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Analysis of Variance
Chapter 15
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15.1 Introduction
Analysis of variance compares two or more
populations of interval data.
Specifically, we are interested in determiningwhether differences exist between the population
means.
The procedure works by analyzing the samplevariance.
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The analysis of variance is a procedure thattests to determine whether differences exitsbetween two or more population means.
To do this, the technique analyzes the sample
variances
15.2 One Way Analysis of
Variance
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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 products low price.
One Way Analysis of Variance
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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
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One Way Analysis of Variance
Convnce Quality Price529 804 672658 630 531793 774 443514 717 596
663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691
663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532
See file
Xm15 -01
Weekly
sales
Weekly
sales
Weekly
sales
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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
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H0: m1 = m2= m3
H1: At least two means differ
To build the statistic needed to test thehypotheses use the following notation:
Solution
Defining the Hypotheses
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Independent samples are drawn from k populations (treatments).
1 2 k
X11
x21...
Xn1,1
1
1
x
n
X12x22...
Xn2,2
2
2
x
n
X1kx2k...
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
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Terminology
In the context of this problem
Response variable weekly sales
Responses actual sale valuesExperimental 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.
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Two types of variability are employed when
testing for the equality of the population
means
The rationale of the test statistic
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Graphical demonstration:
Employing two types of variability
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20
25
30
1
7
Treatment 1 Treatment 2 Treatment 3
10
12
19
9
Treatment 1 Treatment 2 Treatment 3
20
16
1514
11109
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.
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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 samplemeans.
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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 ofSquares forTreatmentsSST
In our example treatments are
represented by the different
advertising strategies.
Variability between sample means
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2k
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.
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Solution continued
Calculate SST
2k
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...xnxnX
Sum of squares for treatments (SST)
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Is SST = 57,512.23 large enough to
reject H0
in favor of H1?
See next.
Sum of squares for treatments (SST)
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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
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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 forError
SSEIn our example this is thesum of all squared differences
between sales in city j and the
sample mean of city j (over all
the three cities).
Within samples variability
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Solution continued
Calculate SSE
Sum of squares for errors (SSE)
k
j
jij
n
i
xxSSE
sss
j
1
2
1
23
22
21
)(
24.670,811,238,700.775,10
(n1 - 1)s12 + (n2 -1)s2
2 + (n3 -1)s32
= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24
= 506,983.50
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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)
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To perform the test we need to calculate
the mean squaresas follows:
The mean sum of squares
Calculation ofMST -Mean Square forTreatments
12.756,28
13
23.512,57
1
k
SSTMST
Calculation ofMSEMean Square forError
45.894,8
360
50.983,509
kn
SSEMSE
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Calculation of the test statistic
23.3
45.894,812.756,28
MSE
MSTF
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.
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And finally the hypothesis test:
H0: m1 = m2= =mk
H1: At least two means differ
Test statistic:
R.R: F>Fa,k-1,n-k
M SE
M STF
The F test rejection region
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The F test
Ho: m1 = m2= m3
H1: At least two means differ
Test statistic F= MST MSE= 3.23
15.3FFF:.R.R 360,13,05.0knk a 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
MSTF
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-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
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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 7238.11Price 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
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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.
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Factor ALevel 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)
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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
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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 thelevels studied.
Random effects
If the levels included in our analysis represent a random
sample of all the possible levels, we have a random-effectANOVA.
The conclusion of the random-effect ANOVA applies to all the
levels (not only those studied).
Models ofFixed and Random Effects
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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 ofFixed and Random Effects.
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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.
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Treatment 4
Treatment 3
Treatment 2
Treatment 1
Block 1Block3 Block2
Block all the observations with some
commonality across treatments
Randomized Blocks
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TreatmentBlock 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
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The sum of square total is partitioned into three
sources of variation
Treatments
Blocks
Within samples (Error)
SS(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
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Calculating the sums of squares
Formulai for the calculation of the sums of squares
TreatmentBlock 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 x2
1 X)]T[x(b
...X)]T[x(b2
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=
...)()(...
)()(...)()()(
22
21
222
212
221
211
XxXX
XxXxXxXxTotalSS
kk
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Calculating the sums of squares
Formulai for the calculation of the sums of squares
TreatmentBlock 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 x2
1 X)]T[x(b
...X)]T[x(b2
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
22kk2
21kk1
22222
21212
22121
21111
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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
SSTMST
1b
SSBMSB
1bkn
SSEMSE
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Test statistics for the randomized block
design ANOVA
MSEMSTF
MSE
MSBF
Test statistic for treatments
Test statistic for blocks
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Testing the mean responses for treatments
F > Fa,k-1,n-k-b+1
Testing the mean response for blocks
F> Fa,b-1,n-k-b+1
The F test rejection regions
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Example 15.2 Are there differences in the effectiveness of cholesterol
reduction drugs?
To answer this question the following experiment wasorganized:
25 groups of men with high cholesterol were matched by ageand 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 aredifferences in mean cholesterol reduction among the fourdrugs?
Randomized Blocks ANOVA - Example
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Solution
Each drug can be considered a treatment.
Each 4 records (per group) can be blocked, becausethey are matched by age and weight.
This procedure eliminates the variability in
cholesterol reductionrelated to differentcombinations of age and weight.
This helps detect differences in the mean cholesterol
reduction attributed to the different drugs.
Randomized Blocks ANOVA - Example
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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
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Analysis of Variance
Chapter 15 - continued
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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
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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
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Solution
The hypotheses tested are:
H0: m1= m2= m3= m4= m5= m6
H1: At least two means differ.
Attempting one-way ANOVA
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City1 City2 City3 City4 City5 City6Convnce 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 marketingemphasis and media usage is employed.
Solution
Attempting one-way ANOVA
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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 City6Convnce Convnce Quality Quality Price Price
TV Paper TV Paper TV Paper
Solution
Xm15-03
Attempting one-way ANOVA
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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
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The current experimental design cannot provide
answers to these questions.
A new experimental design is needed.
Two-way ANOVA (two factors)
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Two-way ANOVA (two factors)
City 1sales
City3sales
City 5sales
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:
Adv
ertisingmedia
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Test whether mean sales of Convenience, Quality,and Price significantly differ from one another.
H0: mConv.= mQuality = mPrice
H1: At least two means differ
Calculations are
based on the sum of
square for factor A
SS(A)
Two-way ANOVA (two factors)
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Two-way ANOVA (two factors)
City 1sales
City 3sales
City 5sales
City 2sales
City 4sales
City 6sales
Factor A: Marketing strategy
FactorB:
Adv
ertisingmedia
Are there differences in the mean sales
caused by different advertising media?
TV
Newspapers
Convenience Quality Price
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Test whether mean sales of the TV, and Newspapers
significantly differ from one another.
H0: mTV = mNewspapersH
1
: The means differ
Calculations are based on
the sum of square for factor B
SS(B)
Two-way ANOVA (two factors)
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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
Factor
B:
Advertising
media
Are there differences in the mean sales
caused by interaction between marketing
strategy and advertising medium?
City 3
sales
TV
Quality
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Test whether mean sales of certain cellsare different than the level expected.
Calculation are based on the sum of square forinteraction SS(AB)
Two-way ANOVA (two factors)
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Graphical description of the possible
relationships between factors A and B.
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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 Re e
sa pn o
nse
M Re e
sa pn o
ns
e
M Re e
sa pn o
nse
M Re e
sa pn o
nse
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Sums of squares
a
1i
2i )x]A[x(rb)A(SS })()()){(2(10(
222. xxxxxx pricequalityconv
b
1j
2j )x]B[x(ra)B(SS })()){(3)(10(
22xxxx 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
)][(
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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 > Fa,a-1 ,n-ab F > Fa, b-1, n-ab
Test for interaction between factors A and BF=
MS(AB)
MSE
Rejection region: F > Fa(a-1)(b-1),n-ab
SS(A)/(a-1) SS(B)/(b-1)
SS(AB)/(a-1)(b-1)
SSE/(n-ab)
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Required conditions:
1. The response distributions is normal
2. The treatment variances are equal.
3. The samples are independent.
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Example 15.3 continued( Xm15-03)
F tests for the Two-way ANOVA
C on ven ien ce Qu al ity Pr ice
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 584Newspaper 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
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Example 15.3 continued
Test of the difference in mean sales between the three marketing
strategies
H0: mconv. = mquality = mpriceH1: 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
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Example 15.3 continued
Test of the difference in mean sales between the three
marketing strategies
H0
: mconv.
= mquality
= mprice
H1: At least two mean sales are different
F = MS(Marketing strategy)/MSE = 5.33
Fcritical = Fa,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
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Example 15.3 - continued Test of the difference in mean sales between the two
advertising media
H0: mTV. = mNespaperH1: 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 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17
Within 501136.7 54 9280.3
Total 614757.0 59
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Example 15.3 - continued
Test of the difference in mean sales between the two
advertising media
H0: mTV. = mNespaper
H1: The two mean sales differ
F = MS(Media)/MSE = 1.42
Fcritical = Faa-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
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Example 15.3 - continued
Test for interaction between factors A and B
H0: mTV*conv. = mTV*quality==mnewsp.*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 3.17Within 501136.7 54 9280.3
Total 614757.0 59
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Example 15.3 - continued
Test for interaction between factor A and B
H0
: mTV*conv.
= mTV*quality
==mnewsp.*price
H1: At least two means differ
F = MS(Marketing*Media)/MSE = .09
Fcritical = Fa(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
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15.7 Multiple Comparisons
When the null hypothesis is rejected, it may bedesirable to find which mean(s) is (are) different,and at what ranking order.
Three statistical inference procedures, geared atdoing this, are presented: Fishers least significant difference (LSD) method
Bonferroni adjustment
Tukeys multiple comparison method
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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 populationmean.
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
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Fisher Least Significant Different (LSD) Method
This method builds on the equal variances t-test of thedifference between two means.
The test statistic is improved by using MSE rather than sp2.
We can conclude that mi and mj differ (at a% significancelevel if |mi - mj| > LSD, where
kn.f.d
)n1
n1(MSEtLSD
ji2
a
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Experimentwise Type I error rate (aE)(the effective Type I error)
The Fishers method may result in an increased probability ofcommitting a type I error.
The experimentwise Type I error rate is the probability of
committing at least one Type I error at significance level ofa Itiscalculated byaE = 1-(1a)
C
where C is the number of pairwise comparisons (I.e.C = k(k-1)/2
The Bonferroni adjustment determines the required Type I errorprobability per pairwise comparison (a) ,to secure a pre-determined overall aE
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The procedure:
Compute the number of pairwise comparisons (C)
[C=k(k-1)/2], where k is the number of populations.
Set a = aE/C, where aE is the true probability of making atleast one Type I error (called experimentwise Type I error).
We can conclude that mi and mj differ (at a/C% significance
level if
kn.f.d
)n
1
n
1(M SEt
ji
)C2(ji
mm a
Bonferroni Adjustment
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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 Fishers method, and the Bonferroni adjustment method
Solution (the Fishers method)
The sample mean sales were 577.55, 653.0, 608.65.
Then,
71.59)20/1()20/1(8894t
)n
1
n
1(M SEt
2/05.
ji
2
a
Fisher and Bonferroni Methods
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Solution (the Bonferroni adjustment)
We calculate C=k(k-1)/2 to be 3(2)/2 = 3.
We set a = .05/3 = .0167, thus t.01672, 60-3 = 2.467 (Excel).
54.73)20/1()20/1(8894467.2
)n
1
n
1(M SEt
ji
2
a
Again, the significant difference is between m1 and m2.
35.4465.6080.653xx
10.3165.60855.577xx
45.750.65355.577xx
32
31
21
Fisher and Bonferroni Methods
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The test procedure:
Find a critical number w as follows:
gn
M SE),k(q w
a
k = the number of samples
=degrees of freedom = n - kng = number of observations per sample
(recall, all the sample sizes are the same)
a = significance level
qa(k,) = a critical value obtained from the studentized range table
Tukey Multiple Comparisons
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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. k21n1...n1n1
kgn
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 toconclude that mmax > mmin .
minmax xx
w minmax xx
Tukey Multiple Comparisons
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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, = n-k = 60-3 = 57,MSE = 8894.
minmax xx
70.71
20
8894)57,3(.q
n
MSE),k(q 05
g
w a
Take q.05(3,60) from the table.
Population
Sales - City 1
Sales - City 2
Sales - City 3
Mean
577.55
653
698.65
w minmax xx
Tukey Multiple Comparisons
Excel Tukey and Fisher LSD method
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Excel Tukey and Fisher LSD method
Xm15-01
Fishers LDS
Bonferroni adjustments
a = .05
a = .05/3 = .0167
Multiple Comparisons
LSD Omega
Treatment Treatment Difference Alpha = 0.05 Alpha = 0.05
Convenience Quality -75.45 59.72 71.70
Price -31.1 59.72 71.70
Quality Price 44.35 59.72 71.70
Multiple Comparisons
LSD Omega
Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05
Convenience Quality 75 45 73 54 71 70
http://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xlsRecommended