One Way ANOVA: Description
One Way ANOVA: Description
One Way ANOVA: Description
3.2.3.1.OneWayANOVA
3.ProductionProcessCharacterization
3.2.Assumptions/Prerequisites
3.2.3.AnalysisofVarianceModels(ANOVA)
3.2.3.1. OneWayANOVA
Description
Aonewaylayoutconsistsofasinglefactorwithseverallevelsandmultipleobservationsat
eachlevel.Withthiskindoflayoutwecancalculatethemeanoftheobservationswithin
eachlevelofourfactor.Theresidualswilltellusaboutthevariationwithineachlevel.We
canalsoaveragethemeansofeachleveltoobtainagrandmean.Wecanthenlookatthe
deviationofthemeanofeachlevelfromthegrandmeantounderstandsomethingaboutthe
leveleffects.Finally,wecancomparethevariationwithinlevelstothevariationacross
levels.Hencethenameanalysisofvariance.
Model
Itiseasytomodelallofthiswithanequationoftheform:
yij = m + a i + ij
Theequationindicatesthatthejthdatavalue,fromleveli,isthesumofthreecomponents:
thecommonvalue(grandmean),theleveleffect(thedeviationofeachlevelmeanfromthe
grandmean),andtheresidual(what'sleftover).
Estimation
clickhereto
seedetails
ofoneway
value
splitting
ANOVA
tablefor
oneway
case
Estimationfortheonewaylayoutcanbeperformedoneoftwoways.First,wecancalculate
thetotalvariation,withinlevelvariationandacrosslevelvariation.Thesecanbe
summarizedinatableasshownbelowandtestscanbemadetodetermineifthefactorlevels
aresignificant.Thevaluesplittingexampleillustratesthecalculationsinvolved.
Ingeneral,theANOVAtablefortheonewaycaseisgivenby:
Source
Factor
Residual
Corr.Total
SumofSquares
DoF
MeanSquare
F0
I1
MSF=SS F/(I1)
MSF/MSE
I(J1)
MSE=SS E/(I(J1))
IJ1
)
S S F = J (y
y
i.
..
)
S S E = (yij y
i.
)
S S T = (yij y
..
where
y
=
i.
1
J
yij
j=1
and
y
=
..
1
IJ
yij
i=1
j=1
Therowlabeled,"Corr.Total",intheANOVAtablecontainsthecorrectedtotalsumof
squaresandtheassociateddegreesoffreedom(DoF).
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3.2.3.1.OneWayANOVA
Leveleffects
mustsumto
zero
ThesecondwaytoestimateeffectsisthroughtheuseofCLMtechniques.Ifyoulookatthe
modelaboveyouwillnoticethatitisintheformofaCLM.Theonlyproblemisthatthe
modelissaturatedandnouniquesolutionexists.Weovercomethisproblembyapplyinga
constrainttothemodel.Sincetheleveleffectsarejustdeviationsfromthegrandmean,they
mustsumtozero.Byapplyingtheconstraintthattheleveleffectsmustsumtozero,wecan
nowobtainauniquesolutiontotheCLMequations.Mostanalysisprogramswillhandlethis
foryouautomatically.SeethechapteronProcessModelingforamorecompletediscussion
onestimatingthecoefficientsforthesemodels.
Testing
Wearetestingtoseeiftheobserveddatasupportthehypothesisthatthelevelsofthefactor
aresignificantlydifferentfromeachother.Thewaywedothisisbycomparingthewithin
levelvariancstothebetweenlevelvariance.
Ifweassumethattheobservationswithineachlevelhavethesamevariance,wecan
calculatethevariancewithineachlevelandpoolthesetogethertoobtainanestimateofthe
overallpopulationvariance.Thisworksouttobethemeansquareoftheresiduals.
Similarly,iftherereallywerenoleveleffect,themeansquareacrosslevelswouldbean
estimateoftheoverallvariance.Therefore,iftherereallywerenoleveleffect,thesetwo
estimateswouldbejusttwodifferentwaystoestimatethesameparameterandshouldbe
closenumerically.However,ifthereisaleveleffect,thelevelmeansquarewillbehigher
thantheresidualmeansquare.
Itcanbeshownthatgiventheassumptionsaboutthedatastatedbelow,theratioofthelevel
meansquareandtheresidualmeansquarefollowsanFdistributionwithdegreesoffreedom
asshownintheANOVAtable.IftheF0valueissignificantatagivensignificancelevel
(greaterthanthecutoffvalueinaFtable),thenthereisaleveleffectpresentinthedata.
Assumptions
Forestimationpurposes,weassumethedatacanadequatelybemodeledasthesumofa
deterministiccomponentandarandomcomponent.Wefurtherassumethatthefixed
(deterministic)componentcanbemodeledasthesumofanoverallmeanandsome
contributionfromthefactorlevel.Finally,itisassumedthattherandomcomponentcanbe
modeledwithaGaussiandistributionwithfixedlocationandspread.
Uses
TheonewayANOVAisusefulwhenwewanttocomparetheeffectofmultiplelevelsofone
factorandwehavemultipleobservationsateachlevel.Thefactorcanbeeitherdiscrete
(differentmachine,differentplants,differentshifts,etc.)orcontinuous(differentgasflows,
temperatures,etc.).
Example
Let'sextendthemachiningexamplebyassumingthatwehavefivedifferentmachines
makingthesamepartandwetakefiverandomsamplesfromeachmachinetoobtainthe
followingdiameterdata:
Analyze
Machine
3
0.125
0.118
0.123
0.126
0.118
0.127
0.122
0.125
0.128
0.129
0.125
0.126
0.120
0.124
0.125
0.124
0.126
0.127
0.127
0.120
0.128
0.119
0.126
0.129
0.121
UsingANOVAsoftwareorthetechniquesofthevaluesplittingexample,wesummarizethe
datainanANOVAtableasfollows:
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3.2.3.1.OneWayANOVA
Source
SumofSquares
DegreesofFreedom
MeanSquare
F0
Factor
0.000137
0.000034
4.86
Residual
CorrectedTotal
0.000132
0.000269
20
24
0.000007
Test
Bydividingthefactorlevelmeansquarebytheresidualmeansquare,weobtainanF0value
of4.86whichisgreaterthanthecutoffvalueof2.87fromtheFdistributionwith4and20
degreesoffreedomandasignificancelevelof0.05.Therefore,thereissufficientevidenceto
rejectthehypothesisthatthelevelsareallthesame.
Conclusion
Fromtheanalysisofthesedatawecanconcludethatthefactor"machine"hasaneffect.
Thereisastatisticallysignificantdifferenceinthepindiametersacrossthemachineson
whichtheyweremanufactured.
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