商务统计学最新英文版教学课件第8章

ConfidenceIntervalEstimationChapter8ObjectivesInthischapter,youlearn:ToconstructandinterpretconfidenceintervalestimatesforthepopulationmeanandthepopulationproportionTodeterminethesamplesizenecessarytodevelopaconfidenceintervalforthepopulationmeanorpopulationproportionChapterOutlineContentofthischapterConfidenceIntervalsforthePopulationMean,μwhenPopulationStandardDeviationσisKnownwhenPopulationStandardDeviationσisUnknownConfidenceIntervalsforthePopulationProportion,πDeterminingtheRequiredSampleSizePointandIntervalEstimatesApointestimateisasinglenumber,aconfidenceintervalprovidesadditionalinformationaboutthevariabilityoftheestimatePointEstimateLowerConfidenceLimitUpperConfidenceLimitWidthofconfidenceintervalDCOVAWecanestimateaPopulationParameter…PointEstimateswithaSampleStatistic(aPointEstimate)MeanProportionpπXμDCOVAConfidenceIntervalsHowmuchuncertaintyisassociatedwithapointestimateofapopulationparameter?AnintervalestimateprovidesmoreinformationaboutapopulationcharacteristicthandoesapointestimateSuchintervalestimatesarecalledconfidenceintervalsDCOVAConfidenceIntervalEstimateAnintervalgivesarangeofvalues:TakesintoconsiderationvariationinsamplestatisticsfromsampletosampleBasedonobservationsfrom1sampleGivesinformationaboutclosenesstounknownpopulationparametersStatedintermsoflevelofconfidencee.g.95%confident,99%confidentCanneverbe100%confidentDCOVAConfidenceIntervalExampleCerealfillexamplePopulationhasµ=368andσ=15.Ifyoutakeasampleofsizen=25youknow368±1.96*15/=(362.12,373.88).95%oftheintervalsformedinthismannerwillcontainµ.Whenyoudon’tknowµ,youuseXtoestimateµIfX=362.3theintervalis362.3±1.96*15/=(356.42,368.18)Since356.42≤µ≤368.18theintervalbasedonthissamplemakesacorrectstatementaboutµ.Butwhatabouttheintervalsfromotherpossiblesamplesofsize25?2525DCOVAConfidenceIntervalExample(continued)Sample#XLowerLimitUpperLimitContainµ?1362.30356.42368.18Yes2369.50363.62375.38Yes3360.00354.12365.88No4362.12356.24368.00Yes5373.88368.00379.76YesDCOVAConfidenceIntervalExampleInpracticeyouonlytakeonesampleofsizenInpracticeyoudonotknowµsoyoudonotknowiftheintervalactuallycontainsµHoweveryoudoknowthat95%oftheintervalsformedinthismannerwillcontainµThus,basedontheonesample,youactuallyselectedyoucanbe95%confidentyourintervalwillcontainµ(thisisa95%confidenceinterval)(continued)Note:95%confidenceisbasedonthefactthatweusedZ=1.96.DCOVAEstimationProcess(mean,μ,isunknown)PopulationRandomSampleMeanX=50SampleIam95%confidentthatμisbetween40&60.DCOVAGeneralFormulaThegeneralformulaforallconfidenceintervalsis:PointEstimate±(CriticalValue)(StandardError)Where:•PointEstimateisthesamplestatisticestimatingthepopulationparameterofinterest•CriticalValueisatablevaluebasedonthesamplingdistributionofthepointestimateandthedesiredconfidencelevel•StandardErroristhestandarddeviationofthepointestimateDCOVAConfidenceLevelConfidencetheintervalwillcontaintheunknownpopulationparameterApercentage(lessthan100%)DCOVAConfidenceLevel,(1-)Supposeconfidencelevel=95%Alsowritten(1-)=0.95,(so=0.05)Arelativefrequencyinterpretation:95%ofalltheconfidenceintervalsthatcanbeconstructedwillcontaintheunknowntrueparameterAspecificintervaleitherwillcontainorwillnotcontainthetrueparameterNoprobabilityinvolvedinaspecificinterval(continued)DCOVAConfidenceIntervalsPopulationMeanσUnknownConfidenceIntervalsPopulationProportionσKnownDCOVAConfidenceIntervalforμ(σKnown)AssumptionsPopulationstandarddeviationσisknownPopulationisnormallydistributedIfpopulationisnotnormal,uselargesample(n30)Confidenceintervalestimate:whereisthepointestimateZα/2isthenormaldistributioncriticalvalueforaprobabilityof/2ineachtailisthestandarderrornσ/2ZXαXnσ/DCOVAFindingtheCriticalValue,Zα/2Considera95%confidenceinterval:Zα/2=-1.96Zα/2=1.960.05so0.951αα0.0252α0.0252αPointEstimateLowerConfidenceLimitUpperConfidenceLimitZunits:Xunits:PointEstimate01.96/2ZαDCOVACommonLevelsofConfidenceCommonlyusedconfidencelevelsare90%,95%,and99%ConfidenceLevelConfidenceCoefficient,Zα/2value1.281.6451.962.332.583.083.270.800.900.950.980.990.9980.99980%90%95%98%99%99.8%99.9%1DCOVAμμxIntervalsandLevelofConfidenceConfidenceIntervalsIntervalsextendfromto(1-)100%ofintervalsconstructedcontainμ;()100%donot.SamplingDistributionoftheMeannσ2/αZXnσ2/αZXxx1x2/2/21DCOVAExampleAsampleof11circuitsfromalargenormalpopulationhasameanresistanceof2.20ohms.Weknowfrompasttestingthatthepopulationstandarddeviationis0.35ohms.Determinea95%confidenceintervalforthetruemeanresistanceofthepopulation.DCOVA2.40681.99320.20682.20)11(0.35/1.962.20nσ/2ZXμαExampleAsampleof11circuitsfromalargenormalpopulationhasameanresistanceof2.20ohms.Weknowfrompasttestingthatthepopulationstandarddeviationis0.35ohms.Solution:(continued)DCOVAInterpretationWeare95%confidentthatthetruemeanresistanceisbetween1.9932and2.4068ohmsAlthoughthetruemeanmayormaynotbeinthisinterval,95%ofintervalsformedinthismannerwillcontainthetruemeanDCOVAConfidenceIntervalsPopulationMeanσUnknownConfidenceIntervalsPopulationProportionσKnow