On the Analysis of Crossover Designs Dallas E. Johnson Professor Emeritus Kansas State University

On the Analysis of Crossover Designs

Dallas E. Johnson

Professor Emeritus

Kansas State University

[email protected]

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

SEQ 1 SEQ 2 Period 1 A B

Period 2 B A

We have n1 subjects randomly assigned tosequence 1, and n2 subjects randomlyassigned to sequence 2.

Suppose nj subjects are assigned to thejth sequence of treatments. Let yijk denotethe observed response from the kth subjectin 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), andall jk’s and ijk’s are independent.

Main Advantage:

Each subject serves as his/her owncontrol.

Most appropriate for conditions thatreoccur.

Disadvantages:

The designs cannot be used for treatmentcomparisons when the condition beingtreated is cured during the firstperiod.

The two period/two treatment crossoverdesign should not be used when carryovereffects exist unless one can include a“washout period” prior to administeringa treatment in the second period.

IDEAL TREATMENT STRUCTURE MODEL

ModelParameters SEQ 1 SEQ 2 Period 1

Period 2

IDEAL TREATMENT STRUCTURE MODEL


Period 2

CARRYOVER MODEL:


Period 2

NOTE:NOTE:


Period 2


Period 2

Therefore

,

,

.

Therefore

,

,

.

If carry-over exists, the difference intreatment effects can only be obtained fromPeriod 1 data. Then

.

If carry-over exists, the difference intreatment effects can only be obtained fromPeriod 1 data. Then

.

Usual Analysis

SV df MS F Hyp TstdSeq 1 S12 S1

2/S22

Sub(Seq) n1+n2-2 S22

Trt 1 S32 S32/S52

Per 1 S42 S42/S52

Error n1+n2-2 S52

Usual Analysis

SV df MS F Hyp TstdSeq 1 S12 S1

2/S22

Sub(Seq) n1+n2-2 S22

Trt 1 S32 S32/S52

Per 1 S42 S42/S52

Error n1+n2-2 S52

Question: What if the treatments have differentvariances?

Let yjk be the vector of responses for the kthsubject in the jth sequence of treatments, andsuppose we assume that

yjk ~ N( j, ) where

and where and * are determined by i and j.

1

2

jj

j

2 * 2 ** *2 11 1211 12 11 12

* * 2 * 2 *21 2221 22 21 22

1 11 1

ij i jS

Two period/two treatment crossover designswhere the treatments have unequalvariances.

Here we assume that

for the AB sequence

and that

for the BA sequence.

Two period/two treatment crossover designswhere the treatments have unequalvariances.

Here we assume that

for the AB sequence

and that

for the BA sequence.

A A B

A B B

2

2

B A B

A B A

2

2

Note that

2 211 21 1 2

1

~ ( , 2 ),

1, 2,...,

k k A B A A B By y N

k n

2 222 12 2 1

2

~ ( , 2 ),

1, 2,...,

k k A B A A B By y N

k n

and

Shanga (2003) showed that in the twoperiod/two treatment crossover designwithout carryover effects, the estimatedstandard errors for testing both betweensubject and within subject contrasts of the ij’s when the treatments have unequalvariances are exactly the same as they arewhen the treatments have equal variances. Thus if one is only interested in thedifference between the two treatments, thenthere is no need to worry about unequalvariances in this design.

Other questions:

Are there any consequences to ignoringcarryover and/or unequal variances in thetwo period/two treatment crossover design? In particular - does unequal variances showup as carryover or does carryover show upas 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) 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).

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;

N= 6, =.5, B=2

Analysis Assumptions

Simulation C0V0 C0V1 C1V0 C1V1

C0V0 =.040 (1) =.87 =.040 (1) =.87 = .050 (1) =.38 =.050 (1)=.38C0V1 =.045 (1) =.43 =.045 (1) =.43 = .050 (1) =.18 =.046 (1)=.17C1V0 =.124 (1) =.66 =.124 (1) =.66 = .050 (1) =.38 =.050 (1)=.38C1V1 =.066 (1) =.26 =.066 (1) =.26 = .050 (1) =.18 =.046 (1)=.17

N= 12, =.5, B=2 C0V0 =.048 (1) =1.0 =.048 (1) =1.0 = .055 (1) =.68 =.055 (1)=.66C0V1 =.055 (1) =.79 =.055 (1) =.80 = .055 (1) =.32 =.054 (1)=.31C1V0 =.214 (1) =.95 =.214 (1) =.95 = .055 (1) =.67 =.055 (1)=.66C1V1 =.102 (1) =.54 =.102 (1) =.54 = .055 (1) =.32 =.054 (1)=.31

Tests for equal treatment effects.

N= 18, =.5, B=2



C0V0 =.046 (1) =1.0 =.046 (1) =1.0 = .045 (1) =.83 =.045 (1)=.83C0V1 =.040 (1) =.92 =.040 (1) =.92 = .034 (1) =.47 =.034 (1)=.46C1V0 =.297 (1) =.99 =.297 (1) =.99 = .045 (1) =.83 =.045 (1)=.83C1V1 =.117 (1) =.69 =.117 (1) =.69 = .034 (1) =.47 =.034 (1)=.46

N= 30, =.5, B=2 C0V0 =.051 (1) =1.0 =.051 (1) =1.0 = .061 (1) =.96 =.061 (1)=.96C0V1 =.055 (1) =.99 =.055 (1) =.99 = .057 (1) =.67 =.055 (1)=.67C1V0 =.507 (1) =1.0 =.508 (1) =1.0 = .061 (1) =.96 =.061 (1)=.96C1V1 =.230 (1) =.92 =.230 (1) =.92 = .054 (1) =.67 =.054 (1)=.67


NOTE: Failing to assume carryover when carryover exists invalidatesthe tests for equal treatment effects and the invalidation generally gets worse as the

THREE TREATMENT - THREE PERIODCROSSOVER DESIGN

SEQUENCE

1 2 3 4 5 6 Period 1 A A B B C C

Period 2 B C A C A B

Period 3 C B C A B A

Carryover Case:

Table 1. Model parameters for the 3 period/3 treatmentcrossover design with carryover.

Sequence

Per 1 2 3 4 5 6

1 + 1+ A + 1+ A + 1+ B + 1+ B + 1+ C + 1+ C

2 + 2+ B+A + 2+ C+A + 2+ A+B + 2+ C+B + 2+ A+C + 2+ B+C

3 + 3+ C+B + 3+ B+C + 3+ C+A + 3+ A+C + 3+ B+A + 3+ A+B

Let represent the expected

response in Period i of Sequence j. Notethat

Let represent the expected

response in Period i of Sequence j. Notethat

Suppose nj subjects are assigned to the jthsequence of treatments.

Let yijk denote the observed response from thekth subject in the jth sequence during the ithperiod; i=1,2,3; j=1,2,...,6; k=1,2,...,nj.

Means Model:


where represents the expected response

in Period i of Sequence j.

Ideal Conditions:


Means Model:


where represents the expected response

in Period i of Sequence j.

Ideal Conditions:


Remarks:

A crossover experiment is reallya special type of a repeated measuresexperiment where the treatment ischanging over time.

We know that traditional ANOVAanalyses of repeated measuresexperiments are only valid when therepeated measures satisfy compoundsymmetry and tests for differencesacross time points are valid if and onlyif the repeated measures satisfy theHyuhn-Feldt Conditions.

Question:

What do the above remarks have to dowith the validity of our analyses ofcrossover experiments described previously?

The analyses that we have performed upto this point in time using our idealconditions are valid if the vector ofmeasurements on a subject within eachsequence satisfies compound symmetry. Thatis, the variance of the measured responseis the same for each treatment*periodcombination and the correlation betweenmeasurements in different periods is thesame for all pairs of periods.

Again, let yjk be the vector ofresponses for the kth subject in the jthsequence of treatments, and suppose weassume that

yjk ~ N( j, )

where , , and

and where and * are determined by i andj.

Again, let yjk be the vector ofresponses for the kth subject in the jthsequence of treatments, and suppose weassume that

yjk ~ N( j, )

where , , and

and where and * are determined by i andj.

j

j

j

p j

1

2

11 12 1

21 22 2

1 2

p

p

p p pp

ij i jS

Remark:

The covariance matrix is said topossess compound symmetry if

= 2[(1- )I + J]for some 2 and .

Remark:

The covariance matrix is said tosatisfy the H-F conditions if

= I + j’ + j ’ for some and .

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 tt 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 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 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:

2

21

2

A A B A C

B A B B C

C A C B C

Σ

Shanga simulated three period/three treatment crossover experiments satisfying four different conditions:


Each of 1000 sets of data under each of these conditions was analyzed four different ways assuming:


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;

RUN;

N= 6 =.5 B=2 C=4



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



N= 12 =.5B=2 C=4



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


N= 18 =.5 B=2 C=4



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

Simulation C1V0 C1V1N= 6A=1B=1 C=1

= .043 (.5) =.52(1) =.99

=.040 (.5)=.53(1) =.99

N= 6A=1B=.5 C=.25

= .054 (.5) =.86(1) =1.0

=.044 (.5)=.99(1) =1.0

N= 6A=1B=2 C=4

= .048 (.5) =.10(1) =.25

=.044 (.5)=.13(1) =.35

Tests for Carryover

Simulation C1V0 C1V1

N= 12 A=1B=1 C=1

= .040 (.5) =.85(1) =1.0

=.042 (.5)=.85(1) =1.0

N= 12 A=1B=.5 C=.25

= .047(.5) =1.0(1) =1.0

=.046 (.5)=1.0(1) =1.0

N= 12 A=1B=2 C=4

= .064 (.5) =.15(1) =.45

=.056 (.5)=.20(1) =.62

Tests for Carryover

In the three treatment/three period/six sequence crossover design, Shanga also considered testing

2 2 20 : A B CH

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.

That’s All For Now!

On the Analysis of Crossover Designs Dallas E. Johnson Professor Emeritus Kansas State University

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