4.5. Two-way ANOVA

Example. (Moore/McCabe Example 13.34)

Red palm oil, due to its high content of vi-

tamin A, is thought to reduce the occurence

and severity of malarity for young children.

To investigate whether this is indeed the case,

a supplement will be prepared that contains

either a placebo, a low dose, or a high dose

of red palm oil. Because boys and girls may

differ in exposure to malaria and the response

to the red palm oil supplement, we consider a

two-way ANOVA, that takes also gender into

account. Suppose we recruit 75 boys and 75

girls to the study. We will then randomly as-

sign 25 of each gender to each of the red

palm oil levels:

GenderRed palm oil Girls Boys TotalPlacebo 25 25 50Low dose 25 25 50High dose 25 25 50Total 75 75 150

165

This example illustrates several advantagesof two-way designs. The first is efficiency.With a two-way design, we can use the sametool to answer two questions: 1. Does redpalm oil/ 2. Does gender have an impactupon prevalence of malaria? That is, we gettwo one-way ANOVAs for the price of one.

The second and more important advantage isthat we may investigate interactions betweenfactors. Suppose that boys and girls reactindeed differently to the red palm oil supple-ment. That piece of information would notcome out in one-way ANOVAs neither for redpalm oil nor for gender.

The final advantage is increased power of thetests. Suppose we would run a one-way de-sign and there are indeed differences betweenboys and girls. The one-way ANOVA wouldassign this variation to the residual (withingroups) part of the model. In the two-wayANOVA, gender is included in the fit (be-tween groups) part of the model, which in-creases the power of the tests.

166

Assumptions for Two-Way ANOVA

The assumptions for two-way ANOVA arethe same as for one-way ANOVA, just thatwe have now 2 instead of only 1 factor:

We have independent random samples of sizenij from each of I J normal populations.The population means ij may differ, but allpopulations have the same variance 2.

Let xijk represent the kth observation fromthe population having factor A at level i andfactor B at level j. The statistical model is

Xijk = ij + ijk, where

i = 1, . . . , I; j = 1, . . . , J; k = 1, . . . , nij; andijk N(0, 2).

We estimate ij by xij =1

nij

nijk=1

xijk,

and 2 by (using s2ij =nij

k=1(xijkxij)2/(nij1)):

s2 =

ij(nij 1)s2ijN IJ =

SSE

DFE= MSE,

where DFE = observationsgroups = NIJ.167

Main effects and interactions

We shall now explore in detail the FIT part of themodel, represented by the population means ij.

Because we have independent samples from each group,

we can think of the problem initially as a one-way

ANOVA with IJ groups, that is SST = SSM + SSE,

where

SSM =I

i=1

Jj=1

nijk=1

(xij x)2

with DFM = groups1 = IJ 1.In two-way ANOVA, the terms SSM and DFM are

broken down into a main effect for A, a main effect

for B, and an AB interaction as follows:

SSM = SSA + SSB + SSABDFM = DFA + DFB + DFAB

SSA represents variation among the means for the

different levels of A. Because there are I such means,

SSA has DFA= I 1 degress of freedom. Similarly,DFB= J 1. For DFAB we obtain by subtraction:

DFAB = DFMDFADFB= (IJ 1) (I 1) (J 1)= (I 1)(J 1).

168

The two-way ANOVA model

Taking the split up of the population meansij into main effects and an interaction effectinto account, the two-way ANOVA modelmay be restated as

Xijk=

ij +i+j+()ij +ijk, ijkN(0, 2),

where is the overall mean; i is the effectof level i (i = 1, . . . , I) of factor A; j is theeffect of level j (j = 1, . . . , J) of factor B;()ij is the interaction effect of levels i andj; and ijk is the error associated with thekth data point from level i of factor A andlevel j of factor B.

Note that since i, j and ()ij are devi-ations from the overall mean , in the fixedeffects model the sums of all these deviationsare zero:

i

i =j

j =i,j

()ij = 0.

169

The Hypothesis Tests in Two-Way ANOVA

Factor A main-effects test:

H0 : i = 0 for all i = 1, . . . , IH1 : Not all i are zero

F = MSA/MSE F (I 1, N IJ).

Factor B main-effects test:

H0 : j = 0 for all j = 1, . . . , JH1 : Not all j are zero

F = MSB/MSE F (J 1, N IJ).

Test for AB interactions:

H0 : ()ij = 0 i=1,. . .,I; j=1,. . .,JH1 : Not all ()ij are zero

F = MSAB/MSE F ((I1)(J1), NIJ).

Note: Finding a significant interaction effectsubstantially reduces the usefulness of factorA/B main effects, because then they applyonly on average, but not individually for eachlevel of the other factor.

170

Example: (Azcel, 4th edition.)

The brightness of films produced by 3 differ-ent manufacturers has been compared using3 different development processes:

All p-values are below 0.01%, so both man-ufacturer and development method have animpact.

There is also an interaction effect:

F = 61.756/5.811 = 10.627Degrees of freedom:(3 1)(3 1) = 4 and 45 3 3 = 36Critical value: F0.05(4,36) = 2.63

This implies that main effects must be checkedfor each level of the other factor separately.

171

Analyze => ANOVA => Linear Models ...

Model: Process, Film => Factorial

Model Options => Sum of Squares to show: Type III

Plots => Custom list of Plots => Interaction Plot

Linear Models Dependent Variable: Brightness Brightness

Source DF Sum of

Squares Mean Square F Value Pr > F

Model 8 1776.044444 222.005556 38.20 F

Process 2 165.644444 82.822222 14.25

Two-Way ANOVA with 1 Observation/Cell

The case of one data point in every cell presentsa problem in two-way ANOVA because then

DFE = observations groups = 0.However, if we may assume that there are nointeractions, then SSAB is only due to errorand contains no other information. In thatcase we may use SSAB and its associateddegrees of freedom DFAB= (I 1)(J 1) inplace of SSE and its degrees of freedom.

We can thus conduct the tests for the maineffects by dividing MSA (MSB) by MSABwhen testing for factor A (factor B) maineffects. The resulting F -statistics has I 1and (I1)(J1) degrees of freedom for fac-tor A, and J 1 and (I 1)(J 1) degreesof freedom for factor B.

The fixed effects two-way ANOVA model withone observation per cell reduces then to:

Xijk = + i + j + ijk, ijk N(0, 2).175

Randomized Complete Block Design

When we want to compare the means of kpopulation means while controlling extrane-ous variables, a procedure known as blockingis used. A block (lohko) is a collection of kexperimental units that are as nearly alike aspossible with respect to the extraneous vari-ables. (A block could be the same persontrying k different products.)

Each treatment is randomly assigned to 1unit within each block (random order in try-ing the products). Since the effect of theextraneous variables is controlled by match-ing like experimental units, any differences inresponse are attributed to treatment effects.

This randomized complete block design (alsocalled one-way ANOVA with repetition) maybe regarded as a two-way ANOVA with oneitem per cell because the blocks may be viewedas one factor and the treatment levels as theother. In the randomized block design, how-ever, we are only interested in the treatmentlevels and not in the blocks.

176

Because the randomized complete block de-

sign is just a two-way ANOVA with one ob-

servation per cell, we may test the null hy-

pothesis

H0 : 1 = 2 = = k.with an F -test of the form

F =MSTR

MSE F (k 1, (k 1)(b 1)),

where k denotes the number of treatments

and b denotes the number of blocks.

Note that the randomized complete block de-

sign assumes that there is no interaction be-

tween treatments and blocks!

177

Example: (Cochran & Snedecor)

Below are percentages of soya beans fail-

ing to sprout, planted under 5 individually

identical conditions (blocks) under 5 differ-

ent treatments, with ANOVA output from

Excel:

Degrees of freedom:

Treatment: 5 1 = 4Block: 5 1 = 4Error: (5 1)(5 1) = 16

There is a significant main effect for treat-

ments, but not for blocks.

178

Further Two-Way Analysis

Post hoc tests and contrasts work in muchthe same way as in one-way ANOVA. We onlyneed to replace MSE from one-way ANOVAwith the new MSE for two-way ANOVA asthe new estimator s2 for the variance 2.

Example: (soya beans continued.)Consider using the Bonferroni method to com-pare all

(52

)= 542 = 10 pairs of treatment.

The combined significance level at 5% is

= 0.05/10 = 0.005.

The standard error for differences in treat-ment means is

SEB =

s2( 1ni

+1

nj

)=

5.41 2

5= 1.471.

The degrees of freedom are

df = (k 1)(b 1) = 4 4 = 16,so the smallest significant difference becomes

t0.005/2(16) SEB = 3.252 1.471 = 4.78.Only the difference of the pair FermateNoneexceeds this value and is therefore significant.

179

Analyze => ANOVA => Mixed Models... Dependent variable: Failures Classification variables: Treatment, Block => Fixed Effect Model: Treatment, Block => Main => Fixed Effect Model Options:

=> Least Squares Post Hoc Tests => Add:

Mixed Models Analysis of: Failures

Covariance Parameter Estimates

Cov Parm Estimate

Residual 5.4100

Type 3 Tests of Fixed Effects

Effect Num

DFDen DF F Value Pr > F

Treatment 4 16 3.87 0.0219

Block 4 16 2.30 0.1032

Least Squares Means

Effect Treatment EstimateStandard

Error DF t Value Pr > |t|

Treatment 1 6.2000 1.0402 16 5.96

Example: (film processing continued.)

Consider testing the following two contrasts:

H0(1): A =12(B + C), H0(2): B = C.

A B C2i

Xi 33.867 30.733 29.267i(1) 1 -0.5 -0.5 1.5i(2) 0 1 -1 2

The contrast values and standard errors are:

L1 = 3.867, s(L1) =

5.811 1.5/15 = 0.762L2 = 1.467, s(L2) =

5.811 2/15 = 0.880

after inserting MSW=5.811 and n=15 into

s(L) =

MSE

2i /n.

The t-statistics for the two contrasts are:

t1 =L1

s(L1)= 5.07 and t2 =

L2s(L2)

= 1.67

with df = N IJ = 453 3 = 36, such thatt0.025(36) = 2.03; p1 = 0.000, p2 = 0.104.

So we reject H0(1) and accept H0(2).

182

Analyze => ANOVA => Linear Models ... Post Hoc Tests => Add Class effects to use: Process: True Show p-values for differences: All pairwise differences Preview Code => Insert Code

OK => OK => Run

Linear Models

Least Squares Means

ProcessBrightness

LSMEANLSMEAN Number

1 33.8666667 1

2 30.7333333 2

3 29.2666667 3

Least Squares Means for effect Process Pr > |t| for H0: LSMean(i)=LSMean(j)

Dependent Variable: Brightness

i/j 1 2 3

1 0.0011