实验设计与数据处理：Design of Experiments and Analysis of Variance

Chapter10DesignofExperimentsandAnalysisofVarianceOne-WayANOVAF-TestTypesof

RegressionModelsExperimentalDesignsOne-WayAnovaCompletelyRandomizedRandomizedBlockTwo-WayAnovaFactorialOne-WayANOVAF-Test1. TeststheEqualityof2orMore(p)PopulationMeans2. VariablesOneNominalScaledIndependentVariable2orMore(p)TreatmentLevelsorClassificationsOneIntervalorRatioScaledDependentVariable3. UsedtoAnalyzeCompletelyRandomizedExperimentalDesignsOne-WayANOVAF-TestAssumptions1. Randomness&IndependenceofErrorsIndependentRandomSamplesareDrawnforeachcondition2. NormalityPopulations(foreachcondition)areNormallyDistributed3. HomogeneityofVariancePopulations(foreachcondition)haveEqualVariancesOne-WayANOVAF-TestHypothesesH0:

1=

2=

3=...=

pAllPopulationMeansareEqualNoTreatmentEffectHa:NotAll

jAreEqualAtLeast1Pop.MeanisDifferentTreatmentEffectNOT

1

2

...

pOne-WayANOVAF-TestHypothesesH0:

1=

2=

3=...=

pAllPopulationMeansareEqualNoTreatmentEffectHa:NotAll

jAreEqualAtLeast1Pop.MeanisDifferentTreatmentEffectNOT

1

2

...

pXf(X)

1

=

2

=

3Xf(X)

1

=

2

3WhyVariances?ObserveonesamplefromeachtreatmentgroupTheirmeansmaybeslightlydifferentHowdifferentisenoughtoconcludepopulationmeansaredifferent?DependsonvariabilitywithineachpopulationHighervarianceinpopulationhighervarianceinmeansStatisticaltestsareconductedbycomparingvariabilitybetweenmeanstovariabilitywithineachsampleTwoPossible

ExperimentOutcomesSametreatmentvariationDifferentrandomvariationACan’trejectequalityofmeans!Rejectequalityofmeans!TwoMorePossible

ExperimentOutcomesSametreatmentvariationDifferentrandomvariationABDifferenttreatmentvariationSamerandomvariationCan’trejectequalityofmeans!RejectReject1. Compares2TypesofVariationtoTestEqualityofMeans2. ComparisonBasisIsRatioofVariances3. IfTreatmentVariationIsSignificantlyGreaterThanRandomVariationthenMeansAreNotEqual4. VariationMeasuresAreObtainedby‘Partitioning’TotalVariationOne-WayANOVA

BasicIdeaOne-WayANOVA

PartitionsTotalVariationOne-WayANOVA

PartitionsTotalVariationTotalvariationOne-WayANOVA

PartitionsTotalVariationVariationduetotreatmentTotalvariationOne-WayANOVA

PartitionsTotalVariationVariationduetotreatmentVariationduetorandomsamplingTotalvariationOne-WayANOVA

PartitionsTotalVariationVariationduetotreatmentVariationduetorandomsamplingTotalvariationSumofSquaresAmongSumofSquaresBetweenSumofSquaresTreatmentAmongGroupsVariationOne-WayANOVA

PartitionsTotalVariationVariationduetotreatmentVariationduetorandomsamplingTotalvariationSumofSquaresWithinSumofSquaresError(SSE)WithinGroupsVariationSumofSquaresAmongSumofSquaresBetweenSumofSquaresTreatment(SST)AmongGroupsVariationTotalVariation

XGroup1Group2Group3Response,XTreatmentVariation

X

X3

X2

X1Group1Group2Group3Response,XRandom(Error)Variation

X2

X1

X3Group1Group2Group3Response,XSS=SSE+SSTButThus,SS=SSE+SSTOne-WayANOVAF-Test

TestStatistic1. TestStatisticF=MST/MSE

MSTIsMeanSquareforTreatmentMSEIsMeanSquareforError2. DegreesofFreedom

1=p-1

2=n-pp=#Populations,Groups,orLevelsn=TotalSampleSizeOne-WayANOVA

SummaryTableSourceofVariationDegreesofFreedomSumofSquaresMeanSquare(Variance)FTreatmentp-1SSTMST=SST/(p-1)MSTMSEErrorn-pSSEMSE=SSE/(n-p)Totaln-1SS(Total)=SST+SSETheFdistributionTwoparametersincreasingeitheronedecreasesF-alpha(exceptforv2<3)I.e.,thedistributiongetssmushedtotheleft

SeeSection9.5

F

v1v2(,)0FOne-WayANOVAF-TestCriticalValue

Ifmeansareequal,F=MST/MSE

1.OnlyrejectlargeF!AlwaysOne-Tail!Fapnp(,)

10RejectH0DoNotRejectH0F©1984-1994T/MakerCo.One-WayANOVAF-TestExampleAsproductionmanager,youwanttoseeif3fillingmachineshavedifferentmeanfillingtimes.Youassign15similarlytrained&experiencedworkers,5permachine,tothemachines.Atthe.05level,isthereadifferenceinmeanfillingtimes?

Mach1 Mach2

Mach3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40F03.89One-WayANOVAF-Test

SolutionH0:

1=

2=

3Ha:NotAllEqual

=.05

1=2

2=12CriticalValue(s):TestStatistic:Decision:Conclusion:Rejectat

=.05ThereIsEvidencePop.MeansAreDifferent

=.05FMSTMSE

2358209211256...SummaryTable

SolutionFromComputerSourceofVariationDegreesofFreedomSumofSquaresMeanSquare(Variance)FTreatment(Machines)3-1=247.164023.582025.60Error15-3=1211.0532.9211Total15-1=1458.2172Reminder:AssumptionsforEqualityofMeansTestIndependentrandomsamplesfromeachpopulationAllpopulationprobabilitiesarenormallydistributedAllpopulationshaveequalvariances (Teststartswithassumptionofequalmeansaswell,butthatmayberejectedasaresultofthetest)Exercise10.26

|Summaryofvaluecondition|MeanStd.Dev.Freq.------------+------------------------------------1|30.6420.035438502|26.21428623.701946423|15.1276615.70324947------------+------------------------------------Total|24.05755420.878451139Exercise10.26

AnalysisofVarianceSourceSSdfMSFProb>F------------------------------------------------------------------------Betweengroups6109.714123054.857057.690.0007Withingroups54045.8255136397.395776------------------------------------------------------------------------Total60155.5396138435.909707Bartlett'stestforequalvariances:chi2(2)=7.1931Prob>chi2=0.027One-WayANOVAF-Test

ThinkingChallengeYou’reatrainerforMicrosoftCorp.Isthereadifferenceinmeanlearningtimesof12peopleusing4differenttrainingmethods(

=.05)?

M1

M2

M3

M4

10 11 13 18

9 16 8 23

5 9 9 25Usethefollowingtable. ©1984-1994T/MakerCo.SummaryTable

(PartiallyCompleted)SourceofVariationDegreesofFreedomSumofSquaresMeanSquare(Variance)FTreatment(Methods)348Error80TotalF04.07One-WayANOVAF-Test

Solution*H0:

1=

2=

3=

4Ha:NotAllEqual

=.05

1=3

2=8CriticalValue(s):TestStatistic:Decision:Conclusion:Rejectat

=.05ThereIsEvidencePop.MeansAreDifferent

=.05FMSTMSE

11610116.SummaryTable

Solution*SourceofVariationDegreesofFreedomSumofSquaresMeanSquare(Variance)FTreatment(Methods)4-1=334811611.6Error12-4=88010Total12-1=1142810.26:condition1vs.2Two-samplettestwithequalvariances------------------------------------------------------------------------------Group|ObsMeanStd.Err.Std.Dev.[95%Conf.Interval]---------+--------------------------------------------------------------------1|5030.642.83343920.0354424.9459936.334012|4226.214293.6572923.7019518.8282433.60033---------+--------------------------------------------------------------------combined|9228.619572.27025321.775524.1099933.12914---------+--------------------------------------------------------------------diff|4.4257144.559216-4.63196313.48339------------------------------------------------------------------------------Degreesoffreedom:90Ho:mean(1)-mean(2)=diff=0Ha:diff<0Ha:diff!=0Ha:diff>0t=0.9707t=0.9707t=0.9707P<t=0.8329P>|t|=0.3343P>t=0.167110.26condition2vs.3Two-samplettestwithequalvariances------------------------------------------------------------------------------Group|ObsMeanStd.Err.Std.Dev.[95%Conf.Interval]---------+--------------------------------------------------------------------2|4226.214293.6572923.7019518.8282433.600333|4715.127662.29055415.7032510.5170119.73831---------+--------------------------------------------------------------------combined|8920.359552.17653320.5333716.0341524.68495---------+--------------------------------------------------------------------diff|11.086634.2207672.69739419.47586------------------------------------------------------------------------------Degreesoffreedom:87Ho:mean(2)-mean(3)=diff=0Ha:diff<0Ha:diff!=0Ha:diff>0t=2.6267t=2.6267t=2.6267P<t=0.9949P>|t|=0.0102P>t=0.0051MultipleComparisonsProblemP{Atleastoneofpintervalsfailstocontainthetruedifference}=1–P{Allcintervalscontainthetruedifferences}=1–(1-alpha)c>alphaIfcomparingmanypairs,needgreaterconfidenceforanyoneofthemthanyouwouldforrejectingequalityofanyonepairMultipleComparisonsProcedure1. TellsWhichPopulationMeansAreSignificantlyDifferentExample:

1=

2

32. PostHocProcedureDoneAfterRejection

ofEqualMeansin

ANOVAOutputFromManyStatisticalcomputerPrograms–variousversions(Tukey,Bonferroni,etc.)10.26MultipleComparisons

(Bonferroni)RowMean-|ColMean|12---------+----------------------2|-4.42571|0.872|3|-15.5123-11.0866|0.0010.029RandomizedBlockDesignTypesof

RegressionModelsExperimentalDesignsOne-WayAnovaCompletelyRandomizedRandomizedBlockTwo-WayAnovaFactorialRandomizedBlockDesign1. ExperimentalUnits(Subjects)AreAssignedRandomlytoBlocksBlocksareAssumedHomogeneous2. OneFactororIndependentVariableofInterest2orMoreTreatmentLevelsorClassifications3.OneBlockingFactorRandomizedBlockDesignFactorLevels:(Treatments)A,B,C,D

ExperimentalUnits

TreatmentsarerandomlyassignedwithinblocksBlock1ACDBBlock2CDBABlock3BADC

...............BlockbDCABRandomizedBlockF-Test1. TeststheEqualityof2orMore(p)PopulationMeans2. VariablesOneNominalScaledIndependentVariable2orMore(p)TreatmentLevelsorClassificationsOneNominalScaledBlockingVariableOneIntervalorRatioScaledDependentVariable3. UsedwithRandomizedBlockDesignsRandomizedBlockF-TestAssumptions1. NormalityProbabilityDistributionofeachBlock-TreatmentcombinationisNormal2. HomogeneityofVarianceProbabilityDistributionsofallBlock-TreatmentcombinationshaveEqualVariancesRandomizedBlockF-TestHypothesesH0:

1=

2=

3=...=

pAllPopulationMeansareEqualNoTreatmentEffectHa:NotAll

jAreEqualAtLeast1Pop.MeanisDifferentTreatmentEffect

1

2

...

pIsWrong

RandomizedBlockF-TestHypothesesH0:

1=

2=

3=...=

pAllPopulationMeansareEqualNoTreatmentEffectHa:NotAll

jAreEqualAtLeast1Pop.MeanisDifferentTreatmentEffect

1

2

...

pIsWrong

Xf(X)

1

=

2

=

3Xf(X)

1

=

2

3TheFRatioforRandomizedBlockDesignsSS=SSE+SSB+SSTRandomizedBlockF-Test

TestStatistic1. TestStatisticF=MST/MSEMSTIsMeanSquareforTreatmentMSEIsMeanSquareforError2. DegreesofFreedom

1=p-1

2=n–b–p+1p=#Treatments,b=#Blocks,n=TotalSampleSizeRandomizedBlockF-TestCriticalValue

Ifmeansareequal,F=MST/MSE

1.OnlyrejectlargeF!AlwaysOne-Tail!Fapnp(,)

10RejectH0DoNotRejectH0F©1984-1994T/MakerCo.RandomizedBlockF-TestExampleYouwishtodeterminewhichoffourbrandsoftireshasthelongesttreadlife.Yourandomlyassignoneofeachbrand(A,B,C,andD)toatirelocationoneachof5cars.Atthe.05level,isthereadifferenceinmeantreadlife?

TireLocationBlockLeftFrontRightFrontLeftRearRightRearCar1A:42,000C:58,000B:38,000D:44,000Car2B:40,000D:48,000A:39,000C:50,000Car3C:48,000D:39,000B:36,000A:39,000Car4A:41,000B:38,000D:42,000C:43,000Car5D:51,000A:44,000C:52,000B:35,000F03.49RandomizedBlockF-Test

SolutionH0:

1=

2=

3=

4Ha:NotAllEqual

=.05

1=3

2=12CriticalValue(s):TestStatistic:Decision:Conclusion:Rejectat

=.05ThereIsEvidencePop.MeansAreDifferent

=.05F=11.9933Exercise10.47

Whatisthepurposeofblockingonweeksinthisstudy?c.Arethemeannumberofwalkersdifferentamongthepromptingconditions?d.Whichpairwisemeansaresignificantlydifferent?e.Whatassumptionsarerequiredfortheanalysisincandd?FactorialExperimentsTypesof

RegressionModelsExperimentalDesignsOne-WayAnovaCompletelyRandomizedRandomizedBlockTwo-WayAnovaFactorialFactorialDesign1. ExperimentalUnits(Subjects)AreAssignedRandomlytoTreatmentsSubjectsareAssumedHomogeneous2. TwoorMoreFactorsorIndependentVariablesEachHas2orMoreTreatments(Levels)3. AnalyzedbyTwo-WayANOVAFactorialDesign

Example

TreatmentFactor2(TrainingMethod)FactorLevelsLevel1Level2Level3Level119hr.

20hr.

22hr.

Factor

1(High)11hr.

17hr.

31hr.

(Motivation)Level227hr.

25hr.

31hr.

(Low)29hr.

30hr.

49hr.

Advantages

ofFactorialDesigns1. SavesTime&Efforte.g.,CouldUseSeparateCompletelyRandomizedDesignsforEachVariable2. ControlsConfoundingEffectsbyPuttingOtherVariablesintoModel3. CanExploreInteractionBetweenVariablesTwo-WayANOVATypesof

RegressionModelsExperimentalDesignsOne-WayAnovaCompletelyRandomizedRandomizedBlockTwo-WayAnovaFactorialTwo-WayANOVA1. TeststheEqualityof2orMorePopulationMeansWhenSeveralIndependentVariablesAreUsed2. SameResultsasSeparateOne-WayANOVAonEachVariableButInteractionCanBeTested3. UsedtoAnalyzeFactorialDesignsTwo-WayANOVAAssumptions1. NormalityPopulationsareNormallyDistributed2. HomogeneityofVariancePopulationshaveEqualVariances3. IndependenceofErrorsIndependentRandomSamplesareDrawnTwo-WayANOVA

DataTableXijk