Topic 28: Unequal
Replication in Two-Way
ANOVA
Outline
• Two-way ANOVA with unequal
numbers of observations in the cells
– Data and model
– Regression approach
– Parameter estimates
• Previous analyses with constant n
just special case
Data for two-way ANOVA
• Y is the response variable
• Factor A with levels i = 1 to a
• Factor B with levels j = 1 to b
• Yijk is the kth observation in cell (i,j)
• k = 1 to nij and nij may vary
Recall Bread Example
• KNNL p 833
• Y is the number of cases of bread sold
• A is the height of the shelf display, a=3
levels: bottom, middle, top
• B is the width of the shelf display, b=2:
regular, wide
• n=2 stores for each of the 3x2
treatment combinations (BALANCED)
Regression Approach
• Create a-1 dummy variables to represent
levels of A
• Create b-1 dummy variables to represent
levels of B
• Multiply each of the a-1 variables with
b-1 variables for B to get variables for
AB
LET’S LOOK AT THE RELATIONSHIP
AMONG THESE SETS OF VARIABLES
Common Set of Variables
   0,    0,
 (  )  0, (  )
data a2;
0
set a1;
X1 = (height eq 1) - (height eq 3);
X2 = (height eq 2) - (height eq 3);
X3 = (width eq 1) - (width eq 2);
X13 = X1*X3;
X23 = X2*X3;
i
i
i
j
ij
j
j
ij
Run Proc Reg
proc reg data=a2;
model sales= X1 X2 X3 X13 X23
/ XPX I;
height: test X1, X2;
width: test X3;
interaction: test X13, X23;
run;
X′X Matrix
Model Crossproducts X'X X'Y Y'Y
Variable Intercept X1 X2 X3 X13 X23 Sets of
Intercept
12 0 0 0
0
0 variables
X1
0 8 4 0
0
0 orthogonal
X2
0
4
8
0
0
X3
0
0
0 12
0
X13
0
0
0
0
8
0 Crossproducts
0
between
4 sets is 0
X23
0
0
0
0
4
8
Orthogonal X’s
• Order in which the variables are fit in
the model does not matter
– Type I SS = Type III SS
• Order of fit not mattering is true for
all choices of restrictions when nij is
constant
• Orthogonality lost when nij are not
constant
KNNL Example
• KNNL p 954
• Y is the change in growth rates for
children after a treatment
• A is gender, a=2 levels: male, female
• B is bone development, b=3 levels:
severely, moderately, or mildly
depressed
• nij=3, 2, 2, 1, 3, 3 children in the groups
Read and check the data
data a3;
infile 'c:\...\CH23TA01.txt';
input growth gender bone;
proc print data=a1;
run;
Obs
1
2
3
4
5
6
7
8
9
10
11
12
13
14
growth
1.4
2.4
2.2
2.1
1.7
0.7
1.1
2.4
2.5
1.8
2.0
0.5
0.9
1.3
gender
1
1
1
1
1
1
1
2
2
2
2
2
2
2
bone
1
1
1
2
2
3
3
1
2
2
2
3
3
3
Common Set of Variables
   0,    0,
 (  )  0, (  )
data a3;
0
set a3;
X1 = (bone eq 1) - (bone eq 3);
X2 = (bone eq 2) - (bone eq 3);
X3 = (gender eq 1) - (gender eq 2);
X13 = X1*X3;
X23 = X2*X3;
i
i
i
j
ij
j
j
ij
Run Proc Reg
proc reg data=a3;
model growth= X1 X2 X3 X13 X23
/ XPX I;
run;
X′X Matrix
Model Crossproducts X'X X'Y Y'Y
Variable Intercept X1 X2 X3 X13 X23 CrossIntercept
14 -1 0 0
3
0 product
X1
-1 9 5 3
1 -1 terms no
longer 0
X2
0 5 10 0 -1 -2
Order of
X3
0 3 0 14 -1
0
fit
X13
3 1 -1 -1
9
5 matters
X23
0 -1
-2
0
5
10
How does this impact the
analysis?
• In regression, this happens all the
time (explanatory variables are
correlated)
– t tests look at significance of
variable when fitted last
• When looking at comparing means
order of fit will alter null hypothesis
Prepare the data for a plot
data a1; set a1;
if (gender eq 1)*(bone
then gb='1_Msev ';
if (gender eq 1)*(bone
then gb='2_Mmod ';
if (gender eq 1)*(bone
then gb='3_Mmild';
if (gender eq 2)*(bone
then gb='4_Fsev ';
if (gender eq 2)*(bone
then gb='5_Fmod ';
if (gender eq 2)*(bone
then gb='6_Fmild';
eq 1)
eq 2)
eq 3)
eq 1)
eq 2)
eq 3)
Plot the data
title1 'Plot of the data';
symbol1 v=circle i=none;
proc gplot data=a1;
plot growth*gb;
run;
Find the means
proc means data=a1;
output out=a2 mean=avgrowth;
by gender bone;
run;
Plot the means
title1 'Plot of the means';
symbol1 v='M' i=join c=blue;
symbol2 v='F' i=join c=green;
proc gplot data=a2;
plot avgrowth*bone=gender;
run;
Plot of the means
avgrowth
2.4 F
Interaction?
2.2
F
2.0 M
M
1.8
1.6
1.4
1.2
1.0
M
F
0.8
1
2
3
bone
gender
M M M 1
F F F 2
Cell means model
• Yijk = μij + εijk
– where μij is the theoretical mean or
expected value of all observations
in cell (i,j)
– the εijk are iid N(0, σ2)
– Yijk ~ N(μij, σ2), independent
Estimates
• Estimate μij by the mean of the
observations in cell (i,j), Yij.
Yij.  (  k Yijk ) / n ij
• For each (i,j) combination, we can get
an estimate of the variance
s   k ( Yijk  Yij. ) /( n ij  1)
2
ij
2
• We pool these to get an estimate of σ2
Pooled estimate of
2
σ
• In general we pool the sij2, using
weights proportional to the df, nij -1
• The pooled estimate is
s2 = (Σ (nij-1)sij2) / (Σ(nij-1))
Nothing different in terms
of parameter estimates
from balanced design
Run proc glm
proc glm data=a1;
class gender bone;
model growth=gender|bone/solution;
means gender*bone;
run;
Shorthand way to write main
effects and interactions
Parameter Estimates
• Solution option on the model statement
gives parameter estimates for the glm
parameterization
• These constraints are
– Last level of main effect is zero
– Interaction terms with a or b are zero
• These reproduce the cell means in the
usual way
Parameter Estimates
Parameter
Intercept
gender 1
bone
1
bone
2
gender*bone 1 1
gender*bone 1 2
Estimate
0.90000000
-0.00000000
1.50000000
1.20000000
-0.40000000
-0.20000000
B
B
B
B
B
B
Standard
Error t Value Pr > |t|
0.2327373
3.87 0.0048
0.3679900 -0.00 1.0000
0.4654747
3.22 0.0122
0.3291403
3.65 0.0065
0.5933661 -0.67 0.5192
0.5204165 -0.38 0.7108
Example : ˆ 22  0.90  1.20  2.10
Output
Source
Model
Error
Sum of
DF Squares
5 4.4742857
8 1.3000000
Mean
Square F Value Pr > F
0.89485714
5.51 0.0172
0.16250000
Corrected Total 13 5.7742857
Note DF and SS add as usual
Output Type I SS
Source
gender
DF Type I SS Mean Square F Value Pr > F
1 0.0028571
0.00285714
0.02 0.8978
bone
2 4.3960000
2.19800000
13.53 0.0027
gender*bone
2 0.0754286
0.03771429
0.23 0.7980
SSG+SSB+SSGB=4.47429
Output Type III SS
Source
gender
DF Type III SS Mean Square F Value Pr > F
1 0.12000000
0.12000000
0.74 0.4152
bone
2 4.18971429
2.09485714
12.89 0.0031
gender*bone
2 0.07542857
0.03771429
0.23 0.7980
SSG+SSB+SSGB=4.38514
Type I vs Type III
• SS for Type I add up to model SS
• SS for Type III do not necessarily add up
• Type I and Type III are the same for the
interaction because last term in model
• The Type I and Type III analysis for the
main effects are not necessarily the same
• Different hypotheses are being examined
Type I vs Type III
• Most people prefer the Type III
analysis
• This can be misleading if the cell
sizes differ greatly
• Contrasts can provide some insight
into the differences in hypotheses
Contrast for A*B
• Same for Type I and Type III
• Null hypothesis is that the profiles are
parallel; see plot for interpretation
• μ12 - μ11 = μ22 - μ21 and
μ13 - μ12 = μ23 - μ22
• μ11 - μ12 - μ21 + μ22 = 0 and
μ12 - μ13 - μ22 + μ23 = 0
A*B Contrast statement
contrast 'gender*bone
Type I and III'
gender*bone 1 -1 0 -1 1 0,
gender*bone 0 1 -1 0 -1 1;
run;
Type III Contrast for gender
• (1) μ11 = (1)(μ + α1 + β1 + (αβ)11)
• (1) μ12 = (1)(μ + α1 + β2 + (αβ)12)
• (1) μ13 = (1)(μ + α1 + β3 + (αβ)13)
• (-1) μ21 = (-1)(μ + α2 + β1 + (αβ)21)
• (-1) μ22 = (-1)(μ + α2 + β2 + (αβ)22)
• (-1) μ23 = (-1)(μ + α2 + β3 + (αβ)23)
L
= 3α1 – 3α2 + (αβ)11 + (αβ)12 + (αβ)13 –
(αβ)21 – (αβ)22 – αβ23
Contrast statement
Gender Type III
contrast 'gender Type III'
gender 3 -3
gender*bone 1 1 1 -1 -1 -1;
Type I Contrast for gender
• (3) μ11 = (3)(μ + α1 + β1 + (αβ)11)
• (2) μ12 = (2)(μ + α1 + β2 + (αβ)12)
• (2) μ13 = (2)(μ + α1 + β3 + (αβ)13)
• (-1) μ21 = (-1)(μ + α2 + β1 + (αβ)21)
• (-3) μ22 = (-3)(μ + α2 + β2 + (αβ)22)
• (-3) μ23 = (-3)(μ + α2 + β3 + (αβ)23)
L
= (7α1 – 7α2 )+(2β1 – β2 – β3)+3(αβ)11
+2(αβ)12 +2(αβ)13 –1(αβ)21 –3(αβ)22 –3(αβ)23
Contrast statement
Gender Type I
contrast 'gender Type I'
gender 7 -7
bone 2 -1 –1
gender*bone 3 2 2 -1 -3 -3;
Contrast output
Contrast
DF Contrast SS
gender Type III 1 0.12000000
gender Type I
1 0.00285714
bone Type III
2 4.18971429
gender*bone
Type I and III
2 0.07542857
Summary
• Type I and Type III F tests test
different null hypotheses
• Should be aware of the differences
• Most prefer Type III as it follows logic
similar to regression analysis
• Be wary, however, if the cell sizes
vary dramatically
Comparing Means
• If interested in Type III hypotheses,
need to use LSMEANS to do
comparisons
• If interested in Type I hypotheses,
need to use MEANS to do
comparisons.
• We will show this difference via the
ESTIMATE statement
SAS Commands
• Will use earlier contrast code to set
up the ESTIMATE commands
estimate 'gender Type III' gender 3 -3
gender*bone 1 1 1 -1 -1 -1 / divisor=3;
estimate 'gender Type I' gender 7 -7
bone 2 -1 -1 gender*bone 3 2 2 -1 -3 -3 /
divisor=7;
MEANS OUPUT
Level of ------------growth----------gender N
Mean
Std Dev
1
2
7
7
1.65714286
1.62857143
0.62411843
0.75655862
Diff = 0.0286
LSMEANS OUPUT
gender
1
2
growth
LSMEAN
1.60000000
1.80000000
Diff = -0.20
Estimate output
Parameter
Estimate
gender Type III -0.200
gender Type I
0.029
Std Err
0.2327
0.2155
Notice that these two estimates
agree with the difference of
estimates for LSMEANS or MEANS
Analytical Strategy
• First examine interaction
• Some options when the interaction is
significant
– Interpret the plot of means
– Run A at each level of B and/or B
at each level of A
– Run as a one-way with ab levels
– Use contrasts
Analytical Strategy
• Some options when the interaction is
not significant
– Use a multiple comparison
procedure for the main effects
– Use contrasts for main effects
– If needed, rerun without the
interaction
Example continued
proc glm data=a3;
class gender bone;
model growth=gender bone/
solution;
For Type I
hypotheses means gender bone/
tukey lines;
run;
Pool here because
small df error
Output
Source
Model
Sum of
DF Squares Mean Square F Value Pr > F
3 4.3988571 1.46628571
10.66 0.0019
Error
10 1.3754286
Corrected Total
13 5.7742857
0.13754286
Output Type I SS
Source
gender
bone
DF Type I SS Mean Square F Value Pr > F
1 0.00285714
0.00285714
0.02 0.8883
2 4.39600000
2.19800000
15.98 0.0008
Output Type III SS
Source
gender
bone
DF Type III SS Mean Square F Value Pr > F
1 0.09257143
0.09257143
0.67 0.4311
2
4.39600000
2.19800000
15.98 0.0008
Although different null
hypothesis for gender, both
Type I and III tests are
not found significant
Tukey comparisons
Group
Mean
N
bone
A
A
A
2.1000
4
1
2.0200
5
2
B
0.9000
5
3
Tukey Comparisons
• Why don’t we need a Tukey
adjustment for gender?
• Means statement does provide mean
estimates so you know directionality
of F test but that is all the statement
provides you
Last slide
• Read KNNL Chapter 23
• We used program topic28.sas to
generate the output for today