应用统计学_卡方检验ppt课件

BEO2255AppliedStatisticsforBusiness,WeekSixAnalyzingcategoricaldata:Chi-squaredtests,1,Thisweeklecturewillcover.,Analysingcategoricaldata(nominal)Chi-squaretestofdifferencesbetweenproportionsChi-squaretestofindependence,2,SPSS单样本非参数检验,总体分布的chi-square检验(1)目的:根据样本数据推断总体的分布与某个已知分布是否有显著差异-吻合性检验。适用于分类资料的统计推断,3,SPSS单样本非参数检验,总体分布的chi-square检验(2)基本假设:H0:总体分布与理论分布无显著差异(3)基本方法根据已知总体的构成比计算出样本中各类别的期望频数，计算实际观察频数与期望频数的差距,即：计算卡方值卡方值较小,则实际频数和期望频数相差较小.如果P大于a,不能拒绝H0,认为总体分布与已知分布无显著差异.反之,4,SPSS单样本卡方检验,总体分布的chi-square检验(4)基本操作步骤:菜单:analyze-nonparametrictest-chisquare选定待检验变量入testvariablelist框确定待检验个案的取值范围(expectedrange)getfromdata:全部样本usespecifiedrange:用户自定义个案范围指定期望频数(expectedvalues)allcategoriesequal:所有类别有相同的构成比value:用户自定义构成比,5,Categoricalvariable,VariablesthatdescribecategoriesofentitiesDealingwiththemallthetimeinstatisticsMakingcomparisonsamongvariablesForexample,whetherconsumerspreferaparticularbrandofaproductamongothercompetingbrands.CheckingwhetherthereisarelationshipbetweentwocategoricalvariablesGenderandpreferenceforaproduct,whetherthepreferenceforaproductisindependentfromgender,6,Chi-squaretestfordifferencesbetweenproportions,ThistestinvolveswithnominaldataproducedbymultinomialexperimentItisageneralisationofabinomialexperimentThesetestthenullhypothesisthatdatainthetargetpopulationhasaparticularprobabilitydistribution.Example1Wemighttestwhetherconsumersareindifferenttowhichoffourmaterials(glass,plastic,steeloraluminium)thatcouldbeusedtomakesoftdrinkcontainers.Thenullhypothesisisthattheyareindifferent(orthatequalnumberspreferglass,plastic,steelandaluminium).,7,Example1,DataLetpGbetheprobabilitythatanindividualselectedatrandomwillnominateglassashis/herpreferenceifrequiredtomakeachoice.SimilarlyforpP(plastic),pS(steel)andpA(aluminium)HypothesesHO:pG=pP=pS=pA=0.25.HA:atleastonepi0.25.Thealternativeisthatatleastonematerialismorepreferred(orlesspreferred)thantheothers.,8,Example1cont.,Procedure:Selectarandomsampleof,say,100consumersanddeterminetheirpreferences.UnderthenullhypothesisWeexpect25consumerstonominateglass,25tonominateplastic,25tonominatesteeland25tonominatealuminiumThesearetheexpectedfrequencies,Ei.Ei=npi.Wecomparetheexpectedfrequencieswiththesampleresultsortheobservedfrequencies,Oi.Iftheyareapproximatelythesamewewouldconcludethatthenullhypothesisistrue.OiEiHOisprobablytrue.,9,Example1cont.,Chisquare,Werequireateststatistictodecidewhetherthedifferenceislargeenoughtorejectthenullhypothesis.WeusechisquarewithG-1degreesoffreedomwhereGisthenumberofgroups.,Supposeinourexample,39preferglass,16preferplastic,20prefersteeland25preferaluminium.Recallthattheexpectedfrequencieswereall25.,10,ObtainthecriticalvalueofchisquareCritical23=7.82.Obtainthecriticalvalueat5%significancelevelat3d.f.,(TableE4,page742,Berensonet.al.2013)i.e.thereisonlya5percentchanceorlessthat237.82ifHOistrue.Comparisonofchisquarevalues23=12.087.82rejectHO.Conclusion:atthe5%significancelevelthereissufficientevidencetorejectthenullhypothesis.Atleastoneoftheprobabilities(pi)isdifferent.Thesampleresultsindicatethatthematerialsarenotequallypreferredbyconsumersinthetargetpopulation.Thus,atleastpreferencesfortwomaterialsaredifferent.,11,ChisquaretestusingSPSS,Example:Supposethatwewanttotestwhetherornotcustomershaveacolourpreferenceforpackaging.Threedifferentcolours,Blue,Green不同层为水平数积.(5)是否显示各分组的棒图(displayclusteredbarcharts),20,产生交叉列联表,进一步计算cells选项:选择在频数分析表中输出各种百分比.row:行百分比(Rowpct);column:列百分比(Colpct);total:总百分比(Totpct);,21,分析列联表中变量间的关系,目的：通过列联表分析，检验行列变量之间是否独立。方法：卡方检验：对品质数据的相关性进行度量,22,分析列联表中变量间的关系,卡方检验年龄与工资收入交叉列联表低中高青40000中05000老00600低中高青00500中06000老40000,23,分析列联表中变量间的关系,卡方检验基本步骤(1)H0:行列变量之间无关联或相互独立(2)构造卡方统计量统计量服从(r-1)*(c-1)个自由度的卡方分布count:观察(实际)频数expectedcount:期望频数(期望频数反映的是H0成立情况下的数据分布特征)Residual:剩余(观察频数-期望频数),24,1、列联表,2、三维柱形图,3、二维条形图,从三维柱形图能清晰看出各个频数的相对大小。,从二维条形图能看出，吸烟者中患肺癌的比例高于不患肺癌的比例。,通过图形直观判断两个分类变量是否相关：,25,Testsofindependencecont,Example2Supposeweinterviewed400people&askedthemwhichofthreeagegroupstheyarein(under25,25to60,andover60).Wealsoasktheirresponsetothestatementthat“Allimportsofautomobilesshouldbebannedinordertoprotectthelocalindustry”(agree,novieweitherway,disagree).,attitudestowardsbanningimportsagreenoviewdisagreeTotalagegroupunder251953259725-60469447187over60305630116Total95203102400,26,Testsofindependencecont,Example2cont.Nullhypothesis:Thenullhypothesisisthatanswerstothetwoquestionsareindependent.Underthenull:Probover60andagree=Probover60ProbagreeMultiplicationruleforindependenteventsExpectedfrequency=Probover60Probagreesamplesize.,27,ProcedureWesetupacross-tabulationshowingtheobservedfrequenciesofanswerstothetwoquestions.Wecalculatetheexpectedfrequencies.TestOurtestisbasedonacomparisonoftheobservedandexpectedfrequencies.Short-cutforexpectedfrequencies,28,Age*attitudetobanningimportsCrosstabulation,19.0,53.0,25.0,97.0,23.0,49.2,24.7,96.9,46.0,94.0,47.0,187.0,44.4,94.9,47.7,187.0,30.0,56.0,30.0,116.0,27.6,58.9,29.6,116.1,95.0,203.0,102.0,400.0,95.0,203.0,102.0,400.0,Count,ExpectedCount,Count,ExpectedCount,Count,ExpectedCount,Count,ExpectedCount,Under25,25-60,Over60,Age,Group,Total,Agree,Noview,Disagree,Attitudetobanimports,Total,Calculationforexpectedfrequencyofagreeandover60,95116/400,29,Age*attitudetobanningimportsCrosstabulation,19.0,53.0,25.0,97.0,23.0,49.2,24.7,96.9,46.0,94.0,47.0,187.0,44.4,94.9,47.7,187.0,30.0,56.0,30.0,116.0,27.6,58.9,29.6,116.1,95.0,203.0,102.0,400.0,95.0,203.0,102.0,400.0,Count,ExpectedCount,Count,ExpectedCount,Count,ExpectedCount,Count,ExpectedCount,Under25,25-60,Over60,Age,Group,Total,Agree,Noview,Disagree,Attitudetobanimports,Total,Thecount(observed)andtheexpectedaredifferent,butdifferentenoughtorejectthenull?,30,Chi-squaredtestforindependence,Rationale:OijEijHOisprobablytrue.TeststatisticWerequireateststatistictodecidewhetherthedifferenceislargeenoughtorejectthenullhypothesis.,31,Chi-SquareTests,1.438,a,4,.837,1.517,4,.805,1.307,1,.758,400,PearsonChi-Square,LikelihoodRatio,Linear-by-Linear,Association,NofValidCases,Value,df