六西格玛绿带培训3

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wrapupDay3:Multipleregression和modelbuildingIntroductiontomultipleregressionmodelBuildingamodelFittingthemodelwithleastsquaresapproachAssumptionsformodelUsefulnessofamodelAnalysisofvarianceUsingthemodelforestimation和predictionPitfallsinpredictionmodel--------------------------------------------------------Break-----------------------------------------------------------------Multipleregressionexercise(statapult)Presentationformultipleregressionexercise--------------------------------------------------------Lunch------------------------------------------------------------------Qualitativedata和dummyvariablesModelswith2ormorequantitativeindependentvariablesTestingthemodelModelswithonequalitativeindependentvariableComparingslopes和responsecurve--------------------------------------------------------Break-----------------------------------------------------------------ModelbuildingexampleStepwiseregression–anapproachtoscreenoutfactorsDay3wrapupDay4: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confirmationrunDay4wrapupDay5: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)OutofcontrolnotnecessarilyisbadDay5wrapupRecapofStatisticalTerminologyDistributionsdiffersinlocationDistributionsdiffersinspreadDistributionsdiffersinshapeNormalDistribution-6-5-4-3-2-10123456------------------------------99.9999998%-------------------------------------------99.73%--------------------95.45%---------68.27%--±3

variationiscallednaturaltoleranceAreaunderaNormalDistribution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,Cpk<Cp;then流程isnotcenteredCpk=Cp;then流程iscenteredUSL-Y3Y-LSL3Cpk=min,The流程ofcollecting,presenting和describingsampledata,usinggraphical工具和numbers.ParetoChartPopulationmeanHistogramPopulation标准偏差DescriptiveStatisticsEstimatesforDescriptiveStatisticsThe流程ofestimatingthepopulationparametersfromsample(s)thatwastakenfromthepopulation.Samplemean,XPopulationmean,mSample标准偏差,SPopulation标准偏差,(whensamplesize,n>20)Estimated标准偏差,R/d2

Population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samplingdistribution

Forlargesamplesize(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标准偏偏差isunknownPopulationdistributionisatleastapproximatelynormallydistributedt(a,u)aAreaunderthecurvePercentilesoftheNormalDistribution/ZDistributionZaAreaunderthecurveWhereby,Shadedarea=one-tailedprobabilityofoccurencea=1––ShadedareaStudenttDistrbutionexampleFDArequirespharmaceuticalcompaniestoperformextensivetestsonallnewdrugsbeforetheycanbemarketedtothepublic.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:标准偏偏差,Sforsmallsamplesizeisprobablynotaccurate标准偏偏差,SforsmallsamplesizeisprobablytoooptimisticWiderintervalisthereforenecessarytoachievetherequiredconfidencelevelSummaryofformulaforconfidencelimit6Sigma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%----------------LCLUCLOutofcontroldatapointsOurExplanationMostfrequentlyusedsamplesizeforSPCinindustryis3to5unitspersampling.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).DesignatedasH1ElementsofHypothesisTesting(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.Thereforeifmpismorethan500by

1.645stderror,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importantInferencesbasedonasinglesample“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.m

X=1.05msec; s

S=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质量量ofallpotentialcandidateshead-huntedbythe招聘聘Agency.Themanagementwantstoknowtheeffectivenessofthe招聘聘Agency,asitwasbelievedthattheaveragetestscoreforalltheidentifiedcandidatesshouldbeequalormorethan90inordertoreducetheriskofassigningthewrongcandidatesforthetask.WhenAmyreviewsthetestsresultofaparticularbatchof20candidates,shefindsthatthemeanscoreis84和the标准准偏偏差差is11.Asthisisaverycritical招聘聘project,Amywantstobemorereservewithheranalysis,和decidedtobemorebiastowardsprovingthatthepopulationmeanislesserthan90.Asaresult,aconfidencelevelof90%willbeusedinheranalysis.“Smallsampletestofhypothesisaboutapopulationmean”(con’’t)Solution:Sincethesamplesizeisrelativelysmall(i.e.<30),weshouldusetstatistic.m

X=84; s

S=11; n=20; a=0.1(1tail)Degreeoffreedom,df=19NullhypothesisH0:m=mH0(90)AlternatehypothesisH1:m<mH0(90)mH0=900.1ofarea(a)RejectArea1.3277stderrorX=84Fromstudent‘‘t’’distributiontable,Thetvaluewith19dfcorrespondingto0.1tailareais=1.3277.HowmanystderrorisXlesserthan90?ThereforeXis–2.439stderrorsawayfrom90和intherejectarea.Amywillrejectthenullhypothesis和concludethattheaveragescoreislessthan90marks.2.439stderrorAcceptanceArea“Largesampletestofhypothesisaboutapopulationproportion”Amethodcurrentlyusedbydoctorstoscreenforpossiblestomachulcerfailstodetecttheulcerin20%ofthepatientswhoactuallyhavethedisease.Supposeanewmethodhasbeendevelopedthatresearchershopewilldetectstomachulcermoreaccurately.Thisnewmethodwasusedtoscreenarandomsampleof140patientsknowntohavestomachulcer.Ofthese,thenewmethodfailedtodetectulcerin12ofthepatients.Using95%confidencelevel,doesthissample提供供evidencethatthefailurerateofthenewmethoddiffersfromtheonecurrentlyinuse?Solution:Lettheprobabilityofsuccessinmisseddetectionasptherefore; H0:p=0.2(i.e.pH0) H1:p<>0.2Samplesize,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).m1

ControlbyaControlbybPowerofaHypothesisTesting(con’’t)Assuchthereisatradeoffbetweena和b.Asadecreases,bincreases和viceversa.mH0(a/2)RejectArea(a/2)RejectAream1

75%(a=0.25)95%(a=0.05)ConfidenceIntervalbform1whena=0.05bform1whena=0.25PowerofaHypothesisTesting(con’t)Ahospitaluseslargequantitiesofpackageddosesofaparticulardrug.Theindividualdoseofthisdrugis100cc.Theactionofthedrugissuchthatthebodywillharmlesslypassoffexcessivedoses.Ontheotherhand,insufficientdoses(i.e.99.6cc和below)donotproducethedesiredmedicaleffect,和theyinterferewithpatienttreatment.Thehospitalhaspurchaseditsrequirementsofthisdrugfromthesamemanufacturerforanumberofyears和knowsthatthepopulation标准偏差差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(nullhypothesisthatme