Other hypotheses of interest (cont’d)
• In addition to the simple null hypothesis of no treatment
effects, we might wish to test other hypothesis of the general
form (examples follow):
H0 : Ck×g βg×p = 0, Comparisons among treatments
H0 : βg×pMp×q = 0, Comparisons across traits
H0 : Ck×g βg×pMp×q = 0, A combination of both.
• For example, suppose that we measure p = 2 traits on each
unit in g = 3 treatment groups, and we set τ30 = [τ31 τ32]0 =
0. In this case,


µ1 µ2


β =  τ11 τ12 
τ21 τ22
389
Other hypotheses of interest (cont’d)
• Comparisons of treatment means:
"
Cβ =
0 1 0
0 1 −1
#


#
"
µ1 µ2
τ11
τ12


.
=
τ
τ
 11 12 
τ11 − τ21 τ12 − τ22
τ21 τ22
• Under the restriction τ13 = τ23 = 0, the first row of β is the
expected value of units in treatment 3.
• First row of C provides differences between mean resposnes
for treatments 3 and 1 for both traits.
• Second row of C provides differences between mean responses
for treatments 1 and 2 for both traits.
390
Other hypotheses of interest (cont’d)
• Comparison of trait means with each treatment group:


µ1 µ2


βM =  τ11 τ12 
τ21 τ22
"
1
−1
#


µ1 − µ2


=  τ11 − τ12  .
τ21 − τ22
• The first row compares mean responses for trait 1 and
trait 2 for units receiving treatment 3.
• The second row compares mean responses for trait 1 and
trait 2 for units receiving treatment 1.
• The third row compares mean responses for trait 1 and
trait 2 for units receiving treatment 2.
391
Other hypotheses of interest (cont’d)
• One might be interested in comparing the effects of the
treatments on just the first trait:
"
CβM =
0 1 0
0 1 −1
#


"
#
µ1 µ2 " #
τ11

 1
=
.
 τ11 τ12 
0
τ11 − τ21
τ21 τ22
• Row 1 is the difference of the effects of treatments 3 and 1
on mean responses for trait 1.
• Row 2 is the difference of the effect of treatments 1 and 2
on mean responses for trait 1.
392
Other hypotheses of interest (cont’d)
• One might be interested in interaction contrasts :
"
CβM =
0 1 0
0 1 −1
#


#
"
#
µ1 µ2 "
1
τ11 − τ21


=
.
τ
τ
 11 12 
−1
(τ11 − τ21) − (τ12 − τ22)
τ21 τ22
• The first is the difference in mean responses to treatments
3 and 1 for trait 1 minus the difference in mean responses
to treatments 3 and 1 for trait 2.
• The second is the difference in mean responses to treatments
1 and 2 for trait 1 minus the difference in mean responses
to treatments 1 and 2 for trait 2.
393
F approximation to the sampling distribution of
Wilk’s Criterion
C. R. Rao (1951) Bulletin Int. Stat. Inst. 33(2), 177-180.
• Coincides with exact F-distribution for cases described on a
previous slide
• More accurate than large sample chi-square approximation
• Similar F-approximations are used by SAS for Pillai’s trace
and the Lawley-Hotelling trace.
394
F approximation to the sampling distribution of
Wilk’s Criterion
• When H0 : Ck×r βr×pMp×u = 0k×u is true
1 − Λ1/b ab − c
∼ F(ku,ab−c)
1/b
uk
Λ
where
a = (n − r) − (u − k + 1)/2
v
u
u u2k 2 − 4
b = t 2
u + k2 − 5
c = (uk − 2)/2
395
Example: One-way MANOVA
• Populations: g = 3 types of students
– (k = 1) Technical school students (n1 = 23)
– (k = 2) Architecture students (n2 = 38)
– (k = 3) Medical technology students (n3 = 21)
• Response variables: p = 4 test scores
– aptitude test
– mathematics test
– language test
– general knowledge
396
Example (cont’d)
One-way MANOVA model



Xij = 

Xij1
Xij2
Xij3
Xij4






=


µ1
µ2
µ3
µ4








+


τi1
τi2
τi3
τi4






+


ij1
ij2
ij3
ij4





where

ij = 


ij1
ij2
ij3
ij4


 ∼ N ID(0,

Σ)
and use the SAS constraints τ31 = τ32 = τ33 = τ34 = 0
397
Example (cont’d)
• Test the null hypothesis that the mean vectors for the four
traits are the same for all three types of students
• Write the one-way MANOVA model in matrix form
Xn×p = An×r βr×p + n×p
where


µ1 µ2 µ3 µ4


β3×4 =  τ11 τ12 τ13 τ14 
τ21 τ22 τ23 τ24
398
Example (cont’d)
The null hypothesis that the mean vectors for the four traits are
the same for all three types of students can be written as
"
H0 : Cβ =
0 1 0
0 0 1
#


"
#
µ1 µ2 µ3 µ4
0 0 0 0


=
τ
τ
τ
τ
 11 12 13 14 
0 0 0 0
τ21 τ22 τ23 τ24
Here k = 2 rows in C
r = 3 rows in β
p = 4 response variables
M = I4×4 so u = p = 4
n = n1 + n2 + n3 = 23 + 38 + 21 = 82
399
Example (cont’d)
|W |
The value of Wilks Lambda is |B+W | = 0.544
a = (82 − 3) − (4 − 2 + 1)/2 = 77.5
v
u
u 4222 − 4
b = t 2
=2
2
4 +2 −5
c = ((4)(2) − 2)/2 = 3
and
F =
√ ! 1− Λ
ab − c
√
= 6.76
uk
Λ
on (uk, ab−c) = (8, 152) degrees of freedom and p-value < .0001
Conclude that the means scores are different for at least two
types of students for at least one the the four response variables
400
Example (Cont’d)
• Can test the null hypothesis of equal response means across
populations by running a one way ANOVA for each of the
four response variables.
• SAS code (in morel.sas)
PROC GLM DATA=SET1;
CLASS GROUP;
MODEL X1-X4 = GROUP / Solution;
MANOVA H=group /PRINTH PRINTE;
RUN;
401
Example (Cont’d)
MANOVA Test Criteria and F Approximations for
the Hypothesis of No Overall group Effect
H = Type III SSCP Matrix for group
E = Error SSCP Matrix
S=2
Statistic
Wilks’ Lambda
Pillai’s Trace
Hotelling-Lawley Trace
Roy’s Greatest Root
M=0.5
Value
0.543448
0.492698
0.773588
0.675059
N=37
F Value
6.77
6.29
7.29
12.99
Num DF
8
8
8
4
Den DF
152
154
106.27
77
p-value
<.0001
<.0001
<.0001
NOTE: F Statistic for Roy’s Greatest Root is an upper bound.
NOTE: F Statistic for Wilks’ Lambda is exact.
402
Example (Cont’d): morel.R
> morel[,1] <- as.factor(morel[,1])
> fit <- manova(as.matrix(morel[,-1])~morel[,1])
> summary(fit, test="Wilks")
Df
Wilks approx F num Df den Df
Pr(>F)
morel[, 1] 2 0.54345
6.7736
8
152 1.384e-07 ***
Residuals 79
--Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
> summary(fit, test="Pillai")
Df Pillai approx F num Df den Df
Pr(>F)
morel[, 1] 2 0.4927
6.2923
8
154 4.824e-07 ***
Residuals 79
--403
Bonferroni t-tests and intervals
• Can also examine Bonferroni simultaneous t-tests and CIs
• For the difference in treatment means for the i-th response
(µi + τki) − (µi + τ`i) = τki − τ`i,
we need the variance of X̄ki = X̄`i:
!
Var(τ̂ki − τ̂`i) = Var(X̄ki − X̄`i) =
1
1
+
σii,
nk
n`
wii
where σii is estimated as spooled,ii = n−g
, with wii the ith
diagonal element of the within-groups SSP matrix W .
404
Bonferroni t-tests and intervals
• If we wish to carry out all pairwise comparisons, there will be
pg(g − 1)/2 of them.
• To maintain a simultaneous type I error level of no more than
α we can use
tn−g (
α
)
2m
where
m=
pg(g − 1)
.
2
• Reject H0 : τki = τ`i if
t = r
|X̄ki = X̄`i|
1 + 1 s
pooled,ii
nk
n`
> tn−g (
α
)
2m
405
Bonferroni simultaneous intervals
• We have three groups and four variables, for a total of
4×3×2/2 = 12 comparisons, three for each response variable.
• From the output:
w11 = 55036.1955
w33 = 4759.8655
w22 = 14588.9983
w44 = 7040.6248
with n1 = 23, n2 = 37, n3 = 20 and n − g = 82 − 3 = 79.
406
Bonferroni simultaneous intervals
• A 95% confidence interval for the true difference between
technical school and architecture students on mathematics
is
v
u
α u
w22
t
x̄12 − x̄22 ± tn−g (
)
2m
n−g
s
1
1
+
n1
n2
!
14588.9983 1
1
47.3913 − 51.1842 ± t79(0.05/24)
+
79
23
37
√
−3.7929 ± 2.951 184.671 × 0.0705
⇒
(−14.441, 6.855).
Since the interval includes 0, we conclude that there is
insufficient evidence to reject the null hypothesis of equal
mathematics means scores for technical and architecture
students.
407
Bonferroni simultaneous intervals
• We carry out similar calculations for other types of students.
Compare technical school to medical technology students:
v
u
α u
w
x̄12 − x̄32 ± tn−g (
)t 22
2m
n−g
s
1
1
+
n1
n2
!
14588.9983 1
1
+
47.3913 − 38.0952 ± t79(0.05/24)
79
23
20
√
9.296 ± 2.951 184.671 × 0.09348
⇒
(−2.965, 21.577).
408
Bonferroni simultaneous intervals
• Architecture versus medical technology students:
v
u
w
α u
)t 22
x̄22 − x̄32 ± tn−g (
1
1
+
n1
n2
!
2m
n−g
√
13.089 ± 2.951 184.671 × 0.0713
⇒
(2.381, 23.797).
• Repeat these calculations for each of the other three
response variables to construct 12 confidence intervals
with simultaneous 95% confidence.
409
Test for Parallel Profiles
• Are the differences in mean scores for each pair of tests
consistent across all 3 student groups?
• Are the differences in mean scores between two student
groups consistent across all response variables?
• Parallel profiles are equivalent to no group × test interaction
410
Test for Parallel Profiles
• The no group × test interaction null hypothesis is expressed
as
[(µ1 + τ11) − (µ2 + τ12)] − [µ1 − µ2] = 0
[(µ2 + τ12) − (µ3 + τ13)] − [µ2 − µ3] = 0
[(µ3 + τ13) − (µ4 + τ14)] − [µ3 − µ4] = 0
[(µ1 + τ21) − (µ2 + τ22)] − [µ1 − µ2] = 0
[(µ2 + τ22) − (µ3 + τ23)] − [µ2 − µ3] = 0
[(µ3 + τ23) − (µ4 + τ24)] − [µ3 − µ4] = 0
411
Test for Parallel Profiles
• This is equivalent to
H0 : Ck×r βr×pMp×u
"
=
"
=
0 1 0
0 0 1
#
0 0 0
0 0 0
#

1
0
0
µ1 µ2 µ3 µ4

1
0
  −1

 τ11 τ12 τ13 τ14  
 0 −1
1
τ21 τ22 τ23 τ24
0
0 −1







where k = 2 rows in C
r = 3 rows in β
p = 4 response variables
u=3
n = n1 + n2 + n3 = 23 + 38 + 21 = 82
412
Test for Parallel Profiles
|W |
The value of Wilks Lambda is |B+W | = 0.569
a = (82 − 3) − (3 − 2 + 1)/2 = 78
v
u
u 3222 − 4
b = t 2
=2
3 + 22 − 5
c = ((3)(2) − 2)/2 = 2
√ ! 1− Λ
ab − c
√
F =
= 8.36
uk
Λ
on (uk, ab−c) = (6, 154) degrees of freedom and p-value < .0001
Conclude that the mean score profiles are not parallel. For at
least one pair of response variables, the difference in the mean
responses is not the same for all types of students.
413
Testing for parallel profiles using R
> M <- matrix(c(1, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, -1), ncol = 4)
> fitp <- manova((as.matrix(morel[, -1]) %*% t(M)) ~ morel[, 1])
> summary(fitp, test="Wilks")
Df Wilks approx F num Df den Df
Pr(>F)
morel[, 1] 2 0.5689
8.3624
6
154 7.461e-08 ***
Residuals 79
--Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1
1
414
MANOVA Using the (preferred) car package
> morel$studentgroup <- as.factor(morel$studentgroup)
> library(car)
> fit.lm <- lm(cbind(aptitude, mathematics, language,
generalknowledge)~studentgroup , data = morel)
> fit.manova <- Manova(fit.lm)
> summary(fit.manova)
Type II MANOVA Tests:
Sum of squares and products for error:
aptitude mathematics language generalknowledge
aptitude
55036.195
8140.0376 5569.9774
5490.1278
mathematics
8140.038 14588.9983 2619.2097
-128.0217
language
5569.977
2619.2097 4759.8655
-102.4044
generalknowledge 5490.128
-128.0217 -102.4044
7040.6248
415
Term: studentgroup
Sum of squares and products for the hypothesis:
aptitude mathematics language generalknowledge
aptitude
24600.207
6832.584 5709.5104
-2040.4327
mathematics
6832.584
2329.599 1570.1805
-1064.7222
language
5709.510
1570.181 1325.6955
-455.5712
generalknowledge -2040.433
-1064.722 -455.5712
743.4849
Multivariate Tests:
Df
Pillai
2
Wilks
2
Hotelling-Lawley 2
Roy
2
studentgroup
test stat approx F num Df den Df
Pr(>F)
0.4926982 6.292329
8
154 4.8238e-07 ***
0.5434483 6.773565
8
152 1.3843e-07 ***
0.7735884 7.252392
8
150 4.0901e-08 ***
0.6750591 12.994889
4
77 3.9151e-08 ***
416
> fit$SSPE
aptitude mathematics language generalknowledge
aptitude
55036.195
8140.0376 5569.9774
5490.1278
mathematics
8140.038 14588.9983 2619.2097
-128.0217
language
5569.977
2619.2097 4759.8655
-102.4044
generalknowledge 5490.128
-128.0217 -102.4044
7040.6248
> C <- matrix(c(0, 1, 0, 0, 1, -1), ncol = 3, by = T)
> M <- matrix(c(1, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, -1), nrow = 4,
by = T)
> newfit <- linearHypothesis(model = fit.lm, hypothesis.matrix = C)
> print(newfit)
Multivariate Tests:
Df test stat approx F num Df den Df
Pr(>F)
Pillai
2 0.4926982 6.292329
8
154 4.8238e-07 ***
Wilks
2 0.5434483 6.773565
8
152 1.3843e-07 ***
Hotelling-Lawley 2 0.7735884 7.252392
8
150 4.0901e-08 ***
Roy
2 0.6750591 12.994889
4
77 3.9151e-08 ***
417
> newfit <- linearHypothesis(model = fit.lm, hypothesis.matrix = C,
P = M)
> print(newfit)
Response transformation matrix:
[,1] [,2] [,3]
aptitude
1
0
0
mathematics
-1
1
0
language
0
-1
1
generalknowledge
0
0
-1
418
Sum of squares and products for the hypothesis:
[,1]
[,2]
[,3]
[1,] 13264.6375 363.6553 5115.040
[2,]
363.6553 514.9337 853.636
[3,] 5115.0403 853.6360 2980.323
Sum of squares and products for error:
[,1]
[,2]
[,3]
[1,] 53345.119 -9399.728 -2667.382
[2,] -9399.728 14110.444 -2115.038
[3,] -2667.382 -2115.038 12005.299
Multivariate Tests:
Df test stat approx F num Df den Df
Pillai
2 0.4539686 7.634505
6
156
Wilks
2 0.5689034 8.362413
6
154
Hotelling-Lawley 2 0.7175639 9.089142
6
152
Roy
2 0.6563063 17.063963
3
78
Pr(>F)
3.3709e-07
7.4605e-08
1.7098e-08
1.2962e-08
419
***
***
***
***