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Chapter 15

# OMIS 2010 Chapter Notes - Chapter 15: Operations Management, Analysis Of Variance, Multiple Comparisons Problem

by OC9945

School

York UniversityDepartment

Operations Management and Information SystemCourse Code

OMIS 2010Professor

Alan MarshallChapter

15This

**preview**shows pages 1-3. to view the full**18 pages of the document.**Chapter 15: Analysis of Variance

15.1 Introduction

In this chapter, we introduced the analysis of variance technique, which deals with problems whose

objective is to compare two or more populations of quantitative data. You are expected to learn how to

do the following:

l. Recognize when the analysis of variance is to be employed.

2. Recognize which of the three models introduced in this chapter is to be used.

3. How to interpret the ANOVA table.

4. How to perform Fisher's LSD method, the Bonferonni adjustment and Tukey’s multiple

comparison procedure.

5. How to conduct Bartlett's test.

15.2 Single-Factor (One-Way) Analysis of Variance: Independent

Samples

When the k samples are drawn independently of each other, we partition the total sum of squares

into two sources of variability: sum of squares for treatment and sum of squares for error. The F-test is

then used to complete the technique.

The important formulas are

SS(Total) = (xij !x)2

i"1

nj

#

j"1

k

#

SST =

!!

nj(x j!x)2

j"1

k

#"n1(x

1!x)2$n2(x

2!x)2$"$nk(x

k!x)2

SSE = (xij !x

j)2

i"1

nj

#

j"1

k

#

=

!!

(xi1!x

1)2

i"1

n

1

#$(xi2!x

2)2

i"1

n2

#$"$(xik !x

k)2

i"1

nk

#

= !!

(n1!1)s1

2$(n2!1)s2

2

$

"

$

(nk

!

1) sk

2

MST = SST

k

!1

MSE = SSE

n!

k

179

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

F = MST

MSE

The ANOVA table for the completely randomized design is shown below.

Source d.f. Sums of Squares Mean Squares F-ratio

Treatments k – 1 SST MST F

Error n – k SSE MSE

Total n – 1 SS(Total)

Example 15.1

A major computer manufacturer has received numerous complaints concerning the short life of its

disk drives. Most need repair within two years. Since the cost of repairs often exceeds the cost of a new

disk drive, the manufacturer is concerned. In his search for a better disk drive, he finds three new prod-

ucts. He decides to test these three plus his current disk drive to determine if differences in lifetimes

exist among the products. He takes a random sample of five disk drives of each type and links it with a

computer. The number of weeks until the drive breaks down is recorded and is shown below. Do these

data allow us to conclude at the 5% significance level that there are differences among the disk drives?

Type 1 Type 2 Type 3 Current Product

78 125 143 101

92 110 125 96

101 116 133 88

105 88 108 125

98 128 121 128

x

1"94.8

x

2"113.4

x

3

"

126.0

x

4

"

107.6

s1

2"110.7 s2

2"252.8 s3

2

"

172.0 s4

2

"

320.3

Solution

The problem objective is to compare the populations of lifetimes of the four disk drives, and the

data are quantitative. Because the samples are independent, the appropriate technique is the completely

randomized design of the analysis of variance. The null and alternative hypotheses automatically follow.

H0: %&

1"%

2"%

3"%

4

180

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

H1: At least two means differ.

The rejection region is

F >

F

',k!1, n!k"

F

.05, 3, 16 "3.24

The test statistic is computed as follows:

n

1 = n2 = n3 = n4 = 5

x = 5(94.80) $5(113.4) $5(126.0)

$

5(107.6 )

20 "2,209

20 "110.45

SST = 5(94.8 – 110.45)2 + 5(113.4 – 110.45)2 + 5(126.0 – 110.45)2 + 5(107.6 – 110.45)2

= 2,517.75

SSE = 4(110.7) + 4(252.8) + 4(172.0) + 4(320.3) = 3,423.2

MST = 2,517.75

3"839.25

MSE = 3,423.2

16 "213.95

F = 839.25

213.95 "3.92

The complete ANOVA table follows.

Source d.f. Sums of Squares Mean Squares F-ratio

Treatments 3 2,517.75 839.25 3.92

Error 16 3,423.2 213.95

Total 19 5,940.95

Since the F-ratio (3.92) exceeds

F

',k!1, n!k (3.24), we reject H0 and conclude that at least two means

differ.

181

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