商务统计学课件英文版BSFC7e-CH10

Two-SampleTestsandOne-WayANOVAChapter10ObjectivesInthischapter,youlearn:

HowtousehypothesistestingforcomparingthedifferencebetweenThemeansoftwoindependentpopulationsThemeansoftworelatedpopulationsTheproportionsoftwoindependentpopulationsThevariancesoftwoindependentpopulationsThemeansofmorethantwopopulationsTwo-SampleTestsTwo-SampleTestsPopulationMeans,IndependentSamplesPopulationMeans,RelatedSamplesPopulationVariancesGroup1vs.Group2Samegroupbeforevs.aftertreatmentVariance1vs.Variance2Examples:PopulationProportionsProportion1vs.Proportion2DCOVADifferenceBetweenTwoMeansPopulationmeans,independentsamplesGoal:Testhypothesisorformaconfidenceintervalforthedifferencebetweentwopopulationmeans,μ1–μ2

ThepointestimateforthedifferenceisX1–X2*σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalDCOVADifferenceBetweenTwoMeans:IndependentSamplesPopulationmeans,independentsamples*UseSptoestimateunknownσ.UseaPooled-Variance

ttest.σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalUseS1andS2toestimateunknownσ1andσ2.UseaSeparate-variancettestDifferentdatasourcesUnrelatedIndependentSampleselectedfromonepopulationhasnoeffectonthesampleselectedfromtheotherpopulationDCOVAHypothesisTestsfor

TwoPopulationMeansLower-tailtest:H0:μ1

μ2H1:μ1<μ2i.e.,H0:μ1–μ2

0H1:μ1–μ2

<0Upper-tailtest:H0:μ1≤μ2H1:μ1

>

μ2i.e.,H0:μ1–μ2

≤0H1:μ1–μ2

>0Two-tailtest:H0:μ1=μ2H1:μ1

≠

μ2i.e.,H0:μ1–μ2

=0H1:μ1–μ2

≠0TwoPopulationMeans,IndependentSamplesDCOVATwoPopulationMeans,IndependentSamplesLower-tailtest:H0:μ1–μ2

0H1:μ1–μ2

<0Upper-tailtest:H0:μ1–μ2

≤0H1:μ1–μ2

>0Two-tailtest:H0:μ1–μ2

=0H1:μ1–μ2

≠0aa/2a/2a-ta-ta/2tata/2RejectH0iftSTAT<-taRejectH0iftSTAT>taRejectH0iftSTAT<-ta/2

ortSTAT>ta/2

Hypothesistestsforμ1–μ2

DCOVAPopulationmeans,independentsamplesHypothesistestsforµ1-µ2withσ1andσ2unknownandassumedequalAssumptions:

SamplesarerandomlyandindependentlydrawnPopulationsarenormallydistributedorbothsamplesizesareatleast30Populationvariancesareunknownbutassumedequal*σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalDCOVAPopulationmeans,independentsamplesThepooledvarianceis:Theteststatisticis:WheretSTAThasd.f.=(n1+n2–2)(continued)*σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalHypothesistestsforµ1-µ2withσ1andσ2unknownandassumedequalDCOVAPopulationmeans,independentsamplesTheconfidenceintervalfor

μ1–μ2is:Wheretα/2hasd.f.=n1+n2–2*Confidenceintervalforµ1-µ2withσ1andσ2unknownandassumedequalσ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalDCOVAPooled-VariancetTestExampleYouareafinancialanalystforabrokeragefirm.IsthereadifferenceindividendyieldbetweenstockslistedontheNYSE&NASDAQ?Youcollectthefollowingdata:

NYSE

NASDAQ

Number2125Samplemean 3.272.53Samplestddev 1.301.16Assumingbothpopulationsareapproximatelynormalwithequalvariances,is

thereadifferenceinmean

yield(

=0.05)?DCOVAPooled-VariancetTestExample:CalculatingtheTestStatisticTheteststatisticis:(continued)H0:μ1-μ2=0i.e.(μ1=μ2)H1:μ1-μ2≠0i.e.(μ1≠μ2)DCOVAPooled-VariancetTestExample:HypothesisTestSolutionH0:μ1-μ2=0i.e.(μ1=μ2)H1:μ1-μ2≠0i.e.(μ1≠μ2)

=0.05df=21+25-2=44CriticalValues:t=±2.0154TestStatistic:Decision:Conclusion:RejectH0ata=0.05Thereisevidenceofadifferenceinmeans.t02.0154-2.0154.025RejectH0RejectH0.0252.040DCOVAPooled-VariancetTestExample:ConfidenceIntervalforµ1-µ2SincewerejectedH0canwebe95%confidentthatµNYSE>µNASDAQ?95%ConfidenceIntervalforµNYSE-µNASDAQSince0islessthantheentireinterval,wecanbe95%confidentthatµNYSE>µNASDAQDCOVAPopulationmeans,independentsamplesHypothesistestsforµ1-µ2withσ1andσ2unknown,notassumedequalAssumptions:

SamplesarerandomlyandindependentlydrawnPopulationsarenormallydistributedorbothsamplesizesareatleast30Populationvariancesareunknownandcannotbeassumedtobeequal*σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalDCOVAPopulationmeans,independentsamples(continued)*σ1andσ2unknown,assumedequalσ1andσ2unknown,notassumedequalHypothesistestsforµ1-µ2withσ1andσ2unknownandnotassumedequalDCOVATheformulaeforthistestarenotcoveredinthisbook.Seereference8fromthischapterformoredetails.ThistestutilizestwoseparatesamplevariancestoestimatethedegreesoffreedomforthettestSeparate-VariancetTestExampleYouareafinancialanalystforabrokeragefirm.IsthereadifferenceindividendyieldbetweenstockslistedontheNYSE&NASDAQ?Youcollectthefollowingdata:

NYSE

NASDAQ

Number2125Samplemean 3.272.53Samplestddev 1.301.16Assumingbothpopulationsareapproximatelynormalwithunequalvariances,is

thereadifferenceinmean

yield(

=0.05)?DCOVASeparate-VariancetTestExample:CalculatingtheTestStatistic(continued)H0:μ1-μ2=0i.e.(μ1=μ2)H1:μ1-μ2≠0i.e.(μ1≠μ2)DCOVASeparate-VariancetTestExample:HypothesisTestSolutionH0:μ1-μ2=0i.e.(μ1=μ2)H1:μ1-μ2≠0i.e.(μ1≠μ2)

=0.05df=40CriticalValues:t=±2.021TestStatistic:Decision:Conclusion:FailToRejectH0ata=0.05Thereisinsufficientevidenceofadifferenceinmeans.t02.021-2.021.025RejectH0RejectH0.0252.019DCOVARelatedPopulations

ThePairedDifferenceTest

TestsMeansof2RelatedPopulations PairedormatchedsamplesRepeatedmeasures(before/after)Usedifferencebetweenpairedvalues:EliminatesVariationAmongSubjectsAssumptions:DifferencesarenormallydistributedOr,ifnotNormal,uselargesamplesRelatedsamplesDi=X1i-X2iDCOVARelatedPopulations

ThePairedDifferenceTestTheith

paireddifferenceisDi,whereRelatedsamplesDi=X1i-X2i

ThepointestimateforthepaireddifferencepopulationmeanμDisD:nisthenumberofpairsinthepairedsampleThesamplestandarddeviationisSD(continued)DCOVATheteststatisticforμD

is:PairedsamplesWheretSTAThasn-1d.f.ThePairedDifferenceTest:

FindingtSTATDCOVALower-tailtest:H0:μD

0H1:μD<0Upper-tailtest:H0:μD≤0H1:μD

>0Two-tailtest:H0:μD=0H1:μD

≠0PairedSamplesThePairedDifferenceTest:PossibleHypothesesaa/2a/2a-ta-ta/2tata/2RejectH0iftSTAT<-taRejectH0iftSTAT>taRejectH0iftSTAT<-ta/2

ortSTAT>ta/2

WheretSTAThasn-1d.f.DCOVATheconfidenceintervalforμDisPairedsampleswhereThePairedDifferenceConfidenceIntervalDCOVAAssumeyousendyoursalespeopletoa“customerservice”trainingworkshop.Hasthetrainingmadeadifferenceinthenumberofcomplaints?Youcollectthefollowingdata:PairedDifferenceTest:Example

NumberofComplaints:

(2)-(1)Salesperson

Before(1)

After(2)

Difference,

DiC.B. 6

4-2T.F. 20

6-14M.H. 3

2-1R.K. 0

00M.O. 4

0

-4 -21

D=Din

=-4.2DCOVAHasthetrainingmadeadifferenceinthenumberofcomplaints(atthe0.01level)?

-4.2D=H0:μD=0H1:

μD

0TestStatistic:t0.005=±4.604

d.f.=n-1=4Reject

/2

-4.6044.604Decision:

DonotrejectH0(tstatisnotintherejectionregion)Conclusion:

Thereisinsufficientofachangeinthenumberofcomplaints.PairedDifferenceTest:SolutionReject

/2

-1.66

=.01DCOVATheconfidenceintervalforμDis:Sincethisintervalcontains0youare99%confidentthatμD=0ThePairedDifferenceConfidenceInterval--ExampleDCOVAD=-4.2,SD=5.67TwoPopulationProportionsGoal:testahypothesisorformaconfidenceintervalforthedifferencebetweentwopopulationproportions, π1–π2

ThepointestimateforthedifferenceisPopulationproportionsAssumptions:

n1π1

5,n1(1-π1)5n2π2

5,n2(1-π2)5

DCOVATwoPopulationProportionsPopulationproportionsThepooledestimatefortheoverallproportionis:whereX1andX2arethenumberofitemsofinterestinsamples1and2Inthenullhypothesisweassumethenullhypothesisistrue,soweassumeπ1=π2andpoolthetwosampleestimatesDCOVATwoPopulationProportionsPopulationproportionsTheteststatisticforπ1–π2isaZstatistic:(continued)whereDCOVAHypothesisTestsfor

TwoPopulationProportionsPopulationproportionsLower-tailtest:H0:π1

π2H1:π1<π2i.e.,H0:π1–π2

0H1:π1–π2

<0Upper-tailtest:H0:π1≤π2H1:π1

>

π2i.e.,H0:π1–π2

≤0H1:π1–π2

>0Two-tailtest:H0:π1=π2H1:π1

≠

π2i.e.,H0:π1–π2

=0H1:π1–π2

≠0DCOVAHypothesisTestsfor

TwoPopulationProportionsPopulationproportionsLower-tailtest:H0:π1–π2

0H1:π1–π2

<0Upper-tailtest:H0:π1–π2

≤0H1:π1–π2

>0Two-tailtest:H0:π1–π2

=0H1:π1–π2

≠0aa/2a/2a-za-za/2zaza/2RejectH0ifZSTAT<-ZaRejectH0ifZSTAT>ZaRejectH0ifZSTAT<-Za/2

orZSTAT>Za/2

(continued)DCOVAHypothesisTestExample:

TwopopulationProportionsIsthereasignificantdifferencebetweentheproportionofmenandtheproportionofwomenwhowillvoteYesonPropositionA?Inarandomsample,36of72menand35of50womenindicatedtheywouldvoteYesTestatthe.05levelofsignificanceDCOVAThehypothesistestis:H0:π1–π2

=0(thetwoproportionsareequal)H1:π1–π2

≠0(thereisasignificantdifferencebetweenproportions)Thesampleproportionsare:Men: p1=36/72=0.50Women: p2=35/50=0.70Thepooledestimatefortheoverallproportionis:HypothesisTestExample:

TwopopulationProportions(continued)DCOVATheteststatisticforπ1–π2is:HypothesisTestExample:

TwoPopulationProportions(continued).025-1.961.96.025-2.20Decision:

RejectH0Conclusion:

Thereisevidenceofasignificantdifferenceintheproportionofmenandwomenwhowillvoteyes.RejectH0RejectH0CriticalValues=±1.96For=.05DCOVAConfidenceIntervalfor

TwoPopulationProportionsPopulationproportionsTheconfidenceintervalfor

π1–π2is:DCOVAConfidenceIntervalforTwoPopulationProportions--ExampleThe95%confidenceintervalforπ1–π2is:Sincethisintervaldoesnotcontain0canbe95%confidentthetwoproportionsaredifferent.DCOVATestingfortheRatioOfTwoPopulationVariancesTestsforTwoPopulationVariancesFteststatisticH0:σ12=σ22H1:σ12≠σ22H0:σ12≤σ22H1:σ12>σ22*Hypotheses FSTATS12/S22S12=Varianceofsample1(thelargersamplevariance)n1=samplesizeofsample1S22=Varianceofsample2(thesmallersamplevariance)n2=samplesizeofsample2n1–1=numeratordegreesoffreedomn2–1=denominatordegreesoffreedomWhere:DCOVATheFcriticalvalue

isfoundfromtheFtableTherearetwodegreesoffreedomrequired:numeratoranddenominatorThelargersamplevarianceisalwaysthenumeratorWhenIntheFtable,numeratordegreesoffreedomdeterminethecolumndenominatordegreesoffreedomdeterminetherowTheFDistributiondf1=n1–1;df2=n2–1DCOVAFindingtheRejectionRegionH0:σ12=σ22H1:σ12≠σ22H0:σ12≤σ22H1:σ12>σ22F

0

Fα

RejectH0DonotrejectH0RejectH0ifFSTAT>FαF

0

/2RejectH0DonotrejectH0Fα/2

RejectH0ifFSTAT>Fα/2DCOVAFTest:AnExampleYouareafinancialanalystforabrokeragefirm.YouwanttocomparedividendyieldsbetweenstockslistedontheNYSE&NASDAQ.Youcollectthefollowingdata:

NYSE

NASDAQ

Number 21 25Mean 3.27 2.53Stddev 1.30 1.16Isthereadifferenceinthe variancesbetweentheNYSE &NASDAQatthe

=

0.05level?DCOVAFormthehypothesistest:H0:σ21=σ22(thereisnodifferencebetweenvariances)H1:σ21≠σ22(thereisadifferencebetweenvariances)FTest:ExampleSolutionFindtheFcriticalvaluefor

=0.05:Numeratord.f.=n1–1=21–1=20Denominatord.f.=n2–1=25–1=24Fα/2=F.025,20,24=2.33DCOVATheteststatisticis:0

/2=.025F0.025=2.33RejectH0DonotrejectH0H0:σ12=σ22H1:σ12

≠

σ22FTest:ExampleSolutionFSTAT=1.256isnotintherejectionregion,sowedonotrejectH0(continued)Conclusion:Thereisnotsufficientevidenceofadifferenceinvariancesat=.05F

DCOVAGeneralANOVASettingInvestigatorcontrolsoneormorefactorsofinterestEachfactorcontainstwoormorelevelsLevelscanbenumericalorcategoricalDifferentlevelsproducedifferentgroupsThinkofeachgroupasasamplefromadifferentpopulationObserveeffectsonthedependentvariableArethegroupsthesame?Experimentaldesign:theplanusedtocollectthedataDCOVACompletelyRandomizedDesignExperimentalunits(subjects)areassignedrandomlytogroupsSubjectsareassumedhomogeneousOnlyonefactororindependentvariableWithtwoormorelevelsAnalyzedbyone-factoranalysisofvariance(ANOVA)DCOVAOne-WayAnalysisofVarianceEvaluatethedifferenceamongthemeansofthreeormoregroupsExamples:Numberofaccidentsfor1st,2nd,and3rdshiftExpectedmileageforfivebrandsoftiresAssumptionsPopulationsarenormallydistributedPopulationshaveequalvariancesSamplesarerandomlyandindependentlydrawnDCOVAHypothesesofOne-WayANOVA

Allpopulationmeansareequali.e.,nofactoreffect(novariationinmeansamonggroups)

Atleastonepopulationmeanisdifferenti.e.,thereisafactoreffectDoesnotmeanthatallpopulationmeansaredifferent(somepairsmaybethesame)DCOVAOne-WayANOVAWhenTheNullHypothesisisTrueAllMeansarethesame:(NoFactorEffect)DCOVAOne-WayANOVAWhenTheNullHypothesisisNOTtrueAtleastoneofthemeansisdifferent(FactorEffectispresent)or(continued)DCOVAPartitioningtheVariationTotalvariationcanbesplitintotwoparts:SST=TotalSumofSquares

(Totalvariation)SSA=SumofSquaresAmongGroups

(Among-groupvariation)SSW=SumofSquaresWithinGroups

(Within-groupvariation)SST=SSA+SSWDCOVAPartitioningtheVariationTotalVariation=theaggregatevariationoftheindividualdatavaluesacrossthevariousfactorlevels(SST)Within-GroupVariation=variationthatexistsamongthedatavalueswithinaparticularfactorlevel(SSW)Among-GroupVariation=variationamongthefactorsamplemeans(SSA)SST=SSA+SSW(continued)DCOVAPartitionofTotalVariationVariationDuetoFactor(SSA)VariationDuetoRandomError(SSW)TotalVariation(SST)=+DCOVATotalSumofSquaresWhere:

SST=Totalsumofsquares c=numberofgroupsorlevels

nj=numberofvaluesingroupj

Xij=ithobservationfromgroupj

X=grandmean(meanofalldatavalues)SST=SSA+SSWDCOVATotalVariation(continued)DCOVAAmong-GroupVariationWhere:

SSA=Sumofsquaresamonggroups c=numberofgroups

nj=samplesizefromgroupj

Xj=samplemeanfromgroupj

X=grandmean(meanofalldatavalues)SST=SSA+SSWDCOVAAmong-GroupVariationVariationDuetoDifferencesAmongGroupsMeanSquareAmong=SSA/degreesoffreedom(continued)DCOVAAmong-GroupVariation(continued)DCOVAWithin-GroupVariationWhere:

SSW=Sumofsquareswithingroups c=numberofgroups

nj=samplesizefromgroupj

Xj=samplemeanfromgroupj

Xij=ithobservationingroupjSST=SSA+SSWDCOVAWithin-GroupVariationSummingthevariationwithineachgroupandthenaddingoverallgroupsMeanSquareWithin=SSW/degreesoffreedom(continued)DCOVAWithin-GroupVariation(continued)DCOVAObtainingtheMeanSquaresTheMeanSquaresareobtainedbydividingthevarioussumofsquaresbytheirassociateddegreesoffreedomMeanSquareAmong(d.f.=c-1)MeanSquareWithin(d.f.=n-c)MeanSquareTotal(d.f.=n-1)DCOVAOne-WayANOVATableSourceofVariationSumOfSquaresDegreesofFreedomMeanSquare(Variance)AmongGroupsc-1MSA=WithinGroupsSSWn-cMSW=TotalSSTn–1SSAMSAMSWFc=numberofgroupsn=sumofthesamplesizesfromallgroupsdf=degreesoffreedomSSAc-1SSWn-cFSTAT=DCOVAOne-WayANOVA

FTestStatisticTeststatistic

MSAismeansquaresamonggroups MSWismeansquareswithingroupsDegreesoffreedomdf1=c–1(c=numberofgroups)df2=n–c(n=sumofsamplesizesfromallpopulations)H0:μ1=μ2=…

=μcH1:AtleasttwopopulationmeansaredifferentDCOVAInterpretingOne-WayANOVA

FStatisticTheFstatisticistheratiooftheamongestimateofvarianceandthewithinestimateofvarianceTheratiomustalwaysbepositivedf1=c-1willtypicallybesmall

df2=n-cwilltypicallybelargeDecisionRule:RejectH0ifFSTAT>Fα,otherwisedonotrejectH00

RejectH0DonotrejectH0FαDCOVAOne-WayANOVA

FTestExampleYouwanttoseeifthreedifferentgolfclubsyielddifferentdistances.Yourandomlyselectfivemeasurementsfromtrialsonanautomateddrivingmachineforeachclub.Atthe0.05significancelevel,isthereadifferenceinmeandistance?

Club1

Club2

Club3

254 234 200

263 218 222

241 235 197

237 227 206

251 216 204DCOVA•••••One-WayANOVAExample:

ScatterPlot270260250240230220210200190••••••••••Distance

Club1

Club2

Club3

254 234 200

263 218 222

241 235 197

237 227 206

251 216 204Club123DCOVAOne-WayANOVAExampleComputations

Club1

Club2

Club3

254 234 200

263 218 222

241 235 197

237 227 206

251 216 204X1=249.2X2=226.0X3=205.8X=227.0n1=5n2=5n3=5n=15c=3SSA=5(249.2–227)2+5(226–227)2+5(205.8–227)2=4716.4SSW=(254–249.2)2+(263–249.2)2+…+(204–205.8)2=1119.6MSA=4716.4/(3-1)=2358.2MSW=1119.6/(15-3)=93.3DCOVAOne-WayANOVAExampleSolutionH0:μ1=μ2=μ3H1:μjnotallequal

=0.05df1=2df2=12TestStatistic:Decision:Conclusion:RejectH0at

=0.05Thereisevidencethatatleastoneμjdiffersfromtherest0

=.05F0.05=3.89RejectH0DonotrejectH0CriticalValue:Fα

=3.89DCOVASUMMARYGroupsCountSumAverageVarianceClub151246249.2108.2Club25113022677.5Club351029205.894.2ANOVASourceofVariationSSdfMSFP-valueFcritBetweenGroups4716.422358.225.2750.00003.89WithinGroups1119.61293.3Total5836.014

One-WayANOVAExcelOutputDCOVAOne-WayANOVA

MinitabOutputOne-wayANOVA:DistanceversusClubSourceDFSSMSFPClub24716.42358.225.280.000Error121119.693.3Total145836.0S=9.659R-Sq=80.82%R-Sq(adj)=77.62%Individual95%CIsForMeanBasedonPooledStDevLevelNMeanStDev-------+---------+---------+---------+--15249.2010.40(-----*-----)25226.008.80(-----*-----)35205.809.71(-----*-----)-------+---------+---------+---------+--208224240256PooledStDev=9.66DCOVAANOVAAssumptionsRandomnessandIndependenceSelectrandomsamplesfromthecgroups(orrandomlyassignthelevels)NormalityThesamplevaluesforeachgrouparefromanormalpopulationHomogeneityofVarianceAllpopulationssampledfromhavethesamevarianceCanbetestedwithLevene’sTestDCOVAANOVAAssumptions

Levene’sTestTeststheassumptionthatthevariancesofeachpopulationareequal.First,definethenullandalternativehypotheses:H0:σ21=σ22=…=σ2cH1:Notallσ2jareequalSecond,computetheabsolutevalueofthedifferencebetweeneach