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Chapter 15

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

Department
Operations Management and Information System
Course Code
OMIS 2010
Professor
Alan Marshall
Chapter
15

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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

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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 nk 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
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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