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Two-Way ANOVA
Quantitative Research Methods II
EDMS 646
Hong Jiao
Fall 2012
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Identify α, 1-α, β, 1-β under different
hypothesis testing (Practice)1. Null hypothesis: H0: μ1- μ2 = 0
Alternative hypothesis: H1
: μ1
- μ2
≠ 0
2. Null hypothesis: H0: μ1- μ2 ≤ 0
Alternative hypothesis: H1: μ1- μ2 > 0
3. Null hypothesis: H0: μ1- μ2 ≥ 0
Alternative hypothesis: H1: μ1- μ2 < 0
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µ 0 Xcrit
Null distribution
Rejection regionNon-Rejection regionRejection
region
Non-Null distribution
H 0: μ1-μ2 =0
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µ 0 Xcrit
µ A
Null distribution Alternative distribution
Acceptanceregion
Rejectionregion
H 0: μ1-μ2 =<0
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µ 0Xcrit
µ A
Non-rejection regionRejection region
Null distribution
Alternative distribution
H 0: μ1-μ2 =>0
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Review-One Way ANOVA• The analysis of variance tests for differences between
– groups – individuals
• One can reasonably conclude there is no difference between the means when
– the observed F ratio is smaller than the critical value
– the observed F ratio is greater than the critical value – the MS-within is smaller than MS-between
– the SS-within is greater than SS-between
• Rejection of the null hypothesis in the one-way ANOVAleads to the conclusion that
– each population mean is different from every other populationmean
– all of the sample means are significantly different
– not all population means are equal to each other
– the alternate hypothesis is also rejected
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Outline• Two-way ANOVA
– Purpose
– Interaction
– Model setup
– Hypothesis testing
– Test statistics
– Additivity
– Assumptions
– Variance decomposition
– Types of sum of squares – SPSS runs
– Power analysis
– Measures of association
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Scenario 1
A researcher is interested in whether graduatestudents with low, medium, and high IQ
differed in terms of their achievement. He
classified a random sample of 300 graduate
students into low, medium, and high IQ groups
and test them on an achievement test. Which
statistical procedure should the researcher
employ to answer his research question?
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Scenario 2• A researcher is interested in investigating the effect of different levels of
familiarity of text and reading perception on the level of students readingcomprehension. To investigate the research question, the researcher randomly selected 25 students and randomly assigns them to one of the four treatment conditions.
• Familiar material which the teacher tells the student is difficult tounderstand (F/D)
• Familiar material which the teacher tells the student is easy tounderstand (F/E)
• Unfamiliar material which the teacher tells the student is difficult tounderstand (UF/D)
• Unfamiliar material which the teacher tells the student is easy to
understand (UF/E)• As a result of experimental mortality, the number of students within each of
the treatment differs. Following the implementation of the treatments, thestudents’ comprehension of the passage that they read was assessed. Thefour variables in the data file are the reading comprehension scores for the
four groups.
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Reanalyzing Scenario 2• A researcher is interested in investigating the effect of different levels of
familiarity of text and reading perception on the level of students readingcomprehension. To investigate the research question, the researcher randomly selected 25 students and randomly assigns them to one of the four treatment conditions.
• Familiarity of material• Familiar material
• Unfamiliar material• Easy/difficult told by teacher
• the teacher tells the student is difficult to understand (F/D)• the teacher tells the student is easy to understand (F/E)
• As a result of experimental mortality, the number of students within each of the treatment differs. Following the implementation of the treatments, thestudents’ comprehension of the passage that they read was assessed. Thefour variables in the data file are the reading comprehension scores for thefour groups.
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Purpose of two-way ANOVA
• Test equality of means related to each of twoIVs
• Test the presence of interaction effects
between two IVs• Such effect deals with the relative magnitude
of cell means
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Two-Way ANOVA (2X2)
B Margin Meansof A
1 2
A 1 μ11 μ12 μ1.
2 μ21 μ22 μ2.
MarginMeans of B
μ.1 μ.2 μ..
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Example
• Investigate whether either of 2 factors affects the DV-post
operative recovery rate of children who receive tonsillectomies.• Two IVs:
– A: nature of pre-operative program with two levels
• 1). Anxiety reduction
• 2). Procedure orientation
– B: time of pre-operative program also with two levels• 1). Early treatment
• 2). Delayed treatment
• Children are randomly assigned to one of the four joint levels of the two IVs.
– A=1, B=1: anxiety reduction/early treatment
– A=1, B=2: anxiety reduction/delayed treatment
– A=2, B=1: procedure orientation/early treatment
– A=2, B=2: procedure orientation/delayed treatment
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Use Two-Way ANOVA• Mean post operative recovery rates under two pre-
operative programs – Main effect A hypothesis: H0: μ1.= μ2.
• Mean post operative recovery rates under two timeframes for pre-operative programs
– Main effect B hypothesis: H0: μ.1= μ.2
• Mean postoperative recovery rates under various four joint levels of pre-operative program and time of pre-operative program (interaction hypothesis betweentreatments A and B H0: μ jk – μ j.- μ.k + μ.. =0 for all
levels of j and k.
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Interaction effects
• When the values of the cells means can not be
defined in terms of corresponding marginal
means and grand means μ jk ≠ μ j. + μ.k - μ..
• When the mean differences among the levels of
one independent variable A differ from one
level of a second independent variable B to
another.
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Presence of InteractionB Margin Means
of A
1 2
A 1 5 7 6
2 3 11 7
MarginMeans of B
4 9 6.5
Difference between the mean of level one of A and level two
of A at level one of B is equal to μ11-μ21=5-3=2
Difference between the mean of level one of A and level two
of A at level two of B is equal to μ12-μ22=7-11=-4
There is interaction.
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No Presence of InteractionB Margin Means
of A
1 2
A 1 10 2 6
2 11 3 7
MarginMeans of B
10.5 2.5 6.5
Difference between the mean of level one of A and level two of A
at level one of B is equal to μ11-μ21=10-11=-1
Difference between the mean of level one of A and level two of A
at level two of B is equal to μ12-μ22=2-3=-1
Both are equal to difference in marginal means μ1. -μ2. =6-7=-1
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Identification of Main Effects and Interaction Effects-Lomax p88 (2X2)
a c b e
d f g h
A1 A2
B1
B2
A1 A1 A1
A1 A1 A1 A1
A2 A2 A2
A2 A2 A2 A2
B1B1
B1
B1
B1
B1 B1
B2
B2B2
B2B2 B2 B2
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Types of Interaction
B
1 2 3
A 1 10 30 52 30 90 25
Mean at level one of A is smaller than the mean at level two of A
for each of the levels of B- the order of the difference are the sameAt level one of B: 10<30
At level two of B: 30<90
At level three of B: 5<25
1. Ordinal interaction: the order of the mean magnitude on one
IV does not change across the various levels of the 2nd IV
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Graphical representation of ordinal interaction
Ordinal Interactio
0
20
40
60
80
100
B1 B2 B3
IV - B
D V - A A1
A2
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Types of Interaction
B
1 2 3
A 1 10 70 20
2 30 50 40
Mean at level 1 of A is smaller than the mean at level 2 of A for
level 1 and 3 of B but there is a reversal in the order of mean
magnitude for level 2 of B
At level one of B: 10<30
At level two of B: 70>50
At level three of B: 20<40
2. Disordinal interaction: the order of the mean magnitude on
one IV changes across the various levels of the 2nd IV
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Graphical representation of disordinal interaction
0
10
20
30
40
50
6070
80
B1 B2 B3
D V
- A
IV - B
Disordinal Interaction
A1
A2
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Graphical representation of disordinal interaction
Change from Disordinal to Ordinal Interaction
0
10
20
30
40
50
60
70
80
A1 A2
IV-A
D
B1
B2
B3
IV, which is placed on the baseline in the graph may affect
whether the line segments cross.
If variable A is put on the baseline, the interaction changed
from disordinal to ordinal
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Which IV should be placed on the baseline of the
graph for describing the interaction effects?
• If both IV are manipulated variables, either IV
may be appropriately be placed on the baseline
• If one of the IVs is not a manipulated variablesuch as gender, the non-manipulated IV should
be placed on the baseline of the graph
• Factor with most levels put on x axis
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No InteractionB Margin
Means of A
1 2 3
A 1 40 20 60 40
2 70 50 90 70
Margin
Means of B
55 35 75 55
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No Interaction
If there is no interaction, graphs of cell means will result in parallel
line segments, no matter which IV is placed on the baseline
Every cell mean may be defined in terms of its marginal means and
the grand mean: μ jk = μ j. + μ.k - μ..
μ 11= μ1. + μ.1- μ.. =40+55-55=40
μ 12= μ1. + μ.2- μ.. =40+35-55=20
μ 13= μ1. + μ.3- μ.. =40+75-55=60
μ 21= μ2. + μ.1- μ.. =70+55-55=70
μ 22= μ2. + μ.2- μ.. =70+35-55=50
μ 23= μ2. + μ.3- μ.. =70+75-55=90
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Graphical representation of no interaction
No Interactio
0
20
40
60
80
100
B1 B2 B3
IV - B
D V - A A1
A2
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Categories of two-way ANOVA
• Two-way factorial ANOVA:
– Both IVs are manipulated by the researcher
• Two-way randomized blocks ANOVA:
– One of the IVs is not manipulated by the
researcher.
– The non-manipulated IV is called blockingvariable
h A O A
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Why use two-way ANOVA
instead of two one-way ANOVA?
• Possible to increase power by reducing error variance – Score variability on DV with cells will be less than the
score variability within each row or column
•Economy of efforts – Simultaneous consideration of two IVs with the same
subjects is more economical than using separate samplesof subjects for each IV
• Possible to investigate interaction effects which can
not be assessed with one-way ANOVA• When on IV is a blocking variable, may be able to
increase the level of comparability of subjectsacross the levels of the manipulated IV
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Notations used in Two-Way ANOVA
IV B
1 2 … K • jn • jY • j µ
1
111Y
211Y
1111nY
112Y
212Y
1212nY
k Y 11
k Y 21
k n k Y 11
•1n •1Y
•1 µ
2
1 2 1Y
221Y
2121nY
122Y
222Y
2222nY Cell(j,k)
k Y 12
k Y 22
k n k Y 22 •.2n •2Y •2 µ
…
ijk Y
jk Y
jk µ
2
jk σ
jk n
IV A
J
11 J Y
12 J Y
11 J n J Y
21 J Y
22 J Y
22 J n J Y
Jk Y 1
Jk Y 2
Jk n Jk Y
• J n
• J Y • J
µ
k n• 1•
n 2•n K n• N
k Y •
1•Y 2•Y K Y • ••Y
k • µ 1• µ
2• µ K •
µ ••
µ
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ijk jk k jijk Y ε αβ β α µ ++++= •••• )(
where• Y ijk is the score for the ith experimental unit in the jk th
treatment combination,• µ.. is the grand mean of the scores,
• α j. is the treatment effect of the jth level of the first factor,• β.k is the treatment effect of the k th level of the second
factor,
• (αβ) jk is the interaction effect of α. j and βk ., and
• εijk is the error effect for Y ijk . εijk ~ N (0, σ2
ε).
i = (1, …. , n); j = (1, … , J ); k = (1, …. , K )
Two-way ANOVA model
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jk ijk ijk Y µ ε −=
µ µ µ µ αβ +−−= •• k j jk jk )(
•••• −= µ µ β k k
•••• −= µ µ α j j
Computationally
• Usually, both α j and βk are fixed effects.
• Accordingly, (αβ) jk is a fixed effect as well.
St ti ti l H th i T ANOVA
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Statistical Hypotheses in Two-way ANOVA
1. Marginal means of A are equal,
H0: μ1.= μ2. =, …, = μJ. or H 0: α1 = α2 = …. = α j = 0 for all j
H1: μ j. ≠ μ j.’ for at least one pair of levels of A
Or H 1: α j 0 for at least one j .
2. Marginal means of B are equal,
H0: μ.1= μ.2 =, …, = μ.K or H 0: β1 = β2 = …. = βk = 0 for all k
H1: μ.k ≠ μ.k ’ for at least one pair of levels of B
Or H 1: β.
k ≠ 0 for at least one k .
3. Cell means may be defined in terms of marginal means and grand means,
H0: μ jk.= μ j. + μ.k - μ.. for all j, k
H 1: μ jk. ≠ μ j. + μ.k - μ.. for at least one cell jk
H0: μ jk.- μ j. - μ.k + μ.. =0 for all j and k
H 1: μ jk. -μ j. - μ.k +μ.. ≠ 0 for at least one cell jk
H0: (αβ) jk.= 0 for all j and k
H 1: (αβ) jk ≠. 0 for at least one cell jk
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Expected values of Mean Squares in Two-way ANOVA
1)( 1
2
2
−+=
∑=
•
J
n
MS E
J
j
j j
A
α
σ ε
1)(1
2
2
−+=
∑=
•
K
n
MS E
K
k
k k
B
β
σ ε
)1)(1(
)(
)(1
2
12
−−
+=∑∑
= =
K J
n
MS E
J
j
jk jk
K
k
AB
αβ
σ ε
2)( ε σ =with MS E
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F -statistic for testing different hypotheses
with
A
MS
MS F =
1. Main effect A
Used for assessing H0: α j=0 for all j
2. Main effect B
with
B
MS
MS F = Used for assessing H0: βk =0 for all k
3. Interaction effect AB
with
AB
MS
MS F = Used for assessing H0: (αβ) jk =0 for all j, k
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Elements in F statistics
Assessed effects df Sum of Squares Mean Squares
Main Effect A
Main effect B
Interaction AB
Within cell
Total
Adf
withdf
ASS
BSS
total df total SS
Bdf
ABdf ABSS
withSS
A MS
B MS
total MS
with MS
AB MS
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Definitional formulas for the elements in F statistics
1−= J df A
JK N df with −=
2
1
)(
∑=••••
−= J
j j j A
Y Y nSS
1−= N df total
1−= K df B
)1)(1( −−= K J df AB
2
1
)(∑=
•••• −= K
k
k k B Y Y nSS
2
11
)(∑∑=
••••
=
+−−= J
j
k j jk jk
K
k
AB Y Y Y Y nSS
2
1 11
)(∑∑∑= ==
−= J
j
jk ijk
n
i
K
k
with Y Y SS jk
2
1 11
)(∑∑∑=
••
==
−= J
j
ijk
n
i
K
k
total Y Y SS jk
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• Sum of squares and degree of freedoms are additive
when n jk =n j’k’ for all j, j’, k, k’ or more generally
when the cell sample sizes are the same or
proportional.
• Cell sample sizes are proportional when
n jk =(n j.*n.k )/N for all j, k total with AB B A SS SS SS SS SS =+++
total with AB B A df df df df df =+++
However, when the cell sample sizes are not proportional, the
additivity will not hold for sums of squares
No independence among sums of squares. Thus, F statistics are
not independent of one another.
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Necessary conditions required for SSA,
SSB, and SSAB to be independent
• The number of elements in each cell, n jk ,
must be proportional by row and bycolumn, or the relative magnitude of
corresponding cell sample sizes must be thesame, or proportionality is present if andonly if
• n jk =(n j.*n.k )/N for all j,k
• When all the cell sizes are equal, this is aspecial case in which proportionality is present
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B
1 2
A 1 n11=5 n12=20 n1.=25
2 n21=20 n22=80 n2.=100
3 n31=15 n32=60 n3.=75
n.1=40 n.2=160 N=200
Proportional cell sample sizes
B
1 2
A 1 n11=15 n12=20 n1.=35
2 n21=5 n22=70 n2.=75
3 n31=20 n32=70 n3.=90
n.1=40 n.2=160 N=200
when proportionality is present or not present
Disproportional cell sample sizes, n jk
n32=(75*160)/200=60 n32≠(90*160)/200
SStotal ≠ SSA+SSB+SSAB+SSwith
Because these terms are not independent
SStotal=SSA+SSB+SSAB+SSwith
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Assumptions for Two-way ANOVA
1. Observations on DV are random and independent both
within and across treatments• F-test is not robust to violation of this assumption
2. For each joint level of IVs A and B, scores on the DV is
normally distributed
• F -test is robust to the violation of this assumption if the
sample size for each cell is large, n jk ≥20 for all j, k, or with
balanced or nearly balanced design
3. For each joint level of IVs A and B, scores on the DV
have equal variance-homogeneity of variance• H 0: σ11
2 = σ122 = …. = σ jk
2 = σε2
• Robust if the sample sizes for each cell are equal or nearly
equal or with large n
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Example 1-one way ANOVA:
– Research interest: investigating whether three different behavior modification techniques differ in terms of their effectiveness for
reducing the level of ‘acting out’ behavior manifested byhyperactive children.
– Three modification procedures:• Punishment
• Negative reinforcement
• Positive reinforcement
– To investigate the research question, the researcher randomlyselected 12 hyperactive children who are having problems with‘acting out’ behavior and randomly assigned 4 to each of thethree behavior modification treatment regimens
– Following the implementation of the treatment which lasted for 6 weeks, the researcher observed the amount of ‘acting out’
behavior which each child manifested during a 3 hour controlled observation period
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Example 1-Two Way ANOVA: – Research interest: investigating whether three different behavior modification
techniques and difference in child age differ in terms of their effectiveness for reducingthe level of ‘acting out’ behavior manifested by hyperactive children.
– Variable A: Three modification procedures:
Level 1: Positive reinforcement
Level 2: Negative reinforcement
Level 3: Punishment
– Variable B: Age of child
Level 1: four years of age
Level 2: eight years of age
– To investigate the research question, the researcher randomly selected 12 hyperactivechildren who are having problems with ‘acting out’ behavior and randomly assigned 4to each of the three behavior modification treatment regimens based on their age levels
– Following the implementation of the treatment which lasted for 6 weeks, the researcher observed the amount of ‘acting out’ behavior which each child manifested during a 3
hour controlled observation period
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Data layout
IV B
1 2 • jn
• jY
1
01
5.011 =Y
12
5.112 =Y 4 •1Y =1
2
21
5.121 =Y
12
5.122 =Y 4 •2Y =1.5IV A
3
33
0.331 =Y
44
0.432 =Y 4 • J Y =3.5
k n• 6 6 N=12
k Y • 1•Y =1.667 2•Y =2.333 ••Y =2
Computation of the statistics
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Computation of the statistics
2131 =−=−= J df A
62312=×−=−=
JK N df with
14)25.3(4)25.1(4)21(4
)(
222
2
1
=−+−+−=
−= ∑=
••••
J
j
j j A Y Y nSS
111121 =−=−= N df total
121 −=−= K df B
2)12)(13()1)(1( =−−=−−= K J df AB
333.1)2333.2(6)2667.1(6
)(
22
2
1
=−+−=
−=∑=
••••
K
k
k k B Y Y nSS
667.0)2333.25.34(2)2667.15.33(2
)2333.25.15.1(2)2667.15.15.1(2)2667.115.1(2
)2667.115.0(2)(
22
222
2
2
11
=+−−++−−+
+−−++−−++−−+
+−−=+−−= ∑∑=
••••
=
J
j
k j jk jk
K
k
AB Y Y Y Y nSS
2)44()44()5.12(
)5.11()5.12()5.11()33()33()5.11(
)5.12()5.01()5.00()(
222
222222
222
2
1 11
=−+−+−+
−+−+−+−+−+−+
−+−+−=−= ∑∑∑= ==
J
j
jk ijk
n
i
K
k
with Y Y SS jk
18)24(2)22()21()22()21()23(2
)21()22()21()20()(
222222
2222
2
1 11
=−+−+−+−+−+−+
−+−+−+−=−= ∑∑∑=
••
==
J
j
ijk
n
i
K
k
total Y Y SS jk
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ANOVA Summary Table
Source df SS MS F-observed F-critical Sig
A 2 14 7 21 .99F2,6=10.92 0.002
B 1 1.333 1.333 4 .99F1,6=13.75 0.092
AB 2 0.667 0.333 1 .99F2,6=10.92 0.422
Within 6 2 0.333
Total 11 18
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Finding Critical Statistics
• df for numerator • df for denominator
• Significance level
• F table – Lomax Page 472
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Making decision
• Compare F-observed and F-critical – Reject the null when F-observed is equal to
larger than F-critical
– Otherwise fail to reject the null• Use Sig. value
– Reject a null when the Sig value is equal to or
smaller than α
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Decision for the example
• Reject the null for the main effect for variable A because – F-observed=21, > F-critical=10.92 – Sig=0.002, < 0.01
– Rejections implies that at least two of the marginal
means of IV A differ from each other – The population marginal means of acting out behavior
differ for at least two types of behavior modificationreinforcement procedures
– Can not tell which of these three population means differ
based on the F test• Fail to reject the null for main effect B and
interaction between A and B
I i f l
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Interpretation of results• No significant interaction effects,
– main effects are additive – Main effects are statistically independent of one another
– Can make a statement about the constant added benefitsof A1 over A2 regardless of the levels of B
• Significant interaction effects, – main effects are not additive – Main effects are not statistically independent of one
another
– Warning one can not generalize statements about a maineffect for A over all levels of B
– Can not make a blanket statement about the constantadded benefits of A1 over A2, because relationshipamong the levels of factor A depends on the level of factor B
Comparison of statistics for one-way and two-way
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p y y
ANOVA
Source df Sum of Squares Mean Squares F
Between groups 2 14 7 15.75
Within groups 9 4 0.4444
Total 11 18
Source df SS MS F-observed
A 2 14 7 21
B 1 1.333 1.333 4
AB 2 0.667 0.333 1
Within 6 2 0.333
Total 11 18
Comparison of statistics for one-way and two-way
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ANOVA
Sum of Squares
One-way Two-waySS betw = SSA
=with
AB
B
with
SS SS
SS
SS
SStotal = SStotal
SS-within for two-way ANOVA is often smaller than that for one-way
ANOVA.
The larger the difference, the greater the relative power of the two-way
ANOVA for testing the hypothesis regarding the main effect A
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Comparison of statistics for one-way and two-way
ANOVADegree of freedom
One-way Two-waydf betw = df A
=with
AB
B
with
df df
df
df
df total = df total
df-within for one-way ANOVA is always larger than that for two-wayANOVA. This affects both the magnitude of MS-within and the critical
value of F.
The larger the difference, the greater the relative power of the two-way
ANOVA for testing the hypothesis regarding the main effect A
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Type of Sum of Squares How A Main Effect is Tested How B Main Effect is Tested How AB Interaction is Tested
Type I Alone Controlling for A Controlling for A and B
Type II Controlling for B Controlling for A Controlling for A and B
Type III Controlling for B and AB Controlling for A and AB Controlling for A and B
Type IV Controlling for B and AB Controlling for A and AB Controlling for A and B
• Type I SS should only be used in circumstances where theeffects can be ordered in terms of importance.
• In Type II SS, the interaction is tested first. Only if theinteraction is NOT significant can one go on to test the maineffects.
• Type III SS is always recommended because its control of
interaction effects in testing main effects• Type IV SS is recommended when there are missing values
in some cells
Properties for each sums of squares
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Properties for each sums of squares
1. The values for SSAB and SSwithin are the same for the four
different types of sums of squares. Thus the values of theF-statistics for interaction will be the same
2. Whenever all cell sample sizes are proportional
• SS-Type I = SS-Type II
• SS-Type III = SS-Type IV
3. Whenever all cell sample sizes are equal
• All 4 types of SS are the same
• Main effect hypotheses that are tested deal with
unweighted population means
T ANOVA l d l
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Two-way ANOVA: example data layout
T ANOVA l d l
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Two-way ANOVA: example data layout
T ANOVA l d t l t
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Two-way ANOVA: example data layout
Between-S ubjects Factors
4
4
4
age 4 6age 8 6
1.00
2.00
3.00
mod_A
1.002.00
age_B
Value Label N
Descriptive Statistics
Dependent Variable: act_out
.5000 .70711 2
1.5000 .70711 2
1.0000 .81650 4
1.5000 .70711 2
1.5000 .70711 2
1.5000 .57735 4
3.0000 .00000 2
4.0000 .00000 23.5000 .57735 4
1.6667 1.21106 6
2.3333 1.36626 6
2.0000 1.27920 12
age_B
age 4
age 8
Total
age 4
age 8
Total
age 4
age 8Total
age 4
age 8
Total
mod_A
1.00
2.00
3.00
Total
Mean Std. Deviation N
T ANOVA l d t l t
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Two-way ANOVA: example data layout
Estimated Marginal Means
1. mod_A
Dependent Variable: act_out
1.000 .289 .294 1.706
1.500 .289 .794 2.2063.500 .289 2.794 4.206
mod_A1.00
2.003.00
Mean Std. Error Lower Bound Upper Bound
95% Confidence Interval
2. age_B
Dependent Variable: act_out
1.667 .236 1.090 2.243
2.333 .236 1.757 2.910
age_B
age 4
age 8
Mean Std. Error Lower Bound Upper Bound
95% Confidence Interval
3. mod_A * age_B
Dependent Variable: act_out
.500 .408 -.499 1.499
1.500 .408 .501 2.499
1.500 .408 .501 2.499
1.500 .408 .501 2.499
3.000 .408 2.001 3.999
4.000 .408 3.001 4.999
age_Bage 4
age 8
age 4
age 8
age 4
age 8
mod_A1.00
2.00
3.00
Mean Std. Error Lower Bound Upper Bound
95% Confidence Interval
T ANOVA l t t
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Two-way ANOVA: example outputTests of Between-Subjects Effects
De pe nd en t Var iab le: act_o ut
16.000a 5 3.20 0 9 .6 0 0 .0 0 8
4 8 .0 00 1 4 8 .0 0 0 1 44 .0 00 .0 0 0
1 4 .0 00 2 7 .0 0 0 21 .0 0 0 .0 0 21 .333 1 1 .333 4 .0 0 0 .0 9 2
.66 7 2 .333 1 .0 0 0 .422
2.00 0 6 .333
6 6 .0 00 12
1 8 .0 00 1 1
Source
Corre cted Mode l
Intercept
mo d _ Aa g e _ B
mod_A * age_B
Error
Total
Co rrected Tota l
Typ e III Su m
o f Sq u a re s d f Mea n Squ a re F S ig .
R Sq uare d = .889 (Ad justed R Squa red = .796)a .
T ANOVA l d t l t
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Two-way ANOVA: example data layout
T ANOVA l d t l t
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Two-way ANOVA: example data layout
G*P M i Eff t df J 1
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G*Power-Main Effect df=J-1
G*Power-Interaction Effect df=(J-1)(K-1)
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G Power Interaction Effect df (J 1)(K 1)
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Measure of Association
Effect Size
Two way ANOVA
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• Dependent variable: Attitude toward minority group following
the course.• Factor A: Type of beat ( J = 3)
1 = upper class; 2 = middle class; 3 = inner city• Factor B: Length of the course ( K = 3)
1 = 5 hours; 2 = 10 hours; 3 = 15 hours
Data Summary:
1 2 3 mean
1 33.00 35.00 38.00 35.33
2 30.00 31.00 36.00 32.33
3 20.00 40.00 52.00 37.33
mean 27.67 35.33 42.00 35.00
Factor B
Factor A
Two-way ANOVA
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• ANOVA table
S S d f M S F p - va l u
A 1 9 0 . 0 0 0 2 9 5 . 0 0 0 1 . 5 2 0 . 2 3 2
B 1 5 4 3 . 3 3 3 2 7 7 1 . 6 6 61 2 . 3 5 0 . 0 0 0A x B 1 2 3 6 . 3 3 3 4 3 0 9 . 1 6 74 . 9 5 0 . 0 0 3
W i t h i n c e l l2 2 5 0 . 0 0 03 6 6 2 . 5 0 0
T o t a l 5 2 2 0 . 0 0 04 4
Effect Sizes for two way ANOVA design
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Effect Sizes for two-way ANOVA design
(assume α and β are fixed effects)
nK
MS MS J with A J
j
j
))(1(ˆ
1
2 −−=∑
=
α
For the effect of each factor, use
nJ
MS MS K with B K
k
k
))(1(ˆ
1
2 −−=∑
=
β
n
MS MS K J with AB J
j
jk
K
k
))(1)(1()(
1
2
1
−−−=∑∑= =
αβ
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333.415
65
)3(5
)5.620.95)(13(ˆ
1
2==
−−=∑
=
J
j
jα
56.94)3(5
)5.62667.771)(13(ˆ
1
2=
−−=∑
=
K
k
k β
33.1975
)5.62167.309)(13)(13()(
1 1
2 =−−−=∑∑= =
J
j
K
k
jk αβ
Effect Size (partial omega-squared)
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Effect Size (partial omega-squared)
02.03/333.45.62
3/333.4
/ˆˆ
/ˆ
ˆ
1
22
1
2
2=
+=
+
=
∑
∑
=
=
J
j
j
J
j
j
A
J
J
α σ
α
ω
ε
34.03/56.945.62
3/56.94
/ˆˆ
/ˆ
ˆ
1
22
1
2
2 =+
=+
=
∑
∑
=
= K
k
k
K
k
k
B
K
K
β σ
β
ω
ε
26.0)3(3/33.1975.62
)3(3/33.197
/)(ˆ
/)(
ˆ
1 1
22
1 1
2
2=
+=
+
=
∑∑
∑∑
= =
= =
J
j
K
k
jk
J
j
K
k
jk
AB
JK
JK
αβ σ
αβ
ω
ε
22
22
α ε
α
σ σ
σ ω
+=
Simplified formula for estimating ω2 for two-way ANOVA
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p g y
Cohen’s guidelines:
ω 2 = 0.010 is a small association
ω 2 = 0.059 is a medium association
ω 2 = 0.138 is a large association
withtot
withbet
MS SS
MS J SS
+
−−=
)1(ˆ 2ω
withtot
with A A
MS SS
MS J SS
+
−−=
)1(ˆ 2ω
withtot
with B B
MS SS
MS J SS
+
−−
=
)1(ˆ 2ω
withtot
with AB AB
MS SS
MS J SS
+
−−
=
)1(ˆ 2ω
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Highlights• Interaction
• Variance decomposition compared with one-way
ANOVA
d i bl l
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Dependent Variable: Current Salary
Source Sum of Squares
df Mean Square F
Gender ? 1 110 11
Degree ? 3 250 ?
Gender *Degree
300 ? ? ?
Error ? 45 ?
D d V i bl C S l
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Dependent Variable: Current Salary
Source Sum of Squares
df Mean Square F
Degree ? 3 250 ?
Error ? 45 ?
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