two way anova 4 uploaded

7/31/2019 Two Way ANOVA 4 Uploaded

http://slidepdf.com/reader/full/two-way-anova-4-uploaded 1/75

1

Two-Way ANOVA

Quantitative Research Methods II

EDMS 646

Hong Jiao

Fall 2012



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



µ 0 Xcrit

Null distribution

Rejection regionNon-Rejection regionRejection

region

Non-Null distribution

H 0: μ1-μ2 =0



µ 0 Xcrit

µ A

Null distribution Alternative distribution

Acceptanceregion

Rejectionregion

H 0: μ1-μ2 =<0



µ 0Xcrit

µ A

Non-rejection regionRejection region

Null distribution

Alternative distribution

H 0: μ1-μ2 =>0



6

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



7

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



8

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?



9

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.



10

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.



11

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



12

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..



13

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



14

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.



15

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.



16

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.



17

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



18

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



19

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



20

Graphical representation of ordinal interaction

Ordinal Interactio

0

20

40

60

80

100

B1 B2 B3

IV - B

D V - A A1

A2



21

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



22

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



23

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



24

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



25

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



26

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



27

Graphical representation of no interaction

No Interactio

0

20

40

60

80

100

B1 B2 B3

IV - B

D V - A A1

A2



28

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



29

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



30

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



31

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



32

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



33

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



34

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



35

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



36

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



37

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



38

• 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.



39

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



40

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



41

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



42

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



43

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



44

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



45

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



46

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



47

Finding Critical Statistics

• df for numerator • df for denominator

• Significance level

• F table – Lomax 2



48

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



49

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



50

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



51

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




52

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



53


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



54

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



55

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



56

Two-way ANOVA: example data layout

T ANOVA l d l



57


T ANOVA l d t l t



58


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



59


Estimated Marginal Means

1. mod_A


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


1.667 .236 1.090 2.243

2.333 .236 1.757 2.910

age_B

age 4

age 8



3. mod_A * age_B


.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



T ANOVA l t t



60

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



61


T ANOVA l d t l t



62


G*P M i Eff t df J 1



63

G*Power-Main Effect df=J-1

G*Power-Interaction Effect df=(J-1)(K-1)



64

G Power Interaction Effect df (J 1)(K 1)



65

Measure of Association

Effect Size

Two way ANOVA



66

• 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



67

• 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



68

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



69

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)



70

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



71

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



72

Highlights• Interaction

• Variance decomposition compared with one-way

ANOVA

d i bl l



73

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



74

Dependent Variable: Current Salary

Source Sum of Squares

df Mean Square F

Degree ? 3 250 ?

Error ? 45 ?

two way anova 4 uploaded

Documents