn through a scatter plot. Here, since we will use this line to predict y on the basis of x, we call ŷ the predicted value of y when the value of the independent variable is x. In addition, b0 is the y-intercept and b1 is the slope of the estimated regression line. When we determine numerical values for b0 and b1, these values will be the point estimates of the y-intercept β0 and the slope β1 of the line of means. To explain which estimated regression line is the least squares line, we begin with the fuel consumption situation. Figure __13.6__ shows an estimated regression line drawn through a scatter plot of the fuel consumption data. In this figure the red dots represent the eight observed fuel consumptions and the black squares represent the eight predicted fuel consumptions given by the estimated regression line. Furthermore, the line segments drawn between the red dots and black squares represent __ residual__s, which are the

The least squares line

is the line that minimizes the **sum of squared residuals**. That is, the least squares line is the line positioned on the scatter plot so as to minimize the sum of the squared vertical distances between the observed and predicted fuel consumptions.

Figure 13.6: An Estimated Regression Line Drawn Through the Fuel Consumption Scatter Plot

To define the least squares line in a general situation, consider an arbitrary observation (xi, yi) in a sample of n observations. For this observation, the predicted value of the dependent variable y given by an estimated regression line is

ŷ = b0 + b1xi

Furthermore, the difference between the observed and predicted values of y, yi − ŷi is the **
residual
** for the observation, and the sum of squared residuals for all n observations is

The least squares line is the line that minimizes SSE. To find this line, we find the values of the y-intercept b0 and slope b1 that give values of ŷi = b0 + b1xi that minimize SSE. These values of b0 and b1 are called the **
least squares point estimates
** of β0 and β1. Using calculus, it can be shown that these estimates are calculated as follows:3

The Least Squares Point Estimates

For the simple linear regression model

:

1 The **least squares point estimate of the slope β1** is where

2 The **least squares point estimate of the y-intercept β0** is where

Here n is the number of observations (an observation is an observed value of x and its corresponding value of y).

The following example illustrates how to calculate these point estimates and how to use these point estimates to estimate mean values and predict individual values of the dependent variable. Note that the quantities SSxy and SSxx used to calculate the least squares point estimates are also used throughout this chapter to perform other important calculations.

EXAMPLE 13.3: The Fuel Consumption Case

Part 1: Calculating the least squares point estimates

Again consider the fuel consumption problem. To compute the least squares point estimates of the regression parameters β0 and β1 we first calculate the following preliminary summations:

Using these summations, we calculate SSxy and SSxx as follows.

It follows that the least squares point estimate of the slope β1 is

Furthermore, because

the least squares point estimate of the y-intercept β0 is

Since b1 = −.1279, we estimate that mean weekly fuel consumption decreases (since b1 is negative) by .1279 MMcf of natural gas when average hourly temperature increases by 1 degree. Since b0 = 15.84, we estimate that mean weekly fuel consumption is 15.84 MMcf of natural gas when average hourly temperature is 0°F. However, we have not observed any weeks with temperatures near 0, so making this interpretation of b0 might be dangerous. We discuss this point more fully after this example.

The least squares line

ŷ = b0 + b1x = 15.84 − .1279x

is sometimes called the **least squares prediction equation**. In Table 13.3 (on the next page) we summarize using this prediction equation to calculate the predicted fuel consumptions and the residuals for the eight weeks of fuel consumption data. For example, since in week 1 the average hourly temperature was 28°F, the predicted fuel consumption for week 1 is

Table 13.3: Calculation of SSE Obtained by Using the Least Squares Point Estimates

ŷ1 = 15.84 −.1279(28) = 12.2588

It follows, since the observed fuel consumption in week 1 was y1 = 12.4, that the residual for week 1 is

y1 − ŷ1 = 12.4 − 12.2588 = .1412

If we consider all of the residuals in Table 13.3 and add their squared values, we find that SSE, the sum of squared residuals, is 2.568. This SSE value will be used throughout this chapter. Figure 13.7 gives the MINITAB output of the least squares line. Note that this output gives the least squares estimates b0 = 15.84 and b1 = −0.1279. In general, we will rely on MINITAB, Excel, and MegaStat to compute the least squares estimates (and to perform many other regression calculations).

Figure 13.7: The MINITAB Output of the Least Squares Line

Part 2: Estimating a mean fuel consumption and predicting an individual fuel consumption

We define the ** experimental region** to be the range of the previously observed values of the average hourly temperature x. Referring to Table 13.3, we see that the experimental region consists of the range of average hourly temperatures from 28°F to 62.5°F. The simple linear regression model relates weekly fuel consumption y to average hourly temperature x for values of x that are in the experimental region. For such values of x, the least squares line is the estimate of the line of means. It follows that the point on the least squares line corresponding to an average hourly temperature of x

ŷ = b0 + b1x

is the point estimate of β0 + β1x the mean fuel consumption for all weeks that have an average hourly temperature of x. In addition, we predict the error term ε to be 0. Therefore, ŷ is also the **point prediction** of an individual value y = β0 + β1x + ε, which is the amount of fuel consumed in a single week that has an average hourly temperature of x. Note that the reason we predict the error term to be zero is that, because of several regression assumptions to be discussed in the next section, has a 50 percent chance of being positive and a 50 percent chance of being negative.

For example, suppose a weather forecasting service predicts that the average hourly temperature in the next week will be 40°F. Because 40°F is in the experimental region

is

1 The point estimate of the mean fuel consumption for all weeks that have an average hourly temperature of 40°F, and

2 The point prediction of the amount of fuel consumed in a single week that has an average hourly temperature of 40°F.

Figure 13.8 illustrates ŷ = 10.72 as a square on the least squares line.

Figure 13.8: Point Estimation and Point Prediction in the Fuel Consumption Case

To conclude this example, note that Figure 13.9 illustrates the potential danger of using the least squares line to predict outside the experimental region. In the figure, we extrapolate the least squares line far beyond the experimental region to obtain a prediction for a temperature of −10°F. As shown in Figure 13.7 (page 524) for values of x in the experimental region the observed values of y tend to decrease in a straight-line fashion as the values of x increase. However, for temperatures lower than 28°F the relationship between y and x might become curved. If it does, extrapolating the straight-line prediction equation to obtain a prediction for x = −10 might badly underestimate mean weekly fuel consumption (see Figure 13.9).

Figure 13.9: The Danger of Extrapolation Outside the Experimental Region

The previous example illustrates that when we are using a least squares regression line, we should not estimate a mean value or predict an individual value unless the corresponding value of x = 0 is in the **
experimental region
**—the range of the previously observed values of x. Often the value x = 0 is not in the experimental region. In such a situation, it would not be appropriate to interpret the y-intercept b0 as the estimate of the mean value of y when x equals 0. For example, consider the fuel consumption problem. Figure 13.9 illustrates that the average hourly temperature 0°F is not in the experimental region. Therefore, it would not be appropriate to use b0 = 15.84 as the point estimate of the mean weekly fuel consumption when average hourly temperature is 0. Because it is not meaningful to interpret the y-intercept in many regression situations, we often omit such interpretations.

We now present a general procedure for estimating a mean value and predicting an individual value:

Point Estimation and Point Prediction in Simple Linear Regression

Let b0 and b1 be the least squares point estimates of the y-intercept β0 and the slope β1 in the **simple linear regression model,** and suppose that x0, a specified value of the independent variable x, is inside the experimental region. Then

ŷ = b0 + b1X0

1 is the point estimate of the **mean value of the dependent variable** when the value of the independent variable is x0.

2 is the point prediction of an **individual value of the dependent variable** when the value of the independent variable is x0. Here we predict the error term to be 0.

EXAMPLE 13.4: The QHIC Case

Consider the simple linear regression model relating yearly home upkeep expenditure, y, to home value, x. Using the data in Table 13.2 (page 520), we can calculate the least squares point estimates of the y-intercept β0 and the slope β1 to be b0 = −348.3921 and b1 = 7.2583. Since b1 = 7.2583, we estimate that mean yearly upkeep expenditure increases by $7.26 for each additional $1,000 increase in home value. Consider a home worth $220,000, and note that x0 = 220 is in the range of previously observed values of x: 48.9 to 286.18 (see Table 13.2 on page 520). It follows that

is the point estimate of the mean yearly upkeep expenditure for all homes worth $220,000 and is the point prediction of a yearly upkeep expenditure for an individual home worth $220,000.

The marketing department at QHIC wishes to determine which homes should be sent advertising brochures promoting QHIC’s products and services. If the marketing department has decided to send an advertising brochure to any home that has a predicted yearly upkeep expenditure of at least $500, then a home worth $220,000 would be sent an advertising brochure. This is because the predicted yearly upkeep expenditure for such a home is (as calculated above) $1,248.43. Other homes can be evaluated in a similar fashion.

Exercises for Section 13.1

CONCEPTS

13.1 When does the scatter plot of the values of a dependent variable y versus the values of an independent variable x suggest that the simple linear regression model

y = β0 + β1x + ε

might appropriately relate y to x?

13.2 What is the least squares regression line, and what are the least squares point estimates?

13.3 How do we obtain a **point estimate of the mean value of the dependent variable** and a **point prediction of an individual value** of the dependent variable?

13.4 Why is it dangerous to extrapolate outside the experimental region?

METHODS AND APPLICATIONS

In Exercises 13.5 through 13.9 we present five data sets involving a dependent variable y and an independent variable x. For each data set, assume that the simple linear regression model

y = β0 + β1x + ε

relates y to x.

13.5 ** THE STARTING SALARY CASE** StartSal

The chairman of the marketing department at a large state university undertakes a study to relate starting salary (y) after graduation for marketing majors to grade point average (GPA) in major courses. To do this, records of seven recent marketing graduates are randomly selected, and the data shown below on the left are obtained. The MINITAB output obtained by fitting a least squares regression line to the data is below on the right.

a Find the least squares point estimates b0 and b1 on the computer output and report their values. Interpret b0 and b1. Does the interpretation of b0 make practical sense?

b Use the least squares line to compute a point estimate of the mean starting salary for allmarketing graduates having a grade point average of 3.25 and a point prediction of thestarting salary for an individual marketing graduate having a grade point average of 3.25.

13.6 ** THE SERVICE TIME CASE** SrvcTime

Accu-Copiers, Inc., sells and services the Accu-500 copying machine. As part of its standard service contract, the company agrees to perform routine service on this copier. To obtain information about the time it takes to perform routine service, Accu-Copiers has collected data for 11 service calls. The data and Excel output from fitting a least squares regression line to the data follow on the next page.

a Find the least squares point estimates b0 and b1 on the computer output and report their values. Interpret b0 and b1. Does the interpretation of b0 make practical sense?

b Use the least squares line to compute a point estimate of the mean time to service four copiers and a point prediction of the time to service four copiers on a single call.

13.7 ** THE FRESH DETERGENT CASE** Fresh

Enterprise Industries produces Fresh, a brand of liquid laundry detergent. In order to study the relationship between price and demand for the large bottle of Fresh, the company has gathered data concerning demand for Fresh over the last 30 sales periods (each sales period is four weeks). Here, for each sales period,

y = demand for the large bottle of Fresh (in hundreds of thousands of bottles) in the sales period, and

x = the difference between the average industry price (in dollars) of competitors’ similar detergents and the price (in dollars) of Fresh as offered by Enterprise Industries in the sales period.

The data and MINITAB output from fitting a least squares regression line to the data follow below.

a Find the least squares point estimates b0 and b1 on the computer output and report their values. Interpret b0 and b1. Does the interpretation of b0 make practical sense?

b Use the least squares line to compute a point estimate of the mean demand in all sales periods when the price difference is .10 and a point prediction of the actual demand in an individual sales period when the price difference is .10.

13.8 ** THE DIRECT LABOR COST CASE** DirLab

An accountant wishes to predict direct labor cost (y) on the basis of the batch size (x) of a product produced in a job shop. Data for 12 production runs are given in the table below, along with the Excel output from fitting a least squares regression line to the data.

a By using the formulas illustrated in Example 13.3 (see page 523) and the data provided, verify that (within rounding) b0 = 18.488 and b1 = 10.146, as shown on the Excel output.

b Interpret the meanings of b0 and b1. Does the interpretation of b0 make practical sense?

c Write the least squares prediction equation.

d Use the least squares line to obtain a point estimate of the mean direct labor cost for all batches of size 60 and a point prediction of the direct labor cost for an individual batch of size 60.

13.9 ** THE REAL ESTATE SALES PRICE CASE** RealEst

A real estate agency collects data concerning y = the sales price of a house (in thousands of dollars), and x = the home size (in hundreds of square feet). The data are given in the table below. The MINITAB output from fitting a least squares regression line to the data is on the next page.

a By using the formulas illustrated in Example 13.3 (see page 523) and the data provided, verify that (within rounding) b0 = 48.02 and b1 = 5.700, as shown on the MINITAB output.

b Interpret the meanings of b0 and b1. Does the interpretation of b0 make practical sense?

c Write the least squares prediction equation.

d Use the least squares line to obtain a point estimate of the mean sales price of all houses having 2,000 square feet and a point prediction of the sales price of an individual house having 2,000 square feet.

13.2: Model Assumptions and the Standard Error

Model assumptions

In order to perform hypothesis tests and set up various types of intervals when using the simple linear regression model

we need to make certain assumptions about the error term ε. At any given value of x, there is a population of error term values that could potentially occur. These error term values describe the different potential effects on y of all factors other than the value of x. Therefore, these error term values explain the variation in the y values that could be observed when the independent variable is x. Our statement of the simple linear regression model assumes that μy, the mean of the population of all y values that could be observed when the independent variable is x, is β0 + β1x. This model also implies that ε = y − (β0 + β1x), so this is equivalent to assuming that the mean of the corresponding population of potential error term values is 0. In total, we make four assumptions—called the **regression assumptions—about** the simple linear regression model. These assumptions can be stated in terms of potential y values or, equivalently, in terms of potential error term values. Following tradition, we begin by stating these assumptions in terms of potential error term values:

The Regression Assumptions

1 At any given value of x, the population of potential error term values has a **mean equal to 0**.

2 **Constant Variance Assumption**

At any given value of x, the population of potential error term values has a variance that does not depend on the value of x. That is, the different populations of potential error term values corresponding to different values of x have **equal variances**. We denote the **constant variance as σ2**.

3 **Normality Assumption**

At any given value of x, the population of potential error term values has a normal distribution.

4 **Independence Assumption**

Any one value of the error term ε is statistically independent of any other value of ε. That is, the value of the error term ε corresponding to an observed value of y is statistically independent of the value of the error term corresponding to any other observed value of y.

Taken together, the first three assumptions say that, at any given value of x, the population of potential error term values is **normally distributed** with **mean zero** and a **variance σ2 that does not depend on the value of x.** Because the potential error term values cause the variation in the potential y values, these assumptions imply that the population of all y values that could be observed when the independent variable is x is normally distributed with **mean β0 + β1 x
** and

Figure 13.10: An Illustration of the Model Assumptio

The **independence assumption**

is most likely to be violated when time series data are being utilized in a regression study. Intuitively, this assumption says that there is no pattern of positive error terms being followed (in time) by other positive error terms, and there is no pattern of positive error terms being followed by negative error terms. That is, there is no pattern of higher-than-average y values being followed by other higher-than-average y values, and there is no pattern of higher-than-average y values being followed by lower-than-average y values.

It is important to point out that the regression assumptions very seldom, if ever, hold exactly in any practical regression problem. However, it has been found that regression results are not extremely sensitive to mild departures from these assumptions. In practice, only pronounced departures from these assumptions require attention. In optional Section 13.8 we show how to check the regression assumptions. Prior to doing this, we will suppose that the assumptions are valid in our examples.

In Section 13.1 we stated that, when we predict an individual value of the dependent variable, we predict the error term to be 0. To see why we do this, note that the regression assumptions state that, at any given value of the independent variable, the population of all error term values that can potentially occur is normally distributed with a mean equal to 0. Since we also assume that successive error terms (observed over time) are statistically independent, each error term has a 50 percent chance of being positive and a 50 percent chance of being negative. Therefore, it is reasonable to predict any particular error term value to be 0.

The **mean square error** and the **standard error**

To present statistical inference formulas in later sections, we need to be able to compute point estimates of σ2 and σ, the constant variance and standard deviation of the error term populations. The point estimate of σ2 is called the mean square error and the point estimate of σ is called the **standard error.** In the following box, we show how to compute these estimates:

The Mean Square Error and the Standard Error

If the regression assumptions are satisfied and SSE is the sum of squared residuals:

1 The point estimate of σ is the mean square error

2 The point estimate of σ2 is the standard error

In order to understand these point estimates, recall that σ2 is the variance of the population of y values (for a given value of x) around the mean value μy. Because ŷ is the point estimate of this mean, it seems natural to use

SSE = Σ (yi − ŷi)2

to help construct a point estimate of σ2. We divide SSE by n − 2 because it can be proven that doing so makes the resulting s2 an unbiased point estimate of σ2. Here we call n − 2 the **number of degrees of freedom** associated with SSE.

EXAMPLE 13.5: The Fuel Consumption Case

Consider the fuel consumption situation, and recall that in Table 13.3 (page 524) we have calculated the sum of squared residuals to be SSE = 2.568. It follows, because we have observed n = 8 fuel consumptions, that the point estimate of σ2 is the mean square error

This implies that the point estimate of σ is the standard error

As another example, it can be verified that the standard error for the simple linear regression model describing the QHIC data is s = 146.8970.

To conclude this section, note that in optional Section 13.9 we present a shortcut formula for calculating SSE. The reader may study Section 13.9 now or at any later point.

Exercises for Section 13.2

CONCEPTS

13.10 What four assumptions do we make about the simple linear regression model?

**13.11** What is estimated by the mean square error, and what is estimated by the standard error?

METHODS AND APPLICATIONS

**13.12** THE STARTING SALARY CASE StartSal

When a least squares line is fit to the 7 observations in the starting salary data, we obtain SSE = 1.438. Calculate s2 and s.

**13.13** THE SERVICE TIME CASE SrvcTime

When a least squares line is fit to the 11 observations in the service time data, we obtain SSE = 191.7017. Calculate s2 and s.

**13.14** THE FRESH DETERGENT CASE Fresh

When a least squares line is fit to the 30 observations in the Fresh detergent data, we obtain SSE = 2.806. Calculate s2 and s.

**13.15** THE DIRECT LABOR COST CASE DirLab

When a least squares line is fit to the 12 observations in the labor cost data, we obtain SSE = 746.7624. Calculate s2 and s.

**13.16** THE REAL ESTATE SALES PRICE CASE Real Est

When a least squares line is fit to the 10 observations in the real estate sales price data, we obtain SSE = 896.8. Calculate s2 and s.

**13.17** Ten sales regions of equal sales potential for a company were randomly selected. The advertising expenditures (in units of $10,000) in these 10 sales regions were purposely set during July of last year at, respectively, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14. The sales volumes (in units of $10,000) were then recorded for the 10 sales regions and found to be, respectively, 89, 87, 98, 110, 103, 114, 116, 110, 126, and 130. Assuming that the simple linear regression model is appropriate, it can be shown that b0 = 66.2121, b1 = 4.4303, and SSE = 222.8242. Calculate s2 and s. SalesVol

13.3: Testing the Significance of the Slope and y-Intercept

Testing the significance of the slope

A simple linear regression model is not likely to be useful unless there is a **significant relationship between y and x.** In order to judge the significance of the relationship between y and x, we test the null hypothesis

**
H0**: β1 = 0

which says that there is no change in the mean value of y associated with an increase in x, versus the alternative hypothesis

Ha: β1 ≠ 0

which says that there is a (positive or negative) change in the mean value of y associated with an increase in x. It would be reasonable to conclude that x is significantly related to y if we can be quite certain that we should reject H0 in favor of Ha.

In order to test these hypotheses, recall that we compute the least squares point estimate b1 of the true slope β1 by using a sample of n observed values of the dependent variable y. Different samples of n observed y values would yield different values of the least squares point estimate b1. It can be shown that, if the regression assumptions hold, then the population of all possible values of b1 is normally distributed with a mean of β1 and with a standard deviation of

The standard error s is the point estimate of σ, so it follows that a point estimate of σ b 1 is

which is called the **standard error of the estimate b1.** Furthermore, if the regression assumptions hold, then the population of all values of

has a t distribution with n−2 degrees of freedom. It follows that, if the null hypothesis H0: β1 = 0 is true, then the population of all possible values of the test statistic

has a t distribution with n − 2 degrees of freedom. Therefore, we can test the significance of the regression relationship as follows:

Testing the Significance of the Regression Relationship: Testing the Significance of the Slope

Define the test statistic

and suppose that the regression assumptions hold. Then we can test H0: β1 − 0 versus a particular alternative hypothesis at significance level α (that is, by setting the probability of a Type I error equal to α) by using the appropriate critical value rule, or, equivalently, the corresponding p-value.

Here tα/2, tα, and all p-values are based on n − 2 degrees of freedom. **If we can reject H0: β1 = 0 at a given value of α, then we conclude that the slope (or, equivalently, the regression relationship) is significant at the α level.**

We usually use the two-sided alternative Ha: β1 ≠ 0 for this test of significance. However, sometimes a one-sided alternative is appropriate. For example, in the fuel consumption problem we can say that if the slope β1 is not 0, then it must be negative. A negative β1 would say that mean fuel consumption decreases as temperature x increases. Because of this, it would be appropriate to decide that x is significantly related to y if we can reject H0: β1 = 0 in favor of the one-sided alternative Ha: β1 < 0. Although this test would be slightly more effective than the usual two-sided test, there is little practical difference between using the one-sided or two-sided alternative. Furthermore, computer packages (such as MINITAB and Excel) present results for testing a two-sided alternative hypothesis. For these reasons we will emphasize the two-sided test.

It should also be noted that

1 **If we can decide that the slope is significant at the.05 significance level,** then we have concluded that x is significantly related to y by using a test that allows only a.05 probability of concluding that x is significantly related to y when it is not. **This is usually regarded as strong evidence that the regression relationship is significant.**

2 **If we can decide that the slope is significant at the.01 significance level, this is usually regarded as very strong evidence that the regression relationship is significant.**

3 The smaller the significance level α at which H0 can be rejected, the stronger is the evidence that the regression relationship is significant.

EXAMPLE 13.6: The Fuel Consumption Case

Again consider the fuel consumption model

y = β0 + β1x + ε

For this model SSxx = 1,404.355, b1 = −.1279, and s = .6542 [see Examples 13.3 (page 523) and 13.5 (page 532)]. Therefore

and

To test the significance of the slope we compare | t | with tα/2 based on n − 2 = 8 − 2 = 6 degrees of freedom. Because

| t | = 7.33 > t.025 = 2.447

we can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 and conclude that the slope (regression relationship) is significant at the .05 level.

The p-value for testing H0 versus Ha is twice the area to the right of | t | = 7.33 under the curve of the t distribution having n − 2 = 6 degrees of freedom. Since this p-value can be shown to be .0003, we can reject H0 in favor of Ha at level of significance .05, .01, or .001. We therefore have extremely strong evidence that x is significantly related to y and that the regression relationship is significant.

Figure 13.11 presents the MINITAB and Excel outputs of a simple linear regression analysis of the fuel consumption data. Note that b0 (labeled as on the outputs), b1 (labeled ), s (labeled ), (labeled ), and t (labeled ) are given on each of these outputs. Also note that each output gives the p-value related to t = −7.33 (labeled ). Excel tells us that this p-value equals 0.0003, while MINITAB has rounded this p-value to 0.000 (which means less than .001). Other quantities on the MINITAB and Excel outputs will be discussed later.

Figure 13.11: MINITAB and Excel Output of a Simple Linear Regression Analysis of the Fuel Consumption Data

In addition to testing the significance of the slope, it is often useful to calculate a **confidence interval** for β1. We show how this is done in the following box:

A Confidence Interval for the Slope

If the regression assumptions hold,

**a 100(1 − α) percent confidence interval for the true slope β1** is [b1 ± tα / 2s b 1 ]. Here tα/2 is based on n − 2 degrees of freedom.

EXAMPLE 13.7: The Fuel Consumption Case

The MINITAB output in Figure 13.11(a) tells us that b1 = −.12792 and s b 1 = .01746. Thus, for instance, because t.025 based on n − 2 = 8 − 2 = 6 degrees of freedom equals 2.447, a 95 percent confidence interval for β1 is

This interval says we are 95 percent confident that, if average hourly temperature increases by one degree, then mean weekly fuel consumption will decrease (because both the lower bound and the upper bound of the interval are negative) by at least .0852 MMcf of natural gas and by at most .1706 MMcf of natural gas. Also, because the 95 percent confidence interval for β1 does not contain 0, we can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 at level of significance .05. Note that the 95 percent confidence interval for β1 is given on the Excel output but not on the MINITAB output (see Figure 13.11).

EXAMPLE 13.8: The QHIC Case

Figure 13.12 presents the MegaStat output of a simple linear regression analysis of the QHIC data. We summarize some important quantities from the output as follows (we discuss the other quantities later): b0 = −348.3921, b1 = 7.2583, s = 146.897, s b 1 = .4156, and t = b1 / s b 1 = 17.466. Since the p-value related to t = 17.466 is less than.001 (see the MegaStat output), we can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 at the .001 level of significance. It follows that we have extremely strong evidence that the regression relationship is significant. The MegaStat output also tells us that a 95 percent confidence interval for the true slope β1 is [6.4170, 8.0995]. This interval says we are 95 percent confident that mean yearly upkeep expenditure increases by between $6.42 and $8.10 for each additional $1,000 increase in home value.

Figure 13.12: MegaStat Output of a Simple Linear Regression Analysis of the QHIC Data

Testing the significance of the y-intercept

We can also test the significance of the y-intercept β0. We do this by testing the null hypothesis H0: β0 = 0 versus the alternative hypothesis Ha: β0 ≠ 0. **If we can reject H0 in favor of Ha by setting the probability of a Type I error equal to α, we conclude that the intercept β0 is significant at the α level.** To carry out the hypothesis test, we use the test statistic

Here the critical value and p-value conditions for rejecting H0 are the same as those given previously for testing the significance of the slope, except that t is calculated as b0s b 0. For example, if we consider the fuel consumption problem and the MINITAB output in Figure 13.11, we see that b0 = 15.8379, s b 0 = .8018, t = 19.75, and p-value = .000. Because t = 19.75 > t.025 = 2.447 and p-value < .05, we can reject H0: β0 = 0 in favor of Ha: β0 ≠ 0 at the .05 level of significance. In fact, since the p-value < .001, we can also reject H0 at the .001 level of significance. This provides extremely strong evidence that the y-intercept β0 does not equal 0 and thus is significant. Therefore, we should include β0 in the fuel consumption model.

In general, if we fail to conclude that the intercept is significant at a level of significance of .05, it might be reasonable to drop the y-intercept from the model. However, remember that β0 equals the mean value of y when x equals 0. If, logically speaking, the mean value of y would not equal 0 when x equals 0 (for example, in the fuel consumption problem, mean fuel consumption would not equal 0 when the average hourly temperature is 0), it is common practice to include the y-intercept whether or not H0: β0 = 0 is rejected. In fact, experience suggests that it is definitely safest, when in doubt, to include the intercept β0.

Exercises for Section 13.3

CONCEPTS

**13.18** What do we conclude if we can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 by setting

a α equal to .05?

b α equal to .01?

**13.19** Give an example of a practical application of the confidence interval for β1.

METHODS AND APPLICATIONS

In Exercises **13.20** through **13.24**, we refer to MINITAB, MegaStat, and Excel output of simple linear regression analyses of the data sets related to the five case studies introduced in the exercises for Section 13.1. Using the appropriate output for each case study,

a Find the least squares point estimates b0 and b1 of β0 and β1 on the output and report their values.

b Find SSE and s on the computer output and report their values.

c Find s b 1 and the t statistic for testing the significance of the slope on the output and report their values. Show how t has been calculated by using b1 and s b 1 from the computer output.

d Using the t statistic and appropriate critical value, test H0: β1 = 0 versus Ha: β1 ≠ 0 by setting α equal to .05. Is the slope (regression relationship) significant at the .05 level?

e Using the t statistic and appropriate critical value, test H0: β1 = 0 versus Ha: β1 ≠ 0 by setting α equal to .01. Is the slope (regression relationship) significant at the .01 level?

f Find the p-value for testing H0: β1 = 0 versus Ha: β1 ≠ 0 on the output and report its value. Using the p-value, determine whether we can reject H0 by setting a equal to .10, .05, .01, and .001. How much evidence is there that the slope (regression relationship) is significant?

g Calculate the 95 percent confidence interval for β1 using numbers on the output. Interpret the interval.

h Calculate the 99 percent confidence interval for β1 using numbers on the output.

i Find s b 0 and the t statistic for testing the significance of the y intercept on the output and report their values. Show how t has been calculated by using b0 and s b 0 from the computer output.

j Find the p-value for testing H0: β0 = 0 versus Ha: β0 ≠ 0. Using the p-value, determine whether we can reject H0 by setting a equal to .10, .05, .01, and .001. What do you conclude about the significance of the y intercept?

k Using the appropriate data set and s from the computer output, hand calculate SSxx, s b 0 and s b 1.

13.20 THE STARTING SALARY CASE StartSal

The MINITAB output of a simple linear regression analysis of the data set for this case (see Exercise 13.5 on page 527) is given in Figure 13.13. Recall that a labeled MINITAB regression output is on page 535.

Figure 13.13: MINITAB Output of a Simple Linear Regression Analysis of the Starting Salary Data

**13.21** THE SERVICE TIME CASE SrvcTime

The MegaStat output of a simple linear regression analysis of the data set for this case (see Exercise 13.6 on pages 527 and 528) is given in Figure 13.14. Recall that a labeled MegaStat regression output is on page 537.

Figure 13.14: MegaStat Output of a Simple Linear Regression Analysis of the Service Time Data

**13.22** THE FRESH DETERGENT CASE Fresh

The MINITAB output of a simple linear regression analysis of the data set for this case (see Exercise 13.7 on page 528) is given in Figure 13.15. Recall that a labeled MINITAB regression output is on page 535.

Figure 13.15: MINITAB Output of a Simple Linear Regression Analysis of the Fresh Detergent Demand Data

**13.23** THE DIRECT LABOR COST CASE DirLab

The Excel and MegaStat output of a simple linear regression analysis of the data set for this case (see Exercise 13.8 on page 529) is given in Figure 13.16. Recall that labeled Excel and MegaStat regression outputs are on pages 535 and 537.

Figure 13.16: Excel and MegaStat Output of a Simple Linear Regression Analysis of the Direct Labor Cost Data

13.24 THE REAL ESTATE SALES PRICE CASE RealEst

The MINITAB output of a simple linear regression analysis of the data set for this case (see Exercise 13.9 on page 529) is given in Figure 13.17 on page 540. Recall that a labeled MINITAB regression output is on page 535.

Figure 13.17: MINITAB Output of a Simple Linear Regression Analysis of the Real Estate Sales Price Data

**13.25** Find and interpret a 95 percent confidence interval for the slope β1 of the simple linear regression model describing the sales volume data in Exercise 13.17 (page 533). SalesVol

**13.26** **THE FAST-FOOD RESTAURANT RATING CASE** FastFood

In the early 1990s researchers at The Ohio State University studied consumer ratings of six fast-food restaurants: Borden Burger, Hardee’s, Burger King, McDonald’s, Wendy’s, and White Castle. Each of 406 randomly selected individuals gave each restaurant a rating of 1, 2, 3, 4, 5, or 6 on the basis of taste, and then ranked the restaurants from 1 through 6 on the basis of **overall** preference. In each case, 1 is the best rating and 6 the worst. The mean ratings given by the 406 individuals are given in the following table:

Figure 13.18 gives the Excel output of a simple linear regression analysis of this data. Here, mean preference is the dependent variable and mean taste is the independent variable. Recall that a labeled Excel regression output is given on page 535.

Figure 13.18: Excel Output of a Simple Linear Regression Analysis of the Fast-Food Restaurant Rating Data

a Find the least squares point estimate b1 of β1 on the computer output. Report and interpret this estimate.

b Find the 95 percent confidence interval for β1 on the output. Report and interpret the interval.

13.4: Confidence and Prediction Intervals

We have seen that

ŷ = b0 + b1x0

**is the point estimate of the mean value of y**

when the value of the independent variable x is x0. We have also seen that ŷ is the point prediction of an individual value of y

when the value of the independent variable x is x0. In this section we will assess the accuracy of ŷ as both a point estimate and a point prediction. To do this, we will find a confidence interval for the mean value of y

and a prediction interval for an individual value of y.

Because each possible sample of n values of the dependent variable gives values of b0 and b1 that differ from the values given by other samples, different samples give different values of ŷ = b0 + b1x0. A confidence interval for the mean value of y is based on the estimated standard deviation of the population of all possible values of ŷ. This estimated standard deviation is called the **standard error of ŷ
** and is denoted sŷ If the regression assumptions hold, the formula for is sŷ

Here, s is the standard error (see Section 13.2), is the average of the n previously observed values of x, and .

As explained above, a confidence interval for the mean value of y is based on the standard error sŷ. A prediction interval for an individual value of y is based on a more complex standard error: the estimated standard deviation of the population of all possible values of y − ŷ the prediction error obtained when predicting y by ŷ. We refer to this estimated standard deviation as the **standard error of y − ŷ
** and denote it as s(y − ŷ). If the regression assumptions hold, the formula for s(y − ŷ) is

Intuitively, the “extra 1” under the radical in the formula for s(y − ŷ) accounts for the fact that there is more uncertainty in predicting an individual value y = β0 + β1×0 + ε than in estimating the mean value β0 + β1×0 (because we must predict the error term ε when predicting an individual value). Therefore, as shown in the following summary box, the prediction interval for an individual value of y is longer than the confidence interval for the mean value of y.

A Confidence Interval and a Prediction Interval

If the regression assumptions hold,

1 A **100(1 − α) percent confidence interval for the mean value of y
** when x equals x0 is

2 A **100(1 − α) percent prediction interval for an individual value of y
** when x equals x0 is

Here, tα/2 is based on (n − 2) degrees of freedom.

The summary box tells us that both the formula for the confidence interval and the formula for the prediction interval use the quantity We will call this quantity the **
distance value,** because it is a measure of the distance between x0 the value of x for which we will make a point estimate or a point prediction, and , the average of the previously observed values of x. The farther that x0 is from which represents the center of the experimental region, the larger is the distance value, and thus the longer are both the confidence interval and the prediction interval . Said another way, when x0 is farther from the center of the data, ŷ = b0 + b1x0 is likely to be less accurate as both a point estimate and a point prediction.

EXAMPLE 13.9: The Fuel Consumption Case

In the fuel consumption problem, recall that a weather forecasting service has predicted that in the next week the average hourly temperature will be x0 = 40°F. Also, recall that

is the point estimate of the mean fuel consumption for all weeks that have an average hourly temperature of 40°F and is the point prediction of the fuel consumption in a single week that has an average hourly temperature of 40°F. Using the information in Example 13.3 (page 523), we compute

Since s = .6542 (see Example 13.5 on page 532) and since tα/2 = t.025 based on n − 2 = 8 − 2 = 6 degrees of freedom equals 2.447, it follows that a 95 percent confidence interval for the mean fuel consumption when x equals 40°F is

This interval says we are 95 percent confident that the mean fuel consumption for all weeks that have an average hourly temperature of 40°F is between 10.13 MMcf of natural gas and 11.31 MMcf of natural gas.

If the natural gas company bases its transmission nomination for next week on the point prediction ŷ = 10.72, it will order 10.72 MMcf of natural gas. To evaluate the accuracy of the point prediction ŷ = 10.72, we can calculate the following 95 percent prediction interval for the fuel consumption in a single week that has an average hourly temperature of 40°F:

This interval says that we are 95 percent confident that the fuel consumption in a single week that has an average hourly temperature of 40°F will be between 9.01 MMcf of natural gas and 12.43 MMcf of natural gas. Furthermore, note that the half-length of the 95 percent prediction interval is (12.43 − 9.01)/2 = 1.71, which is (1.71/10.72)100% = 15.91% of the transmission nomination 10.72. It follows that we are 95 percent confident that the actual amount of natural gas that will be used by the city next week will differ from the natural gas company’s transmission nomination by no more than 15.91 percent. That is, we are 95 percent confident that the natural gas company’s percentage nomination error will be less than or equal to 15.91 percent. Although this does not imply that the natural gas company is likely to make a terribly inaccurate nomination, we are not confident that the company’s percentage nomination error will be within the 10 percent allowance granted by the pipeline transmission system. Therefore, the natural gas company may be assessed a transmission fine. In Chapter 14 we use a multiple regression model to substantially reduce the natural gas company’s percentage nomination errors.

Below we repeat the bottom of the MINITAB output in Figure 13.11(a) on page 535. This output gives the point estimate and prediction ŷ = 10.72 the 95 percent confidence interval for the mean value of y when x equals 40, and the 95 percent prediction interval for an individual value of y when x equals 40.

Predicted Values for New Observations

Although the MINITAB output does not directly give the distance value, it does give under the heading “SE Fit.” A little algebra shows that this implies that the distance value equals (sŷ/s)2. Specifically, because sŷ = .241 and s = .6542, the distance value equals (.241/.6542)2 = .1357. This distance value is (within rounding) equal to the distance value that we hand calculated earlier.

To conclude this example, note that Figure 13.19 illustrates the MINITAB output of the 95 percent confidence and prediction intervals corresponding to all values of x in the experimental region. Here = 43.98 can be regarded as the center of the experimental region. Notice that the farther x0 is from = 43.98, the larger is the distance value and, therefore, the longer are the 95 percent confidence and prediction intervals. These longer intervals are undesirable because they give us less information about mean and individual values of y.

Figure 13.19: MINITAB Output of 95% Confidence and Prediction Intervals for the Fuel Consumption Case

In general, the prediction interval is useful if, as in the fuel consumption problem, it is important to predict an individual value of the dependent variable. A confidence interval is useful if it is important to estimate the mean value. Although it is not important to estimate a mean value in the fuel consumption problem, it is important to estimate a mean value in other situations. To understand this, recall that the mean value is the average of all the values of the dependent variable that could potentially be observed when the independent variable equals a particular value. Therefore, it might be important to estimate the mean value if we will observe and are affected by a very large number of values of the dependent variable when the independent variable equals a particular value. We illustrate this in the following example.

EXAMPLE 13.10: The QHIC Case

Consider a home worth $220,000. We have seen that the predicted yearly upkeep expenditure for such a home is

This predicted value is given at the bottom of the MegaStat output in Figure 13.12, which we repeat here:

Predicted values for: Upkeep

In addition to giving ŷ = 1,248.43 the MegaStat output also tells us that the distance value, which is given under the heading “Leverage” on the output, equals .042. Therefore, since s equals 146.897 (see Figure 13.12 on page 537), it follows that a 95 percent prediction interval for the yearly upkeep expenditure of an individual home worth $220,000 is calculated as follows:

Here t.025 is based on n − 2 = 40 − 2 = 38 degrees of freedom. Note that this interval is given on the MegaStat output.

Because there are many homes worth roughly $220,000 in the metropolitan area, QHIC is more interested in the mean upkeep expenditure for all such homes than in the individual upkeep expenditure for one such home. The MegaStat output tells us that a 95 percent confidence interval for this mean upkeep expenditure is [1187.79, 1309.06]. This interval says that QHIC is 95 percent confident that the mean upkeep expenditure for all homes worth $220,000 is at least $1,187.79 and is no more than $1,309.06. Furthermore, suppose that a special, more expensive advertising brochure will be sent only to homes that QHIC is very sure have a mean yearly upkeep expenditure that exceeds $1,000. Because the lower end of the 95 percent confidence interval for $220,000 homes is above $1,000, the special brochure will be sent to homes worth $220,000.

Exercises for Section 13.4

CONCEPTS

**13.27** What is the difference between a confidence interval and a prediction interval?

**13.28** What does the distance value measure? How does the distance value affect a confidence or prediction interval?

METHODS AND APPLICATIONS

**13.29** THE STARTING SALARY CASE StartSal

The following partial MINITAB regression output for the starting salary data relates to predicting the starting salary of a marketing graduate having a grade point average of 3.25.

Predicted Values for New Observations

a Report (as shown on the computer output) a point estimate of and a 95 percent confidence interval for the mean starting salary of all marketing graduates having a grade point average of 3.25.

b Report (as shown on the computer output) a point prediction of and a 95 percent prediction interval for the starting salary of an individual marketing graduate having a grade point average of 3.25.

c Remembering that s = .536321 and that the distance value equals (sŷ/s)2, use sŷ from the computer output to hand calculate the distance value when x = 3.25.

d Remembering that for the starting salary data n = 7, b0 = 14.816 and b1 = 5.7066 hand calculate (within rounding) the confidence interval of part (a) and the prediction interval of part (b).

**13.30** THE SERVICE TIME CASE SrvcTime

The following partial MegaStat regression output for the service time data relates to predicting service times for 1, 2, 3, 4, 5, 6, and 7 copiers.

Predicted values for: Minutes (y)

a Report (as shown on the computer output) a point estimate of and a 95 percent confidence interval for the mean time to service four copiers.

b Report (as shown on the computer output) a point prediction of and a 95 percent prediction interval for the time to service four copiers on a single call.

c For this case: n = 11, b0 = 11.4641, b1 = 24.6022, and s = 4.615. Using this information and a distance value from the MegaStat output, hand calculate (within rounding) the confidence interval of part (a) and the prediction interval of part (b).

d If we examine the service time data, we see that there was at least one call on which Accu-Copiers serviced each of 1, 2, 3, 4, 5, 6, and 7 copiers. The 95 percent confidence intervals for the mean service times on these calls might be used to schedule future service calls. To understand this, note that a person making service calls will (in, say, a year or more) make a very large number of service calls. Some of the person’s individual service times will be below, and some will be above, the corresponding mean service times. However, since the very large number of individual service times will average out to the mean service times, it seems fair to both the efficiency of the company and to the person making service calls to schedule service calls by using estimates of the mean service times. Therefore, suppose we wish to schedule a call to service five copiers. Examining the MegaStat output, we see that a 95 percent confidence interval for the mean time to service five copiers is [130.753, 138.197]. Since the mean time might be 138.197 minutes, it would seem fair to allow 138 minutes to make the service call. Now suppose we wish to schedule a call to service four copiers. Determine how many minutes to allow for the service call.

**13.31** THE FRESH DETERGENT CASE Fresh

The following partial MINITAB regression output for the Fresh detergent data relates to predicting demand for future sales periods in which the price difference will be .10 (see New Obs 1) and .25 (see New Obs2).

Predicted Values for New Observations

a Report (as shown on the computer output) a point estimate of and a 95 percent confidence interval for the mean demand for Fresh in all sales periods when the price difference is .10.

b Report (as shown on the computer output) a point prediction of and a 95 percent prediction interval for the actual demand for Fresh in an individual sales period when the price difference is .10.

c Remembering that s = .316561 and that the distance value equals (sŷ/s)2, use sŷ from the computer output to hand calculate the distance value when x = .10.

d For this case: n = 12, b0 = 18.4875, b1 = 10.1463, and s = 8.6415. and Using this information, and your result from part (c), find 99 percent confidence and prediction intervals for mean and individual demands when x = .10.

e Repeat parts (a), (b), (c), and (d) when x = .25.

**13.32** THE DIRECT LABOR COST CASE DirLab

The following partial MegaStat regression output for the direct labor cost data relates to predicting direct labor cost when the batch size is 60.

Predicted values for: LaborCost (y)

a Report (as shown on the MegaStat output) a point estimate of and a 95 percent confidence interval for the mean direct labor cost of all batches of size 60.

b Report (as shown on the MegaStat output) a point prediction of and a 95 percent prediction interval for the actual direct labor cost of an individual batch of size 60.

c For this case: n = 12, b0 = 18.4875, b1 = 10.1463, and s = 8.6415. Use this information and the distance value from the MegaStat output to compute 99 percent confidence and prediction intervals for the mean and individual labor costs when x = 60.

**13.33** THE REAL ESTATE SALES PRICE CASE RealEst

The following partial MINITAB regression output for the real estate sales price data relates to predicting the sales price of a home having 2,000 square feet.

Predicted Values for New Observations

a Report (as shown on the MINITAB output) a point estimate of and a 95 percent confidence interval for the mean sales price of all houses having 2,000 square feet.

b Report (as shown on the MINITAB output) a point prediction of and a 95 percent prediction interval for the sales price of an individual house having 2,000 square feet.

c If you were purchasing a home having 2,000 square feet, which of the above intervals would you find to be most useful? Explain.

13.5: Simple Coefficients of Determination and Correlation

The __simple coefficient of determination__

**The simple coefficient of determination is** a measure of the usefulness of a simple linear regression model. To introduce this quantity, which is denoted **
r2** (pronounced

r

Figure 13.20: The Reduction in the Prediction Errors Accomplished by Employing the Predictor Variable X

Next, suppose we decide to employ the predictor variable x and observe the values x1, x2, …, xn corresponding to the observed values of y. In this case the prediction of yi is

ŷi = b0 + b1xi

and the error of prediction is yi − ŷi. For example, Figure 13.20(b) illustrates the prediction errors obtained in the fuel consumption problem when we use the predictor variable x. Together, Figures 13.20(a) and (b) show the reduction in the prediction errors accomplished by employing the predictor variable x (and the least squares line).

Using the predictor variable x decreases the prediction error in predicting yi from (yi − ) to (yi − ŷi), or by an amount equal to

(yi − ) − (yi − ŷi) = (ŷi − )

It can be shown that in general

The sum of squared prediction errors obtained when we do not employ the predictor variable x, Σ (yi − )2, is called the **total variation.** Intuitively, this quantity measures the total amount of variation exhibited by the observed values of y. The sum of squared prediction errors obtained when we use the predictor variable x, Σ (yi − ŷi)2, is called the **un explained variation (this is another name for **

SSE

). Intuitively, this quantity measures the amount of variation in the values of y that is not explained by the predictor variable. The quantity Σ (ŷi − )2 is called the explained variation. Using these definitions and the above equation involving these summations, we see that

**Total variation** − **Unexplained variation** = **Explained variation**

It follows that the explained variation is the reduction in the sum of squared prediction errors that has been accomplished by using the predictor variable x to predict y. It also follows that

**Total variation = Explained variation** + Unexplained variation

Intuitively, this equation implies that the explained variation represents the amount of the total variation in the observed values of y that is explained by the predictor variable x (and the simple linear regression model).

We now define the **
simple coefficient of determination
** to be

That is, r2 is the proportion of the total variation in the n observed values of y that is explained by the simple linear regression model. Neither the explained variation nor the total variation can be negative (both quantities are sums of squares). Therefore, r2 is greater than or equal to 0. Because the explained variation must be less than or equal to the **total variation,** r2 cannot be greater than 1. The nearer r2 is to 1, the larger is the proportion of the total variation that is explained by the model, and the greater is the utility of the model in predicting y. If the value of r2 is not reasonably close to 1, the independent variable in the model does not provide accurate predictions of y. In such a case, a different predictor variable must be found in order to accurately predict y. It is also possible that no regression model employing a single predictor variable will accurately predict y. In this case the model must be improved by including more than one independent variable. We see how to do this in Chapter 14.

In the following box we summarize the results of this section:

**The Simple Coefficient of Determination, r2**

For the simple linear regression model

1 Total variation = Σ(yi − )2

2 Explained variation = Σ(ŷi − )2

3 Unexplained variation = Σ(yi − ŷi)2

4 Total variation = Explained variation + Unexplained variation

5 The simple coefficient of determination is

6 r2 is the proportion of the total variation in the n observed values of the dependent variable that is explained by the simple linear regression model.

EXAMPLE 13.11: The Fuel Consumption Case

For the fuel consumption data (see Table 13.1 on page 517) we have seen that = (12.4 + 11.7 + … + 7.5)/8 = 81.7/8 = 10.2125. It follows that the total variation is

Furthermore, we found in Table 13.3 (page 524) that the unexplained variation is SSE = 2.568. Therefore, we can compute the explained variation and r2 as follows:

This value of r2 says that the regression model explains 89.9 percent of the total variation in the eight observed fuel consumptions.

EXAMPLE 13.12: The QHIC Case

In the QHIC case, it can be shown that: Total variation = 7,402,755.2399; Explained variation = 6,582,759.6972; SSE = Unexplained variation = 819,995.5427; and

This value of r2 says that the simple linear regression model that employs home value as a predictor variable explains 88.9 percent of the total variation in the 40 observed home upkeep expenditures.

In optional Section 13.9 we present some shortcut formulas for calculating the total, explained, and unexplained variations. Finally, for those who have already read Section 13.3, r2, the explained variation, the unexplained variation, and the total variation are calculated by MINITAB, Excel, and MegaStat. These quantities are identified on the MINITAB and Excel outputs of Figure 13.11 (page 535) and on the MegaStat output of Figure 13.12 (page 537) by, respectively, the labels ,,, and. These outputs also give an “adjusted r2.” We explain the meaning of this quantity in Chapter 14.

The __simple correlation coefficient__, r

People often claim that two variables are correlated. For example, a college admissions officer might feel that the academic performance of college students (measured by grade point average) is correlated with the students’ scores on a standardized college entrance examination. This means that college students’ grade point averages are related to their college entrance exam scores. One measure of the relationship between two variables y and x is the **
simple correlation coefficient.** We define this quantity as follows:

The Simple Correlation Coefficient

The **simple correlation coefficient between**

y

and **
x
**, denoted by

r

, is

where b1 is the slope of the least squares line relating y to x. This correlation coefficient **measures the strength of the linear relationship between**

y

and

x

.

Because r2 is always between 0 and 1, the correlation coefficient r is between −1 and 1. A value of r near 0 implies little linear relationship between y and x. A value of r close to 1 says that y and x have a strong tendency to move together in a straight-line fashion with a positive slope and, therefore, that y and x are highly related and **positively correlated.** A value of r close to −1 says that y and x have a strong tendency to move together in a straight-line fashion with a negative slope and, therefore, that y and x are highly related and **negatively correlated.** Figure 13.21 illustrates these relationships. Notice that when r = 1, y and x have a perfect linear relationship with a positive slope, whereas when r = −1, y and x have a perfect linear relationship with a negative slope.

Figure 13.21: An Illustration of Different Values of the Simple Correlation Coefficient

EXAMPLE 13.13: The Fuel Consumption Case

In the fuel consumption problem we have previously found that b1 = −.1279 and r2 = .899. It follows that the simple correlation coefficient between y (weekly fuel consumption) and x (average hourly temperature) is

This simple correlation coefficient says that x and y have a strong tendency to move together in a linear fashion with a negative slope. We have seen this tendency in Figure 13.2 (page 517), which indicates that y and x are negatively correlated.

If we have computed the least squares slope b1 and r2, the method given in the previous box provides the easiest way to calculate r. The simple correlation coefficient can also be calculated using the formula

Here SSxy and SSxx have been defined in Section 13.1 on page 522, and SSyy denotes the total variation, which has been defined in this section. Furthermore, this formula for r automatically gives r the correct (+ or −) sign. For instance, in the fuel consumption problem, SSxy = −179.6475, SSxx = 1,404.355, and SSyy = 25.549 (see Examples 13.3 on page 523 and 13.11 on page 549). Therefore

It is important to make two points. First, **the value of the simple correlation coefficient is not the slope of the least squares line.** If we wish to find this slope, we should use the previously given formula for b1.4 Second, **high correlation does not imply that a cause-and-effect relationship exists.** When r indicates that y and x are highly correlated, this says that y and x have a strong tendency to move together in a straight-line fashion. The correlation does not mean that changes in x cause changes in y. Instead, some other variable (or variables) could be causing the apparent relationship between y and x. For example, suppose that college students’ grade point averages and college entrance exam scores are highly positively correlated. This does not mean that earning a high score on a college entrance exam causes students to receive a high grade point average. Rather, other factors such as intellectual ability, study habits, and attitude probably determine botha student’s score on a college entrance exam and a student’s college grade point average. In general, while the simple correlation coefficient can show that variables tend to move together in a straight-line fashion, scientific theory must be used to establish cause-and-effect relationships.

Exercises for Section 13.5

CONCEPTS

**13.34** Discuss the meanings of the total variation, the unexplained variation, and the explained variation.

**13.35** What does the simple coefficient of determination measure?

METHODS AND APPLICATIONS

In Exercises **13.36** through **13.40**, we give the total variation, the unexplained variation (SSE), and the least squares point estimate b1 that are obtained when simple linear regression is used to analyze the data set related to each of five previously discussed case studies. Using the information given in each exercise, find the explained variation, the simple coefficient of determination (r2), and the simple correlation coefficient (r). Interpret r2.

13.36 THE STARTING SALARY CASE StartSal

Total variation = 61.380; SSE = 1.438; b1 = 5.7066

**13.37** THE SERVICE TIME CASE SrvcTime

Total variation = 20,110.5455; SSE = 191.7017; b1 = 24.6022

**13.38** THE FRESH DETERGENT CASE Fresh

Total variation = **13.45**9; SSE = 2.806; b1 = 2.6652

**13.39** THE DIRECT LABOR COST CASE DirLab

Total variation = 1,025,339.6667; SSE = 746.7624; b1 = 10.1463

13.40 THE REAL ESTATE SALES PRICE CASE RealEst

Total variation = 7447.5; SSE = 896.8; b1 = 5.7003

**13.41** THE FAST-FOOD RESTAURANT RATING CASE FastFood

Note: This exercise is only for those who have previously read Section 13.3. Use the Excel output of Figure 13.18 (page 540) to find and report each of the following: explained variation, unexplained variation, total variation, r2. Interpret r2.

13.6: Testing the Significance of the Population Correlation Coefficient (Optional)

We have seen that the simple correlation coefficient measures the linear relationship between the observed values of x and the observed values of y that make up the sample. A similar coefficient of linear correlation can be defined for the population of all possible combinations of observed values of x and y. We call this coefficient the **population correlation coefficient** and denote it by the symbol **
ρ
** (pronounced

H0:

ρ = 0, which says there is no linear relationship between x and y,

H

ρ ≠ 0, which says there is a positive or negative linear relationship between x and y.

and is based on the assumption that the population of all possible observed combinations of values of x and y has a **bivariate normal probability distribution.** See Wonnacott and Wonnacott (1981) for a discussion of this distribution. It can be shown that the preceding test statistic t and the p-value used to test H0: ρ = 0 versus Ha: ρ ≠ 0 are equal to, respectively, the test statistic t = b1 / s b 1 and the p-value used to test H0: β1 = 0 versus Ha: β1 ≠ 0, where β1 is the slope in the simple linear regression model. Keep in mind, however, that although the mechanics involved in these hypothesis tests are the same, these tests are based on different assumptions (remember that the test for significance of the slope is based on the regression assumptions). If the bivariate normal distribution assumption for the test concerning ρ is badly violated, we can use a nonparametric approach to correlation. One such approach is **Spearman’s rank correlation coefficient.** This approach is discussed in optional Section 18.5.

EXAMPLE 13.14: The Fuel Consumption Case

Again consider testing the significance of the slope in the fuel consumption problem. Recall that in Example 13.6 (page 534) we found that t = −7.33 and that the p-value related to this t statistic is .0003. We therefore (if the regression assumptions hold) can reject H0: β1 = 0 at level of significance .05, .01, or .001, and we have extremely strong evidence that x is significantly related to y. This also implies (if the population of all possible observed combinations of x and y has a bivariate normal probability distribution) that we can reject H0: ρ = 0 in favor of Ha: ρ ≠ 0 at level of significance .05, .01, or .001. It follows that we have extremely strong evidence of a linear relationship, or correlation, between x and y. Furthermore, because we have previously calculated r to be −.9482, we estimate that x and y are negatively correlated.

Exercises for Section 13.6

CONCEPTS

**13.42** Explain what is meant by the population correlation coefficient ρ.

**13.43** Explain how we test H0: ρ = 0 versus Ha: ρ ≠ 0. What do we conclude if we reject H0: ρ = 0?

METHODS AND APPLICATIONS

**13.44** **The starting salary case** StartSal

Consider testing H0: β1 = 0 versus Ha: β1 ≠ 0. Figure 13.13 (page 538) tells us that t = 14.44 and that the related p-value is less than .001. Assuming that the bivariate normal probability distribution assumption holds, test H0: ρ = 0 versus Ha: ≠ 0 by setting α equal to .05, .01, and .001. What do you conclude about how x and y are related?

13.45 **The service Time Case** SrvcTime

Consider testing H0: β1 = 0 versus Ha: β1 ≠ 0. Figure 13.14 (page 538) tells us that t = 30.580 and that the related p-value is less than .001. Assuming that the bivariate normal probability distribution assumption holds, test H0: ρ = 0 versus Ha: ρ ≠ 0 by setting α equal to .05, .01, and .001. What do you conclude about how x and y are related?

13.7: An F Test for the Model

In this section we discuss an F test that can be used to test the significance of the regression relationship between x and y. Sometimes people refer to this as testing the significance of the simple linear regression model. For simple linear regression, this test is another way to test the null hypothesis H0: β1 = 0 (the relationship between x and y is not significant) versus Ha: β1 ≠ 0 (the relationship between x and y is significant). If we can reject H0 at level of significance α, we often say that **the simple linear regression model is significant at level of significance α
**.

An F Test for the Simple Linear Regression Model

Suppose that the regression assumptions hold, and define the overall

F

statistic to be

Also define the p-value related to F(model) to be the area under the curve of the F distribution (having 1 numerator and n − 2 denominator degrees of freedom) to the right of F(model)—see Figure 13.22(b).

Figure 13.22: An F Test for the Simple Linear Regression Model

We can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 at level ofsignificance α if either of the following equivalent conditions hold:

1 F(model) > Fα

2 p-value < α

Here the point Fα is based on 1 numerator and n − 2 denominator degrees of freedom.

The first condition in the box says we should reject H0: β1 = 0 (and conclude that the relationship between x and y is significant) when F(model) is large. This is intuitive because a large overall F statistic would be obtained when the explained variation is large compared to the unexplained variation. This would occur if x is significantly related to y, which would imply that the slope β1 is not equal to 0. Figure 13.22(a) illustrates that we reject H0 when F(model) is greater than Fα. As can be seen in Figure 13.22(b), when F(model) is large, the related p-value is small. When the p-value is small enough [resulting from an F(model) statistic that is large enough], we reject H0. Figure 13.22(b) illustrates that the second condition in the box (p-value < α) is an equivalent way to carry out this test.

EXAMPLE 13.15: The Fuel Consumption Case

Consider the fuel consumption problem and the following partial MINITAB output of the simple linear regression analysis relating weekly fuel consumption y to average hourly temperature x:

Analysis of Variance

Looking at this output, we see that the explained variation is 22.981 and the unexplained variation is 2.568. It follows that

Note that this overall F statistic is given on the MINITAB output and is also given on the following partial Excel output:

The p-value related to F(model) is the area to the right of 53.69 under the curve of the F distribution having 1 numerator and 6 denominator degrees of freedom. This p-value is given on both the MINITAB output (labeled “p”) and the Excel output (labeled “Significance F”) and is less than .001. If we wish to test the significance of the regression relationship with level of significance α = .05, we use the critical value F.05 based on 1 numerator and 6 denominator degrees of freedom. Using Table A.6 (page 867), we find that F.05 = 5.99. Since F(model) = 53.69 > F.05 = 5.99, we can reject H0: β1 = 0 in favor of Ha: β1 ≠ 0 at level of significance .05. Alternatively, since the p-value is smaller than .05, .01, and .001, we can reject H0 at level of significance .05, .01, or .001. Therefore, we have extremely strong evidence that H0: β1 = 0 should be rejected and that the regression relationship between x and y is significant. That is, we might say that we have extremely strong evidence that the simple linear model relating y to x is significant.

As another example, consider the following partial MegaStat output:

This output tells us that for the QHIC simple linear regression model, F(model) is 305.06 and the related p-value is less than .001. Because the p-value is less than .001, we have extremely strong evidence that the regression relationship is significant.

Testing the significance of the regression relationship between y and x by using the overall F statistic and its related p-value is equivalent to doing this test by using the t statistic and its related p-value. Specifically, it can be shown that (t)2 = F(model) and that (tα/2)2 based on n − 2 degrees of freedom equals Fα based on 1 numerator and n − 2 denominator degrees of freedom. It follows that the critical value conditions

| t | > tα/2 and F(model) > Fα

are equivalent. Furthermore, the p-values related to t and F(model) can be shown to be equal. Because these tests are equivalent, it would be logical to ask why we have presented the F test. There are two reasons. First, most standard regression computer packages include the results of the F test as a part of the regression output. Second, the F test has a useful generalization in multiple regression analysis (where we employ more than one predictor variable). The F test in multiple regression is not equivalent to a t test. This is further explained in Chapter 14.

Exercises for Section 13.7

CONCEPTS

**13.46** What are the null and alternative hypotheses for the F test in simple linear regression?

**13.47** The F test in simple linear regression is equivalent to what other test?

METHODS AND APPLICATIONS

In Exercises **13.48** through **13.53**, we give MINITAB, MegaStat, and Excel outputs of simple linear regression analyses of the data sets related to six previously discussed case studies. Using the appropriate computer output,

a Use the explained variation and the unexplained variation as given on the computer output to calculate the F(model) statistic.

b Utilize the F(model) statistic and the appropriate critical value to test H0: β1 = 0 versus Ha: β1 ≠ 0 by setting α equal to .05. What do you conclude about the regression relationship between y and x?

c Utilize the F(model) statistic and the appropriate critical value to test H0: β1 = 0 versus Ha: β1 ≠ 0 by setting α equal to .01. What do you conclude about the regression relationship between y and x?

d Find the p-value related to F(model) on the computer output and report its value. Using the p-value, test the significance of the regression model at the .10, .05, .01, and .001 levels of significance. What do you conclude?

e Show that the F(model) statistic is (within rounding) the square of the t statistic for testing H0: β1 = 0 versus Ha: β1 ≠ 0. Also, show that the F.05 critical value is the square of the t.025 critical value.

Note that in the lower right hand corner of each output we give (in parentheses) the number of observations, n, used to perform the regression analysis and the t statistic for testing H0: β1 = 0 versus Ha: β1 ≠ 0.

13.48 THE STARTING SALARY CASE StartSal

Analysis of Variance

**13.49** THE SERVICE TIME CASE SrvcTime

**13.50** THE FRESH DETERGENT CASE Fresh

Analysis of Variance

**13.51** THE DIRECT LABOR COST CASE DirLab

**13.52** THE REAL ESTATE SALES PRICE CASE RealEst

Analysis of Variance

13.53 THE FAST-FOOD RESTAURANT RATING CASE FastFood

13.8: Residual Analysis (Optional)

__Chapter 16__

In this section we explain how to check the validity of the regression assumptions. The required checks are carried out by analyzing the **regression residuals.** The residuals are defined as follows:

For any particular observed value of y, the corresponding

residual

is

e = y − ŷ = (observed value of y − predicted value of y)

where the predicted value of y is calculated using the least squares prediction equation

ŷ = b0 + b1x

The linear regression model y = β0 + β1x + ε implies that the error term ε is given by the equation ε = y (β0 + β1x). Since ŷ in the previous box is clearly the point estimate of β0 + β1x, we see that the residual e = y − ŷ is the point estimate of the error term ε. If the regression assumptions are valid, then, for any given value of the independent variable, the population of potential error term values will be normally distributed with mean 0 and variance σ2 (see the regression assumptions in Section 13.2 on page 530). Furthermore, the different error terms will be statistically independent. Because the residuals provide point estimates of the error terms, it follows that

If the regression assumptions hold, the residuals should look like they have been randomly and independently selected from normally distributed populations having mean 0 and variance σ2.

In any real regression problem, the regression assumptions will not hold exactly. In fact, it is important to point out that mild departures from the regression assumptions do not seriously hinder our ability to use a regression model to make statistical inferences. Therefore, we are looking for pronounced, rather than subtle, departures from the regression assumptions. Because of this, we will require that the residuals only approximately fit the description just given.

Residual plots

One useful way to analyze residuals is to plot them versus various criteria. The resulting plots are called **
residual plots.** To construct a residual plot, we compute the residual for each observed y value. The calculated residuals are then plotted versus some criterion. To validate the regression assumptions, we make residual plots against (1) values of the independent variable x; (2) values of ŷ, the predicted value of the dependent variable; and (3) the time order in which the data have been observed (if the regression data are time series data).

We next look at an example of constructing residual plots. Then we explain how to use these plots to check the regression assumptions.

EXAMPLE 13.16: The QHIC Case

Figure 13.23 gives the QHIC upkeep expenditure data and a scatterplot of the data. If we use a simple linear regression model to describe the QHIC data, we find that the least squares point estimates of β0 and β1 are b0 = −348.3921 and b1 = 7.2583. The MegaStat output in Figure 13.24(a) presents the predicted home upkeep expenditures and residuals that are given by the simple linear regression model. Here each residual is computed as

Figure 13.23: The QHIC Upkeep Expenditure Data and a Scatterplot of the Data QHIC

Figure 13.24: MegaStat and MINITAB Output of the Residuals and Residual Plots for the QHIC Simple Linear Regression Model

e = y = ŷ = y − (b0 + b1x) = y − (−348.3921 + 7.2583x)

For instance, for the first observation (home) when y = 1,412.08 and x = 237.00 (see Figure 13.23), the residual is

The MINITAB output in Figure 13.24(b) and (c) gives plots of the residuals for the QHIC simple linear regression model against values of x and ŷ. To understand how these plots are constructed, recall that for the first observation (home) y = 1,412.08, x = 237.00, ŷ = 1,371.816 and the residual is 40.264. It follows that the point plotted in Figure 13.24(b) corresponding to the first observation has a horizontal axis coordinate of the x value 237.00 and a vertical axis coordinate of the residual 40.264. It also follows that the point plotted in Figure 13.24(c) corresponding to the first observation has a horizontal axis coordinate of the ŷ value 1,371.816, and a vertical axis coordinate of the residual 40.264. Finally, note that the QHIC data are cross-sectional data, not time series data. Therefore, we cannot make a residual plot versus time.

The constant variance assumption

To check the validity of the constant variance assumption, we examine plots of the residuals against values of x, ŷ, and time (if the regression data are time series data). When we look at these plots, the pattern of the residuals’ fluctuation around 0 tells us about the validity of the constant variance assumption. A residual plot that “fans out” [as in Figure 13.25(a)] suggests that the error terms are becoming more spread out as the horizontal plot value increases and that the constant variance assumption is violated. Here we would say that an **increasing error variance** exists. A residual plot that “funnels in” [as in Figure 13.25(b)] suggests that the spread of the error terms is decreasing as the horizontal plot value increases and that again the constant variance assumption is violated. In this case we would say that a **decreasing error variance** exists. A residual plot with a “horizontal band appearance” [as in Figure 13.25(c)] suggests that the spread of the error terms around 0 is not changing much as the horizontal plot value increases. Such a plot tells us that the constant variance assumption (approximately) holds.

Figure 13.25: Residual Plots and the Constant Variance Assumption

As an example, consider the QHIC case and the residual plot in Figure 13.24(b). This plot appears to fan out as x increases, indicating that the spread of the error terms is increasing as x increases. That is, an increasing error variance exists. This is equivalent to saying that the variance of the population of potential yearly upkeep expenditures for houses worth x (thousand dollars) appears to increase as x increases. The reason is that the model y = β0 + β1x + ε says that the variation of y is the same as the variation of ε. For example, the variance of the population of potential yearly upkeep expenditures for houses worth $200,000 would be larger than the variance of the population of potential yearly upkeep expenditures for houses worth $100,000. Increasing variance makes some intuitive sense because people with more expensive homes generally have more discretionary income. These people can choose to spend either a substantial amount or a much smaller amount on home upkeep, thus causing a relatively large variation in upkeep expenditures.

Another residual plot showing the increasing error variance in the QHIC case is Figure 13.24(c). This plot tells us that the residuals appear to fan out as ŷ (predicted y) increases, which is logical because ŷ is an increasing function of x. Also, note that the scatter plot of y versus x in Figure 13.23 shows the increasing error variance—the y values appear to fan out as x increases. In fact, one might ask why we need to consider residual plots when we can simply look at scatter plots of y versus x. One answer is that, in general, because of possible differences in scaling between residual plots and scatter plots of y versus x, one of these types of plots might be more informative in a particular situation. Therefore, we should always consider both types of plots.

When the constant variance assumption is violated, we cannot use the formulas of this chapter to make statistical inferences. Later in this section we discuss how we can make statistical inferences when a nonconstant error variance exists.

The assumption of correct functional form

If the functional form of a regression model is incorrect, the residual plots constructed by using the model often display a pattern suggesting the form of a more appropriate model. For instance, if we use a simple linear regression model when the true relationship between y and x is curved, the residual plot will have a curved appearance. For example, the scatter plot of upkeep expenditure, y, versus home value, x, in Figure 13.23 (page 557) has either a straight-line or slightly curved appearance. We used a simple linear regression model to describe the relationship between y and x, but note that there is a “dip,” or slightly curved appearance, in the upper left portion of each residual plot in Figure 13.24. Therefore, both the scatter plot and residual plots indicate that there might be a slightly curved relationship between y and x. Later in this section we discuss one way to model curved relationships.

The normality assumption

If the normality assumption holds, a histogram and/or stem-and-leaf display of the residuals should look reasonably bell-shaped and reasonably symmetric about 0. Figure 13.26(a) gives the MINITAB output of a stem-and-leaf display of the residuals from the simple linear regression model describing the QHIC data. The stem-and-leaf display looks fairly bell-shaped and symmetric about 0. However, the tails of the display look somewhat long and “heavy” or “thick,” indicating a possible violation of the normality assumption.

Figure 13.26: Stem-and-Leaf Display and Normal Plots of the Residuals from the Simple Linear Regression Model Describing the QHIC Data

Another way to check the normality assumption is to construct a **
normal plot
** of the residuals. To make a normal plot, we first arrange the residuals in order from smallest to largest. Letting the ordered residuals be denoted as e(1), e(2),…, e(n) we denote the ith residual in the ordered listing as e(i). We plot e(i) on the vertical axis against a point called z(i) on the horizontal axis. Here z(i) is defined to be the point on the horizontal axis under the standard normal curve so that the area under this curve to the left of z(i) is (3i − 1)/(3n + 1). For example, recall in the QHIC case that there are n = 40 residuals given in Figure 13.24(a). It follows that, when i = 1, then

Therefore, z(1) is the normal point having an area of .0165 under the standard normal curve to its left. Thus, as illustrated in Figure 13.26(b), z(1) equals −2.13. Because the smallest residual in Figure 13.24(a) is −289.044, the first point plotted is e(1) = −289.044 on the vertical scale versus z(1) = −2.13 on the horizontal scale. When i = 2, it can be verified that (3i − 1)/(3n + 1) equals .0413 and thus that z(2) = −1.74. Therefore, because the second-smallest residual in Figure 13.24(a) is −259.958, the second point plotted is e(2) = −259.958 on the vertical scale versus z(2) = −1.74 on the horizontal scale. This process is continued until the entire normal plot is constructed. The MegaStat output of this plot is given in Figure 13.26(c).

An equivalent plot is shown in Figure 13.26(d), which is a MINITAB output. In this figure, we plot the percentage p(i) of the area under the standard normal curve to the left of z(i) on the vertical axis. Thus, the first point plotted in this normal plot is e(1) = −289.044 on the horizontal scale versus p(1) = (.0165)(100) = 1.65 on the vertical scale, and the second point plotted is e(2) = −259.958 on the horizontal scale versus p(2) = (.0413)(100) = 4.13 on the vertical scale. It is important to note that the scale on the vertical axis does not have the usual spacing between the percentages. The spacing reflects the distance between the z-scores that correspond to the percentages in the standard normal distribution. Hence, if we wished to create the plot in Figure 13.26(d) by hand, we would need special graphing paper with this vertical scale.

It can be proven that, if the normality assumption holds, then the expected value of the ith ordered residual e(i) is proportional to z(i). Therefore, a plot of the e(i) values on the horizontal scale versus the z(i) values on the vertical scale (or equivalently, the e(i) values on the horizontal scale versus the p(i) values on the vertical scale) should have a straight-line appearance. That is, if the normality assumption holds, then the normal plot should have a straight-line appearance. A normal plot that does not look like a straight line (admittedly, a subjective decision) indicates that the normality assumption is violated. Since the normal plots in Figure 13.26 have some curvature (particularly in the upper right portion), there is a possible violation of the normality assumption.

It is important to realize that violations of the constant variance and correct functional form assumptions can often cause a histogram and/or stem-and-leaf display of the residuals to look nonnormal and can cause the normal plot to have a curved appearance. Because of this, it is usually a good idea to use residual plots to check for nonconstant variance and incorrect functional form before making any final conclusions about the normality assumption. Later in this section we discuss a procedure that sometimes remedies simultaneous violations of the constant variance, correct functional form, and normality assumptions.

We have concluded that the QHIC data may violate the assumptions underlying a simple linear regression model because the relationship between x and y may not be linear and because the errors may not be normally distributed with constant variance. However, the fanning out seen in the residual plots in Figure 13.24(a) and (b) and the slight curvature seen in Figure 13.23 are not extreme. Also, the heavy-tailed nature of the stem-and-leaf display of the residuals and the nonlinearity of the normal probability plots in Figure 13.26 are not pronounced. In optional Section 15.6 we will discuss procedures for transforming data that do not satisfy the regression assumptions into data that do. When we use these procedures to fit a new model to the QHIC data, we will find the expenditure predictions given by the transformed regression model do not differ much from the predictions given by the simple linear regression model of this chapter. This is good evidence that the model of the current chapter does allow QHIC managers to make reasonable decisions about which homeowners should be sent brochures. Note that optional Section 15.6 can be read now without loss of continuity.

The independence assumption

The independence assumption is most likely to be violated when the regression data are time series data—that is, data that have been collected in a time sequence. For such data the time-ordered error terms can be **autocorrelated**. Intuitively, we say that error terms occurring over time have **
positive autocorrelation
** if a positive error term in time period i tends to produce, or be followed by, another positive error term in time period i + k (some later time period) and if a negative error term in time period i tends to produce, or be followed by, another negative error term in time period i + k. In other words, positive autocorrelation exists when positive error terms tend to be followed over time by positive error terms and when negative error terms tend to be followed over time by negative error terms. Positive autocorrelation in the error terms is depicted in Figure 13.27(a), which illustrates that

Figure 13.27: Positive and Negative Autocorrelation

Intuitively, error terms occurring over time have **
negative autocorrelation
** if a positive error term in time period i tends to produce, or be followed by, a negative error term in time period i + k and if a negative error term in time period i tends to produce, or be followed by, a positive error term in time period i + k. In other words, negative autocorrelation exists when positive error terms tend to be followed over time by negative error terms and negative error terms tend to be followed over time by positive error terms. An example of negative autocorrelation in the error terms is depicted in Figure 13.27(b), which illustrates that

The independence assumption basically says that the time-ordered error terms display no positive or negative autocorrelation. This says that **the error terms occur in a random pattern over time.** Such a random pattern would imply that the error terms (and their corresponding y values) are statistically independent.

Because the residuals are point estimates of the error terms, a residual plot versus time is used to check the independence assumption. If a residual plot versus the data’s time sequence has a cyclical appearance, the error terms are positively autocorrelated, and the independence assumption is violated. If a plot of the time-ordered residuals has an alternating pattern, the error terms are negatively autocorrelated, and again the independence assumption is violated. However, if a plot of the time-ordered residuals displays a random pattern, the error terms have little or no autocorrelation. In such a case, it is reasonable to conclude that the independence assumption holds. Note that a statistical test for autocorrelation is presented in Section 15.7. This test is called the **Durbin-Watson test,** and you are prepared to read about it now if you wish to do so.

EXAMPLE 13.17

Figure 13.28(a) on the next page presents data concerning weekly sales at Pages’ Bookstore (Sales), Pages’ weekly advertising expenditure (Adver), and the weekly advertising expenditure of Pages’ main competitor (Compadv). Here the sales values are expressed in thousands of dollars, and the advertising expenditure values are expressed in hundreds of dollars. Figure 13.28(a) also gives the residuals that are obtained when MegaStat is used to perform a simple linear regression analysis relating Pages’ sales to Pages’ advertising expenditure. These residuals are plotted versus time in Figure 13.28(b). We see that the residual plot has a cyclical pattern. This tells us that the error terms for the model are positively autocorrelated and the independence assumption is violated. Furthermore, there tend to be positive residuals when the competitor’s advertising expenditure is lower (in weeks 1 through 8 and weeks 14, 15, and 16) and negative residuals when the competitor’s advertising expenditure is higher (in weeks 9 through 13). Therefore, the competitor’s advertising expenditure seems to be causing the positive autocorrelation.

Figure 13.28: Pages’ Bookstore Sales and Advertising Data, and Residual Analysis

To conclude this example, note that the simple linear regression model relating Pages’ sales to Pages’ advertising expenditure has a standard error, s, of 5.038. The MegaStat residual plot in Figure 13.28(b) includes grid lines that are placed one and two standard errors above and below the residual mean of 0. All MegaStat residual plots use such grid lines to help better diagnose potential violations of the regression assumptions.

When the independence assumption is violated, various remedies can be employed. One approach is to identify which independent variable left in the error term (for example, competitors’ advertising expenditure) is causing the error terms to be autocorrelated. We can then remove this independent variable from the error term and insert it directly into the regression model, forming a **multiple regression model.** (Multiple regression models are discussed in Chapter 14.)

Exercises for Section 13.8

CONCEPTS

**13.54** In a regression analysis, what variables should the residuals be plotted against? What types of patterns in residual plots indicate violations of the regression assumptions?

**13.55** In regression analysis, how do we check the normality assumption?

**13.56** What is one possible remedy for violations of the constant variance, correct functional form, and normality assumptions?

METHODS AND APPLICATIONS

**13.57** **THE FUEL CONSUMPTION CASE** FuelCon1

Recall that Table 13.3 gives the residuals from the simple linear regression model relating weekly fuel consumption to average hourly temperature. Figure 13.29(a) gives the Excel output of a plot of these residuals versus average hourly temperature. Describe the appearance of this plot. Does the plot indicate any violations of the regression assumptions?

Figure 13.29: Residual Diagnostics for Exercises 13.57 and **13.58**

13.58 THE FRESH DETERGENT CASE Fresh

Figure 13.29(b) gives the MINITAB output of residual diagnostics that are obtained when the simple linear regression model is fit to the Fresh detergent demand data. Interpret the diagnostics and determine if they indicate any violations of the regression assumptions.

**13.59** THE SERVICE TIME CASE SrvcTime

The MegaStat output of the residuals given by the service time model is given in Figure 13.30(a), and MegaStat output of residual plots versus x and ŷ is given in Figures 13.30(b) and (c). Do the plots indicate any violations of the regression assumptions?

Figure 13.30: MegaStat Output of Residual Analysis for the Service Time Model (for Exercise 13.59)

**13.60** THE SERVICE TIME CASE SrvcTime

Figure 13.30(a) gives the MegaStat output of the residuals from the simple linear regression model describing the service time-data in Exercise 13.6.

a In this exercise we construct a normal plot of the residuals from the simple linear regression model. To construct this plot, we must first arrange the residuals in order from smallest to largest. These ordered residuals are given in Table 13.4 on the next page. Denoting the ith ordered residual as e(i) (i = 1, 2,…, 11), we next compute for each value of i the point z(i). These computations are summarized in Table 13.4. Show how z(4) = −.46 and z(10) = 1.05 have been obtained.

Table 13.4: Ordered Residuals and Normal Plot Calculations for Exercise 13.60(a)

b The ordered residuals (the e(i)’s) are plotted against the z(i)’s on the MegaStat output of Figure 13.31. Does this figure indicate a violation of the normality assumption?

Figure 13.31: Normal Plot of the Residuals for Exercise 13.60(b)

**13.61** A simple linear regression model is employed to analyze the 24 monthly observations given in Table 13.5. Residuals are computed and are plotted versus time. The resulting residual plot is shown in Figure 13.32. Discuss why the residual plot suggests the existence of positive autocorrelation. SalesAdv

Table 13.5: Sales and Advertising Data for Exercise 13.61 SalesAdv

Figure 13.32: Residual Plot for Exercise 13.61

**13.62** **THE UNEQUAL VARIANCES SERVICE TIME CASE** SrvcTime2

Figure 13.33(a) presents data concerning the time, y, required to perform service and the number of microcomputers serviced, x, for 15 service calls. Figure 13.33(b) gives a plot of y versus x, and Figure 13.33(c) gives the Excel output of a plot of the residuals versus x for a simple linear regression model. What regression assumption appears to be violated?

Figure 13.33: The Data, Data Plot, and Residual Plot for Exercise 13.62 SrvcTime2

13.9: Some Shortcut Formulas (Optional)

Calculating the sum of squared residuals

A shortcut formula for the sum of squared residuals is

where

For example, consider the fuel consumption case. If we square each of the eight observed fuel consumptions in Table 13.1 (page 517) and add up the resulting squared values, we find that . We have also found in Example 13.3 (page 523) that Σyi = 81.7, SSxy = −179.6475 and SSxx = 1,404.355. It follows that

and

Finally, note that equals b1SSxx. However, we recommend using the first of these expressions, because doing so usually gives less round-off error.

Calculating the total, explained, and unexplained variations

The unexplained variation is SSE, and thus the shortcut formula for SSE is a shortcut formula for the unexplained variation. The quantity SSyy defined on page 567 is the total variation, and thus the shortcut formula for SSyy is a shortcut formula for the total variation. Lastly, it can be shown that the expression equals the explained variation and thus is a shortcut formula for this quantity.

Chapter Summary

This chapter has discussed **simple linear regression analysis,** which relates a dependent variable to a single independent (predictor) variable. We began by considering the simple linear regression model, which employs two parameters: the slope and

y

intercept. We next discussed how to compute the least squares point estimates of these parameters and how to use these estimates to calculate a point estimate of the mean value of the dependent variable and a point prediction of an individual value of the dependent variable. Then, after considering the assumptions behind the simple linear regression model, we discussed **testing the significance of the regression relationship (slope),** calculating a confidence interval for the mean value of the dependent variable, and calculating a prediction interval for an individual value of the dependent variable. We next explained several measures of the utility of the simple linear regression model. These include the simple coefficient of determination and an

F

**test for the simple linear model.** We concluded this chapter by giving an optional discussion of using **residual analysis** to detect violations of the regression assumptions.

Glossary of Terms

**cross-sectional data:**

Data that are observed at a single point in time. (page 520)

**dependent variable:**

The variable that is being described, predicted, or controlled. (page 515)

**distance value:**

A measure of the distance between a particular value x0 of the independent variable x and , the average of the previously observed values of x (the center of the experimental region). (page 541)

**error term:**

The difference between an individual value of the dependent variable and the corresponding mean value of the dependent variable. (page 519)

**experimental region:**

The range of the previously observed values of the independent variable. (page 526)

**independent variable:**

A variable used to describe, predict, and control the dependent variable. (page 515)

**least squares point estimates:**

The point estimates of the slope and y intercept of the simple linear regression model that minimize the sum of squared residuals. (pages 521–522)

**negative autocorrelation:**

The situation in which positive error terms tend to be followed over time by negative error terms and negative error terms tend to be followed over time by positive error terms. (page 562)

**normal plot:**

A residual plot that is used to check the normality assumption. (page 560)

**positive autocorrelation:**

The situation in which positive error terms tend to be followed over time by positive error terms and negative error terms tend to be followed over time by negative error terms. (page 562)

**residual:**

The difference between the observed value of the dependent variable and the corresponding predicted value of the dependent variable. (pages 522, 556)

**residual plot:**

A plot of the residuals against some criterion. The plot is used to check the validity of one or more regression assumptions. (page 557)

**simple coefficient of determination:**

The proportion of the total variation in the observed values of the dependent variable that is explained by the simple linear regression model. (page 548)

**simple correlation coefficient:**

A measure of the linear association between two variables. (page 549)

**simple linear regression model:**

An equation that describes the straight-line relationship between a dependent variable and an independent variable. (page 519)

**slope (of the simple linear regression model):**

The change in the mean value of the dependent variable that is associated with a one-unit increase in the value of the independent variable. (page 519)

**time series data:**

Data that are observed in time sequence. (page 520)

y-intercept (of the simple linear regression model):

The mean value of the dependent variable when the value of the independent variable is 0. (page 519)

Important Formulas and Tests

**Simple linear regression** model: page 519

Least squares point estimates of β0 and β1: pages 521–522

Least squares line (prediction equation): page 522

The predicted value of y: page 521

The residual: pages 522 and 556

Sum of squared residuals: pages 522 and 548

Mean square error: page 532

Standard error: page 532

Standard error of the estimate b1: page 533

Testing the significance of the slope: page 534

Testing the significance of the y-intercept: page 536

Confidence interval for the slope: page 536

Point estimate of a mean value of y: page 541

Point prediction of an individual value of y: page 541

Standard error of ŷ: page 541

Confidence interval for a mean value of y: page 541

Prediction interval for an individual value of y: page 541

Explained variation: page 548

Unexplained variation: page 548

Total variation: page 548

Simple coefficient of determination: page 548

Simple correlation coefficient: page 549

Testing the significance of the population correlation coefficient: page 552

An F test for the simple linear regression model: page 553

Normal plot calculations: page 560

Supplementary Exercises

**13.63** Consider the following data concerning the demand (y) and price (x) of a consumer product. Demand

a Plot y versus x. Does it seem reasonable to use the simple linear regression model to relate y to x?

b Calculate the least squares point estimates of the parameters in the simple linear regression model.

c Write the least squares prediction equation. Graph this equation on the plot of y versus x.

d Test the significance of the regression relationship between y and x.

e Find a point prediction of and a 95 percent prediction interval for the demand corresponding to each of the prices $2.10, $2.75, and $3.10.

**13.64** In an article in Public Roads (1983), Bissell, Pilkington, Mason, and Woods study bridge safety (measured in accident rates per 100 million vehicles) and the difference between the width of the bridge and the width of the roadway approach (road plus shoulder):5 AutoAcc

The MINITAB output of a simple linear regression analysis relating accident to width difference is as follows:

Using the MINITAB output

a Identify and interpret the least squares point estimate of the slope of the simple linear regression model.

b Identify and interpret the p-value for testing H0: β1 = 0 versus Ha: β1 ≠ 0.

c Identify and interpret r2.

**13.65** The data in Table 13.6 concerning the relationship between smoking and lung cancer death are presented in a course of The Open University, Statistics in Society, Unit C4, The Open University Press, Milton Keynes, England, 1983. The original source of the data is Occupational Mortality: The Registrar General’s Decennial Supplement for England and Wales, 1970–1972, Her Majesty’s Stationery Office, London, 1978. In the table, a smoking index greater (less) than 100 indicates that men in the occupational group smoke more (less) than average when compared to all men of the same age. Similarly, a lung cancer death index greater (less) than 100 indicates that men in the occupational group have a greater (less) than average lung cancer death rate when compared to all men of the same age. In Figure 13.34 we present a portion of a MINITAB output of a simple linear regression analysis relating the lung cancer death index to the smoking index. In Figure 13.35 we present a plot of the lung cancer death index versus the smoking index. Smoking

Table 13.6: The Smoking and Lung Cancer Death Data Smoking

Figure 13.34: MINITAB Output of a Simple Linear Regression Analysis of the Data in Table 13.6

Figure 13.35: A Plot of the Lung Cancer Death Index versus the Smoking Index

a Although the data do not prove that smoking increases your chance of getting lung cancer, can you think of a third factor that would cause the two indexes to move together?

b Does the slope of the hypothetical line relating the two indexes when the smoking index is less than 100 seem to equal the slope of the hypothetical line relating the two indexes when the smoking index is greater than 100? If you wish, use simple linear regression to make a more precise determination. What practical conclusion might you make?

**13.66** On January 28, 1986, the space shuttle Challenger exploded soon after takeoff, killing all eight astronauts aboard. The temperature at the Kennedy Space Center at liftoff was 31°F. Before the launch, several scientists argued that the launch should be delayed because the shuttle’s O-rings might harden in the cold and leak. Other scientists used the data plot in Figure 13.36 to argue that there was no relationship between temperature and O-ring failure. On the basis of this figure and other considerations, Challenger was launched to its disastrous, last flight.

Figure 13.36: A Data Plot Based on Seven Launches

Scientists using the data plot in Figure 13.36 made a horrible mistake. They relied on a data plot that was created by using only the seven previous launches where there was at least one O-ring failure. A plot based on all 24 previous launches—17 of which had no O-ring failures—is given in Figure 13.37 on the next page.

Figure 13.37: A Data Plot Based on All 24 Launches

a Intuitively, do you think that Figure 13.37 indicates that there is a relationship between temperature and O-ring failure? Use simple linear regression to justify your answer.

b Even though the figure using only seven launches is incomplete, what about it should have cautioned the scientists not to make the launch?

**13.67** In an article in the Journal of Accounting Research, Benzion Barlev and Haim Levy consider relating accounting rates on stocks and market returns. Fifty-four companies were selected. For each company the authors recorded values of x, the mean yearly accounting rate for the period 1959 to 1974, and y, the mean yearly market return rate for the period 1959 to 1974. The data in Table 13.7 were obtained. Here the accounting rate can be interpreted to represent input into investment and therefore is a logical predictor of market return. Use the simple linear regression model and a computer to do the following: AcctRet

Table 13.7: Accounting Rates on Stocks and Market Returns for 54 Companies AcctRet

a Find a point estimate of and a 95 percent confidence interval for the mean market return rate of all stocks having an accounting rate of 15.00.

b Find a point prediction of and a 95 percent prediction interval for the market return rate of an individual stock having an accounting rate of 15.00.

**13.68** In New Jersey, banks have been charged with withdrawing from counties having a high percentage of minorities. To substantiate this charge, P. D’Ambrosio and S. Chambers (1995) present the data in Table 13.8 concerning the percentage, x, of minority population and the number of county residents, y, per bank branch in each of New Jersey’s 21 counties. If we use Excel to perform a simple linear regression analysis of this data, we obtain the output given in Figure 13.38. NJBank

Table 13.8: The New Jersey Bank Data NJBank

Figure 13.38: Excel Output of a Simple Linear Regression Analysis of the New Jersey Bank Data

a Determine if there is a significant relationship between x and y.

b Describe the exact nature of any relationship that exists between x and y. (Hint: Estimate β1 by a point estimate and a confidence interval.)

**13.69** In analyzing the stock market, we sometimes use the model y = β0 + β1x + ε to relate y, the rate of return on a particular stock, to x, the rate of return on the overall stock market. When using the preceding model, we can interpret β1 to be the percentage point change in the mean (or expected) rate of return on the particular stock that is associated with an increase of one percentage point in the rate of return on the overall stock market.

If regression analysis can be used to conclude (at a high level of confidence) that β1 is greater than 1 (for example, if the 95 percent confidence interval for β1 were [1.1826, 1.4723]), this indicates that the mean rate of return on the particular stock changes more quickly than the rate of return on the overall stock market. Such a stock is called an aggressive stock because gains for such a stock tend to be greater than overall market gains (which occur when the market is bullish). However, losses for such a stock tend to be greater than overall market losses (which occur when the market is bearish). Aggressive stocks should be purchased if you expect the market to rise and avoided if you expect the market to fall.

If regression analysis can be used to conclude (at a high level of confidence) that β1 is less than 1 (for example, if the 95 percent confidence interval for β1 were [.4729, .7861]), this indicates that the mean rate of return on the particular stock changes more slowly than the rate of return on the overall stock market. Such a stock is called a defensive stock. Losses for such a stock tend to be less than overall market losses, whereas gains for such a stock tend to be less than overall market gains. Defensive stocks should be held if you expect the market to fall and sold off if you expect the market to rise.

If the least squares point estimate b1 of β1 is nearly equal to 1, and if the 95 percent confidence interval for β1 contains 1, this might indicate that the mean rate of return on the particular stock changes at roughly the same rate as the rate of return on the overall stock market. Such a stock is called a neutral stock.

In a 1984 article in Financial Analysts Journal, Haim Levy considers how a stock’s value of β1 depends on the length of time for which the rate of return is calculated. Levy calculated estimated values of β1 for return length times varying from 1 to 30 months for each of 38 aggressive stocks, 38 defensive stocks, and 68 neutral stocks. Each estimated value was based on data from 1946 to 1975. In the following table we present the average estimate of β1 for each stock type for different return length times:

Let y = average estimate of β1 and x = return length time, and consider relating y to x for each stock type by using the simple linear regression model

Here and are regression parameters relating y to x. We use the asterisks to indicate that these regression parameters are different from β0 and β1. Calculate a 95 percent confidence interval for for each stock type. Carefully interpret the meaning of each interval.

13.70: Internet Exercise US News

Graduate business schools use a variety of factors to guide the selection of applicants for admission to MBA programs. Among the key indicators are undergraduate GPA and Graduate Management Admissions Test (GMAT) score. Is there a statistically significant relationship between the average GMAT score and the average undergraduate GPA for admitted MBA program applicants? Can we develop a statistical model describing the relationship? How reliable are the predictions of the model? The U.S. News website ranks the top MBA programs and provides some interesting data on these questions—the average undergraduate GPA and the average GMAT score of admitted applicants for its 50 top-ranked MBA programs. Note that the data are somewhat limited in scope in that they reflect only the experience of top 50 programs.

Go to the U.S. News website and retrieve the data for the 50 top-ranked MBA programs, http://www.usnews.com. Click on (in turn) .edu : Business: Top Business Schools or go directly to http://www.usnews.com/usnews/edu/beyond/gradrank/mba/gdmbat1.htm. To capture the data—select the entire data table, copy and paste it into Excel or MINITAB, add your own variable labels, and clean up the data as necessary. Excel and MINITAB data files are also included on the CD-ROM (USNews.xls and USNews.mtw). Construct a scatter plot of GMAT versus GPA. Describe any apparent relationship between the two variables. Develop a simple linear regression model expressing GMAT as a linear function of GPA. Identify and interpret the key summary measures—R2, the standard error, and the F-statistic from the ANOVA table. Identify and interpret the estimated regression coefficients. Suppose that the average undergraduate GPA for a particular program is 3.50. Use your regression model to predict the average GMAT score for the program. Prepare a brief report summarizing your analysis and conclusions.

Appendix 13.1: Simple Linear Regression Analysis Using MINITAB

The instruction blocks in this section each begin by describing the entry of data into the MINITAB data window. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.1 for further information about entering data, saving data, and printing results when using MINITAB.

Simple linear regression of the fuel consumption data in Figure 13.11(a) on page 535 (data file: FuelCon1.MTW):

Appendix 13.2: Simple Linear Regression Analysis Using Excel

The instruction blocks in this section each begin by describing the entry of data into an Excel spreadsheet. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.2 for further information about entering data, saving data, and printing results when using Excel.

Simple linear regression in Figure 13.11(b) on page 535 (data file: FuelCon1.xlsx):

• Enter the fuel consumption data from Table 13.1 (page 517) with the temperatures in column A with label Temp and the fuel consumptions in column B with label FuelCons.

• Select **Data : Data Analysis : Regression** and click OK in the Data Analysis dialog box.

• In the Regression dialog box:

Enter B1.B9 into the “Input Y Range” box.

Enter A1.A9 into the “Input X Range” box.

• Place a checkmark in the Labels checkbox.

• Be sure that the “Constant is Zero” checkbox is NOT checked.

• Select the “New Worksheet Ply” option.

• Click OK in the Regression dialog box to obtain the regression results in a new worksheet.

**To produce residual plots** similar to Figures 13.24 (page 559) and 13.26 (page 561):

• In the Regression dialog box, place a checkmark in the Residuals check box to request predicted values and residuals.

• Place a checkmark in the Residual Plots check box.

• Place a checkmark in the Normal Probability Plots check box.

• Click OK in the Regression dialog box.

• Move the plots to chart sheets to format them for effective viewing. Additional residual plots—residuals versus predicted values and residuals versus time—can be produced using the Excel charting features.

**To compute a point prediction for fuel consumption when temperature is 40°F** (data file: FuelCon1.xlsx):

• The Excel Analysis ToolPak does not provide an option for computing point or interval predictions. A point prediction can be computed from the regression results using Excel cell formulas.

• In the regression output, the estimated intercept and slope parameters from cells A17.B18 have been copied to cells D2.E3 and the predictor value 40 has been placed in cell E5.

• In cell E6, enter the Excel formula = E2 + E3*E5 (= 10.7210) to compute the prediction.

Appendix 13.3: Simple Linear Regression Analysis Using MegaStat

The instructions in this section begin by describing the entry of data into an Excel worksheet. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.2 for further information about entering data, saving data, and printing results in Excel. Please refer to Appendix 1.3 for more information about using MegaStat.

Simple linear regression for the service time data in Figure 13.14 on page 538 (data file: SrvcTime.xlsx):

• Enter the service time data (page 528) with the numbers of copiers serviced in column A with label Copiers and with the service times in column B with label Minutes.

• Select **Add-Ins : MegaStat : Correlation/Regression : Regression Analysis.**

• In the Regression Analysis dialog box, click in the Independent Variables window and use the autoexpand feature to enter the range A1. A12.

• Click in the Dependent Variable window and use the autoexpand feature to enter the range B1.B12.

• Check the appropriate Options and Residuals checkboxes as follows:

1 Check “Test Intercept” to include a y-intercept and to test its significance.

2 Check “Output Residuals” to obtain a list of the model residuals.

3 Check “Plot Residuals by Observation,” and “Plot Residuals by Predicted Y and X” to obtain residual plots versus time, versus the predicted values of y, and versus the values of the independent variable.

4 Check “Normal Probability Plot of Residuals” to obtain a normal plot.

5 Check “Durbin-Watson” for the Durbin-Watson statistic (to be explained in Chapter 15).

To obtain a point prediction of y when four computers will be serviced (as well as a confidence interval and prediction interval):

• Click on the drop-down menu above the Predictor Values window and select “Type in predictor values.”

• Type the value of the independent variable for which a prediction is desired (here equal to 4) into the Predictor Values window.

• Select a desired level of confidence (here 95%) from the Confidence Level drop-down menu or type in a value.

• Click OK in the Regression Analysis dialog box.

**To compute several point predictions** of y—say, when 1, 2, 3, and 4 computers will be serviced—(and corresponding confidence and prediction intervals):

• Enter the values of x for which predictions are desired into a column in the spreadsheet—these values can be in any column. Here we have entered the values 1, 2, 3, and 4 into cells A15 through A18.

• Click on the drop-down menu above the Predictor Values box and select “Predictor values from spreadsheet cells.”

• Enter the range A15.A18 into the Predictor Values box.

• Click OK in the Regression Analysis dialog box.

1 Generally, the larger the sample size is—that is, the more combinations of values of y and x that we have observed—the more accurately we can describe the relationship between y and x. Therefore, as the natural gas company observes values of y and x in future weeks, the new data should be added to the data in Table 13.1.

2 As implied by the discussion of Example 13.1, if we have not observed any values of x near 0, this interpretation is of dubious practical value.

3 In order to simplify notation we will often drop the limits on summations in this and subsequent chapters. That is, instead of using the summation we will simply write Σ.

4 Essentially, the difference between r and b1 is a change of scale. It can be shown that b1 and r are related by the equation b1 = (SSyy/SSxx)1/2 r.

5 Source: H. H. Bissell, G. B. Pilkington II, J. M. Mason, and D. L. Woods, “Roadway Cross Section and Alignment,” Public Roads 46 (March 1983), pp. 132–41.

(Bowerman 514)

Bowerman, Bruce L. Business Statistics in Practice, 5th Edition. McGraw-Hill Learning Solutions, 022008.

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