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MANOVA: Part 2 Slide 1 of 75

MANOVA: Part 2Profile Analsysis and 2–way MANOVA

Edps/Soc 584 and Psych 594Applied Multivariate Statistics

Carolyn J. AndersonDepartment of Educational Psychology

I L L I N O I SUNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN

c©Board of Trustees, University of Illinois

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

Example: Distributed vs

Massed Practice

Validity of Assumptions

Unbalanced Designs

Conclusions


Outline

Extensions of 1–Way MANOVA:

■ Profile Analysis◆ Two groups◆ Two or more groups

■ 2–way MANOVA◆ Review 2-way ANOVA◆ 2-way MANOVA◆ Example

■ Validity of Assumptions■ Unbalanced Designs

Reading: Johnson & Wichern pages xx–xx.

● Outline

Profile Analysis

● Profile Analysis

● Example of Profiles for Two

Groups

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Profile Analysis

Profile analysis is an extension of 1–way MANOVA involving presponse variables administered to g groups of individuals orcases.

e.g., A battery of personality tests (sub-scales) where the p testscores are measured in the same units (or at least similar orcommensurate units).

Profile Analysis gives you more specific hypotheses to testthan the standard

Ho : µ1 = µ2 = · · · = µg.

Consider p× 1 mean vector for each group

µ′l = (µl1, µl2, . . . , µlp)

and plot the means as “profiles” for each group.

● Outline

Profile Analysis

● Profile Analysis

● Example of Profiles for Two

Groups

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Example of Profiles for Two Groups

Variable

One Two Three . . .p

Mea

nV

alue

5

10

15

q

q

q

q Group 1q

q

q

q

Group 2

● Outline

Profile Analysis

2 Groups

● Profile Analysis: 2 Groups

● Question 2

● Question 3

● Testing Question 1: Parallel

● Question 1 continued

● Finishing Question 1

● Testing Question 2:

Coincident profiles


● Finishing Up Question 2

● Question 3: Flat



● Finishing up Question 3

● Example: WAIS data

● WAIS Profiles

● WAIS: Are Profiles Parallel?


● WAIS: Are Profiles Parallel?● WAIS: Are Profiles

Coincident?● WAIS: Are Profiles

Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Profile Analysis: 2 GroupsThe usual null hypothesis is Ho : µ1 = µ2, but in profileanalysis, we’ll also consider both a more general and morerestrictive hypotheses.

The Three Questions (Steps):

Question 1:

Are the profiles parallel? orAre successive (adjacent)differences between meansequal?i.e., Ho1 :

µ1i − µ1,i−1 = µ2i − µ2,i−1

for all i = 2, . . . , p

Variables

1 2 3 4 5M

ean

Val

ue

Group 1

r

r

r

r

r

µ11

µ12

Group 2

r

r

r

r

r

µ21

µ22

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 2

Assuming that profiles areparallel, are they coincident?or

Are the population meansequal for the two groups?

i.e.,

Ho2 : µ1i = µ2i

for all i = 1, . . . p; that is,

Ho2:µ1 = µ2

Variables

1 2 3 4 5

Mea

nV

alue

Group 1

r

r

r

r

r

µ11

µ12

Group 2

r

r

r

r

r

µ21 =

µ22 =

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 3

Assuming that the profilesare coincident, are they Level(flat)?or

Are all the means equal?

(over groups and variables).

Ho3 : µ11 = µ12 = · · · = µ1p

= µ21 = µ22 = · · · = µ2p

Variables

1 2 3 4 5

Mea

nV

alue

Group 1r r r r rµ1i

Group 2r r r r

rµ2i =

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Testing Question 1: ParallelAssuming

X1j ∼ N (µ1,Σ) and X2j ∼ N (µ2,Σ)

for j = 1, . . . , nl, and independent.

Question 1: Are the profiles parallel?

We can write the hypothesis are Ho1 : Cµ1 = Cµ2 whereC(p−1)×p contrast matrix. e.g.,

C(p−1)×p =

−1 1 0 0 · · · 0 0

0 −1 1 0 · · · 0 0...

......

.... . . · · · · · ·

0 0 0 0 · · · −1 1

What we’re doing is linearly transforming our original pvariables into (p− 1) new variables.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 1 continuedWhen we take

CX =

−X1 +X2

−X2 +X3

...−Xp−1 +Xp

,

which are measured on both groups (populations, etc).

Random variable CX1j ∼ Np−1(Cµ1,CΣC ′),andRandom variable CX2j ∼ Np−1(Cµ2,CΣC ′).

To estimate the covariance matrix Σ, use Spool,

Spool(n1 − 1)S1 + (n2 − 1)S2

n1 + n2 − 2

and the estimate of CΣC ′ equals CSpoolC′.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Finishing Question 1To test Ho1 : Cµ1 −Cµ2 = C(µ1 − µ2) = 0, we can just useHotelling’s T 2 for Two independent samples.

Reject Ho1 if

T 2 = (X̄1−X̄2)′C ′

[(1

n1+

1

n2

)

CSpoolC′]−1

C(X̄1−X̄2) > c2

where

c2 =(n1 + n2 − 2)(p− 1)

n1 + n2 − pF(p−1),(n1+n2−p)(α)

If you reject Ho1 −→ STOP.. . . You can do any follow-up to examine differences.

If you retain Ho2 −→ Conclude profiles are parallel andPROCEED to next question.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Testing Question 2: Coincident profilesAssuming that the the profiles are parallel, are the profilescoincident?

If profiles are parallel, then one will be “above” the other for alli = 1, . . . , p; that is,

µ1j > µ2j for all i = 1, . . . , p

or

µ1j < µ2j for all i = 1, . . . , p

So, profiles will be coincident only if the total “heights” areequal

(µ11 + µ12 + · · ·+ µ1p) = (µ21 + µ22 + · · ·+ µ2p)

1′pµ1 = 1

′pµ2

where 1p is a (p× 1) vector of ones.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 2 continuedThe null hypothesis for question 2 is

Ho2 : 1′µ1 = 1′µ2

We are forming a new variable, 1′X, which is a simple sumand test whether this variable is equal for the two groups.

1′X1j ∼ N1(1

′µ1,1′Σ1) j = 1, . . . , n1

1′X2j ∼ N1(1

′µ2,1′Σ1) j = 1, . . . , n2

We estimate

1′µ1 by 1

′x̄1

1′µ2 by 1

′x̄2

and

1′Σ1 by 1

′Spool1

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Finishing Up Question 2To test

Ho2:1′µ1 = 1

′µ2,

we can do a simple univariate 2 independent sample t-test.

We will reject Ho2 at the α-level if

t =1′(x̄1 − x̄2)

√(

1n1

+ 1n2

)

1′Spool1

> tn1+n2−2(α)

If Reject Ho2 −→ STOP and conclude that the profiles areparallel but not coincident.

If Retain Ho2 −→ PROCEED and test whether profiles are flat.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 3: FlatIf the profiles are coincident, do they all have the same mean?i.e.,

Ho3 : µ11 = µ12 = · · · = µ1p = µ21 = µ22 = · · · = µ2p

We can test this using a contrast matrix C, such as the one weused for Question 1 (testing parallel profiles)

Ho3 : C(µ1+µ2) = 0

Note that we are adding rather than subtracting as we did inHo1.

This testsCµ1 +Cµ2 = 0

(µ12 − µ11) + (µ22 − µ21)

(µ13 − µ12) + (µ23 − µ22)...

(µ1p − µ1,p−1) + (µ2p − µ2,p−1)

= 0

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 3 continuedTesting

Cµ1 +Cµ2 = 0

(µ12 − µ11) + (µ22 − µ21)

(µ13 − µ12) + (µ23 − µ22)...

(µ1p − µ1,p−1) + (µ2p − µ2,p−1)

= 0

Is the same as this, which is what we want to test,

(µ12 + µ22) = (µ11 + µ21)

(µ13 + µ23) = (µ12 + µ22)

......

(µ1p + µ2p) = (µ1,p−1 + µ2,p−1)

Because If the profiles are coincident, then µ1i = µ2i for alli = 1, . . . , p. So if Ho3 is true, then C(µ1 + µ2) = 0.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Question 3 continuedFor Ho3 : C(µ1 + µ2) = 0, flat profiles, we are taking linearcombinations in two ways simultaneously

■ Over variables via the use of C.

■ Over groups via an additive function.

The mean C(µ1 + µ2) is estimated by taking the grand mean

X̄ =

∑n1

j=1 X1j +∑n2

j=1X2j

n1 + n2=

n1X̄1 + n2X̄2

n1 + n2

and the distribution is of CX̄ is

CX̄ ∼ Np−1

(

C(µ1 + µ2),CΣC′(

1

n1 + n2

))

How do we get this?

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Finishing up Question 3

Reject Ho3 : C(µ1 + µ2) = 0 (i.e., that profiles are flat) atα-level if

(n1+n2)X̄′C ′(CSC)−1CX̄ >

(n1 + n2 − 1)(p− 1)

n1 + n2 − p+ 1F(p−1),(n1+n2−p+1(α)

where S is the total sample covariance matrix

S =1

n1 + n2 − 1

n1∑

j=1

(X1j − X̄)(X1j − X̄)′ +n2∑

j=1

(X2j − X̄)(X2j − X̄)′

Note that we use the total sample mean in computing S.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Example: WAIS dataThis example is from Morrison (2005).

Fourty nine elderly men participating in an interdisciplinarystudy of human aging were classified into the diagnosticcategories “senile factor present” and “no senile factor” on thebasis of an intensive psychiatric examination. The WechslerAdult Intelligence Scale (WAIS) was administered to allsubjects by an independent investigator.

Below are mean scores by group on some of the WAISsubtests.

Not Senile (n = 37) Senile (n = 12)

Sub-Test x̄ std dev x̄ std dev

Information 12.566 3.387 8.750 3.251

Similarities 9.486 3.380 5.333 4.271

Arithmetic 11.514 3.363 8.500 3.631

Picture 7.973 1.922 4.750 3.571

Note:

My results differ slightly from Morrison’s. There is either a typoin the text or in the data set. (no way to find out which).

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS Profiles

WAIS Sub-Tests

Information

Similarities

ArithmeticPicture completion

Sub

-tes

tGro

upM

ean

5

10

15

q

q

q

q No Senile Factorq

q

q

q

Senile Factor

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS: Are Profiles Parallel?The within group (pooled) sample covariance matrix:

Spool =1

n1 + n2 − 2((n1 − 1)S1 + (n2 − 1)S2)

=

11.262 8.995 7.164 3.379

8.995 13.019 7.037 2.308

7.164 7.037 11.750 2.639

3.379 2.308 2.639 5.813

Test for parallel profiles: The hypothesis is thatHo : µ1i − µ1,i−1 = µ2,i − µ2,i−1 for i = 2, 3, 4; that is,simultaneously, the following three qualities hold in thepopulation,

µ12 − µ11 = µ22 − µ21

µ13 − µ12 = µ23 − µ22

µ14 − µ13 = µ24 − µ23

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS: Are Profiles Parallel?To test this hypothesis, we need a (3× 4) contrast matrix C

C =

−1 1 0 0

0 −1 1 0

0 0 −1 1

Our hypothesis can now be,expressed as Ho : Cµ1 = Cµ2, orHo : C(µ1 − µ2) = 0. This hypothesis is tested by Hotelling’sT 2, and we need

CSpoolC′ =

6.291 −4.150 −0.944−4.151 10.694 −4.382−0.944 −4.382 12.286

,

(x̄n − x̄y)′ = (3.818, 4.153, 3.014, 3.223),

and(C(x̄n − x̄y))

′ = (0.336,−1.140, 0.209)

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS: Are Profiles Parallel?Putting all of this into our equation for T 2 gives us

T 2 =n1n2

n1 + n2(x̄1 − x̄2)

′C ′(CSpoolC′)−1C(x̄1 − x̄2)

= 1.224

and compare the following statistic to the F distribution,

F =n1 + n2 − p

(n1 + n2 − 2)(p− 1)T 2 =

45

(47)(3)1.224 = 0.391

Since F3,45(α = .05) = 2.8115 is greater than our observedstatistic (or the p-value of F equals .76), we do not reject thehypothesis that the profiles are parallel.

Since we Retained the null, we will PROCEED to test whetherthe profiles are coincident.

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS: Are Profiles Coincident?Test for Equal (coincident) Profiles.

We now test Ho : µ1i = µ2i for i = 1, 2, 3, 4 variables. Since weconcluded that the profiles are parallel, then we can test thishypothesis by testing whether

(µ11 + µ12 + µ13 + µ14) = (µ21 + µ22 + µ23 + µ24)

1′µ1 = 1

′µ2

where 1 is a (4× 1) vector of ones. This is just a univariate, 2independent sample t.

To test Ho : 1′(µ1 − µ2) = 0, we need for each group the sumsof the means over all variables, which are

1′x̄1 = 41.541 and 1

′x̄2 = 27.333

We also need an estimate of the variance of the differencebetween means,

1′Spool1 = 104.891

● Outline

Profile Analysis

2 Groups


● Question 2

● Question 3





Coincident profiles








● WAIS Profiles





Coincident?

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


WAIS: Are Profiles Coincident?

Putting all of these statistics together for our test statisticsgives us

t =41.541− 27.333

√

104.890(

112 + 1

37

) = 4.176

Since t47(α = .025) = 2.012 (or the p-value of t47 = 4.15 is< .001), we reject the null hypothesis. The profiles are notcoincident.

At this point we STOP and should not go on to test whetherprofiles are level (flat).

● Outline

Profile Analysis

2 Groups

Two or More Groups

● Profiles Analysis for g ≥ 2

Groups

● Two or More: Parallel

● GLM: Parallel and WAIS

● Two or More: Coincident?

● Two or More: Level Profiles

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Profiles Analysis for g ≥ 2 GroupsTo deal with two or more groups, we follow the same logic:

1. Using assumptions (including results from earlier tests),determine what the hypothesis implies for populationparameters. This has implications for data.

2. Compute a statistic that reflects the implications for data.

3. Find a transformation of the statistic with a known samplingdistribution.

It is easier to do profile analysis using the GLM framework. Wewill use

X = AB + ǫ,

and test hypotheses of the form

Ho : LBM = 0

Appropriate definitions of L and M will lead to tests forparallelism, coincidence, and level means.

● Outline

Profile Analysis

2 Groups

Two or More Groups


Groups





MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Two or More: ParallelParallel profiles imply

Ho1 : (µ1i − µ1,i−1) = (µ2i − µ2,i−1) = · · · = (µgi − µg,i−1)

for all i = 2, . . . , p. This will be a MANOVA on differences.

This hypothesis takes linear combinations of the means withingroups and then compare the groups. For the lineartransformations of the regression parameters (i.e., the τ orµ’s), define

Mp×(p−1) =

−1 0 · · · 0

1 −1 · · · 0

0 1 · · · 0...

.... . .

...0 0 · · · −10 0 · · · 1

For the comparisons of groups on these “new” variables, defineL = C(g−1)×g contrast matrix.

● Outline

Profile Analysis

2 Groups

Two or More Groups


Groups





MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


GLM: Parallel and WAISFor the WAIS example, Ho1 is

LBM = (0, 1,−1)

β01 β02 β03 β04

β11 β12 β13 β14

β21 β22 β23 β24

−1 0 0

1 −1 0

0 1 −10 0 1

= 0

= (0, 1,−1)

(β01 − β02) (β02 − β03) (β03 − β04)

(β11 − β12) (β12 − β13) (β13 − β14)

(β21 − β22) (β22 − β23) (β23 − β24)

= 0

= [(β11 − β12)− (β21 − β22), (β12 − β13)− (β22 − β23),

(β13 − β14)− (β23 − β24)] = (0, 0)

So we have(

(β11 − β12)− (β21 − β22)

(β12 − β13)− (β22 − β23)

)

=⇒(

(µ11 − µ12)− (µ21 − µ22)

(µ12 − µ13)− (µ22 − µ23)

)

=

(

0

0

)

● Outline

Profile Analysis

2 Groups

Two or More Groups


Groups





MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Two or More: Coincident?

Assuming parallel, then coincidence implies

Ho2 : 1′µ1 = 1′µ2 = · · · = 1

′µg

So we’re testing equivalence of a single variable over g groups:1-way ANOVA.The linear transformation of the variables is given by

Mp×1 =

1

1...1

For the groups,L = I

See SAS for WAIS example

● Outline

Profile Analysis

2 Groups

Two or More Groups


Groups





MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


Two or More: Level ProfilesIf profiles are coincidence, then flat or level profiles means

Ho3 : µ11 = µ12 = · · · = µ1p = µ21 = · · · = µgp

That is, sums of two variables are the same for pairs of groups.The required linear transformation,

M (p×(p−1) =

1 0 0 · · · 0

1 1 0 · · · 0

0 1 1 · · · 0...

......

. . ....

0 0 0 · · · 1

The required contrasts of the groups,

L = C(g−1)×g

which is the same one used for testing parallel profiles.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

● MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions


MANOVAMini Outline:1. Review 2–way ANOVA2. 2–way MANOVA for balanced data3. Example: Distributed versus Massed Practice/instruction4. Unbalanced designs.5. Multivariate GLM and further extensions (MANCOVA,

longitudinal)

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA

● ANOVA Model for Two

Factors● Examples: Interaction & No

Interaction● Observation as Sum of

Means● Decomposition of Sums of

Squares

● Sums of Squares

Decomposition

● ANOVA Summary Table

● Hypothesis Tests

Two-Way MANOVA


Massed Practice


Unbalanced Designs

ConclusionsMANOVA: Part 2 Slide 31 of 75

Two-Way ANOVA“2–way” ANOVA −→ 2 Factors (qualitative variables).

Notation:Factor B

1 · · · k · · · b

1 X11r · · · X1kr · · · X1br ←− n observations in...

......

... each of the gb

Factor A l Xl1r · · · Xlkr · · · Xlbr combinations of levels...

......

... of the factors.g Xgkr · · · Xgkr · · · Xgbr

Observation: Xlkr = the rth observation at level l of Factor Aand level k of Factor B.

Factor A: l = 1, . . . , g

Factor B: k = 1, . . . , b

Replications: r = 1, . . . , n For now, a balanced design.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


ANOVA Model for Two FactorsXlkr = µ

︸︷︷︸

overalllevel

+ τl︸︷︷︸

fixed effectfactor Aat level l

+ βk︸︷︷︸

fixed effectfactor Bat level k

+ γlk︸︷︷︸

interactionbetween

factors A &Bat levelsl & , k

+ ǫlkr︸︷︷︸

residual

whereg∑

l=1

τg =b∑

k=1

βk +

g∑

l=1

γlk =b∑

k=1

γlk = 0

and ǫlkr ∼ N (0, σ2) and all independent.

Model for the mean:

E(Xlkr) = µlk = µ+ τl + βk + γlk

TThe effects are not additive; there is an interaction.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


Examples: Interaction & No Interaction

Factor A

a1 a2 a3 a4

Cel

lMea

n

5

10

15Factor B

p

p

p

p b1p

p

p

p b2

p

p

p

p b3

Factor A

a1 a2 a3 a4

5

10

15

Factor B

p

p

p

p b1p

p

p

p b2

p

p

p

p b3

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


Observation as Sum of Means

xlkr = x̄︸︷︷︸

µ̂

+(x̄l�� − x̄)︸︷︷︸

τ̂l

+(x̄�k� − x̄)

︸︷︷︸

β̂k

+(x̄lk� − x̄l�� − x̄�k� − x̄)

︸︷︷︸

γ̂lk

+(xlkr − x̄lk�)︸︷︷︸

ǫ̂lkr

where

x̄ = overall sample mean.

x̄l�� = mean for lth level of Factor A (or “row” mean).

x̄�k� = mean for kth level of Factor B (or “column” mean).

x̄lk� = mean for the lth level of Factor A and kth level of FactorB (or the “cell” mean).

The decomposition of the observations above shows theestimates of the population effects.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


Decomposition of Sums of Squaresxlkr = x̄

︸︷︷︸

µ̂

+(x̄l�� − x̄)︸︷︷︸

τ̂l

+(x̄�k� − x̄)

︸︷︷︸

β̂k

+(x̄lk� − x̄l�� − x̄�k� − x̄)

︸︷︷︸

γ̂lk

+(xlkr − x̄lk�)︸︷︷︸

ǫ̂lkr

To get the sums of squares:

1. Subtract x̄ from both sides.2. Square both sides.3. Sum over observations within cells, levels of Factor A, and

levels of Factor B.

4. After a little algebra,g∑

l=1

b∑

k=1

n∑

r=1

(xlkr − x̄)2 =

g∑

l=1

n(x̄l�� − x̄)2 +b∑

k=1

gn(x̄�k� − x̄)2

+

g∑

l=1

b∑

k=1

n(x̄lk� − x̄l�� − x̄�k� + x̄)2

+

g∑

l=1

b∑

k=1

n∑

r=1

(xlkr − x̄lk�)2

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


Sums of Squares Decomposition

We get an orthogonal decomposition of the sums of squares:

SStotal = SSA + SSB + SSAB + SSresidual

Note: SStotal is corrected for the grand mean.

Also, we get decomposition of degrees of freedom,

(gbn− 1)︸︷︷︸

total

= (g − 1)︸︷︷︸

Factor A

+ (b− 1)︸︷︷︸

Factor B

+(g − 1)(b− 1)︸︷︷︸

Interaction

+ gb(n− 1)︸︷︷︸

Residual

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


ANOVA Summary Table

Source of Degrees SSvariation of freedom Sums of Squared

Factor A (g − 1) SSA =∑g

l=1 n(x̄l�� − x̄)2

Factor B (b− 1) SSB =∑b

k=1 gn(x̄�k� − x̄)2

Interaction (g − 1)(b− 1) SSAB =∑g

l=1

∑bk=1 n(x̄lk� − x̄l�� − x̄

�k� + x̄)2

Residual gb(n− 1) SSres =∑g

l=1

∑bk=1

∑nr=1(xlkr − x̄lk�)

2

(error)

Total gbn− 1 SStotal =∑g

l=1

∑bk=1

∑nr=1(xlkr − x̄)2

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

● Two-Way ANOVA





Squares

● Sums of Squares

Decomposition



Two-Way MANOVA


Massed Practice


Unbalanced Designs


Hypothesis Tests

Interaction: Ho : γ11 = γ12 = · · · = γgb = 0. If the null is trueand all assumptions valid,

F =SSAB/((g − 1)(b− 1))

SSres/(gb(n− 1))∼ F(g−1)(b−1),gb(n−1)

Factor A: Ho : τ1 = τ2 = . . . = τg. If the null is true and allassumptions valid,

F =SSA/(g − 1)

SSres/(gb(n− 1))∼ F(g−1),gb(n−1)

Factor B: Ho : β1 = β2 = · · · = βb. If the null is true and allassumptions valid,

=SSB/(b− 1)

SSres/(gb(n− 1))∼ F(b−1),gb(n−1)

We should start with the Interaction before consideringwhether to test the main effects.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA

● Decomposition of

Observation

● Decomposition of SSCP

● Decomposition of SSCP● Hypothesis Testing:

Interaction First● Distribution of Wilk’s Lambda

Λ∗

● Hypothesis Testing:

Interaction First● Hypothesis Testing: Main

Effects

● Following Rejection


Massed Practice


Unbalanced Designs


Two-Way MANOVAFor p variables, X lkr is a p× 1 vector of measurements on pvariables.

The modelXlkr = µ+ τ l + βk + γlk + ǫlkr

and ǫlkr ∼ Np(0,Σ) and independent.

where

l = 1, . . . , g

k = 1, . . . , b

r = 1, . . . , nlk

And for identification

g∑

l=1

τ l =b∑

k=1

βk =

g∑

l=1

γlk =b∑

k=1

γlk = 0

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Decomposition of ObservationWe decompose the observation vector (xlkr − x̄) into sums ofvarious vectors,

(xlkr − x̄) = (x̄l�� − x̄)︸︷︷︸

τ̂ l

+(x̄�k� − x̄)

︸︷︷︸

ˆβk

+(x̄lk� − x̄l�� − x̄�k� + x̄)

︸︷︷︸

γ̂lk

+(xlkr − x̄lk�)︸︷︷︸

ǫ̂lkr

If we take the sums of squares and cross-products of this, weobtain the SSCP decomposition,

∑

l

∑

k

∑

r

(x̄l�� − x̄)(x̄l�� − x̄)′

︸︷︷︸

SSCPtotal

= . . .

which will take more space than is left on this slide

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Decomposition of SSCP

∑

l

∑

k

∑

r

(x̄l�� − x̄)(x̄l�� − x̄)′

︸︷︷︸

SSCPtotal

=∑

l

bn(x̄l�� − x̄)(x̄l�� − x̄)′

︸︷︷︸

SSCPFactorA

+∑

k

gn(x̄�k� − x̄)(x̄

�k� − x̄)′

︸︷︷︸

SSCPFactorB

+∑

l

∑

k

n(x̄lk� − x̄l�� − x̄�k� + x̄)(x̄lk� − x̄l�� − x̄

�k� + x̄)′

︸︷︷︸

SSCPinteraction

+∑

l

∑

k

∑

r

(xlkr − x̄lk�)(xlkr − x̄lk�)′

︸︷︷︸

SSCPresidual

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Decomposition of SSCPSo,

SSCPtotal = SSCPFactorA+SSCPFactorB+SSCPInteraction+SSCPresidual

which is orthogonal so long as the design is balanced (i.e.,nlk = n) or proportional. . . what to do with unbalanced isdiscussed a bit later.

And

dftotal = dfFactorA + dfFactorB + dfInteraction + dfresidual

(gbn− 1) = (g − 1) + (b− 1) + (g − 1)(b− 1) + gb(n− 1)

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Hypothesis Testing: Interaction First

Ho : γ11 = γ12 = · · · = γgp

Test Statistic is Wilk’s Lambda

Λ∗ =det(SSCPresidual)

det(SSCPresidual + SSCPinteraction)=

det(E)

det(E +H)

νh = νinteraction = (g − 1)(n− 1)

νe = νresidual + gb(n− 1)

Distribution of Λ∗ are given on the following slide for variouscases.

If you don’t have p ≤ 2 or νh ≤ 2, but you have large n thenthe approximate sampling distribution of

−(

νe −p+ 1− νh

2

)

ln(Λ∗) is χ2pνh

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Distribution of Wilk’s Lambda Λ∗

Wilk’s Λ∗ =|SSCPe|

|SSCPe + SSCPh|Number df forvariables Hypothesis Sampling distribution for multivariate data

p = 1 νh ≥ 1(

νe

νh

)(1−Λ∗

Λ∗

)

∼ Fνh,νe

p = 2 νh ≥ 1(

νe−1νh

)(1−

√Λ∗√

Λ∗

)

∼ F2νh,2(νe−1)

p ≥ 1 νh = 1(

νe+νh−pp

)(1−Λ∗

Λ∗

)

∼ Fp,(νe+νh−p)

p ≥ 2 νh = 2(

νe+νh−p−1p

)(1−

√Λ∗√

Λ∗

)

∼ F2p,2(νe+νh−p−1)

where νh = degrees of freedom for hypothesis, andνe = degrees of freedom for error (residual).

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Hypothesis Testing: Interaction FirstYou should test the interaction first.

■ If Ho for the interaction is rejected, then you don’t so tests formain effects because the Factor main effects do not have aclear interpretation.

■ J&W recommend doing p-univariate 2-way ANOVAs to findout “where” the interactions exist. (i.e., are there interactionsfor just some of the p variables and not others?).

■ For variables with no interaction, you can interpret maineffects.

■ Note: In some cases you can go on to test main effects eventhough the interaction is significant (we’ll do this in ourexample).

■ In general, if you retain the null hypothesis for interaction, wecontinue and test main effects. . .

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Hypothesis Testing: Main EffectsFor Factor A:Ho : τ 1 = τ 2 = · · · = τ g = 0 versus Ha : at least one τ l 6= 0

Λ∗A =

det(SSCPresidual)

det(SSCPresidual + SSCPFactorA)=

E

E +HA

and νh = g − 1. For p ≤ 2 or νA ≤ 2 (i.e., g ≤ 3), use the exactdistribution given in the table.

For Factor B:Ho : β1 = β2 = · · · = βb = 0 versus Ha : at least one βk 6= 0

Λ∗B =

det(SSCPresidual)

det(SSCPresidual + SSCPFactorB)=

E

E +HB

and νh = b− 1. For p ≤ 2 or νB ≤ 2 (i.e., v ≤ 3), use the exactdistribution given in the table.

For Large n, you can use

−(

νe −p+ 1− νh

2

)

ln(Λ∗h) ≈ χ2

pνh

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA

● Two-Way MANOVA


Observation




Λ∗



Effects



Massed Practice


Unbalanced Designs


Following RejectionIf a null hypothesis is rejected, you should perform additionalanalyses to figure out where the effects are. These couldinclude■ Multivariate T 2; that is, generalized squared distances

D2(X)lk|l′k′ = (X̄ lk − X̄ l′k′)′S−1pool(X̄ lk − X̄ l′k′)

■ Simultaneous confidence statements/intervals.■ When interactions are not significant, concentrate on

contrasts of levels of Factor A and of Factor B (if they’resignificant). These are the same as those described in 1-wayMANOVA notes.

For example,

(τli − τmi) or (µli − µmi)

Use Bonferroni if planned.■ 2-way ANOVAs with multiple comparisons.■ Discriminant Analysis.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice● Example: Distributed vs

Massed Practice

● The Design and Data

● Cell Means & 95%

Confidence Ellipses

● Factor Means & 95%

Confidence Ellipses

● Test for Interaction

● Test for Method Main Effect

● Test for Condition Main Effect

● Interpretation & Conclusion


Unbalanced Designs

Conclusions


Example: Distributed vs Massed Practice1–Way MANOVA: Data from Tatsuoka (1988), MultivariateAnalysis: Techniques for Educational and Psychological Research, pp273–279.

An experiment was conducted for comparing 2 methods (A &B) of teaching shorthand computer programing to 60 femaleseniors in a vocational high school (a dated example). Also ofinterest were the effects of distributed versus massed practice

C1: 2 hours of instruction/day for 6 weeksC2: 3 hours of instruction/day for 4 weeksC3: 4 hours of instruction/day for 3 weeks

So each subject received a total of 12 hours of instruction.Note: nl = 10 per cell of the design

Two variables (dependent measures):X1 = speedX2 = accuracy

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


The Design and DataVarious Means:

C1 C2 C3

X1 X2 X1 X2 X1 X2 X1 X2

A 34.30 21.80 32.50 17.90 29.60 14.10 32.13 17.93

B 42.80 25.60 35.50 18.59 27.00 11.60 35.10 18.57

38.55 23.79 34.00 18.20 28.30 12.85 33.62 18.25

Notes:■ Cell means (black) based on nlk = 10.■ Means for method (green) are based on nl� = 30

■ Means for condition (blue) are based on n�k = 20.

■ Total sample means (red) are based on n��= 60.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Cell Means & 95% Confidence Ellipses

X1 (Speed)5 10 15 20 25 30 35 40

X2

(Acc

urac

y)

5

10

15

20

25

30

35

40

r

x̄AC1

r

x̄AC2

r

x̄AC3

rx̄BC1

r

x̄BC2

r

x̄BC3

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Factor Means & 95% Confidence Ellipses

X1 (Speed)5 10 15 20 25 30 35 40

X2

(Acc

urac

y)

5

10

15

20

25

30

35

40

b

x̄A b x̄B

b x̄C1

b

x̄C2

b

x̄C3

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Test for Interaction

Ho : γAC1= γAC2

= γAC3= γBC1

= γBC2= γBC3

= 0

Ha : not all γlk = 0

W = E =

(

1333.10 211.20

211.20 184.70

)

HAB =

(

308.04 174.82

174.82 99.23

)

Λ∗ =det(E)

det(E +HAB)=

201.618

316.957= .6361

and

F =

(53

2

)(

1−√.6361√

.6361

)

= 6.73

Comparing F = 6.73 to the F4,106 distribution, F has p < .001.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Test for Method Main Effect. . . For purposes of illustration. . .

Ho : τA = τB = 0 versus Ha : not all γl = 0

W = E =

(

1333.10 211.20

211.20 184.70

)

HMethod =

(

132.02 28.18

28.18 6.02

)

Λ∗ =det(E)

det(E +HMethod)=

201.618

222.13= .9077

and

F =

(53

1

)(

1−√.9077√

.9077

)

= 2.63

Comparing F = 2.63 to the F2,106 distribution, F has p = .08.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Test for Condition Main Effect

Ho : βC1= βC2

= βC3= 0 versus Ha : not all βk = 0

W = E =

(

1333.10 211.20

211.20 184.70

)

HCondition =

(

1055.03 1111.55

1111.55 1177.30

)

Λ∗ =det(E)

det(E +HCondition)=

201.618

1502.966= .1341

and

F =

(53

2

)(

1−√.1341√

.1341

)

= 45.85

Comparing F = 45.85 to the F4,106 distribution, F has p < .001.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA



Massed Practice



Confidence Ellipses


Confidence Ellipses






Unbalanced Designs

Conclusions


Interpretation & Conclusion■ The effectiveness of the teaching methods (A and B)

depends on which of the three conditions of distributedpractice that a student used.

(i.e., There is a significant interaction).

■ The method by condition cell means indicate that method Bis more effective in terms of both speed and accuracy underconditions C1 and C2, but A is more effective in terms ofspeed and accuracy under condition C3.

■ Statistical tests of these conclusions should be performed infollow up analyses. Potentially useful supplement analysesinclude ANOVAs for each of the dependent variables,simultaneous confidence intervals for various treatmenteffects, and discriminant analysis.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


● Validity of Assumptions

● Shorthand Example: Scatter

Plot● Shorthand Example: QQ Plot

1● Shorthand Example: QQ Plot

2● Equality of Covariance

Matrices● Testing Equality of

Covariance Matrices● Example: Box’s Test of

Equality of Σlk● Example of Box’s Test

continued● Box’s Test— the easier

method● Checking Independence

AssumptionMANOVA: Part 2 Slide 56 of 75

Validity of AssumptionsThe assumptions are

■ Multivariate normality.■ Equality of Σ’s.■ Independence between and within groups.

Checks on multivariate normality:

■ Scatter plots of residuals.■ QQ plots of residuals.■ Scatter & QQ plots of principal components.■ Tests of univariate normality for each residuals.■ Tests of multivariate normality (see text).

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Shorthand Example: Scatter Plot

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Shorthand Example: QQ Plot 1

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Shorthand Example: QQ Plot 2

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Equality of Covariance Matrices■ When there are equal nlk’s (balanced), violating equality ofΣlk’s probably doesn’t hurt too much.

■ When there are unequal nlk ’s and det(Σlk|’s differsubstantially, then will tend to make more errors.

Where you make more Type I or Type II errors depends onhow different the nlk ’s and |Σlk|’s are from each other.

■ Test of Equality of Σ’s — this gets into ideas of tests andinference for covariance matrices, which leads into topics oncanonical correlation and discriminate analysis.

Assume:

X lj ∼ Np(µl,Σl) for l = 1, . . . , g and j = 1, . . . , nl

and independent over the g groups and with groups.The null hypothesis:

Ho : Σ1 = Σ2 = · · · = Σg ≡ Σ

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Testing Equality of Covariance MatricesIf Ho : Σ1 = Σ2 = · · · = Σg ≡ Σ is true, then

Spool = S =

(1

∑gl=1 νl

) g∑

l=1

νl ln(|Si|)

where νl = (ni − 1) = df(Si). When Ho is true, then Spool is anunbiased estimate of Σ.

Box’s test (1949ish):

M =

(g∑

l=1

νl

)

ln(| S︸︷︷︸

Spool

|)−g∑

l=1

νl ln(|Sl|)

Test statistic is[

1− 2p2 + 3p− 1

6(p+ 1)(g − 1)

(g∑

l=1

(1

νl

)

− 1∑g

l=1 νl

)]

︸︷︷︸

Correction factor

M ≈ χ2(g−1)p(p+1)/2

i.e., asymptotically

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Example: Box’s Test of Equality of Σlk

Ho : ΣA,C1= ΣA,C2

= ΣA,C3= ΣB,C1

= ΣB,C2= ΣB,C3

versus Ha : not all Σlk are equal.

The method by condition sample covariance matrices:

SA,C1=

(

34.68 8.40

8.40 5.07

)

SB,C1=

(

29.73 1.47

1.47 2.93

)

SA,C2=

(

38.50 4.17

4.17 7.43

)

SB,C2=

(

14.50 3.72

3.72 1.83

)

SA,C3=

(

19.16 3.78

3.78 1.66

)

SB,C3=

(

11.56 2.33

2.33 1.60

)

Pooled Covariance Matrix:

Spool =

(

25.69 3.91

3.91 3.42

)

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Example of Box’s Test continued

M = (∑

l,k

(nlk − 1)) ln |Spool| −∑

l,k

(nlk − 1)|Slk|

= 6(10− 1) ln(69.14)− (10− 1)(ln(105.14 + ln(268.82) + ln(20.30)

+ ln(85.07) + ln(12.73) + ln(13.04))

= 228.75− 205.34

= 23.41

Since nlk = 10 for all K = (2)(3) = 6 combinations of methodand condition (i.e., equal cell sizes), the correction factorsimplifies to

1− (2p2 + 3p− 1)(g + 1)

6(p+ 1)gn= 1− (2(2)2 + 3(2)− 1)(6 + 1)

6(2 + 1)10(6)= .9157

Test statistic = .9158(23.41) = 21.44 withdf = (K − 1)p(p+ 1)/2 = (6− 1)2(2 + 1)/2 = 15, which usingthe χ2 distribution has p-value = .123. Retain null hypothesis;that is, the equal covariance matrix assumption is reasonable.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Box’s Test— the easier methoddata shorthand;input speed accuracy method $ conditin $;if conditin=’C1’ and method=’A’ then group=1;else if conditin=’C1’ and method=’B’ then group=2;else if conditin=’C2’ and method=’A’ then group=3;else if conditin=’C2’ and method=’B’ then group=4;else if conditin=’C3’ and method=’A’ then group=5;else if conditin=’C3’ and method=’B’ then group=6;

proc discrim simple pool=test Wcov Pcov list;class group;var speed accuracy;run;

. . . and the output. . .

Notation:

K = Number of Groups

P = Number of Variables

N = Total Number of Observations - Number of Groups

N(i) = Number of Observations in the i’th Group - 1

__ N(i)/2|| |Within SS Matrix(i)|

V = -----------------------------------N/2

|Pooled SS Matrix|

_ _ 2| 1 1 | 2P + 3P - 1

RHO = 1.0 - | SUM ----- - --- | -------------|_ N(i) N _| 6(P+1)(K-1)

DF = .5(K-1)P(P+1)_ _

| PN/2 || N V |

Under the null hypothesis: -2 RHO ln | ------------------ || __ PN(i)/2 ||_ || N(i) _|

is distributed approximately as Chi-Square(DF).

Chi-Square DF Pr > ChiSq

21.218963 15 0.1300

Since the Chi-Square value is not significant at the 0.1 level, a pooledcovariance matrix will be used in the discriminant function.Reference: Morrison, D.F. (1976) Multivariate Statistical Methodsp252

Note: There is a more recent edition of Morri-son. . . 2007 (I think).

64-1

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice













Checking Independence Assumption■ MANOVA is generally relatively robust to violations of

multivariate normality and equal covariance matrices;however, it is not robust to violations of◆ Independence of observations within groups.◆ Independence of observations between groups.

■ Violation of independence generally increases the Type Ierror rate (i.e., it’s higher than what you think it is, andpossibly much larger).

■ You can check for possible violation by computing theintra-class correlation for each of the variables using

rintra =MSbetween −MSerror

MSbetween + (n− 1)MSerror

■ You can also consider the data collection procedure/method,because this is where dependence can slip in.

■ It’s the independence within that is most likely to be violated.In some cases this can be dealt with.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

● Unbalanced Designs

● Order in which SSCP are

extracted● Type I SSCP: “Sequential

SSCP”● Example Type I SSCP:

“Sequential SSCP”

● Type I: “Sequential SSCP”

continued

● Type II or “Partial SSCP”

● Example: Type II or “Partial

SSCP”

● Type III & Type IV SSCP● Type IV and When to Use

Which

Conclusions


Unbalanced DesignsSimilar to 1-way ANOVA.

With balanced designs (or proportional)

SSCPtotal = SSCPA + SSCPB + SSCPAB + SSCPresidual

orSSCPmodel = SSCPA + SSCPB + SSCPAB

and interpretation is straight forward.

With unbalanced designs, the partitioning of the is not longerunique. It depends on

■ The model.

■ Various sub-models of model as specified by the order inwhich various SSCP are extracted.. . .

These notes are based on Khattree & Naik.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Order in which SSCP are extractedSuppose

Xlkr = µ+ τ l + βk + γlk + ǫlkr

where l = 1, . . . , g, k = 1, . . . , b, and r = 1, . . . , nlk.

We could partition total SSCP as

SSCPtotal corrected = SSCPA|µ+SSCPB|µ,A+SSCPAB|µ,A,B

or

SSCPtotal corrected = SSCPB|µ+SSCPA|µ,B+SSCPAB|µ,A,B

When the design is unbalanced,

SSCPA|µ 6= SSCPA|µ,B

There are 4 different ways of computing SSCP’s, which canlead to different results and different interpretation.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Type I SSCP: “Sequential SSCP”The partitioning of the model SSCP into component SSCP dueto each variable or effect (including interactions) as it’s addedsequentially to the model as given in the model statement inproc glm in SAS.

e.g.,

model x1 x2 = A B C A ∗B A ∗ C B ∗ C A ∗B ∗ CThat is all main effects, 2–way interactions and a 3-wayinteraction.

The SSCP for A ∗ C = SSCP (A ∗ C|µ,A,B,C,A ∗B); that is,SSCP is adjusted for all previously entered terms,

model x1 x2 = A B C A ∗B︸︷︷︸

already entered

A ∗ C B ∗ C A ∗B ∗ C

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Example Type I SSCP: “Sequential SSCP”To demonstrate this, I deleted some observations from theshorthand data set to make it a bit unbalanced. The samplesizes are now

Practice ConditionMethod C1 C2 C3 Total

A 5 10 8 23B 10 7 6 23

Total 15 17 14 46

Below are the hypothesis SSCP matrices for two differentorders:

Order: M C C ∗M C M C ∗M

Hm =

(

164.54 83.22

83.22 42.09

)

Hm =

(

43.09 0.72

0.72 0.01

)

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Type I: “Sequential SSCP” continued

■ The sum of Hm +Hc will be equivalent regardless ofwhether method order condition was entered first.

■ With Type I SSCP, the sum of SSCP matrices of all theeffects in the model will sum to the total for the model,regardless of the order in which they are entered into themodel.

■ The total SSCP is completely partitioned.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Type II or “Partial SSCP”With Type II, an SSCP matrix equals the increase in modelSSCP for a particular variable (effect).

e.g.,

model x1 x2 = A B C A ∗B A ∗ C B ∗ C A ∗B ∗ CThat is all main effects, 2–way interactions and a 3-wayinteraction.

Consider the SSCP for A ∗ C. The SSCP for A ∗ C is theincrease in model SSCP by adding A ∗ C to the model that hasA, B, A ∗ C and B ∗ C in it.

The three-way interaction A ∗B ∗C∗ contains A ∗C in it, so theSSCP for A ∗ C is “in the model”. The effects A ∗B ∗ C andA ∗ C are correlated.

With Type II, the sum of the SSCP’s do not equal the totalmodel SSCP.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Example: Type II or “Partial SSCP”Below are the SSCP matrices for Type I and Type II and thetotals:

Type I: Sequential Type II: Partial

M C C ∗M C M C ∗M

Hm

164.54 83.22

83.22 42.09

43.09 0.72

0.72 0.01

43.09 0.72

0.72 0.01

Hc

886.58 925.59

925.59 966.37

1008.02 1008.08

1008.08 1008.45

886.58 925.59

925.59 966.37

Hcm

280.31 137.06

37.06 67.64

280.31 137.06

37.06 67.64

280.31 137.06

37.06 67.64

Sum

1331.43 1145.87

1145.87 1076.10

1209.98 1063.38

1063.38 1034.03

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs







continued



SSCP”


Which

Conclusions


Type IV and When to Use Which

Type IV are designed for situations where there are emptycells.

■ For balanced designs, the 4 ways to compute SSCP yieldthe same results.

■ For unbalanced designs, the 4 ways can lead to differentconclusions.

■ Type III is appropriate when interest is in comparing theeffects of experimental variables.

■ For model building purposes (finding a good model forprediction), Types I and/or II might be preferable.

■ If there are design parameters, make sure that tests do notinvolve these.

■ The choice should be made on a case-by-case basis withthought given to what it is you want to test.

● Outline

Profile Analysis

2 Groups

Two or More Groups

MANOVA

Two-Way ANOVA

Two-Way MANOVA


Massed Practice


Unbalanced Designs

Conclusions

● Conclusions


Conclusions■ Any design that you can form for univariate case can be

formed for the multivariate situation.

■ This can be done by creating an appropriate design matrix inthe general linear model context.

■ Models can include fixed (as we have done), but alsorandom effects (as in the univariate case).

■ It is possible to reformulated MANOVA as a linear mixedmodel, and hence provides a natural way to handle missingobservations (be sure there are systematic reasons formissingness). . . . see Goldstein.

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