2. 15.1 Introduction
• Analysis of variance compares two or more
populations of interval data.
• Specifically, we are interested in determining
whether differences exist between the population
means.
• The procedure works by analyzing the sample
variance.
3. • The analysis of variance is a procedure that
tests to determine whether differences exits
between two or more population means.
• To do this, the technique analyzes the sample
variances
15.2 One Way Analysis of
Variance
4. • Example 15.1
– An apple juice manufacturer is planning to develop a new
product -a liquid concentrate.
– The marketing manager has to decide how to market the
new product.
– Three strategies are considered
• Emphasize convenience of using the product.
• Emphasize the quality of the product.
• Emphasize the product’s low price.
One Way Analysis of Variance
5. • Example 15.1 - continued
– An experiment was conducted as follows:
• In three cities an advertisement campaign was launched .
• In each city only one of the three characteristics
(convenience, quality, and price) was emphasized.
• The weekly sales were recorded for twenty weeks
following the beginning of the campaigns.
One Way Analysis of Variance
7. • Solution
– The data are interval
– The problem objective is to compare sales in three
cities.
– We hypothesize that the three population means are
equal
One Way Analysis of Variance
8. H0: µ1 = µ2= µ3
H1: At least two means differ
To build the statistic needed to test the
hypotheses use the following notation:
• Solution
Defining the Hypotheses
9. Independent samples are drawn from k populations (treatments).
1 2 k
X11
x21
.
.
.
Xn1,1
1
1
x
n
X12
x22
.
.
.
Xn2,2
2
2
x
n
X1k
x2k
.
.
.
Xnk,k
k
k
x
n
Sample size
Sample mean
First observation,
first sample
Second observation,
second sample
X is the “response variable”.
The variables’ value are called “responses”.
Notation
10. Terminology
• In the context of this problem…
Response variable – weekly sales
Responses – actual sale values
Experimental unit – weeks in the three cities when we
record sales figures.
Factor – the criterion by which we classify the populations
(the treatments). In this problems the factor is the marketing
strategy.
Factor levels – the population (treatment) names. In this
problem factor levels are the marketing trategies.
11. Two types of variability are employed when
testing for the equality of the population
means
The rationale of the test statistic
13. 20
25
30
1
7
Treatment 1 Treatment 2 Treatment 3
10
12
19
9
Treatment 1Treatment 2Treatment 3
20
16
15
14
11
10
9
10x1 =
15x2 =
20x3 =
10x1 =
15x2 =
20x3 =
The sample means are the same as before,
but the larger within-sample variability
makes it harder to draw a conclusion
about the population means.
A small variability within
the samples makes it easier
to draw a conclusion about the
population means.
14. The rationale behind the test statistic – I
• If the null hypothesis is true, we would expect all
the sample means to be close to one another
(and as a result, close to the grand mean).
• If the alternative hypothesis is true, at least
some of the sample means would differ.
• Thus, we measure variability between sample
means.
15. • The variability between the sample means is
measured as the sum of squared distances
between each mean and the grand mean.
This sum is called the
Sum of Squares for Treatments
SST
In our example treatments are
represented by the different
advertising strategies.
Variability between sample means
16. 2
k
1j
jj
)xx(nSST ∑=
−=
There are k treatments
The size of sample j The mean of sample j
Sum of squares for treatments (SST)
Note: When the sample means are close to
one another, their distance from the grand
mean is small, leading to a small SST. Thus,
large SST indicates large variation between
sample means, which supports H1.
17. • Solution – continued
Calculate SST
2
k
1j
jj
321
)xx(nSST
65.608x00.653x577.55x
∑=
−=
===
= 20(577.55 - 613.07)2
+
+ 20(653.00 - 613.07)2
+
+ 20(608.65 - 613.07)2
=
= 57,512.23
The grand mean is calculated by
k21
kk2211
n...nn
xn...xnxn
X
+++
+++
=
Sum of squares for treatments (SST)
18. Is SST = 57,512.23 large enough to
reject H0 in favor of H1?
See next.
Sum of squares for treatments (SST)
19. • Large variability within the samples weakens the
“ability” of the sample means to represent their
corresponding population means.
• Therefore, even though sample means may
markedly differ from one another, SST must be
judged relative to the “within samples variability”.
The rationale behind test statistic – II
20. • The variability within samples is measured by
adding all the squared distances between
observations and their sample means.
This sum is called the
Sum of Squares for Error
SSEIn our example this is the
sum of all squared differences
between sales in city j and the
sample mean of city j (over all
the three cities).
Within samples variability
21. • Solution – continued
Calculate SSE
Sum of squares for errors (SSE)
∑∑= =
−=
===
k
j
jij
n
i
xxSSE
sss
j
1
2
1
2
3
2
2
2
1
)(
24.670,811,238,700.775,10
= (n1 - 1)s1
2
+ (n2 -1)s2
2
+ (n3 -1)s3
2
= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24
= 506,983.50
22. Is SST = 57,512.23 large enough
relative to SSE = 506,983.50 to reject
the null hypothesis that specifies that
all the means are equal?
Sum of squares for errors (SSE)
23. To perform the test we need to calculate
the mean squaresmean squares as follows:
The mean sum of squares
Calculation of MST -
Mean Square for Treatments
12.756,28
13
23.512,57
1
=
−
=
−
=
k
SST
MST
Calculation of MSE
Mean Square for Error
45.894,8
360
50.983,509
=
−
=
−
=
kn
SSE
MSE
24. Calculation of the test statistic
23.3
45.894,8
12.756,28
=
=
=
MSE
MST
F
with the following degrees of freedom:
v1=k -1 and v2=n-k
Required Conditions:
1. The populations tested
are normally distributed.
2. The variances of all the
populations tested are
equal.
25. And finally the hypothesis test:
H0: µ1 = µ2 = …=µk
H1: At least two means differ
Test statistic:
R.R: F>Fα,k-1,n-k
MSE
MST
F=
The F test rejection region
26. The F test
Ho: µ1 = µ2= µ3
H1: At least two means differ
Test statistic F= MST/ MSE= 3.23
15.3FFF:.R.R 360,13,05.0knk ≈=> −−−,−,α 1
Since 3.23 > 3.15, there is sufficient evidence
to reject Ho in favor of H1,and argue that at least one
of the mean sales is different than the others.
23.3
17.894,8
12.756,28
MSE
MST
F
=
=
=
27. -0.02
0
0.02
0.04
0.06
0.08
0.1
0 1 2 3 4
• Use Excel to find the p-value
– fx Statistical FDIST(3.23,2,57) = .0467
The F test p- value
p Value = P(F>3.23) = .0467
28. Excel single factor ANOVA
SS(Total) = SST + SSE
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Convenience 20 11551 577.55 10775.00
Quality 20 13060 653.00 7238.11
Price 20 12173 608.65 8670.24
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 57512 2 28756 3.23 0.0468 3.16
Within Groups 506984 57 8894
Total 564496 59
Xm15-01.xls
29. 15.3 Analysis of Variance
Experimental Designs
• Several elements may distinguish between one
experimental design and others.
– The number of factors.
• Each characteristic investigated is called a factor.
• Each factor has several levels.
30. Factor A
Level 1Level2
Level 1
Factor B
Level 3
Two - way ANOVA
Two factors
Level2
One - way ANOVA
Single factor
Treatment 3 (level 1)
Response
Response
Treatment 1 (level 3)
Treatment 2 (level 2)
31. • Groups of matched observations are formed into
blocks, in order to remove the effects of
“unwanted” variability.
• By doing so we improve the chances of
detecting the variability of interest.
Independent samples or blocks
32. • Fixed effects
– If all possible levels of a factor are included in our analysis we
have a fixed effect ANOVA.
– The conclusion of a fixed effect ANOVA applies only to the
levels studied.
• Random effects
– If the levels included in our analysis represent a random
sample of all the possible levels, we have a random-effect
ANOVA.
– The conclusion of the random-effect ANOVA applies to all the
levels (not only those studied).
Models of Fixed and Random Effects
33. • In some ANOVA models the test statistic of the fixed
effects case may differ from the test statistic of the
random effect case.
• Fixed and random effects - examples
– Fixed effects - The advertisement Example (15.1): All the
levels of the marketing strategies were included
– Random effects - To determine if there is a difference in the
production rate of 50 machines, four machines are randomly
selected and there production recorded.
Models of Fixed and Random Effects.
34. 15.4 Randomized Blocks (Two-way)
Analysis of Variance
• The purpose of designing a randomized block
experiment is to reduce the within-treatments
variation thus increasing the relative amount of
between treatment variation.
• This helps in detecting differences between the
treatment means more easily.
35. Treatment 4
Treatment 3
Treatment 2
Treatment 1
Block 1Block3 Block2
Block all the observations with some
commonality across treatments
Randomized Blocks
36. Treatment
Block 1 2 k Block mean
1 X11 X12 . . . X1k
2 X21 X22 X2k
.
.
.
b Xb1 Xb2 Xbk
Treatment mean
1]B[x
2]B[x
b]B[x
1]T[x 2]T[x k]T[x
Block all the observations with some
commonality across treatments
Randomized Blocks
37. • The sum of square total is partitioned into three
sources of variation
– Treatments
– Blocks
– Within samples (Error)
SS(Total) = SST + SSB + SSESS(Total) = SST + SSB + SSE
Sum of square for treatments Sum of square for blocks Sum of square for error
Recall.
For the independent
samples design we have:
SS(Total) = SST +
SSE
Partitioning the total variability
38. Calculating the sums of squares
• Formulai for the calculation of the sums of squares
Treatment
Block 1 2 k Block mean
1 X11 X12 . . . X1k
2 X21 X22 X2k
.
.
.
b Xb1 Xb2 Xbk
Treatment mean
1]B[x
2]B[x
1]T[x 2]T[x k]T[x x
2
1 X)]T[x(b
− ...X)]T[x(b
2
2 +
−+
2
k X)]T[x(b
−+SST =
+
−
2
1 X)]B[x(k
+
−
2
2 X)]B[x(k
2
k X)]B[x(k
−
SSB=
=+−+−+
+−+−++−+−=
...)()(...
)()(...)()()(
2
2
2
1
2
22
2
12
2
21
2
11
XxXX
XxXxXxXxTotalSS
kk
39. Calculating the sums of squares
• Formulai for the calculation of the sums of squares
Treatment
Block 1 2 k Block mean
1 X11 X12 . . . X1k
2 X21 X22 X2k
.
.
.
b Xb1 Xb2 Xbk
Treatment mean
1]B[x
2]B[x
1]T[x 2]T[x k]T[x x
2
1 X)]T[x(b
− ...X)]T[x(b
2
2 +
−+
2
k X)]T[x(b
−+SST =
+
−
2
1 X)]B[x(k
+
−
2
2 X)]B[x(k
2
k X)]B[x(k
−
SSB=
...)X]B[x]T[xx()X]B[x]T[xx(
...)X]B[x]T[xx()X]B[x]T[xx(
...)X]B[x]T[xx()X]B[x]T[xx(SSE
2
2kk2
2
1kk1
2
2222
2
1212
2
2121
2
1111
++−−++−−
++−−++−−
++−−++−−=
40. To perform hypothesis tests for treatments and blocks we
need
• Mean square for treatments
• Mean square for blocks
• Mean square for error
Mean Squares
1k
SST
MST
−
=
1b
SSB
MSB
−
=
1bkn
SSE
MSE
+−−
=
41. Test statistics for the randomized block
design ANOVA
MSE
MST
F =
MSE
MSB
F =
Test statistic for treatments
Test statistic for blocks
42. • Testing the mean responses for treatments
F > Fα,k-1,n-k-b+1
• Testing the mean response for blocks
F> Fα,b-1,n-k-b+1
The F test rejection regions
43. • Example 15.2
– Are there differences in the effectiveness of cholesterol
reduction drugs?
– To answer this question the following experiment was
organized:
• 25 groups of men with high cholesterol were matched by age
and weight. Each group consisted of 4 men.
• Each person in a group received a different drug.
• The cholesterol level reduction in two months was recorded.
– Can we infer from the data in Xm15-02 that there are
differences in mean cholesterol reduction among the four
drugs?
Randomized Blocks ANOVA - Example
44. • Solution
– Each drug can be considered a treatment.
– Each 4 records (per group) can be blocked, because
they are matched by age and weight.
– This procedure eliminates the variability in
cholesterol reduction related to different
combinations of age and weight.
– This helps detect differences in the mean cholesterol
reduction attributed to the different drugs.
Randomized Blocks ANOVA - Example
45. BlocksTreatments b-1 MST / MSE MSB / MSE
Conclusion: At 5% significance level there is sufficient evidence
to infer that the mean “cholesterol reduction” gained by at least
two drugs are different.
K-1
Randomized Blocks ANOVA - Example
ANOVA
Source of Variation SS df MS F P-value F crit
Rows 3848.7 24 160.36 10.11 0.0000 1.67
Columns 196.0 3 65.32 4.12 0.0094 2.73
Error 1142.6 72 15.87
Total 5187.2 99
47. 15.5 Two-Factor Analysis of Variance -
• Example 15.3
– Suppose in Example 15.1, two factors are to be
examined:
• The effects of the marketing strategy on sales.
– Emphasis on convenience
– Emphasis on quality
– Emphasis on price
• The effects of the selected media on sales.
– Advertise on TV
– Advertise in newspapers
48. • Solution
– We may attempt to analyze combinations of levels, one
from each factor using one-way ANOVA.
– The treatments will be:
• Treatment 1: Emphasize convenience and advertise in TV
• Treatment 2: Emphasize convenience and advertise in
newspapers
• …………………………………………………………………….
• Treatment 6: Emphasize price and advertise in newspapers
Attempting one-way ANOVA
49. • Solution
– The hypotheses tested are:
H0: µ1= µ2= µ3= µ4= µ5= µ6
H1: At least two means differ.
Attempting one-way ANOVA
50. City1 City2 City3 City4 City5 City6
Convnce Convnce Quality Quality Price Price
TV Paper TV Paper TV Paper
– In each one of six cities sales are recorded for ten
weeks.
– In each city a different combination of marketing
emphasis and media usage is employed.
• Solution
Attempting one-way ANOVA
51. • The p-value =.0452.
• We conclude that there is evidence that differences
exist in the mean weekly sales among the six cities.
City1 City2 City3 City4 City5 City6
Convnce Convnce Quality Quality Price Price
TV Paper TV Paper TV Paper
• Solution
Xm15-03
Attempting one-way ANOVA
52. • These result raises some questions:
– Are the differences in sales caused by the different
marketing strategies?
– Are the differences in sales caused by the different
media used for advertising?
– Are there combinations of marketing strategy and
media that interact to affect the weekly sales?
Interesting questions – no answers
53. • The current experimental design cannot provide
answers to these questions.
• A new experimental design is needed.
Two-way ANOVA (two factors)
54. Two-way ANOVA (two factors)
City 1
sales
City3
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
TV
Newspapers
Convenience Quality Price
Are there differences in the mean sales
caused by different marketing strategies?
Factor A: Marketing strategy
FactorB:
Advertisingmedia
55. Test whether mean sales of “Convenience”, “Quality”,
and “Price” significantly differ from one another.
H0: µConv.= µQuality = µPrice
H1: At least two means differ
Calculations are
based on the sum of
square for factor A
SS(A)
Two-way ANOVA (two factors)
56. Two-way ANOVA (two factors)
City 1
sales
City 3
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
Factor A: Marketing strategy
FactorB:
Advertisingmedia
Are there differences in the mean sales
caused by different advertising media?
TV
Newspapers
Convenience Quality Price
57. Test whether mean sales of the “TV”, and “Newspapers”
significantly differ from one another.
H0: µTV = µNewspapers
H1: The means differ
Calculations are based on
the sum of square for factor B
SS(B)
Two-way ANOVA (two factors)
58. Two-way ANOVA (two factors)
City 1
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
TV
Newspapers
Convenience Quality Price
Factor A: Marketing strategy
FactorB:
Advertisingmedia
Are there differences in the mean sales
caused by interaction between marketing
strategy and advertising medium?
City 3
sales
TV
Quality
59. Test whether mean sales of certain cells
are different than the level expected.
Calculation are based on the sum of square for
interaction SS(AB)
Two-way ANOVA (two factors)
60. Graphical description of the possible
relationships between factors A and B.
Graphical description of the possible
relationships between factors A and B.
61. Levels of factor A
1 2 3
Level 1 of factor B
Level 2 of factor B
1 2 3
1 2 31 2 3
Level 1and 2 of factor B
Difference between the levels of factor A
No difference between the levels of factor B
Difference between the levels of factor A, and
difference between the levels of factor B; no
interaction
Levels of factor A
Levels of factor A Levels of factor A
No difference between the levels of factor A.
Difference between the levels of factor B
Interaction
M R
e e
s
a p
n o
n
s
e
M R
e e
s
a p
n o
n
s
e
M R
e e
s
a p
n o
n
s
e
M R
e e
s
a p
n o
n
s
e
62. Sums of squares
∑=
−=
a
1i
2
i )x]A[x(rb)A(SS })()()){(2(10( 222
. xxxxxx pricequalityconv −+−+−
∑=
−=
b
1j
2
j )x]B[x(ra)B(SS })()){(3)(10( 22
xxxx NewspaperTV −+−
∑∑
==
+−−=
b
1j
2
jiij
a
1i
)x]B[x]A[x]AB[x(r)AB(SS
∑∑∑ ===
−=
r
k
ijijk
b
j
a
i
ABxxSSE
1
2
11
)][(
63. F tests for the Two-way ANOVA
• Test for the difference between the levels of the main
factors A and B
F=
MS(A)
MSE
F=
MS(B)
MSE
Rejection region: F > Fα,a-1 ,n-ab F > Fα, b-1, n-ab
• Test for interaction between factors A and B
F=
MS(AB)
MSE
Rejection region: F > Fα,(a-1)(b-1),n-ab
SS(A)/(a-1) SS(B)/(b-1)
SS(AB)/(a-1)(b-1)
SSE/(n-ab)
64. Required conditions:
1. The response distributions is normal
2. The treatment variances are equal.
3. The samples are independent.
65. • Example 15.3 – continued( Xm15-03)
F tests for the Two-way ANOVA
Convenience Quality Price
TV 491 677 575
TV 712 627 614
TV 558 590 706
TV 447 632 484
TV 479 683 478
TV 624 760 650
TV 546 690 583
TV 444 548 536
TV 582 579 579
TV 672 644 795
Newspaper 464 689 803
Newspaper 559 650 584
Newspaper 759 704 525
Newspaper 557 652 498
Newspaper 528 576 812
Newspaper 670 836 565
Newspaper 534 628 708
Newspaper 657 798 546
Newspaper 557 497 616
Newspaper 474 841 587
66. • Example 15.3 – continued
– Test of the difference in mean sales between the three marketing
strategies
H0: µconv. = µquality = µprice
H1: At least two mean sales are different
F tests for the Two-way ANOVA
ANOVA
Source of Variation SS df MS F P-value F crit
Sample 13172.0 1 13172.0 1.42 0.2387 4.02
Columns 98838.6 2 49419.3 5.33 0.0077 3.17
Interaction 1609.6 2 804.8 0.09 0.9171 3.17
Within 501136.7 54 9280.3
Total 614757.0 59
Factor A Marketing strategies
67. • Example 15.3 – continued
– Test of the difference in mean sales between the three
marketing strategies
H0: µconv. = µquality = µprice
H1: At least two mean sales are different
F = MS(Marketing strategy)/MSE = 5.33
Fcritical = Fα,a-1,n-ab = F.05,3-1,60-(3)(2)= 3.17; (p-value = .0077)
– At 5% significance level there is evidence to infer that
differences in weekly sales exist among the marketing
strategies.
F tests for the Two-way ANOVA
MS(A)/MSE
68. • Example 15.3 - continued
– Test of the difference in mean sales between the two
advertising media
H0: µTV. = µNespaper
H1: The two mean sales differ
F tests for the Two-way ANOVA
Factor B = Advertising media
ANOVA
Source of Variation SS df MS F P-value F crit
Sample 13172.0 1 13172.0 1.42 0.2387 4.02
Columns 98838.6 2 49419.3 5.33 0.0077 3.17
Interaction 1609.6 2 804.8 0.09 0.9171 3.17
Within 501136.7 54 9280.3
Total 614757.0 59
69. • Example 15.3 - continued
– Test of the difference in mean sales between the two
advertising media
H0: µTV. = µNespaper
H1: The two mean sales differ
F = MS(Media)/MSE = 1.42
Fcritical = Fα,a-1,n-ab = F.05,2-1,60-(3)(2)= 4.02 (p-value = .2387)
– At 5% significance level there is insufficient evidence to infer
that differences in weekly sales exist between the two
advertising media.
F tests for the Two-way ANOVA
MS(B)/MSE
70. • Example 15.3 - continued
– Test for interaction between factors A and B
H0: µTV*conv. = µTV*quality =…=µnewsp.*price
H1: At least two means differ
F tests for the Two-way ANOVA
Interaction AB = Marketing*Media
ANOVA
Source of Variation SS df MS F P-value F crit
Sample 13172.0 1 13172.0 1.42 0.2387 4.02
Columns 98838.6 2 49419.3 5.33 0.0077 3.17
Interaction 1609.6 2 804.8 0.09 0.9171 3.17
Within 501136.7 54 9280.3
Total 614757.0 59
71. • Example 15.3 - continued
– Test for interaction between factor A and B
H0: µTV*conv. = µTV*quality =…=µnewsp.*price
H1: At least two means differ
F = MS(Marketing*Media)/MSE = .09
Fcritical = Fα,(a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2)= 3.17 (p-value= .9171)
– At 5% significance level there is insufficient evidence to infer
that the two factors interact to affect the mean weekly sales.
MS(AB)/MSE
F tests for the Two-way ANOVA
72. 15.7 Multiple Comparisons
• When the null hypothesis is rejected, it may be
desirable to find which mean(s) is (are) different,
and at what ranking order.
• Three statistical inference procedures, geared at
doing this, are presented:
– Fisher’s least significant difference (LSD) method
– Bonferroni adjustment
– Tukey’s multiple comparison method
73. • Two means are considered different if the difference
between the corresponding sample means is larger than
a critical number. Then, the larger sample mean is
believed to be associated with a larger population
mean.
• Conditions common to all the methods here:
– The ANOVA model is the one way analysis of variance
– The conditions required to perform the ANOVA are satisfied.
– The experiment is fixed-effect
15.7 Multiple Comparisons
74. Fisher Least Significant Different (LSD) Method
• This method builds on the equal variances t-test of the
difference between two means.
• The test statistic is improved by using MSE rather than sp
2
.
• We can conclude that µi and µj differ (at α% significance
level if |µi - µj| > LSD, where
kn.f.d
)
n
1
n
1
(MSEtLSD
ji
2
−=
+= α
75. Experimentwise Type I error rate (αE)
(the effective Type I error)
• The Fisher’s method may result in an increased probability of
committing a type I error.
• The experimentwise Type I error rate is the probability of
committing at least one Type I error at significance level of α. It
is calculated by
αE = 1-(1 – α)C
where C is the number of pairwise comparisons (I.e.
C = k(k-1)/2
• The Bonferroni adjustment determines the required Type I error
probability per pairwise comparison (α) ,to secure a pre-
determined overall αE.
76. • The procedure:
– Compute the number of pairwise comparisons (C)
[C=k(k-1)/2], where k is the number of populations.
– Set α = αE/C, where αE is the true probability of making at
least one Type I error (called experimentwise Type I error).
– We can conclude that µi and µj differ (at α/C% significance
level if
kn.f.d
)
n
1
n
1
(MSEt
ji
)C2(ji
−=
+>µ−µ α
Bonferroni Adjustment
77. 35.4465.6080.653xx
10.3165.60855.577xx
45.750.65355.577xx
32
31
21
=−=−
=−=−
=−=−
• Example 15.1 - continued
– Rank the effectiveness of the marketing strategies
(based on mean weekly sales).
– Use the Fisher’s method, and the Bonferroni adjustment method
• Solution (the Fisher’s method)
– The sample mean sales were 577.55, 653.0, 608.65.
– Then,
71.59)20/1()20/1(8894t
)
n
1
n
1
(MSEt
2/05.
ji
2
≈+
=+α
Fisher and Bonferroni Methods
78. • Solution (the Bonferroni adjustment)
– We calculate C=k(k-1)/2 to be 3(2)/2 = 3.
– We set α = .05/3 = .0167, thus t.0167/2,60-3 = 2.467 (Excel).
54.73)20/1()20/1(8894467.2
)
n
1
n
1
(MSEt
ji
2
=+
=+α
Again, the significant difference is between µ1 and µ2.
35.4465.6080.653xx
10.3165.60855.577xx
45.750.65355.577xx
32
31
21
=−=−
=−=−
=−=−
Fisher and Bonferroni Methods
79. • The test procedure:
– Find a critical number ω as follows:
g
n
MSE
),k(q ν=ω α
k = the number of samples
ν =degrees of freedom = n - k
ng = number of observations per sample
(recall, all the sample sizes are the same)
α = significance level
qα(k,ν) = a critical value obtained from the studentized range table
Tukey Multiple Comparisons
80. If the sample sizes are not extremely different, we can use the
above procedure with ng calculated as the harmonic mean of
the sample sizes. k21 n1...n1n1
k
gn
+++
=
• Repeat this procedure for each pair of samples.
Rank the means if possible.
• Select a pair of means. Calculate the difference
between the larger and the smaller mean.
• If there is sufficient evidence to
conclude that µmax > µmin .
minmax xx −
ω>− minmax xx
Tukey Multiple Comparisons
81. City 1 vs. City 2: 653 - 577.55 = 75.45
City 1 vs. City 3: 608.65 - 577.55 = 31.1
City 2 vs. City 3: 653 - 608.65 = 44.35
• Example 15.1 - continued We had three populations
(three marketing strategies).
K = 3,
Sample sizes were equal. n1 = n2 = n3 = 20,
ν = n-k = 60-3 = 57,
MSE = 8894.
minmax xx −
70.71
20
8894
)57,3(.q
n
MSE
),k(q 05
g
==ν=ω α
Take q.05(3,60) from the table.
Population
Sales - City 1
Sales - City 2
Sales - City 3
Mean
577.55
653
698.65
ω>− minmax xx
Tukey Multiple Comparisons
