Hence P 4 5 x 8 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 con t CentralLimitTheorem Exampleofsamplingdistribution Thepopulationdistributionofannualincomeofengineersisskewednegatively Thisdistributionhasameanof 48 000 和a标准偏差of 5000 Ifwedrawasampleof100engineers whatistheprobabilitythattheiraverageannualincomeis 48700和more Exampleofsamplingdistribution con t 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 con t Applyingtheconfidenceintervalsequation Upperconfidencelimit 27 75 1 96 0 6 28 926Lowerconfidencelimit 27 75 1 96 0 6 26 574 Assuch thereis95 confidencelevelthattheaverageusefullifeofthemagneticmediatofallbetween26 574和28 926months ConfidenceintervalfordiscretedataBreakinto4or5teams combinedalltheM M sinoneteam和calculatetheconfidenceintervalsforeachcolortype usingbelowtable Combineallthedatafromthe4teams whatarethechangestotheconfidenceintervals Determinethesamplesizeforconfidenceintervals continuousdata GivenSamplemean 21S 6Whatisthesamplesizerequired suchthatthereisa95 confidencelevelthattheaveragevaluewillfallwithin 1 176fromitsmean Ans Applyingtheequation n 1 96x6 1 176 2 10 2 100Thesamplesizemustbe100 Determinethesamplesizeforconfidenceintervals discretedata Given p 0 4 Proportionagree q 0 6 Proportionnotagree Whatisthesamplesizerequired suchthatthereisa99 confidencelevelthattheproportionagreewillfallwithin 0 146fromp Ans Applyingtheequation n 2 58 2 0 4 0 6 0 146 2 74 95 75Thesamplesizemustbe75 Revision 1 00Date June2001 第二天 TestsofHypothesesWeek1recapofStatisticsTerminologyIntroductiontoStudentTdistributionExampleinusingStudentTdistributionSummaryofformulaforConfidenceLimitsIntroductiontoHypothesisTestingTheelementsofHypothesisTesting Break LargesampleTestofHypothesisaboutapopulationmeanp Values theobservedsignificancelevelsSmallsampleTestofHypothesisaboutapopulationmeanMeasuringthepowerofhypothesistestingCalculatingTypeIIErrorprobabilitiesHypothesisExerciseI Lunch HypothesisExerciseIPresentationComparing2populationMeans IndependentSamplingComparing2populationMeans PairedDifferenceExperimentsComparing2populationProportions F Test Break HypothesisTestingExerciseII paperclip HypothesisTestingPresentation第一天wrapup