anova (1)
TRANSCRIPT
Analysis of VarianceAnalysis of Variance
Chapter 15
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.
• 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
• 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
• 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
One Way Analysis of Variance
Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532
Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 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
• 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
H0: 1 = 2= 3
H1: At least two means differ
To build the statistic needed to test thehypotheses use the following notation:
• Solution
Defining the Hypotheses
Independent samples are drawn from k populations (treatments).
1 2 kX11
x21
.
.
.Xn1,1
1
1x
n
X12
x22
.
.
.Xn2,2
2
2x
n
X1k
x2k
.
.
.Xnk,k
k
kx
n
Sample sizeSample mean
First observation,first sample
Second observation,second sample
X is the “response variable”.The variables’ value are called “responses”.
Notation
Terminology
• In the context of this problem…Response variable – weekly salesResponses – 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.
Two types of variability are employed when testing for the equality of the population
means
The rationale of the test statistic
Graphical demonstration:Employing two types of variability
20
25
30
1
7
Treatment 1 Treatment 2 Treatment 3
10
12
19
9
Treatment 1Treatment 2Treatment 3
20
161514
1110
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 conclusionabout the population means.
A small variability withinthe samples makes it easierto draw a conclusion about the population means.
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.
• 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
SSTIn our example treatments arerepresented by the differentadvertising strategies.
Variability between sample means
2k
1jjj )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 toone 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.
• Solution – continuedCalculate SST
2k
1jjj
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...nnxn...xnxn
X
Sum of squares for treatments (SST)
Is SST = 57,512.23 large enough to reject H0 in favor of H1?
See next.
Sum of squares for treatments (SST)
• 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
• 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 differencesbetween sales in city j and thesample mean of city j (over all the three cities).
Within samples variability
• Solution – continuedCalculate SSE
Sum of squares for errors (SSE)
k
jjij
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
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)
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
12.756,2813
23.512,571
k
SSTMST
Calculation of MSEMean Square for Error
45.894,8360
50.983,509
kn
SSEMSE
Calculation of the test statistic
23.3
45.894,8
12.756,28
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.
And finally the hypothesis test:
H0: 1 = 2 = …=k
H1: At least two means differ
Test statistic:
R.R: F>F,k-1,n-k
MSEMST
F
The F test rejection region
The F test
Ho: 1 = 2= 3
H1: At least two means differ
Test statistic F= MST MSE= 3.2315.3FFF:.R.R 360,13,05.0knk 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.
23.317.894,812.756,28
MSEMST
F
-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
Excel single factor ANOVA
SS(Total) = SST + SSE
Anova: Single Factor
SUMMARYGroups Count Sum Average Variance
Convenience 20 11551 577.55 10775.00Quality 20 13060 653.00 7238.11Price 20 12173 608.65 8670.24
ANOVASource of Variation SS df MS F P-value F crit
Between Groups 57512 2 28756 3.23 0.0468 3.16Within Groups 506984 57 8894
Total 564496 59
Xm15-01.xls
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.
Factor ALevel 1Level2
Level 1
Factor B
Level 3
Two - way ANOVATwo factors
Level2
One - way ANOVASingle factor
Treatment 3 (level 1)
Response
Response
Treatment 1 (level 3)
Treatment 2 (level 2)
• 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
• 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
• 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.
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.
Treatment 4
Treatment 3
Treatment 2
Treatment 1
Block 1Block3 Block2
Block all the observations with some commonality across treatments
Randomized Blocks
TreatmentBlock 1 2 k Block mean
1 X11 X12 . . . X1k2 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
• 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
Calculating the sums of squares
• Formulai for the calculation of the sums of squares
TreatmentBlock 1 2 k Block mean
1 X11 X12 . . . X1k2 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(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=
...)()(...
)()(...)()()(
22
21
222
212
221
211
XxXX
XxXxXxXxTotalSS
kk
Calculating the sums of squares
• Formulai for the calculation of the sums of squares
TreatmentBlock 1 2 k Block mean
1 X11 X12 . . . X1k2 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(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
22kk2
21kk1
22222
21212
22121
21111
To perform hypothesis tests for treatments and blocks we need
• Mean square for treatments• Mean square for blocks• Mean square for error
Mean Squares
1kSST
MST
1bSSB
MSB
1bknSSE
MSE
Test statistics for the randomized block design ANOVA
MSEMST
F
MSEMSB
F
Test statistic for treatments
Test statistic for blocks
• Testing the mean responses for treatments
F > F,k-1,n-k-b+1
• Testing the mean response for blocks
F> F,b-1,n-k-b+1
The F test rejection regions
• 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
• 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
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
ANOVASource of Variation SS df MS F P-value F critRows 3848.7 24 160.36 10.11 0.0000 1.67Columns 196.0 3 65.32 4.12 0.0094 2.73Error 1142.6 72 15.87
Total 5187.2 99
Analysis of VarianceAnalysis of Variance
Chapter 15 - continued
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
• 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
• Solution– The hypotheses tested are:
H0: 1= 2= 3= 4= 5= 6
H1: At least two means differ.
Attempting one-way ANOVA
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 marketing emphasis and media usage is employed.
• Solution
Attempting one-way ANOVA
• 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
• 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
• The current experimental design cannot provide answers to these questions.
• A new experimental design is needed.
Two-way ANOVA (two factors)
Two-way ANOVA (two factors)
City 1sales
City3sales
City 5sales
City 2sales
City 4sales
City 6sales
TV
Newspapers
Convenience Quality Price
Are there differences in the mean sales caused by different marketing strategies?
Factor A: Marketing strategy
Fact
or B
: Ad
verti
sing
med
ia
Test whether mean sales of “Convenience”, “Quality”, and “Price” significantly differ from one another.
H0: Conv.= Quality = Price
H1: At least two means differ
Calculations are based on the sum of square for factor ASS(A)
Two-way ANOVA (two factors)
Two-way ANOVA (two factors)
City 1sales
City 3sales
City 5sales
City 2sales
City 4sales
City 6sales
Factor A: Marketing strategy
Fact
or B
: Ad
verti
sing
med
ia
Are there differences in the mean sales caused by different advertising media?
TV
Newspapers
Convenience Quality Price
Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another.
H0: TV = Newspapers
H1: The means differ
Calculations are based onthe sum of square for factor BSS(B)
Two-way ANOVA (two factors)
Two-way ANOVA (two factors)
City 1sales
City 5sales
City 2sales
City 4sales
City 6sales
TV
Newspapers
Convenience Quality Price
Factor A: Marketing strategy
Fact
or B
: Ad
verti
sing
med
ia
Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium?
City 3sales
TV
Quality
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)
Graphical description of the possible relationships between factors A and B.Graphical description of the possible relationships between factors A and B.
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 ANo difference between the levels of factor B
Difference between the levels of factor A, anddifference between the levels of factor B; nointeraction
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 n s e
M Re e sa pn o n s e
M Re e sa pn o n s e
M Re e sa pn o n s e
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( 22 xxxx NewspaperTV
b
1j
2jiij
a
1i
)x]B[x]A[x]AB[x(r)AB(SS
r
kijijk
b
j
a
i
ABxxSSE1
2
11
)][(
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 > F,a-1 ,n-ab F > F, b-1, n-ab
• Test for interaction between factors A and B
F= MS(AB)MSE
Rejection region: F > Fa-1)(b-1),n-ab
SS(A)/(a-1) SS(B)/(b-1)
SS(AB)/(a-1)(b-1)
SSE/(n-ab)
Required conditions:
1. The response distributions is normal2. The treatment variances are equal.3. The samples are independent.
• Example 15.3 – continued( Xm15-03)
F tests for the Two-way ANOVA
Convenience Quality Price
TV 491 677 575TV 712 627 614TV 558 590 706TV 447 632 484TV 479 683 478TV 624 760 650TV 546 690 583TV 444 548 536TV 582 579 579TV 672 644 795
Newspaper 464 689 803Newspaper 559 650 584Newspaper 759 704 525Newspaper 557 652 498Newspaper 528 576 812Newspaper 670 836 565Newspaper 534 628 708Newspaper 657 798 546Newspaper 557 497 616Newspaper 474 841 587
• Example 15.3 – continued– Test of the difference in mean sales between the three marketing
strategiesH0: conv. = quality = price
H1: At least two mean sales are different
F tests for the Two-way ANOVA
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
Factor A Marketing strategies
• Example 15.3 – continued– Test of the difference in mean sales between the three
marketing strategiesH0: conv. = quality = price
H1: At least two mean sales are different
F = MS(Marketing strategy)/MSE = 5.33
Fcritical = F,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
• Example 15.3 - continued– Test of the difference in mean sales between the two
advertising mediaH0: TV. = Nespaper
H1: The two mean sales differ
F tests for the Two-way ANOVA
Factor B = Advertising media
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
• Example 15.3 - continued– Test of the difference in mean sales between the two
advertising mediaH0: TV. = Nespaper
H1: The two mean sales differ
F = MS(Media)/MSE = 1.42 Fcritical = Fa-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
• Example 15.3 - continued– Test for interaction between factors A and B
H0: TV*conv. = TV*quality =…=newsp.*price
H1: At least two means differ
F tests for the Two-way ANOVA
Interaction AB = Marketing*Media
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
• Example 15.3 - continued– Test for interaction between factor A and B
H0: TV*conv. = TV*quality =…=newsp.*price
H1: At least two means differ
F = MS(Marketing*Media)/MSE = .09
Fcritical = Fa-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
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
• 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
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 sp2.
• We can conclude that i and j differ (at % significance level if |i - j| > LSD, where
kn.f.d
)n1
n1
(MSEtLSDji
2
Experimentwise Type I error rate (E)(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 It iscalculated by
E = 1-(1 – )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 () , to secure a pre-determined overall E
• The procedure:– Compute the number of pairwise comparisons (C)
[C=k(k-1)/2], where k is the number of populations.– Set = E/C, where E is the true probability of making at
least one Type I error (called experimentwise Type I error).– We can conclude that i and j differ (at /C% significance
level if
kn.f.d
)n1
n1
(MSEtji
)C2(ji
Bonferroni Adjustment
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
)n1
n1
(MSEt
2/05.
ji
2
Fisher and Bonferroni Methods
• Solution (the Bonferroni adjustment)– We calculate C=k(k-1)/2 to be 3(2)/2 = 3.– We set = .05/3 = .0167, thus t.01672, 60-3 = 2.467 (Excel).
54.73)20/1()20/1(8894467.2
)n1
n1
(MSEtji
2
Again, the significant difference is between 1 and 2.
35.4465.6080.653xx
10.3165.60855.577xx
45.750.65355.577xx
32
31
21
Fisher and Bonferroni Methods
• The test procedure:– Find a critical number as follows:
gnMSE
),k(q
k = the number of samples =degrees of freedom = n - kng = number of observations per sample (recall, all the sample sizes are the same) = significance levelq(k,) = a critical value obtained from the studentized range table
Tukey Multiple Comparisons
If the sample sizes are not extremely different, we can use the above procedure with ng calculated as the harmonic mean ofthe sample sizes. k21 n1...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 to conclude that max > min .
minmax xx
minmax xx
Tukey Multiple Comparisons
City 1 vs. City 2: 653 - 577.55 = 75.45City 1 vs. City 3: 608.65 - 577.55 = 31.1City 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
70.7120
8894)57,3(.q
nMSE
),k(q 05g
Take q.05(3,60) from the table.
Population
Sales - City 1Sales - City 2Sales - City 3
Mean
577.55653698.65
minmax xx
Tukey Multiple Comparisons
Excel – Tukey and Fisher LSD method
Xm15-01
Fisher’s LDS
Bonferroni adjustments
= .05
= .05/3 = .0167
Multiple Comparisons
LSD OmegaTreatment Treatment Difference Alpha = 0.05 Alpha = 0.05Convenience Quality -75.45 59.72 71.70
Price -31.1 59.72 71.70Quality Price 44.35 59.72 71.70
Multiple Comparisons
LSD OmegaTreatment Treatment Difference Alpha = 0.0167 Alpha = 0.05Convenience Quality -75.45 73.54 71.70
Price -31.1 73.54 71.70Quality Price 44.35 73.54 71.70