Chi-Square Applications ppt课件


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Chi-Square Applications ppt课件内容预览:Apply the chi-square distribution to:Goodness-of-fit testsTests of independence between 2variablesTests comparing proportions from multiple populationsTests of a single population variance.Chapter 13- KeyTermsObserved versus expected frequenciesNumber of parameters estimated,mNumber of categories used,kContingency tableIndependent variablesGoodness-of-FitTestsTheQuestion:Does the distribution of sample data resemble a specified probability distribution,such as:the binomial,hypergeometric,or Poisson discrete distributions.pptthe uniform,normal,or exponential continuous distributions.a predefined probability distribution.Hypotheses:H0:pi =values expected H1:pi 1values expectedwhereGoodness-of-FitTestsRejectionRegion:Degrees of Freedom =k –1– mwhere k =of categories,m =of parametersUniform Discrete:m =0so df =k –1Binomial: m =0when p is known,so df =k –1m =1when p is unknown,so df =k –2Poisson:m =1since μusually estimated,df =k –2Normal:m =2when μand s estimated,df =k –3Exponential:m =1since μusually estimated,df =k –2Goodness-of-FitTestsGoodness-of-Fit: AnExampleProblem 13.20: In a study of vehicle ownership,it has been found that 13.516035350f U.S.households do not own a vehicle,with 33.720100240wning 1vehicle,33.5554wning 2vehicles,and 19.34574570wning 3or more vehicles. The data for a random sample of 100households in a resort community are summarized below. At the 0.05level of significance,can we reject the possibility that the vehicle-ownership distribution in this community differs from that of the nation as a whole?Vehicles Owned Households0 201 352 233or more 22Goodness-of-Fit: AnExampleVehicles Oj Ej [Oj– Ej ]2Ej0 20 13.53.12961 35 33.70.05012 23 33.53.29103+ 22 19.30.3777Sum =6.8484I. H0:p0=0.135,p1=0.337,p2=0.335,p3+=0.193Vehicle-ownership distribution in this community is the same as it is in the nation as a whole.H1: At least one of the proportions does not equal the stated value. Vehicle-ownership distribution in this community is not the same as it is in the nation as a whole.Goodness-of-Fit: AnExampleII. RejectionRegion:a =0.05df =k –1– m =4–1–0=3III. TestStatistic:c2=6.8484IV. Conclusion: Since the test statistic of c2=6.8484falls below the critical value of c2=7.815,we do not reject H0with at least 95耾nfidence.V. Implications: There is not enough evidence to show that vehicle ownership in this community differs from that in the nation as a whole.Chi-Square Tests of Independence Between TwoVariables
课件关键字:Chi-Square Applications ,Appli。

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