6sigma统计概念培训.ppt

Hence,P(4.5x8.5)=0.9864-0.3745=0.6119(approxequalto0.6123bybinomialdistribution),天马行空官方博客：,IntroductiontoSampling,Whatispopulationinstatistic?Apopulationinstatisticreferstoallitemsthathavebeenchosenforstudy.,Whatisasampleinstatistic?Asampleinstatisticreferstoaportionchosenfromapopulation,bywhichthedataobtaincanbeusedtoinferontheactualperformanceofthepopulation,Population,Sample2,Sample6,Sample8,Sample1,Sample3,Sample7,Sample4,Sample5,Samplingdistribution-adistributionofsamplemeansIfyoutake10samplesoutofthesamepopulations,youwillmostlikelyendupwith10differentsamplemeans和sample标准偏差s.ASamplingdistributiondescribestheprobabilityofallpossiblemeansofthesamplestakenfromthesamepopulation.,Whensamplesizeincreases,thestandarderror(orthestddeviationofsamplingdistribution)willgetsmaller.,Samplingdistribution(cont),CentralLimitTheorem,Exampleofsamplingdistribution,Thepopulationdistributionofannualincomeofengineersisskewednegatively.Thisdistributionhasameanof$48,000,和a标准偏差of$5000.Ifwedrawasampleof100engineers,whatistheprobabilitythattheiraverageannualincomeis$48700和more.,Exampleofsamplingdistribution(cont),Therefore,mean=48000sigma=500X=50000Z=(48700-48000)/500=700/500=1.4,Fromthestandardizednormaldistributiontable,P(X$48700)=0.9192Therefore;P(X$48700)=1-0.9192=0.0808Thus,wehavedeterminedthatithasonly8.08%chancefortheaverageannualincomeof100engineerstobemorethan$48700.,CentralLimitTheoremExercise,Breakinto4groupsasbelow:Group1:Thepopulationgroup.Group2to4:Thesamplesub-groupThepopulationgroupwillhave3oftheirmembersthrowingasingledices60timeseach.Atotalof180throwswillberecorded和thisdatawillbethepopulationdata.Eachsamplesub-groupwillhave3oftheirmembersthrowing5dicesatonetime,和collectthesum和averagevalueoftheparticularthrow.Eachmemberistoconduct20throws和obtainthesamplemeanofeachthrow.Attheendoftheexercise,atotalof180samplemeanswillbecollectedfromthe3sub-groups.Fromthearriveddata,plotthehistogram和commentonthedistributionofboththepopulation和thesamplings.,Thefinitepopulationmultiplier,Previouslywesaythat:,Finitepopulationmultiplierwithrespecttopopulation和samplesize,RuleofthumbsThefinitepopulationmultiplierneedonlybeincludedifpopulationsizetosamplesizeratioislessthan25.,ConfidenceInterval,“Pointestimates”Apointestimateisasinglenumberthatisusedtoestimateanunknownpopulationparameter.,Whatdoes95%confidenceintervalsmeans?Itdefines95%ofthetime,theaveragevalueofarandomsamplingwillfallwithinavaluerangewhichis+/-1.96standarderrorfromthesamplemean.,为什么1.96standarderror?Referringtothestandardizednormaldistributiontable,whenz=1.96,theassociatedprobabilityis0.975(or97.5%)asbelow:,Butthisisa2tailsinterval(i.e.+/-1.96),henceweneedtominusoffanother2.5%fromtheotherend,givingatotalcoverageof95%.,EquationforcomputingconfidenceintervalsForcontinuousdata:,Fordiscretedata:,ConfidenceintervalforcontinuousdataAlargediskdrivemanufacturerneedsanestimateofthemeanlifeitcanexpectfromthemagneticmediabyreciprocallyswitchingitsdigitalstateat1MHzfrequency.Thedevelopmentteamhasdeterminedpreviouslythatthevarianceofthepopulationlifeis36months,和hadconductedareliabilitytestingfor100samples,collectingdataofitsusefullifeasbelow:,Fromtheabovedata,whatisthe95%confidenceintervalfortheusefulmeanlifeofthemagneticmediainadiskdrive?Whatdoesthismean?,Confidenceintervalforcontinuousdata(cont),Applyingtheconfidenceintervalsequation:Upperconfidencelimit=27.75+1.96(0.6)=28.926Lowerconfidencelimit=27.751.96(0.6)=26.574,Assuch,thereis95%confidencelevelthattheaverageusefullifeofthemagneticmediatofallbetween26.574和28.926months.,ConfidenceintervalfordiscretedataBreakinto4or5teams,combinedalltheMthen流程isnotcentered,Cpk=Cp;then流程iscentered,The流程ofcollecting,presenting和describingsampledata,usinggraphical工具和numbers.ParetoChartPopulationmeanHistogramPopulation标准偏差,DescriptiveStatistics,ProbabilityTheory,Probabilityisthechanceforaneventtooccur.Statisticaldependence/independencePosteriorprobabilityRelativefrequencyMakedecisionthroughprobabilitydistributions(i.e.Binomial,Poisson,Normal),InferentialStatistics,The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),CentralLimitTheorem,某些takeawaysforsamplesize和samplingdistribution,PercentilesofthetDistribution,Whereby,df=Degreeoffreedom=n(samplesize)1Shadedarea=one-tailedprobabilityofoccurencea=1ShadedareaApplicablewhen:Samplesize450).DesignatedasH1,ElementsofHypothesisTesting(cont),Exampleifwewanttotestwhetherapopulationmeanisequalto500,wewouldtranslateitto:NullHypothesis,H0:mp=500和consideralternatehypothesisas:AlternateHypothesis,H1:mp500;(2tailstest),Rememberconfidenceinterval,at95%confidencelevelstatesthat:95%ofthetimethemeanvaluewillfluctuatewithintheconfidenceinterval(limit)5%chancethatthemeanisnaturalfluctuation,butwethinkitisnotalpha(a)probability,TypeIIErrorAcceptinganullhypothesis(H0),whenitisfalse.Probabilityofthiserrorequalsb,TypeIErrorRejectingthenullhypothesis(H0),whenitistrue.Probabilityofthiserrorequalsa,Usethestderrorobservedfromthesampletosetconfidencelimiton500(mH0).TheassumptionismH0hasthesamevarianceasmp.,ElementsofHypothesisTesting(cont),Otherpossiblealternatehypothesisare:AlternateHypothesis,H1:mp500;(1tailtest)AlternateHypothesis,H1:mp500For95%confidencelevel,a=0.05.SinceH1isonetailtest,rejectareadoesnotneedtobedividedby2.,某些hypothesistestingsthatareapplicabletoengineers:Theimpactonresponsemeasurementwithnew和old流程parameters.Comparisonofanewvendorsparts(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”(cont),Fromstandardnormaldistributiontable,TheZvaluecorrespondingto0.005tailareais2.58.,a=0.01(2tails),since2tailstest,thereforetailarea=a/2=0.005;,“Largesampletestofhypothesisaboutapopulationmean”(cont),Whatdoes99%confidencelevelmeansintheaboveexample?Itdefinesthelimitswhereby99%oftheaveragesamplingvalueshouldfallwithin,giventhedesirable(hypothesised)meanasmH0.AnyvaluefalloutsidethisconfidencelimitindicatesthesamplemeanissignificantlydifferentfrommH0.Inotherwords,wewillonlyconcludethealternatehypothesisH1(thatthemeansaredifferent)ifwearemorethan99%sure.,TheObservedSignificancelevel,p-value,p-valueistheprobabilityforconcludingthenullhypothesisH0thatbothpopulationmeansareequalwiththeobservedsampledata.Hence1pvaluewillbetheconfidencelevelwehaveonthealternatehypothesis.,Usingthethrottlequestionasanexample:Weknowthatthemeanresponsetimeis3standarderrorawayfrom1.2msec(mH0),thereforeZ=3.Sincethisisa2tailstest,p-value=P(Z3)=2P(Z3)Fromstandardnormaldistributiontable,P(Z=3)=0.9987P(Z3)=10.9987=0.0013p-value=2P(Z3)=0.0026,Instatisticalterm,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”(cont),Fromstudenttdistributiontable,Thetvaluewith19dfcorrespondingto0.1tailareais=1.3277.,“Largesampletestofhypothesisaboutapopulationproportion”,Amethodcurrentlyusedbydoctorstoscreenforpossiblestomachulcerfailstodetecttheulcerin20%ofthepatientswhoactuallyhavethedisease.Supposeanewmethodhasbeendevelopedthatresearchershopewilldetectstomachulcermoreaccurately.Thisnewmethodwasusedtoscreenarandomsampleof140patientsknowntohavestomachulcer.Ofthese,thenewmethodfailedtodetectulcerin12ofthepatients.Using95%confidencelevel,doesthissample提供evidencethatthefailurerateofthenewmethoddiffersfromtheonecurrentlyinuse?,Solution:Lettheprobabilityofsuccessinmisseddetectionasptherefore;H0:p=0.2(i.e.pH0)H1:p0.2Samplesize,n=140(i.e.usestandardnormalzastheteststatistic)Computethestandarderrorfornullhypothesis(i.e.whenp=0.2),TestifpH03stderrorwillgivereasonablevalue(i.e.between0to1)pH03stderror0.23(0.034)=(0.166,0.234),“Largesampletestofhypothesisaboutapopulationproportion”(cont),Calculatethenumberofstandarderrorsbetweenthesampled和hypothesisedvalue:,Conclusions:Sincep=0.086isinrejectarea,werejectnullhypothesis和concludethatthenewscreenmethodissignificantlydifferentthantheoldscreenmethodwith95%confidencelevel.With3.36stderrorfrompH0(0.2),thep-valueiscalculatedtobe0.00078,hencethereisa99.922%confidenceinthealternatehypothesis.Itappearsthatthenewscreenmethodwillgivelessermissdetectionforstomachulcer.,PowerofaHypothesisTesting,TypeIErrorRejectingthenullhypothesis(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).,PowerofaHypothesisTesting(cont),Assuchthereisatradeoffbetweena和b.Asadecreases,bincreases和viceversa.,mH0,(a/2)RejectArea,(a/2)RejectArea,PowerofaHypothesisTesting(cont),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(nullhypothesisthatmeandosagefromshippmentis100