管理统计学ppt课件 65页

SlidesPreparedbyJOHNS.LOUCKSSt.Edward’sUniversityUpdatedbyChenWeihua\nChapter8IntervalEstimationLearningobjectivesKnowhowtoconstructandinterpretanintervalestimateofapopulationmeanand/orapopulationproportion.Understandtheconceptofasamplingerror.Beabletouseknowledgeofasamplingdistributiontomakeprobabilitystatementsaboutthesamplingerror.Understandandbeabletocomputethemarginoferror.Learnaboutthetdistributionanditsuseinconstructinganintervalestimateforapopulationmean.Beabletodeterminethesizeofasimplerandomsamplenecessarytoestimateapopulationmeanand/orapopulationproportionwithaspecifiedlevelofprecision.\n[------------------------------------------][------------------------------------------][------------------------------------------]Chapter8IntervalEstimationIntervalEstimationofaPopulationMean:Large-SampleCaseIntervalEstimationofaPopulationMean:Small-SampleCaseDeterminingtheSampleSizeIntervalEstimationofaPopulationProportion\nIntroductionConsiderthefollowingstatements:x=550Asingle-valuedestimatethatconveyslittleinformationabouttheactualvalueofthepopulationmean.Weare99%confidentthatisintheinterval[449,551]Anintervalestimatewhichlocatesthepopulationmeanwithinanarrowinterval,withahighlevelofconfidence.Weare90%confidentthatisintheinterval[400,700]Anintervalestimatewhichlocatesthepopulationmeanwithinabroaderinterval,withalowerlevelofconfidence.\n抽样推断抽样推断的概念：抽样推断是在抽样调查的基础上，利用样本的实际资料计算样本指标，并据以推算总体相应数量特征的一种统计分析方法\n抽样推断抽样推断的内容：参数估计假设检验\n抽样推断抽样推断的特点：抽样推断是由部分推断整体的一种认识方法抽样推断是建立在随机取样的基础上抽样推断是运用概率估计的方法，而不是确定的数学分析法抽样推断的误差可以事先计算并加以控制\nPointEstimateAsingle-valuedestimate.Asingleelementchosenfromasamplingdistribution.Conveyslittleinformationabouttheactualvalueofthepopulationparameter,abouttheaccuracyoftheestimate.ConfidenceIntervalorIntervalEstimateAnintervalorrangeofvaluesbelievedtoincludetheunknownpopulationparameter.Associatedwiththeintervalisameasureoftheconfidencewehavethattheintervaldoesindeedcontaintheparameterofinterest.TypesofEstimators\n参数估计的基本知识参数估计的条件：要有合适的统计量作为估计量要有合理的允许误差范围（准确性）抽样误差抽样极限误差误差率估计精度要有可接受的置信度（可靠性）估计的置信度概率度\n参数估计点估计以样本指标直接估计总体参数。评价准则的数学期望等于总体参数，即该估计量称为无偏估计。无偏性有效性当为的无偏估计时，方差越小，无偏估计越有效。一致性对于无限总体，如果对任意则称的一致估计。是充分性一个估计量如能完全地包含未知参数信息，即为充分量估计量\nAconfidenceintervalorintervalestimateisarangeorintervalofnumbersbelievedtoincludeanunknownpopulationparameter.Associatedwiththeintervalisameasureoftheconfidencewehavethattheintervaldoesindeedcontaintheparameterofinterest.Aconfidenceintervalorintervalestimatehastwocomponents:ArangeorintervalofvaluesAnassociatedlevelofconfidenceConfidenceIntervalorIntervalEstimate\nConfidenceIntervalorIntervalEstimateTwoformulas:P(x1≤x≤x2)=1-\nSamplingErrorTheabsolutevalueofthedifferencebetweenanunbiasedpointestimateandthepopulationparameteritestimatesiscalledthesamplingerror.Forthecaseofasamplemeanestimatingapopulationmean,thesamplingerrorisSamplingError=LetE(Δ)=themaximumsamplingerrormentionedintheprecisionstatement.Eistheamountaddedtoandsubtractedfromthepointestimatetoobtainanintervalestimate.Eisoftenreferredtoasthemarginoferror.Wehave\nProbabilityStatementsAbouttheSamplingErrorKnowledgeofthesamplingdistributionofenablesustomakeprobabilitystatementsaboutthesamplingerroreventhoughthepopulationmeanisnotknown.Aprobabilitystatementaboutthesamplingerrorisaprecisionstatement.\nPrecisionStatementThereisa1-probabilitythatthevalueofasamplemeanwillprovideasamplingerroroforless.ProbabilityStatementsAbouttheSamplingError/2/21-ofallvaluesSamplingdistributionof\nConfidenceIntervalorIntervalEstimateXf(X)1-aa/2a/2(x1)(x2)P(x1≤x≤x2)=1-\nConfidenceIntervalorIntervalEstimateWhen△isknown，P=?compute，△isknownZF（z）Refertotable（x1,x2）P=2F（z）\nConfidenceIntervalorIntervalEstimateWhen1-isknown,（x1,x2）?compute，1-isknownRefertotable（x1,x2）(1-a)/2Z/2\nIntervalEstimateofaPopulationMean:Large-SampleCase(n>30)WithKnownwhere:isthesamplemean1-istheconfidencecoefficientz/2isthezvalueprovidinganareaof/2intheuppertailofthestandardnormalprobabilitydistributionsisthepopulationstandarddeviationnisthesamplesize\nIntervalEstimateofaPopulationMean:Large-SampleCase(n>30)WithUnknownInmostapplicationsthevalueofthepopulationstandarddeviationisunknown.Wesimplyusethevalueofthesamplestandarddeviation,s,asthepointestimateofthepopulationstandarddeviation.\nIfthepopulationdistributionisnormal,thesamplingdistributionofthemeanisnormal.Ifthesampleissufficientlylarge,regardlessoftheshapeofthepopulationdistribution,thesamplingdistributionisnormal(CentralLimitTheorem).43210-1-2-3-40.40.30.20.10.0zf(z)StandardNormalDistribution:95%IntervalConfidenceIntervalforWhenIsKnown\nConfidenceIntervalforwhenisKnown(Continued)\nA95%confidenceintervalforwhenisknownandsamplingisdonefromanormalpopulation,oralargesampleisused:Thequantityisoftencalledthemarginoferrororthesamplingerror.Forexample,if:n=25=20=122A95%confidenceinterval:The95%ConfidenceIntervalform\nWedefineasthezvaluethatcutsoffaright-tailareaofunderthestandardnormalcurve.(1-)iscalledtheconfidencecoefficient.iscalledtheerrorprobability,and(1-)100%iscalledtheconfidencelevel.PzzPzzPzzzzn>æèç=<-æèç=-<<æèç=-±aaaaaaas222221()(1-)100%ConfidenceInterval:x543210-1-2-3-4-50.40.30.20.10.0Zf(z)StandardNormalDistributionA(1-a)100%ConfidenceIntervalforma/2a/2æèçæèçæèç\n0.990.0052.5760.980.0102.3260.950.0251.9600.900.0501.6450.800.1001.282CriticalValuesofzandLevelsofConfidence543210-1-2-3-4-50.40.30.20.10.0Zf(z)StandardNormalDistribution\nWhensamplingfromthesamepopulation,usingafixedsamplesize,thehighertheconfidencelevel,thewidertheconfidenceinterval.543210-1-2-3-4-50.40.30.20.10.0Zf(z)StandardNormalDistribution543210-1-2-3-4-50.40.30.20.10.0Zf(z)StandardNormalDistributionTheLevelofConfidenceandtheWidthoftheConfidenceInterval\nTheSampleSizeandtheWidthoftheConfidenceIntervalWhensamplingfromthesamepopulation,usingafixedconfidencelevel,thelargerthesamplesize,n,thenarrowertheconfidenceinterval.0.90.80.70.60.50.40.30.20.10.0xf(x)SamplingDistributionoftheMean95%ConfidenceInterval:n=400.40.30.20.10.0xf(x)SamplingDistributionoftheMean95%ConfidenceInterval:n=20\nPopulationconsistsoftheFortune500Companies(FortuneWebSite),asrankedbyRevenues.YouaretryingtotofindouttheaverageRevenuesforthecompaniesonthelist.Thepopulationstandarddeviationis$15,056.37.Arandomsampleof30companiesobtainsasamplemeanof$10,672.87.Givea95%and90%confidenceintervalfortheaverageRevenues.Example\nNationalDiscounthas260retailoutletsthroughouttheUnitedStates.Nationalevaluateseachpotentiallocationforanewretailoutletinpartonthemeanannualincomeoftheindividualsinthemarketingareaofthenewlocation.SamplingcanbeusedtodevelopanintervalestimateofthemeanannualincomeforindividualsinapotentialmarketingareaforNationalDiscount.Asampleofsizen=36wastaken.Thesamplemean,,is$21,100andthesamplestandarddeviation,s,is$4,500.Wewilluse.95astheconfidencecoefficientinourintervalestimate.Example:NationalDiscount,Inc.\nPrecisionStatementThereisa.95probabilitythatthevalueofasamplemeanforNationalDiscountwillprovideasamplingerrorof$1,470orless…….determinedasfollows:95%ofthesamplemeansthatcanbeobservedarewithin+1.96ofthepopulationmean.If,then1.96=1,470.Example:NationalDiscount,Inc.\nExample:NationalDiscount,Inc.IntervalEstimateofthePopulationMean:UnknownIntervalEstimateofis:$21,100+$1,470or$19,630to$22,570Weare95%confidentthattheintervalcontainsthepopulationmean.\nIntervalEstimationofaPopulationMean:Small-SampleCase(n<30)PopulationisNotNormallyDistributedTheonlyoptionistoincreasethesamplesizeton>30andusethelarge-sampleinterval-estimationprocedures.PopulationisNormallyDistributedandisKnownThelarge-sampleinterval-estimationprocedurecanbeused.PopulationisNormallyDistributedandisUnknownTheappropriateintervalestimateisbasedonaprobabilitydistributionknownasthetdistribution.\ntDistributionThetdistributionisafamilyofsimilarprobabilitydistributions.Aspecifictdistributiondependsonaparameterknownasthedegreesoffreedom.Asthenumberofdegreesoffreedomincreases,thedifferencebetweenthetdistributionandthestandardnormalprobabilitydistributionbecomessmallerandsmaller.Atdistributionwithmoredegreesoffreedomhaslessdispersion.Themeanofthetdistributioniszero.Thestandarddeviationofthetdistributionismorethan1.\nIntervalEstimationofaPopulationMean:Small-SampleCase(n<30)withUnknownIntervalEstimatewhere1-=theconfidencecoefficientt/2=thetvalueprovidinganareaof/2intheuppertailofatdistributionwithn-1degreesoffreedoms=thesamplestandarddeviation\nThetisafamilyofbell-shapedandsymmetricdistributions,oneforeachnumberofdegreeoffreedom.Theexpectedvalueoftis0.Fordf>2,thevarianceoftisdf/(df-2).Thisisgreaterthan1,butapproaches1asthenumberofdegreesoffreedomincreases.Thetisflatterandhasfattertailsthandoesthestandardnormal.ThetdistributionapproachesastandardnormalasthenumberofdegreesoffreedomincreasesIfthepopulationstandarddeviation,,isnotknown,replacewiththesamplestandarddeviation,s.Ifthepopulationisnormal,theresultingstatistic:hasatdistributionwith(n-1)degreesoffreedom.Standardnormalt,df=20t,df=10ConfidenceIntervalorIntervalEstimateforWhenIsUnknown-ThetDistribution\nA(1-)100%confidenceintervalforwhenisnotknown(assuminganormallydistributedpopulation):whereisthevalueofthetdistributionwithn-1degreesoffreedomthatcutsoffatailareaoftoitsright.ConfidenceIntervalsforwhenisUnknown-ThetDistribution\ndft0.100t0.050t0.025t0.010t0.005-------------------------------13.0786.31412.70631.82163.65721.8862.9204.3036.9659.92531.6382.3533.1824.5415.84141.5332.1322.7763.7474.60451.4762.0152.5713.3654.03261.4401.9432.4473.1433.70771.4151.8952.3652.9983.49981.3971.8602.3062.8963.35591.3831.8332.2622.8213.250101.3721.8122.2282.7643.169111.3631.7962.2012.7183.106121.3561.7822.1792.6813.055131.3501.7712.1602.6503.012141.3451.7612.1452.6242.977151.3411.7532.1312.6022.947161.3371.7462.1202.5832.921171.3331.7402.1102.5672.898181.3301.7342.1012.5522.878191.3281.7292.0932.5392.861201.3251.7252.0862.5282.845211.3231.7212.0802.5182.831221.3211.7172.0742.5082.819231.3191.7142.0692.5002.807241.3181.7112.0642.4922.797251.3161.7082.0602.4852.787261.3151.7062.0562.4792.779271.3141.7032.0522.4732.771281.3131.7012.0482.4672.763291.3111.6992.0452.4622.756301.3101.6972.0422.4572.750401.3031.6842.0212.4232.704601.2961.6712.0002.3902.6601201.2891.6581.9802.3582.6171.2821.6451.9602.3262.57600.40.30.20.10.0tf(t)tDistribution:df=10Area=0.10}Area=0.10}Area=0.025}Area=0.025}1.372-1.3722.228-2.228Wheneverisnotknown(andthepopulationisassumednormal),thecorrectdistributiontouseisthetdistributionwithn-1degreesoffreedom.Note,however,thatforlargedegreesoffreedom,thetdistributionisapproximatedwellbytheZdistribution.ThetDistribution\nAstockmarketanalystwantstoestimatetheaveragereturnonacertainstock.Arandomsampleof15daysyieldsanaverage(annualized)returnofandastandarddeviationofs=3.5%.Assuminganormalpopulationofreturns,givea95%confidenceintervalfortheaveragereturnonthisstock.Thecriticalvalueoftfordf=(n-1)=(15-1)=14andaright-tailareaof0.025is:Thecorrespondingconfidenceintervalorintervalestimateis:Exampledft0.100t0.050t0.025t0.010t0.005-------------------------------13.0786.31412.70631.82163.657..................131.3501.7712.1602.6503.012141.3451.7612.1452.6242.977151.3411.7532.1312.6022.947..................\nWheneverisnotknown(andthepopulationisassumednormal),thecorrectdistributiontouseisthetdistributionwithn-1degreesoffreedom.Note,however,thatforlargedegreesoffreedom,thetdistributionisapproximatedwellbytheZdistribution.dft0.100t0.050t0.025t0.010t0.005-------------------------------13.0786.31412.70631.82163.657..................1201.2891.6581.9802.3582.6171.2821.6451.9602.3262.576LargeSampleConfidenceIntervalsforthePopulationMean\nExample:Aneconomistwantstoestimatetheaverageamountincheckingaccountsatbanksinagivenregion.Arandomsampleof100accountsgivesx-bar=$357.60ands=$140.00.Givea95%confidenceintervalfor,theaverageamountinanycheckingaccountatabankinthegivenregion.LargeSampleConfidenceIntervalsforthePopulationMean\nExample:ApartmentRentsIntervalEstimationofaPopulationMean:Small-SampleCase(n<30)withUnknownAreporterforastudentnewspaperiswritinganarticleonthecostofoff-campushousing.Asampleof10one-bedroomunitswithinahalf-mileofcampusresultedinasamplemeanof$550permonthandasamplestandarddeviationof$60.Letusprovidea95%confidenceintervalestimateofthemeanrentpermonthforthepopulationofone-bedroomunitswithinahalf-mileofcampus.We’llassumethispopulationtobenormallydistributed.\ntValueAt95%confidence,1-=.95,=.05,and/2=.025.t.025isbasedonn-1=10-1=9degreesoffreedom.Inthetdistributiontableweseethatt.025=2.262.Example:ApartmentRents\nIntervalEstimationofaPopulationMean:Small-SampleCase(n<30)withUnknown550+42.92or$507.08to$592.92Weare95%confidentthatthemeanrentpermonthforthepopulationofone-bedroomunitswithinahalf-mileofcampusisbetween$507.08and$592.92.Example:ApartmentRents\nsummaryClassification：Populationisnormallydistributed，andσisknown（usenormaldistributiontosolve）Wedon’tknowifpopulationisnormallydistributed，butσisknown（whenlargesample，usenormaldistributiontosolve）σisnotknownandn<30（usetdistributiontosolve）σisnotknownandn>=30（usetdistributionornormaldistributiontosolve）\nLetE=themaximumsamplingerrormentionedintheprecisionstatement.Eistheamountaddedtoandsubtractedfromthepointestimatetoobtainanintervalestimate.Eisoftenreferredtoasthemarginoferror.WehaveSolvingfornwehaveSampleSizeforanIntervalEstimateofaPopulationMean\nButwhensamplingwithoutreplacementforfinitepopulationWehaveSolvingfornwehaveSampleSizeforanIntervalEstimateofaPopulationMean\nHowclosedoyouwantyoursampleestimatetobetotheunknownparameter?(Whatisthedesiredbound,B?)Whatdoyouwantthedesiredconfidencelevel(1-)tobesothatthedistancebetweenyourestimateandtheparameterislessthanorequaltoB?Whatisyourestimateofthevariance(orstandarddeviation)ofthepopulationinquestion?Beforedeterminingthenecessarysamplesize,threequestionsmustbeanswered:}Bound,BSample-SizeDetermination\nStandarderrorofstatisticSamplesize=nSamplesize=2nStandarderrorofstatisticThesamplesizedeterminestheboundofastatistic,sincethestandarderrorofastatisticshrinksasthesamplesizeincreases:SampleSizeandStandardError\nAmarketingresearchfirmwantstoconductasurveytoestimatetheaverageamountspentonentertainmentbyeachpersonvisitingapopularresort.Thepeoplewhoplanthesurveywouldliketodeterminetheaverageamountspentbyallpeoplevisitingtheresorttowithin$120,with95%confidence.Frompastoperationoftheresort,anestimateofthepopulationstandarddeviationiss=$400.Whatistheminimumrequiredsamplesize?Sample-SizeDetermination:Example\nExample:NationalDiscount,Inc.SampleSizeforanIntervalEstimateofaPopulationMeanSupposethatNational’smanagementteamwantsanestimateofthepopulationmeansuchthatthereisa.95probabilitythatthesamplingerroris$500orless.Howlargeasamplesizeisneededtomeettherequiredprecision?\nExample:NationalDiscount,Inc.SampleSizeforIntervalEstimateofaPopulationMeanAt95%confidence,z.025=1.96.Recallthat=4,500.SolvingfornwehaveWeneedtosample312toreachadesiredprecisionof+$500at95%confidence.\nIntervalEstimationofaPopulationProportionIntervalEstimatewhere:1-istheconfidencecoefficientz/2isthezvalueprovidinganareaof/2intheuppertailofthestandardnormalprobabilitydistributionisthesampleproportion\nLarge-SampleConfidenceIntervalsforthePopulationProportion,p\nLarge-SampleConfidenceIntervalsforthePopulationProportion,p\nAmarketingresearchfirmwantstoestimatethesharethatforeigncompanieshaveintheAmericanmarketforcertainproducts.Arandomsampleof100consumersisobtained,anditisfoundthat34peopleinthesampleareusersofforeign-madeproducts;therestareusersofdomesticproducts.Givea95%confidenceintervalfortheshareofforeignproductsinthismarket.Thus,thefirmmaybe95%confidentthatforeignmanufacturerscontrolanywherefrom24.72%to43.28%ofthemarket.Large-SampleConfidenceIntervalforthePopulationProportion,p\nThewidthofaconfidenceintervalcanbereducedonlyatthepriceof:alowerlevelofconfidence,oralargersample.90%ConfidenceIntervalSampleSize,n=200LowerLevelofConfidenceLargerSampleSizeReducingtheWidthofConfidenceIntervals-TheValueofInformation\nExample:PoliticalScience,Inc.IntervalEstimationofaPopulationProportionPoliticalScience,Inc.(PSI)specializesinvoterpollsandsurveysdesignedtokeeppoliticalofficeseekersinformedoftheirpositioninarace.Usingtelephonesurveys,interviewersaskregisteredvoterswhotheywouldvoteforiftheelectionwereheldthatday.Inarecentelectioncampaign,PSIfoundthat220registeredvoters,outof500contacted,favoredaparticularcandidate.PSIwantstodevelopa95%confidenceintervalestimatefortheproportionofthepopulationofregisteredvotersthatfavorsthecandidate.\nIntervalEstimateofaPopulationProportionwhere:n=500,=220/500=.44,z/2=1.96.44+.0435PSIis95%confidentthattheproportionofallvotersthatfavorsthecandidateisbetween.3965and.4835.Example:PoliticalScience,Inc.\nsummaryconditionConfidenceintervalInfinitepopulation，Finitepopulationsamplingwithreplacementnp>5nq>5Populationproportion（p）Finitepopulationsamplingwithoutreplacement，np>5nq>5\nLetE=themaximumsamplingerrormentionedintheprecisionstatement.WehaveSolvingfornwehaveSampleSizeforanIntervalEstimateofaPopulationProportion\nButwhensamplingwithoutreplacementforfinitepopulationWehaveSolvingfornwehaveSampleSizeforanIntervalEstimateofaPopulationProportion\nThemanufacturersofasportscarwanttoestimatetheproportionofpeopleinagivenincomebracketwhoareinterestedinthemodel.Thecompanywantstoknowthepopulationproportion,p,towithin0.01with99%confidence.Currentcompanyrecordsindicatethattheproportionpmaybearound0.25.Whatistheminimumrequiredsamplesizeforthissurvey?Sample-SizeforProportion:Example\nSampleSizeforanIntervalEstimateofaPopulationProportionSupposethatPSIwouldlikea.99probabilitythatthesampleproportioniswithin+.03ofthepopulationproportion.Howlargeasamplesizeisneededtomeettherequiredprecision?Example:PoliticalScience,Inc.\nSampleSizeforIntervalEstimateofaPopulationProportionAt99%confidence,z.005=2.576.Note:Weused.44asthebestestimateofpintheaboveexpression.Ifnoinformationisavailableaboutp,then.5isoftenassumedbecauseitprovidesthehighestpossiblesamplesize.Ifwehadusedp=.5,therecommendednwouldhavebeen1843.Example:PoliticalScience,Inc.\nEndofChapter8