ts, course 02411, imm, dtuhs/engnote.pdf · 31 2.4 f ractional 2 k factorial design 34 c hs. design...

DesignandAnalysisafExperiments

withkFactorshavingpLevels

HenrikSpliid

LecturenotesintheDesignandAnalysisofExperiments

1stEnglishedition2002

InformaticsandMathematicalModelling

TechnicalUniversityofDenmark,DK{2800Lyngby,Denmark

0

c hs.

DesignofExperiments,Course02411,IMM,DTU

1

Foreword

Thesenoteshavebeenpreparedforuseinthecourse02411,StatisticalDesignofEx-

periments,attheTechnicalUniversityofDenmark.Thenotesareconcernedsolelywith

experimentsthathavekfactors,whichalloccuronplevelsandarebalanced.Suchex-

perimentsaregenerallycalledpk

factorialexperiments,andtheyareoftenusedinthe

laboratory,whereitiswantedtoinvestigatemanyfactorsinalimited-perhapsasfew

aspossible-numberofsingleexperiments.

Readersareexpectedtohaveabasicknowledgeofthetheoryandpracticeofthedesign

andanalysisoffactorialexperiments,or,inotherwords,tobefamiliarwithconcepts

andmethodsthatareusedinstatisticalexperimentalplanningingeneral,includingfor

example,analysisofvariancetechnique,factorialexperiments,blockexperiments,square

experiments,confounding,balancingandrandomisationaswellastechniquesforthecal-

culationofthesumsofsquaresandestimatesonthebasisofaveragevaluesandcontrasts.

ThepresentversionisarevisedEnglishedition,whichinrelationtotheDanishhasbeen

improvedasregardscontents,layout,notationand,inpart,organisation.Substantial

partsofthetexthavebeenrewrittentoimprovereadabilityandtomakethevarious

methodseasiertoapply.Finally,theexamplesonwhichthenotesarelargelybasedhave

beendrawnupwithagreaterdegreeofdetailing,andnewexampleshavebeenadded.

Sincethepresentversionisthe�rstinEnglish,errorsinformulationanspellingmay

occur.

HenrikSpliid

IMM,March2002

April2002:SincetheversionofMarch2002afewcorrectionshavebeenmadeonthe

pages21,25,26,40,68and82.

Lecturenotesforcourse02411.IMM-DTU.

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Contents

1

6

1.1

Introduction...................................

6

1.2

Literaturesuggestionsconcerningthedrawingupandanalysisoffactorial

experiments...................................

7

2

2k{factorialexperiment

9

2.1

Complete2kfactorialexperiments.......................

9

2.1.1

Factors..................................

9

2.1.2

Design..................................

9

2.1.3

Modelforresponse,parametrisation..................10

2.1.4

E�ectsin2k{factorexperiments....................11

2.1.5

Standardnotationforsingleexperiments

..............11

2.1.6

Parameterestimates..........................12

2.1.7

Sumsofsquares.............................13

2.1.8

Calculationmethodsforcontrasts

..................13

2.1.9

Yates'algorithm

............................14

2.1.10Replicationsorrepetitions.......................15

2.1.1123factorialdesign............................16

2.1.122kfactorialexperiment

........................19

2.2

Blockconfounded2kfactorialexperiment..................20

2.2.1

Constructionofaconfoundedblockexperiment...........25

2.2.2

Aone-factor-at-a-timeexperiment..................27

2.3

Partiallyconfounded2kfactorialexperiment.................28

2.3.1

Somegeneralisations..........................31

2.4

Fractional2kfactorialdesign

.........................34

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2.5

Factorson2and4levels............................43

3

Generalmethodsforpk-factorialdesigns

48

3.1

Completep

kfactorialexperiments.......................48

3.2

CalculationsbasedonKempthorne'smethod

................57

3.3

Generalformulationofinteractionsandarti�ciale�ects

..........60

3.4

Standardisationofgenerale�ects.......................62

3.5

Block-confoundedpkfactorialexperiment..................65

3.6

Generalisationofthedivisionintoblockswithseveralde�ningrelations..70

3.6.1

Constructionofblocksingeneral...................74

3.7

Partialconfounding...............................78

3.8

Constructionofafractionalfactorialdesign.................86

3.8.1

Resolutionforfractionalfactorialdesigns

..............90

3.8.2

Practicalandgeneralprocedure....................91

3.8.3

Aliasrelationswith1=pq�pkexperiments..............95

3.8.4

Estimationandtestingin1=pq�pkfactorialexperiments......101

3.8.5

Fractionalfactorialdesignlaidoutinblocks.............105

Index

.....

116

Myownnotes

.....

118

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Tabels

2.1

Asimpleweighingexperimentwith3items..................34

2.2

A1/4�25factorialexperiment.........................40

2.3

A2�4experimentin2blocks.........................44

2.4

Afractional2�2�4factorialdesign......................45

3.1

MakingaGraeco-Latinsquareina32factorialexperiment.........50

3.2

Latincubesin33experiments.........................53

3.3

EstimationandSSQinthe32-factorialexperiment..............58

3.4

Indexvariationwithinversionofthefactororder

..............61

3.5

Generalisedinteractionsandstandardisation.................62

3.6

Latinsquaresin23factorialexperimentsandYates'algorithm

.......63

3.7

23factorialexperimentin2blocksof4singleexperiments

.........65

3.8

32factorialexperimentin3blocks.......................66

3.9

Divisionofa23factorialexperimentinto22blocks..............69

3.10Dividinga33factorialexperimentinto9blocks...............71

3.11Divisionofa25experimentinto23blocks

..................72

3.12Divisionof3kexperimentsinto33blocks...................73



3.15Partiallyconfounded23factorialexperiment.................78

3.16Partiallyconfounded32factorialexperiment.................82

3.17FactorexperimentdoneasaLatinsquareexperiment............86

3.18Confoundingsina3�

1

�33factorialexperiment,aliasrelations......88

3.19A2�

2

�25factorialexperiment........................92

3.20Constructionof3�

2�35factorialexperiment................96

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3.21Estimationina3�

1

�33-factorialexperiment................101

3.22TwoSASexamples...............................104

3.23A3�

2

�35factorialexperimentin3blocksof9singleexperiments

....106

3.24A2�

4

�28factorialin2blocks........................110

3.25A2�

3

�27factorialexperimentin4blocks..................114

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1 1.1

Introduction

Theselecturenotesareconcerned

with

theconstructionofexperimentaldesignswhich

are

particularlysuitablewhenitiswantedtoexaminealargenumberoffactorsandoftenunder

laboratoryconditions.

Thecomplexityoftheproblem

canbeillustratedwiththefactthatthenumberofpossiblefactor

combinationsinamulti-factorexperimentistheproductofthelevelsofthesinglefactors.If,

forexample,oneconsiders10factors,eachononly2levels,thenumberofpossibledi�erent

experimentsis2�

2�

:::�

2=

2k=

1024.Ifitiswantedtoinvestigatethefactorson3levels,

thisnumberincreasesto310=

59049singleexperiments.Ascanbeseen,thenumberofsingle

experimentsrapidlyincreaseswiththenumberoffactorsandfactorlevels.

Forpracticalexperimentalwork,thisimpliestwomainproblems.

First,itquicklybecomes

impossibletoperform

allexperimentsinwhatiscalledacompletefactorstructure,andsecond,it

isdiÆculttokeeptheexperimentalconditionsunchangedduringalargenumberofexperiments.

Doingtheexperiments,forexample,necessarilytakesalongtime,useslargeamountsoftest

material,usesalargenumberofexperimentalanimals,orinvolvesmanypeople,allofwhich

tendtoincreasetheexperimentaluncertainty.

Thesenoteswillintroducegeneralmodelsforsuchmulti-factorexperimentswhereallfactors

areonplevels,andwewillconsiderfundamentalmethodstoreducetheexperimentalworkvery

considerablyinrelationtothecompletefactorialexperiment,andtogroupsuchexperimentsin

smallblocks.Inthisway,bothsavingsintheexperimentalworkandmoreaccurateestimates

areachieved.

Ane�orthasbeenmadetokeepthenotesas"non-mathematical"aspossible,forexampleby

showingthevarioustechniquesintypicalexamplesandgeneralisingonthebasisofthese.Onthe

otherhand,thishasthedisadvantagethatthetextisperhapssomewhatlongerthanapurely

mathematicalstatisticalrun-throughwouldneed.

Generally,extensivenumericalexamplesarenotgivennorexamplesofthedesignofexperiments

forspeci�cproblem

complexes,butthewholediscussioniskeptonsuchagenerallevelthat

experimentaldesignersfrom

di�erentdisciplinesshouldhavereasonablepossibilitiestobene�t

from

themethodsdescribed.Asmentionedintheforeword,itisassumedthatthereaderhasa

certainfundamentalknowledgeofexperimentalworkandstatisticalexperimentaldesign.

Finally,Ithinkthat,onthebasisofthesenotes,apersonwouldbeabletounderstandtheidea

intheexperimentaldesignsshown,andwouldalsobeabletodraw

upandanalyseexperimen-

taldesignsthataresuitableingivenproblem

complexes.However,thismustnotpreventthe

designerofexperimentsfrom

consultingtherelevantspecialistliteratureonthesubject.Here

canbefoundmanynumericalexamples,bothdetailedandrelevant,andinmanycases,alter-

nativeanalysismethodsaresuggested,whichcanbeveryusefulintheinterpretationofspeci�c

experimentresults.Below,afew

examplesof"classical"literatureinthe�eldarementioned.

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1.2

Literaturesuggestionsconcerningthedrawingupandana-

lysisoffactorialexperiments

. Box,G.E.P.,Hunter,W.G.andHunter,J.S.:StatisticsforExperimenters,Wiley,1978.

Chapter10introduces2k

factorialexperiments.Chapter11showsexamplesoftheiruseand

analysis.Inparticular,section10.9showsamethodofanalysingexperimentswithmanye�ects,

whereonedoesnothaveanexplicitestimateofuncertainty.Themethodusesthetechnique

from

thequantilediagram

(Q-Q

plot)andisbothsimpleandillustrativefortheuser.Anumber

ofstandardblockexperimentsaregiven.

Chapter12introducesfractionalfactorialdesigns

andchapter13givesexamplesofapplications.

Thebookcontainsmanyexamplesthatare

completelycalculated-althoughonthebasisofquitemodestamountofdata.Ingenerala

highlyrecommendablebookforexperimenters.

Davies,O.L.andothers:TheDesignandAnalysisofExperiments,OliverandBoyd,1960(1st

edition1954).

Chapters7,8,9and10dealwithfactorialexperimentswithspecialemphasison2k

and3k

factorialexperiments.A

largenumberofpracticalexamplesaregivenbasedonrealproblems

withachemical/technicalbackground.Eventhoughthebookisalittleold,itishighlyrecom-

mendableasabasisforconductinglaboratoryexperiments.Italsocontainsagoodchapter(11)

aboutexperimentaldeterminationofoptimalconditionswherefactorialexperimentsareused.

Fisher,R.A.:TheDesignofExperiments,OliverandBoyd,1960(1stedition1935)

A

classic(perhaps"theclassic"),writtenbyoneofthefoundersofstatistics.Chapters6,7and

8introducenotationandmethodsfor2kand3kfactorialexperiments.Veryinterestingbook.

Johnson,N.L.andLeone,F.C,:StatisticsandExperimentalDesign,VolumeII,Wiley1977.

Chapter15givesapracticallyorientatedandquitecondensedpresentationof2k

factorialex-

perimentsforuseinengineering.WithVolumeI,thisisagoodgeneralbookaboutengineering

statisticalmethods.

Kempthorne,O.:TheDesignandAnalysisofExperiments,Wiley1973(1stedition1952).

Thiscontainsthemathematicalandstatisticalbasisforpk

factorialexperimentswithwhich

thesenotesareconcerned(chapter17).Inadditionitdealswithanumberofspeci�cproblems

relevantformulti-factorialexperiments,forexampleexperimentswithfactorsonboth2and

3levels(chapter18).

Itisbasedon

agriculturalexperimentsinparticular,butisactually

completelygeneralandhighlyrecommended.

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Montgomery,D.C.:DesignandAnalysisofExperiments,Wiley1997(1stedition1976).

Thelatestedition(5th)isconsiderablyimprovedinrelationtothe�rsteditions.Thebook

givesagood,thoroughandrelevantrun-throughofmanyexperimentaldesignsandmethods

foranalysingexperimentalresults.Chapters7,8and9dealwith2kfactorialexperimentsand

chapter10dealswith3kfactorialexperiments.Anexcellentmanualand,uptoapoint,suitable

forself-tuition.

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2

2k{factorialexperiment

Chapter2discussessomefundamentalexperimentalstructuresformulti-factorexperi-

ments.Here,forthesakeofsimplicity,weconsideronlyexperimentswhereallfactors

occurononly2levels.Theselevelsforexamplecanbe\low"/"high"foranamountof

additiveor\notpresent"/"present"foracatalyst.

Aspecialnotationisintroducedandanumberoftermsandmethods,whicharegenerally

applicableinplanningexperimentswithmanyfactors.Thischaptershouldthusbeseen

asanintroductiontothemoregeneraltreatmentofthesubjectthatfollowslater.

2.1

Complete2k

factorialexperiments

2.1.1

Factors

Thename,2kfactorialexperiments,referstoexperimentsinwhichitiswishedtostudyk

factorsandwhereeachfactorcanoccurononly2levels.Thenumberofpossibledi�erent

factorcombinationsisprecisely2k,andifonechoosestodotheexperimentsothatall

thesecombinationsaregonethroughinarandomiseddesign,theexperimentiscalleda

complete2kfactorialexperiment.

Inthissection,themainpurposeistointroduceageneralnotation,sowewillonly

consideranexperimentwithtwofactors,eachhavingtwolevels.Thisexperimentisthus

calleda22factorialexperiment.

ThefactorsintheexperimentarecalledAandB,anditispractical,nottosayrequired,

alwaystousethesenames,evenifitcouldperhapsbewishedtouse,forexample,Tfor

temperatureorVforvolumeformnemonicreasons.

Inaddition,thefactorsareorganisedsothatAisalwaysthe�rstfactorandBisthe

secondfactor.

2.1.2

Design

Foreachcombinationofthetwofactors,weimaginethatanumber(r)ofmeasurements

aremade.Therandomerroriscalled(generally)E.Theresultofasingleexperiment

withacertainfactorcombinationisoftencalledtheresponse,andthisterminologyis

alsousedforthesumoftheresultsobtainedforthegivenfactorcombination.

Thisdesignisasfollowswheretherearerrepetitionsperfactorcombinationinacom-

pletelyrandomisedsetup:

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9

B=0

B=1

Y001

Y011

A=0

:

:

Y00r

Y01r

Y101

Y111

A=1

:

:

Y10r

Y11r

Ifforexampleweinvestigatehowtheoutputfromaprocessdependsonpressureand

temperature,thetwolevelsoffactorAcanrepresenttwovaluesofpressurewhilethetwo

levelsoffactorBrepresenttwotemperatures.Themeasuredvalue,Yij�,thengivesthe

resultofthe�'thmeasurementwiththefactorcombination(Ai,Bj).

2.1.3

Modelforresponse,parametrisation

Itisassumed,asmentioned,thattheexperimentisdoneasacompletelyrandomised

experiment,thatis,thatthe2�2�robservationsaremade,forexample,incompletely

randomorderorrandomlydistributedovertheexperimentalmaterialwhichmaybeused

intheexperiment.

Themathematicalmodelfortheyieldofthisexperiment(theresponse)is,inthatfactor

Aisstillthe�rstfactorandfactorBisthesecondfactor:

Yij�=�+Ai+Bj+ABij+Eij�,wherei=(0;1);j=(0;1);�=(1;2;::;r)

wheretheususalrestrictionsapply

1 X i=0

Ai=0

;

1 X j=0

Bj=0

;

1 X i=0

ABij=0

;

1 X j=0

ABij=0

Theserestrictionsimplythat

A0=�A1

;

B0=�B1

;

AB00=�AB10=�AB01=+AB11

Therefore,inreality,thereareonly4parametersinthismodel,namelytheexperiment

level�andthefactorparametersA1,B1andAB11,ifone,(asusual)referstothe\high"

levelsofthefactors.

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2.1.4

E�ectsin2k{factorexperiments

Ina2-levelfactorialexperiment,oneoftenspeaksofthe"e�ects"ofthefactors.Bythis

isunderstoodinthisspecialcasethemeanchangeoftheresponsethatisobtainedby

changingafactorfromits"low"toits"high"level.

Thee�ectsinanexperimentwherethefactorshaveprecisely2levelsarethereforede�ned

inthefollowingmanner:

A=A1�A0=2A1

,andlikewiseB=2B1

;

AB=2AB11

Inotherfactorialexperiments,oneoftenspeaksmoregenerallyaboutfactore�ectsas

expressionsoftheactionofthefactorsontheresponse,withouttherebyreferringtoa

de�niteparameterform.

2.1.5

Standardnotationforsingleexperiments

Inthetheoreticaltreatmentofthisexperiment,itispracticaltointroduceastandard

notationfortheexperimentalresultsinthesamewayasforthee�ectsinthemathematical

model.

Fortheexperimentsthataredoneforexamplewiththefactorcombination(A1;B0),the

sumoftheresultsoftheexperimentisneeded.Thissumiscalleda,thatis

a=

r X �=1

Y10�

wherethissumisthesumofalldatawithfactorAonthehighlevelandtheotherfactors

onthelowlevel.Asmentioned,aisalsocalledtheresponseofthefactorcombinationin

question.

Inthesameway,thesumfortheexperimentswiththefactorcombination(A0;B1)is

calledb,whilethesumfor(A1;B1)iscalled"ab".Finally,thesumfor(A0;B0)iscalled

"(1)".

Inthedesignabove,cellsumsarethusfoundasinthefollowingtable

B=0

B=1

A=0

(1)

b

A=1

a

ab

Somepresentationsusenamesthatdirectlyrefertothefactorlevelsasforexample:

B=0

B=1

A=0

00

01

A=1

10

11

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11

Whenoneworkswiththesecellsums,theyaremostpracticallyshownintheso-called

standardorderforthe22experiment:

(1);a;b;ab

Itisimportanttokeepstrictlytotheintroducednotation,i.e.upper-caseletterforparam-

etersinthemodelandlower-caselettersforcellsums,andthattheorderofparameters

aswellasdata,iskeptasshown.Ifnot,thereisaconsiderableriskofmakingamessof

it.

2.1.6

Parameterestimates

Wecannowformulatetheanalysisoftheexperimentinmoregeneralterms.

We�ndthefollowingestimatesfortheparametersofthemodel:

^�=[(1)+a+b+ab]=(4�r)=[(1)+a+b+ab]=(2

k�r)

wherek=2,asmentioned,givesthenumberoffactorsinthedesignandristhenumber

ofrepetitionsofthesingleexperiments.

Furtherwe�nd:

b A 1=�b A 0=[�(1)+a�b+ab]=(2

k�r)

b B 1=�b B 0=[�(1)�a+b+ab]=(2

k�r)

d AB11=�d AB10=�d AB01=d AB00=[(1)�a�b+ab]=(2

k�r)

IfwealsowanttoestimateforexampletheA-e�ect,i.e.thechangeinresponsewhen

factorAischangedfromlow(i=0)tohigh(i=1)level,we�nd

b A=b A 1�b A 0=2b A 1=[�(1)+a�b+ab]=(2

k�

1�r)

Theparenthesis[�(1)+a�b+ab]givesthetotalincreaseinresponse,whichwasfound

bychangingthefactorAfromitslowleveltoitshighlevel.Thisamountiscalledthe

A-contrast,andiscalled[A].Therefore,inthecaseofthefactorA,wehaveinsummary

theequations:

[A]=[�(1)+a�b+ab]

;

b A 1=�b A 0=[A]=(2

k�r)

;

b A=2b A 1

andcorrespondinglyfortheothertermsinthemodel.Speci�callyforthetotalsumof

observations,thenotation[I]=[(1)+a+b+ab]isused.Thisquantitycanbecalledthe

pseudo-contrast.

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2.1.7

Sumsofsquares

Further,wecanderivethesumsofsquaresforalltermsinthemodel.Thiscanbedone

withordinaryanalysisofvariancetechnique.Forexample,thisgivesinthecaseoffactor

A:

SSQA

=[A]2=(2

k�r)

Correspondingexpressionsapplyforalltheotherfactore�ectsinthemodel.

Thesumsofsquaresforthesefactore�ectsallhave1degreeoffreedom.

Iftherearerepeatedmeasurementsforthesinglefactorcombinations,i.e.r>1,wecan

�ndtheresidualvariationasthevariationwithinthesinglecellsinthedesignintheusual

manner:

SSQresid=

1 X i=0

1 X j=0

([r X �

=1

Y2 ij�]�T

2 ij�=r)

;

where

Tij�

=

r X �=1

Yij�

isthesum(thetotal)incell(i;j).

Wecansummarisetheseconsiderationsinananalysisofvariancetable:

Sourceof

Sumofsquares

Degreesof

S2

F-value

variation

=SSQ

freedom=f

=SSQ/f

A

[A]2=(2

k�r)

1

S2 A

FA

=S

2 A=S

2 resid

B

[B]2=(2k�r)

1

S2 B

FB

=S

2 B=S

2 resid

AB

[AB]2=(2k�r)

1

S2 AB

FAB

=S

2 AB=S

2 resid

Residual

SSQresid

2k�(r�1)

S2 resid

Totalt

SSQtot

r�2k�1

Inthetable,forexample,FA

iscomparedwithanFdistributionwith(1;2k�(r�1))

degreesoffreedom.

2.1.8

Calculationmethodsforcontrasts

Thesalientpointintheaboveanalysisisthecalculationofthecontrasts.Variousmethods,

somemorepracticalthanothers,canbegiventosolvethisproblem.

Mathematically,thecontrastscanbecalculatedbythefollowingmatrixequation:

2 6 6 6 4I A B A

B

3 7 7 7 5=2 6 6 6 41

1

1

1

�1

1

�1

1

�1

�1

1

1

1

�1

�1

13 7 7 7 52 6 6 6 4(1)

a b ab

3 7 7 7 5

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13

Onenotesthatbothcontrastsandcellsumsaregiveninstandardorder.Inadditionit

canbeseenthattherowforexamplefortheA-contrastcontains+1foraandab,where

factorAisatitshighlevel,but-1for(1)andb,wherefactorAisatitslowlevel.Finally,

itisnoticedthattherowforAB

foundbymultiplyingtherowsforAandB

byeach

other.

Insomepresentations,thematrixexpresssionshownisgivenjustas+and-signsina

table:

(1)

a

b

ab

I

+

+

+

+

A

�

+

�+

B

�

�+

+

AB

+

��+

2.1.9

Yates'algorithm

Finallywegiveacalculationalgorithm

whichisnamedaftertheEnglishstatistician

FrankYatesandiscalledYates'algorithm.Data,i.e.thecellsums,arearrangedin

standardorderinacolumn.Thenthesearetakeninpairsandsummed,andafterthat

thesamevaluesaresubtractedfromeachother.Thesumsareputatthetopofthenext

columnfollowedbythedi�erences.Whenformingthedi�erences,theuppermostvalue

issubtractedfromthebottomone(mnemonicrule:Ascomplicatedaspossible).The

operationisrepeatedasmanytimesastherearefactors.Herethiswouldbek=2times:

Cellsums

1sttime

2ndtime

=

Contrasts

SumofSq.

(1)

(1)+a

(1)+a+b+ab

=

[I]

[I]2=(2

k�r)

a

b+ab

�(1)+a�b+ab

=

[A]

[A]2=(2k�r)

b

�(1)+a

�(1)�a+b+ab

=

[B]

[B]2=(2k�r)

ab

�b+ab

(1)�a�b+ab

=

[AB]

[AB]2=(2k�r)

Wegiveanumericalexamplewherethedataareshowninthefollowingtable:

B=0

B=1

A=0

12.1

19.8

14.3

21.0

A=1

17.9

24.3

19.1

23.4

One�nds(1)=12:1+14:3=26:4,a=17:9+19:1=37:0,b=19:8+21:0=40:8and

ab=24:3+23:4=47:7.

Yates=algorithmnowgives

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14

Cellsums

1sttime

2ndtime

=

Contrasts

SumsofSquares

(1)=26.4

63.4

151.9

=

[I]

[I]2=(2k�r)=2884.20

a=37.0

88.5

17.5

=

[A]

[A]2=(2k�r)=38.28

b=40.8

10.6

25.1

=

[B]

[B]2=(2k�r)=78.75

ab=47.7

6.9

-3.7

=

[AB]

[AB]2=(2

k�r)=1.71

Inthisexperimentr=2,andSSQresid

canbefoundasthesumofsquareswithinthe

singlefactorcombinations.

SSQresid=(12:12+14:3

2�(12:1+14:3)2=2)

+(17:92+19:1

2�(17:9+19:1)2=2)

+(19:82+21:0

2�(19:8+21:0)2=2)

+(24:32+23:4

2�(24:3+23:4)2=2)=2:42+0:72+0:72+0:41=4:27

ANOVA

Sourceofvariation

SSQ

df

s2

F-value

Amaine�ect

38.28

2�1=1

38.28

35.75

Bmaine�ect

78.75

2�1=1

78.75

73.60

ABinteraction

1.71

(2�1)(2�1)=1

1.71

1.60

Residualvariation

4.27

4(2�1)=4

1.07

Totalt

123.01

8�1=7

Asweshallsee,Yates'algorithmisgenerallyapplicabletoall2kfactorialexperiments

andforexamplecanbeeasilyprogrammedonacalculator.Thealgorithmalsoappears

insignalanalysisunderthename\fastFouriertransform".

Thelastcolumninthealgorithmgivesthecontraststhatareusedfortheestimationas

wellasthecalculationofthesumsofsquaresforthefactore�ects.

2.1.10

Replicationsorrepetitions

Beforewemoveontoexperimentswith3ormorefactors,letuslookatthefollowing

experiment

B=0

B=1

A=0

Y001

Y011

A=1

Y101

Y111

Dayno.1

,

B=0

B=1

A=0

Y002

Y012

A=1

Y102

Y112

Dayno.2

,��,

B=0

B=1

A=0

Y00R

Y01R

A=1

Y10R

Y11R

Dayno.R

c hs.


15

thatis,a2�2,replicatedR

times.Themathematicalmodelforthisexperimentis

notidenticalwiththemodelpresentedonpage10thebeginningofthischapter.The

experimentisnotcompletelyrandomisedinthatrandomisationisdonewithindays.

Anexperimentalcollectionofsingleexperimentsthatcanberegardedashomogeneous

withrespecttouncertainty,suchasthedaysintheexample,isgenerallycalledablock.

Ifitisassumedthatthecontributionfromthedayscanbedescribedbyanadditivee�ect,

correspondingtoageneralincreaseorreductionoftheresponseonthesingledays(block

e�ect),areasonablemathematicalmodelwouldbe:

Yij�=�+Ai+Bj+ABij+D�+Fij�

;

i=(0;1);j=(0;1);�=(1;2;:::;R);

whereD�givesthecontributionfromthe�'thdag,andFij�givesthepurelyrandomerror

withindays.

Wewillsaythatthe22experimentisreplicatedRtimes.

Thisisessentiallydi�erentfromthecasewhereforexample2�2�rmeasurementsare

madeinacompletelyrandomiseddesignasonpage10.

Ifoneisinthepracticalsituationofhavingtochoosebetweenthetwodesigns,anditis

assumedthatbothexperiments(becauseofthetimeneeded)mustextendoverseveral

days,thelatterdesignispreferable.Inthe�rstdesigntherandomisationisdoneacross

dayswithrrepetitions,andtheexperimentaluncertainty,Eij�

willalsocontainthe

variationbetweendays.

OnecanregardD�,i.e.thee�ectfromthe�'thday,asarandomlyvaryingamountwith

thevariance�

2 D,whileFij�,i.e.theexperimentalerrorwithinoneday,isassumedtohave

thevariance�

2 F.FromthiscanbederivedthatEij�,i.e.thetotalexperimentalerrorin

acompletelyrandomiseddesignoverseveraldays,hasthevariance

�2 E

=�

2 D

+�

2 F

Theexampleillustratestheadvantageofdividingone'sexperimentintosmallerhomo-

geneousblocksasdistinctfromcompleterandomisation.Italsoshowsthatthereisa

fundamentaldi�erencebetweentheanalysisofanexperimentwithrrepetitionsina

completelyrandomiseddesignandarandomiseddesignreplicatedRtimes.

2.1.11

23

factorialdesign

Wenowstatethedescribedtermsforthe23factorialexperimentwithaminimumof

comments.

ThefactorsarenowA,B,andCwithindicesi,jandk,respectively.Thefactorsare

againorderedsoAisthe�rstfactor,BthesecondandCthethirdfactor.

c hs.


16

Themathematicalmodelwithrrepetitionspercellinacompletelyrandomiseddesignis:

Yijk�=�+Ai+Bj+ABij+Ck+ACik+BCjk+ABCijk+Eijk�

wherei;j;k=(0;1)and�=(1;::;r).

Theusualrestrictionsare:

1 X i=0

Ai=

1 X j=0

Bj=

1 X i=0

ABij=

1 X j=0

ABij=

1 X k=0

Ck=��=

1 X k=0

ABCijk=0

whichimpliesthat

A1=

�

A0

;

B1=

�

B0

;

AB11=

�

AB10=

�

AB01=

AB00

;

C1=

�

C0

;

�

�

�

;

(andfurtheronuntil)

AB

C000=

�

AB

C100=

�

AB

C010=

AB

C110=

�

AB

C001=

AB

C101=

AB

C011=

�

AB

C111

Thee�ectsoftheexperiment(whichgivethedi�erenceinresponsewhenafactoris

changedfrom\low"levelto\high"level,cf.page11)are

A=2A1

;

B=2B1

;

AB=2AB11

;

C=2C1

;��;ABC=2ABC111

Thestandardorderforthe23=8di�erentexperimentalconditions(factorcombinations)

is:

(1)

,

a

,

b

,

ab

,

c

,

ac

,

bc

,

abc

wheretheintroductionofthefactorCisdonebymultiplyingcontothetermsforthe22

experimentandaddingtheresultingtermstothesequence:(1);a;b;ab;((1);a;b;ab)c=

(1);a;b;ab;c;ac;bc;abc.

c hs.


17

[I]

=

[+(1)+a+b+ab+c+ac+bc+abc]

[A]

=

[�(1)+a�b+ab�c+ac�bc+abc]

[B]

=

[�(1)�a+b+ab�c�ac+bc+abc]

[AB]

=

[+(1)�a�b+ab+c�ac�bc+abc]

[C]

=

[�(1)�a�b�ab+c+ac+bc+abc]

[AC]

=

[+(1)�a+b�ab�c+ac�bc+abc]

[BC]

=

[+(1)+a�b�ab�c�ac+bc+abc]

[ABC]

=

[�(1)+a+b�ab+c�ac�bc+abc]

orinmatrixformulation

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I A B A

B C AC

BC

ABC

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1)

a b ab c a

c bc abc

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

Yates'algorithmisperformedasabove,buttheoperationonthecolumnsshouldnowbe

done3timesasthereare3factors.Ifonewritesindetailwhathappens,onegets:

response

1sttime

2ndtime

3rdtime

contrasts

(1)

(1)+a

(1)+a+b+ab

(1)+a+b+ab+c+ac+bc+abc

[I]

a

b+ab

c+ac+bc+abc

�(1)+a�

b+ab�

c+ac�

bc+abc

[A]

b

c+ac

�(1)+a�

b+ab

�(1)�

a+b+ab�

c�

ac+bc+abc

[B]

ab

bc+abc

�c+ac�

bc+abc

(1)�

a�

b+ab+c�

ac�

bc+abc

[AB]

c

�(1)+a

�(1)�

a+b+ab

�(1)�

a�

b�

ab+c+ac+ab+abc

[C]

ac

�b+ab

�c�

ac+bc+abc

(1)�

a+b�

ab�

c+ac�

bc+abc

[AC]

bc

�c+ac

(1)�

a�

b+ab

(1)+a�

b�

ab�

c�

ac+bc+abc

[BC]

abc

�bc+abc

c�

ac�

bc+abc

�(1)+a+b�

ab+c�

ac�

bc+abc

[ABC]

Parameterestimatesare,withk=3:

^�=

[I]

2k�r

;

b A 1=

[A]

2k�r

;b B 1=

[B]

2k�r

;:::;

dABC111=

[ABC]

2k�r

c hs.


18

Correspondingly,thee�ectestimatesare:

b A=2b A 1;

b B=2b B 1;

��;

dABC=2dABC111

Thesumsofsquaresare,forexample:

SSQA

=

[A]2

2k�r

;SSQB

=

[B]2

2k�r

;SSQABC

=[ABC]2

2k�r

Thevariancesofthecontrastsarefound,with[A]asexample,as

Varf[A]g=Varf�(1)+a�b+ab�c+ac�bc+abcg=2k�r��

2

;

wherek=3here.

Theresultisseenbynotingthatthereare2kterms,whichallhavethesamevariance,

whichforexampleis

Varfabg=Varf

r X �=1

Y110�g=r��

2

Further,itisnowfound,that

Varfb A 1g=Varf[A]=(2

k�r)g=�

2=(2

k�r)

Varfb Ag=Varf2b A 1g=�

2=(2

k�

2�r)

2.1.12

2k

factorialexperiment

Thestatedequationsaregeneraliseddirectlytofactorialexperimentswithkfactors,each

on2levels,withrrepetitionsinarandomiseddesign.Writingupthemathematical

model,namesforcellsums,calculationofcontrastsetc.aredoneinexactlythesameway

asdescribedabove.Forestimatesandsumsofsquares,thengenerally

Parameterestimate=(Contrast)/(2k�r)

E�ectestimate=2�Parameterestimate

Sumofsquares(SSQ)=(Contrast)

2=(2

k�r)

c hs.


19

Regardingtheconstructionofcon�denceintervalsfortheparametersande�ects,the

varianceoftheestimatescanbederived.One�nds

VarfContrastg=�

2�2k�r

VarfParameterestimateg=VarfContrastg/(2k�r)2=�

2=(2k�r)

VarfE�ectestimateg=22�

2=(2k�r)=�

2=(2k�

2�r)

Thecon�denceintervalsforparametersore�ectscanbeconstructedifonehasanestimate

of�

2.Supposethatonehassuchanestimate,^�

2

=s2,andthatithasfdegreesof

freedom.If(1��)con�denceintervalsarewanted,onetherebygets

I 1�

�(parameter)=Parameterestimate�s�t(f) (1�

�=2)=

p 2k�r

I 1�

�(e�ekt)=E�ectestimate�2�s�t(f) (1�

�=2)=

p2k�r

wheret(f) (1�

�=2)

denotesthe(1��=2)-fractileinthet-distributionwithfdegreesof

freedom.

2.2

Blockconfounded2k

factorialexperiment

Inexperimentswithmanyfactors,thenumberofsingleexperimentsquicklybecomes

verylarge.Forpracticalexperimentalwork,thismeansthatitcanbediÆculttoensure

homogeneousexperimentalconditionsforallthesingleexperiments.

Agenerallyoccurringproblemisthatinaseriesofexperiments,rawmaterialisused

thattypicallycomesintheformofbatches,i.e.homogeneousshipments.Aslongaswe

performtheexperimentsonrawmaterialfromthesamebatch,theexperimentswillgive

homogeneousresults,whileresultsofexperimentsdoneonmaterialfromdi�erentbatches

willbemorenon-homogeneous.Thebatchesofrawmaterialinthiswayconstituteblocks.

Inthesameway,itwilloftenbethecasethatexperimentsdoneclosetogetherintime

aremoreuniformthanexperimentsdonewithalongtimebetweenthem.

Inaseriesofexperimentsonewilltrytodoexperimentsthataretobecomparedonthe

mostuniformbasispossible,sincethatgivesthemostexactevaluationofthetreatments

thatarebeingstudied.Forexample,onewilltrytodotheexperimentonthesamebatch

andwithinasshortaspaceoftimeaspossible.Butthisofcourseisaproblemwhenthe

numberofsingleexperimentsislarge.

Letusimaginethatwewanttodoa23factorialexperiment,i.e.anexperimentwith8

singleexperiments,correspondingtothe8di�erentfactorcombinations.Supposefurther

c hs.


20

thatitisnotpossibletodoallthese8singleexperimentsonthesameday,butperhaps

onlyfourperday.

Anobviouswaytodistributethe8singleexperimentsoverthetwodayscouldbetodraw

lots.Weimaginethatthisdrawinglotsresultsinthefollowingdesign:

day1

day2

(1)

c

abc

a

bc

ac

ab

b

Forthisdesign,wegetforexampletheA-contrast:

[A]=[�(1)+a�b+ab�c+ac�bc+abc]

Aslongasthetwodaysgiveresultswithexactlythesamemeanresponse,thisestimate

will,inprinciple,bejustasgoodasiftheexperimentshadbeendoneonthesameday.

(howeverthevarianceisgenerallyincreasedwhenexperimentsaredoneovertwodays

insteadofononeday).

Butifontheotherhandthereisacertainunavoidabledi�erenceinthemeanresponse

onthetwodays,weobviouslyhaveariskthatthisa�ectstheestimates.Asasimple

modelforsuchadi�erenceinthedays,wecanassumethattheresponseonday1is1g

undertheideal,whileitis2govertheidealonday2.Ane�ectofthistypeisablock

e�ect,andthedaysconstitutetheblocks.Onesaysthattheexperimentislaidoutintwo

blockseachwith4singleexperiments.

FortheA-contrast,itisshownbelowhowtheseunintentional,butunavoidable,e�ects

ontheexperimentalresultsfromthedayswilla�ecttheestimation,as1gissubtracted

fromalltheresultsfromday1and2gisaddedtoalltheresultsfromday2:

[A]=[�((1)�1g)+(a�1g)�(b+2g)+(ab+2g)�(c�1g)+(ac+2g)�(bc+2g)+(abc�1g)]

=[�(1)+a�b+ab�c+ac�bc+abc]+[1�1�2+2+1+2�2�1]g

=[�(1)+a�b+ab�c+ac�bc+abc]

Thus,adi�erenceinlevelontheresultsfromthetwodays(blocks)willnothaveanye�ect

ontheestimateforthemaine�ectoffactorA.Inotherwords,factorAisinbalance

withtheblocks(thedays).

Ifwerepeattheprocedureforthemaine�ectoffactorB,weget

[B]=[�((1)�1g)�(a�1g)+(b+2g)+(ab+2g)�(c�1g)�(ac+2g)+(bc+2g)+(abc�1g)]

c hs.


21

=[�(1)�a+b+ab�c�ac+bc+abc]+[1+1+2+2+1�2+2�1]g

=[�(1)+a�b+ab�c+ac�bc+abc]+6g

TheestimatefortheBe�ect(i.e.thedi�erenceinresponsewhenBischangedfromlow

tohighlevel)istherebyonaverage(6g=4)=1:5ghigherthantheidealestimate.

Ifwelookbackatthedesign,thisisbecausefactorBwasmainlyat"highlevel"onday

2,wheretheresponseonaverageisalittleabovetheideal.

ThesamedoesnotapplyinthecaseoffactorA.Thishasbeenat\highlevel"twotimes

eachdayandlikewiseat\lowlevel"twotimeseachday.ThesameappliesforfactorC.

ThusfactorsAandCareinbalanceinrelationtotheblocks(thedays),whilefactorB

isnotinbalance.

Anoverallevaluationofthee�ectoftheblocks(thedays)ontheexperimentcanbeseen

fromthefollowingmatrixequation

2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1)�1g

a�1g

b+2g

ab+2g

c�1g

ac+2g

bc+2g

abc�1g

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I+4g

A B+6g

AB�6g

C AC

BC�6g

ABC�6g

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

ItcanbeseenthatallcontraststhatonlyconcernfactorsAandCarefoundcorrectly,

becausethetwofactorsareinbalanceinrelationtotheblocksinthedesign,whileall

contraststhatalsoconcernBarea�ectedbythe(unintentional,butunavoidable)e�ect

fromtheblocks.

Whatwenowcanaskiswhetheritispossibleto�ndadistributionoverthetwodaysso

thatthein uencefromtheseiseliminatedtothegreatestpossibleextent.

Wecannotethatitisthedi�erencebetweenthedaysthatisimportantfortheestimates

ofthee�ectsofthefactors,whilegenerallevelofthedaysisabsorbedinthecommon

averageforalldata.

Ifweoncemoreregardthecalculationofthecontrast[A],wecandrawupthefollowing

table,whichshowshowthein uenceofthedaysisweightedintheestimate:

Contrast[A]

Response

(1)

a

b

ab

c

ac

bc

abc

Weight

�

+

�

+

�+

�

+

Day

1

1

2

2

1

2

2

1

c hs.


22

Wenotethatday1entersanequalnumberoftimeswith+andwith�,andday2as

well.Ifwelookatoneofthecontrastswherethedaysdonotcancel,e.g.[B],wegeta

tablelikethefollowing:

Contrast[B]

Response

(1)

a

b

ab

c

ac

bc

abc

Weight

�

�+

+

��

+

+

Day

1

1

2

2

1

2

2

1

wherethebalanceisobviouslynotpresent.

Theconditionthatisnecessarysothatane�ectisnotin uencedbythedaysisobviously

thatthereisabalanceasdescribed.Thepossibilitiesforcreatingsuchabalanceare

linkedtothematrixofonesintheestimation:

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I A B A

B C AC

BC

ABC

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1)

a b ab c a

cbc a

bc3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

Thismatrixhasthespecialcharacteristicthattheproductsumofanytworowsiszero.

Ifoneforexampletakestherowsfor[A]and[B],onegets(-1)(-1)+(+1)(-1)+...+

(+1)(+1)=0.Thetwocontrasts[A]and[B]arethusorthogonalcontrasts(linearly

independent).

IfonethereforechoosesforexampleadesignwherethedaysfollowfactorB,itisabsolutely

certainthatinanycasefactorAwillbeinbalanceinrelationtothedays.Thisdesign

wouldbe:

day1

day2

(1)

a

c

ac

b

ab

bc

abc

Thein uencefromthedayscannowbecalculatedbyadding�1gtoalldatafromday1

andadding+2gtoalldatafromday2:

c hs.


23

2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1)�1g

a�1g

b+2g

ab+2g

c�1g

ac�1g

bc+2g

abc+2g

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I+4g

A B+12g

AB

C AC

BC

ABC

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

Onecanseethatnow,becauseofthedescribedattributeofthematrix,itisonlytheB

contrastandtheaveragethatarea�ectedbythedistributionoverthetwodays.

Ofcoursethisdesignisnotveryusefulifwealsowanttoestimatethee�ectoffactor

B,aswecannotunequivocallyconcludewhetheraB-e�ectfoundcomesfromfactorB

orfromdi�erencesintheblocks(thedays).Ontheotherhand,alltheothere�ectsare

clearlyfreefromtheblocke�ect(thee�ectofthedays).

Onesaysthatmaine�ectoffactorBisconfoundedwiththee�ectoftheblocks(the

word\confound"isfromLatinandmeansto\mixup").

Thelastexampleshowshowwe(byfollowingthe+1and�1variationforthecorrespond-

ingcontrast)candistributethe8singleexperimentsoverthetwodayssothatprecisely

oneofthee�ectsofthemodelisconfoundedwithblocks,andnomorethantheone

chosen.Onecanshowthatthiscanalwaysjustbedone.

If,forexample,wechoosetodistributeaccordingtothethree-factorinteractionABC,it

canbeseenthattherowfor[ABC]has+1fora,b,cogabc,but�1for(1),ab,acogbc.

Onecanalsofollowthe+and�signsinthefollowingtable:

(1)

a

b

ab

c

ac

bc

abc

I

+

+

+

+

+

+

+

+

A

�

+

�+

�+

�

+

B

�

�+

+

��

+

+

AB

+

��+

+

�

�

+

C

�

��+

+

+

+

AC

+

�+

��+

�

+

BC

+

+

��

+

+

ABC

�

+

+

�+

�

�

+

Thisgivesthefollowingdistribution,aswenowingeneraldesignatethedaysasblocks

andletthesehavethenumbers0and1:

block0

block1

(1)

ab

ac

bc

a

b

c

abc

c hs.


24

Theblockthatcontainsthesingleexperiment(1)iscalledtheprincipalblock.The

practicalmeaningofthisisthatonecanmakeastartinthisblockwhenconstructingthe

design.

2.2.1

Constructionofaconfoundedblockexperiment

Theexperimentdescribedaboveiscalledablockconfounded(orjustconfounded)23

factorialexperiment.Thechosenconfoundingisgivenwiththeexperiment's

de�ningrelation:I=

ABC

AndinthisconnectionABCiscalledthede�ningcontrast.

Aneasywaytocarryoutthedesignconstructionistoseeifthesingleexperimentshavean

evenoranunevennumberoflettersincommonwiththede�ningcontrast.Experiments

withanevennumberincommonshouldbeplacedintheoneblockandexperimentswith

anunevennumberincommonshouldgointheotherblock.

Alternativelyonemayusethefollowingtabularmethodwherethecolumnfor'Block'

isfoundbymultiplyingtheA,BandCcolumns:

A

B

C

code

Block=ABC

�1

�1

�1

(1)

�1

+1

�1

�1

a

+1

�1

+1

�1

b

+1

+1

+1

�1

ab

�1

�1

�1

+1

c

+1

+1

�1

+1

ac

�1

�1

+1

+1

bc

�1

+1

+1

+1

abc

+1

Theexperimentisanalysedexactlyasanordinary23factorialexperiment,butwiththe

exceptionthatthecontrast[ABC]cannotunambiguouslybeattributedtothefactorsin

themodel,butisconfoundedwiththeblocke�ect.

Onecanaskwhetheritispossibletodotheexperimentin4blocksof2singleexperiments

inareasonableway.Thishasgeneralrelevance,sincepreciselytheblocksize2(which

naturallyisthesmallestimaginable)occursfrequentlyinpracticalinvestigations.

Onecouldimaginethatthe8observationswereputintoblocksaccordingtotwocriteria,

i.e.bychoosingtwode�ningrelationsthatforexamplecouldbe:

c hs.


25

I 1=

(1)

ab

ac

bc

ABC

c

abc

a

b

I 2=AB

block(0,0)�1

block(0,1)�2

block(1,0)�3

block(1,1)�4

Onenoticesforexamplethattheexperimentsinblock(0,1)haveanevennumberof

lettersincommonwithABCandanunevennumberoflettersincommonwithAB.

Thetabularmethodgives

A

B

C

code

B1=ABC

B2=AB

Blockno.

�1

�1

�1

(1)

�1

+1

1�(0,0)

+1

�1

�1

a

+1

�1

4�(1,1)

�1

+1

�1

b

+1

�1

4�(1,1)

+1

+1

�1

ab

�1

+1

1�(0,0)

�1

�1

+1

c

+1

+1

3�(1,0)

+1

�1

+1

ac

�1

�1

2�(0,1)

�1

+1

+1

bc

�1

�1

2�(0,1)

+1

+1

+1

abc

+1

+1

3�(1,0)

Inthe�gure,thereisa2�2blocksystem,correspondingtothegroupingaccordingto

ABCandAB.OnecannotethatthefactorsAandBarebothon\high"aswellas\low"

levelinall4blocks.Thesefactorsareobviouslyinbalanceinrelationtotheblocks.

However,thisdoesnotapplytofactorC.Itisat\high"levelintwooftheblocksandat

\low"levelintheothertwo.Ifitissounfortunatethatthetwoblocksdesignated(0,0)

and(1,1)togetherresultinahigherresponsethantheothertwoblocks,wewillgetan

undervaluationofthee�ectoffactorC.ThusfactorCisconfoundedwithblocks.

Tobeabletoforeseethis,onecanperceiveABC

andAB

asfactorsandthenwitha

formalcalculation�ndtheinteractionbetweenthem:

Blocke�ect=Blocklevel+ABC+AB+(ABC�AB)

Forthee�ectthuscalculated(ABC�AB)=A

2B

2C,thearithmeticruleisintroduced

thatinthe2kexperiment,theexponentsarereducedmodulo2.Thus(ABC�AB)=

A2B

2C

�!A

0B

0C

�!C

.Therebyonegetstheformalexpressionfortheblock

confounding:

Blocke�ect=Blocklevel+ABC+AB+C

whichtellsusthatitispreciselythethreee�ectsABC,ABandCthatbecomeconfounded

withtheblocksinthegivendesign.

c hs.


26

Ifonewantstoestimatethemaine�ectofC,thisdesignisthereforeunfortunate.A

betterdesignwouldbe:

I 1=

(1)

abc

ac

b

AC

c

ab

a

bc

I 2=AB

block0,0

block0,1

block1,0

block1,1

Since(AC�AB)=A

2BC=BC,thein uenceoftheblocksinthedesignisformally

givenby

Blocks=Blocklevel+AC+AB+BC

andthede�ningrelation:I=AB=AC=BC.

Thethreee�ectsAB,ACandBCareconfoundedwithblocks.Allothere�ectscanbe

estimatedwithoutin uencefromtheblocks.Takespecialnotethatthemaine�ectsA,

BandCallappearatbothhighandlowlevelsinall4blocks.Thethreefactorsarethus

allinbalanceinrelationtotheblocks.

Thedesignshownisthebestexistingdesignforestimatingthemaine�ectsof3factors

inminimalblocks,thatis,with2experimentsineach.Sinceminimalblocksatthesame

timeresultinthemostaccurateexperiments,thedesignisparticularlyimportant.

Thedesigndoesnotgivethepossibilityofestimatingthetwo-factorinteractionsAB,AC

andBC.

2.2.2

A

one-factor-at-a-timeexperiment

Itcouldbeinterestingtocomparethedesignshownwiththefollowingone-factor-at-a-

timeexperiment,whichisalsocarriedoutinblocksofsizeof2:

(1)a

(1)b

(1)c

thatis3blocks,wherethefactorsareinvestigatedeachinoneblock.

Theexperimentcouldbeaweighingexperiment,whereonewantstodeterminetheweight

ofthreeitems,A,BandC.Themeasurement(1)correspondstothezeropointreading,

whileagivesthereadingwhenitemAis(alone)ontheweightandcorrespondinglyforb

andc.

Inthisdesign,anestimateforexampleoftheAe�ectisfoundas

c hs.


27

b A=[�(1)+a]withvariance2�

2

whereitishereassumedthatr=1.Intheprevious23designin2�2blocks,itwasfound

b A=[�(1)+a�:::+abc]=(2

3�

1)withvariance�

2=2

Ifoneistoachieveanaccuracyasgoodasthe\optimal"designwithrepeateduseofthe

one-factor-at-a-timedesign,ithastoberepeated(2��

2=(�

2=2)=4times.Thus,there

willbeatotalof4�6=24singleexperimentsincontrasttothe8thatareusedinthe

\optimal"design.

Anotherone-factor-at-a-timein2blocksof2singleexperimentsisthefollowingexperi-

ment:

(1)

a

b

c

block0

block1

Whyisthisahopelessexperiment?Whatcanoneestimatefromtheexperiment?

2.3

Partiallyconfounded2k

factorialexperiment

Wewillagainconsiderthe2�2experimentwiththetwofactorsAandB:

B=0

B=1

A=0

(1)

b

A=1

a

ab

Supposethatthisexperimentistobedoneinblocksofthesize2.Theblockscan

correspondforexampletobatchesofrawmaterialthatarenolargerthanatmost2

experimentsperbatchcanbedone.Bychoosingthede�ningcontrastasI=AB,the

followingblockgroupingisobtained:

Experiment1:

(1) 1

ab 1

a2

b 2

batch1

batch2

I=AB

Themathematicalmodeloftheexperimentisthefollowing:

Yij�=�+Ai+Bj+ABij+Eij�,wherei=(0;1);j=(0;1);�=1

c hs.


28

buttheABinteractione�ectisconfoundedwithblocks.

Supposenowthatwefurtherwanttoestimateand/ortesttheinteractioncontribution

ABij.Thisofcoursecanonlybedonebydoingyetanotherexperiment,inwhichABis

notconfoundedwithblocks.Therearetwosuchexperiments,onewhereAisconfounded

withblocksandonewhereBisconfoundedwithblocks.Asexamplewechoosethelatter:

Experiment2:

(1) 3

a3

b 4

ab 4

batch3

batch4

I=B

Fromthetwoexperimentsshown(eachwithtwoblocksandtwosingleexperimentsin

eachblock)wewillnowestimatethevariouse�ects.Themaine�ectforfactorAcanbe

estimatedbothinthetwo�rstblocksandinthetwolastblocksandatotalAcontrastis

foundas:

[A] total=[A] 1+[A] 2

;

thatis,thesumoftheAcontrastsinboththetwoexperimentalparts:

[A] 1=�(1) 1+a2�b 2+ab 1

(from

experiment1)

[A] 2=�(1) 3+a3�b 4+ab 4

(from

experiment2)

astheindexofthe8singleexperimentscorrespondstotheblock(batch)inwhichthe

singleexperimentsweremade.Theindexofthecontrastgiveswhetheritisthe�rstor

thesecondexperimentalpartitiscalculatedin.

Further,wecannow�ndacontrastforthemaine�ectB,butonlyfromthe�rstexperi-

ment:

[B] 1=�(1) 1�a2+b 2+ab 1

Finallyacontrastfortheinteractione�ectABisfound,butnowfromtheotherexperiment

whereitisnotconfoundedwithblocks:

[AB] 2=+(1) 3�a3�b 4+ab 4

SincethetwoAcontrastsarebothfreeofblocke�ects,inadditiontotheirsumwecan

�ndtheirdi�erence:

[A] di�erence=[A] 1�[A] 2

c hs.


29

Thisamountmeasuresthedi�erencebetweentheAestimatesinthetwopartsofthe

experiment.Thisdi�erence,astheexperimentislaidout,canonlybeduetoexperimental

uncertainty,andcanthusbeinterpretedasanexpressionoftheexperimentaluncertainty,

thatis,theresidualvariation.

Thetwocontrasts

[B] 2=�(1) 3�a3+b 4+ab 4

(from

experiment2)

[AB] 1=+(1) 1�a2�b 2+ab 1

(from

experiment1)

arebothconfoundedwithblocks.

Asexpressionsoftheexperimentallevelsinthetwopartsoftheexperimentwe�nd

[I] 1=+(1) 1+a2+b 2+ab 1

(from

experiment1)

[I] 2=+(1) 3+a3+b 4+ab 4

(from

experiment2)

Thisresultsin

[I] total=[I] 1+[I] 2

[I] di�erence=[I] 1�[I] 2

Thequantity[I] totalandthecontrast[I] di�erencemeasurethelevelofthewholeex-

perimentandthedi�erenceinlevelbetweenthe�rstandsecondpartoftheexperiment,

respectively.

Onecaninvestigatewhetherthequantitiesdrawnupareorthogonalcontrastsbylooking

atthefollowingmatrixexpression:

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4[I]

[A] total

[B] 1

[AB] 2

[A] 1�[A] 2

[B] 2

[AB] 1

[I] 1�[I] 2

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

0

0

0

0

0

0

0

0

1

�1

�1

1

�1

1

�1

1

1

�1

1

�1

0

0

0

0

�1

�1

1

1

1

�1

�1

1

0

0

0

0

1

1

1

1

�1

�1

�1

�1

3 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1) 1

a2 b 2 a

b 1(1) 3

a3 b 4 a

b 43 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

c hs.


30

Oneseesthatallthecontrastsaremutuallyorthogonalandthatthereareexactly7

contrastsandthesum(thepseudocontrast).Theycanthereforeasawholedescribeall

thevariationbetweenthe8singleexperimentsthathavebeencarriedout.

Btmeansofthegeneralformulaforthesumsofsquaresandforcontrastsinparticular,

wecanthen�ndallthesumsofsquares.

SSQA

=

[A]2 total

RA�2k�r

,

whereRA

=2andk=2

SSQB

=

[B]2 1

RB�2k�r

,

whereRB

=1andk=2

SSQAB

=

[AB]2 2

RAB�2k�r

,

whereRAB

=1andk=2

SSQA;uncertainty

=

[A]2 1+[A]2 2

2k�r

�[A]2 total

RA�2k�r

,

whereRA

=2

SSQB;blocks

=

[B]2 2

RB;blocks�

2k�r

,

whereRB;blocks=1

SSQAB;blocks

=

[AB]2 1

RAB;blocks�

2k�r

,

whereRAB;blocks=1

SSQleveldi�erence

=

[I]2 1+[I]2 2

2k�r

�[I]2 total

R�2k�r

,

whereR=2

Allthesesumsofsquareshave1degreeoffreedom,andwecannowdrawupananalysis

ofvariancetablebasedon:

SSQA

,

fA

=1

,

(precision=1)

SSQB

,

fB

=1

,

(precision=1/2)

SSQAB

,

fAB

=1

,

(precision=1/2)

SSQblocks=SSQB;blocks+SSQAB;blocks+SSQleveldi�erence;fblocks=3

SSQuncertainty=SSQA;uncertainty

;

funcertainty=1

2.3.1

Somegeneralisations

Asshown,allthedrawnupcontrastsaremutuallyorthogonal.

c hs.


31

Thismeansthatthesumofsquaresforthe7contrastsmakeupthetotalsumofsquares.

Thisisalsosynonymouswiththestatementthatthee�ectswehaveintheexperiment

areallmutuallybalanced.

Onecanalsonotethatthenumberofdegreesoffreedomisprecisely8�1=7,namely

oneforeachofthecontrasts,towhichcomes1forthetotalsum[I].

Onecanthentesttheindividuale�ectsofthemodelagainsttheestimateofuncertainty.

Intheexample,thisestimatehasonly1degreeoffreedom,andofcoursethisdoesnot

giveareasonablystrongtest.

ThefactthatoneseeminglycomestothesameFtestforthee�ectofA,Baswellas

ABisnotsynonymouswiththestatementthattheA,BandABe�ectsareestimated

equallyprecise.Forexamplethiscanbeseenbycalculatingthevariancesintheparameter

estimates: V

ar(b A 1)=Varf

[A]

RA

�2k�rg=�

2�

RA

�2k�r

(RA

�2k�r)2

=

�2

RA

�2k�r

;RA

=2

Var(b B 1)=Varf

[B] 1

RB

�2k�rg=�

2�

RB

�2k�r

(RB

�2k�r)2

=

�2

RB

�2k�r

;RB

=1

Var(d AB11)=Varf

[AB] 2

RAB

�2k�rg=�

2�

RAB

�2k�r

(RAB

�2k�r)2

=

�2

RAB

�2k�r

;RAB

=1

sothatthevariancesoftheBandABestimatesaredoublethevarianceoftheAestimate.

ThisofcourseisduetothefactthattheAestimateisbasedontwiceasmanyobservations

astheotherestimates(RA=RB

=2andRA=RAB

=2).

Thedi�erencebetweenthetestsofthethreee�ectsistheirpower.ThetestoftheA

e�ecthasgreaterpowerthantheothertwotests(forthesametestlevel�).

Onecangenerallywrite

VarfParameterestimateg=Varf[Contrast]=(R�2

k�r)g=�

2=(R�2

k�r)

;

VarfE�ectestimateg=Varf2�[Contrast]=(R�2

k�r)g=�

2�4=(R�2

k�r)

;

whereRgivesthenumberof2kfactorialexperimentsonwhichtheestimateisbased,

andrgivesthenumberofrepetitionsforthesinglefactorcombinationsinthesefactorial

c hs.


32

experiments.FortheAe�ectintheexample,R=2,whileR=1forboththeBandthe

ABe�ect.

Ifonehasrepeateda2kfactorialexperimentRtimes,whereane�ectcanbeestimated,

onecan�ndthevariationbetweentheseestimatesinasimilarwayasshownfortheA

e�ectintheexample.Ifwesuppose,forthesakeofsimplicity,thatitistheAe�ect

thatcanbeestimatedintheseRdi�erent2kexperimentswithrrepetitionsperfactor

combination,wecangenerally�ndanestimateofuncertaintyasthesquaresum:

SSQA;uncertainty=

[A]2 1+[A]2 2+��+[A]2 R

2k�r

�([A] 1+[A] 2+��+[A] R)2

R�2k�r

whichwillhaveR�1degreesoffreedom.Theamount[A] �givestheAcontrastinthe

�'thfactorialexperiment.

OnenotesthatthissumofsquaresisexactlythevariationbetweentheRestimatesfor

theAe�ect.Ifonehasseverale�ectswhichinthiswayareestimatedseveraltimes,all

theiruncertaintycontributionscanbesummedupinacommonuncertaintyestimate,

whichcanbeusedfortesting.

Estimationafblocke�ects

Insomeconnections,itcanbeofinteresttoestimatespeci�cblockdi�erences.Ifwe

againtakethetwoconfounded2kexperimentsthatformthebasisforthissection,we

couldforexamplebeinterestedinestimatingthedi�erencebetweenblock0andblock

1.Anestimateforthisdi�erencecanbederivedbyrememberingthatthedi�erence

betweenblocks0and1isconfoundedwiththeABe�ect,andthatthepureABe�ect

canbeestimatedinthesecondexperimentalhalf.Inotherwords,wecandrawupthe

contrast

[AB] 1�[AB] 2=[(1)1�a2�b 2+ab 1]�[(1)3�a3�b 4+ab 4]

Thisquantityhas2�(di�erencebetweenblock0andblock1)asitsexpectedvalue,and

onecanthereforeusetheestimateX1=([AB] 1�[AB] 2)=2astheestimatefortheblock

di�erenceBlock1-Block2.

Inthisestimationofblocke�ects,theprincipleisthesimpleonethatoneestimatesthe

e�ectsthattheblocksareconfoundedwithandthenbreakstheconfoundingwiththese

estimates.

Wewillnotgofurtherherewiththeseideas,butonlypointoutthegeneralpossibilities

thatlieinusingpartialconfounding.

c hs.


33

Itbecomespossibletotestandestimateallfactore�ectsinfactorialexperimentswith

smallblocks,justasitbecomespossibletoextractblocke�ectswiththeoutlinedestima-

tiontechnique.

2.4

Fractional2k

factorialdesign

Inthissection,wewillintroduceaspecialandveryimportanttypeofexperiment,which

undercertainassumptionscanhelptoreduceexperimentalworkgreatlyincomparison

withcompletefactorialexperiments.

Example2.1:

A

simpleweighingexperimentwith3items

Supposewewanttodeterminetheweightofthreeitems,A,BandC.Aweighingresult

canbedesignatedinthesamewayasdescribedabove.Forexample"a"designatesthe

resultoftheweighingwhereitemAisontheweightalone,while(1)designatesweighing

withoutanyitembeingontheweight,i.e.,thezeropointadjustment.

Thesimplestexperimentconsistsindoingthefollowing4singleexperiments:

(1)

a

b

c

thatis,thatonemeasurementobtainedwithoutanyitemontheweightisobtained�rst,

andthethreeitemsareweighedseparately.

Theestimatesfortheweightofthethreeitemsare:

b A=[�(1)+a]

;

b B=[�(1)+b]

;

b C=[�(1)+c]

Thiskindofdesignisprobablyfrequently(butunfortunately)usedinpractice.Itcanbe

brie ycharacterisedas\one-factor-at-a-time".

Onecandirectly�ndthevarianceintheestimates:

Varfb Ag=Varfb Bg=Varfb Cg=2��

2

Abasiccharacteristicofgoodexperimentaldesignsisthatalldataareusedinestimates

foralle�ects.Thisisseennottobethecasehere,andonecanaskifonecouldpossibly

�ndanexperimentaldesignthatismore\eÆcient"thantheoneshown.

Theexperimentsthatcanbecarriedoutare:

(1)

a

b

ab

c

ac

bc

abc

c hs.


34

Thecompletefactorialexperimentofcourseconsistsofdoingall8singleexperiments,

andtheestimatesforthee�ectsarefoundaspreviouslyshown.InthecaseoffactorA,

wegetthee�ectestimate: b A=

[�(1)+a�b+ab�c+ac�bc+abc]=4

whichisthustheestimatefortheweightofitemA.Thevarianceforthisestimateis�

2=2

(namely8�

2=42).

OnecaneasilyconvinceoneselfthattheweightofitemBanditemCarebalancedout

oftheestimateforA.Thesameappliestoapossiblezeropointerrorinthescaleofthe

weight(�).

Asanalternativetothesetwoobviousexperiments,wecanconsiderthefollowingexper-

iment:

(1)

ab

bc

ac

Theexperimentthusconsistsofweighingtheitemstogethertwobytwo.Forexamplethe

estimatefortheweightofAis:

b A=[A]=2=[�(1)+ab+ac�bc]=2

Varfb Ag=�

2

Notethatthezeropoint�aswellastheweightsofitemsBandCareeliminatedinthis

estimate.

Onealsonotesthatinrelationtotheprimitive\one-factor-at-a-time"experiment,inthis

designwecanuseall4observationstoestimatetheAe�ect,thatis,theweightofitem

A.ThesameobviouslyappliestotheestimatesfortheBandCe�ects.Inaddition,

thevarianceoftheestimatehereisonlyhalfthevarianceoftheestimatesinthe\one-

factor-at-a-time"experiment.Theexperimentisthereforeappreciablybetterthanthe

\one-factor-at-a-time"experiment.

Theexperimentiscalleda

1 2�23factorialexperimentora23

�

1

factorialex-

periment,asitconsistspreciselyofhalfthecomplete23factorialexperiment.

Finallyasmallnumericalexample:

(1)=6:78g

ab=28:84g

ac=20:66g

bc=18:12g

c hs.


35

b A=

(�6:78+28:84+20:66�18:12)=2

=

12.30g

b B=

(�6:78+28:84�20:66+18:12)=2

=

9.76g

b C=

(�6:78�28:84+20:66+18:12)=2

=

1.58g

Letussupposethatthemanufacturerhasstatedthattheweighthasanaccuracycorre-

spondingtothestandarddeviation�=0:02g.WiththisisfoundVarfb Ag=4�0:02

2=22=

0:02

2g2.ThestandarddeviationoftheestimatedAweightisthus0.02g.Thesamestan-

darddeviationisfoundfortheweightsofBandC.

A95%con�denceintervalfortheweightofAis12.30�2�0.02g=[12.26,12.34]g.

Endofexample2.1

Wewillnowdiscusswhatcangenerallybeestimatedinanexperimentasdescribedinthe

aboveexample.Ifonecanassumethatitisonlythemaine�ectsthatareimportantin

theexperiments,therearenoproblemsestimatingthese.Intheexample,onecantakeit

thattheweightofthetwoitemsisexactlythesumoftheweightsofthetwoitems,which

correspondstosayingthatthereisnointeraction.

Alternatively,wenowimaginethatthefollowinggeneralmodelappliesforthedescribed

experimentwiththethreefactors,A,BandC:

Yijk�=�+Ai+Bj+ABij+Ck+ACik+BCjk+ABCijk+Eijk�

wherei;j;k=(0;1)and�=(1;:::;r)withtheusualrestrictions.Completerandomisa-

tionisassumed.

ThequantityEijk�

designatestheexperimentalerrorinthe�'threpetitionofthesingle

experimentindexedby(i;j;k).

Forthesingleexperiment"(1)"inthedescribedexperiment,allindicesareonlevel\0",

anditsexpectedvalueis:

Ef(1)g=�+A0+B0+AB00+C0+AC00+BC00+ABC000

ByusingthefactthatA0=�A1

andcorrespondinglyfortheothertermsofthemodel,

we�nd

Ef(1)g=��A1�B1+AB11�C1+AC11+BC11�ABC111

Efabg=�+A1+B1+AB11�C1�AC11�BC11�ABC111

c hs.


36

Efacg=�+A1�B1�AB11+C1+AC11�BC11�ABC111

Efbcg=��A1+B1�AB11+C1�AC11+BC11�ABC111

Inthisway,fortheAcontrastwecannow�nd

Ef[A]g=Ef�(1)+ab+ac�bcg=4(A1�BC11)

ThismeansthatifthefactorsBandCinteract,soBC116=0,theestimateforthemain

e�ectoffactorAwillbea�ectedinthishalfexperiment.Thee�ectsAandBCare

thereforeconfoundedintheexperiment.Itholdstruegenerallyinthisexperimentthat

thee�ectsareconfoundedingroupsoftwo.

Thisisformallyexpressedthroughthealiasrelation"A=BC".Therelationexpresses

thatthee�ectsAandBCactsynchronouslyintheexperimentandthattheytherefore

areconfounded.TheAandBCe�ectscannotbedestinguishedfromeachotherinthe

experiment.

Thealiasrelationsforthewholeexperimentare

I

=

ABC

A

=

BC

B

=

AC

C

=

AB

wherethe�rstrelation,I=ABC,iscalledthede�ningrelationoftheexperimentand

ABCcalledthede�ningcontrast-inthesamewayasintheconstructionofaconfounded

blockexperiment(cf.page25).Thisexpressesthatthethree-factor-interactionABC

doesnotvaryintheexperiment,buthasthesamelevelinallthesingleexperiments

(namely�ABC111).

Theotheraliasrelationsaresimplyderivedbymultiplyingbothsidesofthede�ning

relationwiththee�ectsofinterest,andthenreducingtheexponentsmodulo2.For

example,thealiasrelationfortheAe�ectisfoundasA�I=A�ABC

i.e.A

=

A2BC

�!BC,where"I"isheretreatedasa\one"andthe2-exponentinA

2BC

is

reducedto0(modulo2reduction).

Ifwerecalltheconfoundedblockexperiment,whereacomplete23factorialexperiment

couldbelaidoutintwoblocksaccordingtothede�ningrelationI=ABC,weseethat

ourexperimentispreciselytheprincipalblockinthatexperiment.Ifitisacaseofa

1 2�2kfactorialexperiment,thefractionthatcontains"(1)"canbecalledtheprincipal

fraction.

c hs.


37

Wecancheckwhetherfromtheotherhalfofthecompleteexperimentonecould�nd

estimatesthatarejustasgoodasinthehalfwetreatedinourexample.Theexperiment

is

a

b

c

abc

[A]=[a�b�c+abc]

Ef[A]g=Efa�b�c+abcg=4(A1+BC11)

Notethattheconfoundinghastheoppositesigncomparedwithearlier.Ifoneaddsthe

twocontrasts,thatis

[�(1)+ab+ac�bc]+[a�b�c+abc]

one�ndspreciselytheAcontrastforthecompleteexperiment,whilesubtractingthem,

thatis

�[�(1)+ab+ac�bc]+[a�b�c+abc]

�ndspreciselytheBCcontrast.

Thetwoalternativehalfexperimentsarecalledcomplementaryfractionalfactorials,as

togethertheyformthecompletefactorialexperiment.

Wewillnowshowhowonechoosesforexamplea

1 2�23factorialexperimentinpractice.

Wenotethata

1 2�23factorialexperimentconsistsof22measurements.Theexperiment

thatistobederivedcanthereforebeunderstoodasa22experimentwithanextrafactor

putin.Letusthereforeconsiderthecomplete22experimentwiththefactorsAandB.

Themathematicalmodelforthisexperimentis:

Yij�=�+Ai+Bj+ABij+Eij�

;

i=(0;1);j=(0;1);�=(1;2;:::;r)

Ifwesuspectthatall4parametersinthismodelcanbeimportant,furtherfactorscannot

beputintotheexperiment,butifweassumethattheinteractionABisnegligible,asin

theweighingexperiment,wecanintroducefactorC,sothatitisconfoundedwithjust

AB.

c hs.


38

WethereforechoosetoconfoundCwithAB,thatisusingthealiasrelationC=AB.

Thisaliasrelationcan(only)bederivedfromthede�ningrelationI=ABC,which

canbeseenbymultiplyingthealiasrelationC=ABonbothsideswithC(orABfor

thatmatter)andreducingallexponentsmodulo2.

C=AB

=)

I

=

ABC(thede�ningrelation)

A

=

BC

B

=

AC

AB

=

C

(thegeneratorequation)

WeshallcallC=ABthegeneratorequationsinceitisthealiasrelationfromwhich

thedesignisgenerated.

Theprincipalfractionismadeupofallsingleexperimentsthathaveanevennumber

oflettersincommonwithABC,i.e.,theexperiments(1),ab,ac,bc.Alternatively,the

complementaryfractioncouldbechosen,whichcontainsallsingleexperimentsthathave

anunevennumberoflettersincommonwithABC,i.e.,a,b,candabc.

Withthislastmethod,wherethestartingpointisthecompletefactorialexperimentforthe

two(�rst)factorsAandB,itissaidthattheseformanunderlyingcompletefactorial

forthefractionalfactorialdesign.Wewillreturntothisimportantconceptlater.

Letusnowsupposethatwechoosetheexperimentcorrespondingto\uneven":

a

b

c

abc

To�ndthesignfortheconfoundings,itisenoughtoconsideroneofthealiasrelations,

forexampleC=ABandcomparethiswithoneoftheexperimentsthatistobedone,

forexampletheexperiment\a".

Fortheexperiment"a",thee�ectChasthevalueC0

(sincefactorCison0level),and

thee�ectABhasthevalueAB10.TheconfoundingisthereforeC0

=AB10.Sincewe

calculateonthebasisofthe\high"levelsC1andAB11,theseareputin.

SinceC0=�C1

andAB10=�AB11,we�nallygetthatthealiasrelationisC1=AB11.

Therestofthealiasrelationsgetthesamesignwhentheyareexpressedinthehighlevels.

ForexampleonegetsA1=BC11.

Onewritesforexample

+C=+AB

=)

+I

=

+ABC(thede�ningrelation)

+A

=

+BC

+B

=

+AC

+AB

=

+C

(thegeneratorequation)

c hs.


39

Whethertheconstructedexperimentisasuitableexperimentdependsonwhetherthe

aliasrelationstogethergivesatisfactorypossibilitiesforestimatingthee�ects,which,a

priori,areconsideredinteresting.

Example2.2:

A

1/4�2

5

factorialexperiment

.

We�nishthissectionbyshowinghow,withthehelpoftheintroducedideas,onecan

constructa1=4�25factorialexperiment,i.e.,anexperimentthatconsistsonlyof23=8

measurements,butincludes5factors.Thesearecalled(always)A,B,C,DandE(for

1st,2nd,3rd,4thand5thfactor).

Thecompletefactorialexperimentwith3factorscontainsinadditiontothelevel�the

e�ectsA,B,AB,C,AC,BC,andABC.

SupposenowthatitcanbeassumedthatfactorsBandCdonotinteract,i.e.that

BC=0.AreasonableinferencefromthiscouldbethatalsoABC=0.Therebyitwould

benaturaltochoosetwogeneratorequations,namelyD

=

BC

andE

=

ABC.

ThesegiveI 1=BCDandI 2=ABCE,respectively.Theprincipalfractionconsistsofthe

singleexperimentsthathaveanevennumberoflettersincommonwithboththede�ning

contrastsBCDandABCE.Thesesingleexperimentsare:

(1)

ae

bde

abd

cde

acd

bc

abce

Adirectandeasymethodtoconstructthisexperimentistowriteoutatableasfollows:

A

B

C

D=�BC

E=ABC

Code

�1

�1

�1

�1

�1

(1)

+1

�1

�1

�1

+1

ae

�1

+1

�1

+1

+1

bde

+1

+1

�1

+1

�1

abd

�1

�1

+1

+1

+1

cde

+1

�1

+1

+1

�1

acd

�1

+1

+1

�1

�1

bc

+1

+1

+1

�1

+1

abce

NotethatforthefactorsA,BandCtheorderingofthelevelscorrespondtothestandard

order:(1),a,b,ab,c,ac,bc,abc,asusedinYatesalgorithm,forexample.

TheminussigninD=�BCensuresthattheexperiment(1)isobtainedasthe�rstone,

iftheprincipalfractioniswanted.

Thistabularmethodofwritingouttheexperimentcanbeusedquitegenerallyaswillbe

demonstratedinthefollowing.Afurtheradvantageisthatthesignsoftheconfoundings

areobtaineddirectly.

Analternativeexperimentisfoundbyconstructingoneoftheother\fractions".Iffor

c hs.


40

exampleonewantsanexperimentthatcontainsthesingleexperiment"a",thecorre-

spondingfractioncanbefoundbymultiplyingtheprincipalfractionthroughwith\a"

andreducingtheexponentsmodulo2.Inthiswayonegets:

a

e

abde

bd

acde

cd

abc

bce

ThesameexperimentwouldhavebeenobtainedbychangingthesigninE=ABCso

thatE=�ABCisusedintheabovetabularmethod.

Now,to�ndthealiasrelationsintheexperiment,wewillagainusethetwode�ning

relations.

Theinteractionofthetwode�ningcontrastsisfoundbymultiplyingthemtogetherand

againreducingallexponentsmodulo2:

D=BC,E=ABC=)I 1=BCD,I 2=ABCE

andI 3=I 1�I 2=AB

2

C2DE!ADE

sothatthede�ningrelationandthealiasrelations(withoutsigns)oftheexperimentare:

I

=

BCD

=

ABCE

=

ADE

A

=

ABCD

=

BCE

=

DE

B

=

CD

=

ACE

=

ABDE

AB

=

ACD

=

CE

=

BDE

C

=

BD

=

ABE

=

ACDE

AC

=

ABD

=

BE

=

CDE

BC

=

D

=

AE

=

ABCDE

ABC

=

AD

=

E

=

BCDE

Roughlyspeaking,theexperimentisonlyagoodexperimentifonecanassumethatthe

interactionsarenegligible(inrelationtothemaine�ects).

Thesignsfortheconfoundingscanagainbefoundbyconsideringanaliasrelation,e.g.

A=ABCD=BCE=DE,togetherwithoneofthesingleexperimentsthatarepartof

thechosenexperimentaldesign.

Forexample"a"isintheexperimentanditcorrespondstoasingleexperimentwith

indices(1,0,0,0,0)forthefactorsA,B,C,DandE,respectively.Thus

A1=ABCD1000=BCE000=DE00()+A1=�ABCD1111=�BCE111=+DE11

Thissignpatternisrepeatedinallthealiasrelations:

c hs.


41

+I

=

�BCD

=

�ABCE

=

+ADE

+A

=

�ABCD

=

�BCE

=

+DE

+B

=

�CD

=

�ACE

=

+ABDE

+AB

=

�ACD

=

�CE

=

+BDE

+C

=

�BD

=

�ABE

=

+ACDE

+AC

=

�ABD

=

�BE

=

+CDE

+BC

=

�D

=

�AE

=

+ABCDE

+ABC

=

�AD

=

�E

=

+BCDE

Weneednowto�ndestimatesandsumsofsquares.Thiscanbedonebyagainusing

thefactthattheexperimentisformedonthebasisofthecompleteunderlyingfactorial

structurecomposedoffactorsA,BandC.Inthisstructurewenowestimateallthee�ects

correspondingtothethreefactors.

Inordertosubsequently�ndtheDe�ectweonlyneedtolookuptheBCrow,wherethe

De�ectappearswiththeoppositesign.Dataaregroupedinstandardorderaccordingto

thefactorsA,BandC.Thisisdonebyignoring\d"and\e".Thenthecontrastscanbe

calculated,withtheuseofYates'algorithm,forexample.Onegets:

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I

=

�BCD

=

�ABCE

=

ADE

A

=

�ABCD

=

�BCE

=

DE

B

=

�CD

=

�ACE

=

ABDE

AB

=

�ACD

=

�CE

=

BDE

C

=

�BD

=

�ABE

=

ACDE

AC

=

�ABD

=

�BE

=

CDE

BC

=

�D

=

�AE

=

ABCDE

ABC

=

�AD

=

�E

=

BCDE

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

=2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4e a bd

abde

cdacde

bce

abc

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

Notethattherowe;a;bd;abde;cd;acde;bce;abc,becomestherow(1);a;b;ab,c;ac;bc;abc,

ifoneleavesoutdande,i.e.thestandardorderforthecomplete23factorialexperiment

forA,BandC.

Theexperimentwhichwe�ndbyignoringthefactorsDandE,i.e.thecomplete23

factorialexperimentincludingA,BandC,isagainanunderlyingcompletefactorial

experimentandA,BandCconstituteanunderlyingcompletefactorialstructure.

c hs.


42

WecaneasilycheckthatforexampletheABCe�ectisconfoundedwith-E.Onewayto

dothisistoconsidertheABCcontrast:

[ABC]=�e+a+bd�abde+cd�acde�bce+abc

wherewenotethatalldatawithEatthehighlevel,i.e.,e,abde,acdeandbce,appear

with-1ascoeÆcient,whiletheremainder,i.e.a,bd,cdandabcappearwith+1.The

contrastthereforecontainsacontributionof�4(E1)+4(�E1)=�8E1fromthefactorE.

Thesuggestedexperimentcouldbedoneintwoblocksof4byforexampleconfounding

theABinteractionwithblocks.Thatwouldgivethegrouping:

abc

e

cd

abde

a

bce

bd

acde

block0

block1

Theconfoundingsinthisexperimentwouldbe:

I

=

�BCD

=

�ABCE

=

ADE

A

=

�ABCD

=

�BCE

=

DE

B

=

�CD

=

�ACE

=

ABDE

AB

=

�ACD

=

�CE

=

BDE

=

Blocks

C

=

�BD

=

�ABE

=

ACDE

AC

=

�ABD

=

�BE

=

CDE

BC

=

�D

=

�AE

=

ABCDE

ABC

=

�AD

=

�E

=

BCDE

wherethecontrastsarecalculatedaspreviously,butwherethecontrastthatappearsin

theABrownowcontainspossiblefactore�ectsaswellastheblocke�ects.

Endofexample2.2

2.5

Factorson2and4levels

Inmanycaseswhereseveralfactorsareanalysed,itcanbedesirableandperhapseven

necessaryforsinglefactorsthattheycanappearon3orperhaps4levelstogetherwith

the2levelsoftheotherfactors.Incasethereisaneedforamixtureof2and3levels,itis

diÆculttoconstructgoodexperimentaldesigns,butinthetextbookbyOscarKempthorne

(1952):TheDesignandAnalysisofExperiments,Wiley,NewYork,therearehowever

somesuggestionsforthis.

If4levelsareused,onecanusetheprocedurebelow,whichisdemonstratedwiththehelp

oftwoexamples,sothepresentationisnottoocomplicated.

c hs.


43

Example2.3:

A

2�4experimentin2blocks

Supposethatinafactorialexperimenttwofactorsaretobeanalysed,namelyafactor

Athatappearson2levelsandafactorGthatappearson4levels.Themathematical

modelfortheexperimentis

Yil�=�+Ai+G`+AGi`+Ei`�

wherei=(0;1),`=(0;1;2;3)and�=(1;2;:::;r).

Usualparameterrestrictionsareused

1 X i=0

Ai=

3 X `=0

G`=

1 X i=0

AGi`=

3 X `=0

AGi`=0

Toreformulatethemodeltoa2kfactorialstructure,twonewfactorsareintroduced,B

andC,asreplacementsforG.

G=0

G=1

G=2

G=3

B=0,C=0

B=1,C=0

B=0,C=1

B=1,C=1

Yij�=�+Ai+Bj+ABij+Ck+ACik+BCjk+ABCijk+Eijk�

wheretheindexj=

remainderof(`=2)andk=

integerpartof(`=2).Inversely,

`=j+2k.

ThecorrespondencebetweenfactorGandthetwoarti�cialfactorsBandCisthat

G`=Bj+Ck+BCjk

;

`=j+2k

Supposenowthatonewantstodoacompletefactorialexperimentwiththetwofactors

AandG,i.e.a2�4experiment,oratotalof8singleexperiments.

Supposefurtherthatonewantstodotheexperimentin2blockswith4singleexperiments.

Inthereformulatedmodel,wherefactorGisreplacedbyBandC,weseethatthemain

e�ectoffactorGisgivenasB+C+BC.Thee�ectsB,CandBCmustthereforebe

estimatedandcannotbeusedasde�ningcontrastwhendividingintoblocks.

FortheinteractionbetweenfactorAandfactorG,itholdstruethat

AGi`=ABij+ACik+ABCijk

;

`=j+2k

c hs.


44

andoneofthesethreee�ects(itdoesnotmatterwhich)canbereasonablyusedasde�ning

contrast.WechooseforexampletheABe�ect.

Theexperimenttherebyis

Block1

Block2

(1)

ab

c

abc

a

b

ac

bc

orconvertedtofactorsAandG:

Block1

Block2

(1)

ab

c

abc

a

b

ac

bc

A=0

A=1

A=0

A=1

A=1

A=0

A=1

A=0

G=0

G=1

G=2

G=3

G=0

G=1

G=2

G=3

TheexperimentcanbeanalysedwithYates'algorithm,andonegetsatableofanalysis

ofvariancewhich(inoutline)isbuiltupasthefollowing:

Sourceof

Sumof

Degreesof

S2

F-value

variation

Squares

dom

A

SSQA

1

S2 A

G

SSQB

+SSQC

+SSQBC

3

S2 G

AG-unconfounded

SSQAC

+SSQABC

2

S2 AG

Blocks+AG

SSQAB

1

S2 AG+blocks

Possibleresidual

frompreviousexp.

Total

OneseesthatsomeofthevariationarisingfromtheAGinteractioncanbetakenout

andtested,whiletheremainingpartisconfoundedwithblocks.Ontheotherhand,one

cannotestimatespeci�cAGinteractione�ects,sincethepartdescribedbytheABpart

cannotbeestimated(AG=AB+AC+ABC).

Endofexample2.3

Example2.4:

A

fractional2�2�4factorialdesign

IfforexampleonewantstoevaluatethreefactorsA,BandGwith2,2,and4levels

respectively,twonewarti�cialfactorsareintroduced,CandD,sothatG=C+D+CD,

andinthiscaseonemustkeepthethreee�ectsC,DandCDclearofconfoundings.It

couldbewishedtodosucha2�2�4designwithatotalof16possiblesingleexperiments

asa

1 2�24experiment,usingonly8singleexperiments.

c hs.


45

IfitisassumedthatthethreefactorsA,BandGdonotinteract,theexperimentcanbe

constructedsothatthee�ectsA,B,C,DandCDcanbefound(asG=C+D+CD).

Onecanusethefollowingde�ningcontrastandaliasrelations:

I

=

ABCD

A

=

BCD

B

=

ACD

AB

=

CD

C

=

ABD

AC

=

BD

BC

=

AD

ABC

=

D

wheree�ectsthatareinterestingareunderlined,whilee�ectsconsideredtobewithout

interestarewrittennormally.

Therelationbetweenthefactorsisthat

G=C+D+CD

,

AG=AC+AD+ACD

,

BG=BC+BD+BCD

and

ABG=ABC+ABD+ABCD

Theexperimentwantedcouldbethefollowing(trytoconstructityourself!):

a

b

c

abc

d

abd

acd

bcd

orconvertedtothelevelsofthefactors,astheindexforthefactorGisk+2`,wherek

istheindexforCwhile`istheindexforD:

a

b

c

abc

d

abd

acd

bcd

A=1

A=0

A=0

A=1

A=0

A=1

A=1

A=0

B=0

B=1

B=0

B=1

B=0

B=1

B=0

B=1

G=0

G=0

G=1

G=1

G=2

G=2

G=3

G=3

Ofcourse,onecouldalsohaveusedthecomplementaryexperimentasthestartingpoint:

(1)

ab

ac

bc

ad

bd

cd

abcd

TrytowritethecorrespondingexperimentoutinfactorsA,BandG.

Iftheconstructedexperimentshouldbelaidoutintwoblocksof4singleexperiments,

onecoulduseeitherACorBCasde�ningcontrast.IfACisused,onegetsthedesign:

c hs.


46

Block1

Block2

b

abc

d

acd

a

c

abd

bcd

A=0

A=1

A=0

A=1

A=1

A=0

A=1

A=0

B=1

B=1

B=0

B=0

B=0

B=0

B=1

B=1

G=0

G=1

G=2

G=3

G=0

G=1

G=2

G=3

Note,forexample,thatallthreefactorsA,BandGarebalancedwithinbothblocks.

Endofexample2.4

c hs.


47

3

Generalmethodsforpk-factorialdesigns

Inthischapterwewillintroducegeneralmethodsforfactorialexperimentsinwhichthere

arekfactors,eachonplevels.Thepurposeofthisistogeneralizetheconceptsand

methodsthatwerediscussedinthepreviouschapter,whereweconsideredkfactorsof

whicheachwasononlyp=2levels.

Inparticular,wewilllookatexperimentswithmanyfactorsthathavetobeevaluatedon

2or3levels,whicharemostrelevantinpractice.

Ingeneral,noproofsaregiven,butthesubjectispresentedthroughexamplesanddirect

demonstrationinspeci�ccases.

ThemethodwewilldealwithisoftencalledKempthorne'smethod,andtheinterested

readerisreferredtothetextbookbyOscarKempthorne(1952):TheDesignandAnalysis

ofExperiments,Wiley,NewYork.Thisbookhasasomewhatmoremathematicalreview

oftheexperimentalstructuresandmodelsthatwewilldealwithhere.Infairness,it

shouldbesaidthatitwasactuallyR.A.Fischerandotherswho,around1935,formulated

importantpartsofthebasisforKempthorne'spresentation.

3.1

Completepk


Wenowconsiderexperimentswithkfactorseachonplevels,wherepiseverywhereas-

sumedtobeaprimenumber.Inaddition,completerandomisationisgenerallyassumed.

Incaseswhereexperimentsarediscussedinwhichthereisusedblocking,completeran-

domisationisassumedwithinblocks.

ThefactorsarealwayscalledA,B,C,etc.FactorAisthe�rstfactor,Bthesecondfactor

etc.Inaddition(tothegreatestpossibleextent)weusetheindicesi,j,k,etc.forthe

factorsA,B,C,etc.,respectively.

Theexperimentisgenerallycalledapkfactorialexperiment,andthenumberofpossible

di�erentfactorcombinationsispreciselyp�p�:::�p=pk.

Foranexperimentwith3factors,A,B,andC,thestandardmathematicalmodelis:

Yijk�=�+Ai+Bj+ABi;j+Ck+ACi;k+BCj;k+ABCi;j;k+Eijk�

wherei;j;k=(0;1;::;p�1)and�=(1;2;::;r).

Theindex�=(1;2;::;r)givesthenumberofrepetitionsofeachsingleexperimentinthe

experiment.Theotherindicesassumethevalues(0,1,2,..,p�1).Itshouldbenoted

thattheindexalwaysrunsfrom0uptoandincludingp�1.

c hs.


48

Forsuchexperimentsweintroduceastandardnotationforthesingleexperimentsinthe

samewayaswiththe2kexperiment.Inthecasewherep=

3andk=

3wehave

thefollowingtable,whichshowsallthesingleexperimentsinthecomplete33factorial

experiment:

A

A

A

0

1

2

0

1

2

0

1

2

B=0

(1)

a

a2

c

ac

a2c

c2

ac2

a2c2

B=1

b

ab

a2b

bc

abc

a2bc

bc2

abc2

a2bc2

B=2

b2

ab2

a2b2

b2c

ab2c

a2b2c

b2c2

ab2c2

a2b2c2

C=0

C=1

C=2

Aspreviously,weuseoneoftheseexpressionsasthetermforacertain"treatment"or

factorcombinationinasingleexperiment,aswellasforthetotalresponsefromthesingle

experimentsdonewiththisfactorcombination.Thus,forexample

ab2c=

r X �=1

Y121�=T121�

orjustT121

Forthe2kfactorialexperiment,wearrangedthesetermsinwhatwascalledastandard

order.Wecanalsodothisforthepkexperimentingeneral.Thesestandardordersare:

2k:(1);a;b;ab;c;ac;bc;abc;d;ad;bd;abd;cd;:::

3k:(1);a;a

2;b;ab;a

2b;b2;ab2;a

2b2;c;ac;a

2c;::;a

2b2c2;d;:::

5k:(1);a;a

2;a

3;a

4;b;ab;a

2b;a

3b;a

4b;b2;ab2;a

2b2;a

3b2;a

4b2;:::;a

4b4;

c;ac;a

2c;:::;a

4b4c4;d;ad;:::

7k:(1);a;a

2;:::;a

6;b;ab;a

2b;:::;a

6b6;c;ac;:::;a

6b6c6;d;ad;:::

Forexampleinthe3kfactorialexperiment,anewfactorisaddedbymultiplyingallthe

termsuntilnowwiththefactorinthe�rstpowerandinthesecondpowerandadding

boththesenewrowstotheoriginalorder.

Theseterms,ofcourse,canperfectlywellbeusedasnamesforfactorcombinationsin

completelygeneralfactorialexperiments,buttheresultswewillshowareonlygenerally

applicabletoexperimentsthatcanbeformulatedaspk

factorialexperimentswherepis

aprimenumber.

Beforewecontinue,itwouldbeusefultolookmorecloselyata32factorialexperiment

andshowhowthetotalvariationinthisexperimentcanbedescribedandfoundwiththe

helpofaGraeco-Latinsquare.Inadditionwewillintroducesomemnemonictermsfor

newarti�ciale�ects,whichwilllaterprovetobepracticalintheconstructionofmore

sophisticatedexperimentaldesigns.

c hs.


49

Example3.1:

MakingaGraeco-Latinsquareina32factorialexperiment

Theexperimenthas3�3=9di�erentsingleexperiments:

A=0

A=1

A=2

B=0

(1)

a

a2

B=1

b

ab

a2b

B=2

b2

ab2

a2b2

Themathematicalmodelfortheexperimentis

Yij�=�+Ai+Bj+ABi;j+Eij�

;i=(0;1;2);j=(0;1;2);�=(1;:::;r)

2 X i=0

Ai=

2 X j=0

Bj=

2 X i=0

ABi;j=

2 X j=0

ABi;j=0

Inthisexperimentwecanintroducetwoarti�cialfactors,whichwecancallXandZ.We

letthesefactorshaveindicessandt,respectively,whichwedeterminewith

s=(i+j)3

andt=(i+2j) 3

wherethedesignation(:) 3nowstandsfor"modulo3",i.e."remainderof(.)afterdivision

by3".

Wewillnowseehowtheindicessandtforthede�nednewe�ectsXandZvarythroughout

theexperimentwiththeindicesiandjofthetwooriginalfactorsAandB.

Thisisshowninthetablebelow,asi+jandi+2jarestillcalculated"modulo3".

i

j

s=(i+j) 3

t=(i+2j) 3

0

1

2

0

0

0

0

1

2

0

1

2

0

1

2

1

1

1

1

2

0

2

0

1

0

1

2

2

2

2

2

0

1

1

2

0

Ai

Bj

Xi+j

Zi+2j

Wenotethatifwe�xoneofthelevelsforoneofthe4indices,eachoftheother3indices

appearspreciselywiththevalues0,1and2withinthislevel.Asanexampleofthis,we

considerthesingleexperimentswhereZ'sindext=(i+2j) 3=1:

i

j

s=(i+j) 3

t=(i+2j) 3

1

0

1

1

2

1

0

1

0

2

2

1

Ai

Bj

Xi+j

Zi+2j

c hs.


50

Inthedesignshown,the4factorsA,B,XandZareobviouslyinbalanceinrelationto

eachother.ThedesignisaGraeco-LatinsquarewiththeintroducedfactorsXandZ

insidethesquareandwiththefactorsAandBatthesides.Sincethevariationbetween

the9singleexperimentsor"treatments"intheexperimenthasatotalof9-1degrees

offreedom,andthe4factorsareinbalance,asdescribed,itcanbeshownthatthese4

factorscandescribethewholevariationbetweenthesingleexperiments.

AandBareidenticalwiththeoriginalmaine�ects,anditcanbeshownthatXand

ZtogetherpreciselymakeuptheinteractiontermABi;jinthe"natural"mathematical

modeloftheexperiment.

Wewillnotprovethisresult,butonlyillustrateitwithanexample,whereweimagine

thata32experimentwithoneobservationpercellhasresultedinthefollowingdata:

A=0

A=1

A=2

sum-B

sum-X

sum-Z

B=0

(1)=10

a=15

a2=18

43

X=0

35

Z=0

33

B=1

b=8

ab=12

a2b=16

36

X=1

34

Z=1

36

B=2

b2=5

ab2=9

a2b2=11

25

X=2

35

Z=2

35

sum-A

23

36

45

104

104

104

Theusualtwo-sidedanalysisofvarianceforthesedatawiththefactorsAandBgives:

ANOVA

Sourceof

Sumof

Degreesof

F-value

variation

squares

freedom

A

81.556

(3-1)=2

B

54.889

(3-1)=2

AB

1.778

(3-1)(3-1)=4

Residual

0.000

0

Total

138.223

(9-1)=8

Totherightofthedatatablearesumsforthetwoarti�cialfactorsXandZ.Forexample

the(X=0)sumisfoundas10+16+9=35,i.e.thesumofthedatawheretheindex

(i+j)3=0asXhastheindex=(i+j)3.

Fromthiswe�ndthefollowingsumsofsquaresanddegreesoffreedom,wherethe4

factorsA,B,XandZconstituteaGraeco-Latinsquare:

SSQ(treatments)

=

102+15

2+18

2+82+::+11

2�1042=9

=

138.222

,

f=9-1

SSQ(A)

=

(232+36

2+45

2)=3�1042=9

=

81.556

,

f=2

SSQ(B)

=

(252+36

2+43

2)=3�1042=9

=

54.889

,

f=2

SSQ(X)

=

(352+34

2+35

2)=3�1042=9

=

0.222

,

f=2

SSQ(Z)

=

(332+36

2+35

2)=3�1042=9

=

1.556

,

f=2

SSQ(A)+SSQ(B)+SSQ(X)+SSQ(Z)

=

138.223

,

f=8

c hs.


51

Itisseen(exceptfortherounding)thatfortheinteractionABandcorrespondingdegrees

offreedomwehave:

SSQ(AB-interaction)=SSQ(X)+SSQ(Z),andf(AB-interaction)=f(X)+f(Z)

Further,itcanbegenerallyshownthatfortheinteractiontermitappliesthat

ABi;j=Xi+j+Zi+2j

;

i=(0;1;2);j=(0;1;2)

Thiscanbeillustratedby�ndingtheestimatesfortheinteractiontermsaswellasfor

thearti�ciale�ectsXandZ.Asanexamplewecan�ndtheinteractionestimatefor

(A=1,B=2),i.e.AB1;2.

^�=104=9=11:556

;

b A 1=36=3�104=9=0:444

;

b B 2=25=3�104=9=�3:222

=)d AB1;2=Y1;2�^��b A 1�b B 2=9:000�11:556�0:444�(�3:222)=0:222

c X 1+2=c X 3!c X 0=35=3�104=9=0:111

b Z 1+2�2=

b Z 5!b Z 2=35=3�104=9=0:111

sothatc X 1+2+b Z 1+2�2=d AB1;2,aspostulated(rememberthatindicesarestillcalculated

"modulo3").TrytoworkoutwhetheritiscorrectthatAB2;2

=X2+2+Z2+2�2,when

theseareestimated.

Inordertousetheresultsoftheexamplegenerally,itispracticaltointroducesomemore

mnemonicnamesforthetwointroducede�ectsXandZ.Wethusset

Xi+j=ABi+j

andZi+2j=AB

2 i+2j

Correspondingly,wewritetheoriginalmodelontheform

Yij�=�+Ai+Bj+ABi+j+AB

2 i+2j+Eij�

where

ABi+j+AB

2 i+2j=ABi;j

;

i=(0;1;2);j=(0;1;2)

Itappliesthatwiththisnewformalnotation:

c hs.


52

2 X i=0

Ai=

2 X j=0

Bj=

2 X r=0

ABr=

2 X s=0

AB

2 s=0

;

r=(i+j)3

;s=(i+2j) 3

whereallindicesarestillcalculated\modulo3".

Thetwoe�ectsABi+jandAB

2 i+2jinthiswaydesignatethearti�ciallyintroducede�ects,

whichenableadecompositionoftheusualinteractiontermABi;j

fromthetraditional

modelformulation.Theexponent"2"onAB

2

shouldonlybeconsideredasamnemonic

helpandnotasanexpressionofraisingtoapowerof2.

Finallynotehowtheindicesandnamesfortheintroducedarti�ciale�ectsmatch.

Endofexample3.1

Sucharti�ciale�ectscanbede�nedforgeneralpk

factorialexperiments.Tobeableto

keepthee�ectsinorder,weintroducea"standardorder"fore�ects.Foranexperiment

whereallfactorshaveplevels,thearti�ciale�ectswilllikewiseallhaveplevels:

2k:I;A;B;AB;C;AC;BC;ABC;D;AD;BD;ABD;CD;ACD;BCD;ABCD;E;AE;::

3k:I;A;B;AB;AB2;C;AC;AC2;BC;BC2;ABC;ABC2;AB2C;AB2C2;D;AD;AD2,

BD;BD2;:::;AB2C2D2;E;:::

5k:I;A;B;AB;AB2;AB3;AB4;C;AC;AC2;:::;BC;:::;AB4C;:::;AB4C4;D;:::

Thesee�ectshaveindicesaccordingtothesamerulesthatwereusedintheprevious

example.Thatis,forexample,thatinthe5kexperimentwithfactorsA,BandC,each

with5levels,thee�ectAB

3Chasindex=(i+3j+k) 5,i.e.(i+3j+k)modulo5.

FactorAisthe�rstfactor,BthesecondfactorandCthethirdfactor.

Notethatthisstandardordercanbederivedfromthestandardorderforsingleexperi-

mentsbychangingtoupper-caselettersandleavingoutthetermswheretheexponent

onthe�rstfactorinthee�ectisgreaterthan1.Forexample,AB

3Cshouldbeincluded,

whileforexampleB

2CDshouldbeleftout.

Example3.2:

Latincubesin33experiments

Lettherebeacompletelyrandomised33experimentwithrrepetitionsofeachsingle

experiment.Wehaveintheusualmodelformulation:


wherei=(0;1;2);j=(0;1;2);k=(0;1;2)and�=(1;2;:::;r)

Thecellsoftheexperimentorsingleexperimentsmakeupacube,thelengthofitsedge

being3.

c hs.


53

Itthuslookslikethis:

A

A

A

0

1

2

0

1

2

0

1

2

B=0

(1)

a

a2

c

ac

a2c

c2

ac2

a2c2

B=1

b

ab

a2b

bc

abc

a2bc

bc2

abc2

a2bc2

B=2

b2

ab2

a2b2

b2c

ab2c

a2b2c

b2c2

ab2c2

a2b2c2

C=0

C=1

C=2

Withthestatedarithmeticrulesforindicesusedonthestandardorderfortheintroduced

arti�ciale�ectsfora33factorialexperiment,wecannow�ndtheindexvaluesforallthe

e�ects.Theindexvaluesarefoundasshowninthefollowingtables:

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

0

1

2

0

1

2

indexfor

j=1

0

1

2

0

1

2

0

1

2

Ai

j=2

0

1

2

0

1

2

0

1

2

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

0

0

0

0

0

0

0

0

indexfor

j=1

1

1

1

1

1

1

1

1

1

Bj

j=2

2

2

2

2

2

2

2

2

2

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

0

1

2

0

1

2

indexfor

j=1

1

2

0

1

2

0

1

2

0

ABi+j

j=2

2

0

1

2

0

1

2

0

1

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

0

1

2

0

1

2

indexfor

j=1

2

0

1

2

0

1

2

0

1

AB2 i+2j

j=2

1

2

0

1

2

0

1

2

0

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

0

0

1

1

1

2

2

2

indexfor

j=1

0

0

0

1

1

1

2

2

2

Ck

j=2

0

0

0

1

1

1

2

2

2

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

1

2

0

2

0

1

indexfor

j=1

0

1

2

1

2

0

2

0

1

ACi+k

j=2

0

1

2

1

2

0

2

0

1

k=0

k=1

k=2

c hs.


54

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

2

0

1

1

2

0

indexfor

j=1

0

1

2

2

0

1

1

2

0

AC2 i+2k

j=2

0

1

2

2

0

1

1

2

0

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

0

0

1

1

1

2

2

2

indexfor

j=1

1

1

1

2

2

2

0

0

0

BCj+k

j=2

2

2

2

0

0

0

1

1

1

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

0

0

2

2

2

1

1

1

indexfor

j=1

1

1

1

0

0

0

2

2

2

BC2 j+2k

j=2

2

2

2

1

1

1

0

0

0

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

1

2

0

2

0

1

indexfor

j=1

1

2

0

2

0

1

0

1

2

ABCi+j+k

j=2

2

0

1

0

1

2

1

2

0

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

2

0

1

1

2

0

indexfor

j=1

1

2

0

0

1

2

2

0

1

ABC2 i+j+2k

j=2

2

0

1

1

2

0

0

1

2

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

1

2

0

2

0

1

indexfor

j=1

2

0

1

0

1

2

1

2

0

AB2Ci+2j+k

j=2

1

2

0

2

0

1

0

1

2

k=0

k=1

k=2

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

0

1

2

2

0

1

1

2

0

indexfor

j=1

2

0

1

1

2

0

0

1

2

AB2C2 i+2j+2k

j=2

1

2

0

0

1

2

2

0

1

k=0

k=1

k=2

Forexample,wecanlookatthetermBCj+k

andnotewhereithastheindexvalue1.

Thisisstatedbelow:

c hs.


55

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

1

1

1

indexfor

j=1

1

1

1

BCj+k

j=2

1

1

1

k=0

k=1

k=2

Foranyotherarbitrarilychosenterm,therewillbeanequalnumberof0's,1'sand2'sin

these9places.Asanexample,wetakethetermABC

2,wherethecorrespondingplaces

areshownbelow:

i=0

i=1

i=2

i=0

i=1

i=2

i=0

i=1

i=2

j=0

2

0

1

indexfor

j=1

1

2

0

ABC2 i+j+2k

j=2

0

1

2

k=0

k=1

k=2

Wehavethusconstructed13e�ects,eachwith3levels(0,1,and2),whichareinbalance

witheachotherinthesamewayaswiththeGraeco-Latinsquareintheexamplementioned

earlier.

Itcanbederivedfromthisthatwecanre-writeouroriginalmodelwiththehelpofthe

newarti�ciale�ects:

Yijk�=�+Ai+Bj+ABi+j+AB

2 i+2j+Ck+ACi+k+AC

2 i+2k+BCj+k

+BC

2 j+2k+ABCi+j+k+ABC

2 i+j+2k+AB

2Ci+2j+k+AB

2C

2 i+2j+2k+Eijk�

wherethetermsinthemodelaredecompositionsoftheconventionalmodelterms:

Ai

)

Ai

Bj

)

Bj

ABi;j

)

ABi+j+AB

2 i+2j

Ck

)

Ck

ACi;k

)

ACi+k+AC

2 i+2k

BCj;k

)

BCj+k+BC

2 j+2k

ABCi;j;k

)

ABCi+j+k+ABC

2 i+j+2k+AB

2Ci+2j+k+AB

2C

2 i+2j+2k

withtheusualmeaningtotheleftandthearti�ciale�ectstotheright.

Inthe33experiment,thereare27cellsorsingleexperiments.Todescribethemeanvalues

inthesecells,27parametersshouldbeused,ofwhichoneis�,sothatthereshouldbe

27�1=26degreesoffreedomforfactore�ect.

c hs.


56

The13termsinthestandardorderallhave3levels,whichsumupto0.Thus3�1=2

freeparameters(degreesoffreedom)areconnectedtoeachofthe13terms,oratotalof

13�(3�1)=26freeparameters(degreesoffreedom).

Itisfurtherseenthat,becauseofthebalance,allparametersareestimatedbyforming

theaverageinthesamewayasforthemaine�ectsandcorrectingwiththetotalaverage.

Forexample:

d AC2 0

=

133�

1X i

X j

X k

� Yijk:�Æ i+2k;0�� Y::::;

where

Æ r;s=

( 1

forr=s

0

forr6=s

astheindicatorÆ i+2k;0

pointsoutthedatawheretheindexforAC

2

is0(zero),i.e.

(i+2k) 3=0,while� Yijk:givestheaverageresponseincells(i;j;k),and� Y::::givesthe

averageresponseforthewholeexperiment.

Thus,inordertoestimateAC

2 0

thecellswherethecorrespondingindex,namely,i+2k

modulo3is0(zero)areincluded.Correspondingly,i+2khastobe1,respectively2to

gointotheestimatesforAC

2 1

andAC

2 2,respectively:

d AC2 0

=1 9(� Y000:+� Y010:+� Y020:+� Y101:+� Y111:+� Y121:+� Y202:+� Y212:+� Y222:)�� Y::::

d AC2 1

=1 9(� Y100:+� Y110:+� Y120:+� Y201:+� Y211:+� Y221:+� Y002:+� Y012:+� Y022:)�� Y::::

d AC2 2

=1 9(� Y200:+� Y210:+� Y220:+� Y001:+� Y011:+� Y021:+� Y102:+� Y112:+� Y122:)�� Y::::

Endofexample3.2

3.2

CalculationsbasedonKempthorne'smethod

Wehaveseenwithexamplesthattheintroducednewe�ects/parameters,whichobviously

donotreferdirectlyforexampletocertaintreatments(apartfromthemaine�ects),give

risetoamathematicaldecompositionoftheinteractions.Thisprocedureandthe

methodsderivedfromitaregenerallycalled"Kempthorne's"method,afterthenameof

thestatisticiantowhomitsoriginisoftenascribed,andwhohasdescribedit(cf.thelist

ofliteraturesuggestionsatthebeginningofthesenotes).

c hs.


57

Wehaveillustratedthatthecorrespondingestimatesareindependentbecauseofthe

describedbalance,and�nallywesawinexample3.1,page50,thatwecanalsoform

sumsofsquare,whichareindependentandsumtothetotalsumofsquares.

Wenowconsideranarbitrarilychosene�ectinthestandardorder.Wegenerallycallthis

e�ectF:

Ft=A

�B

�:::C

t

wheretheindexis

t=i��+j��+:::+k� ;

modulop

Withthenotationintroduced,whereYij:::k�

givestheresponseinthesingleexperiment

no.�withthefactorcombination(ij:::k),wehavethat

Tij:::k=a

i bj:::c

k=

r X �=1

Yij:::k�

andestimatesare:

b F l=P ij

:::kTij:::k�Æ l;t

N=p

�P ij

:::kTij:::k

N

forl=(0;1;:::;p�1);wheret=(i��+j��+:::+k� ) pandN=r�p

k

SSQ(F)=

P p�1

l=0

� Pij:::kTij:::k�Æ l;t

� 2

N=p

�� P

ij:::kTij:::k

� 2

N

=(N=p)�p

�

1 X l=0

b F2 l

wherewestillusetheindicator Æ r

;s=

( 1

forr=s

0

forr6=s

;

Thee�ectestimatecanbeexpressedinwords

b F l=sumofdata,wheret=l

numberofdata,wheret=l�averageafalldata;l=(0;1;:::;p�1)

Example3.3:

EstimationandSSQ

inthe3

2-factorialexperiment

c hs.


58

Ifweagainconsiderthe32experimentinexample3.1page50,whereweletTijgivethe

sumofdataincells(i;j)forexample,we�nd:

b A i=

Ti0+Ti1+Ti2

r�32�

1

�T::

r�32

SSQ(A)=r�3

2�

1

2 X i=0

b A2 i

b B j=

T0j+T1j+T2j

r�32�

1

�T::

r�32

d AB0=

T00+T21+T12

r�32�

1

�T::

r�32

d AB1=

T10+T01+T22

r�32�

1

�T::

r�32

d AB2=

T20+T11+T02

r�32�

1

�T::

r�32

SSQ(AB)=r�3((d AB0)2+(d AB1)2+(d AB2)2)

d AB20=

T00+T11+T22

r�32�

1

�T::

r�32

d AB21=

T10+T21+T02

r�32�

1

�T::

r�32

d AB22=

T20+T01+T12

r�32�

1

�T::

r�32

SSQ(AB

2)=r�3((d AB20)2+(d AB21)2+(d AB22)2)

wheretheinnermost2exponentissymbolic-mnemonic,whiletheoutermosthereisthe

usuallysquaring.

Endofexample3.3

c hs.


59

3.3

Generalformulationofinteractionsandarti�ciale�ects

ConsiderapkfactorialexperimentwithfactorsA,B,...,C,andpisaprimenumber.

Theinteractionbetweenanyfactorsisdecomposedaswehaveseeninthepreviousex-

amples:

ABi;j=ABi+j+AB

2 i+2j+:::+AB

p�

1

i+(p�

1)j

Ageneralnotationcanbeintroducedfortheinteractione�ectsinafactorstructureby

introducinganoperator"�"inthefollowingway

A�B=AB+AB

2+:::+AB

p�

1

andA�I=A.Laterwewillneedthefurtherarithmeticrulethat

(A+B)�=A

�+B

�

sothatforexample

(A�B)�=(AB+AB

2+:::+AB

p�

1)�=(AB)�+(AB

2)�+:::+(AB

p�

1)�

Inadditionanevenmoregeneraloperator"�"canbeintroduced,workinginthefollowing

way:

A�B=A+B+A�B

Inthiswaytheoperator"�"generatesallthetermsinthestandardorderforthefactors

onwhichitworks.

Foranycompletepkfactorialexperiment,thefactormodelcanthenbewritten

Y=�+A�B�:::�C+E

=�+A+B+A�B+:::+C+A�C+B�C+:::+A�B�:::�C+E

Ifp=3,thedecompositionisaspreviously.Forexample

A�B

=

AB+AB

2

A�B�C

=

(AB+AB

2)�C=ABC+ABC

2+AB

2C+AB

2C

2

c hs.


60

SupposeonebeginswithBasthe�rstfactor.Inthe32casethiswouldgivethedecom-

position

BAj;i=B�A=BAj+i+BA

2 j+2i

butthenitemergesthattheindicesforthissetofarti�ciale�ectsvarysynchronously

withtheindicesforthee�ectsABi+jandAB

2 i+2j.

Wecanillustratethiswiththefollowing

Example3.4:

Indexvariationwithinversionofthefactororder

Takea32experimentandconsiderthefollowingtable:

A

B

AB

AB

2

BA

BA

2

i

j

i+j

i+2j

j+i

j+2i

0

0

0

0

0

0

1

0

1

1

1

2

2

0

2

2

2

1

0

1

1

2

1

1

1

1

2

0

2

0

2

1

0

1

0

2

0

2

2

1

2

2

1

2

0

2

0

1

2

2

1

0

1

0

NotethatfortheindicesofthetwotermsAB

2

andBA

2,itappliesthat

(i+2j) 3=0()(j+2i) 3=(2i+j)3=0

(i+2j) 3=1()(j+2i) 3=(2i+j)3=2

(i+2j) 3=2()(j+2i) 3=(2i+j)3=1

aswestillcalculate"modulo3".

ThismeansthattheindicesforAB

2

andBA

2

varysynchronouslysothatAB

2 0

�BA

2 0,

AB

2 1

�BA

2 2,andAB

2 2

�BA

2 1orsaidinanotherway:Inordertoextractthepropersum

ofsquaresweonlyneedoneofthem,ofwhichwehavechosenAB

2.

Endofexample3.4

Theexampleillustratestherulethatweshouldonlyincludee�ectswheretheexponent

onthe�rstfactorinthee�ectequals1.Incertainsituations,however,onecancometo

c hs.


61

e�ectsthatdonotful�lthiscondition.Butfortunatelyitiseasyto�ndthee�ectfrom

thestandardorderwithwhichitcanbereplaced.

3.4

Standardisationofgenerale�ects

Weconsiderageneralnon-standardisede�ect.Asexamplewecantakeane�ectsuchas

CA

2B

4D

3froma3kfactorialexperiment.To�ndthee�ectfromthestandardorderwith

anindexvariationthatvariessynchronouslywiththis,oneproceedsasfollows:

1.ArrangethefactorsinthefactororderA,B,C,D,...etc

CA

2B

4D

3�!A

2B

4CD

3

2.Reduceallexponentsmodulop

A2B

4CD

3�!A

2BCD

0�!A

2BC(p=3her)

3.Iftheexponentonthe�rstfactorinthee�ect(hereA)is1,weare�nished.Oth-

erwise,thewholee�ectisliftedtothesecondpowerandtheexponentsareagain

reducedmodulop:

A2BC�!A

4B

2C

2�!AB

2C

2

4.Step3isrepeateduntilthe�rstfactorinthee�ecthastheexponent1.

Withthehelpofthisalgorithm,onecanalwaysmaketheexponentonthe�rstfactor

inaKempthornee�ectbe1,andbyusinga�xedorderoffactors,onegetsanunequiv-

ocalstandardorder.Forexampleweseethattheindexforthee�ectCA

2B

4D

3

varies

synchronouslywiththeindexforthee�ectAB

2C

2

-calculatedmodulo3.

Example3.5:

Generalisedinteractionsandstandardisation

Ifwehavetwoe�ectsina35experiment,forexampleABC

2

andABDE,we�ndtheir

generalisedinteractionas(wherep=3)

ABC

2�ABDE=ABC

2(ABDE)+ABC

2(ABDE)2=ABCD

2E

2+CDE

wherewehavealsousedtheabove-mentionedsquaringmethodtogettheexponent1on

the�rstfactorinthee�ect.

Inthisconnectionitcanbeusefultorememberthatinasquareexperiment,onefactor

canbemoved"outofthesquare"andouttotheedge,wherebytheedgeinquestion

"movesintothesquare".Fromthe32experimentdealtwithinexample3.1page50:

c hs.


62

i!

0

1

2

0

1

2

0

1

2

0

1

2

j

0

0

1

2

0

0

0

0

1

2

0

1

2

#

1

0

1

2

1

1

1

1

2

0

2

0

1

2

0

1

2

2

2

2

2

0

1

1

2

0

Ai

Bj

ABi+j

AB

2 i+2j

Themaine�ectsAiandBjconstitutetheedgesinthesquareandABi+jandAB

2 i+2jare

"insidethesquare".IfwenowmoveABi+jandAB

2 i+2jouttothesides,weseethatAi

andBjmove"intothesquare":

i+j!

0

1

2

0

1

2

0

1

2

0

1

2

i+2j

0

0

1

2

0

0

0

0

2

1

0

2

1

#

1

0

1

2

1

1

1

2

1

0

1

0

2

2

0

1

2

2

2

2

1

0

2

2

1

0

ABi+j

AB

2 i+2j

Ai

Bj

Examples:(i+j=1andi+2j=2))(i=0andj=1),(i+j=2andi+2j=0))

(i=1andj=1).

AB�AB

2=AB(AB

2)+AB(AB

2)2=A

2B

3+A

3B

5=A

2+B

2=A+B

Therefore,byusingtherulesabove,wecouldhaveforeseenthatAandB

wouldcome

intothesquareasgeneralisedinteractionsforthearti�ciale�ectsABandAB

2.

Thefoure�ectsA,B,ABandAB

2togetherformtheelementsinwhatiscalledagroup.

Inbrief,itdistinguishesitselfbythefactthatwiththeintroducedarithmeticruleswe

cancreatenewelementsfromotherelements,andallelementscreatedwillbelongtothe

group.

Endofexample3.5

Example3.6:Latinsquaresin23factorialexperimentsandYates'algorithm

Inchapter2wewentthroughthe2kexperiment,whilehere-exempli�edwiththe3k

experiment-wehaveintroducedmoregeneralpkexperiments.Wewillnowshowbrie y

howtheintroducedmoregeneralmethodslookina2kexperiment.

Withthreefactors,A,B,andC,themathematicalmodelintheintroducedformulation,

withp=2,is:

Yijk�=�+Ai+Bj+ABi+j+Ck+ACi+k+BCj+k+ABCi+j+k+Eijk�

c hs.


63

wherei;j;k=(0;1)and�=(1;::;r).

Theusualrestrictionsare:

1 X i=0

Ai=

1 X j=0

Bj=

1 Xi+j=0

ABi+j=

1 X k=0

Ck=

1 Xi+k=0

ACi+k=

1 Xj+k=0

B

Cj+k=

1 Xi+j+k=0

AB

Ci+j+k=

0

Theconnectionbetweenthetraditionalmodelformulationandtheformulationintroduced

hereis(asalsoshownonpage56forthe3kexperiment)that(withtheusualformulation

totheleftandthenewformulationtotheright):

Ai

)

Ai

Bj

)

Bj

ABi;j

)

ABi+j

Ck

)

Ck

ACi;k

)

ACi+k

BCj;k

)

BCj+k

ABCi;j;k

)

ABCi+j+k

IfdataareanalysedwiththehelpofYates'algorithm,onemustensurethatthee�ect

estimatesgetthecorrectsign.Yates'algorithmalwaysgivesestimatescorrespondingto

thelevelwhereallfactorsintheparameterareonlevel"1".FortheABCinteraction,

Yates'algorithmgivesthat

dABC1;1;1=[ABCkontrast]=(2

k�r)

Ifthedataareanalysedaccordingtotheintroducedmodel,itisfoundthat

dABC1;1;1)dABC1+1+1!dABC3!dABC1

thatis,thatthealgorithm�ndstheABCi+j+k-parameterlevel"1".

Ifontheotherhand,theinteractionABisconsidered,Yates'algorithm�nds

d AB1;1)d AB1+1!d AB2!d AB0=�d AB1

thatisplusABi+jparameterlevel"0"orminusitslevel"1".

ItthusgenerallyappliesthatYates'algorithm

usedfora2kfactorialexperimentfor

theintroducedgenerale�ectswithanunevennumberoffactorsgivestheparameters'

c hs.


64

level"1",whilethealgorithmforparameterswithanevennumberoffactorsgivesthe

parameters'level"0".

Endofexample3.6

3.5

Block-confoundedpk

factorialexperiment

Inthissectionwewillgeneralisethemethodsthatwereintroducedinsection2.2page20,

andwestartwiththefollowing

Example3.7:

23factorialexperimentin2blocksof4singleexperiments

Weconsidera23factorialexperimentwithfactorsA,BandCandwithrrepetitionsper

factorcombination.

Thetraditionalmathematicalmodelforthisexperimentis


wherei;j;k=(0;1)and�=(1;::;r)

Wehavepreviouslyseenthatsuchanexperimentcanbelaidoutintwoblocksbychoosing

toconfoundoneofthefactore�ectswithblocks,andwehaveseenthatthisisformalised

bychoosingade�ningcontrast.Thee�ectcorrespondingtothiswillbeconfoundedwith

blocks.Inordertousetheintroducedmethodforanalysisofp

kfactorialexperiments,we

willwritethemodelonthegeneralform,whichforp=2is:


wherei;j;k;=(0;1)and�=(1;:::;r)

Todividetheexperimentintotwoparts,wenowchoosea

De�ningrelation:I=ABC

where,asanexample,wechoosetoconfoundthe3-factorinteractionABCwithblocks.

Thise�ecthasindex=(i+j+k) 2,whichthustakesthevalues0or1.Welettheblock

numberfollowthisindex,i.e.thatinBlock0areplacedtheexperimentswhereitapplies

that(i+j+k) 2=0.Correspondingly,experimentswhere(i+j+k) 2=1areputin

block1.To�ndtheprincipalblock,wemustinotherwords�ndallthesolutionstothe

equation:

c hs.


65

(i+j+k)(modulo2)

=0

Wetry:

i=0;j=0=)k=0:experiment=(1)

i=1;j=0=)k=1:experiment=ac

i=0;j=1=)k=1:experiment=bc

i=1;j=1=)k=0:experiment=ab

Thelastsolutioncouldbefoundbyaddingthetwoprevioussolutionstoeachother:

i

j

k

1

+

0

+

1

=

2

!

0

ac

0

+

1

+

1

=

2

!

0

bc

1

+

1

+

2!0

=

2

!

0

ac�bc=ab

Onenotesthatthisindexadditioncorrespondsto"multiplying"thetwosolutionsacand

bcbyeachother.

Theotherblockisconstructedby�ndingthesolutionsto(i+j+k) 2=1.Thesesolutions

are(a;b;c;abc).

Inthiswaytheblockingisfound:

Block0

Block1

(1)

ab

ac

bc

a

b

c

abc

Wenotethatthissolutionisexactlythesameastheonewefoundinsection2.2using

forexamplethetabularmethod.

Endofexample3.7

WehavenowseenasimpleexampleoftheuseofKempthorne'smethodtomakeblock

experiments.Theprincipleisstillthatwelettheblockvariablevarysynchronouslywith

thelevelsforthefactore�ectthatwewillconfoundwithblocks.

Example3.8:

32factorialexperimentin3blocks

Supposewehavetheexperiment

A=0

A=1

A=2

B=0

(1)

a

a2

B=1

b

ab

a2b

B=2

b2

ab2

a2b2

c hs.


66

whereweagainwritethemodelonthegeneralform,whichforp=3is:

Yij�=�+Ai+Bj+ABi+j+AB

2 i+2j+Eij�

Wenowwanttocarryouttheexperimentbein3blocks,eachwith3singleexperiments.

Forthatpurposewecanlettheblockindexfollowtheindexforthearti�ciale�ectABi+j,

wherebyitisstillpossibletoestimatethetwomaine�ectsAandB:

De�ningrelation:I=AB

Index=i+j:

i=0

i=1

i=2

j=0

0

1

2

j=1

1

2

0

j=2

2

0

1

Block0

Block1

Block2

(1)

ab2

a2b

a

b

a2b2

a2

b2

ab

Block0isgivenwithallsolutionstotheequation(i+j)3=0.Theothertwoblocksare

givenwith(i+j)3

=1and(i+j)3

=2respectively.

Thedesigncouldhavebeencomputeddirectlyusingthefollowingtabularmethod:

i

j

code

Block=(i+j)3

0

0

(1)

0

1

0

a

1

2

0

a2

2

0

1

b

1

1

1

ab

2

2

1

a2b

0

0

2

b2

2

1

2

ab2

0

2

2

a2b2

1

Ifweonlywantedtheprincipalblockwecanusethemethodshownintheprevious

example,whichconsistsofsolvingtheequation:

(i+j)modulo3=0

i=0

)

j=0

Experiment:(1)

i=1

)

j=�1!�1+3

=2

Experiment:ab2

i=2

)

j=�2!�2+3

=1

Experiment:a

2b

c hs.


67

Itisseenthattheprincipalblockhastheappearance:

Block0

(1)

x

x2

wherex=ab2)x

2=(ab2)2=a

2b4=a

2b

wherexcanrepresentanychosensolutionto(i+j)3

=0except(i;j)=(0;0).

Theothertwoblocksarecanbefoundbytheabovetabularmethodor,equivalently,by

solvingtheequations(i+j)3=1and(i+j)3

=2respectively.Forexample,theterma

isthesolutionto(i+j)3

=1inthati=1,andj=0correspondtoa.

Theblockthatcontainstheexperimentacanbeconstructedby"multiplying"aonthe

principalblockfound:

Block0

Block1

a�(1)

ab2

a2b

=)

a

a2b2

b

principalblock

Thelastblockisconstructedby�ndingasolutiontotheequation(i+j)3=2,forexample

a2

andmultiplyingthisontheprincipalblock:

Block0

Block

a2�(1)

ab2

a2b

=)

a2

b2

ab

principalblock

Itiseasytoshowthatwiththisblocking,theonlye�ectinourmodelthatisconfounded

withtheblocke�ectispreciselytheABi+je�ect.

Ifwewantedanalternativeblockgrouping,wheretheAB

2

e�ectwasconfoundedwith

blocks,wewouldusethede�ningrelationI=AB

2anddeterminetheprincipalblockby

solvingtheindexequation(i+2j)=0.Onesolutionisab,andtheprincipalblockis

therefore(1),ab,(ab)

2

=

(1),ab,a

2b2.Afterthisone�ndstheblocking

Block0

Block1

Block2

(1)

ab

a2b2

a

a2b

b2

a2

b

ab2

principalblock

(i+2j) 3=0

Tryityourself!

Endofexample3.8

c hs.


68

Wehaveseenhowapkfactorialexperimentcanbedividedintopblockssothatane�ect

choseninadvanceisconfoundedwithblocks.

Wecangeneralisethismethodtoadivisionintopqblocks,whereq<k.Todothis,we

startbydividinga23experimentinto2�2=22=4blocks.

Example3.9:

Divisionofa23factorialexperimentinto22blocks

Lettherebea23factorialexperimentwithfactorsA,BandC.Withtheintroduced

formulationthemodelis,inthatp=2:


whereallindicesi;j;k=(0;1)and�=(1;::;r)

Tode�ne4(=2�2)blocks,weuse2de�ningrelations,forexample

I 1=AB

and

I 2=AC

aspreviouslyshownonpage27.

Thestructureofthe4blockscanbeillustrated

I 1=AB

i+j=0

i+j=1

I 2=AC

i+k=0

Block(0,0)

Block(1,0)

i+k=1

Block(0,1)

Block(1,1)

IftheindexforbothABi+jandACi+kis0forexample,thesingleexperimentsareplaced

inblock(0,0).

Inthisway,orbyusingthetabularmethod,one�ndstheblocking:

I 1=AB

i+j=0

i+j=1

I 2=AC

i+k=0

(1)

abc

b

ac

i+k=1

ab

c

a

bc

Iftheblocke�ectsaremodelledasa2�2design,wecanwritethattheblockscontribute

with

Blocks=�+Ff+Gg+FGf+g

;

wheref=(i+j)2

andg=(i+k) 2

c hs.


69

Itisclearthatthee�ectABvariessynchronouslywithFandthatthetwoe�ectsare

confounded.Correspondingly,ACisconfoundedwithG.Thatpartoftheblockvariation,

whichisherecalledFG,hastheindex(f+g) 2=((i+j)+(i+k))2=(j+k) 2,whichis

preciselytheindexforthetermBCinthemodelfortheresponseoftheexperiment.

Thereforeitcanbeconcludedthatthee�ectBCwillalsobeconfoundedwithblocks,

whichcanalsobeseenfromthefollowingtable,wheretheindexoftheBCe�ectis0on

onediagonaland1ontheotherone:

I 1=AB

i+j=0

i+j=1

I 2=AC

i+k=0

j+k=0

j+k=1

i+k=1

j+k=1

j+k=0

Moreformallywecanwrite:

Blocks=AB+AC+AB�AC=AB+AC+BC

Endofexample3.9

3.6

Generalisationofthedivisionintoblockswithseveralde�n-

ingrelations

LetI 1=A�B�:::C

denoteade�ningrelation,thatdividesapk

factorialexperiment

intopblocks.Further,letI 2=A

aB

b:::C

cdenoteade�ningrelationthatlikewisedivides

thepkexperimentintopblocks.

Inthedivisionoftheexperimentintop�pblocksonthebasisofthesede�ningrelations,

bothe�ects

I 1=A�

B�...C

andI 2=AaBb...Cc

willbeconfoundedwithblocks.Inaddition,theirgeneralisedinteractionwillbecon-

foundedwithblockssothatbesidesI 1andI 2thee�ectsgivenintheexpression:

I 1�I 2=(A

�

B�...C

)�(A

aB

b...C

c)

willbeconfoundedwithblocks.Alltermsintheexpression:

I 1�I 2=I 1+I 2+I 1�I 2=I 1+I 2+I 1I 2+I 1(I2)2+...++I 1(I2)(p�

1)

c hs.


70

areconfoundedwithblocks.

Wecangenerallywriteuptheconfoundingsforanydivisionofap

kfactorialexperiment

inpqblocks.

Ifwehavethecorrespondingde�ningrelationsgivenbyI 1,I 2,I 3,..,I q,allthee�ectsin

theequation

I 1�I 2�I 3�...�I q=I 1+I 2+I 1�I 2+I 3+...+I 1�I 2�I 3�...�I q

willbeconfoundedwithblocks.Theoperators"�"and"�"workasstatedinsection3.3

page60

Example3.10:

Dividinga33factorialexperimentinto9blocks

Lettherebea33factorialexperimentwithfactorsA,BandC.

Asanexamplewedividetheexperimentinto3�3blocksusing

I 1=ABC

2

andI 2=AC

Thereby,ABC

2

andACtogetherwiththeirgeneralisedinteractionareconfoundedwith

blocks,thatis,allthee�ectsintheexpression(wherep=3):

ABC

2

�AC=ABC

2

+AC+ABC

2

�AC

=ABC

2

+AC+ABC

2(AC)+ABC

2(AC)2=ABC

2

+AC+AB

2

+BC

Intheanalysisofvariancetable,thesumsofsquaresforABC

2,AC,AB

2andBCthere-

forealsocontainpossibleblocke�ectsandthustheycannotbeinterpretedasexpressing

factore�ectsalone.

Theprincipalblockinthisexperimentisfoundbysolvingtheequations(p=3):

(i+j+2k) 3=0and(i+k) 3=0

One�ndsforexamplei=1)k=2)j=1,thatisabc

2.Theprincipalblockcontains

33=32=3singleexperiments.Thismeansthatitsatis�esto�ndonesolutioninorder

todeterminetheblock.Ifthissolutioniscalled"x",theprincipalblockexperimentsare

(1),xandx

2.

c hs.


71

Inourcase,wethenforx=abc

2getthethreeexperiments(1),abc

2and(abc

2)2=a

2b2c.

Onecancheckthat(i=2;j=2;c=1)isalsoasolutiontothetwoindexequations.

Theotherblocksarefoundby�ndingthesolutionstotheindexequationsfortheright-

handsidesequalto(0,1,2)inthecaseofbothequations,i.e.atotalof9di�erentcases,

correspondingtothe3�3blocks.

Foraanyoneoftheseblocks,itappliesthattheycanbefoundwhenjustoneexperiment

isfoundintheblock.Bymultiplyingthisexperimentontheprincipalblock,thewhole

blockisdetermined.

Endofexample3.10

Example3.11:

Divisionofa25experimentinto23blocks

LetthefactorsbeA,B,C,DandE,whichallappearon2levels.Todividetheexperiment

into2�2�2blocks,3de�ningrelationsareused,f.ex.

I 1=ABC,I 2=BDEandI 3=ABE

Therebyalle�ectsinthefollowingexpressionareconfoundedwithblocks:

I 1�I 2�I 3=I 1+I 2+I 1�I 2+I 3+I 1�I 3+I 2�I 3+I 1�I 2�I 3

Thatis,inadditiontoABC,BDEandABE,thefollowingterms(sincep=2):

(I1

�I 2)

=

ABCBDE

=

ACDE

(I1

�I 3)

=

ABCABE

=

CE

(I2

�I 3)

=

BDEABE

=

AD

(I1

�I 2�I 3)

=

ABCBDEABE

=

BCD

Thedesigncanbefoundbythetabularmethod(sef.ex.page25).

Ifonlytheprincipalblockiswantedwecansolvetheequations

(i+j+k) 2=0

;

(j+l+m) 2=0

;

(i+j+m) 2=0

Oneblockcontains25=23=22=4singleexperiments.Therefore2solutionshavetobe

found.Ifthesesolutionsarecalledxandy,theprincipalblockis:(1),x,yandxy

Aningeniouswayto�ndthesesolutionsistotrywith(i=1,j=0)and(i=0,j=1),

whichcorrespondtotheelementsaandbinthefactorstructureforfactorsAandB.The

methodworksifthemaine�ectsAandBarenotconfoundedwitheachotherorwith

blocks.

c hs.


72

We�nd (i

=1;j=0)

)

k=1,

m=1,

l=1,

theexperimentisx=acde

(i=0;j=1)

)

k=1,

m=1,

l=0,

theexperimentisy=bce

Theprincipalblockistherefore

Principalblock

Block(0,0,0)

(1)

x

y

xy

=

(1)

acde

bce

abd

Notethatallexperimentsintheprincipalblockhaveanevennumberoflettersincommon

withthe3de�ningcontrasts,ABC,BDEandABE.Theremainingblockscannowbe

foundbymultiplyingwithelementsthatarenotintheprincipalblock.

Fortheblockcorrespondingtotheequations

(i+j+k) 2=1

;

(j+l+m) 2=0

;

(i+j+m) 2=0

thatisblock(1,0,0),thereisasolution:(i;j;k;l;m)=(1;0;0;1;1)=ade(startwith

(i;j)=(1;0),whichistheeasiestmethod).Therestoftheblockisfoundbymultiplying

thissolutionontotheprincipalblock:

Principalblock

Block(1,0,0)

ade�(1)

acde

bce

abd

=)

ade

c

abcd

be

Theremaining6blockscanbefoundbysettingtheright-handsidesoftheequationsto

(0,1,0),(1,1,0),(0,0,1),(1,0,1),(0,1,1)and(1,1,1),respectively.

Endofexample3.11

Example3.12:

Divisionof3kexperimentsinto33blocks

Lettherebea3kfactorialexperimentandsupposethatI 1,I 2andI 3de�neadivisionof

theexperimentinto3�3�3=27blocks.

Theconfoundingistherebygivenwith

I 1�I 2�I 3=I 1+I 2+I 1�

I 2+I 3+I 1�

I 3+I 2�

I 3+I 1�

I 2�

I 3

=I 1+I 2+I 1I 2+I 1I2 2

+I 3+I 1I 3+I 1I2 3

+I 2I 3+I 2I2 3

+I 1(I2

�

I 3)+I 1(I2

�

I 3)2

whereexponentsarereducedmodulo3.Forthetwolasttermswehave:

c hs.


73

I 1(I2

�

I 3)=I 1(I2

I 3+I 2I2 3)=I 1I 2I 3+I 1I 2I2 3

and

I 1(I2

�

I 3)2=I 1(I2

I 3+I 2I2 3)2=I 1I2 2

I2 3

+I 1I2 2

I4 3

=I 1I2 2

I2 3

+I 1I2 2

I 3

Thetermsfoundwillallbeconfoundedwithblocks.Eachofthecorrespondinge�ects

hasprecisely3levels,i.e.thevariationbetweenthese3levelshas2degreesoffreedom.A

totalof13termswith2degreesoffreedomarefound,i.e.atotalof26degreesoffreedom,

whichcorrespondexactlytothevariationbetween27blocks.

Thedesigncanbefoundbythetabularmethod(sef.ex.page67).

Endofexample3.12

3.6.1

Constructionofblocksingeneral

Wehaveseenabovethatinapkfactorialexperiment,onede�ningrelation,I 1,dividesthe

experimentintopblocks,whileqrelations,forexampleI 1,...,I q,dividetheexperiment

intopqblockseachcontainingpreciselypk�

qsingleexperiments.

Thiscorrespondstothefactthatinoneblockarejustasmanysingleexperimentsas

thereareinacompletep

k�

qexperiment,thatis,anexperimentwithk�qfactorseach

onplevels.

We�rstconstructtheprincipalblockamongthesepqblocks,and,onthebasisofthis,

theremainingblockscanbedetermined,asisshownintheexamples.

Themethodwewilluseisinbrief:

1)::Lettherebeqde�ningcontrastsI 1,I 2,...,I q,andagainletallthesingleexperi-

mentsbedesignatedwith(1);a;a

2;:::;ap�

1;b;ab;:::;ap�

1b;b2;ab2;:::;ap�

1bp

�

1,...,

etc.

2):Determinek�qofthesesingleexperiments,whichareintheprincipalblock.For

exampletheycanbe:f1,f2,...,fk�

q.Itisrequiredforthesesingleexperimentsthat

thecorrespondingsolutionstotheindexequationsarelinearlyindependent.

3):Theprincipalblockcannowbeconstructedbyusingthesesingleexperimentsas

"basicexperiments"andmakingacompletepk�

q

factorialexperiment,i.e.the

standardorderforthesingleexperimentsonthebasisoff1,,f2,...,fk�

q:

(1);f1;f1

2;:::;f1

p�

1;f2;f1f2;f1

2f2;:::;f1

p�

1f2

p�

1;:::;f1

p�

1f2

p�

1��fk�

qp�

1

c hs.


74

Thiscollectionofsingleexperimentsthenmakesuptheprincipalblock.Duringthe

formation,allexponentsarereducedmodulop.

4):Changethelevelforoneoftheindexequationsandthereby�ndanewsingleex-

perimentthatisnotintheprincipalblockandmultiplytheexperimentonallthe

experimentsintheprincipalblock.Inthiswayanewblockisformed.

5):Continuewith4)untilallblocksareformed.

Inordertoassurethatallthesingleexperimentsintheprincipalblockaredi�erent,we

mustrequirefortheoriginal(k�q)solutionsthattheyarelinearlyindependent(where

allarestillcalculatedmodulop).

Otherwisetheprincipalblockwillnotbecompletelydetermined,andthesamesingle

experimentswillbefoundseveraltimeswhentryingto�ndtheexperimentsintheblock.

Withthesamekindofargumentation,itcanbeshownhow,onthebasisofoneexperiment

belongingtoanalternativeblock,therestofthatblockcanbeformedbymultiplyingit

ontotheprincipalblock.

Forexampleif,withthehelpofaspreadsheetoracomputerprogram,onewantsto

�ndablockdistribution,thesimplestmethodistorunthroughallsingleexperiments

instandardorderandforeachsingleexperimentcalculatethevalueoftheindicesofthe

de�ningcontrasts,thatistousethetabularmethod.

Example3.13:


Letthenotationbeasusual.Thesingleexperimentsaregivenby:

(1);a;a

2

;b;ab;a

2b;b2;ab2;a

2b2;:::;a

2b2c2d

2

Takeforexamplethede�ningrelations:

I 1=ABi+jandI 2=BCD

2 j+k+2l

Theprincipalblockconsistsoftheexperimentswhereboth(i+j)3=0and(j+k+2l) 3=0.

Inoneblockthereare34

�

2

=32singleexperiments.

Thecompletedesigncanbeconstructedandwrittenoutbymeansofthetabularmethod

(seepage67).

Ifwewanttheprincipalprincipalblock,forexample,wemustjustdeterminetwo"linearly

independent"singleexperimentsandfromthatformtherestasa32experiment.

Thus:Findtwolinearlyindependentsolutionsto:

c hs.


75

i+j=0andj+k+2l=0

Trywithi=0)j=0)k+2l=0andchoose(k=1;l=1),forexample,giving

(i;j;k;l)=(0;0;1;1)asausablesolution.Theexperimentiscd.

Thentryforexamplewithi=1)j=2,(j+k+2l)=0)(k+2l) 3=(�2)3

=

(�2+3)3=1whereweforexamplechoosel=0andk=1.Notethatonecanalways

addanarbitrarymultipleof"3"toa(negative)numberwhenonehasto�nd"modulo

3"ofthenumber.Thatistosaythatgenerally(x) p=(x+kp) pwhere(:) pheredenotes

"(.)modulop".

Thus(i;j;k;l)=(1;2;1;0)isausablecombinationandtheexperimentis"ab2c".

Checktheindependencebyverifyingthatcd(ab2c)�6=(1)forall�(therelevant�'sare1

and2):OK.

Nowcallf1=cdandf2=ab2c.Theprincipalblockthenis

(1)

f1

f2 1

(1)

cd

(cd)2

f2

f1f2

f2 1f2

=

ab2c

cdab2c

(cd)2ab2c

f2 2

f1f

2 2

f2 1f

2 2

(ab2c)

2

cd(ab2c)

2

(cd)2(ab2c)

2

byorderingtheelements,multiplyingoutandreducingallexponentsmodulo3,theblock

isfound:

(1)

cd

c2d

2

ab2c

ab2c2d

ab2d

2

a2bc

2

a2bd

a2bcd

2

To�ndanalternativeblock,welookforasingleexperimentthatisnotintheblock

alreadyfound.Wecanforexampletake"a".

Thenewblockisthen: (1

)

cd

c2d

2

a

acd

ac2d

2

a�

ab2c

ab2c2d

ab2d

2

=)

a2b2c

a2b2c2d

a2b2d

2

a2bc

2

a2bd

a2bcd

2

bc2

bd

bcd

2

orbymultiplyingwithb:

(1)

cd

c2d

2

b

bcd

bc2d

2

b�

ab2c

ab2c2d

ab2d

2

=)

ac

ac2d

ad

2

a2bc

2

a2bd

a2bcd

2

a2b2c2

a2b2d

a2b2cd

2

c hs.


76

Endofexample3.13

Example3.14:


A53experimentconsistsofatotalof125singleexperiments.Withthedivisioninto5

blocks,thereare25singleexperimentsineachblock.

ThefactorsareA,BandC,andasde�ningrelationwechooseforexample

I=ABC

3 i+j+3k

Intheprincipalblock,wherep=5,itappliesthat

i+j+3k=0

(modulo5)

Sincethesizeoftheblockis5�5=52,wehaveto�nd2linearlyindependentsolutions

tothisequation.Forexample,

(i;j;k)=(1;0;3)�ac3and(i;j;k)=(0;1;3)�bc

3

canbeused.Asastart,theprincipalblockisthereby

(1)

ac3

a2c6

a3c9

a4c1

2

bc3

abc

6

a2bc

9

a3bc

12

a4bc

15

b2c6

ab2c9

a2b2c1

2

a3b2c1

5

a4b2c1

8

b3c9

ab3c1

2

a2b3c1

5

a3b3c1

8

a4b3c2

1

b4c1

2

ab4c1

5

a2b4c1

8

a3b4c2

1

a4b4c2

4

andafterreductionoftheexponentsmodulo5,one�nallygets

(1)

ac3

a2c

a3c4

a4c2

bc3

abc

a2bc

4

a3bc

2

a4b

b2c1

ab2c4

a2b2c2

a3b2

a4b2c3

b3c4

ab3c2

a2b3

a3b3c3

a4b3c

b4c2

ab4

a2b4c3

a3b4c1

a4b4c4

Itcanbeinterestingtonotethatthisisa5�5Latinsquare,whichwithCinsidethe

squareis:

c hs.


77

A=0

A=1

A=2

A=3

A=4

B=0

0

3

1

4

2

B=1

3

1

4

2

0

B=2

1

4

2

0

3

B=3

4

2

0

3

1

B=4

2

0

3

1

4

whichforinstanceshowsthatthethreefactorsaremutuallybalancedwithintheblock

found.Thesamewillnaturallyapplywithintheother4blocksintheexperiment.One

oftheseblockscanbeeasilyconstructedforexamplebymultiplyingtheprincipalblock

withanexperimentthatisnotincludedintheprincipalblock.Bymultiplyingwitha,

forexample,we�nd

a

a2c3

a3c

a4c4

c2

abc

3

a2bc

a3bc

4

a4bc

2

b

ab2c1

a2b2c4

a3b2c2

a4b2

b2c3

ab3c4

a2b3c2

a3b3

a4b3c3

b3c

ab4c2

a2b4

a3b4c3

a4b4c1

b4c4

whichisthusalsoaLatinsquare.

Theremainingblockscanbefoundinthesameway,butnaturallyonecanalsoleta

programconstructalltheblocksbycalculatingthevalueoftheindex(i+j+3k)forall

singleexperimentsandplacingtheexperimentsaccordingtowhether(i+j+3k)(modulo

5)is0,1,2,3or4,thatisbythetabularmethod.

Endofexample3.14

3.7

Partialconfounding

Partialconfoundingin2kfactorialexperimentswasintroducedinsection2.3page28.

Wewillgiveanotherexampleofpartialconfoundingina2kexperiment,wherewenow

forthesakeofillustrationuseKempthorne'smethodtoformtherelevantblocks.

Example3.15:

Partiallyconfounded2

3

factorialexperiment

Againweconsideranexperimentwith3factorsA,BandC,eachon2levels.Weassume

thattheexperimentscanonlybedoneinblockswhicheachcontain4singleexperiments.

Tobeabletoestimateallthee�ectsinthemodel

Yijk=�+Ai+Bj+ABi+j+Ci+ACi+k+BCj+k+ABCi+j+k+E

c hs.


78

itisnecessarytodoapartiallyconfoundedfactorialexperiment.

Supposethatinthe�rstexperimentalserieswechoosetoconfoundthethree-factorin-

teractionABC.

Todividetheexperimentinto2blocks,wehaveto�nd2solutionstotheindexequation

sincetheblocksizeis23

�

1=2�2.

Thereforewehaveto�nd2solutionstotheequation(i+j+k) 2=0.

Bytrialanderror,we�ndforexamplex=acandy=bc.

Theprincipalblockisthen

block1

(1)

x

y

xy

=

(1)

ac

bc

ab

(i+j+k) 2=0

Bymultiplyingwitha,wegettheotherblock,whichofcourseconsistsoftheremaining

singleexperimentsinthecomplete23factorialexperiment:

block2

a�(1)

x

y

xy

=

a

c

abc

b

(i+j+k) 2=1

Analysisofthis�rstblock-confoundedexperimentcanbedonewithYates'algorithm,

whichgivesaresultthatcanalsobeexpressedinmatrixformintheusualway:

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I (1)

A(1)

B(1)

AB(1)

C(1)

AC(1)

BC(1)

ABC(1)=blocks

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5=2 6 6 6 6 6 6 6 6 6 6 6 6 6 41

1

1

1

1

1

1

1

�1

1

�1

1

�1

1

�1

1

�1

�1

1

1

�1

�1

1

1

1

�1

�1

1

1

�1

�1

1

�1

�1

�1

�1

1

1

1

1

1

�1

1

�1

�1

1

�1

1

1

1

�1

�1

�1

�1

1

1

�1

1

1

�1

1

�1

�1

13 7 7 7 7 7 7 7 7 7 7 7 7 7 52 6 6 6 6 6 6 6 6 6 6 6 6 6 4(1)

a b ab c a

c bc abc

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

Theindex(:) (1)onthecontrastsreferstothis�rstexperiment.

Wethendoanotherexperiment,butthistimewechoosetoconfoundthee�ectAB.The

blockingoftheexperimentisthus:

block3

block4

(1)

ab

abc

c

and

a

b

ac

bc

(i+j)2=0

(i+j)2=1

c hs.


79

Forthisexperimentwecan�ndcontrastsinthesamewayasinthe�rstexperiment.And

�nallywewillcombinethetwoexperiments.Wehavethefollowingsourcesofvariation:

1)

Factore�ects

whicharenotconfounded

2)

Factore�ects

whicharepartiallyconfounded

3)

Blocke�ects,

i.e.variationbetweenthetotalsofthe4blocks

4)

Residualvariation

Analysisofthetwoexperimentsgivesrespectively:

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I (1)

A(1)

B(1)

AB(1)

C(1)

AC(1)

BC(1)

ABC(1)=blocks

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5and

2 6 6 6 6 6 6 6 6 6 6 6 6 6 4I (2)

A(2)

B(2)

AB(2)=blocks

C(2)

AC(2)

BC(2)

ABC(2)

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

whereindex(:) (2)correspondstothesecondexperiment.

Wecan�ndsumsofsquaresanddegreesoffreedomcorrespondingtothefoursourcesof

variation:

c hs.


80

1)

Unconfoundedfactore�ects

SSQA

=

([A(1)]+[A(2)])

2=(2�2

3),

f=1

SSQB

=

([B(1)]+[B(2)])

2=(2�2

3),

f=1

SSQC

=

([C(1)]+[C(2)])

2=(2�2

3),

f=1

SSQAC

=

([AC(1)]+[AC(2)])

2=(2�2

3),

f=1

SSQBC

=

([BC(1)]+[BC(2)])

2=(2�2

3),

f=1

2)

Partiallyconfounded

factore�ects

SSQAB

(halfprecision)

=

[AB(1)]2=(2

3),

f=1

SSQABC

(halfprecision)

=

[ABC(2)]2=(2

3),

f=1

3)

Blocke�ectsandconfounded

factore�ects

Betweenexperiments

=

([I (1)]�[I(2)])

2=(2�2

3),

f=1

SSQAB+blocks(3-4)

=

[AB(2)]2=(2

3),

f=1

SSQABC+blocks(1-2)

=

[ABC(1)]2=(2

3),

f=1

4)

Residualvariation:

BetweenA-estimates(SSQA,Uncert.)

=

([A(1)]�[A(2)])

2=(2�2

3),

f=1

BetweenB-estimates(SSQB,Uncert.)

=

([B(1)]�[B(2)])

2=(2�2

3),

f=1

BetweenC-estimates(SSQC,Uncert.)

=

([C(1)]�[C(2)])

2=(2�2

3),

f=1

BetweenAC-estimates(SSQAC,Uncert.)

=

([AC(1)]�[AC(2)])

2=(2�2

3),

f=1

BetweenBC-estimates(SSQBC,Uncert.)

=

([BC(1)]�[BC(2)])

2=(2�2

3),

f=1

5)

Totalvariation

=

SSQtotwithdegreesoffreedom

f=15

Generally,thevariationcanbecalculatedbetweenforexampleRA

Aestimatesthatare

allbasedoncontrastsfrom(RA

di�erent)2kexperimentswithrrepetitions(inwhich

theyareallunconfounded)withtheexpression:

SSQA,Uncert.=

[A(1)]2+:::+[A(RA)]2

2k�r

�([A(1)]+:::+[A(RA)])

2

RA

�2k�r

Intheexample,RA

=2,k=3andr=1.Forothernon-confoundedestimates,naturally,

correspondingexpressionsarefound.See,too,page31.

Wehavetherebyaccountedforallthevariationinthetwoexperimentscollectively.Note

thatwehavecalculatedsumsofsquarescorrespondingtoatotalof15sourcesofvariance,

eachwithonedegreeoffreedom.Thiscorrespondspreciselytothetotalvariationbetween

the16singleexperiments,whichgivesriseto(16�1)=15degreesoffreedom.

Iftherearerrepetitionsforeachofthesingleexperiments,alltheSSQ'shavetobedivided

c hs.


81

byr.Inthiscase,onecan,ofcourse,�ndvariationwithineachfactorcombination(a

totalof8+8singleexperimentswith(r�1)degreesoffreedom)andusethem

to

calculateaseparateestimatefortheremaindervariation.Thisestimatecan,ifnecessary,

becomparedwiththementionedestimate,whichwascalculatedabove.

Endofexample3.15

Theexampleshownillustratestheprinciplesforcombiningseveralexperimentswithdif-

ferentconfoundings.Thewholeanalysiscanbesummarisedtothefollowing.Suppose

that,inall,experimentsaremadeinRblocksmednblocksingleexperimentsineachblock.

Thevariationcanthenbedecomposedinthefollowingcontributions,whereTblockigives

thetotalinthei'thblock:

SSQblocks

=

(T2 block1+T

2 block2+:::+T

2 blockR)=nblock�(T

2 tot)=(R�nblock)

=

variationbetweenblocktotals

SSQe�ects

=

SSQforallfactore�ectsbasedonexperimentsinwhich

thee�ectsarenoterconfoundedwithblocks

SSQresid

=

variationbetweene�ectestimatesfromexperiments,where

thee�ectsarenotconfounded

SSQuncertainty

=

variationbetweenrepeatedsingleexperimentswithinblocks

The�rstcontribution,SSQblocksalsocontains,inadditiontothetotalvariationbetween

blocks,thevariationfromconfoundedfactore�ects.

Example3.16:

Partiallyconfounded32factorialexperiment

We�nishthissectionbyshowingtheprinciplesfortheconstructionandanalysisofa

partiallyconfounded32factorialexperiment.Thisexperimentispossiblylittleusedin

practice,butitillustratesthegeneralprocedurewell.Anditshowshowallthemain

e�ectsandinteractionsina3�3experimentcanbedetermined,eventhoughthesizeof

theblockisonly3.

Experiment1:Dividethe32experimentinto3blocksof3accordingtoI=ABi+j

One�nds:

block1

:

(1) (1)

ab2 (1)

a2b (1)

total=T1

block2

:

a(1)

a2b2 (1)

b (1)

total=T2

block3

:

a2 (1)

b2 (1)

ab (1)

total=T3

c hs.


82

Index(1)indicatesthatthisisexperiment1.

Experiment2:NowdivideaccordingtoI=AB

2 i+2j.Thisgivestheblocking:

block4

:

(1) (2)

ab (2)

a2b2 (2)

total=T4

block5

:

a(2)

a2b (2)

b2 (2)

total=T5

block6

:

a2 (2)

b (2)

ab2 (2)

total=T6

Index(2)indicatesthatthisisexperiment2.

Wecan�ndthesumofsquaresbetweenthe6blocks,inthatTtot=T1+T2+:::+T6

:

SSQblocks

=

(T2 1

+T

2 2

+:::+T

2 6)=3�T

2 tot=18

Wethenhave

TA0

=

(1) (1)+b (1)+b2 (1)+(1) (2)+b (2)+b2 (2)

TA1

=

a(1)+ab (1)+ab2 (1)+a(2)+ab (2)+ab2 (2)

TA2

=

a2 (1)+a

2b (1)+a

2b2 (1)+a

2 (2)+a

2b (2)+a

2b2 (2)

Thatis,thatforexampleTA0

=thesumofallthemeasurementswherefactorAhas

beenonlevel"0"intheRA

experimentswheree�ectAisnotconfoundedwithblocks,

andcorrespondinglyforlevels"1"and"2".WithTtot;A

=TA0

+TA1

+TA0

wegetquite

generally:

SSQA

=

(T2 A

0

+T

2 A1

+T

2 A2)=(RA

�3k

�

1)�T

2 tot;A=(RA

�3k)

withf=(3�1)=2degreesoffreedom.InourexampleRA

=2andk=2.

TB0

=

(1) (1)+a(1)+a

2 (1)+(1) (2)+a(2)+a

2 (2)

TB1

=

b (1)+ab (1)+a

2b (1)+b (2)+ab (2)+a

2b (2)

TB2

=

b2 (1)+ab2 (1)+a

2b2 (1)+b2 (2)+a

2b2 (2)+a

2b2 (2)

c hs.


83

SSQB

=

(T2 B

0

+T

2 B1

+T

2 B2)=(RB

�3k

�

1)�T

2 tot;B=(RB

�3k)

withf=(3�1)=2degreesoffreedomandRB

=2andk=2.

TAB0

=

(1) (2)+ab2 (2)+a

2b (2)

TAB1

=

a(2)+a

2b2 (2)+b (2)

TAB2

=

a2 (2)+b2 (2)+ab (2)

thatis,sumsfromtheRAB

experimentsinwhichthearti�ciale�ectABi+jisnotcon-

foundedwithblocks,i.e.theexperimentconsistingofblocks4,5and6.WithTtot;AB

=

T4+T5+T6,one�nds

SSQAB

=

(T2 AB0

+T

2 AB1

+T

2 AB2)=(RAB

�3k

�

1)�T

2 tot;AB=(RAB

�3k)

withf=(3�1)=2degreesoffreedomandRAB

=1andk=2.

Finallyone�nds

TAB2 0

=

(1) (1)+ab (1)+a

2b2 (1)

TAB2 1

=

a(1)+a

2b (1)+b2 (1)

TAB2 2

=

a2 (1)+b (1)+ab2 (1)

thatis,sumsfromtheexperimentsinwhichthearti�ciale�ectAB

2 i+jisnotconfounded

withblocks,i.e.theexperimentconsistingofblocks1,2and3.WithTtot;AB2

=T1+T2+T3

one�nds

SSQAB2

=

(T2 AB20

+T

2 AB21

+T

2 AB22)=(RAB2

�3k

�

1)�T

2 tot;AB2=(RAB2

�3k)

withf=(3�1)=2degreesoffreedomandRAB2

=1andk=2.

Theresidualvariationisfoundasthevariationbetweenestimatesfore�ectsthatarenot

confoundedwithblocks.

FromtheAe�ectone�nds,whereSSQA(bothexperiments)istheabovecalculatedsum

ofsquaresfore�ectA,whileSSQA(experiment1)andSSQA(experiment2)arethesums

ofsquaresfore�ectAcalculatedseparatelyforthetwoexperiments:

c hs.


84

SSQUA

=

SSQA(experiment1)+SSQA(experiment2)-SSQA(bothexperiments)

withf=4�2=2degreesoffreedom.

FromtheBe�ectone�ndscorrespondingly

SSQUB

=

SSQB(experiment1)+SSQB(experiment2)-SSQB(bothexperiments)

likewisewithf=4�2=2degreesoffreedom.

Sincetheremaininge�ectsABi+jandAB

2 i+2jareonly"purely"estimatedonetimeeach,

wedonotgetanycontributiontotheresidualvariationfromthesee�ects.

Insummary,wegetthefollowingvariancedecomposition:

Blocksand/orconfounded

factore�ects

SSQblocks

5

Maine�ectAi

SSQA

2

Maine�ectBj

SSQB

2

InteractionABi+j

SSQAB

2

(halfpr�cision)

InteractionAB

2 i+j

SSQAB2

2

(halfpr�cision)

Residualvariation

SSQUA

+SSQUB

2+2

Total

SSQtotal

17

Notethatwehavederivedvariationcorrespondingto17=18�1degreesoffreedom.

Inconclusionwecangiveestimatesforallthee�ectsinthisexperiment:

c �2 resid=(SSQUA

+SSQUB)=4

(^ A0;^ A1;^ A2)=(T

A0

6

�Ttot

18;

TA1

6

�Ttot

18;

TA2

6

�Ttot

18)

andthemaine�ectBisfoundcorrespondingly.

c hs.


85

Fortheinteraction,theestimatesarefoundintheblockswheretheyarenotconfounded:

(d AB0;d AB1;d AB2)=(T

AB0

3

�Ttot;2

9

;TAB1

3

�Ttot;2

9

;TAB2

3

�Ttot;2

9

)(fromexperiment2)

(d AB2 0;d AB2 1;d AB2 2)=(T

AB2 0

3

�Ttot;1

9

;TAB2 1

3

�Ttot;1

9

;TAB2 2

3

�Ttot;1

9

)(fromexperiment1)

andwiththehelpoftherelationABi;j=ABi+j+AB

2 i+2jonecan�nally�ndtheparameter

estimatesinthetraditionalmodelformulaYi;j=�+Ai+Bj+ABi;j+E.

Endofexample3.16

3.8

Constructionofafractionalfactorialdesign

WewillnowconcernourselveswithconstructingdesignswherethefactorsformLatin

squares/cubes.Thepresentationisageneralisationoftheresultsinsection2.4,wherewe

introducedfractional2kfactorialdesigns.Wewilllimitthediscussiontoexperimentswith

factorson2oron3levels,sincethesearetheexperimentsthatareusedmostfrequently

inpractice.

Asbefore,mainlyexamplesareusedtoshowthedi�erenttechniques.

Example3.17:

FactorexperimentdoneasaLatinsquareexperiment

Letusassumethatwehavethreefactors,A,BandCandthatwewanttoevaluatethese

eachon3levels.WechoosetomakeaLatinsquareexperimentwiththefactorCinside

thesquare.Accordingtothesameprincipledescribedintheprevioussection,wecanfor

exampleletChaveindexk=(i+j)3.Thismeansthatthemaine�ectCwillhavethe

sameindexastheKempthornee�ectABwithindex(i+j)3.

TheexperimentaldesignwheretheindexforfactorCisinsidethesquareisthus:

A=0

A=1

A=2

B=0

0

1

2

B=1

1

2

0

B=2

2

0

1

IndexforC

orequivalently:

(1)

ac

a2c2

bc

abc

2

a2b

b2c2

ab2

a2b2c

Insteadofthe3�3�3singleexperimentsinthecompletefactorexperiment,wechoose

todoonlythe3�3singleexperimentsintheLatinsquare.

c hs.


86

Notethatifx=acandy=bcwhichbothcorrespondtoasolutiontotheindexequation

k=(i+j)3,theexperimentcanbewritten:

(1)

x

x2

y

xy

x2y

y2

xy

2

x2y

2

Finally,wecanalsobymeansofthetabularmethodwriteoutthedesign:

Experiment

Experiment

Alevel

Blevel

Clevel

code

no.

sequence

i

j

(i+j)3

1

3

0

0

0

(1)

2

7

1

0

1

ac

3

1

2

0

2

a2c2

4

9

0

1

1

bc

5

5

1

1

2

abc

2

6

6

2

1

0

a2b

7

2

0

2

2

b2c2

8

4

1

2

0

ab2

9

8

2

2

1

a2b2c

Inthepracticalexecutionoftheexperiment,theorderisrandomised,asexempli�edin

thetable.

ThevariationofthemeanvalueduetothetwofactorsAandBcanbewritten

Ai+Bj+ABi;j=Ai+Bj+ABi+j+AB

2 i+2j

wheretheleft-handsideistheconventionalmeaning,whiletheright-handsideisthe

formulationaccordingtoKempthorne'smethod.

TheintroductionoffactorC,asmentioned,wasdonebygivingCtheindexvaluek=

(i+j)3.IfCispurelyadditive,i.e.ifCdoesnotinteractwiththeotherfactors,the

followingmodelwilldescribetheresponse,wherethein uencefromCisputin:


2 i+2j+Ck=i+j+Eijk�

whereindex�correspondstopossiblerepetitionsofthe9singleexperiments.

IfthereisinteractionbetweenthetwofactorsAandB,thatpartoftheinteraction

describedbythearti�ciale�ectABi+jcannotberegardedasnegligible(thesameistrue

ofcourseforAB

2 i+2j).

c hs.


87

IfwenowtrytoestimatetheCe�ect,wecannotavoidhavingtheABi+jpartoftheAB

interactionconfoundedtheCestimate,preciselybecauseweusedk=(i+j)3

Thee�ectsCkandABi+jinotherwordsareconfoundedintheexperiment.

Wecanalsodemonstratethisbydirectcalculation.Using� Ytotfortheaverageofthe9

measurements,wehave,since(1)=Y0;0;0,a

2b=Y2;1;0,andab2=Y1;2;0:

b C 0=(Y0;0;0+Y2;1;0+Y1;2;0)=3�� Ytot

whichhastheexpectedvalue

Efb C 0g=Ef(Y0;0;0+Y2;1;0+Y1;2;0)=3�� Ytotg=C0+AB0

Furtheritisfound


and


InthisexperimentwehaveacertainpossibilityofevaluatingwhethertheABinteraction

canberegardedasnegligible,becausewecanexaminetheAB

2 i+2je�ect.Ifthisisneg-

ligible,onecanperhapsallowoneselftoconcludethattheABe�ectasawholecanbe

zero.

Insummary,onecanseethatonlyifthefactorsAandBdonotinteract,istheexperiment

suitableforestimatingC.

Asweshallseebelow,weneedtoassumethatalltwo-factorinteractionsarezeroinorder

toestimatethemaine�ectsA,BandCinthedescribed(1=3)�33=33

�

1

experiment.

Endofexample3.17

Wehaveseenaboveandpreviouslyinsection2.4thatitisnotwithoutproblemsto

putfurtherfactorsintoanexperimentintheform

ofasquareexperiment.Butwith

appropriateassumptionsaboutthelackofinteractions,itcanbedone.Inthefollowing,

wewilltrytoshowhowitisdoneinpractice.

Example3.18:Confoundingsina3

�

1�3

3factorialexperiment,aliasrelations

Consideragaintheaboveexample.Ifweshouldhavedoneanordinary33factorial

experiment,itwouldhaveconsistedof3�3�3=27singleexperiments.Theexperiment

wedidisonly1/3ofthis,namelyatotalof33=3=33

�

1=9singleexperiments.

Ifingeneralthereareinteractionsbetweenallthefactors,thee�ectsoftheexperiment

willbeconfoundedwitheachotheringroupsof3e�ects,analogouswithforexamplethe

c hs.


88

23�

1experiment,wheretheywereconfoundedingroupsof2.Thecompletemathematical

modelforthe33factorialexperimentcanbewritten,onceagain:


2 i+2j+Ck+ACi+k+AC

2 i+2k+BCj+k

+BC

2 j+2k+ABCi+j+k+ABC

2 i+j+2k+AB

2Ci+2j+k+AB

2C

2 i+2j+2k+Eijk�

Forthe9singleexperimentsweconsideredinthepreviousexample,weusedk=i+j,

correspondingtotheconfoundingCk=ABi+j.

Thisgeneratorequationcanbechangedtoade�ningrelationbymultiplyingonboth

sidesoftheequationsignwithC

2

,andthenreorganisingtheexpressionsandreducing

theexponentsmodulo3.TheresultisC

2C=C

2AB�!I=ABC

2.Thuswehavethe

De�ningrelation:I=ABC

2

Ifonenowwantswhiche�ectsinthegeneralmodelanarbitrarye�ectisconfoundedwith,

thede�ningrelationcanbeused.Itismultipliedwiththee�ectinquestionin�rstand

secondpower(becausethefactorsareon3levelsandaccordingtotheruleslayedoutin

section3.3).One�ndsfortheAe�ect:

A�(I=ABC

2)�!A=(A)(ABC

2)=(A)2(ABC

2)

SincenowA(ABC

2)=A

2BC

2!A

4B

2C

4!AB

2Cand(A)2(ABC

2)=A

3BC

2!BC

2,

itisfoundthat

A=AB

2C=BC

2

Onecanbeconvincedthatindicesforthesethreee�ectsvarysynchronouslythroughout

theexperiment,becauseitisrequiredthatk=(i+j)3."Modulo3"calculationgives

(tryityourself):

E�ects

A

AB

2C

BC

2

Indices

i

(i+2j+k) 3

(j+2k) 3

0)

0

0

1)

2

2

2)

1

1

Ofcoursethesamecalculationscanbemadeforalle�ectsintheexperimentandonecan

beconvincedthatitwillgenerallyholdtruethatalle�ectsareconfoundedingroupsof

3.c hs.


89

Thecompletesetofaliasrelationsisfoundonthebasisofthede�ningrelationbymul-

tiplyingwiththee�ectsintheunderlyingfactorstructurein�rstandsecondpower:

GeneratorC=AB=)

De�ningrelationI=ABC

2

Aliasrelations

A

=

AB

2C

=

BC

2

B

=

AB

2C

2

=

AC

2

AB

=

ABC

=

C

AB

2

=

AC

=

BC

ForexampleAB

2�(I=ABC

2)�!AB

2=AC=BC.

Rememberagain,thatthefactorsAandBconstituteanunderlyingcompletefactor

structureandthatfactorCisintroducedintothisstructurebymeansofthegenerator

equationC

=AB.Thisexampli�esthegeneralmethodofconstructionoffractional

factorials.

Thealiasrelationsaremostusefullywrittenupwithonerelationpere�ectintheunder-

lyingfactorstructureandinstandardorder,asisshowninthetable.

Endofexample3.18

3.8.1

Resolutionforfractionalfactorialdesigns

Thetermresolutiondescribeswhichordersofe�ectsareconfoundedwitheachother.

Correspondingtotheexamplepage34wherefractional2kfactorialdesignswereintro-

duced,page37showsaliasrelationsfora23

�

1

factorialexperiment.Itcanbeseenhere

thatthemaine�ects(�rstordere�ects)areconfoundedwith2-factorinteractions(second

ordere�ects).SuchanexperimentiscalledaresolutionIIIexperiment.Itshouldbe

notedthatprecisely3factorsarepresentinthede�ningrelation(I=ABC)forthe

experiment.

Intheaboveexamplea33

�

1

factorialdesignisdescribedwiththede�ningrelationI=

ABC

2.ThisexperimenttooiscalledaresolutionIIIexperiment,sincemaine�ectsare

confoundedwithtwo-factorinteractions(orhigher).Thede�ningrelationinvolvesat

least3factors.

Ifallmaine�ectsinafractionalfactorialdesignareconfoundedwithe�ectsofatleastsec-

ondorder(2-factorinteractions),theexperimentiscalledaresolutionIIIexperiment.

Thiscorrespondstothefactthatnoe�ectinthede�ningrelationoftheexperimentisof

alowerorderthan3.

c hs.


90

Ifitholdstruethatnoe�ectinthede�ningrelationoftheexperimentisofalowerorder

than4,theexperimentiscalledaresolutionIV

experiment.

InaresolutionIVexperiment,themaine�ectsareallconfoundedwithe�ectsofat

leastthethirdorder,i.e.3-factorinteractions.Two-factorinteractionswillgenerallybe

confoundedwithother2-factorinteractionsand/orinteractionsofahigherorderina

resolutionIVexperiment.

Inmanypracticalcircumstances,onecannotassumeinadvancethatthe2-factorinter-

actionsareunimportantcomparedwiththemainactions.Onewillthereforeoftenwant

anexperimentofresolutionIV-atleast.

InaresolutionVexperiment,themaine�ectsareallconfoundedwithe�ectsofatleast

thefourthorder,i.e.4-factorinteractions.Two-factorinteractionswillgenerallybe

confoundedwith3-factorinteractionsand/orinteractionsofahigherorderinaresolution

Vexperiment.

Asarule,experimentswithahigherresolutionthanVwillnotbeneededtobedone,if

thefactorsinvolvedareofaquantitativenature(temperature,pressure,concentration,

time,densityetc.)wheremaine�ectsand2-factorinteractionsaremostfrequentlyof

considerablygreaterimportancethantheinteractionsofhigherorder.

3.8.2

Practicalandgeneralprocedure

Bymeansofthemethodoutlinedabove,wecannowconstructarbitrary1=pq�pkfactorial

experimentsand�ndtheconfoundings(thealiasrelations)intheexperiment.

kfactorsareconsidered(A,B,C,...,K)andthesefactorsareorderedsothatthe�rst

factorsareaprioriattributedthegreatestimportance.Meaningthattheexperimenter

expectsthatfactorAwillprovetohavethegreatestimportance(e�ect)ontheresponse

Y,andthatBhasthenextgreatestimportanceetc.

Thisorderingofthefactorsbeforetheexperimentisagreathelpbothwithregardto

creatingasuitabledesignandwithregardtoevaluatingtheresultsobtained.

Inthereview,inaddition,onehastomakeadecisionastowhichfactorsthatcouldbe

thoughttointeractandwhichonesthatcanbeassumedtoactadditively.Asthegeneral

rule,interactionsbetweenfactorsthathavealargee�ectwillbelargerthaninteractions

betweenfactorswithmoremoderatee�ects.

Inaddition,onewillgenerallyexpectthatinteractionsofahighorderwillbelessimpor-

tantthaninteractionsofalowerorder.

Inmanycasesoneoftenallowsoneselftoassumethatinteractionsofanorderhigherthan

2(i.e.3-factore�ectssuchasABC,ABD,BCDetc.ande�ectsofevenhigherorder)are

c hs.


91

assumedtohaveconsiderablylessimportancethanthemaine�ects.

Afterthis,theexperimentismostsimplyconstructedbystartingwiththecompletefacto-

rialexperiment,whichismadeupofthe(k�q)�rst(andexpectedtobemostimportant)

factorsandputtingtheremainingqfactorsintothisfactorstructurebyconfoundingwith

e�ectsregardedasnegligible.The�rst(k�q)andoftenmostimportantfactorswill

therebyformtheunderlyingfactorstructureinthe1=pq�pkfactorialexperimentwanted.

Example3.19:

A

2�

2

�2

5

factorialexperiment

Supposethatoneconsiders5factorsA,B,C,DandE,whichonewantstoevaluateeach

on2levelsina1=22�25factorialexperiment,i.e.in25

�

2=23=8singleexperiments.

WeimaginethatacloserevaluationoftheproblemathandindicatesthatfactorsA,B

andCwillhavethegreateste�ect,andwethuslettheunderlyingfactorstructureconsist

ofpreciselythesefactors.

ThedesignisthengeneratedbyconfoundingfactorsDandEwithe�ectsintheunderlying

factorstructure:

Generators

I A B AB C A

CBC

=

E

ABC

=

D

=)

Aliasrelationer

I

=

ABCD

=

BCE

=

ADE

A

=

BCD

=

ABCE

=

DE

B

=

ACD

=

CE

=

ABDE

AB

=

CD

=

ACE

=

BDE

C

=

ABD

=

BE

=

ACDE

AC

=

BD

=

ABE

=

CDE

BC

=

AD

=

E

=

ABCDE

ABC

=

D

=

AE

=

BCDE

TheexperimentisaresolutionIIIexperiment.

Theexperimentcaneasilybewrittenoutusingthetabularmethodasfollows(ifthe

principalfractionwiththeexperiment"(1)"ischosen)

i

j

k

l=(i+j+k) 2

m=(j+k) 2

Code

0

0

0

0

0

(1)

1

0

0

1

0

ad

0

1

0

1

1

bde

1

1

0

0

1

abe

0

0

1

1

1

cde

1

0

1

0

1

ace

0

1

1

0

0

bc

1

1

1

1

0

abcd

c hs.


92

whichisprobablytheeasiestwayto�ndtheexperimentandatthesametimetimewrite

outthewholeplan,forexampleusingasimplespreadsheetprogram.

Wemaywritetheplanas:

D=ABC,E=�BC

l=(i+j+k) 2,m=(j+k) 2

(1)

ad

bde

abe

cde

ace

bc

abcd

wheretheindicesarei,j,k,landm

forfactorsA,B,C,DandE,respectively,andthe

constructionoftheexperimentisgivenwiththepreviouslyintroducedsignnotationfor

2kexperimentsaswellaswiththeindexmethodthatisusedinthepresentchapter.See,

too,theexampleonpage40.

Thereare3alternativepossibilities,namely

D=�ABC,E=�BC

l=(i+j+k+1)2,m=(j+k) 2

d

a

be

abde

ce

acde

bcd

abc

D=+ABC,E=+BC

l=(i+j+k) 2,m=(j+k+1)2

e

ade

bd

ab

cd

ac

bce

abcde

D=�ABC,E=+BC

l=(i+j+k+1)2,m=(j+k+1)2

de

ae

b

abd

c

acd

bcde

abce

Aprerequisiteforobtaininga"good"experimentbydoingoneoftheseexperimentsis

thatfactorsBandCdonotinteractwitheachother(BCandABCunimportant),and

thatfactorsDandEdonotinteractwithotherfactorsatallorwitheachother.Factor

AcanbeallowedtointeractwiththetwofactorsBandC,i.e.ABandACcandi�er

from0.

Ifthesepreconditionscannotberegardedasful�lledtoareasonabledegree,theexperi-

mentwillnotbeappropriatetostudythe5factorssimultaneouslyinafractionalfactorial

designwithonly8singleexperiments.

Thealternativeswillthenbeeithertoexcludeoneofthefactors(bykeepingitconstantin

theexperiment)andbeingcontentwitha1=2�24experimentortoextendtheexperiment

toa1=2�25experiment,i.e.anexperimentwith16singleexperiments.Thesecond

alternativecouldreasonablybeconstructedbyputtingfactorEintothefactorstructure

consistingoffactorsA,B,CandDbytherelationABCD=E,withthefollowingalias

relations:

c hs.


93

I

=

ABCDE

A

=

BCDE

B

=

ACDE

AB

=

CDE

C

=

ABDE

AC

=

BDE

BC

=

ADE

ABC

=

DE

D

=

ABCE

AD

=

BCE

ABD

=

CE

CD

=

ABE

ACD

=

BE

BCD

=

AE

ABCD

=

E

Ifitisassumedthatallinteractionsofanorderhigherthan2areunimportant,onegets

reducedaliasrelations(thefullde�ningrelationisretained)

I

=

ABCDE

A

=

B

=

AB

=

C

=

AC

=

BC

= =

DE

D

=

AD

= =

CE

CD

= =

BE

=

AE

=

E

Onecanseethatall2-factorinteractionscanbetestedinthisdesign.Ifsomeofthese

arereasonablysmall,theirsumsofsquarescouldbepooledintoaresidualsumofsquares

andusedtotesthigherordere�ects.

ThisexperimentisaresolutionVexperiment.

Onecanchooseoneofthetwofollowingcomplementaryexperiments:

E=�ABCD

orm=(i+j+k+l)2

(1)

ae

be

ab

ce

ac

bc

abce

de

ad

bd

abde

cd

acde

bcde

abcd

E=+ABCD

orm=(i+j+k+l+1)2

e

a

b

abe

c

ace

bce

abc

d

ade

bde

abd

cde

acd

bcd

abcde

c hs.


94

Endofexample3.19

3.8.3

Aliasrelationswith1=pq�pk

experiments

Whenqfactorsareputintoafactorstructureconsistingof(k�q)factors,qgenerator

equationsareused.Eachequationgivesrisetoonede�ningrelation.Thatis,one�nds

qde�ningrelationswithqde�ningcontrasts:I 1,I 2,...,I q.Thecompletede�ning

relationcanthensymbolicallybewrittenas

I=I 1�I2�:::�Iq

wheretheoperator"�"isde�nedonpage60.Bycalculatingtheexpressionandreplacing

all"+"with"=",one�ndsthecompletede�ningrelation:

I=I 1=I 2=I 1I 2=I 1I

2 2

=:::=I 1I

p�

1

2

=:::=I 1I

p�

1

2

��I

p�

1

q

correspondingtothe"standardorder"forqfactors,calledI 1,I 2,...,I q.

Thealiasrelationsoftheexperimentdrawnupcanbefoundforanarbitrarye�ect,F,

bycalculatingtheexpression

F=F�I=)F=F�I1�I2�:::�Iq

andFandallthee�ectsemergingontheright-handsideoftheexpressionwillbecon-

foundedwitheachother.

Fromcalculationoftheexpressionandreplacementof"="with"+"subsequently,one

gets:

F=FI 1=:::=FI

p�

1

1

=FI 2=:::=FI

p�

1

2

=:::=F(I1I

p�

1

2

��I

p�

1

q

)p�

1

Duringthecalculationofthesinglee�ectsintheexpression,itcanbehelpfultousethe

factthatfortwoarbitrarye�ectsX

andY,itholdstruethat

XY

�=X

�Y

suchthatinthecalculationofexpressionswithtwoe�ects,itisofteneasiesttoliftupthe

simpleste�ecttothepowerinquestion.Forexampleina32factorialexperiment,both

(AB

2C)2(AB)and(AB

2C)(AB)2becomeBC.Testthis.

c hs.


95

Example3.20:

Constructionof3

�

2

�3

5

factorialexperiment

Lettherebe5factorsA,B,C,DandEeachon3levels.Onewantstodoonly1/9of

thewholeexperiment,i.e.atotalof35

�

2=33=27singleexperiments.

Againwestartwithacompletefactorstructureforthreeofthefactors.Anditisassumed

thatitisreasonabletochooseA,BandC.Inthisfactorstructure,afurther2factorsare

putin,namelyDandE. D

esigngenerators

I A B AB

AB2

C AC

BC

ABC

AB2C

AC2

BC2

=

E

ABC2

AB2C2

=

D

=)

I 1

=

AB

2C

2D

2

I 2

=

BC

2E

2

Otheralternativescanbechosen,forexampletoputbothDandEintothe3-factor

interactionABC(whichisdecomposedin4partseachwith2degreesoffreedom)byfor

exampleD=AB

2C

2andE=ABC

2.(Tryto�ndthecharacteristicsofthisexperiment

(aliasrelations)).

Withtheconfoundingchoseninthetable,one�ndsthede�ningrelation

I=AB

2C

2D

2=BC

2E

2=(AB

2C

2D

2)(BC

2E

2)=(AB

2C

2D

2)(BC

2E

2)2

whichafterreductiongives

I=AB

2C

2D

2=BC

2E

2=ACD

2E

2=ABD

2E

Thealiasrelationsoftheexperimentare

c hs.


96

I

=

AB2C2D2

=BC2E2

=ACD2E2

=ABD2E

=De�ningrelation

A

=

ABCD

=ABC2E2

=AC2DE=AB2DE2

=BCD

=AB2CE=CD2E2

=BD2E

B

=

AC2D2

=BCE=ABCD2E2

=AB2D2E=ABC2D2

=CE=AB2CD2E2

=AD2E

AB

=

ACD

=AB2C2E2

=AB2C2DE=ABDE2

=BC2D2

=ACE=BC2DE=DE2

AB2

=

AB2CD

=AC2E2

=ABC2DE=ADE2

=CD

=ABCE=BCD2E2

=BDE2

C

=

AB2D2

=BE2

=AC2D2E2

=ABCD2E=AB2CD2

=BCE2

=AD2E2

=ABC2D2E

AC

=

ABD

=ABE2

=ACDE=AB2C2DE2

=BC2D

=AB2C2E=DE=BC2D2E

BC

=

AD2

=BE=ABC2D2E2

=AB2CD2E=ABCD2

=CE2

=AB2D2E2

=AC2D2E

ABC

=

AD

=AB2E2

=AB2CDE=ABC2DE2

=BCD2

=AC2E=BDE=CDE2

AB2C

=

AB2D

=AE2

=ABCDE=AC2DE2

=CD2

=ABC2E=BD2E2

=BCDE2

AC2

=

ABC2D

=ABCE2

=ADE=AB2CDE2

=BD

=AB2E=CDE=BCD2E

BC2

=

ACD2

=BC2E=ABD2E2

=AB2C2D2E=ABD2

=E=AB2C2D2E2

=ACD2E

ABC2

=

AC2D

=AB2CE2

=AB2DE=ABCDE2

=BD2

=AE=BCDE=CD2E

AB2C2

=

AB2C2D

=ACE2

=ABDE=ACDE2

=D

=ABE=BC2D2E2

=BC2DE2

Thealiasrelationforexampleofthemaine�ectAisfoundwiththehelpof:

A=A�I1�I2=A�(AB

2C

2D

2)�(BC

2E

2)

whichgives

A=A(AB

2C

2D

2)=A(AB

2C

2D

2)2=A(BC

2E

2)=:::=A(AB

2C

2D

2)2(BC

2E

2)2

TheexpressionsareorganisedinA-B-C-D-Eorderandtheexponentsreducedmodulo3.

Ifnecessarytheexponent1onthe�rstfactorintheexpressionsisfoundbyraisingtothe

powerof2andreducingmodulo3.

Inthesameway,thealiasrelationsforeachoftheothere�ectsarefoundintheunderlying

factorstructureasshowninthetable.

Toelucidatethecharacteristicsoftheexperimentaldesign,alle�ectsconsideredunimpor-

tantcanberemoved.ThisisthecasefortheBCe�ectandallothere�ectsinvolvingmore

than2factors.Forthesakeofclarity,thee�ectsfromtheunderlyingfactorstructureare

retained,butinparenthesisforassuminglyunimportante�ects.

Inthisway,thefollowingtableisfound,whichshowsthat2-factorinteractionsareusually

confoundedwithother2-factorinteractionsorwithmaine�ects.

c hs.


97

Reducedaliasrelations

I

=

AB2C2D2

=BC2E2

=

ACD2E2

=ABD2E

A

=

B

=

CE

AB

=

DE2

AB2

=

CD

C

=

BE2

AC

=

DE

(BC)

=

AD2

=BE=CE2

(ABC)

=

AD

(AB2C)

=

AE2

=CD2

AC2

=

BD

(BC2)

=

E

(ABC2)

=

BD2

=AE

(AB2C2)

=

D

TheexperimentisaresolutionIIIexperiment.Onecanseethatitisnecessarytoassume

thatanumberofthe2-factorinteractionsareunimportantiftheexperimentistobe

suitable.

If,forexample,onecanfurthermoreignoreinteractionsinvolvingfactorsDandE,allelse

canbetestedandestimated.Thisshowstheusefulnessoforderingthefactorsaccording

toimportance(i.e.maine�ectsandthusinteractionsfromDandErelativelysmall).

IfitholdstruethatDandEhaveonlyadditivee�ectsanddonotinteractwiththe

otherfactors,thealiasrelationscanbereducedtothetableshownonpage96,wherethe

experimentwasconstructed.

Thereare3�3=9possibilitiesforimplementingtheexperiment.If,forexample,we

wantthefractionincluding"(1),theexperimentwillbegivenbytheindexrestrictions

(i+2j+2k+2l) 3=0and(j+2k+2m) 3=0.

Ifwewanttowriteoutatableofindicesforthefactors(thatisthedesign),weusethe

tabularmethodandthegeneratorequationsl=i+2j+2kandm=j+2kasfollows:

c hs.


98

Factorsandlevels

A

B

C

D

E

Experiment

i

j

k

l=(i+2j+2k) 3

m

=(j+2k) 3

code

0

0

0

0

0

(1)

1

0

0

1

0

ad

2

0

0

2

0

a2d2

0

1

0

2

1

bd2e

1

1

0

0

1

abe

2

1

0

1

1

a2bde

0

2

0

1

2

b2de2

1

2

0

2

2

ab2d2e2

2

2

0

0

2

a2bd2e

0

0

1

2

2

cd2e2

1

0

1

0

2

ace2

...

...

...

...

...

...

...

...

...

...

...

...

2

2

2

1

0

a2b2c2d

Theexperimentis3�3�3=27singleexperiments.Onecanalsoderivethesingle

experimentsbysolvingindexequations.If3experimentswhich(independently)ful�l

indexequationsarecalledx,yandz,theexperimentwillbe:

(1)

x

x2

y

xy

x2y

y2

xy

2

x2y

2

z

xz

x2z

yz

xyz

x2yz

y2z

xy

2z

x2y

2z

z2

xz

2

x2z

2

yz

2

xyz

2

x2yz

2

y2z

2

xy

2z

2

x2y

2z

2

IntheunderlyingfactorstructureA,BandC,x

0

=a,y

0

=bandz

0

=cwillbesolutions,

andthecorrespondingindexsetsare(i;j;k)=(1;0;0),(i;j;k)=(0;1;0)and(i;j;k)=

(0;0;1).

To�ndthreesolutionsx,yandz,wethereforetrywith

x0

:(i;j;k)=(1;0;0)

=)

(l;m)=(1;0)

=)

x=ad

y0

:(i;j;k)=(0;1;0)

=)

(l;m)=(2;1)

=)

y=bd

2e

z0

:(i;j;k)=(0;0;1)

=)

(l;m)=(2;2)

=)

z=cd

2e2

Theexperimentthenconsistsofthesingleexperimentsbelow:

c hs.


99

(1)

ad

(ad)2

bd2e

ad(bd

2e)

(ad)2(bd

2e)

(bd

2e)

2

ad(bd

2e)

2

(ad)2(bd

2e)

2

cd2e2

adcd

2e2

(ad)2cd

2e2

(bd

2e)cd

2e2

ad(bd

2e)cd

2e2

(ad)2(bd

2e)cd

2e2

(bd

2e)

2cd

2e2

ad(bd

2e)

2cd

2e2

(ad)2(bd

2e)

2cd

2e2

(cd

2e2)2

ad(cd

2e2)2

(ad)2(cd

2e2)2

(bd

2e)(cd

2e2)2

ad(bd

2e)(cd

2e2)2

(ad)2(bd

2e)(cd

2e2)2

(bd

2e)

2(cd

2e2)2

ad(bd

2e)

2(cd

2e2)2

(ad)2(bd

2e)

2(cd

2e2)2

whicharereorganisedandtheexponentsreducedmodulo3:

(1)

ad

a2d

2

bd2e

abe

a2bde

b2de2

ab2d

2e2

a2b2e2

cd2e2

ace

2

a2cde2

bcd

abcd

2

a2bc

b2ce

ab2cde

a2b2cd

2e

c2de

ac2d

2e

a2c2e

bc2e2

abc

2de2

a2bc

2d

2e2

b2c2d

2

ab2c2

a2b2c2d

Inall,thereare9di�erentpossibilitiestoconstructtheexperiment,correspondingtothe

followingtable:

(i+2j+2k+2l)3=

0

(i+2j+2k+2l)3=

1

(i+2j+2k+2l)3=

2

(j+2k+2m

) 3=

0

1=

thedesignshown

2

3

(j+2k+2m

) 3=

1

4

5

6

(j+2k+2m

) 3=

2

7

8

9

Thethreeexperiments"1","4"and"7",forexample,arecomplementarywithregardto

thegeneratorequationBC

2

=E,i.e.thede�ningrelationI 2=BC

2E

2.

Thesameholdstruefor"2","5"and"8",aswellasfor"3","6"and"9".

Ifonecarriesoutoneofthesesetsofcomplementaryexperiments,onebreaksthecon-

foundingsoriginatinginthechoiceofBC

2

=E,andthewholeexperimentwillthenbe

c hs.


100

a1=3�34experimentwiththede�ningrelationI 1=AB

2C

2D

2andthefactorsA,B,C

andEintheunderlyingfactorstructure.

Endofexample3.20

3.8.4

Estimationandtestingin1=p

q�p

k


Theimportantthingtorealiseisthat1=pq�pkfactorialexperimentsareconstructedon

thebasisofacompletefactorstructure,i.e.theunderlyingcompletefactorstructure.

Theanalysisoftheexperimentisthendoneinthefollowingsteps:

1)Forthefractionalfactorialdesign,theunderlyingcompletefactorstructureisiden-

ti�ed.

2)Dataarearrangedinaccordancewiththisunderlyingstructure,andthesumsof

squaresaredeterminedintheusualwayforthefactorsandinteractionsinit.

3)Thealiasrelationsindicatehowallthee�ectsareconfoundedintheexperiment.

Therebythesumsofsquaresarefoundfore�ectsthatarenotintheunderlying

factorstructure.

4)Byconsideringspeci�cfactorcombinationsinasingleexperiment,onecandecide

howtheindexrelationsarebetweenthee�ectsthatarepartofthesamealias

relation.Inthiswayestimatesaredeterminedfortheindividuallevelsforthe

e�ectsthatarenotintheunderlyingstructure.

Asanillustrationofthis,weconsiderthefollowing.

Example3.21:

Estimationina3

�

1

�3

3-factorialexperiment

WehavefactorsA,BandC,allon3levelsandweassumethatA,BandCarepurely

additive,sothatitisrelevanttodoafractionalfactorialexperimentinsteadofacomplete

factorialexperiment.

Astheunderlyingfactorstructure,the(A,B)structureischosen.

Thegeneralised(Kempthorne)e�ectsinthisstructureareA,B,ABandAB

2.Wechoose

toconfoundforexamplewithAB,i.e.ABi+j

=Ck.Thischoiceentailsthefollowing

de�ningrelationandaliasrelations:

I

=

ABC

2

A

=

AB

2C

=

BC

2

B

=

AB

2C

2

=

AC

2

AB

=

ABC

=

C

AB

2

=

AC

=

BC

c hs.


101

Theindexrestrictionontheprincipalfractionoftheexperiment,i.e.thefractionthat

contains"(1)",is(i+j+2k) 3=0,k=(i+j)3.Twolinearlyindependentsolutions

havetobefoundforthisandbystartingin"a"and"b",one�nds:

(i;j)=(1;0)=)k=(1+0)=1:theexperimentisac

(i;j)=(0;1)=)k=(0+1)=1:theexperimentisbc

Onepossibleexperimentistheprincipalfractioninwhich"(1)"isapart:

(1)

ac

(ac)

2

bc

acbc

(ac)

2bc

(bc)

2

ac(bc)2

(ac)

2(bc)

2

=)

(1)

ac

a2c2

bc

abc

2

a2b

b2c2

ab2

a2b2c

Analternativepossibilityistocarryoutoneofthe(two)otherfractions,forexamplethe

oneofwhichthesingleexperimentaispart.Thisfractionisdeterminedby"multiplying"

theprincipalfractionwitha:

a�

(1)

ac

a2c2

bc

abc

2

a2b

b2c2

ab2

a2b2c

=)

a

a2c

c2

abc

a2bc

2

b

ab2c2

a2b2

b2c

Thisexperimenthastheindexrestriction(i+j+2k) 3=1,k=(i+j+2)3.

Thisexperimentischosenhereanddataareorganisedandanalysednowintheusualway

accordingtofactorsAandB(neglectingC):

A=0

A=1

A=2

B=0

c2

a

a2c

or

(1)

a

a2

B=1

b

abc

a2bc

2

withoutc:

b

ab

a2b

B=2

b2c

ab2c2

a2b2

b2

ab2

a2b2

TA0

=c2+b+b2c

,

TA1

=a+abc+ab2c2

,

TA2

=a

2c+a

2bc

2+a

2b2

TB0

=c2+a+a

2c

,

TB1

=b+abc+a

2bc

2

,

TB2

=b2c+ab2c2+a

2b2

TAB0

=c2+a

2bc

2+ab2c2

,

TAB1

=a+b+a

2b2

,

TAB2

=a

2c+abc+b2c

TAB2 0

=c2+abc+a

2b2

,

TAB2 1

=a+a

2bc

2+b2c

,

TAB2 2

=a

2c+b+ab2c2

c hs.


102

SSQ(A)

=

([TA0]2+[TA1]2+[TA2]2)=3r�[Ttot]2=9r

,

f=3�1

SSQ(B)

=

([TB0]2+[TB1]2+[TB2]2)=3r�[Ttot]2=9r

,

f=3�1

SSQ(AB)

=

([TAB0]2+[TAB1]2+[TAB2]2)=3r�[Ttot]

2=9r

,

f=3�1

SSQ(AB

2)

=

� [TAB2 0]2+[TAB2 1]2+[TAB2 2]2

� =3r�[Ttot]

2=9r

,

f=3�1

whererindicatesthatatotalofrsinglemeasurementscouldbemadeforeachsingle

experiment.Inthatcase,itisassumedthattheserrepetitionsarerandomisedoverthe

wholeexperiment.

Finally,thee�ectscanbeestimated:

b A 0=TA0=3r�Ttot=9r,

b A 1=TA1=3r�Ttot=9r,

b A 2=TA2=3r�Ttot=9r

b B 0=TB0=3r�Ttot=9r,

b B 1=TB1=3r�Ttot=9r,

b B 2=TB2=3r�Ttot=9r

d AB0=TAB0=3r�Ttot=9r,

d AB1=TAB1=3r�Ttot=9r,

d AB2=TAB2=3r�Ttot=9r

d AB20=TAB2 0=3r�Ttot=9r,

d AB21=TAB2 1=3r�Ttot=9r,

d AB22=TAB2 2=3r�Ttot=9r

To�ndtheconnectionbetweenCandtheAB

e�ect,theindexrelationisfoundfrom

thespeci�cexperimentbyconsideringtwosingleexperiments,forexamplec2(i=0;j=

0;k=2)anda(i=1;j=0;k=0).One�ndsthatitholdstruethat

Index

ABi+j

0

1

2

Ck

2

0

1

whereindexk=(i+j+2)3.Therefore,b C 0=d AB1,b C 1=d AB2

andb C 2=d AB0.

Themathematicalmodeloftheexperimentcouldbewrittenas

Yijk�=�+Ai+Bj+Ck=i+j+2+Eijk�,where�=1;2:::;r

and,if�>1,andthereisusedcompleterandomisationcorrectly,theresidualsumof

squaresis

SSQresid=

3 X i=1

3 X j=1

" r X �=1

Y2 ijk��r�� Y

2 ijk�

#

c hs.


103

Notethatsumsaremadeoverindicesiandjalone,sinceindexkisofcoursegiven

byiandjinthis33

�

1

factorialexperiment(whichconsistsof9singleexperimentseach

repeatedrtimes).

ThepreconditionofadditivitybetweenthethreefactorsA,BandCcouldbetestedby

testingtheAB

2

e�ectagainstthissumofsquares.

Endofexample3.21

Example3.22:

TwoSASexamples

Thecalculationsshownintheaboveexamplearerelativelyeasytoprogram.Aprogram

canalsobewrittenforthestatisticalpackageSASwhichwilldothework.Thefollow-

ingsmallexamplewithdata(r=1)illustrateshowanalysisofvariancecanbedone

correspondingtofactorsAandBalone,i.e.theunderlyingfactorstructure.

A=0

A=1

A=2

B=0

c2=15.1

a=16.9

a2c=23.0

B=1

b=9.8

abc=12.6

a2bc

2=21.7

B=2

b2c=5.0

ab2c2=10.0

a2b2=12.8

dataexempel1;

input

ABCY;

AB

=mod(A+B,3);

AB2=mod(A+B*2,3);

cards;

00215.1

10016.9

20123.0

0109.8

11112.6

21221.7

0215.0

12210.0

22012.8

; procGLM;

classABABAB2;

model

Y

=ABABAB2;

means

ABABAB2;

run;

Intheexamplestartingonpage96withdatalayoutshownonpage98,aSASjobcould

looklikethefollowing:

dataexempel2;

input

ABCDEY;

c hs.


104

AB

=mod(A+B,3);

AB2=mod(A+B*2,3);

AC=mod(A+C,3);

AC2=mod(A+C*2,3);

BC

=mod(B+C,3);

BC2=mod(B+C*2,3);

ABC=mod(A+B+C,3);

ABC2=mod(A+B+C*2,3);

AB2C=mod(A+B*2+C,3);

AB2C2=mod(A+B*2+C*2,3);

cards;

00000

31.0

10010

16.0

20020

4.0

01001

23.8

11011

23.6

21021

9.7

.....

.....

.....

12200

12.8

22210

15.1

; procGLM;classABABAB2CACAC2BCBC2ABCABC2

AB2CAB2C2

;

model

Y

=ABABAB2CACAC2BCBC2ABCABC2

AB2CAB2C2

;

means

ABABAB2CACAC2BCBC2ABCABC2

AB2CAB2C2

;

run;

Andsumsofsquaresandestimatesofe�ectsoutsidetheunderlyingfactorstructure

(A,B,C)canbedirectlyfoundusingthealiasrelations.

Endofexample3.22

3.8.5

Fractionalfactorialdesignlaidoutinblocks

Afractionalfactorialdesigncanbelaidoutinsmallerblocksbecauseofawishtoincrease

theaccuracyintheexperiment(di�erentbatches,groupsofexperimentalanimals,several

daysetc.).Otherreasonscouldbethatforthesakeofsavingtimeonewantstodothe

singleexperimentsonparallelexperimentalfacilities(severalovens,reactors,set-upsand

such).

Intheorganisationofsuchanexperiment,thefractionalfactorialdesignis�rstsetup

withoutregardtothesepossibleblocks,sinceitisimportant�rstandforemosttohave

anoverviewofwhetheritispossibletoconstructagoodfractionalfactorialdesignand

howthefactore�ectsoftheexperimentwillbeconfounded.

Whenasuitablefractionalfactorialdesignhasbeenconstructed,achoiceismadeofwhich

e�ectore�ectswouldbesuitabletoconfoundwithblocks,andacontrolismadethatall

blockconfoundingsaresensible,perhapsthewholeconfoundingtableisreviewed.Inthe

exampleonpage114,anexampleofthisisshown.

c hs.


105

Bothduringtheconstructionofthefractionalfactorialdesignandinthesubsequent

formationofblocksfortheexperiment,theunderlyingfactorstructureisused,which

mostpracticallyiscomposedofthemostimportantfactors,calledA(�rstfactor),B

(secondfactor)etc.

Inpractice,onecannaturallyimaginealargenumberofvariantsofsuchexperiments,

butthefollowingexamplesillustratethetechniquerathergenerally.

Example3.23:

A

3�

2

�3

5

factorialexperimentin3blocksof9singleexper-

iments

Letusagainconsideranexperimentinwhichthereare5factors:A,B,C,DandE.A

fractionalfactorialdesignconsistsof33=27singleexperiments.Weimaginethatfor

practicalreasons,itcanbeexpedienttodividethese27singleexperimentsinto3blocksof

9;forexampleitcanbediÆculttomaintainuniformexperimentalconditionsthroughout

all27singleexperiments.

The1/32�35factorialexperimentwantedisfoundfromtwogeneratorequations.

Aspreviouslydiscussed,2factors,DandE,areintroducedintoacomplete33factor

structureforthefactorsA,BandC.

Asintheexampleonpage96,wechoosetoputinDandEasinthefollowingtable:

Designgenerators

I A B AB

AB2

C AC

BC

ABC

AB2C

AC2

BC2

=

E

ABC2

AB2C2

=

D

=)

I 1

=

AB2C2D2

I 2

=

BC2E2

Withthisconfounding,onegets(aspreviously)thede�ningrelation

I=AB

2C

2D

2=BC

2E

2=ACD

2E

2=ABD

2E

Constructionoftheexperimentstillfollowstheexampleonpage96,andonecouldperhaps

againchoosetheprincipalfraction(seepage100):

c hs.


106

(1)

ad

a2d

2

bd2e

abe

a2bde

b2de2

ab2d

2e2

a2b2e2

cd2e2

ace

2

a2cde2

bcd

abcd

2

a2bc

b2ce

ab2cde

a2b2cd

2e

c2de

ac2d

2e

a2c2e

bc2e2

abc

2de2

a2bc

2d

2e2

b2c2d

2

ab2c2

a2b2c2d

Tonowdividethisexperimentconsistingofthe27singleexperimentsinto3blocksof9,

onechoosesyetanothergeneratingrelationwhichindicateshowtheblocksareformed.

Whenthisrelationistobechosen,oneagainstartswiththealiasrelationsoftheexper-

imentinsuchareducedformthatonehasanoverviewofhowthemaine�ectsand/or

interactionsofinterestareconfounded.Ifweagainfollowthesameexample,thesere-

ducedaliasrelationscouldbeasshowninthefollowingtable,inwhichwenowalsoput

intheblocks:

I

=

AB2C2D2

=BC2E2

=ACD2E2

=ABD2E

A

=

B

=

CE

AB

=

DE2

AB2

=

CD

C

=

BE2

AC

=

DE

(BC)

=

AD2

=BE=CE2

=blocks

(ABC)

=

AD

(AB2C)

=

AE2

=CD2

AC2

=

BD

(BC2)

=

E

(ABC2)

=

BD2

=AE

(AB2C2)

=

D

Theeasiestwaytowriteoutthethisexperimentisshownonpage110,however

wewilldiscussthedesignalittleindetail.

Thechoiceofconfoundingwithblocksmeansthatalle�ectsintheunderlyingfactor

structurethatarenotconfoundedwithane�ectofinterestcanbeused.Thee�ectBC

couldbesuchane�ect(but,forexample,notBC

2,why?).

Thede�ningcontrastBChastheindexvalue(j+k) 3.Theblockdivisionisthendeter-

minedbywhether(j+k) 3=0,1or2.

c hs.


107

To�ndthe3blocks,onecanagainstartwiththeunderlyingfactorstructureanditcan

beseenthattheblockdivisionissolelydeterminedbyindicesforthefactorsBandC,

namelyjandk.

Aswesawinthepreviousexample,theexperiment,asdescribedabove,wasalsofound

onthebasisoftheunderlyingfactorstructure,andtheblocknumbercorrespondingto

thesingleexperimentsisinsertedinthefollowingtable:

Experiment

block

Experiment

block

Experiment

block

(1)

0

ad

0

a2d

2

0

bd2e

1

abe

1

a2bde

1

b2de2

2

ab2d

2e2

2

a2b2e2

2

cd2e2

1

ace

2

1

a2cde2

1

bcd

2

abcd

2

2

a2bc

2

b2ce

0

ab2cde

0

a2b2cd

2e

0

c2de

2

ac2d

2e

2

a2c2e

2

bc2e2

0

abc

2de2

0

a2bc

2d

2e2

0

b2c2d

2

1

ab2c2

1

a2b2c2d

1

To�ndthethreeblocks,wecouldalsosolvetheequations(modulo3):

Generatorer:

Blocks=BC

D=AB

2C

2

E=BC

2

Block0:

j+k=0

i+2j+2k+2l=0

j+2k+2m=0

Block1:

j+k=1

i+2j+2k+2l=0

j+2k+2m=0

Block2:

j+k=2

i+2j+2k+2l=0

j+2k+2m=0

Forexample2solutionshavetobefoundfor"Block0"whichconsistsof3�3=9single

experiments,andafterthatonefurthersolutionforeachoftheothertwoblocks.

Thestructureofblock0canbeillustrated

(1)

u

u2

v

uv

u2v

v2

uv

2

u2v

2

whereuandvrepresentsolutionstotheequationsforblock0.

Forexample,withi=1andj=0,itisfoundfromj+k=0,thatk=0.Further,

i+2j+2k+2l=0indicatesthatl=1,andfromj+2k+2m=0isfoundthatm=0.

Asolutionistherebyu=ad.

Withi=0,j=1,itisfoundthatk=2,l=0andm=2,fromwhichv=bc

2e2isfound.

c hs.


108

(1)

u=ad

u2=a

2d

2

v=bc

2e2

uv=abc

2de2

u2v=a

2bc

2d

2e2

v2=b2ce

uv

2=ab2cde

u2v

2=a

2b2cd

2e

Anditcanbeseenthatthisispreciselytheblock0foundabove.

To�ndblock1,onesolutionisderivedforj+k=1,i+2j+2k+2l=0andj+2k+2m=0.

Suchasolutionisi=0,j=0,k=1,fromwhichl=2andm=2,correspondingtothe

singleexperimentcd

2e2.

By"multiplying"cd

2e2onthealreadyfoundblock0,block1isformed.Tryityourself.

Block2isfoundbysolvingtheequationsj+k=2,i+2j+2k+2l=0andj+2k+2m=0.

Asolutionisi=0,j=1,k=1,fromwhichl=1andm=0,correspondingtothesingle

experimentbcd.Thissolution"ismultiplied"onblock0,bywhichblock2appears.

Whentheexperimentisanalysed,theblocke�ectisre ectedintheBCe�ecttogether

withtheothere�ectswithwhichBCisconfounded.Inotherwords,theexperimentis

againanalysedonthebasisoftheunderlyingfactorstructuredeterminedbythefactors

A,BandC.

Finallytheexperimentcouldalsobeconstructeddirectlyonthebasisofthegenerators

thatarechosen

I A B AB

AB2

C AC

BC

=

Blocks

ABC

AB2C

AC2

BC2

=

E

ABC2

AB2C2

=

D

andcalculatingthefactorlevelsandblocknumbersasshowninthefollowingtableby

meansofthetabularmethod:

c hs.


109

Experimentaldesign

D=(A+2B+2C) 3,E=(B+2C) 3

andBlock=(B+C) 3

No.

A

B

C

D

E

Block

Experiment

1

0

0

0

0

0

0

(1)

2

1

0

0

1

0

0

ad

3

2

0

0

2

0

0

a2d2

4

0

1

0

2

1

1

bd2e

5

1

1

0

0

1

1

abe

6

2

1

0

1

1

1

a2bde

7

0

2

0

1

2

2

b2de2

8

1

2

0

2

2

2

ab2d2e2

9

2

2

0

0

2

2

a2b2e2

10

0

0

1

2

2

1

cd2e2

11

1

0

1

0

2

1

ace2

12

2

0

1

1

2

1

a2cde2

13

0

1

1

1

0

2

bcd

14

1

1

1

2

0

2

abcd2

15

2

1

1

0

0

2

a2bc

16

0

2

1

0

1

0

b2ce

17

1

2

1

1

1

0

ab2cde

18

2

2

1

2

1

0

a2b2cd2e

19

0

0

2

1

1

2

c2de

20

1

0

2

2

1

2

ac2d2e

21

2

0

2

0

1

2

a2c2e

22

0

1

2

0

2

0

bc2e2

23

1

1

2

1

2

0

abc2de2

24

2

1

2

2

2

0

a2bc2d2e2

25

0

2

2

2

0

1

b2c2d2

26

1

2

2

0

0

1

ab2c2

27

2

2

2

1

0

1

a2b2c2d

Endofexample3.23

Finallytwoexamplesaregiventhatillustratethepracticalprocedureintheconstruction

oftworesolutionIVexperimentsfor8and7factorsrespectively.Theseexperimentsare

ofgreatpracticalrelevance,sincetheyincluderelativelymanyfactorsinrelativelyfew

singleexperiments,namelyonly16.Atthesametime,theexamplesshowdivisioninto2

and4blocks,enablingtheadvantagessuchblockingcanhave.

Example3.24:

A

2�

4

�2

8

factorialin2blocks

Theexperimentcouldbedoneinconnectionwithastudyofthemanufacturingprocess

foradrug,forexample.

Weimaginethatthegivenfactorsandtheirlevelsarecircumstanceswhich,duringman-

ufacture,onenormallyaimstokeepconstant,oratleastwithingivenlimits.Itisthe

e�ectofvariationwithinthesepermittedlimitsthatwewanttostudy.

Eightfactorsarestudiedina2�

4�28factorialintwoblocks.The8factorsare2waiting

c hs.


110

timesduringtwophasesoftheprocess,3temperatures,2pHvaluesandthecontentof

zincinthe�nishedproduct.ThefactorsareorderedsothatfactorAisconsideredthe

mostimportant,whileBisthenextmostimportantetc.

TheexperimentisaresolutionIVexperiment.Undertheassumptionofnegligiblethird

orderinteractions,allmaine�ectscanbeanalysedinthisdesign.

Theexperimentisrandomisedwithintwoblocks,asitisassumedthatitisdoneintwo

facilities(R0

andR1)inparallelexperimentsincompletelyrandomorder.

Theexperimentisconstructedasgiveninthefollowingtables.

Factorsandlevelschosen

Factor

Lowlevel

Highlevel

A:Time Solution1+�ltering

(.)70+30min

(a)30+70min

B:TempMix1

(.)20�1

Æ

C

(b)27�1

Æ

C

C:Time Solution2

(.)30min

(c)100min

D:TempSolution2

(.)5�1

Æ

C

(d)17�1

Æ

C

E:Tempproces

(.)5�1

Æ

C

(e)17�1

Æ

C

F:pHrawproduct1

(.)2.65�0.02

(f)3.25�0.02

G:Zink�nalmix

(.)20.0�g/ml

(g)26.0�g/ml

H:pH�nalmix

(.)7.20�0.02

(h)7.40�0.02

Confoundings

I A B AB C A

CBC

ABC

=

H

D AD

BD

ABD

=

G

CD

ACD

=

F

BCD

=

E

ABCD

=

Blocks

Weusethetabularmethodforcalculatingthelevelsofthefactorsandtheblocknumber

onthebasisoftheunderlyingcompletefactorstructureconsistingoffactorsA,B,Cand

D,asshowninthefollowingtable:

c hs.


111

Experimentaldesign

E=(B+C+D) 2,F=(A+C+D) 2,G=(A+B+D) 2,H=(A+B+C) 2

andFacility=(A+B+C+D) 2

No.

A

B

C

D

E

F

G

H

Experiment

Facility

Randomis.

1

0

0

0

0

0

0

0

0

(1)

R0

9

2

1

0

0

0

0

1

1

1

afgh

R1

4

3

0

1

0

0

1

0

1

1

begh

R1

6

4

1

1

0

0

1

1

0

0

abef

R0

11

5

0

0

1

0

1

1

0

1

cefh

R1

16

6

1

0

1

0

1

0

1

0

aceg

R0

7

7

0

1

1

0

0

1

1

0

bcfg

R0

5

8

1

1

1

0

0

0

0

1

abch

R1

10

9

0

0

0

1

1

1

1

0

defg

R1

12

10

1

0

0

1

1

0

0

1

adeh

R0

3

11

0

1

0

1

0

1

0

1

bdfh

R0

1

12

1

1

0

1

0

0

1

0

abdg

R1

14

13

0

0

1

1

0

0

1

1

cdgh

R0

13

14

1

0

1

1

0

1

0

0

acdf

R1

2

15

0

1

1

1

1

0

0

0

bcde

R1

8

16

1

1

1

1

1

1

1

1

abcdefgh

R0

15

Prescriptionsforthesingleexperiments

Belowareshownthefactorsettingsforthetwo�rstsingleexperimentsandthetwolast

ones.

CarriedoutonfacilityR0

Testnr.FF{1X19

Experiment=bd(fh)

Procesparameter

Levelinexperiment


(.)70+30min

B:TempMix1

(b)27�1

Æ

C

C:Time Solution2

(.)30min

D:TempSolution2

(d)17�1

Æ

C

E:Tempproces

(.)5�1

Æ

C

F:pHrawproduct1

(f)3.25�0.02

G:Zink�nalmix

(.)20.0�g/ml

H:pH�nalmix

(h)7.40�0.02

c hs.


112


Testnr.FF{2X19

Experiment=acd(f)

Procesparameter

Levelinexperiment


(a)30+70min

B:TempMix1

(.)20�1

Æ

C

C:Time Solution2

(c)100min

D:TempSolution2

(d)17�1

Æ

C

E:Tempproces

(.)5�1

Æ

C

F:pHrawproduct1

(f)3.25�0.02

G:Zink�nalmix

(.)20.0�g/ml

H:pH�nalmix

(.)7.20�0.02


Testnr.FF{15X19

Experiment=abcd(efgh)

Procesparameter

Levelinexperiment


(a)30+70min

B:TempMix1

(b)27�1

Æ

C

C:Time Solution2

(c)100min

D:TempSolution2

(d)17�1

Æ

C

E:Tempproces

(e)17�1

Æ

C

F:pHrawproduct1

(f)3.25�0.02

G:Zink�nalmix

(g)26.0�g/ml

H:pH�nalmix

(h)7.40�0.02


Testnr.FF{16X19

Experiment=c(efh)

Procesparameter

Levelinexperiment


(.)70+30min

B:TempMix1

(.)20�1

Æ

C

C:Time Solution2

(c)100min

D:TempSolution2

(.)5�1

Æ

C

E:Tempproces

(e)17�1

Æ

C

F:pHrawproduct1

(f)3.25�0.02

G:Zink�nalmix

(.)20.0�g/ml

H:pH�nalmix

(h)7.40�0.02

Endofexample3.24

c hs.


113

Example3.25:

A

2�

3

�2

7

factorialexperimentin4blocks

Supposethatthereare7factorswhichonewantsstudiedin16singleexperiments.The

�rstfourfactors,A,B,CandDareusedastheunderlyingfactorstructure.Thefactors

E,FandGareputintothisaccordingtothetablebelowinaresolutionIVexperiment.

Atthesametime,onecouldwantthe16singleexperimentsdonein4blocksof4single

experiments.Since4=2�2blockshavetobeused,2de�ningequationsforblockshave

tobechosen.Asuggestionfortheconstructionoftheexperimentaldesigncouldbe:

Generators

I A B AB C A

CBC

ABC

=

blocks

D AD

BD

ABD

=

G

CD

ACD

=

F

BCD

=

E

ABCD

=

blocks

Withthesechoices,thee�ectsABCandABCD,butalsothee�ectABC�ABCDwillbe

confoundedwithblocks.Now,sinceABC�ABCD=D,thisisnotagoodchoice,because

themaine�ectDisobviouslyconfoundedwithblocks.Abetterchoicecouldbe:

Generators

I A B AB C A

CBC

ABC

=

blocks

D AD

BD

ABD

=

G

CD

ACD

=

F

BCD

=

blocks

ABCD

=

E

c hs.


114

ThischoicewillentailthatABC,BCDandABC�BCD=ABwillbeconfoundedwith

blocks.Theexperimentaldesigncanbewrittenoutusingthetabularmethod:

Design

E=(A+B+C+D) 2,F=(A+C+D) 2,G=(A+B+D) 2

andBlock=(A+B+C) 2+2�(B+C+D) 2

Nr

A

B

C

D

E

F

G

Experiment

Block

1

0

0

0

0

0

0

0

(1)

0

2

1

0

0

0

1

1

1

aefg

1

3

0

1

0

0

1

0

1

beg

3

4

1

1

0

0

0

1

0

abf

2

5

0

0

1

0

1

1

0

cef

3

6

1

0

1

0

0

0

1

acg

2

7

0

1

1

0

0

1

1

bcfg

0

8

1

1

1

0

1

0

0

abce

1

9

0

0

0

1

1

1

1

defg

2

10

1

0

0

1

0

0

0

ad

3

11

0

1

0

1

0

1

0

bdf

1

12

1

1

0

1

1

0

1

abdeg

0

13

0

0

1

1

0

0

1

cdg

1

14

1

0

1

1

1

1

0

acdef

0

15

0

1

1

1

1

0

0

bcde

2

16

1

1

1

1

0

1

1

abcdfg

3

Endofexample3.25

c hs.


115

Index

23factorialdesign,16

2kfactorialexperiment,19

3kfactorial,48

aliasrelation,37,41,88,101

confounding,37

generally,95

reduced,94,107

sign,39

balance,21,22,35,78

block,16,70

confounding,20,25,65

construction,25,74,105

e�ect,16,33

level,26

minimal,27

randomisation,48,111

system,26

complementaryexperiment,38,94

completefactorial,9

con�denceinterval,20,36

confounding,24,33,37

block,20,65,70

partial,28,78

contrast,12,13

de�ning,25,65

orthogonal,23

de�ningcontrast,25,37,44,65,73,95

de�ningrelation,25,37,41,70,89

design,9,21

e�ects 2k

experiments,11

generally,11

estimates,12

factore�ects,10

factorialexperiment

2k,9,19

3k,48

pk,48

factors,9

fractional

design,34

factorial,34,86

factorialinblocks,105

generatorequation,39,89,95

Kempthorne,43,48,57

Latinsquare,50,63

modulo2,26,37

modulo3,50

orthogonal,31

contrast,23

pkfactorial,48

parameters,10

partialconfounding,28,78

power(test),32

prime,48,60

principalblock,25,74

principalfraction,37,40,102

pseudocontrast,12

randomisation,36

block,48,111

repetition,16

replication,15

resolution,90

response,9{11

SASexamples,104

signof al

iasrelation,39

standardnotation,11

standardorder,17,62

standardisation,62

tabularmethod,25,40,67,75,87,98,

109,115

underlyingfactorial,39,42,90,92,101

116

Weighingexperiment,34

Yates'algorithm,14,18,42,63

c hs.


117

Myownnotes:

c hs.


118

ts, course 02411, imm, dtuhs/engnote.pdf · 31 2.4 f ractional 2 k factorial design 34 c hs. design...

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