商务与经济统计习题答案（第8版，中文版）SBE8-SM13

13-1Chapter13AnalysisofVarianceandExperimentalDesignLearningObjectives1.Understandhowtheanalysisofvarianceprocedurecanbeusedtodetermineifthemeansofmorethantwopopulationsareequal.2.Knowtheassumptionsnecessarytousetheanalysisofvarianceprocedure.3.UnderstandtheuseoftheFdistributioninperformingtheanalysisofvarianceprocedure.4.KnowhowtosetupanANOVAtableandinterprettheentriesinthetable.5.Beabletouseoutputfromcomputersoftwarepackagestosolveanalysisofvarianceproblems.6.KnowhowtouseFisher’sleastsignificantdifference(LSD)procedureandFisher’sLSDwiththeBonferroniadjustmenttoconductstatisticalcomparisonsbetweenpairsofpopulationsmeans.7.Understandthedifferencebetweenacompletelyrandomizeddesign,arandomizedblockdesign,andfactorialexperiments.8.Knowthedefinitionofthefollowingterms:comparisonwiseTypeIerrorratepartitioningexperimentwiseTypeIerrorrateblockingfactormaineffectlevelinteractiontreatmentreplicationChapter1313-2Solutions:1.a.x=(30+45+36)/3=3721SSTRkjjjnxx=5(30-37)2+5(45-37)2+5(36-37)2=570MSTR=SSTR/(k-1)=570/2=285b.21SSE(1)kjjjns=4(6)+4(4)+4(6.5)=66MSE=SSE/(nT-k)=66/(15-3)=5.5c.F=MSTR/MSE=285/5.5=51.82F.05=3.89(2degreesoffreedomnumeratorand12denominator)SinceF=51.82F.05=3.89,werejectthenullhypothesisthatthemeansofthethreepopulationsareequal.d.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments570228551.82Error66125.5Total636142.a.x=(153+169+158)/3=16021SSTRkjjjnxx=4(153-160)2+4(169-160)2+4(158-160)2=536MSTR=SSTR/(k-1)=536/2=268b.21SSE(1)kjjjns=3(96.67)+3(97.33)+3(82.00)=828.00MSE=SSE/(nT-k)=828.00/(12-3)=92.00c.F=MSTR/MSE=268/92=2.91F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=2.91F.05=4.26,wecannotrejectthenullhypothesis.d.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments53622682.91Error828992Total136411AnalysisofVarianceandExperimentalDesign13-33.a.4(100)6(85)5(79)8715x21SSTRkjjjnxx=4(100-87)2+6(85-87)2+5(79-87)2=1,020MSTR=SSB/(k-1)=1,020/2=510b.21SSE(1)kjjjns=3(35.33)+5(35.60)+4(43.50)=458MSE=SSE/(nT-k)=458/(15-3)=38.17c.F=MSTR/MSE=510/38.17=13.36F.05=3.89(2degreesoffreedomnumeratorand12denominator)SinceF=13.36F.05=3.89werejectthenullhypothesisthatthemeansofthethreepopulationsareequal.d.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1020251013.36Error4581238.17Total1478144.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1200340080Error300605Total150063b.F.05=2.76(3degreesoffreedomnumeratorand60denominator)SinceF=80F.05=2.76werejectthenullhypothesisthatthemeansofthe4populationsareequal.5.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments12026020Error216723Total33674b.F.05=3.15(2numeratordegreesoffreedomand60denominator)F.05=3.07(2numeratordegreesoffreedomand120denominator)Thecriticalvalueisbetween3.07and3.15SinceF=20mustexceedthecriticalvalue,nomatterwhatitsactualvalue,werejectthenullhypothesisthatthe3populationmeansareequal.Chapter1313-46.Manufacturer1Manufacturer2Manufacturer3SampleMean232821SampleVariance6.674.673.33x=(23+28+21)/3=2421SSTRkjjjnxx=4(23-24)2+4(28-24)2+4(21-24)2=104MSTR=SSTR/(k-1)=104/2=5221SSE(1)kjjjns=3(6.67)+3(4.67)+3(3.33)=44.01MSE=SSE/(nT-k)=44.01/(12-3)=4.89F=MSTR/MSE=52/4.89=10.63F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=10.63F.05=4.26werejectthenullhypothesisthatthemeantimeneededtomixabatchofmaterialisthesameforeachmanufacturer.7.SuperiorPeerSubordinateSampleMean5.755.55.25SampleVariance1.642.001.93x=(5.75+5.5+5.25)/3=5.521SSTRkjjjnxx=8(5.75-5.5)2+8(5.5-5.5)2+8(5.25-5.5)2=1MSTR=SSTR/(k-1)=1/2=.521SSE(1)kjjjns=7(1.64)+7(2.00)+7(1.93)=38.99MSE=SSE/(nT-k)=38.99/21=1.86F=MSTR/MSE=0.5/1.86=0.27F.05=3.47(2degreesoffreedomnumeratorand21denominator)SinceF=0.27F.05=3.47,wecannotrejectthenullhypothesisthatthemeansofthethreepopulationsareequal;thus,thesourceofinformationdoesnotsignificantlyaffectthedisseminationoftheinformation.AnalysisofVarianceandExperimentalDesign13-58.MarketingManagersMarketingResearchAdvertisingSampleMean54.56SampleVariance.8.3.4x=(5+4.5+6)/3=5.1721SSTRkjjjnxx=6(5-5.17)2+6(4.5-5.17)2+6(6-5.17)2=7.00MSTR=SSTR/(k-1)=7.00/2=3.521SSE(1)kjjjns=5(.8)+5(.3)+5(.4)=7.50MSE=SSE/(nT-k)=7.50/(18-3)=.5F=MSTR/MSE=3.5/.50=7.00F.05=3.68(2degreesoffreedomnumeratorand15denominator)SinceF=7.00F.05=3.68,werejectthenullhypothesisthatthemeanperceptionscoreisthesameforthethreegroupsofspecialists.9.RealEstateAgentArchitectStockbrokerSampleMean67.7361.1365.80SampleVariance117.72180.10137.12x=(67.73+61.13+65.80)/3=64.8921SSTRkjjjnxx=15(67.73-64.89)2+15(61.13-64.89)2+15(65.80-64.89)2=345.47MSTR=SSTR/(k-1)=345.47/2=172.7421SSE(1)kjjjns=14(117.72)+14(180.10)+14(137.12)=6089.16MSE=SSE/(nT-k)=6089.16/(45-3)=144.98F=MSTR/MSE=172.74/144.98=1.19F.05=3.22(2degreesoffreedomnumeratorand42denominator)Note:Table4doesnotshowavaluefor2degreesoffreedomnumeratorand42denominator.However,thevalueof3.23correspondingto2degreesoffreedomnumeratorand40denominatorcanbeusedasanapproximation.Chapter1313-6SinceF=1.19F.05=3.23,wecannot