lec 1: an introduction to anova - s ugauss.stat.su.se/gu/ep/lec1.pdf · ying li stockholm...
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Lec 1: An Introduction to ANOVA
Ying LiStockholm University
October 31, 2011
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Three end-aisle displays
Which is the best?
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Design of the Experiment
Identify the stores of the similar size and type.
The displays are randomly assigned to use.
First Principle
Randomization
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Design of the Experiment
Identify the stores of the similar size and type.
The displays are randomly assigned to use.
First Principle
Randomization
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Observations
3 levels
5 replicates
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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1 2 3
67
89
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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The Analysis of Variance
a level of the factors (a treatments)
n replicates
N = a× n runs
Completely randomized design
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Statistical Model
yij = µ+ τi + εij
i = 1, · · · a
j = 1, · · · n
µ : overall mean
τi : the effect of ith treatment
εij ∼ N(0, σ2),i .i .d .∑ai τi = 0
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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The Analysis of Variance
We are interested in testing the equality of treatment means
E (yi .) = µ+ τi = µi
i = 1, · · · a The appropriate hypothesis are
H0 : µ1 = µ2 = · · · = µa
H1 : µi 6= µj
for at least one pair (i , j)
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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If H0 is true, the distribution of F0 is Fa−1,a(n−1)
Reject H0 if F0 > Fα,a−1,a(n−1)
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Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Equation for manual calculation
Balanced Data:
SST =a∑
i=1
n∑j=1
y2ij −y2..N
SSTreatment =1
n
a∑i=1
y2i . −y2..N
SSE = SST − SSTreatment
Unbalanced Data
SST =a∑
i=1
ni∑j=1
y2ij −y2..N
SSTreatment =a∑
i=1
y2i .ni− y2..
N
SSE = SST − SSTreatment
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Example
y1. = 5.73,y2. = 6.24, y3. = 8.32,y.. = 6.76SST = 21.93SSTreatment = 18.78SSE = SST − SSTreatment = 3.15
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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DF SS MS F P
Treatment 2 18.78 9.39 35.77 <0.001
Error 12 3.15 0.26
Total 14 21.93
Which ones cause the differ?
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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DF SS MS F P
Treatment 2 18.78 9.39 35.77 <0.001
Error 12 3.15 0.26
Total 14 21.93
Which ones cause the differ?
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Multiple Comparison Methods
We know the end-aisle display different than others. We mightsuspect the first and the second are different. Thus one reasonabletest hypothesis would be
H0 : µ1 = µ2
H1 : µ1 6= µ2
or,H0 : µ1 − µ2 = 0
H1 : µ1 − µ2 6= 0
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Contrasts
In general, a contrast is a linear combination of the parameters ofthe form
Γ =a∑
i=1
ciµi
where ci are the contrast constants, and∑a
i=1 ci = 0.For unbalance design
∑ai=1 cini = 0.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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The hypothesis can be expressed in the terms of contrasts
H0 :a∑
i=1
ciµi = 0
H1 :a∑
i=1
ciµi 6= 0
In our example, c1 = 1,c2 = −1, c3 = 0.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Contrast Tests
Testing hypothesis involving contrast can be done in two basicways
1 t-test:
t0 =
∑ai=1 ci yi .√
MSEn
∑ai=1 c
2i
2 F-test:
F0 =MScMSE
=SSc/1
MSE
where
SSc =(∑a
i=1 ci yi .)2
1n
∑ai=1 c
2i
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Orthogonal Contrasts
Two contrasts∑a
i=1 = ciµi = 0 and∑a
i=1 = diµi = 0are orthogonal if
a∑i=1
cidi = 0
or in unbalanced case if
a∑i=1
nicidi = 0
We can specify a− 1 orthogonal contrasts. Tests performed onorthogonal contrasts are independent.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Example
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Pairwise Comparison
The null hypothesis are
H0 : µi = µj , i 6= j
Tukey’s test
Make use of the distribution of studentized range statistics
q =ymax − ymin√MSE2 ( 1
ni+ 1
nj)
Appendix Table 7 contains upper percentiles for q.The difference d between two averages is significant if
|d | > qα(a, f )√2
√MSE (
1
ni+
1
nj)
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Pairwise Comparison
The null hypothesis are
H0 : µi = µj , i 6= j
Tukey’s test
Make use of the distribution of studentized range statistics
q =ymax − ymin√MSE2 ( 1
ni+ 1
nj)
Appendix Table 7 contains upper percentiles for q.The difference d between two averages is significant if
|d | > qα(a, f )√2
√MSE (
1
ni+
1
nj)
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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The fisher least significant difference method (LSD)
The fisher least significant difference method procedure use T testfor testing
H0 : µi = µj , i 6= j
The test statistics is:
t0 =yi . − yj .√
MSE ( 1ni
+ 1nj
)
The quantity: least significant difference (LSD)
LSD = tα/2,N−a
√MSE (
1
ni+
1
nj)
If|yi . − yj .| > LSD
we conclude that the population means µi and µj differ.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Questions:1
Why in some situation overall F test of ANOVA is significant, butthe pairwise comparison fails to reveal the differences?
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Questions:2
Many tests available:Duncan, Student-Newman-Keuls, REGWQ, Bonferroni, Sidak,Scheff, ·Which pairwise comparison method do I use?
unfortunately, no-clear cut answer.
Fisher’s LSD procedure, only apply if F test is significant
Tukey’s method control overall error rate, many statisticiansprefer.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Questions:2
Many tests available:Duncan, Student-Newman-Keuls, REGWQ, Bonferroni, Sidak,Scheff, ·Which pairwise comparison method do I use?
unfortunately, no-clear cut answer.
Fisher’s LSD procedure, only apply if F test is significant
Tukey’s method control overall error rate, many statisticiansprefer.
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Dunnett’s Test
Often one of the treatments is a control treatment, and we wantto compare the other treatments with the control treatment. Wewant to test a− 1 hypotheses:
H0 : µj = µ1
wherei = 2, · · · , a
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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Dunnett’s Test
For the ith treatment, H0 is rejected if
|yi − y1| > dα(a− 1, f )
√MSE (
1
ni+
1
nj)
Appendix 8 gives upper percentiles for dα(a− 1, f )
Ying Li Stockholm University Lec 1: An Introduction to ANOVA
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One Example
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One Example
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One Example
Ying Li Stockholm University Lec 1: An Introduction to ANOVA