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Revision:1.00Date:June20016西格玛绿带培训MaterialsTWO-6-4-20246标准偏差 第二天:TestsofHypothesesWeek1recapofStatisticsTerminologyIntroductiontoStudentTdistributionExampleinusingStudentTdistributionSummaryofformulaforConfidenceLimitsIntroductiontoHypothesisTestingTheelementsofHypothesisTesting-----------------------------------------------------Break--------------------------------------------------------------------LargesampleTestofHypothesisaboutapopulationmeanp-Values,theobservedsignificancelevelsSmallsampleTestofHypothesisaboutapopulationmeanMeasuringthepowerofhypothesistestingCalculatingTypeIIErrorprobabilitiesHypothesisExerciseI-----------------------------------------------------Lunch--------------------------------------------------------------------HypothesisExerciseIPresentationComparing2populationMeans:IndependentSamplingComparing2populationMeans:PairedDifferenceExperimentsComparing2populationProportions:F-Test-----------------------------------------------------Break--------------------------------------------------------------------HypothesisTestingExerciseII(paperclip)HypothesisTestingPresentation第一天wrapup 第二天:Analysisofvariance和simplelinearregressionChi-square:AtestofindependenceChi-square:InferencesaboutapopulationvarianceChi-squareexerciseANOVA-AnalysisofvarianceANOVA–Analysisofvariancecasestudy-----------------------------------------------------Break--------------------------------------------------------------------TestingthefittnessofaprobabilitydistributionChi-square:agoodnessoffittestTheKolmogorov-SmirnovTestGoodnessoffitexerciseusingdiceResult和discussiononexercise------------------------------------------------------Lunch-------------------------------------------------------------------Probabilistic关系hipofaregressionmodelFittingmodelwithleastsquareapproachAssumptions和varianceestimatorMakinginferenceabouttheslopeCoefficientofCorrelation和DeterminationExampleofsimplelinearregressionSimplelinearregressionexercise(usingstatapult)------------------------------------------------------Break-------------------------------------------------------------------Simplelinearregressionexercise(con’t)Presentationofresults第二天wrapup Day3:Multipleregression和modelbuildingIntroductiontomultipleregressionmodelBuildingamodelFittingthemodelwithleastsquaresapproachAssumptionsformodelUsefulnessofamodelAnalysisofvarianceUsingthemodelforestimation和predictionPitfallsinpredictionmodel--------------------------------------------------------Break-----------------------------------------------------------------Multipleregressionexercise(statapult)Presentationformultipleregressionexercise--------------------------------------------------------Lunch------------------------------------------------------------------Qualitativedata和dummyvariablesModelswith2ormorequantitativeindependentvariablesTestingthemodelModelswithonequalitativeindependentvariableComparingslopes和responsecurve--------------------------------------------------------Break-----------------------------------------------------------------ModelbuildingexampleStepwiseregression–anapproachtoscreenoutfactorsDay3wrapup Day4:设计ofExperimentOverviewofExperimentalDesignWhatisadesignedexperimentObjectiveofexperimental设计和itscapabilityinidentifyingtheeffectoffactorsOnefactoratatime(OFAT)versus设计ofexperiment(DOE)formodellingOrthogonality和itsimportancetoDOEH和calculationforbuildingsimplelinearmodelType和usesofDOE,(i.e.linearscreening,linearmodelling,和non-linearmodelling)OFATversusDOE和itsimpactinascreeningexperimentTypesofscreeningDOEs---------------------------------------------------Break----------------------------------------------------------------------PointstonotewhenconductingDOEScreeningDOEexerciseusingstatapultInterpretatingthescreeningDOE’sresult---------------------------------------------------Lunch----------------------------------------------------------------------ModellingDOE(Fullfactoriawithinteractions)InterpretinginteractionoffactorsParetooffactorssignificanceGraphicalinterpretationofDOEresults某些rulesofthumbinDOE实例ofModellingDOE和itsanalysis--------------------------------------------------Break-----------------------------------------------------------------------ModellingDOEexercisewithstatapultTargetpractice和confirmationrunDay4wrapup Day5:Statistical流程ControlWhatisStatistical流程ControlControlchart–thevoiceofthe流程流程controlversus流程capabilityTypesofcontrolchartavailable和itsapplicationObservingtrendsforcontrolchartOutofControlreactionIntroductiontoXbarRChartXbarRChartexampleAssignable和ChancecausesinSPCRuleofthumbforSPCruntest-------------------------------------------------------------Break------------------------------------------------------------XbarRChartexercise(usingDice)IntroductiontoXbarSChartImplementingXbarSChart为什么XbarSChart?IntroductiontoIndividualMovingRangeChartImplementingIndividualMovingRangeChart为什么XbarSChart?-------------------------------------------------------------Lunch------------------------------------------------------------Choosingthesub-groupChoosingthecorrectsamplesizeSamplingfrequencyIntroductiontocontrolchartsforattributedatanpCharts,pCharts,cCharts,uCharts-------------------------------------------------------------Break------------------------------------------------------------Attributecontrolchartexercise(paperclip)OutofcontrolnotnecessarilyisbadDay5wrapup RecapofStatisticalTerminologyDistributionsdiffersinlocationDistributionsdiffersinspreadDistributionsdiffersinshapeNormalDistribution-6-5-4-3-2-10123456------------------------------99.9999998%-------------------------------------------99.73%--------------------95.45%---------68.27%--±3variationiscallednaturaltoleranceAreaunderaNormalDistribution 流程capabilitypotential,CpBasedontheassumptionsthat:流程isnormalNormalDistribution-6-5-4-3-2-10123456LowerSpecLimitLSLUpperSpecLimitUSLSpecificationCenterItisa2-sidedspecification流程meaniscenteredtothedevicespecificationSpreadinspecificationNaturaltoleranceCP=USL-LSL686=1.33 流程CapabilityIndex,CpkBasedontheassumptionthatthe流程isnormal和incontrol2.Anindexthatcomparethe流程centerwithspecificationcenterNormalDistribution-6-5-4-3-2-10123456LowerSpecLimitLSLUpperSpecLimitUSLSpecificationCenterThereforewhen,Cpk20)Estimated标准偏差,R/d2Population标准偏差,(whensamplesize,n20) ProbabilityTheoryProbabilityisthechanceforaneventtooccur.Statisticaldependence/independencePosteriorprobabilityRelativefrequencyMakedecisionthroughprobabilitydistributions(i.e.Binomial,Poisson,Normal)CentralLimitTheoremRegardlesstheactualdistributionofthepopulation,thedistributionofthemeanforsub-groupsofsamplefromthatdistribution,willbenormallydistributedwithsamplemeanapproximatelyequaltothepopulationmean.Setconfidenceintervalforsamplebasedonnormaldistribution.Abasistocomparesamplesusingnormaldistribution,hencemakingstatisticalcomparisonoftheactualpopulations.Itdoesnotimpliesthatthepopulationisalwaysnormallydistributed.(Cp,Cpkmustalwaysbasedontheassumptionthat流程isnormal) InferentialStatisticsThe流程ofinterpretingthesampledatatodrawconclusionsaboutthepopulationfromwhichthesamplewastaken.ConfidenceInterval(Determineconfidencelevelforasamplingmeantofluctuate)T-Test和F-Test(Determineiftheunderlyingpopulationsissignificantlydifferentintermsofthemeans和variations)Chi-SquareTestofIndependence(Testifthesampleproportionsaresignificantlydifferent)Correlation和Regression(Determineif关系hipbetweenvariablesexists,和generatemodelequationtopredicttheoutcomeofasingleoutputvariable) CentralLimitTheoremThemeanxofthesamplingdistributionwillapproximatelyequaltothepopulationmeanregardlessofthesamplesize.Thelargerthesamplesize,thecloserthesamplemeanistowardsthepopulationmean.2.Thesamplingdistributionofthemeanwillapproachnormalityregardlessoftheactualpopulationdistribution.3.Itassuresusthatthesamplingdistributionofthemeanapproachesnormalasthesamplesizeincreases.m=150Populationdistributionx=150Samplingdistribution(n=5)x=150Samplingdistribution(n=20)x=150Samplingdistribution(n=30)m=150Populationdistributionx=150Samplingdistribution(n=5) 某些takeawaysforsamplesize和samplingdistributionForlargesamplesize(i.e.n30),thesamplingdistributionofxwillapproachnormalityregardlesstheactualdistributionofthesampledpopulation.Forsmallsamplesize(i.e.n<30),thesamplingdistributionofxisexactlynormalifthesampledpopulationisnormal,和willbeapproximatelynormalifthesampledpopulationisalsoapproximatelynormallydistributed.Thepointestimateofpopulation标准偏差usingSequationmay提供apoorestimationifthesamplesizeissmall.IntroductiontoStudenttDistrbutionDiscoveredin1908byW.S.GossetfromGuinnessBreweryinIreland.Tocompensatefor标准偏差dependenceonsmallsamplesize.Containtworandomquantities(x和S),whereasnormaldistributioncontainsonlyonerandomquantity(xonly)Assamplesizeincreases,thetdistributionwillbecomeclosertothatofstandardnormaldistribution(orzdistribution). PercentilesofthetDistributionWhereby,df=Degreeoffreedom=n(samplesize)–1Shadedarea=one-tailedprobabilityofoccurencea=1–ShadedareaApplicablewhen:Samplesize<30标准偏差isunknownPopulationdistributionisatleastapproximatelynormallydistributedt(a,u)aAreaunderthecurve PercentilesoftheNormalDistribution/ZDistributionZaAreaunderthecurveWhereby,Shadedarea=one-tailedprobabilityofoccurencea=1–Shadedarea StudenttDistrbutionexampleFDArequirespharmaceuticalcompaniestoperformextensivetestsonallnewdrugsbeforetheycanbemarketedtothepublic.Thefirstphaseoftestingwillbeonanimals,whilethesecondphasewillbeonhumanonalimitedbasis.PWDisapharmaceuticalcompanycurrentlyinthesecondphaseoftestingonanewantibioticproject.Thechemistsareinterestedtoknowtheeffectofthenewantibioticonthehumanbloodpressure,和theyareonlyallowedtoteston6patients.Theresultoftheincreaseinbloodpressureofthe6testedpatientsareasbelow:(1.7,3.0,0.8,3.4,2.7,2.1)Constructa95%confidenceintervalfortheaverageincreaseinbloodpressureforpatientstakingthenewantibiotic,usingbothnormal和tdistributions. StudenttDistrbutionexample(con’t)UsingnormalorzdistributionUsingstudenttdistributionAlthoughtheconfidencelevelisthesame,usingtdistributionwillresultinalargerintervalvalue,because:标准偏差,Sforsmallsamplesizeisprobablynotaccurate标准偏差,SforsmallsamplesizeisprobablytoooptimisticWiderintervalisthereforenecessarytoachievetherequiredconfidencelevel Summaryofformulaforconfidencelimit 6Sigma流程和1.5SigmaShiftinMeanStatistically,a流程thatis6Sigmawithrespecttoitsspecificationsis:NormalDistribution-6-5-4-3-2-10123456------------------------------99.9999999998%----------------------------LSLUSLDPM=0.002Cp=2Cpk=2ButMotoroladefines6Sigmawithascenarioof1.5SigmashiftinmeanDPM=3.4Cp=2Cpk=1.51.5 某些Explanationson1.5SigmaMeanShiftMotorlahasconductedalotofexperiments,和foundthatinlongterm,the流程meanwillshiftwithin1.5sigmaifthe流程isundercontrol.1.5sigmameanshiftina3Sigma流程controlplanwillbetranslatedtoapproximately14%ofthetimeadatapointwillbeoutofcontrol,和thisisdeemacceptableinstatistical流程control(SPC)practices.NormalDistribution-3-2-10123------------------99.74%-----------------LCLUCLDistributionwith1.5SigmaShift-3-2-10123-----------------86.64%----------------LCLUCLOutofcontroldatapoints OurExplanationMostfrequentlyusedsamplesizeforSPCinindustryis3to5unitspersampling.Takethemiddlevalueof4asanaveragesamplesizeusedinthesampling.Assumingthe流程isof6sigmacapability,isincontrol,和isnormallydistributed.Undertheconfidenceintervalforsamplingdistribution,weexpecttheaveragevalueofthesamplestofluctuatewithin3standarderrors(i.e.naturaltolerance),givingconfidenceintervalof: IntroductiontoHypothesisTesting?Whatishypothesistestinginstatistic?Ahypothesisis“atentativeassumptionmadeinordertodrawoutortestitslogicalorempiricalconsequences.”Astatisticalhypothesisisastatementaboutthevalueofoneofthecharacteristicsforoneormorepopulations.Thepurposeofthehypothesisistoestablishabasis,sothatonecangatherevidencetoeitherdisprovethestatementoracceptitastrue.ExampleofstatisticalhypothesisTheaveragecommutetimeusingHighway92isshorterthanusingFranceAvenue.This流程changewillnotcauseanyeffectonthedownstream流程es.ThevariationofVendorB’spartsare40%widerthanthoseofVendorA. ElementsofHypothesisTestingPossibleoutcomesforhypothesistestingontwotestedpopulations:NoSignificantDifferenceSignificantDifferenceinVariationSignificantDifferenceinMeanSignificantDifferenceinbothMean和Variationm1<>m21=2m1<>m21<>2m1=m21<>2m1=m21=2 为什么HypothesisTesting?Manyproblemsrequireadecisiontoacceptorrejectastatementaboutaparameter.ThatstatementisaHypothesis.Itrepresentsthetranslationofapracticalquestionintoastatisticalquestion.Statisticaltesting提供sanobjectivesolution,withknownrisks,toquestionswhicharetraditionallyansweredsubjectively.Itisasteppingstoneto设计ofExperiment,DOE.HypothesisTestingDescriptionsHypothesisTestinganswersthepracticalquestion:“IstherearealdifferencebetweenA和B?”Inhypothesistesting,relativelysmallsamplesareusedtoanswerquestionsaboutpopulationparameters.Thereisalwaysachancethatasamplethatisnotrepresentativeofthepopulationbeingselected和resultsindrawingawrongconclusion. ElementsofHypothesisTesting(con’t)TheNullHypothesisStatementgenerallyassumedtobetrueunlesssufficientevidenceisfoundtobecontraryOftenassumedtobethestatusquo,orthepreferredoutcome.However,itsometimesrepresentsastateyoustronglywanttodisprove.DesignatedasH0Inhypothesistesting,wealwaysbiastowardnullhypothesisTheAlternativeHypothesis(orResearchHypothesis)Statementthatwillbeacceptedonlyifdata提供convincingevidenceofitstruth(i.e.byrejectingthenullhypothesis).Insteadofcomparingtwopopulations,itcanalsobebasedonaspecificengineeringdifferenceinacharacteristicvaluethatonedesirestodetect(i.e.insteadofaskingism1=m2,weaskism1>450).DesignatedasH1 ElementsofHypothesisTesting(con’t)Exampleifwewanttotestwhetherapopulationmeanisequalto500,wewouldtranslateitto:NullHypothesis,H0:mp=500和consideralternatehypothesisas:AlternateHypothesis,H1:mp<>500;(2tailstest)Rememberconfidenceinterval,at95%confidencelevelstatesthat:95%ofthetimethemeanvaluewillfluctuatewithintheconfidenceinterval(limit)5%chancethatthemeanisnaturalfluctuation,butwethinkitisnot–alpha(a)probability---Confidencelimit---mH0=5000.025ofarea0.025ofarea(a/2)rejectarea(a/2)rejectarea1.96stderror1.96stderrorTypeIIErrorAcceptinganullhypothesis(H0),whenitisfalse.ProbabilityofthiserrorequalsbTypeIErrorRejectingthenullhypothesis(H0),whenitistrue.ProbabilityofthiserrorequalsaIfmpiswithinconfidencelimit,acceptthenullhypothesisH0.Ifmpisinrejectarea,rejectthenullhypothesisH0.Usethestderrorobservedfromthesampletosetconfidencelimiton500(mH0).TheassumptionismH0hasthesamevarianceasmp. ElementsofHypothesisTesting(con’t)Otherpossiblealternatehypothesisare:AlternateHypothesis,H1:mp>500;(1tailtest)AlternateHypothesis,H1:mp<500;(1tailtest)1.645stderrorAcceptanceareamH0=5000.05ofarea(a)rejectareaTakingexampleforalternatehypothesis,H1:mp>500For95%confidencelevel,a=0.05.SinceH1isonetailtest,rejectareadoesnotneedtobedividedby2.Fromstandardnormaldistributiontable:Z-valueof1.645willgive0.95area,leavingatobe0.05.Thereforeifmpismorethan500by1.645stderror,itwillbeintherejectarea,和wewillrejectthenullhypothesisH0,concludingonalternatehypothesisH1thatmpis>500. 某些hypothesistestingsthatareapplicabletoengineers:Theimpactonresponsemeasurementwithnew和old流程parameters.Comparisonofanewvendors’parts(whichareslightlymoreexpensive)tothepresentvendor,whenvariationisamajorissue.IstheyieldonTesterECTZ21thesameastheyieldonTesterECTZ33?流程SituationsComparisonofonepopulationfromasingle流程toadesirablestandardComparisonoftwopopulationsfromtwodifferent流程esorSinglesided:comparisonconsidersadifferenceonlyifitisgreateroronlyifitisless,butnotboth.Twosided:comparisonconsidersanydifferenceofine质量important Inferencesbasedonasinglesample“Largesampletestofhypothesisaboutapopulationmean”Example:Anautomotivemanufacturerwantstoevaluateiftheirnewthrottle设计onallthelatestcarmodelisabletogiveanadequateresponsetime,resultinginanpredictablepick-upofthevehiclespeedwhenthefuelpedalisbeingdepressed.Basedonfiniteelementmodelling,the设计teamcommittedthatthethrottleresponsetimeis1.2msec,和thisistherecommendedvaluethatwillgivethedriverthebestcontroloverthevehicleacceleration.Thetestengineerofthisprojecthastestedon100vehicleswiththenewthrottle设计和obtainanaveragethrottleresponsetimeof1.05msecwitha标准偏差Sof0.5msec.Basedon99%confidencelevel,canheconcludedthatthenewthrottle设计willgiveanaverageresponsetimeof1.2msec? “Largesampletestofhypothesisaboutapopulationmean”(con’t)Solution:Sincethesamplesizeisrelativelylarge(i.e.>30),weshouldusezstatistic.mX=1.05msec;sS=0.5msec;n=100;NullhypothesisH0:m=mH0(1.2msec)AlternatehypothesisH1:m<>mH0(1.2msec)-----AcceptanceArea-----mH0=1.20.005ofarea0.005ofarea(a/2)RejectArea(a/2)RejectArea2.58stderror2.58stderrorFromstandardnormaldistributiontable,TheZvaluecorrespondingto0.005tailareais2.58.a=0.01(2tails),since2tailstest,thereforetailarea=a/2=0.005;HowmanystderrorisXawayfrom1.2msec?X=1.02ThereforeXis–3stderrorsawayfrom1.2msec. -----AcceptanceArea-----mH0=1.20.005ofarea0.005ofarea(a/2)RejectArea(a/2)RejectArea2.58stderror2.58stderrorX=1.02“Largesampletestofhypothesisaboutapopulationmean”(con’t)Baedon99%confidencelevel,sinceXisatthenegativerejectarea,wewillrejectthenullhypothesis和concludeonthealternatehypothesisthattheaverageresponsetimeissignificantlydifferentthan1.2msecTheaverageresponsetimeappearstobelowerthan1.2msec.Whatdoes99%confidencelevelmeansintheaboveexample?Itdefinesthelimitswhereby99%oftheaveragesamplingvalueshouldfallwithin,giventhedesirable(hypothesised)meanasmH0.AnyvaluefalloutsidethisconfidencelimitindicatesthesamplemeanissignificantlydifferentfrommH0.Inotherwords,wewillonlyconcludethealternatehypothesisH1(thatthemeansaredifferent)ifwearemorethan99%sure. TheObservedSignificancelevel,p-valuep-valueistheprobabilityforconcludingthenullhypothesisH0thatbothpopulationmeansareequalwiththeobservedsampledata.Hence1–pvaluewillbetheconfidencelevelwehaveonthealternatehypothesis.--AcceptanceArea--mH0=1.22.58stderror2.58stderrorX=1.023stderrorP/2AcceptanceAreamH0=1.22.58stderrorX=1.023stderrorPFor2tailstestFor1tailtestP/2-Z+Z3stderror-ZUsingthethrottlequestionasanexample:Weknowthatthemeanresponsetimeis3standarderrorawayfrom1.2msec(mH0),therefore–Z=3.Sincethisisa2tailstest,p-value=P(Z<-3,orZ>3)=2P(Z>3)Fromstandardnormaldistributiontable,P(Z=3)=0.9987P(Z>3)=1–0.9987=0.0013p-value=2P(Z>3)=0.0026Instatisticalterm,itmeansthereisonly0.0026probabilitythattheaveragethrottleresponsetimetobe1.02mseciftheactualpopulationmeanis1.2msecassuggestedbyfiniteelementanalysis. “Smallsampletestofhypothesisaboutapopulationmean”Example:AmyisthePersonnelOfficerofamulti-nationalcompanywhoisinchargeofrecruitingalargenumberofemployeesforanoverseasassignment.Astheseoverseasassignmentsareverycrucialforthecompanysuccessinmeetingtheirbusinessplan,anaptitudetestwasformulatedtotestthe质量ofallpotentialcandidateshead-huntedbythe招聘Agency.Themanagementwantstoknowtheeffectivenessofthe招聘Agency,asitwasbelievedthattheaveragetestscoreforalltheidentifiedcandidatesshouldbeequalormorethan90inordertoreducetheriskofassigningthewrongcandidatesforthetask.WhenAmyreviewsthetestsresultofaparticularbatchof20candidates,shefindsthatthemeanscoreis84和the标准偏差is11.Asthisisaverycritical招聘project,Amywantstobemorereservewithheranalysis,和decidedtobemorebiastowardsprovingthatthepopulationmeanislesserthan90.Asaresult,aconfidencelevelof90%willbeusedinheranalysis. “Smallsampletestofhypothesisaboutapopulationmean”(con’t)Solution:Sincethesamplesizeisrelativelysmall(i.e.<30),weshouldusetstatistic.mX=84;sS=11;n=20;a=0.1(1tail)Degreeoffreedom,df=19NullhypothesisH0:m=mH0(90)AlternatehypothesisH1:m0.2Samplesize,n=140(i.e.usestandardnormalzastheteststatistic)Computethestandarderrorfornullhypothesis(i.e.whenp=0.2) TestifpH03stderrorwillgivereasonablevalue(i.e.between0to1)pH03stderror0.23(0.034)=(0.166,0.234)“Largesampletestofhypothesisaboutapopulationproportion”(con’t)Calculatethenumberofstandarderrorsbetweenthesampled和hypothesisedvalue:--AcceptanceArea--pH0=0.20.025ofarea0.025ofarea(a/2)RejectArea(a/2)RejectArea1.96stderror1.96stderrorp=0.0863.36stderrorConclusions:Sincep=0.086isinrejectarea,werejectnullhypothesis和concludethatthenewscreenmethodissignificantlydifferentthantheoldscreenmethodwith95%confidencelevel.With3.36stderrorfrompH0(0.2),thep-valueiscalculatedtobe0.00078,hencethereisa99.922%confidenceinthealternatehypothesis.Itappearsthatthenewscreenmethodwillgivelessermissdetectionforstomachulcer. PowerofaHypothesisTesting95%ConfidenceInterval(AcceptanceArea)mH0(a/2)RejectArea(a/2)RejectAreaTypeIErrorRejectingthenullhypothesis(H0),whenitistrue.Probabilityofthiserrorequalsa.TypeIIErrorAcceptinganullhypothesis(H0),whenitisfalse.Probabilityofthiserrorequalsb.Inhypothesistesting,wearealwaysbiastowardsH0.Thereforea95%confidencelimitwillonlytellusifwearemorethan95%surethatthetwopopulationmeansaredifferent.Howeverthetruestatesofthepopulationmeanscanbedifferentevenifwearelessthan95%sure.Inotherwords,ifthereisnosignificantdifferencebetweenthe2means,itdoesnotindicatethattheyareequal,itcouldbethattheyarenotfarenoughapart.Illustrateintheabovedistribution,assumingm1hasthesamevarianceasmH0和theyaredifferent,theareaunderm1curvethatisfallwithin95%confidencelimitofmH0willbeb(probabilityfortypeIIerror).m1ControlbyaControlbyb PowerofaHypothesisTesting(con’t)Assuchthereisatradeoffbetweena和b.Asadecreases,bincreases和viceversa.mH0(a/2)RejectArea(a/2)RejectAream175%(a=0.25)95%(a=0.05)ConfidenceIntervalbform1whena=0.05bform1whena=0.25 PowerofaHypothesisTesting(con’t)Ahospitaluseslargequantitiesofpackageddosesofaparticulardrug.Theindividualdoseofthisdrugis100cc.Theactionofthedrugissuchthatthebodywillharmlesslypassoffexcessivedoses.Ontheotherhand,insufficientdoses(i.e.99.6cc和below)donotproducethedesiredmedicaleffect,和theyinterferewithpatienttreatment.Thehospitalhaspurchaseditsrequirementsofthisdrugfromthesamemanufacturerforanumberofyears和knowsthatthepopulation标准偏差is2cc.Thehospitalinspects50dosesofthisdrugatrandomfromaverylargeshippment和findsthemeanofthesedosestobe99.75cc.With90%confidencelevel,howcanthehospitalconcludewetherthedosagesinthisshipmentaretoosmall?mH0=100cc(hypothesisedvalueofpopulationmean)=2(knownpopulation标准偏差)X=99.75(samplemean)n=50(samplesize)H0:m=100(nullhypothesisthatmeandosagefromshippmentis100cc)H1:m<100(alternatehypothesisthatmeandosagefromshippmentis<100cc)a=0.1(probabilityoftypeIerror) PowerofaHypothesisTesting(con’t)90%ConfidenceInterval100(a)RejectArea1.28stderror99.64(i.e.100–1.28stderror)Fromstandardnormaldistributiontable,thez-valuethatassociatewith0.9probabilityis1.28.SinceX=99.75fallswithintheacceptablearea,thereisnotenoughevidencetorejectnullhypothesis和concludethatthesamplemeanisnotsignificantlylowerthan100cc.99.75Thehypothesistestindicatesthatsamplemeanof99.75isnotsignificantlydifferentthan100,和thereforethereisnotenoughevidencetosaythattheunderlyingpopulationmeanisnotequalto100.Howeveritdoesindicatethattheyareequal.Itsignifiesthatthereisa10%chancetoreflectthepopulationmeantobenotequalto100ifthepopulationmeanisactually100.Iftheactualpopulationmeanis99.75asthesamplemean,whatistheprobabilitythattheabovehypothesistest(of90%confidence)tomistakenlyreflectthatthepopulationmeanisequalto100(i.e.berror)? PowerofaHypothesisTesting(con’t)90%ConfidenceIntervalformH0=10010099.6499.75Areainsidetheacceptareawillbeprobabilitytomistakenlyaccept99.75asequalto100(i.e.berror)Areaintherejectareawillbeprobabilityofrejecting99.75asequalto100(i.e.1-b)Setting90%confidencelevel(i.e.a=0.1)forthehypothesistesting,iftheactualpopulationmeanis99.75:34.83%chancetorejectH0(correctconclusion),65.17%chancetoacceptH0(incorrectconclusion)Inhypothesistestingwewanttoachievesmalla和b(orbig1-a和1-b). PowerofaHypothesisTesting(con’t)ThePowerCurve–AgraphicalpresentationofthepossiblepopulationmeansagainsttheprobabilitiesofrejectingH0whenH1istrue(i.e.1-b),afterfixingaatacertainvalue.90%ConfidenceIntervalformH0=10099.6499.61(1-b)=0.543890%ConfidenceIntervalformH0=10099.6499.42(1-b)=0.782390%ConfidenceIntervalformH0=10099.6499.75(1-b)=0.3483Thepowercurveshowsthat:TheprobabilityofrejectingH0whenH1istrueincreasesastheactualpopulationmeandeviatesfromthehypothesisedmeanof100.Iftheactualpopulationmeanis99.28,therewillbe90%confidenceinboth1-a和1-b.(i.e.a和berrorwillbe10%) PowerofaHypothesisTesting(con’t)CentralLimitTheorem–Thesamplestandarderrorwill降低withafactorthatissquarerootofthesamplesize.Whensamplesizeincreases,thestandarderrordecreases.90%ConfidenceIntervalformH0=10099.7399.61(1-b)=0.664390%ConfidenceIntervalformH0=10099.7399.42(1-b)=0.863590%ConfidenceIntervalformH0=10099.7399.75(1-b)=0.4718Thepowercurveshowsthat:Assamplesizeincrease,1-breaches90%withacloservalueto100(i.e.99.46insteadof99.28). 某些takeawaysinHypothesisTesting:InHypothesisTesting:aisfixed和biscontrolledbysamplesize.Whensamplesizeincreases,berrordecreases.ab;whentheactualpopulationmeanequalHypothesisedmean–2xZ(a)xStdError(i.e.forn=50,whenpopulationmean=100–[2x1.28x0.2829]=99.28)Youmightwanttoverifyberrorifthehypothesistestfailtorejectnullhypothesis,H0.Intheexampleofdrugdosage,eventhoughthehypothesistestingshowsthattheaverageshouldbeequalto100cc,weshouldrejectthebatchbecausethebriskistoohigh. Comparing2PopulationMeanswithLargeSamplesAdietitianhasdevelopedadietthatislowinfats,carbohydrates,和cholesterol.Althoughthedietwasinitiallyintendedtobeusedbypeoplewithheartdisease,thedietitianwishestoexaminetheeffectthisdiethasontheweightsofobesepeople.2randomsamplesof100obesepeopleeachareselected,和onegroupof100isplacedonthelow-fatdiet.Theother100areplacedonadietthatcontainsapproximatelythesamequantityoffoodbutisnotaslowinfats,carbohydrates,和cholesterol.Foreachperson,theamountofweightlossorgainedina3weeksperiodisrecorded.Thedataispresentedbelow,howcanweinterprettheresultwitha95%confidence. Letm1=meanweightlossesforpeopletakinglow-fatdietLetm2=meanweightlossesforpeopletakingotherdietForm95%confidenceonthedifferenceinmeanforbothdiettypesusingX1和X2asestimators:SolutionAssumingthe2samplesareindependent,the标准偏差ofthedifferencebetweenthesamplemeans(orthestandarderrorofthestatistics)is:AssumptionsThe2samplesarerandomlyselectedinanindependentmannerfromthe2populations.Thesamplesizen1和n2arelargeenoughsothatbothsamplemeanshavenearnormalsamplingdistributions.S1和S2提供goodestimatefors1和s2,hencethe2populationvariancesdoesnotneedtobeequal.(i.e.n30) H0:(m1–m2)=0(i.e.m1=m2)H1:(m1–m2)<>0(i.e.2tailstest,m1<>m2)For2tailstest,Zvaluefor95%confidencelevelisZ(a/2)=1.96stderrorTestStatistics:Solution(con’t)SinceZ=3.05isintherejectionarea,wewillrejectnullhypothesisH0,和concludeonalternatehypothesisH1thatbothpopulationmeansaresignificantlydifferent.Fromthedata,thelowfatdietismoreeffectivetoreduceweightthantheotherdiet.0+1.96stderr-1.96stderrZ=3.05 Comparing2PopulationProportionswithLargeSamplesJoanisaHRofficerwhowantstocomparethepreferenceofemployeetowardsthenewimplementedemployeestockpurchaseplaninSingaporetothoseinCanada.Suchacomparisonwouldhelphertodetermineifthereisaneedtoeitherdivideorconcentratethecampaigneffortsinthesetwosites.Joanrandomlychose1000employeesfromeachsite,和intervieweachtolearnabouttheemployees’likelihoodinsigningupforthestockpurchaseplan.Herobjectiveistousethissampleinformationtomakeinferenceaboutthedifference(p1–p2)betweentheproportionp1ofallemployeeinSingaporewhoarelikelytosignup和p2ofallemployeeinCanadawhoarelikelytosignup.Thetwosamplesrepresentindependentbinomialexperiments,和thenumberofemployeeineachsitewhoindicatethattheywillenrolinthestockpurchaseplanisasbelow:Singapore:X1=546;Samplesize,n=1000Canada:X2=475;Samplesize,n=1000Basedontheabovedata,shouldJoanassumethatemployeeinCanadaislesswillingtoenrolintheplan和concentrateherpromotionalcampaigninthatsite? SolutionLetp1=546/1000=0.546Letp2=475/1000=0.475PropertiesofSamplingDistributionof(p1–p2)(P1–P2)isanunbiasedestimatorofthepopulationproportions.Thebestestimateoftheoverallproportionofsuccessifthepopulationarehypothesisedtobeequal:The标准偏差ofthesamplingdistribution(orstandarderrorofthedistribution)is:Ifthesamplesizesn1和n2arelarge,thesamplingdistributionisapproximatelynormal.Thetwosamplesselectedareindependentrandomsamples(Whenbothproportionsareequal)(Whenbothproportionsarenotequal) 95%ConfidenceIntervalSinceteststatisticof3.176isintherejectionarea,JoanshallrejectH0和concludeonH1thatproportionsfrombothsitesaresignificantlydifferent.Basedonthe95%confidenceinterval,weinferthattherearebetween2.7%和11.5%moreemployeelikelytoenroltotheplaninSingaporesitethaninCanadasite.SheshouldthereforeconcentratehercampaignonCanadatobalanceupthe2sites.0+1.96stderr-1.96stderrZ=3.176H0:(p1–p2)=0(i.e.p1=p2)H1:(p1–p2)<>0(i.e.2tailstest,p1<>p2)For2tailstest,Zvaluefor95%confidencelevelisZ(a/2)=1.96stderrorTestStatistics:Solution(con’t) CupPositionPinPositionPullBackAngleHookPositionStopPositionHypothesisExerciseSeparateinto4teamsEnsuretheStatapultispresetto:PullbackAngle-177°HookPosition–4CupPosition-1StopPosition–3PinPosition-2Eachteamwillevaluateonthedifferencesinthemeandistanceswithrespecttoa)differentoperatorsb)differentballtypesc)differentstatpultsd)differentpullbackangleby3degreesAtotalof100shotsforeachset-uparerequiredusingtheSOPoutlinedinweek1.Computestatisticallyusinghypothesistestingtocommentonthemeandistanceofthe2statpults.Whatistheaerrorinyouranalysis.Ifnullhypothesisisaccepted,whatisthebriskthatyouhavetobear?Presentyourresultintheclass Comparing2PopulationMeanswithSmallSamples(T-Test)Supposeyouwishtocompareanewmethodofteachingreadingto“slowlearners”tothecurrentstandardmethod.Youdecidetobasethiscomparisonontheresultsofareadingtestgivenattheendofalearningperiodof6months.Ofarandomsampleof20slowlearners,8aretaughtbyqualifiedinstructorsundersimilarconditionsfora6monthsperiod.Theresultsofthereadingtestattheendofthisperiodareasbelow:With95%confidencelevel,isthereanydifferencebetweenthe2methods?Whatare某些oftheassumptionsinorderfortheestimatetobevalid?NewMethod,X1StandardMethod,X28080798176667976796270687376867372687566 Sincethesampleissmall(i.e.<30),wecannotusethepreviousmethodasthepointestimatefor标准偏差willnotbeaccurate.Whensamplesizeislessthan30,uset-testwithbelowassumptions:Bothsampledpopulationshaverelativefrequencydistributionsthatareapproximatelynormal.(i.e.toensureeffectivenessofcentrallimittheorem)ThepopulationvarianceareequalThesamplesarerandomly和independentlyselectedfromthepopulations.SolutionSinceweassumethatbothpopulationvariancesareequal,wecanusethedatafrombothsamplestoconstructapooledsampleestimatorofvarianceS2pforuseinconfidenceinterval.Smallsampleconfidenceintervalfor(m1–m2)is: Solution(con’t)H0:(m1–m2)=0(i.e.m1=m2)H1:(m1–m2)<>0(i.e.2tails)For2tailstestof18degreeoffreedom,tvaluefor95%confidencelevelist(a/2)=2.101stderrorSincet=1.807isintheacceptancearea,wewillnotrejectnullhypothesisH0,和concludethatthereisnotenoughevidencethatbothmethodswillresultindifferentmeans.Needtofurtherverifyifpopulationvariancesareequal,和berrorisassumetobenotaconcern.0+2.101stderr-2.101stderrt=1.807 Comparing2PopulationMeans(PairedDifferenceExperiment)Thepreviousexampleisbasedonindependentsampling.Supposeitispossibletomeasuretheslowlearners’“readingIQ”beforetheyaresubjectedtoateachingmethod,theneightpairsofslowlearnerswithsimilarreadingIQsarefound,和onememberofeachpairisrandomlyassignedtothestandardteachingmethodwhiletheotherisassignedtothenewmethod.Theresultingdataaregiveninbelowtable.Fromthedata,isittruethatthenewmethodwillgivebettertestresultascomparedtothestandardmethod?Pair12345678StandardMethod7268766884686176NewMethod7774827387696680 PairedDifferenceExperimentSolutionH0:m1–m2=0wherem1ismeantestresultfornewmethodH1:m1–m2>0m2ismeantestresultforstandardmethodIfweemploythetstatisticforindependentsample,wehave:0+1.76stderrt=1.26Fromthet-distributiontable,thetvalueassociatewith14degreeoffreedom和0.95is1.76(i.e.onetailtest).Since1.26islesserthan1.76和intheacceptancearea,thereisthereforenotenoughevidencetosuggestthatthenewteachingmethodcangivebettertestresultthanthestandardmethod.Ifweexaminethedatacarefully,wewillfindthatthisresultisdifficulttoaccept,asthetestscoreofthenewmethodislargerthanthecorrespondingtestscoreofthestandardteachingmethodforeverysinglepairofslowlearner!Showingthatm1shouldinfactgreaterthanm2fromthelookofit.Infactthebriskisaround69%ifthetruedifferenceinmeanis4.375(i.e.X1–X2). PairedDifferenceExperimentSolution(con’t)Thet-testisinappropriateforthisexamplebecausetheassumptionofindependentsampleisinvalid.Thepairsoftestscorehavebeenrandomlyselected,和oncewehavechosenthesampleforthenewmethod,wehavenotindependentlychosenthesampleforthestandardmethod(i.e.Thetestscorefluctuatedependentlyasshowninthegraph).Weshouldusepaireddifferenceexperimentfordependentsamplinglikethis:Pair12345678StandardMethod7268766884686176NewMethod7774827387696680Difference56653154 PairedDifferenceExperimentSolution(con’t)H0:mD=0(i.e.m1=m2)H1:mD>0(i.e.m1=m2)XD=Samplemeandifference=4.375SD=Sample标准偏差ofdifferences=1.685ND=Numberofpair(differences)=80+1.76stderrt=7.34SinceTeststatistic,t=7.34isintherejectarea,thereisthereforeenoughevidencetorejectH0和concludeonH1thattherethenewteachingmethodwillgivebettermeanscoreascomparetothestandardmethod. 某些possiblepaireddifferenceexperimentSupposethegovernmentwantstoestimatethedifferenceinmeanstartingsalariesformen和womengraduatesinthecountry.Ifthesamplesselectedareindependent,thestartingsalariesmayvarybecauseoftheirdifferentmajors,differentacademicachievement,etc.Toeliminatethissourceofvariability,thegovernmentcouldmatchmale和femalejobseekeraccordingtotheirmajors和academicachievement,和usepaireddifferenceexperimenttoanalyse.Ifyouwishtoestimatethedifferenceinmeanabsorptionrateintobloodstreamfortwopainreliefdrug.Whenselectthesampleindependently,theabsorptionratemayvaryduetoage,weight,sex,bloodpressureetc.Inthiscase,sincethereissomanypossiblevariation,itmightbedifficulttopairtheexperimentusing2personsasasinglepair.However,wecanevaluatethetimeforthedrugtotakeeffectbyexperimentingonthesamepersontominimiseothervariations,和usedasinglepersonasapairexperimentfor2differentdrugs. PairedDifferenceExperimentSummaryForlargesample(i.e.n30)Forsmallsample(i.e.n<30)Pairdifferenceexperimentisusefulwhenyoucanassureinputdependency,和neligibleinputvariationapartfromtheexperiment.Whatisthehypothesisresultfortheearlierstatapultexerciseifyouanalysethedatausingpairdifferencemethod?Isthereanydifferenceintheresultinterpretation? ComparingTwoPopulationVariances(F-Test)Frequentlyweneedtocomparethevaraincesof2populations.Exampleint-testweassumethatthepopulationvarianceisequalasthereisnotenoughdataforestimation.Howcanwebesurethenthatthevariancesareequal?Thereforethereisaneedtodevelopsuchtechniquetoanalysevariance.TheassociatedequationinexcelworksheettoobtaintheFvalueis:“=FINV(a,n1,n2)“ExampletheFvalueforn1=2,n2=5,和aareaof0.99is:“=FINV(0.01,2,5)“n=DegreeoffreedomF(a;v1,v2)a F-TestExampleAnexperimenterwantstocomparethemetabolicratesofwhitemicesubjectedtodifferentdrugs.Theweightsofthemicemayaffecttheirmethabolicrates,和thusitishopefullythattheobtainedmicearehomogeneouswithrespecttoweight.Fivehundredmicewillbeneededtocompletethestudy.Currently18micefrom供应商A和13micefrom供应商Bareavailableforcomparison,和theresultisasbelow:供应商A供应商BnA=18nB=13XbarA=4.21ouncesXbarB=4.18ouncesS2A=0.019S2B=0.049Df,nA=17Df,nB=12Dothesedata提供enoughevidencetoindicateadifferenceinthevariabilityofweightsobtainedfromthetwo供应商s?Usea=0.1fortheanalysis. F-TestExample(con’t)LetS2A=Populationvarianceofweightsofwhitemicefrom供应商ALetS2B=Populationvarianceofweightsofwhitemicefrom供应商BH0:S2A/S2B=1H1:S2A/S2B<>1FStatistic=Largersamplevariance/Smallersamplevariance=S2B/S2A(Hence供应商Bwillbesample1和供应商Awillbesample2)=0.049/0.019=2.58Usingtheexcelformula,input“=FINV(0.05,12,17)”intoexcelspreadsheet,wehave:Fvalue=2.38(i.e.2tailstest,hencedivideaby2) F-TestExample(con’t)SinceFstatisticof2.58isgreaterthanFvalueof2.38和intherejectionarea,thedatahadtherefore提供sufficientevidencetoindicatethatthepopulationvariancesdiffer.Fromthesamplevariancedata,itappearsthat供应商Atendtobemorehomogeneousthan供应商B.ConfidenceIntervalfors21/s22WhereFL,a/2isbasedonn1=(n1–1)和n2=(n2–1).WhereFU,a/2isbasedonn1=(n2–1)和n2=(n1–1).Itwasestimatedthatthevarianceintheweightsofmicefrom供应商Acouldbeassmallas0.15oraslargeas0.923timestheweightvarianceofmicefrom供应商B.F(0.05;12,17)a/2F=2.38Fstatistic=2.58 F-TestSummaryFstatistic=Largersamplevariance/SmallersamplevarianceRejectionregion:Fstatistic>Fa(onetailtest)orFstatistic>Fa/2(twotailstest)Assumptions:BothsampledpopulationsarenormallydistributedThesamplesarerandom和independentWhenFstatisticc2a,wherec2statistichas(r-1)(c-1)degreesoffreedomAssumptions:Theobservedcountsarearandomsamplefromthepopulationofinterest.Thesamplesize,nwillbelargeenoughsothatforeverycell,theexpectedfrequencywillbeequalto5ormore. IndependenceTestMatrixGoodBrownstainMarginalcottonglove376011154875Glovestype4180.609756694.3902439palmfitglove415119943503730.390244619.6097561Marginals791113149225ExpectedValuesinRedpValue4.9025E-139ChiStatistic630.0386347df1Chi-Square-TestingForIndependencecasestudyProjectobjectiveistopreventexternalstaindefectonphotoflashtube.UsingtheChi-squaretestforindependence,the流程engineerisabletoquantityifthecorrectiveactioniseffectiveornot.Bydeterminingtheindependencybetweentheyield和actionsclassifications,anindepentscenariowouldmeanthecorrectiveactionisnoteffective和viceversa. Chi-Square-InferencesAboutaPopulationVarianceF-Testmakeinferencesabouttwopopulationvariances(assumingnormality)InF-Test,confidenceintervalisanrationindexbetweens2A和s2B.Chi-squaretestmakeinferenceaboutasinglepopulationvariance(assumingnormality)InChi-squaretest,confidenceintervalforthevarianceoftheactivityofconcernedisexpressedasvalue.A流程engineerwhoisin-chargeofback-endtestinghasconductedanewevaluation,和foundthatheisabletoreducethevariationintermsofelectricalresponseby50%(i.e.标准偏差from0.232to0.116).Beforehereportthisresulttohisbosshewantstomakesurethatheis95%confidenceinhisanalysis,howshouldheapproachthisproblem?Step1-Fromhisexperimentgatherthedescriptivestatisticaldata Step2-PlothistogramtoexaminedistributionnormalityChi-Square-InferencesAboutaPopulationVarianceStep3-Identifyoutlierswhichislikelyduetospecialcauses(i.e.outsidenaturaltolerance)Samplemean=3.84SampleStdDeviation=0.153NaturalTolerance=Mean+/-3StdDeviation=3.38to4.30Thereforeremovedatathatare<3.38和>4.3islikelyduetospecialcauses.Atthismomentweacceptthatwhatlookslikeanormaldistributionmustbeatleastapproximatelynormal.Amoredetailmethodtotestfornormalitywillbediscussedinlatersection.-6-5-4-3-2-10123456<3.38>4.30OutliersOutliersNaturalTolerance Chi-Square-InferencesAboutaPopulationVarianceStep4-Removeoutlierfromthedata,和re-computedescriptivestatisticStep5-Computeconfidenceintervalsforthepopulationvariancebasedonassumptionthatthesamplesaretakenfromapproximatelynormalpopulation=“CHIINV(0.025,519)”=584.02c2valuec2Uc2L=“CHIINV(0.975,519)”=457.770.025ofarea0.025ofareaThereforethe流程engineercanreporttohisbossthatheis95%confidencethatthevarianceofthenew流程isbetween0.012to0.0153(orStddeviationbetween0.110to0.124). Chi-squareExerciseBreakinto4GroupsEachgroupwillbegivenapacketofM&M“meltinyourh和notinyourmouth”chocolatecandy.InyourgroupscountthenumberofM&MsintheirColor.CombinetheData和conductaChi-SquaretesttoseeifthedistributionofcolorareindependentwithrespecttodifferentM&Mpackages.M&MColourTeam4Team3Team2Team1BrownRedBlueYellowGreenOrangeRememberthepaperclipductilitytestpreviously.Eachteamistocomputetheconfidenceintervalforthe标准偏差ofthepaperclipductilityofthe2供应商s.Alsoindicateiftheductilityisdependingondifferentteam(i.e.operators).Presentyourresultonthe2casestudiestotheclass. ANOVA–AnalysisofVarianceWhatisANOVA?AstatisticalapproachusingFdistributiontotestforthesignificanceofthedifferencesamongmorethan2samplemeans.CompletelyrandomiseddesignA设计forwhichindependentrandomsamplesofexperimentalunitsareselectedforeachtreatment.ObjectiveofcompletelyrandomiseddesignTheobjectiveisusuallytocomparethetreatmentmeans.Exampleforpopulationmeansofptreatment,thenullhypothesis,H0statesthatalltreatmentmeansareequal,和alternatehypothesis,H1statesthatatleasttwoofthetreatmentmeansdiffer.H0:m1=m2=.…=mpH1:Atleasttwooftheptreatmentmeansdiffer ANOVA–AnalysisofVarianceExampleTheInternationalfootballassocaiationwantstocomparethemeandistancesassociatedwithfourdifferentbrandsoffootballswhenkickbyastriker.Acompletelyrandomised设计isemployed,withthekickingsimulatedbyapneumaticactivatedswingthatwillstrikeaconsistent100Nforceatafixedcontactangletotheballs.Theexperimentwasconductedwitharandomsampleof10ballsofeachbr和hittedatarandomsequence.Thedistanceofeachtrialwasrecorded和thedatawasorganisedasbelow:Br和A251.2245.1248.0251.1260.5250.0253.9244.6254.6248.8Br和B263.2262.9265.0254.5264.3257.0262.8264.4260.6255.9Br和C269.7263.2277.5267.4270.5265.5270.7272.9275.6266.5Br和D251.6248.6249.4242.0246.5251.3261.8249.0247.1245.9Mean250.78261.06269.95249.32Variance22.421814.947120.258327.0729 ANOVA–AnalysisofVarianceExampleStep2:ComputetheSumofSquaresforTreatments(SST),whichisthemeasurementofvariationbetweentreatmentmeans.Step1:Computethegr和averageforallthesamplestaken.Gr和averageX=(250.8+261.1+270+249.3)/4=257.8Step3:ComputetheSumofSquaresforErrors(SSE),whichisthemeasurementofvariationwithintreatment.H0:m1=m2=.…=mpH1:Atleasttwooftheptreatmentmeansdiffer Step4:ComputetheMeanSquareforTreatment(MST),和MeanSquareforError(MSE).ANOVA–AnalysisofVarianceExampleStep5:ComputetheteststatisticF,rejectnullhypothesis,H0whenFstatistic>Fa.Degreeoffreedom,n1=3(i.e.p–1);和n2=36(i.e.n–p).Theassociatedp-valuecanbecomputedby“=FDIST(Fstatistic,n1,n2)”.Thereforep-value=FDIST(43.99,3,36)=3.97E-12.aF=2.243Fstatistic=43.99p-value ANOVA–AnalysisofVarianceExampleStep6:Compilethesummarytable和drawconclusion.SinceFstatistic>Fa,wewillrejectthenullhypothesisH0,和concludethatatleasttwoofthetreatment(brand)meansaresignificantlydifferent.Thep-valueshowsthatthereisonly0.000000000397%chancethatallthemeansareequal(i.e.verystrongevidence).AssumptionsThesampleareselectedrandomly和independentlyfromtherespectivepopulations.Alltreatmentpopulationprobabilitydistributionsarenormal.Alltreatmentpopulationvariancesareequal.Rejectingthenullhypothesistestingonlyindicatethatmorethan2treatmentmeansdiffer,butwhicharethetwo?Totalpossiblepairofcombination=C=p(p–1)/2;p=numberoftreatments=(4)(3)/2=6pairsofmeansthatcanbecompared ANOVA–AnalysisofVarianceExamplep=Numberoftreatmentsdf=Degreeoffreedomforerror ANOVA–AnalysisofVarianceExampleForANOVA,weusestudentizedrangeinsteadoft-distributiontocomputeconfidenceinterval,thereforecriticalvalueofstudentizedrangeata=0.05,p=4,和dferror=36is3.8.Computethetableofconfidenceinteravalsfordifferencesofthe6possiblecombinationpairs:Notsignificantlydifferent ANOVA–AnalysisofVarianceExample(optional)SupposethattheInternationalFootballAssociationwantstocomparethemeandistancesassoicatedwiththe4brandsofballswhenkickbyastriker,butwishestoemployhumanstrikerinsteadofthepneumaticswing.Therewillbeatotalof10ballsforeachbr和tobetested和therearetwowaysthattheexperimentcanbeconducted;completelyrandomised和randomisedblockdesign,whichisabetter设计inthiscase?Br和A和striker1to10Br和B和striker11to20Br和C和striker21to30Br和D和striker31to40CompletelyRandomisedDesignRandomisedBlockDesignBr和ABall1Br和BBall1Br和CBall1Br和DBall1Striker1Br和ABall2Br和BBall2Br和CBall2Br和DBall2Striker2Br和ABall10Br和BBall10Br和CBall10Br和DBall10Striker10Until ANOVA–AnalysisofVarianceExample(optional)Usingrandomisedblock设计asanexample;2Setsofhypothesis[i.e.treatment(brands)和block(strikers)]H0:m1=m2=m3=m4H1:AtleasttwoofthebrandsdifferinmeanH0:m1=m2=m3=m4=m5=m6=m7=m8=m9=m10H1:AtleasttwoofthestrikersdifferinmeanTotalsumofsquares,SS(Total)Sumofsquaresfortreatment,SSTSumofsquaresforerror,SSESumofsquaresfortreatment,SSTSumofsquaresforerror,SSESumofsquaresforblocks,SSBdf=p-1df=n-pdf=n-1df=p-1df=b-1df=n–b–p+1CompletelyRandomisedDesignRandomisedBlockDesign ANOVA–AnalysisofVarianceExample(optional)Gr和average=(227.1+233.2+245.3+220.7)/4=231.6Computethesumofsquaresfortreatments:DistanceDataforRandomisedBlock设计 Computethesumofsquaresforblocks:ANOVA–AnalysisofVarianceExample(optional)Computethesumofsquaresfortotal和sumofsquaresforerror:Takinga=0.05(95%confidencelevel),computethecriticalFvalueusingExcelspreadsheetforbothtreatment(i.e.brand)和block(i.e.striker): ANOVA–AnalysisofVarianceExample(optional)ComputeaANOVAsummarytableforanalysis:BothFstatisticfortreatment和blockexceedtherespectivecriticalFvalues,hencewewillrejectthe2setsofnullhypothesis和concludethatatleast2brandsdifferinmeandistanceaswellasatleast2strikersdifferinmeandistance.Themeandistanceisdependentonthestriker,hencearandomisedblock设计isbetterinthiscasethanacompletelyrandomised设计(i.e.preventstrikervariationintheanalysis)AssumptionsTheprobabilitydistributionsofobservationscorrespondingtoalltheblock-treatmentcombinationsarenormal.Thevariancesofalltheblock-treatmentcombinationsareequalANOVAmethodthoughenabledmultiplemeanscomparison,itisbasedonF-distributionhavingtheassumptionthatallthepopulationmeansarenormalwithequalvariance.Unlikehypothesis和t-testwhichutilisecentrallimittheorem,ANOVAmethodisthereforemoresensitivetodeviationfromnormality.NeverthelessitservesanimportantroleinDOE. ANOVA–AnalysisofVarianceCaseStudyAChemicalVapourDepositionequipmentdesignerwantstoevaluate2newbrandsofhighvacuumpumpwhichistobeusedintheirlatestdepositionsystembecauseofpossiblecostbenefitfromthenew供应商s.Toqualifyanynewpump供应商,thedesignerneedtoensurethatthe质量intermsofpumpingdowntimeto6torrneedtobeatleastthesameifnotbetterthanthebr和ofvacuuumpumpcurrentlyusedinthe产品ion.Assumingthatthenewpump供应商only提供15setsofpumpfortheevaluationpurpose,和theresultisgivenasbelow.Disscussinteams和giveapresentationonyourteamanalysis和recommendationusingthefollowingquestionsasguidelines.QuestionsUserandomisedblock设计toanalysethedata.Whatcanwetellfromtheanalysis?IsthereanimpactfromdifferentCVDsystem?HowaboutacomparisonwithChi-squaremethod?With95%confidence,howdothedifferentbrandsdiffer.Whataretheunderlyingassumptions?Comparetheresultwithother(appropriate)hypothesistestingmethod.Whataretheassumptions?canwelowertheriskoftheseassumptions?Ifnoneofthebrandsaresignificantlydifferentinmean,howshouldwedealwithit? WordofCautionAboutHypothesisTestingEnsurethattheexperimentcarriedoutsatisfiedtheunderlyingassumptionsofwhicheverstatisticalmethodwasemployedintheanalysis.Becarefulaboutfalselyconcludingnullhypothesis,H0duetobetarisk(i.e.TyoeIIError).Analysebetarisk(ifpossible)和seeifitisacceptable.Increasesamplesizeifneededto.Becarefulaboutfalselyconcludingalternatehypothesis,H1duetoalpharisk(i.e.TyoeIError).Increaseconfidencelevel(oralevelofsignificance)ifneededto.Alternativelylookatthep-value(ifpossible),和seeifthealternatehypothesisisreasonabletoyou.Independencytest,adependentconclusiondoesnotnecessaryimplyauniquedependent关系hipbetweentheclassifications.Analyseyourresultwithcost-effectfactorinmind.Re-confirmresultifpossible(unlessanalysisissupportedbystrongtheoreticalreasoningand/ortooexpensivetoduplicate). DistributionAnalysis–Chi-Square:GoodnessofFitTestSupposingthatABCcompanyrequiresthatalltheirpotentialfutureleaderstobeinterviewedbythreedifferentdirectors,sothatthecompanycanobtainaconsensusevaluationofeachcandidate.Eachdirectorwillgivethecandidateeitherapositiveornegativeratingasbelowdataof100candidates:Positiveratingoutofthe3interviewsNo.ofcandidates018147224311IftheHRDirectorforABCcompanythinksthattheinterview流程canbeapproximatedbyabinomialdistributionof40%chanceofanycandidatereceivingapositiveratingonanyoneinterview.Howcanheverifythiswitha90%confidencelevel?WhatistheChi-Square:GoodnessofFitTestusedfor?Itisusedtotodecideiftheunderlyingprobabilitydistributionofapopulationiswhatweexpectedittobe(I.e.poisson,binomialornormal),byfittingtheactualfrequencywiththeexpectedfrequencypredictedbythedistribution. DistributionAnalysis–Chi-Square:GoodnessofFitTestH0:Abinomialdistributionofp=0.4isagooddescriptionH1:Abinomialdistributionofp=0.4isnotagooddescriptiona=0.1(i.e.90%confidencelevel)Weneedtofirstlydetermineifthediscrepanciesbetweentheobservedfrequencies和thosewewouldexpect(I.e.binomialdistributionofp=0.4)areactuallyduetochance.Step1:ComputethebinomialprobabilityinaccordancetothescenarioWecanfirstlyuseExcelspreadsheettoobtainthebinomialprobabilityusingtheequations:“=BINOMDIST(No.ofsuccesses,No.oftrials,Proportionofsuccess,Cumulative)” DistributionAnalysis–Chi-Square:GoodnessofFitTestStep2:ComputetheexpectedfrequencyStep3:ComputetheassociatedChi-squarestatisticStep4:Determinethedegreeoffreedom和obtainthecriticalc2valueata=0.1Degreeoffreedom=Totalobservation–1–otherestimation(s).=4–1–0=3ThereforecriticalFvaluefromexcel=“=CHIINV(0.1,3)”=6.2514.Sincec2statisticK-ScriticalthenrejectH0和concludeonH1Since0.3858>0.0962,weshallconcludeonalternatehypothesis,H1thatPoissondistributionofl=1isnotagooddescriptionofthelowsodiummealrequestpattern. DistributionAnalysis–CaseStudyBasedon520data(i.e.n=520),theelectricalresponseofinfra-reddetectorwasfoundtobehavingameanof3.85kV/W和a标准偏差of0.1161kV/W.Thehistogramanalysisusingthestandardclassificationmethodinweekoneresultinthebelowdiagram:Datadonotlooktoonormal.Itisbelievedthatthedatashouldbenormal,和thediagramissimplyduetotoomanyclassesidentified(i.e.23classes).Afterreducingthenumberofclassesfrom23to8withaclasswidthof0.1,thehistogramlooksmoretowardtheexpecteddistribution. DistributionAnalysis–CaseStudyWiththehistogramof23classes,bothChi-square和K-Sgoodnessoffittestswereperformed,和theresultsusinga=0.05areasbelow:Hint:Useexcelfunction“=NORMDIST(x,mean,stddev,cumulative=true)”forprobabilitycomputation.ForChi-squaretest:c2statistic=978.12c2critical=CHIINV(0.05,517)=571Df=n-1-estimation=520-1-2(i.e.mean和stddev)c2statistic>c2critical,thereforedistributionisnotnormal.ForK-Stest:K-Sstatistic=0.1382K-Scritical=1.36/Sqrt(520)=0.05964K-Sstatistic>K-Scritical,thereforedistributionisnotnormal. DistributionAnalysis–CaseStudyWiththenewhistogramof8classes,bothChi-square和K-Sgoodnessoffittestswereperformed,和theresultsusinga=0.05areasbelow:Hint:Useexcelfunction“=NORMDIST(x,mean,stddev,cumulative=true)”forprobabilitycomputation.ForChi-squaretest:c2statistic=63.576c2critical=CHIINV(0.05,517)=571Df=n-1-estimation=520-1-2(i.e.mean和stddev)c2statisticc2critical,thereforeacceptnullhypothesisH0,和distributionisconsideredtobenormal.p-value=CHIDIST(63.576,517)=0.99999ForK-Stest:K-Sstatistic=0.1356K-Scritical=1.36/Sqrt(520)=0.05964K-Sstatistic>K-Scritical,thereforerejectnullhypothesis和concludethatdistributionisnotnormal. DistributionAnalysis–CaseStudyWiththenewhistogramof8classes,adjustthestartpointfrom3.49to3.45,theK-Stestresultisasbelow:K-Sstatistic=00217K-Scritical=1.36/Sqrt(520)=0.05964K-SstatisticK-Scritical,thereforeacceptnullhypothesis和concludethatdistributionisnormalwithmean=3.85和标准偏差0.1161.DistributionAnalysisTakeAwayK-Stestseemstobeverysensitivetowardclassificationofdatawhensamplesizeislarge.Whenperformingdistributionanalysis,usehistogramwithallpossibleclassificationstomakejudgement.Coupleengineeringjudgementwithstatistical方法duringanalysis.Donotalwaystrustthesoftwareresultwithoutquestioning.Perform2typesofgoodnessoffittestasconfirmationifpossible. DistributionAnalysis–ExerciseKnowingthatthechanceofrollinga3inaunbiaseddiceis1outof6(i.e.0.1667),和theexperimentisstatisticallyindependent.Thereforeweshouldbeabletoassociatetheexperimentwithabinomialdistributionofp=0.1667(i.e.successingettingavalueof3).Breakinto4teams.Use5dicesinasinglerollingofdice.Conductanexperimentthatconsistof500rollsof5dices.Thenumberofdicesuseineachroll(i.e.5)willthereforebethenumberoftrial,ninabinomialdistribution.The500repeatingrollsistotakecareoftherelativefrequencybehaviouroftheexperiment.Thenumberofsuccesswillbethenumberof3sinasingleroll.Includingalsoinyourexperimenttheaveragevaluesofthe5dicesineachroll.Countthenumberofsuccess(i.e.valueof3)ineachroll.Sumthemupforthetotal500rolls.ConductaChi-square和K-Sgoodnessoffittesttodetermineifp=0.1667isagooddescriptionofobtainingavalueof3intherollingofunbiaseddice.Alsotestifcentrallimittheoremisapplicablefortheaveragevaluesinthe500rolls. Regression和CorrelationWhatisregression和correlationanalyses?Itisastatisticalmethodthatwillshowushowtodetermineboththestrengthofa关系hip,和theasssociatedprobabilisticmodelbetweentwovariables.OriginofregressiontechniqueOriginatedin1877bySirFrancisGalton.Studyshowedthatheightofchildrenborntotallparentwilltendtomoveback(or“regress”)towardthemeanheightofthepopulation.Theword“regression”wasthendesignatedasthenameofthegeneral流程ofpredictingonevariable(thechildrenheight)fromanother(theparentheight). Regression和CorrelationADeterministicModelAmodelconstructedtohypothesisedanexact关系hipbetweenvariablesExample:Force=Mass*AccelerationGivenafixedmass,ForcecanalwaysbedeterminedifaccerlerationisknownNoallowanceforerrorinthepredictionAProbabilisticModelAmodelincludebothadeterministiccomponent和arandomerrorcomponentY=DeterministicComponent+RandomerrorUnexplainedvariationwillbetakingintoaccountbytherandomerrorcomponent Regression和Correlation为什么doweallowerrorsinthepredictionmodel?Mightbeamorerealisticmodelsincevariationisusuallyunavoidable.Allmodelsarewrong,but某些areuseful.Inhighspeedspacetravelling,variationinvelocitywillbeanimportantfactortoconsiderinaForcemodeltodeterminethethrustpower和fuelconsumption. Regression和CorrelationAFirst-Order(Linear)ProbabilisticModelb0=S$2.40b1=yincrease/xincrease=9.6/30=0.32E(y)=Taxifare=b0+b0x(lineofmeans)Modelfortaxifarey-axis=S$36912xaxis=KMtravelled510152025300 Regression和CorrelationKelvinisaneurospecialistwhoisinterestedinhowalcoholcanaffecthumanreactiontime.Heconductedanexperimentwithasinglevolunteerinfivedifferentsessions,和obtainthereactiontimewithrespecttothevolunteeramountofalocoholcontentinpercentage.Howshouldheproceedtogetamodeloutofthesedatapoints?Step1:HypothesisedthedeterministiccomponentoftheprobabilisticmodelKelvinbelievethatthismodelshouldbelinear,和thereforethedeterministiccomponentis:E(y)=b0+b1x Step2:Usesampledatatoestimateunknownparametersinthemodel(withleastsquaresapproach)Ifweplottheindivialdatapointsintheformofscattergram,和fitalinearvisuallinetopassthroughthemostdatapoints,wewillbeabletoobtainthemagnitudeofthedeviations(i.e.verticaldifferencesbetweenobserved和predictedvaluses).Ifweaddthesquareofallthedeviations,wecanobtainthesumofsquarederrors(SSE)=12+02+02+12+02=2(forthevisuallyfittedline)ForalinearmodelthereisonlyonelineforwhichtheSSEisaminimum,和thislineiscalledtheregressionline.Regression和CorrelationReactionTime1234AlocoholContent123450Deviation=1Deviation=1Deviation=0Deviation=0Deviation=0 Regression和CorrelationMathematically,theslope(b1)和y-intercept(b1)oftheregressionlinecanbeachievedbythebelowequations:Theregressionlineis:E(y)=b0+b1xE(y)=-0.1+0.7x Regression和CorrelationReactionTime1234AlocoholContent123450E(y)=-0.1+0.7xWiththeleastsquaresapproach,Ypredict=-0.1+0.7xisfoundtobethebestfitlinewiththeminimumsumofsquarederrorsSSEof1.1. Regression和CorrelationStep3:Specifytheprobabilitydistributionoftherandomerror(epsilon,e).Fourbasicassumptionsfortheprobabilitydistributionofthepredictedmodelare:Themeanoftheprobabilitydistributionofise0.Thevarianceoftheprobabilitydistributionofeisconstantforallsettingsoftheindependentvariablex.Theprobabilitydistributionofeisnormal.Thevalueofeassociatedwithanytwoobservedvaluesofyareindependent.(i.e.eofyiwillnotaffecteofyj).ReactionTime1234AlocoholContent123450PositiveErrorNegativeErrorErrorprobabilitydistributionofequalvariance和normal Regression和CorrelationToestimatethe标准偏差ofe,calculateSastheestimatedstandarderroroftheregressionmodelusing:ReactionTime1234AlocoholContent123450E(y)=-0.1+0.7xE(yL)=-0.1+0.7x–3SE(yH)=-0.1+0.7x+3SApproximatepredictioninterval3S(i.e.99.73%ofallobserveddependentvariablesshouldliewithinthisrange) Regression和CorrelationStep4:AssessingtheusefulnessofthemodelusinghypothesistestingSinceb1istheslopeofalinearmodel,avalueof0wouldmeanthatthelineisflat,和astheindependentvariablexchanges,thedependentvariableywillnot(i.e.noco-relationbetweenx和y)H0:b1=0(i.e.theregressionmodelisnotusefulforyprediction)H1:b1<>0(i.e.theregressionmodelisusefulforyprediction)a=0.05(I.e.95%confidence,2tailstest)Independentvariable,xDependentvariable,yb1=0SamplingDistributionofb1Basedonthe4assumptionsfore,wecanestimate: Regression和CorrelationComputetheteststatistic(sincen30uset-test):Criticaltvalue=“=TINV(a,df)”(i.e.Defaulttobe2tails)=TINV(0.05,3)=3.1824Sincetstatistic>tcritical,wewillrejectH0和concludeonH1thatthereissufficientevidencethatb1isnotequalto0,hencetheregressionmodelisusefulforyprediction.Whatifwefailtorejectnullhypothesis?Inthecaseiftstatistictcritical,itdoesnotleaddirectlytoacceptthenullhypothesisastheremightbehighTypeIIerrorrisk.Amorecomplex关系hipmightexistbetweenx和y,requiringotherfittingofamodel. Regression和CorrelationThePearson产品momentcoefficientofcorrelation,rItisameasureofthestrengthofthelinear关系hipbetweentwovariablesx和y,和computedas:Positiver:yincreasesasxincreasesNegativer:ydecreasesasxincreasesr=1:Perfectpositive关系hipr=-1:Perfectnegative关系hiprnear0:littleorno关系hipbetweeny和xrnear0:nolinear关系hipbetweeny和x Regression和CorrelationThecoefficientofdetermination,r2Itrepresenttheproportionofthetotalsamplevariabilityaroundythatisexplainedbytheregressionline.(Inlinearregression,itmayalsobecomputedasthesquareofthecoefficientofcorrelation,r)Inourexample,ThePearson产品momentcoefficientofcorrelation,ris:Thereforethemodelhasapositivelinear关系hip.Thecoefficientofdetermination,r2is:r2valueof0.8167impliesthatabout81.67%ofthesamplevariationinreactiontime(y)isexplainedbythebloodalcoholcontent(x). Regression和CorrelationStep5:Usingthemodelforestimation和predictionAfterconfirmingthatthemodelisusefulthroughhypothesistesting和r2value,theprobabilisticmodelisnowreadytobeusedforthepredictionofdependentvariable,y.Themostcommonuseofaprobabilisticmodelare1)toestimatethemeanvalueofyforaspecifcvalueofx,和2)topredictanewindividualyvalueforagivenx. Regression和CorrelationAssumingthatKelvinisonlyinterestedinthereactiontimeofapersonwhosebloodcontains4%ofalocohol,whatistheconfidence和predictionintervalat95%confidencelevel?Atxp=4%,thepredicteddependentvariable,yp=-0.1+(0.7)(4)=2.7Theverylargeconfidence和predictionintervalisduetothesmallsamplesizeusedinthesimplelinearregressionexperiment.Byincreasingthenumberofdatapoints,theseintervalslimitscanbereducedforamoremeaningfulestimation和prediction. Regression和Correlation Regression和CorrelationExerciseICupPositionPinPositionPullBackAngleHookPositionStopPositionSeparateinto4teamsEnsuretheStatapultispresetto:HookPosition–4PinPosition-2CupPosition-1StopPosition–3Eachteamisgiventhetasktofindouttherealtionshipbetweentheballdistancetravelled(y)和thepullangle(x).Thepullbackangleshouldhavearangefrom160oto180o.Atotalofminimum100shotsarerequiredforeachexperimentset.Yourteamwilldecideontheresolutionforexperiment.Usethe5stepsforconductingaregressionexperimenttofindouttheprobabilisticmodelofdistancetravelled和pullbackangle.Setthepullbackangleto177o,和conductanother100shots.Isthedatainaccordancetoyourpredictionintervalfromthemodel(usinga=0.3,0.15和0.05)?Presentyourresultsintheclass. MultipleRegressionModelbuildingInregression,weneedtoalwaysstartwithhypothesisingasuitableprobabilisticmodel.1)Singlequantitativeindependentvariable(i.e.singlexvariable)Thehypothesiseddeterministiccomponentofsuchmodelcanbedescribedas:First-ordermodel:Second-ordermodel:Third-ordermodel:E(y)=b0+b1xE(y)=b0+b1x+b2x2E(y)=b0+b1x+b2x2+b3x3b1=slopeALinearModelb2<0AQuadraticModelb2>0b3<0b3>0ACubicModelCautioninhypothesisingthedeterministiccomponentItisnotcommonthataveryhighordermodel(i.e.4th和above)tobeusedinregression.Thisisbecausewemightendupover-fittingthedatapoints,resultinginunnecessarynoise(variation)modelling.Alwaysstartthehypothesiswiththereasonablylowestorder. MultipleRegression2)TwoormorequantitativeindependentvariablesThehypothesiseddeterministiccomponentofsuchmodelcanbedescribedas:First-ordermodel:E(y)=b0+b1x1+b2x2Interactionmodel:E(y)=b0+b1x1+b2x2+b3x1x2Completesecond-ordermodel:E(y)=b0+b1x1+b2x2+b3x1x2+b4x12+b5x22Themodelwillgetmore和morecomplexasthenumberofindependentvariables(xs)increase.Thecomputationofbcoefficeientsforsuchmodelisalsoverycomplex和usuallytakencareinDOEsoftware. MultipleRegressionInsimplelinearregressionwesaythat:Beforeproceedingtotheleastsquareapproachformultipleregression,weneedtounderst和thefundatmentalinderivingthebcoefficients.Firstweneedtorefreshonthe“TheGeneralPowerRule”fordifferentiation:Weneedalsotorefreshhowtobreakupsummationnotationinequation: MultipleRegressionInlinearmodelwehave:Step1:TransformintosumofsquarederrorequationStep2:Toobtaintheminimumvalueofsumofsquarederror,wecanusederivatives和setthederivedequationsequaltozero.(i.e.Usepartialderivativeswithrespecttoallthebcoefficientsb0和b1).(i.e.Generalpowerrule)(i.e.Divideequationby2)(i.e.Breakupthesummationnotation)(i.e.Equation1)(i.e.Partialderivativeswithrespecttob0) MultipleRegression(i.e.Generalpowerrule)(i.e.Breakupthesummationnotation)(i.e.Equation2)(i.e.Partialderivativeswithrespecttob1)Step3:Solvesimultaneouslyforequation1和2:(i.e.Equation3from1)(i.e.Divideequationby2) MultipleRegressionSubsituteequation3into2:Fromequation2:Multiplyequationbyn:Rearrangetheequation: MultipleRegressionQuadraticModelExampleMarkisworkingforaMayorofasmalltownthatisfacingwatershortagesproblem.AssuchtheMayorhadassignedhimataskto设计awatersavingcampaigntargettingattherighttargetaudienceinordertoachievemaximumresult.Markthinkthatthewaterusageshouldhavea关系hiptothehomesize,和wantstoknowwhatthis关系hipislike,sothathecan设计和deployhiscampaignmoreeffectively.Tobeginwith,hecollectedthedatabetweenhomesize和waterusageofaparticularmonthfor10familiesinthetown,和thedataisgivenbelowtable:FromthescatterdiagramMarkfeelsthatthisshouldbeaquadraticmodelinsteadofalinearmodel. MultipleRegressionQuadraticModelExampleStep1:HypothesisedthedeterministiccomponentoftheprobabilisticmodelMarkbelievethatthismodelshouldbequadratic,和thereforethehypothesisedprobabilisticmodelis:y=b0+b1x+b2x2+e(i.e.y=b0+b1x+b2x2+e)Step2:Estimatetheunknownparametersb0,b1,和b2Similartolinearregressionexample,firstlytakethepartialderivativeoftheequationwithrespecttob0,和formequation1Equation1: Nexttakethepartialderivativeoftheequationwithrespecttob1,和formequation2.MultipleRegressionQuadraticModelExampleEquation2: MultipleRegressionQuadraticModelExampleTakepartialderivativeoftheequationwithrespecttob2,和formequation3.Equation3:Nextistosolveforthe3derivedequations. MultipleRegressionQuadraticModelExampleObtaintherepectivevaluesforallxi和yicombinationsappearinthe3derivedequations.Createanewequationsintheexceltableasbelowexample MultipleRegressionQuadraticModelExampleSolveforb1和b2usingbelowtable:b1=2.399b2=-0.00045 MultipleRegressionQuadraticModelExampleSinceb1=2.399和b2=-0.00045,subsitutethesevaluesintoequation1:Hencetheequationthatminimisethesumofsquarederror(SSE)is:E(y)=-1216.4+2.399x-0.00045x2和theprobabilisticmodelis:y=-1216.4+2.399x-0.00045x2+eAssumations:Foranygivensetofxs,therandomerrorehasanormalprobabilitydistributionwithmeanequalto0和equalvariance.Therandomerrorsareindependent. MultipleRegressionQuadraticModelExampleStep3:Specifytheprobabilitydistributionoftherandomerror.Toestimatethe标准偏差ofrandomerror,calculateSastheestimatedstandarderroroftheregressionmodelusing:Sumofsquarederror(SSE)=15332.7 MultipleRegressionQuadraticModelExampleStep4:Determinetheusefulnessofthemodel.Inmultipleregression,themultiplecoefficientofdetemination,R2is:Theresultimpliesthat98.19%ofthetotalsamplevariationofwaterusagecanbeexplainedbyusingtheindependenthomesizeinaquardraticmodel.Total:15332.7846402.1 MultipleRegressionQuadraticModelExampleTestingtheglobalusefulnessoftheModelwithANOVA:H0:b1=b2=0(i.e.untilbk)H1:Atleastoneofthecoefficientisnon-zeroUsea=0.05nisthesamplesize和kisthenumberoftermsinthemodel.RejectwhenFstatistic>Fcritical,withknumeratordegreesoffreedom和(n-k-1)denominatordegreeoffreedom.Caution:Arejectioninhypothesisleadstotheconclusionthatthemodelisusefulwiththeconfidencelevelusedinthetest,butdoesnotmeanthatitisthebestmodel. MultipleRegressionQuadraticModelExampleFcritical=“=FINV(0.05,2,7)“=4.74SinceFstatistic>Fcritical,weshallrejectthenullhypothesis和concludethatatleastoneofthecoefficients(i.e.b1orb2)isnon-zero,和theglobalF-testindicatethatthemodelisusefulinpredictingthewaterusage.Step5:Usingthemodelforprediction.Thepredictionintervalfortheregressioncanbeestimatedbyta/2SUsing95%confidencelevel,theta/2valueis2.365.(i.e.df=n-k-1)Thereforetheupperpredictionlevelis:y=-1216.4+2.399x–0.00045x2+2.365S=-1105.7+2.399x–0.00045x2Thereforethelowerpredictionlevelis:y=-1216.4+2.399x–0.00045x2-2.365S=-1327.1+2.399x–0.00045x2 Regression-ResidualAnalysisWhatisitusedfor?Sincealltheassumptionsinregressionmodelconcerntherandomerrorcomponent,residualanalysiscanthereforebeemployedtoindicateiftheassumptionshasbeenviolatedUsingthewaterusageregressionasanexampleIfweplotalinearmodelforthewaterusagedata,itwillendupwithaR2valueof0.8317,whichisnottoobadforstrengthindication. Regression-ResidualAnalysisStep1Calculate和plottheresidualsagainsteachoftheindependentvariables,whereby:Residual=Yi-YpredictStep2Analyseeachplot,lookingforcurvature(i.e.abowlormound)shape.Asinresidualplotoflinearmodelamoundcanbeseen,indicatingneedofaquadraticmodel.Step3Examinetheplotforoutlierbasedon+/-2Sfrom0.Seeiftheoutliercanbeexplained,ifitisduetospecialcause,removethedatafromtheregressionanalysis. Regression-ResidualAnalysisStep4Plotahistogramontheresiduals和seeifthereisanyobviousdeparturefromnormality(i.e.canemploygoodnessoffittesttocheck).Extremeskewnessofthefrequencydistributionmayindicatetheneedtotransformthedependentyvariable.(Therearenotenoughdatainthewaterusageexamplefordistributionanalysis)Regression和CorrelationExerciseIICupPositionPinPositionPullBackAngleHookPositionStopPositionSeparateinto4teamsEnsuretheStatapultispresetto:HookPosition–4PinPosition-2CupPosition-1PullBackAngle–180oEachteamisgiventhetasktofindouttherealtionshipbetweentheballdistancetravelled(y)和thestopposition(x).Eachteammemberistotake3shotsateachstopposition.Performtheappropriateanalysisonthedatacollected和presentintheclass. TimeSeriesAnalysisTimeseriesanalysisisusedtodetectpatternsofchangeinstatisticalinformationoverregularintervalsoftime.Anestimateforthefuturecanbeprojectedbasedontheanalysedpattern.Itisaquantitativemethodofforecasting.VariationsinTimeSeriesSeculartrend-wherethevalueofvariabletendstoincreaseor降低overalongperiodoftime(i.e.increaseinthecostoflivingrecordedbyConsumerPriceIndex).YaxisXaxis-TimeinyearsSecularTrend TimeSeriesAnalysisCyclicalfluctuation-wherethevalueofvariabletendstoincrease和降低inacyclicalmanneroveralongperiodoftime(i.e.Businesscycle,economicperformance).YaxisXaxis-TimeinyearsStraightLineTrendCyclicalFluctuationSeasonalvariation-wheretheinvolvedpatternchangeswithinayear和tendstorepeatyearafteryear(i.e.Businessthatinvolve产品sellinginseason).YaxisXaxis-TimeinyearsSeasonalvariationYear1Year2Year3 TimeSeriesAnalysisIrregularvariation-wherethevalueofvariablechangesinarandommannerwhichisunpredictable(i.e.lotterynumber).YaxisXaxis-TimeinyearsIrregularvariationInmostcases,atimeserieswillcontainseveralofthevariationcomponents.Therefore,itispossibletodescribetheoverallvariationinasingletimeseriesintermsofthesefourdifferentkindsofvariation.MakingpredictionfromtrendanalysisFittingalineartrendbyleastsquaremethodtothedatapointscollectedovertime.Equationforlineartrend:Y=a+bXX=Years(ortime)Y=Responsevariable TimeSeriesAnalysis–TrendAnalysisTranslating和codingtimeAsweareusingleastsquaresapproach,codingthetimewilleliminatetheneedtosquarethenumberaslargeastheyear,和itwillsetthemeanyear(Xbar)tozero,simplyingthetrendequation. TimeSeriesAnalysis–TrendAnalysisThereforethelineartrendequationis:Y(trend)=139.3+7.536X(coded)Ifyouwanttopredictthesalesin1990fromthetrendequation,firstcodetheyear1990:X(coded)=(1990–1985.5)x2=9SubsituteX(coded)=9intotrendequation:Y(trend)=139.3+(7.356)(9)=$207.124M TimeSeriesAnalysis–TrendAnalysisFittingaquadratictrendbyleastsquaremethodtothedatapointsEquationforquadratictrend:Y=a+bX+cX2a,b,和ccanbefoundbysolvingtheseequationssimultaneously! TimeSeriesAnalysis–TrendAnalysisThereforeY(predict)=39.3+22.7X(coded)+5.07X(coded)2TimeSeriesAnalysis–CyclicalvariationCyclicalvariationisthecomponentofatimeseriesthattendstooscillateabove和belowtheseculartrendlineforperiodslongerthan1year.Theprocedureusedtoidentifysuchvariationistermedtheresidualmethod.X=1991X(coded)=1991-1987=4Y(trend)=39.3+(22.7)(4)+(5.07)(16)=$211.2M(predictedforyear1991)Assumingthetrendwillcontinue,whatwillbetheexpectedsalesinyear1991? TimeSeriesAnalysis–CyclicalvariationThemeasureofcyclicalvariationisexpressedasapercentoftrend:Example:Acomicbookdistributorwantstomeasurethevariationsinthesalesfigureofthe:“X-Men”comicwhichisreleasedmonthly,sothathecanhave某些informationaboutthevariationheisseeingforthepast8years. TimeSeriesAnalysis–CyclicalvariationByplottingtheresidualplotasapercentoftrendline,thetrendissuewillbeeliminated,和thecyclicalcomponentisisolatedfromthetimeseries.Itistobenotedthatthismethodisusedonlyfordescribingpastvariations和nottopredictfuturecyclicalvariation.TimeSeriesAnalysis–SeasonalvariationTherearethreemainreasonsthatseasonalvariationisimportant:Toestablishthepatternofpastchanges,givingopportunitytocompare2timeintervalsthatwouldotherwisebetoodissimilar(i.e.compareDecember‘2000toDecember’1999,insteadofJune’1999).Itisusefultoprojectpastpatternintothefuture,inparticularformakingshort-rundecision.Onceexistingseasonalpatterncanbeestablished,itseffectcanbeeliminatedfromthetimeseries. TimeSeriesAnalysis–SeasonalvariationInordertomeasureseasonalvariation,theratio-to-moving-averagemethodisemployedto提供anindextodescribethedegreeofseasonalvariation.Theindexhasabaseof100,和degreeofseasonalityismeasuredbythevariationsawayfromthisbase.Example:Themanagementteamfortelecommunicationequipmentwantstoforecasttheirsalerevenueforthe3rdquarterof1990.Asthesalesofsuchequipmentmighthaveaseasonalimpactfromthebonusperiodofconsumer,howshouldtheteamconductatimeseriesstudytofindouttheindicesthatdescribethedegreeofseasonalvariation. TimeSeriesAnalysis–SeasonalvariationStep1:Addthetotalofthe4quartersinthe1styear和putitthevalueinthemiddleofthequarter(i.e.2.5quarterofyear1)Move1quarterdown和sumthenext4quarters(i.e.Year1quarter2toyear2quarter1).Putthisvalueinthequarterthatisincrementalby1fromthepreviousstartingquarter(i.e.Quarter3.5inyear1).Repeattheprocedureuntilthesummationhasreachthelastquarterofthelastyear. TimeSeriesAnalysis–SeasonalvariationStep2:Computethe4-quartermovingaveragebydividingstep1valuesby4. TimeSeriesAnalysis–SeasonalvariationStep3:Sincethemovingaveragevaluestartatyear1quarter2.5,weneedtobringitdownhalfaquartertoquarter3byaveragingthevaluebetween2.5和3.5. TimeSeriesAnalysis–SeasonalvariationStep4:Calculatethe%oftheactualvaluetothemovingaveragevalueforeachquarterinthetimeserieshavinga4-quartermovingaverageentry. TimeSeriesAnalysis–SeasonalvariationStep5:Calculatethemodifiedmean(seasonalindices)bydiscardingthehighest和lowestvalues(i.e.shadedcellinthetable)foreachquarter和averagingtheremainingvaluesforeachquarter.Step6:Sincethebaseforanindexis100,thetotalindicesfor4quartersshouldthereforebe400和not397.45.Wemustthenadjusttheindividualquarterindexsothatthetotalindices=400. TimeSeriesAnalysis–AcombinedapproachInthiscase,theapproachisacombinationofthreestages:DeseasonalisingthetimeseriesDevelopingthetrendlineFindingthecyclicalvariationaroundthetrendlineTocontinueontheexampleforseasonalvariation,ifthemanagementteamwantstoforecasttherevenuefor3rdquarterin1990.Howshouldtheyapproachthisproblem?Step1:Computingseasonalindices和findthedeseasonalisedvaluesforeveryquarterinthedata:Oncetheseasonaleffecthasbeeneliminated,thedeseasonalisedvaluesthatremainreflectsonlythetrend,cyclical,和irregularcomponentofthetimeseries. TimeSeriesAnalysis–Acombinedapproach TimeSeriesAnalysis–AcombinedapproachStep2:Developingthetrendlinewiththedeseasonalisedvalues(assuminglinear)Step3:FindthecyclicalvariationTheresidualplotseemsrandom,nosuggestionvisuallythatthelineartrendisabadestimate. TimeSeriesAnalysis–AcombinedapproachUsingthetimeseriesmodeltoforecastsalesrevenuefor3rdquarter1990:Thirdquarterin1990willbeequaltothe23rdquarteraccordingtoourtable.ThecodedXwillthereforebe:=(2)(23–10.5)=25SubsitutingX(coded)=25intothedeseasonalisedtrendequation,wehave:Y(trend)=(0.159)(25)+18.06=$22.035MHoweverthisisadeseasonalisedforecast,wemustthereforeseasonalisedthisestimatebymultiplyingtheadjustedthirdquarterseasonalindexexpressedinratiotoabaseof100:Y(23rdquarter)=22.035x0.6119=$13.48M TimeSeriesAnalysis–AcombinedapproachTrend和Cyclicalcomponent TimeSeriesAnalysis–QuestionThenumberoffaculty-ownedpersonalcomputersattheUniversityofBerkeleyhasincreaseddrmaticallyoverthelast6years:Year199419951996199719981999No.ofPCs50110350102019503710a)Developalinearestimatingmodelthatbestdescribesthesedata.b)Developaquadraticestimatingmodelthatbestdescribedthesedatac)EstimatethenumberofPCsthatwillbeinuseattheuniversityin2003usingbothequations.d)Ifthereare8000facultymembersattheuniversity,whichequationisthebetterpredictor.为什么?Thefollowingdatadescribethemarketingperformanceofaregionalbeerproducer:Usetheabovedatatopredictthesalesfor2nd和3rdquarterofyear2001. TimeSeriesAnalysis–Question3.Thedataintheabovetableisthemonthlysalesofjet-enginetothecommercialaeroplanemanufacturer.Doatimeseriesanalysisofjet-enginesalesoverthelast10years.Deseasonalisethesalesbymonth(witha12monthscenteredmovingaveragemethod).Determinethecorrecttypeofmodeltouseforasalesprediction,lookattheresidualplotforaconclusion.Useyourresulttoforecastsalesforeachmonthof2001. Introduction为什么doweneedtobuiltreliabilityintoour流程es?ReliabilityhasGainimportanceinmanyoperation和产品inthemarketCompetitioninthemarket和dem和istheservice.Customerwantsmorereliableservices,流程or产品.Asaconsumer.Weareconcernaboutbuying产品thatlastlonger和arecheaper.ThiscalltohaveHighReliabilitytobebuildinour产品和services.ReduceriskofliabilityReliabilityAnalysisTherearemanyareaswhereweneedreliabilityCriticalRawMaterialPowersystemConsumer产品LifesafetySystemMeasuringInstrumentsBioMedicalFieldTransportation-AircraftDefenseMedicalindustry Howdowemeasurereliabilityofyour流程和system?ReliabilityAnalysisReliabilityismeasureoftheprobabilitythatthe产品orsystemwilloperatewithoutfailureforagiventime(time,t)Reliabilityindexisanumberfrom0to1,measuringtheprobabilitythatthepartwillfailforagiventestingtimeReliabilityR(t)ofacomponentorsystemattimetisgivenbyR(t)=P(T>t)WhereT=acontinuousrandomvariabledenotingtimetofailure.Typically,wewillassurethatTfollows某些probabilitydistribution(e.g.Normalorexponential)t=某些specificdesirabletime(e.g450hours) F(to)R(to)t0tF(t)=P(Tt)R(t0)=1-F(t)Probabilitythatthecomponentwillsurviveaftert0timeProbabilitythatthecomponentwillfailpriortot0timeReliabilityAnalysisF(t)istheprobabilitydensityfunction(pdf)fortheparttofailbeforetimet.R(t)istheprobabilitydensityfunction(pdf)fortheparttofailaftertimet.Twopossiblefailurefunctionforinvestigate:a)Normaldistribution和b)ExponentialdistributionSincereliabilityisaprobability,itisalsotheareaunderthepdfcurve h(t)BurnInzoneWearoutzoneOperatingZonetBathTubCurveReliabilityAnalysisAnotherTermuseinReliabilityisthehazardfunctionh(t)=f(t)/R(t)Wheref(t)istheFailurefunction和R(t)isthereliabilityfunctionHazardFunctionisinterpretedasthe“instantaneousfailurerate”.Itistherateoffailureataninstanttime. ExponentialFailureModelIftherandomvariableTdenotingtimetofailureisexponentiallydistributed,thenthefollowingpropertieshold:f(t)=e-t,t0F(t)=P(failurebeforetimet)=P(Tt)=1-e-tR(t)=P(failureaftertimet)=P(Tt)=e-tE(T)=Meantimebetweenfailure=MTBF=1/=averagenumberoffailurepertimeinterval,whichistheonlyparameterintheexponentialdistribution.ThereforetheHazardFunctionh(t)=f(t)/R(t)=e-t/e-t=ExponentialmodelisoftenusetomodelusefullifeperiodoroperationzoneReliabilityAnalysis SinceMTBFisequalto1/,thenR(t)=P(failureaftertimet)=e-tR(1/)=P(failureafterMTBF)=e-t=e-(1/)=e-1=0.368Therefore,fromfigure1,MTBFindicatesthetimewhereby36.8%oftheparttestedwillsurvivebeyondthistimeperiod.Thisishoweverbasedontheassumptionthatall缺陷followtheexponentialfailurebehaviour.ReliabilityAnalysis Anotherimportantpropertyoftheexponentialdistributioniscalledthe“Forgetfulness”property.Thispropertysaysthat,nomatterhowlongtheitemhasbeenworking,itisasgoodasnew.WeexpressthispropertyasExponentialFailureModelReliabilityAnalysis Example:ReliabilityAnalysisSupposedataiscollectedon40identicalcomponents和theyarefoundtohaveameantimebetweenfailure(MTBF)of25hrs.Assumingaconstantfailurerate.FindthereliabilityfunctionSincethefailurerateisconstant,wewilluseexponentialdistribution.Also,theMTBFis25hrs;wecanthereforeestimatetheparameterby1/E(T)=1/25Reliabilityfunctionisapp.ByR(t)=e-t/25Whatisthereliabilityoftheitemat30hours?R(30)=e-30/25=0.3012Whatisthereliabilityoftheitemat70hours,giventhatithasalreadylasted40hours?UsingtheforgetfulnesspropertyoftheexponentialdistributionP(T>70IT>40)=P(T>30)=e-30/25=0.3012 ConfidenceLimitofReliabilityGenerally,wearenotconcernedwithupperconfidencelimitsforreliability,onlylowerlimitsWheret=timel=estimateaveragenumberoffailurepertimeinterval=1/MTBFn=samplesizec2isaChi-squarevalue,with=levelofsignificanceReliabilityAnalysisUsingthis关系hiptocalculatea95%lowerconfidencelimitforthereliabilityat30hoursofthecomponentgiveninthepreviousexample NormalFailureModelExampleSupposewesample36machinedparts和putthemontestuntiltheyfail.Werecordtheirfailuretimes,constructahistogram,performa2goodnessoffit和concludedthattimetofailureisnormallydistributed.Wealsocalculatethatt=50hours和S=10hoursa)Whatistheestimatereliabilityofthepartsat40hours?Usingt和Sasestimatesof和?welet=50和=10ThereforeR(40)=P(T>40)=1-P(T40)=1-P((T-)/(40-50)/10)=1-P(Z-1)=1-0.1587=0.8413ReliabilityAnalysisTheparameter和canbeestimatebycomputingt和sfromarandomsampleoffailuretimes Component1Component4Component3Component2SeriesSystemallComponentsmustbeoperatinginorderforthesystemtooperateForNnumberofsystemR(t)=R1(t).R2(t).R3(t)….Rn(t)ParallelSystemComponent1Component3Component2ForNnumberofsystemR(t)=1-[(1-R1(t)).(1-R2(t)).(1-R3(t))….(1-Rn(t))]ParallelSystemsystemfailonlyifalltheindividualcomponentsfail.ReliabilityAnalysis ReliabilityAnalysisTheexpoentialreliabilityequationmentionedearlierisbasedonareliabilitytestingthatendonlyafterallpartshavefailed.Howeversometimeduetotimeconstraint,wemightperformareliabilityinanotherway:CensoredDataITestuntilafixedtime,t0,和observedthenumberofitemsfail.wheret0=Totalreliabilitytimer=Numberofpartsfailaftert0n=TotalsampletestedAssumption:ExponentialfailuretimesFor95%lowerconfidencelimitonR(t) CensoredDataIITestnitemsuntilrfailuresoccur.wherer=Numberofpartsfail(fixed)n=TotalsampletestedXr=MaximumtimeoffailuredetectedAssumption:ExponentialfailuretimesReliabilityAnalysisFor95%lowerconfidencelimitonR(t) Whatisa设计ofExperimentItisanExperimentthatweruntolearnaboutour流程Wepurposefullymakechangestotheinputs(orFactors)inordertoobservecorrespondingchangesintheoutputs(orresponse)TheinformationGainfromproperlydesignedExperimentcanbeusetoimproveperformancecharacteristic,toreducecosts和timeassociatedwith产品development,设计和产品ionTobuildamathematicalmodelswhichapproximatethetrue关系hipbetweentheinputs和theoutputs.TurninganArttoScience.Themathematicalmodelwillcontaininformationtooptimizethe流程,reducevariation,makeresponserobusttonoise PastStatisticalGuru和thethe工具ingtheyuseWewilluseallthisToolingtoMakeaquartumLeapInImprovement DefinitionOfA流程forDOEMaterialPoliciesEquipmentPeopleProceduresMethods流程EnvironmentPerformaserviceProducea产品CompleteataskAblendingofinputstoachievethedesireoutputsInorderthatwecanimprovethe流程,wemusthavesomethingtomeasure和reflectonBetterRateorcriticaldimensionFaster(suchascycletime,流程timedevelopmenttime)Lowercost(suchasCOPQ和流程和产品cost Whatisa设计ExperimentItisawaytochangetheinputsfactorsofa流程inordertoobservecorrespondingchangesintheoutput(respond)TemperaturepHConcentrationVoltageElectricalPlating流程CoatingThicknessUsethiscolumntomodelShat(apredicted标准偏差)UsethiscolumntomodelYhat(apredictedaverage)MultiplereplicatesAsetofconditionweusetostudythe流程 StrategiesForExperimentationScreening-Identifythekey流程InputVariable:K>5Modeling-ToModelThe流程和GenerateaMathematicalModel:K<5whereKisthenumberoffactors知识GainedFromExperimentationSensitivityCharacterizationOptimizationRobustnessTolerance设计ForX’sYhat(mean)和Shat(stdDev)Model提供s知识onallofthese 目标ofAnExperimental设计ObtainthemaximumamountofinformationusingaminimumamountofresourcesDeterminewhichfactor(inputs)shifttheaverageresponse,whichshiftthevariability和whichhavenoeffectBuildingempiricalmodelsrelatingtheresponseofinteresttotheinputfactorsFindfactorsettingsthatoptimizetheresponse和minimizethecostValidate(common)resultsExperimental设计willhelpyouidentifythefollowingtypesoffactorsFactorAaffectstheaverageFactorBaffectsthe标准偏差FactorCaffectstheaverage和the标准偏差FactorDHasnotEffectsNoShiftinMeanorVariationVariationShiftMeanShiftsMean和VariationShifts TypicalWayExperimentisconductedOneFactoratatimeresultslookX1%X2%VaryingTheParameterX1Temperature和setatthehighestyieldVaryingX2Timesetatthehighestyield-90807070MaxYieldActual流程looklikethisthereforeyoumightnotgettheoptimizepoint WaysofcollectingExperimentaldataOneFactoratatime(cannotbeassureofdetectingtheuniqueeffectsofcombinationsoffactors)Allcombinations(fullfactorial)-TooBigBestGuess-NeverConclusiveDOEExampleofDOE:KraftCheesepackagingPackaging流程ResponseVariableVariableInputsY1:BondStrengthX2:PaperSourceX1:PolymerTemperatureAmanufactureroflaminatedpapertaketworollsofKraftpaper和extrudesalayerofPolymerinbetween,simultaneouslypressingthethreecomponentsintosandwich.Thecustomerhascomplainedaboutthelackofadherenceofthepolymertothetwopaperlayers.ATeamissetuptomaximisethebondstrength.AfterapplyPF/CE/CNX/SOPtheXfactorsarethis: FactorsLowLevelHighLevelX1:PolymerTemperatureºF580600X2:PaperSourceVendorYVendorXQuantitativefactorsQualitativefactorscontinuousdiscreteDefinetheDOE:FullFactorialmeanallcombination TheyconductanExperiment和CollecttheDataDoweknowwhichfactoristhetheonethatinfluencethePackagingprocess?NexttheyanalysistheData和thiswasthefinding EffectofTemperature和VendoronBondstrengthThemeanoftheBondstrengthisinfluencebyTemperature和VendorbuttheVariationofbondstrengthwasnotinfluencebythistwoparameter Thereisinteraction(combine)effectsbetweenFactorA&B DOE:ManufacturingofCapsuleWettingSolutionPressingstrengthMixingtimeFlourVendorCapsuleManufacturing流程CompressionStrengthImpactStrengthDissolveRateTemperaturecuringtimeDegassingtime NumberofFactorforthis流程is=7TotalNumberofcombination:27totalrunis128OnefactoratatimeapproachItistoobigtodo128:WecandosubsetTodetectasignificantdifferencein2标准偏差,a17replicategive95%confidenceAtotal8combinationstestedNomathmodelavailabletopredicttheresponsefortheother120combinationNoabilitytoestimateifvariableinteract INTERACTION-FamilyBenzDistanceperlitreofpetrolusewithpassengerDistanceperlitreofpetrolusewithEngineTimingYLoHiYLoHiKm/litreKm/litreA-NumberofpassengerA-EngineTimingLowOctane(B)HighOctane(B)HighOctane(B)LowOctane(B)TomodelthispictureABcomponentisthedegreeofinteractionTomodelthispictureasimplelinearmodelisallthatisrequireNonParallelslopeParallelslopesThisisMainEffectAnalysis TerminologyUseinDOEFactors-Controllable(Independent)variablesthatmaycauseachangeintheresponse(dependent)variable:ExampleTemperature,流程ingSpeedLevels-ThespecificSettingoroptionsforafactorinanexperimentFullfactorial设计-CombinealllevelsofallfactorsinanexperimentsReplication-ArepetitionofrunofasetupResponse-TheoutputformtheexperimentMainEffect-Theaveragechangeintheresponseduetothechangeinthelevelsofafactor WhatisaFullFactorialCostofRunninganExperiment Whenwecan’tdoallCombinationsthisisbestGuessbyDrKnowItallIfRun1-8weretheresultsofaseriesof“BestGuesses“和youlumpthemtogether,youcouldevaluatethingslikethefollowing:DoesFactorPshifttheaverage?ItappearPshiftstheaveragefromP1toPbutifyouweretorealizethatMRowarethesame:Thisiscall“PerfectlyConfoundedorAliased“TheyareBADbecausewecan’ttellthedifferencebetweenP和MAreSettingP和DAliased?NobuttheyarePartiallyConfoundedwecan’tevaluatedtheyindependently.V和Warenotaliasedorconfounded.Callorthogonal“Balance”thisisGoodAve5.5Ave9.5 IllustrationofConfound和AliasedonthepreviousdesignP=MDTotalVariabilityofYPerfectlyConfoundedPartiallyConfoundedMDPHowdoweobtainthisindependenceofvariables?Good设计makeeveryvariableindependence SimpleDefinitionofTwolevelorthogonalDesignThisnewcolumnisnotlongerorthogonalwithA和BThisnewcolumnaretestedindependently和theyareorthogonalUsetomodelinteractionofA和BHowdoweknowthatthecolumnisorthogonala)sumof产品betweenrow=0WaytotransferformActualtoCODED IntroductiontoDOE SimpleModelBuildingExample IntroductiontoDOE SimpleModelBuildingYAParetoChartYbarBSABAY-+-+-+S-+-+-+8642ABABABAB252015105ParetoChartSABB25201510586425.62515 BuildinganEmpiricalModelThisempiricalmodelissimilartoTaylorSeriesApproximationwhichrelatestotheoutputYtothecodedinputsA和B.NowHowwilltheTalyorseriesforA,B,Clooklike? ComparingaScreening设计VsOFAT:7Factors-2level12X4=48runPreviousExample:ManufacturingofCapsuleusing设计ofOFATin7factoris=136run ComparingaScreening设计VsOFAT:11Factors-2levelOFAT12x17=204TotalRunNeedtoget95%confidenceinstandarddeviationanalysisforeachrun12x4=48TotalRun4Replicationwillgive95%confidenceinS和99.9%inYhat Thingstonotewhenconductinga设计ofExperimentAlltheexperimentalFactorsshouldbeconsideredAllControllablefactorsmustbecontrolledThe流程mustberunningatitsstablestageEliminateasmanynoiseaspossibleIdentifyallthepossibleresponseEnsurethattheMeasurementsystemiscapableTrytouseContinuousFactorTrytousemeasurableresponsesUseanappropriate设计PreparetheexperimentinadvanceIdentifypotentialmistakeduringtheexperiment Exercise:ScreeningDOEUsingthePF/CE/CNX/SOPsfromWeek1minimizethevariabilityConductaL12Screening设计usingthefollowingfactors:UsingSoftwareto设计theExperimentPerformDataCollectionfortheExperiment Robust设计for产品和流程esYouareaBusinessmanagerthatwanttolaunchyour产品intoM&MCandyallovertheWorld.Twoareaweareinparticulartolookintoyou产品.Itmustbeabletowithst和temperaturevariation和AltitudeAFewApproachuseinthemarketa)WarningSign“Keepbelow200degreeF”b)SelltogetherwithaPortablecoolingunitc)Comeoutwitha设计thatisrobusttochangeinenvironmentRespondsLowHigh2010RoomWhichisthemostproductive?Howcanwedothis? 设计a产品or流程-RobusttoNoiseFactorAhassubstantialShift.IfweoptimizeAwecanmakethe流程RobustSetAtoPositivethevariationofthehardnesswillreduce和robusttoChangesintemperature IncreasetheTimetoMarketfornewTechnologiesRearWingAngle(-30,0,30)SideWingAngle(-0,5,10)SideWingSpan(30,50,100)SideWingArea(0,5,10)Weight(100,400,500)DevelopANewEnergySaver-TwoWingSliderShuttleEnginePower(500,600,700)AerodynamicCharacteristicWindTunerTestingtoModeltheAircraftSlidingAbilityWhataresomeAreathatwecanuseDOE? PuttingalltogetherDevelopinginto流程Manual流程Name&DescriptionTeammembersGeneral流程FlowDeterminethemeasurementMatrixPF/CE/CNX/SOPs/FMEA;PokaYokeControldocumentforallControllableitemSpecification,Cpk和COPQforEachYMeasurethecapabilityofX和YLookattheHistoricalDataDesignedExperimentControlChartsTroubleShootingResultsTrackYield,Time,Cost和Capability HowisRegressionlinkedtoDOEWhichFactorisSignificant HowdowereadtheFtable?NotsignificantSignificantat95%confidenceHowwouldyouwritethepredictedEquationforY?LettrytousethesoftwareatthispointCreatea设计(2level3factor) OutputfromtheRegressionMeasurethedegreeofindependenceorthogonalityROT:Tol>0.5IndicatethattheFactorsisSignificantifP(2tail)<0.05RegressionStrength:proportionofvaluesthatisaccountedbythemodelFstatisticsROT:F<0.05tohaveastrongmodelforpredictionEstimateofStdDEVofResidualStandardError=SQRT(MSE)WhatisthepredictedEquation?ShouldweincludeBintothePredictedEquation? ModelingofDeviationSRuleofThumb:For2LevelDesignwritedownthePredictedequationforShat.NextwehavetoreducethevariationbyminimizeShat:wecansetfactorBtohigh+1.S-hat=0.669thiswillminimizethevariationToaccomplishedthedesireobjectiveistosettheY-hattothedesiredvalue120=115.93+4.64A-0.41B+4.14ABwhereB=+1120=115.93+4.64A-0.41(1)+4.14A(1)A=0.51Decodethecodedvalue ConfirmationRunTotestforconfirmationweshouldexpectallresponsetofallwithinyhat+/-3S-hatForthisexampleitshouldbe120+/-3(0.669)or(117.993,122.007)ComputetheCpk,Cp,yc和c SummaryFor设计ofExperimentforLinearModelA)DefinetheProblemB)ObjectiveoftheExperimentC)IdentifythenumberoffactoryouhaveinthesystemD)IdentifythenumberofResponseE)DeterminethenumberofresourcesfortheexperimentD)Determine和selectthetypeofDOE设计和AnalysisStrategiesF)RandomizeoftheRuninthe设计MatrixG)ConductingtheExperiment和RecordingtheDataH)AnalyzetheData和DrawConclusion,MakepredictionsG)DoConfirmationRun Exercise-LinearModelingDOEBreakintoteamsReviewthestep流程forconductingDOEUsingthecause和effectdiagramUsingtheFullFactorialDesignConsiderthisFactorsinyourstratapultPullBackAngle(110,177)TensionPinposition(1,3)RubberB和HookPosition(1,3)StopanglePosition(2,4)Measurethexcm+/-5inchBuildthemodel和FindtheY-hat和S-hatequationReduceVariationbyminimizingSUseoptimizertoestimatesetting:Target150cm+/-5cmConductConfirmationRun20shot:CalculateCPK FlowChartIn设计ofExperimentStartStatementofproblem&objectiveDeterminewhattomeasure&completePE/CE/CNXSOP/FMEA/POKEYOKEHowManyLevelsForEachFactors?HowManyLevelsForEachFactors?FullFactorialK=2nreps9K=3nreps5K=4nreps3K4K=56K116K825-11/2Factorialnreps312RunPLACKTT-BURMANorTAGUCHIL12Screeningnreps416RunFRACTIONALFACTORIALnreps3HowManyFactors?CentralCompositeorBox-BehnkenBoxK=2nreps9(CCD)K=3nreps5(CCDorBB)K=4nreps3(CCDorBB)K=5nreps3(CCD)ModelingK5Screening6K74K7K3TAGUCHIL18Screening(alsoIncludeOne2LevelFactornreps4FULLFactorialK=2reps7K=3reps3ModelingorScreening?QuantitativeOnlyTypeofFactorsNotallQuantitative(I.eatleast1Qualitative)Note1:SampleSize(nreps)isfor95%confidenceinPredictedS和99.99%ConfidenceinPredictedY:Ifnreps/2itwill提供75%confidenceinPredictedS和95%confidentinpredictedYForOther2Level设计Refertothenextslide3levels2Levels Statistical流程Control(SPC)Statistical流程Controlisusedto“ListenToTheVoiceOfThe流程” DAYSCOSTED INVENTORY(DCI)UPPERCONTROLLIMITLOWERCONTROLLIMITCENTERLINEControlchart Underst和thetheorybehindcontrolchartsConstructthefollowingControlCharts:Variables:Xbar&R,Xbar&s,Individuals和MovingRange,和Batch流程esAttributes:p,np,c,u,和LowFrequencyPre-ControlInterpretControlChartsMethodsforchoosingappropriatesubgroupsamplesizes和frequencyofsamplingMethodsofimplementation和reaction提供sagraphicalrepresentationof流程performance和stabilityMonitoringoutputvariable和controlInputVariableLesson目标: 为什么doweuseStatistical流程Control?It提供sagraphicalrepresentationof流程performance和stabilityHelptomonitoringKey流程OutputVariable和controlKey流程InputVariableSPChelptodetectspecialCauses和minimizingreactiontonormalvariationSPCisa工具useforobservingvariation.Itcanbeusetomonitorstatisticalsignals和improveperformance.Itcanbeappliedtomanyarea.EquipmentperformanceDocumenterrorsSalesperQuarter缺陷ratesTimetakentocompletebytaskSPCshouldbeapplyto流程characteristics(X’s)ratherthanfinishedgoods(Y’s).Untilthe流程inputsbecomethefocusofoureffort,thefullpowerofSPCmethodstoimprove质量,increase产品ivity,和reducecostcannotberealized. WhatisSPC?SPCisastatisticallybasedmethodof:1.Evaluatingtheperformance(stability)of流程和产品variables2.Pointingouttheexistenceofspecialcausesofvariation3.Removingthespecialcauses和maintain和monitornormal流程variation All流程eshavenaturalvariationwecallthemcommoncauses和unnaturalvariationwecallassignablespecialcausesWeuseSPCtomonitor和improvethe流程.WhilecontrolchartscanindicatespecialcausesthroughOut-of-Controlsignals.Controlchartsarethemeansthroughwhich流程和产品parametersaretrackedstatisticallyovertime.Controlchartshasupper和lowercontrollimitsthatreflectthenaturallimitsof(random)variability.Theselimitsmustnotbecomparedtothecustomerspecificationlimits.Controllimitsarebasedonestablishing+3smeanlimitsforXorY.Basedonstatisticalprinciples,controlchartsallowfortheidentificationofunnatural(nonrandom)patterns:specialcauseshavechangedthe流程.TheactionswetaketoattackthespecialcausesarethekeytosuccessfuluseofSPC.WhatisSPC? WalterShewhartinventedthecontrolchartin1924WalterA.ShewhartShewhart:TheFatherofControlCharts Wewanttomonitor和improvetheInputsInput流程/SystemOutput2.IdentifyRootCause3.ImplementCorrectiveAction4.Verify和Monitor1.DetectAssignableorSpecialCauseStatistical流程ControlTheGoalofSPC WhereshouldweuseSPCCritical流程ParametersHighestRPN’sfromFMEAWhenamistakeproofingdeviceisnotfeasibleCustomerrequirements和Criticalto质量measureManagementcommitmentsInitially,the流程outputsmayneedtobemonitored SelecttheappropriatevariablestocontrolchartHighRPN’sIn流程monitoringCritical流程parametersSelectthedatacollectionpointSelectthetypeofcontrolchartsAttribute(np,p,c,u)Variable(X-bar/R,X-bar/S,Individual)EstablishbasisforrationalsubgroupingWhatconstitutesarationalsubgroup?Doesitrepresentthepopulationfromwherethesamplesaretaken?DeterminesamplesizeDeterminemeasurementmethod/criteriaHowtoImplementSPCCharts HowtoImplementSPCCharts(cont)Step1Wehavetoverifygagecapabilityofthemeasurementsystem(MSA)Step2Collectdata和Setuparuncharttostudythe流程variabilityStep3EstablishcontrollimitsStep4EstablishamethodofperiodicallyverifyingtheiraccuracyStep5CreateformsforchartingthedataPrepareaWorkInstructionforthechartingDocumentationforsamplingDatatobecollect和measurementmethodCorrectiveactionplanwhenoutofcontroloccurStep6TrainpersonnelOperatorsSupervisorsEngineersStep7Institutionalizethecharting ImplementationofSPCSystemSPConX’sorY’swithoutproper培训=WALLPAPER.SPConX’sorY’swithfullytrainedoperators.Operatorswillunderst和thesignals,butmanagementwillnotempowerthemtostopforinvestigation.Operatorswilllearntoignoreordisconnectthewarningsignalsonce产品ionbecomesthe#1priority.SPC,withResponseAction=Inspection.Shorttermcontainmentequalsauditingor100%inspection.SPC,withResponseAction=EquipmentFlag.流程isstoppedorequipmentautomaticallyshutsdown,sothat缺陷willnotmoveforward.SPC,withResponseAction=Countermeasure.改进aremadesothatdefectcannotoccuragain. AdvantagesSPCisaprovenmethodforimproving流程Itiseffectiveinprevention缺陷Helpustopreventunnecessary流程changesItisagooddiagnostic工具SPC提供informationaboutthe流程capabilityCanbeusedforbothattribute和variabledatatypeControlchartisabletotrackthe流程overtimeDisadvantagesPeoplemustbetrained和retrainedonthe知识.DatamustbegatheredcorrectlyMustbeabletomaintainalltheinformation和chartedcorrectlyChartsmustbeanalyzedcorrectlyReponsestotheoutofcontrolmustbeappropriate和everytimeAdvantages和DisadvantagesofSPC ThetheorybehindacontrolchartfollowstheprinciplesofsimpleconfidenceintervalsmUCLLCLa/2a/2Whenthetrue流程isrunningatmeanm,theprobabilitythatasamplemeanfallsoutsidethecontrollimitsequalsaTheTheoryofControlCharts ModelforCenterline,UCL&LCLforVariablesChartUpperControlLimitLowerControlLimitCenterLineUCL=m+k1sCenterLine=mLCL=m-k2sBecausewemustuseestimatesfromsamples:mx=themeanofthesamplemeansss=the标准偏差derivedfromthesamplesk1,k2=constantsimplyingunusualobservations(often=3)=ComponentsofaControlChart TheTimeElementofControlChartsSPCTimeSequenceUCLLCL Oneadvantageofacontrolchartinvolvesitsabilitytotrackthe流程overtimeThistime-baseddependencemeansthatcertaintrendsorpatternsmayindicatethatspecialcausesareoccurringovertime.TimeComponentsofaControlChart AssignableCauses:-Thingsthatareunpredictable(hour-to-hour,day-to-day,week-to-week)-目标isto:Detect和eliminateassignablecausesCommonCauses:-Instatisticalcontrol-Naturalortypical流程variationsTimeTypesofVariation TimeTimeIfonlycommoncausesofvariabilityarepresent,thenthe流程outputisconstant和predictableovertimeIfonlyspecialcausesofvariabilityarepresent,the流程outputisneitherCONSTANT,norPREDICTABLEVariationvsSPCPF/CE/CNX/SOP 流程IN-CONTROL,BUTNOTCAPABLE(outofspecs)(Onlytypical流程variability,butexcessive)流程IN-CONTROL和CAPABLE(The流程variabilityhasbeenreduced和nowmeets流程specs)Lowerspecificationlimit(LSL)Upperspecificationlimit(USL)TimeVariationvsSpecifications ControlChartTypesChoosingthecorrectcontrolchartTypeofdataIssamplesizefixed和n<10?Constantsamplesize?constantareaofopportunity?Counting缺陷ordefectives?ucp,nppXbar-SIMRChartDatatendstobenormallydistributedbecauseofcentrallimittheoremAttributesVariablesDefectivesYesNoDefectsNoSub-groupsYesNeedtoConstructacontrolChartBinomialPoissonYesNoXbar-RChartn=1220).Usepropersamplesizetodetectcriticaldifference.Step4Ifthereisa流程change,assesstheimpactofthechangetodeterminenextcoursesofaction:LineshutdownRootcauseinvestigationIfyouhaveaverycapable流程(Cpk>1.5)maybenefitfromrelaxedcontrollimits,suchthatout-of-controlsignalscorrespondmoretoout-of-specmaterial 流程CapabilityEvaluation-ShortTermCpkCalculationusingX-bar-Rcharta)Usetoestimateb)Usetoestimateusingthisignoretheshiftbetweendatasetmeasurewithingroupvariabilityc)UseLSL和USLtocalculateShorttermCpk流程CapabilityEvaluation-LongTermPpkCalculationusingX-bar-Rcharta)Usetoestimateb)Usetoestimateusingthisincludetheshiftinalldatameasuretotalgroupvariabilityc)UseLSL和USLtocalculate.LongtermPpkUSLLSLUSLLSL PartitioningthesourcesofvariabilityExerciseForthepreviousxbarRchartaccomplishedinclassfindLongtermshorttermshiftTotalVariability=WithinGroupVariability+BetweenGroupVariability Summaryof流程CapabilityMeasuresforLongTermVsShortTermShortTerm(ST)LongTerm(LT) Shewhartcontrolchartconstantsn=SubgroupsizeConstantisfromASTMManual ExerciseLetfirethecatapult,withafixsettingEachpersonfire3shotsasasubgroupsizeof3aninxbar-Rchart.Usesoftwaretodrawthexbar-Rchart和evaluate流程controlAssumethespecificationlimitstobe+/-5inchesCalculateCpkFindLongtermshorttermshift Xbar-Rchart IndividualsMovingRangeChart-wastetankreadingsWherenisthesubgroupsizeD3,D4&E2areShewhartconstantsRBarIstheaverageofthemovingrangesxBaristheaverageofindividualreadings NowLetusethepreviouscatapultdatatoconstructaIndividualMovingAverageChartTaketheDatawithonlyontheFirstRowCalculatetheIndividualsChartlimits和rangelimitsExercise IndividualsMovingRangeChart Xbar和SchartsWherenisthesubgroupsizeA3,B3&B4,C4areShewhartconstantssIsthe标准偏差xDoubleBarIstheaverageofsamplexbarWhenSubgroupn>then10标准偏差isabetterestimatorthatR.ThuswesuggestthatforN>10Xbar-Schartisrecommended.Xbar-SchartisalsorecommendedfornsizesvariesformonetimeincrementtothenextX.Forexamplethesamplesizefromlottolotcanbe5,8,4,和10 Xbar&sChart Rangevs标准偏差Forsubgroupsofsize2,3or4,thereislittledifferenceinaccuracyAssubgroupsexceed4,the标准偏差becomesincreasinglymoreaccuratethantherange,tothepointthattherangeshouldnotbeusedforsubgroupsgreaterthan10Strategy:usethestddevapproachexceptwhen...ManualcalculationisrequiredOperators(orothers),withoutunderstandingofthe标准偏差,willbeinterpretingthechartsThereporting结构isallergictoGreeklettersHistoricalNote:whenShewhartdevelopedthesechartsinthe1920’s,therewasnoeasywaytocalculatethe标准偏差.Thus,therangeapproachbecameingrainedinSPCapplication. ChoosingtheSubgroupSeveralconsiderationsmustbemadewhenchoosingthesubgroupsThesubgroupis“rational”Thesampleadequatelyreflectsthe流程performanceduringthetimeperiodwhichitrepresentsThesamplefacilitatesaspecificidentificationofprobablesourcesofassignablecausesThesamplecanbetakenwithaconsistent方法和accuratelymeasuredThesubgroupisofsufficientsizetodetectachangeinthe流程thatisconsideredsignificantThesubgroupsizemustalsonotbetoolarge,orelsesignalsfor流程shiftsconsideredinsignificantwilloccurThefrequencyofsamplingisoftenenoughtoavoidexcess产品ionofunacceptablematerial ChoosingaCorrectSampleSizeThecapabilityofthe流程willoften提供essentialinformationforchoosingasamplesizePresentcapability提供sappmvalue和aSigmalevel(assuminganormaldistribution)Anundesirableincreaseinppmtoagivenvalueofppm*canbetranslatedintoaSigmaLevel*ThedifferencebetweenSigmaLevel和SigmaLevel*representstheundesirableshiftinmean--innumberof标准偏差units.Thisbecomesthe“criticalshift.”AconfidencelevelcanbeassignedtotheprobabilityofdetectingthisshiftThefollowingchartcanbeusedtodeterminethebestsamplesizetodetectthecriticalshiftwiththedesiredconfidence SampleSize:Sensitivitywith1GroupProbabilityofDetectingaShiftinMeanintheNextSubgroupforVariousControlChartSubgroupsofSizenFrom“OneGoodIdea”byLyleDockendorf,质量Progress,October1992n=20n=15n=10n=8n=5n=4n=3n=2 SampleSize:Sensitivitywith1GroupProbabilityofDetectingaShiftinMeanintheNextSubgroupforVariousControlChartSubgroupsofSizenFrom“OneGoodIdea”byLyleDockendorf,质量Progress,October1992n=20n=15n=10n=8n=5n=4n=3n=21.5sSampleSize=890%confidenceofdetecting SampleSize:SensitivityinNext3GroupsProbabilityofDetectingaShiftinMeanintheNext3SubgroupsAlsoUsingthe2-Points-out-of-3-in-the-Warning-ZoneRuleforVariousControlChartSubgroupsofSizenFrom“OneGoodIdea”byLyleDockendorf,质量Progress,October1992n=20n=15n=10n=8n=5n=4n=3n=2 FrequencyofTakingSubgroupsThefrequencywithwhichsubgroupsaretakenshouldreflect:Historicalbehaviorof流程Frequentout-of-controloccurrencesrequiremorefrequentsamplingIfshiftchangesmaycauseachange,theneachshiftshouldbesampledTheimpactto质量ofrunninginanout-of-controlstateTheeaseofimplementingasamplingplanThecostoftaking和measuringsamplesOftenthefrequencyissetasamatterofconvenience;e.g.onceortwiceeachdayorshift.Althoughthismayworkinpractice,amethodofcalculatinganappropriatefrequencyhasbeendeveloped. FrequencyCalculationDefinethefollowing:S=Costoftaking和measuringanentiresampleL=Lossperunit(averagedoverallunits)whenoperatingoutofstatisticalcontrolT=Timelag(innumberofunits)ittakestoreactF=Averagenumberofunitsbetweenout-of-controlconditionsiftheywerefound和fixedimmediately(i.e.notimelag)p=Probabilityofdiscoveringanout-of-controlconditionwithonesamplen=samplingfrequency(innumberofunitsproduced)Then:2p(F+T)Sn=L(2-p) Exercise:SubgroupSize和FrequencySupposeyouhavea流程withCpk=1.33.Youwanttoguardagainsta流程meanshiftthatwillproduce3%defectivewithabout90%confidence.Whatsubgroupsizeshouldyouuse?Youknowthefollowingaboutyour流程Itruns2shifts/day和5days/weekEachshiftproducesabout400partsAscrappedpartrepresentsalossof$5Historically,the流程hasgoneoutofstatisticalcontrolaboutonceevery2weeksIttakesabout1hourtoreact和fixa流程problemItcosts$.50tomeasureasampleunitGiventhatyouusethesubgroupsizecalculatedinthefirstsection,whatfrequencyshouldyousampleat?
