Designs for Estimating

Carry-over (or Residual) Effects of Treatments

Ref: “Design and Analysis of Experiments” Roger G. Petersen, Dekker 1985

The Cross-over or Simple Reversal Design An Example• A clinical psychologist wanted to test two drugs,

A and B, which are intended to increase reaction time to a certain stimulus.

• He has decided to use n = 8 subjects selected at random and randomly divided into two groups of four. – The first group will receive drug A first then B, while

– the second group will receive drug B first then A.

To conduct the trial he administered a drug to the individual, waited 15 minutes for absorption, applied the stimulus and then measured reaction time. The data and the design is tabulated below:

Group 1 (Subjects)

Period Drug 1 2 3 4 1 A 30 54 42 56 2 B 28 50 38 49

Group 2 (Subjects)

Period Drug 1 2 3 4 1 B 22 44 18 28 2 A 21 41 17 26

The Switch-back or Double Reversal Design

An Example• A following study was interested in the

effect of concentrate type on the daily production of fat-corrected milk (FCM) .

• Two concentrates were used: – A - high fat; and – B - low fat.

• Five test animals were then selected for each of the two sequence groups – ( A-B-A and B-A-B) in a switch-back design.

The data and the design is tabulated below:

One animal in the first group developed mastitis and was removed from the study.

Group 1 (Test animal) Period Treatment 1 2 3 4 1 A 40.8 21.5 48.4 50.3 2 B 35.2 18.4 44.4 45.7 3 A 30.8 17.8 42.7 43.8

Group 2 (Test animal)

Period Treatment 1 2 3 4 5 1 B 43.3 27.6 57.8 49.4 36.6 2 A 40.9 30.2 53.2 48.5 35.9 3 B 37.6 27.4 45.5 45.5 35.3

The Incomplete Block Switch-back Design

An Example• An insurance company was interested in buying a

quantity of word processing machines for use by secretaries in the stenographic pool.

• The selection was narrowed down to three models (A, B, and C).

• A study was to be carried out , where the time to process a test document would be determined for a group of secretaries on each of the word processing models.

• For various reasons the company decided to use an incomplete block switch back design using n = 6 secretaries from the secretarial pool.

The data and the design is tabulated below:

Table: Time required to process the test document together with word processing model (in brackets) Secretary Period 1 2 3 4 5 6

1 38.7(A) 21.8(B) 48.9(A) 29.9(C) 25.7(B) 22.4(C) 2 37.4(B) 23.9(A) 43.9(C) 35.1(A) 23.1(C) 26.0(B) 3 34.4(A) 21.7(B) 42.0(A) 24.5(C) 23.4(B) 20.9(C)

BIB incomplete block design with t = 3 treatments – A, B and block size k = 2.

A

B

A

C

B

C

Designs for Estimating

Carry-over (or Residual) Effects of Treatments

Ref: “Design and Analysis of Experiments” Roger G. Petersen, Dekker 1985

The Latin Square Change-Over (or Round Robin) Design

Selected Latin Squares Change-Over Designs (Balanced for Residual Effects)

Period = Rows Columns = Subjects

Three Treatments A B C A B C B C A C A B C A B B C A

The Latin Square Change-Over (or Round Robin) Design

Selected Latin Squares Change-Over Designs (Balanced for Residual Effects)

Period = Rows Columns = Subjects

Three Treatments A B C A B C B C A C A B C A B B C A

Four Treatments

An Orthogonal Set (use the complete Set) A B C D A B C D A B C D B A D C C D A B D C B A C D A B D C B A B A D C D C B A B A D C C D A B A Balanced Single Latin Square A B C D B D A C C A D B D C B A

An Example

• An experimental psychologist wanted to determine the effect of three new drugs (A, B and C) on the time for laboratory rats to work their way through a maze.

• A sample of n= 12 test animals were used in the experiment.

• It was decided to use a Latin square Change-Over experimental design.

The data and the design is tabulated below:

Table: Time required for test animals (rats) to negotiate a maze After drug treatment (A, B, or C) Latin Square 1 Latin Square 2 Period\Rat 1 2 3 4 5 6 1 138 A 209 B 224 C 186 A 175 B 201 C 2 125 B 186 C 172 A 176 C 135 A 163 B 3 115 C 139 A 127 B 146 B 134 C 101 A Latin Square 3 Latin Square 4 Period\Rat 7 8 9 10 11 12 1 166 A 194 B 186 C 138 A 164 B 168 C 2 152 B 180 C 130 A 154 C 128 A 150 B 3 137 C 97 A 123 B 129 B 137 C 106 A

Analysis : The Latin Square Change-Over (or Round Robin) Design

Assume that we have q p × p Latin Squares (Balanced for Residual Effects)

Period = Rows Columns = Subjects

e.g. q = 3 4×4 Latin squares A B C D A B C D A B C D B A D C C D A B D C B A C D A B D C B A B A D C D C B A B A D C C D A B

Notation

) treatment the(recieving period the

during square in thesubject theof response )(

thth

ththlijk

lk

ijy

square in the period for the sum )(thth

jlijkik ikyQ

square in thesubject for the sum )(thth

klijkij ijyU

square in the treatment for the sum ththil ilT

square in the treatment e th

following nsobservatio of sum thth

il

il

F

square for sum th

lil

jij

kiki iTUQS

Square 1

Period/subject 11 12 1p Total

1 y111(A) y121(B) y1p1(M) Q11

2 y112 (B) y122(A) Y1p2(D) Q12

p y11p(M) y12p(C) y1pp(A) Q1p

Total U11 U12 U1p S1

Treatment A B M

Total T1A T1B T1M S1

Residual Sum F1A F1B F1M

Square 2

Period/subject 21 22 2p Total

1 y211(B) y221(M) y2p1(A) Q21

2 y212 (C) y212(D) y2p2(M) Q22

p y21p(A) y22p(M) y2pp(E) Q2p

Total U21 U22 U2p S2

Treatment A B M

Total T2A T2B T2M S2

Residual Sum F2A F2B F2M

Square q

Period/subject q1 q2 qp Total

1 yq11(E) yq21(G) yqp1(D) Qq1

2 yq12 (C) yq12(D) yqp2(E) Qq2

p yq1p(F) yq2p(E) yqpp(A) Qqp

Total Uq1 Uq2 Uqp Sq

Treatment A B M

Total TqA TqB TqM Sq

Residual Sum FqA FqB FqM

Some other sumsperiod for the sum th

iikk kQP

period )(last theduring applied was

reatment in which t subjects allfor theof sum th

ijl

p

lUC

treatment for the sum th

iill lTD

treatment thefollowing nsobservatio of sum th

iill lFR

Total Grand k

kPG

Adjusted Effects

effectdirect adjusted

1 12*

pGPCpRDppD llll

effectover -carry adjusted

212*

GppPpCRppDR llll

effectpermanent adjusted

***

lll RDT

ANOVA table entries

2

22

)( qp

GySS

i j klijkTotal

2

22

2

1

qp

GS

pSS

iiSquares

Squarei j

ijSquaresinSubjects SSqp

GU

pSS 2

22

1

2

221

qp

GP

pqSS

kkPeriod

PeriodSquarei k

ikSquarePeriod SSSSqp

GQ

pSS 2

221

2

221

qp

GD

pqSS

llunadjustedTreatment

unadjustedTreatmentSquarei l

ilSquareTreatment SSSSqp

GT

pSS 2

221

SquareTreatment

adjustedresidualunadjustedTreatment

SquarePeriodPeriodSubjectSquareTotalError

SS

SSSS

SSSSSSSSSSSS

lladjustedresidual R

ppqpSS 2*

3 21

1

squares of sumeffect over -carry adjusted

21

1 2*3

l

ladjustedresidual Rppqp

SS

squares of sumeffect Treatment adjusted

121

1 2*2

l

ladjustedTreatment Dppppqp

SS

squares of sumeffect Permanent adjusted

1221

1 2*2

l

ladjustedPermanent Tpppqp

SS

Some additional S.S.

The ANOVA Table

Source df

Total qp2 - 1

Squares q - 1

Subjects in Squares q(p - 1)

Periods p - 1

Period × Square (p - 1)(q - 1)

Treatment(unadjusted) p - 1

Treatment × Square (p - 1)(q - 1)

Residual(adjusted) p - 1

Error (p - 1)(qp -2p - 1)

Treatment(adjusted) p - 1

Permanent(adjusted) p - 1

2

**

21 qp

G

ppqp

Dy l

l

Means and their sample variances after adjustment

*

2*

21

1MSE

ppqp

ppyV l

21

*

ppqp

Rr ll

*

21MSE

ppq

prV l

2

*

21 qp

G

ppqp

Ty lPerm

l

*

2

12MSE

pqp

pyV Perm

l

2. Adjusted residual effect

1. Adjusted treatment mean

3. Permanent treatment mean

An Example

• An experimental psychologist wanted to determine the effect of three new drugs (A, B and C) on the time for laboratory rats to work their way through a maze.

• A sample of n= 12 test animals were used in the experiment.

• It was decided to use a Latin square Change-Over experimental design.

The data and the design is tabulated below:

Table: Time required for test animals (rats) to negotiate a maze After drug treatment (A, B, or C) Latin Square 1 Latin Square 2 Period\Rat 1 2 3 4 5 6 1 138 A 209 B 224 C 186 A 175 B 201 C 2 125 B 186 C 172 A 176 C 135 A 163 B 3 115 C 139 A 127 B 146 B 134 C 101 A Latin Square 3 Latin Square 4 Period\Rat 7 8 9 10 11 12 1 166 A 194 B 186 C 138 A 164 B 168 C 2 152 B 180 C 130 A 154 C 128 A 150 B 3 137 C 97 A 123 B 129 B 137 C 106 A

The ANOVA Table

Source df SS MS F

Total 35 33,624.97

Squares 3 1,738.30 579.43 6.88*

Rats in Squares 8 5,944.67 743.08 8.82**

Periods 2 18,093.56 9,046.78 107.43**

Period × Square 6 1,001.11 168.85 1.98

Drug(unadjusted) 2 5,549.06 2,774.53 32.95**

Drug × Square 6 496.28 82.71 0.98

Carry-over(adjusted) 2 296.72 148.36 1.76

Error 6 505.27 84.21

Drug(adjusted) 2 5,524.63 2,762.32 32.80**

Permanent(adjusted) 2 2,373.23 1,186.62 14.09**

Summary Statistics

period 1 2 3mean 179.1 154.3 124.3

Period Means

Treatment A B C Standard Errormean 134.48 154.95 168.16 2.96

Treatment (Drug) Means

Table: Adjusted mean run times and their standard errors

Drug Direct Residual PermanentA 134.48 -5.54 128.94

B 154.95 0.58 155.53C 168.16 4.96 173.12Standard Error 2.96 3.97 5.92

Next Topic: Orthogonal Linear Contrasts