On the Analysis of Crossover Designs
Dallas E. Johnson
Professor Emeritus
Kansas State University
dejohnsn@ksu.edu
785-532-0510 (Office)
785-539-0137 (Home)
Dallas E. Johnson
1812 Denholm Dr.
Manhattan, KS 66503-2210
CROSSOVER DESIGNS
Two Period - Two Treatment Crossover Design
Period 1
SEQ 1
A
SEQ 2
B
Period 2
B
A
We have n1 subjects randomly assigned to
sequence 1, and n2 subjects randomly
assigned to sequence 2.
Main Advantage:
Each subject serves as his/her own
control.
Most appropriate for conditions that
reoccur.
Disadvantages:
The designs cannot be used for treatment
comparisons when the condition being
treated is cured during the first
period.
The two period/two treatment crossover
design should not be used when carryover
effects exist unless one can include a
“washout period” prior to administering
a treatment in the second period.
IDEAL TREATMENT STRUCTURE MODEL
Model
Parameters
Period 1
Period 2
SEQ 1
SEQ 2
IDEAL TREATMENT STRUCTURE MODEL
Model
Parameters
Period 1
Period 2
NOTE:
SEQ 1
SEQ 2
Therefore
,
, and
.
Suppose nj subjects are assigned to the
jth sequence of treatments. Let yijk denote
the observed response from the kth subject
in the jth sequence during the ith period;
i=1,2; j=1,2; k = 1,2,...,nj.
Means Model:
yijk = ij + jk + ijk for all i,j,k
Suppose nj subjects are assigned to the
jth sequence of treatments. Let yijk denote
the observed response from the kth subject
in the jth sequence during the ith period;
i=1,2; j=1,2; k = 1,2,...,nj.
Means Model:
yijk = ij + jk + ijk for all i,j,k
Ideal Conditions:
jk ~ iid N(0,2),
ijk ~iid N(0,2), and
all jk’s and ijk’s are independent.
Let
be the average of all the
subjects in the (ij)th cell.
.
Then
Computer Analyses
Analysis of Variance Table
SV
Sub
Trt
Per
Error
df
MS
n1+n2-1 S12
1
S32
1
S42
n1+n2-2 S52
F
S32/S52
S42/S52
Hyp Tstd
CARRYOVER MODEL:
Model
Parameters
Period 1
Period 2
SEQ 1
SEQ 2
Model
Parameters
Period 1
Period 2
NOTE:
SEQ 1
SEQ 2
Therefore
,
,
.
If carry-over exists, the difference in
treatment effects can only be obtained from
Period 1 data. Then
.
Let
be the average of all the
subjects in
the (ij)th cell.
If
,
If
.
, then TA - TB must be estimated by
.
REMARKS:
1. In order to eliminate carryover effects, it
is customary to include a "washout" period
between the two treatment periods.
2. If carryover exists, then the two period two treatment crossover design offers no
advantages over a simple one period design.
That is, if carryover exists, the data that
occurs in the second period cannot be used
for estimating treatment differences.
3. The test for carryover effects may lack
power. If one suspects carryover, then one
must include a sufficient number of subjects
in order to have a reasonable chance of
detecting carryover. However, one should not
need any more subjects than would be required
if one were testing for treatment effects in
a one period design.
4. If one expects carryover, then the two period
- two treatment crossover design should
probably not be used as it would waste
resources.
Usual Analysis
SV
Seq
df
1
MS
F
S12 S12/S22
Sub(Seq) n1+n2-2 S22
Trt
1
S32
S32/S52
Per
1
S42
S42/S52
Error
n1+n2-2 S52
Hyp Tstd
Question: What if the ideal conditions are
not satisfied?
Suppose we look at each sequence of
treatments in a crossover experiment as a
vector of repeated measures on each
subject.
Question: What if the treatments have different
variances?
Let yjk be the vector of responses for the kth
subject in the jth sequence of treatments, and
suppose we assume that
yjk ~ N( j,  )
where
 1 j 
 j   
 2j
 *11  *12   2   *11  2   *12   11  12 
1
1


  
 *
*    2
*
2
*  
1
1

  21  22      21     22   21  22 
2
ij     i  S j     

*
and where and 
are determined by i and j.
Two period/two treatment crossover designs
where the treatments have unequal
variances.
Here we assume that
  A2
 
  A B
 A B 

 B2 
for the AB sequence
and that
  B2
 
  A B
 A B 

 A2 
for the BA sequence.
Note that
y11k  y21k ~ N ( A   B  1   2 ,  A2  2  A B   B2 ),
k  1, 2,..., n1
and
y22 k  y12 k ~ N ( A   B   2  1 ,  A2  2  A B   B2 ),
k  1, 2,..., n2
Shanga (2003) showed that in the two
period/two treatment crossover design
without carryover effects, the estimated
standard errors for testing both between
subject and within subject contrasts of the
ij’s when the treatments have unequal
variances are exactly the same as they are
when the treatments have equal variances.
Thus if one is only interested in the
difference between the two treatments, then
there is no need to worry about unequal
variances in this design.
Other questions:
Are there any consequences to ignoring
carryover and/or unequal variances in the
two period/two treatment crossover design?
In particular - does unequal variances show
up as carryover or does carryover show up
as unequal variances?
To answer these kinds of questions, Shanga simulated
two period/two treatment crossover experiments
satisfying four different conditions:
(1) no carryover and equal variances (C0V0),
(2) no carryover and unequal variances(C0V1),
(3) carryover and equal variances (C1V0), and
(4) carryover and unequal variances (C1V1).
Each of 1000 sets of data under each of these
conditions was analyzed four different ways
assuming:
(1)
(2)
(3)
(4)
no carryover and equal variances (C0V0),
no carryover and unequal variances(C0V1),
carryover and equal variances (C1V0), and
carryover and unequal variances (C1V1).
PROC MIXED;
TITLE2 'EQUAL VARIANCES';
CLASSES
SEQ PERIOD TRT PERSON;
MODEL PEF=SEQ TRT PERIOD/DDFM=SATTERTH;
REPEATED TRT/SUBJECT=PERSON(SEQ) TYPE=CS;
LSMEANS TRT /PDIFF;
RUN;
PROC MIXED;
TITLE2 'UNEQUAL VARIANCES';
CLASSES
SEQ PERIOD TRT PERSON;
MODEL PEF=SEQ TRT PERIOD/DDFM=SATTERTH;
REPEATED TRT/SUBJECT=PERSON(SEQ) TYPE=CSH;
LSMEANS TRT /PDIFF;
RUN;
Tests for equal treatment effects.
N= 6, =.5,
B=2
Analysis Assumptions
Simulation
C0V0
C0V1
C1V0
C1V1
C0V0
=.040 (1) =.87
=.040 (1) =.87
= .050 (1) =.38
=.050 (1)=.38
C0V1
=.045 (1) =.43
=.045 (1) =.43
= .050 (1) =.18
=.046 (1)=.17
C1V0
=.124 (1) =.66
=.124 (1) =.66
= .050 (1) =.38
=.050 (1)=.38
C1V1
=.066 (1) =.26
=.066 (1) =.26
= .050 (1) =.18
=.046 (1)=.17
C0V0
=.048 (1) =1.0
=.048 (1) =1.0
= .055 (1) =.68
=.055 (1)=.66
C0V1
=.055 (1) =.79
=.055 (1) =.80
= .055 (1) =.32
=.054 (1)=.31
C1V0
=.214 (1) =.95
=.214 (1) =.95
= .055 (1) =.67
=.055 (1)=.66
C1V1
=.102 (1) =.54
=.102 (1) =.54
= .055 (1) =.32
=.054 (1)=.31
N= 12, =.5,
B=2
Tests for equal treatment effects.
N= 18, =.5,
B=2
Analysis Assumptions
Simulation
C0V0
C0V1
C1V0
C1V1
C0V0
=.046 (1) =1.0
=.046 (1) =1.0
= .045 (1) =.83
=.045 (1)=.83
C0V1
=.040 (1) =.92
=.040 (1) =.92
= .034 (1) =.47
=.034 (1)=.46
C1V0
=.297 (1) =.99
=.297 (1) =.99
= .045 (1) =.83
=.045 (1)=.83
C1V1
=.117 (1) =.69
=.117 (1) =.69
= .034 (1) =.47
=.034 (1)=.46
C0V0
=.051 (1) =1.0
=.051 (1) =1.0
= .061 (1) =.96
=.061 (1)=.96
C0V1
=.055 (1) =.99
=.055 (1) =.99
= .057 (1) =.67
=.055 (1)=.67
C1V0
=.507 (1) =1.0
=.508 (1) =1.0
= .061 (1) =.96
=.061 (1)=.96
C1V1
=.230 (1) =.92
=.230 (1) =.92
= .054 (1) =.67
=.054 (1)=.67
N= 30, =.5,
B=2
NOTE: Failing to assume carryover
when carryover exists invalidates
the tests for equal treatment
effects and the invalidation
generally gets worse as the
THREE TREATMENT - THREE PERIOD
CROSSOVER DESIGN
Period 1
1 2
A A
SEQUENCE
3 4 5
B B C
6
C
Period 2
B C
A
C
A
B
Period 3
C B
C
A
B
A
Let
represent the expected
response in Period i of Sequence j.
that
Note
Suppose nj subjects are assigned to the jth
sequence of treatments.
Let yijk denote the observed response from the
kth subject in the jth sequence during the ith
period;
i=1,2,3; j=1,2,...,6; k=1,2,...,nj.
Means Model:
yijk = ij + jk + ijk for all i,j,k
where
represents the expected response
in Period i of Sequence j.
Ideal Conditions:
jk ~ iid N(0,2),
ijk ~iid N(0,2), and
all jk’s and ijk’s are independent.
No Carryover Case:
=  +  i + Sj +   where the
subscript is determined by i and j.
An ANOVA Table:
Source
Seq
Subj(Seq)
df
5
N-6
MS
S12
S22
F
S12/S22
Trt
Per
Error
2
2
2(N-2)
S32
S42
S52
S32/S52
S42/S52
Let
  y
ij
ij
.
Note that
E ( ij )     i  S j   
E ( i  )     i  S   
E (  j )       S j   
E(
11  12  23  25  34  36
E(
13  14  21  26  32  35
E(
6
)       S   A
6
15  16  22  24  31  33
6
)       S   B
)       S   C
Note:
1. Contrasts in the i  ' s estimate period
effects.
2. Contrasts in the  j ' s estimate
sequence effects.
3. Contrasts in the last three estimate
treatment effects.
TITLE1 'A THREE PERIOD/THREE TRT DESIGN';
TITLE2 'ANALYSIS ASSUMES NO CARRY-OVER';
PROC MIXED;
TITLE3 'ANALYSIS USING SAS-MIXED';
CLASSES SEQ PER TRT SUBJ;
MODEL Y=SEQ TRT PER/DDFM=SATTERTH;
RANDOM SUBJ(SEQ);
LSMEANS TRT PER/PDIFF;
RUN;
Analyses assuming carryover.
Carryover Case:
Table 1. Model parameters for the 3 period/3 treatment
crossover design with carryover.
Sequence
Per
1
2
3
+ 1+ A
+ 1+ B
4
5
+1+B
+1+ C
+2+A+C
1
+ 1+A
2
+2+ B+A
+ 2+C+A
+ 2+A+B
+ 2+C+B
3
+ 3+C+B
+ 3+B+C
+ 3+C+A
+ 3+A+C
+ 3+ B+A
6
+ 1+C
+ 2+B+C
+ 3+A+B
QUESTION: How should the data be analyzed?
QUESTION: How should one define treatment
main effects?
Defn. 1: A Trt A main effect mean could be
defined as the average of all cells that received
Trt A.
A:
B:
C:
Carryover Case:
Table 1. Model parameters for the 3 period/3 treatment
crossover design with carryover.
Sequence
Per
1
2
1
+ 1+A+o
+1+ A+o
2
+ 2+B+A
+2+ C+A
3
+ 3+C+B
+3+ B+C
3
4
5
6
+ 1+ B+o
+1+ C+o
+ 1+ C+o
+ 2+ A+B
+ 2+ C+B
+2+ A+C
+ 2+ B+C
+ 3+ C+A
+ 3+ A+C
+3+ B+A
+ 3+ A+B
+1+B+o
Defn. 1: A Trt A main effect mean could be
defined as the average of all cells that
received Trt A.
A:
B:
C:
Unfortunately, contrasts in these would be
confounded with carryover, so these definitions
don't make much sense. However, if
then treatment main effects could be defined as
above.
If
no carryover effects.
, we say that there are
,
A second definition of treatment main
effects is:
Defn. 2:
A:
B:
C:
Contrasts in Defn. 2 main effect means
are not confounded with carry-over
regardless of whether
or not.
A third possible definition is:
Defn. 3:
A:
B:
C:
Contrasts in Defn. 3 main effect means
are also not confounded with carry-over
regardless of whether
or not.
Question: Are the functions given in Defns.
2 and 3 estimable?
Answer:
Yes provided that there is at
least one subject within each
sequence that is measured in all
three periods.
(1311+1312- 213+ 14- 215+ 16
+421+422 +1023- 224+1025 -226
-531-532+ 433+1334 +435+1336)/72 =
.
Similar results hold for the other two treatment main
effect means in Defn. 2.
Period 1
13
SEQUENCE
1
2
3
4
A 13 A -2 B 1 B
Period 2
4
B
4
C
10
A
-2
C
10
A
-2
B
Period 3
-5
C
-5
B
4
C
13
A
4
B
13
A
-2
5
C
6
1 C
(1311+1312- 213+ 14- 215+ 16
+421+422 +1023- 224+1025 -226
-531-532+ 433+1334 +435+1336)/72 =
.
It is fairly obvious that the functions
in Defn. 3 are estimable.
A:
B:
C:
PROC GLM;
CLASSES SEQ PER TRT PRIORTRT SUBJ;
MODEL Y = SEQ SUBJ(SEQ) TRT PER PRIORTRT/E E3;
LSMEANS TRT PER PRIORTRT/PDIFF STDERR E;
OPTIONS NODATE PAGENO=1;
TITLE1 'CRSOVR EXAMPLE #3 - A THREE PERIOD/THREE TRT
DESIGN';
TITLE2 'ANALYSIS ASSUMES NO CARRY-OVER';
DATA ONE;
INPUT SEQ PER TRT $ PRIORTRT $ @@;
DO N=1 TO 6;
INPUT Y @@; OUTPUT; END;
CARDS;
1 1 A O 20.1 23.3 23.4 19.7 19.2 22.2
1 2 B A 20.3 24.8 24.8 21.3 20.9 22.0
1 3 C B 25.6 28.7 28.3 25.7 25.9 26.2
2 1 A O 24.7 23.8 23.6 20.2 19.8 21.5
2 2 C A 29.4 28.7 26.4 26.2 23.7 25.5
2 3 B C 27.5 24.1 25.0 21.4 23.3 20.8
PROC GLM;
CLASSES SEQ PER TRT PRIORTRT SUBJ;
MODEL Y = SEQ SUBJ(SEQ) TRT PER PRIORTRT/E E3;
LSMEANS TRT PER PRIORTRT/TDIFF PDIFF STDERR E;
RUN;
Incorrect
Source
DF
Type III SS
Mean Square
F Value
Pr > F
5
53.6725833
10.7345167
10.70
<.0001
30
307.7788889
10.2592963
10.23
<.0001
TRT
2
249.7263611
124.8631806
124.51
<.0001
PER
1
15.1250000
15.1250000
15.08
0.0002
PRIORTRT
2
4.4492130
2.2246065
2.22
0.1168
SEQ
SUBJ(SEQ)
Why 1?
Y LSMEAN
LSMEAN
Number
A
Non-est
1
B
Non-est
2
C
Non-est
3
TRT
Least Squares Means for Effect TRT
t for H0: LSMean(i)=LSMean(j) / Pr > |t|
Dependent Variable: Y
i/j
1
1
2
-3.1394
<.0001
3
11.8234
<.0001
2
3
3.139398
<.0001
-11.8234
<.0001
-14.9628
<.0001
14.9628
<.0001
Y LSMEAN
LSMEAN
Number
A
Non-est
1
B
Non-est
2
C
Non-est
3
O
Non-est
4
PRIORTRT
Least Squares Means for Effect PRIORTRT
t for H0: LSMean(i)=LSMean(j) / Pr > |t|
Dependent Variable: Y
i/j
1
1
W
R
G
3
4
-0.77084
<.0001
1.312187
<.0001
.
.
2.083024
<.0001
.
.
2
0.770836
<.0001
3
-1.31219
<.0001
-2.08302
<.0001
4
.
.
.
.
O
N
2
.
.
.
.
PROC MIXED;
CLASSES SEQ PER TRT PRIORTRT SUBJ;
MODEL Y = SEQ SUBJ(SEQ) TRT PER PRIORTRT/
DDFM=SATTERTH;
LSMEANS TRT PER PRIORTRT/PDIFF STDERR;
RUN;
ESTIMATE 'A LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 9 0 0
PRIORTRT 2 2 2 3/DIVISOR=9;
ESTIMATE 'B LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 0 9 0
PRIORTRT 2 2 2 3/DIVISOR=9;
ESTIMATE 'C LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 0 0 9
PRIORTRT 2 2 2 3/DIVISOR=9;
Covariance Parameter Estimates
Cov Parm
Estimate
SUBJ(SEQ)
3.0855
Residual
1.0028
ˆ 2
ˆ 2
Type 3 Tests of Fixed Effects
Num Den
DF DF
F Value
Pr > F
SEQ
5 30.6
1.05
0.4072
TRT
2
66
PER
1
66
15.08
0.0002
PRIORTRT
2
66
2.22
0.1168
Effect
124.51 <.0001
Estimates
Label
Estimate Standard Error
DF
t Value
Pr > |t|
A LSM DFN 2
23.6769
0.3438
45.2
68.87
<.0001
B LSM DFN 2
22.8484
0.3438
45.2
66.46
<.0001
C LSM DFN 2
26.7970
0.3438
45.2
77.94
<.0001
A LSM DFN 3
22.3750
0.3698
57.2
60.50
<.0001
B LSM DFN 3
21.5465
0.3698
57.2
58.26
<.0001
C LSM DFN 3
25.4951
0.3698
57.2
68.94
<.0001
A-B
0.8285
0.2639
66
3.14
0.0025
A-C
-3.1201
0.2639
66
-11.82
<.0001
B-C
-3.9486
0.2639
66
-14.96
<.0001
PER 1 LSM
23.1389
0.3370
42.1
68.66
<.0001
PER 2 LSM
24.6333
0.3370
42.1
73.10
<.0001
PER 3 LSM
25.5500
0.3370
42.1
75.82
<.0001
C_A
25.1556
0.3761
60
66.89
<.0001
C_B
25.4285
0.3761
60
67.62
<.0001
C_C
24.6910
0.3761
60
65.66
<.0001
C_A - C_B
-0.2729
0.3541
66
-0.77
0.4436
C_A - C_C
0.4646
0.3541
66
1.31
0.1940
C_B - C_C
0.7375
0.3541
66
2.08
0.0411
Estimates
Label
Estimate Standard Error
DF
t Value
Pr > |t|
A LSM DFN 2
23.6769
0.3438
45.2
68.87
<.0001
B LSM DFN 2
22.8484
0.3438
45.2
66.46
<.0001
C LSM DFN 2
26.7970
0.3438
45.2
77.94
<.0001
A LSM DFN 3
22.3750
0.3698
57.2
60.50
<.0001
B LSM DFN 3
21.5465
0.3698
57.2
58.26
<.0001
C LSM DFN 3
25.4951
0.3698
57.2
68.94
<.0001
A-B
0.8285
0.2639
66
3.14
0.0025
A-C
-3.1201
0.2639
66
-11.82
<.0001
B-C
-3.9486
0.2639
66
-14.96
<.0001
PER 1 LSM
23.1389
0.3370
42.1
68.66
<.0001
PER 2 LSM
24.6333
0.3370
42.1
73.10
<.0001
PER 3 LSM
25.5500
0.3370
42.1
75.82
<.0001
C_A
25.1556
0.3761
60
66.89
<.0001
C_B
25.4285
0.3761
60
67.62
<.0001
C_C
24.6910
0.3761
60
65.66
<.0001
C_A - C_B
-0.2729
0.3541
66
-0.77
0.4436
C_A - C_C
0.4646
0.3541
66
1.31
0.1940
C_B - C_C
0.7375
0.3541
66
2.08
0.0411
Contrasts
Num
DF
Den
DF
F Value
Pr > F
TRT
2
66
124.51
<.0001
PERIOD
2
66
53.17
<.0001
CARRYOVER
2
66
2.22
0.1168
Label
Remarks:
A crossover experiment is really
a special type of a repeated measures
experiment where the treatment is
changing over time.
We know that traditional ANOVA
analyses of repeated measures
experiments are only valid when the
repeated measures satisfy compound
symmetry and tests for differences
across time points are valid if and only
if the repeated measures satisfy the
Hyuhn-Feldt Conditions.
Question:
What do the above remarks have to do
with the validity of our analyses of
crossover experiments described previously?
The analyses that we have performed up
to this point in time using our ideal
conditions are valid if the vector of
measurements on a subject within each
sequence satisfies compound symmetry. That
is, the variance of the measured response
is the same for each treatment*period
combination and the correlation between
measurements in different periods is the
same for all pairs of periods.
Again, let yjk be the vector of
responses for the kth subject in the jth
sequence of treatments, and suppose we
assume that
yjk ~ N( j,  )
where
 1 j 
 
2j
j   
  
 
  pj 
,
  11  12

 22
21



 

  p1  p 2
  1p 
 2p 
 , and
  

  pp 
ij     i  S j     

*
and where and 
are determined by i and
j.
Remark:
The covariance matrix  is said to
possess compound symmetry if
 = 2[(1- )I +  J]for some 2 and  .
Remark:
The covariance matrix 
satisfy the H-F conditions if
is
said
 = I +  j’ + j ’ for some  and  .
to
Goad and Johnson (2000) showed:
(1)
If  satisfies the H-F conditions, then the
traditional tests for treatment and period
effects are valid for all crossover experiments
both with and without carryover.
(2) There are cases where the ANOVA tests are valid even
when  does not satisfy the H-F conditions.
(a) In the no carryover case, tests for equal treatment
effects are valid for the six sequence three
period/three treatment crossover design when
there are an equal number of subjects assigned
to each sequence.
(b) In the no carryover case, tests for equal
period effects are valid only when the H-F
conditions be satisfied
(b)
The traditional tests for equal treatment effects and
equal period effects are valid for a crossover design
generated by t-1 mutually orthogonal tt Latin
squares when there are equal numbers of subjects
assigned to each sequence.
(c)
The traditional tests for equal treatment effects, equal
period effects, and equal carryover effects are likely
to be invalid in the four period/four treatment design
regardless of whether carryover exists or not.
Cases where the validity of ANOVA tests are still in
doubt.
(4)
When carryover exists, the tests for
equal carryover effects are not valid
unless E satisfies the H-F conditions.
(5)
When there are unequal numbers of
subjects assigned to each sequence, the
ANOVA tests are unlikely to be valid
unless E satisfies the H-F conditions.
Goad and Johnson (2000) provide some
alternative analyses for crossover experiments.
Consider again, the three period/three
treatment crossover design in six sequences.
Question: Suppose the variance of a response depends on
the treatment, but that the correlation is the same between all
pairs of sequence cells. That is, for Sequence 1, the
covariance matrix is:
  A2
Σ1    B A
  
 C A
 A B
 B2
 C B
 A C 
 B C 
 C2 
Shanga simulated three period/three treatment
crossover experiments satisfying four different
conditions:
(1)
(2)
(3)
(4)
no carryover and equal variances (C0V0),
no carryover and unequal variances(C0V1),
carryover and equal variances (C1V0), and
carryover and unequal variances (C1V1).
Each of 1000 sets of data under each of these
conditions was analyzed four different ways
assuming:
(1)
(2)
(3)
(4)
no carryover and equal variances (C0V0),
no carryover and unequal variances(C0V1),
carryover and equal variances (C1V0), and
carryover and unequal variances (C1V1).
TITLE1 'CRSOVR EXAMPLE - A THREE PERIOD/THREE TRT DESIGN';
TITLE2 'ASSUMES CARRYOVER AND UNEQUAL VARIANCES';
PROC MIXED;
CLASSES SEQ PER TRT PRIORTRT SUBJ;
MODEL Y = SEQ TRT PER PRIORTRT/DDFM=KR;
LSMEANS TRT PER PRIORTRT/PDIFF;
REPEATED TRT/SUBJECT=SUBJ TYPE=CSH;
ESTIMATE 'A LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 9 0 0
PRIORTRT 2 2 2 3/DIVISOR=9;
ESTIMATE 'B LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 0 9 0
PRIORTRT 2 2 2 3/DIVISOR=9;
ESTIMATE 'C LSM DFN 2' INTERCEPT 9 PER 3 3 3 TRT 0 0 9
PRIORTRT 2 2 2 3/DIVISOR=9;
Tests for equal treatment effects.
N= 6
=.5
B=2
C=4
Analysis Assumptions
Simulation
C0V0
C0V1
C1V0
C1V1
C0V0
=.053
(1) =1.0
=.057
(1) =1.0
= .057
(1) =1.0
=.051
(1)=1.0
C0V1
=.066
(1) =.50
=.057
(1) =.88
= .049
(1) =.42
=.049
(1)=.67
C1V0
=.138
(1) =1.0
=.149
(1) =1.0
= .057
(1) =1.0
=.051
(1)=1.0
C1V1
=.070
(1) =.32
=.069
(1) =.73
= .049
(1) =.42
=.049
(1)=.67
Tests for equal treatment effects.
N= 12
=.5
B=2
C=4
Analysis Assumptions
Simulation
C0V0
C0V1
C1V0
C1V1
C0V0
=.049
(1) =1.0
=.052
(1) =1.0
= .054
(1) =1.0
=.055
(1)=1.0
C0V1
=.070
(1) =.89
=.053
(1) =.99
= .055
(1) =.77
=.046
(1)=.94
C1V0
=.227
(1) =1.0
=.232
(1) =1.0
= .054
(1) =1.0
=.055
(1)=1.0
C1V1
=.081
(1) =.70
=.100
(1) =.97
= .055
(1) =.77
=.046
(1)=.94
Tests for equal treatment effects.
N= 18
=.5
B=2
C=4
Analysis Assumptions
Simulation
C0V0
C0V1
C1V0
C1V1
C0V0
=.054
(1) =1.0
=.056
(1) =1.0
= .048
(1) =1.0
=.053
(1)=1.0
C0V1
=.071
(1) =.99
=.051
(1) =1.0
= .054
(1) =.91
=.051
(1)=.99
C1V0
=.370
(1) =1.0
=.378
(1) =1.0
= .048
(1) =1.0
=.053
(1)=1.0
C1V1
=.094
(1) =.90
=.125
(1) =1.0
= .054
(1) =.91
=.051
(1)=.99
Tests for Carryover
Simulation
C1V0
C1V1
N= 6
A=1
B=1
C=1
= .043
(.5) =.52
(1) =.99
=.040
(.5)=.53
(1) =.99
N= 6
A=1
B=.5
C=.25
= .054
(.5) =.86
(1) =1.0
=.044
(.5)=.99
(1) =1.0
N= 6
A=1
B=2
C=4
= .048
(.5) =.10
(1) =.25
=.044
(.5)=.13
(1) =.35
Tests for Carryover
Simulation
C1V0
C1V1
N= 12
A=1
B=1
C=1
= .040
(.5) =.85
(1) =1.0
=.042
(.5)=.85
(1) =1.0
N= 12
A=1
B=.5
C=.25
= .047
(.5) =1.0
(1) =1.0
=.046
(.5)=1.0
(1) =1.0
N= 12
A=1
B=2
C=4
= .064
(.5) =.15
(1) =.45
=.056
(.5)=.20
(1) =.62
In the three treatment/three period/six sequence
crossover design, Shanga also considered testing
H 0 :  A2   B2   C2
Shanga claimed that his tests were LRTs, but Jung
(2008) has shown that they are not LRTs.
Nevertheless, Shanga's tests had good power for
detecting unequal variances.