商務(wù)統(tǒng)計(jì)學(xué)（第7版）英文t課件7 Sampling Distributions、8 Confidence Interval Estimation

SamplingDistributionsChapter7ObjectivesInthischapter,youlearn:TheconceptofthesamplingdistributionTocomputeprobabilitiesrelatedtothesamplemeanandthesampleproportionTheimportanceoftheCentralLimitTheoremSamplingDistributionsAsamplingdistributionisadistributionofallofthepossiblevaluesofasamplestatisticforagivensamplesizeselectedfromapopulation.Forexample,supposeyousample50studentsfromyourcollegeregardingtheirmeanGPA.Ifyouobtainedmanydifferentsamplesofsize50,youwillcomputeadifferentmeanforeachsample.WeareinterestedinthedistributionofallpotentialmeanGPAswemightcalculateforanysampleof50students.DCOVADevelopingaSamplingDistributionAssumethereisapopulation…PopulationsizeN=4Randomvariable,X,

isageofindividualsValuesofX:18,20,

22,24(years)ABCDDCOVA.3.2.10

18202224

ABCDUniformDistributionP(x)x(continued)SummaryMeasuresforthePopulationDistribution:DevelopingaSamplingDistributionDCOVA16possiblesamples(samplingwithreplacement)Nowconsiderallpossiblesamplesofsizen=2(continued)DevelopingaSamplingDistribution16SampleMeans1stObs2ndObservation182022241818,1818,2018,2218,242020,1820,2020,2220,242222,1822,2022,2222,242424,1824,2024,2224,24DCOVASamplingDistributionofAllSampleMeans181920212223240.1.2.3P(X)

XSampleMeansDistribution16SampleMeans_DevelopingaSamplingDistribution(continued)(nolongeruniform)_DCOVASummaryMeasuresofthisSamplingDistribution:DevelopingASamplingDistribution(continued)DCOVANote: Herewedivideby16becausethereare16 differentsamplesofsize2.ComparingthePopulationDistribution

totheSampleMeansDistribution181920212223240.1.2.3P(X)X

18

20

22

24AB

C

D0.1.2.3PopulationN=4P(X)X_SampleMeansDistributionn=2_DCOVASampleMeanSamplingDistribution:

StandardErroroftheMeanDifferentsamplesofthesamesizefromthesamepopulationwillyielddifferentsamplemeansAmeasureofthevariabilityinthemeanfromsampletosampleisgivenbytheStandardErroroftheMean:

(Thisassumesthatsamplingiswithreplacementor samplingiswithoutreplacementfromaninfinitepopulation)NotethatthestandarderrorofthemeandecreasesasthesamplesizeincreasesDCOVASampleMeanSamplingDistribution:

IfthePopulationisNormalIfapopulationisnormalwithmeanμandstandarddeviationσ,thesamplingdistributionofisalsonormallydistributedwith and

DCOVAZ-valueforSamplingDistribution

oftheMeanZ-valueforthesamplingdistributionof:where: =samplemean =populationmean =populationstandarddeviation

n=samplesizeDCOVANormalPopulationDistributionNormalSamplingDistribution(hasthesamemean)SamplingDistributionProperties(i.e.isunbiased

)DCOVASamplingDistributionPropertiesAsnincreases,decreasesLargersamplesizeSmallersamplesize(continued)DCOVADeterminingAnIntervalIncludingAFixedProportionoftheSampleMeansFindasymmetricallydistributedintervalaroundμthatwillinclude95%ofthesamplemeanswhenμ=368,σ=15,andn=25.Sincetheintervalcontains95%ofthesamplemeans5%ofthesamplemeanswillbeoutsidetheintervalSincetheintervalissymmetric2.5%willbeabovetheupperlimitand2.5%willbebelowthelowerlimit.Fromthestandardizednormaltable,theZscorewith2.5%(0.0250)belowitis-1.96andtheZscorewith2.5%(0.0250)aboveitis1.96.DCOVADeterminingAnIntervalIncludingAFixedProportionoftheSampleMeansCalculatingthelowerlimitoftheintervalCalculatingtheupperlimitoftheinterval95%ofallsamplemeansofsamplesize25arebetween362.12and373.88(continued)DCOVASampleMeanSamplingDistribution:

IfthePopulationisnotNormalWecanapplytheCentralLimitTheorem:Evenifthepopulationisnotnormal,…samplemeansfromthepopulationwillbe

approximatelynormalaslongasthesamplesizeislargeenough.Propertiesofthesamplingdistribution:andDCOVAn↑CentralLimitTheoremAsthesamplesizegetslargeenough…thesamplingdistributionofthesamplemeanbecomesalmostnormalregardlessofshapeofpopulationDCOVAPopulationDistributionSamplingDistribution(becomesnormalasnincreases)CentralTendencyVariationLargersamplesizeSmallersamplesizeSampleMeanSamplingDistribution:

IfthePopulationisnotNormal(continued)Samplingdistributionproperties:DCOVAHowLargeisLargeEnough?Formostdistributions,n>30willgiveasamplingdistributionthatisnearlynormalForfairlysymmetricdistributions,n>15Foranormalpopulationdistribution,thesamplingdistributionofthemeanisalwaysnormallydistributedDCOVAExampleSupposeapopulationhasmeanμ=8andstandarddeviationσ=3.Supposearandomsampleofsizen=36isselected.Whatistheprobabilitythatthesamplemeanisbetween7.8and8.2?DCOVAExampleSolution:Evenifthepopulationisnotnormallydistributed,thecentrallimittheoremcanbeused(n>30)…sothesamplingdistributionofisapproximatelynormal…withmean=8…andstandarddeviation

(continued)DCOVAExampleSolution(continued):(continued)Z7.88.2-0.40.4SamplingDistributionStandardNormalDistributionPopulationDistribution????????????SampleStandardizeXDCOVAPopulationProportions

π=theproportionofthepopulationhavingsomecharacteristicSampleproportion

(p)providesanestimateofπ:0≤p≤1pisapproximatelydistributedasanormaldistributionwhennislarge (assumingsamplingwithreplacementfromafinitepopulationorwithoutreplacementfromaninfinitepopulation)DCOVASamplingDistributionofpApproximatedbya

normaldistributionif:

where

and(whereπ=populationproportion)SamplingDistributionP(

ps).3.2.100.2.4.681pDCOVAZ-ValueforProportionsStandardizeptoaZvaluewiththeformula:DCOVAExampleIfthetrueproportionofvoterswhosupportPropositionAisπ=0.4,whatistheprobabilitythatasampleofsize200yieldsasampleproportionbetween0.40and0.45?i.e.:ifπ=0.4andn=200,whatis P(0.40≤p≤0.45)?DCOVAExample

ifπ=0.4andn=200,whatis P(0.40≤p≤0.45)?(continued)Find:Converttostandardizednormal:DCOVAExampleZ0.451.440.4251StandardizeSamplingDistributionStandardized

NormalDistribution

ifπ=0.4andn=200,whatis P(0.40≤p≤0.45)?(continued)Utilizethecumulativenormaltable:P(0≤Z≤1.44)=0.9251–0.5000=0.42510.400pDCOVAChapterSummaryInthischapterwediscussed:TheconceptofasamplingdistributionComputingprobabilitiesrelatedtothesamplemeanandthesampleproportionTheimportanceoftheCentralLimitTheoremConfidenceIntervalEstimationChapter8ObjectivesInthischapter,youlearn:

ToconstructandinterpretconfidenceintervalestimatesforthepopulationmeanandthepopulationproportionTodeterminethesamplesizenecessarytodevelopaconfidenceintervalforthepopulationmeanorpopulationproportionChapterOutlineContentofthischapterConfidenceIntervalsforthePopulationMean,μwhenPopulationStandardDeviationσisKnownwhenPopulationStandardDeviationσisUnknownConfidenceIntervalsforthePopulationProportion,πDeterminingtheRequiredSampleSizePointandIntervalEstimatesApointestimateisasinglenumber,aconfidenceintervalprovidesadditionalinformationaboutthevariabilityoftheestimatePointEstimateLowerConfidenceLimitUpperConfidenceLimitWidthofconfidenceintervalDCOVAWecanestimateaPopulationParameter…PointEstimateswithaSample

Statistic(aPointEstimate)MeanProportionpπXμDCOVAConfidenceIntervalsHowmuchuncertaintyisassociatedwithapointestimateofapopulationparameter?AnintervalestimateprovidesmoreinformationaboutapopulationcharacteristicthandoesapointestimateSuchintervalestimatesarecalledconfidenceintervalsDCOVAConfidenceIntervalEstimateAnintervalgivesarangeofvalues:TakesintoconsiderationvariationinsamplestatisticsfromsampletosampleBasedonobservationsfrom1sampleGivesinformationaboutclosenesstounknownpopulationparametersStatedintermsoflevelofconfidencee.g.95%confident,99%confidentCanneverbe100%confidentDCOVAConfidenceIntervalExampleCerealfillexamplePopulationhasμ=368andσ=15.Ifyoutakeasampleofsizen=25youknow368±1.96*15/=(362.12,373.88).95%oftheintervalsformedinthismannerwillcontainμ.Whenyoudon’tknowμ,youuseXtoestimateμIfX=362.3theintervalis362.3±1.96*15/=(356.42,368.18)Since356.42≤μ≤368.18theintervalbasedonthissamplemakesacorrectstatementaboutμ.Butwhatabouttheintervalsfromotherpossiblesamplesofsize25?DCOVAConfidenceIntervalExample(continued)Sample#XLowerLimitUpperLimitContainμ?1362.30356.42368.18Yes2369.50363.62375.38Yes3360.00354.12365.88No4362.12356.24368.00Yes5373.88368.00379.76YesDCOVAConfidenceIntervalExampleInpracticeyouonlytakeonesampleofsizenInpracticeyoudonotknowμsoyoudonotknowiftheintervalactuallycontainsμHoweveryoudoknowthat95%oftheintervalsformedinthismannerwillcontainμThus,basedontheonesample,youactuallyselectedyoucanbe95%confidentyourintervalwillcontainμ(thisisa95%confidenceinterval)(continued)Note:95%confidenceisbasedonthefactthatweusedZ=1.96.DCOVAEstimationProcess(mean,μ,isunknown)PopulationRandomSampleMeanX=50SampleIam95%confidentthatμisbetween40&60.DCOVAGeneralFormulaThegeneralformulaforallconfidenceintervalsis:PointEstimate±(CriticalValue)(StandardError)Where:PointEstimateisthesamplestatisticestimatingthepopulationparameterofinterestCriticalValueisatablevaluebasedonthesamplingdistributionofthepointestimateandthedesiredconfidencelevelStandardErroristhestandarddeviationofthepointestimateDCOVAConfidenceLevelConfidencetheintervalwillcontaintheunknownpopulationparameterApercentage(lessthan100%)DCOVAConfidenceLevel,(1-)Supposeconfidencelevel=95%Alsowritten(1-

)=0.95,(so

=0.05)Arelativefrequencyinterpretation:95%ofalltheconfidenceintervalsthatcanbeconstructedwillcontaintheunknowntrueparameterAspecificintervaleitherwillcontainorwillnotcontainthetrueparameterNoprobabilityinvolvedinaspecificinterval(continued)DCOVAConfidenceIntervalsPopulationMean

σUnknownConfidenceIntervalsPopulationProportion

σKnownDCOVAConfidenceIntervalforμ

(σKnown)AssumptionsPopulationstandarddeviationσ

isknownPopulationisnormallydistributedIfpopulationisnotnormal,uselargesample(n>30)Confidenceintervalestimate:

whereisthepointestimate Zα/2isthenormaldistributioncriticalvalueforaprobabilityof/2ineachtail isthestandarderror

DCOVAFindingtheCriticalValue,Zα/2Considera95%confidenceinterval:Zα/2=-1.96Zα/2=1.96PointEstimateLowerConfidenceLimitUpperConfidenceLimitZunits:Xunits:PointEstimate0DCOVACommonLevelsofConfidenceCommonlyusedconfidencelevelsare90%,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%DCOVAIntervalsandLevelofConfidenceConfidenceIntervals

Intervalsextendfrom

to

(1-

)100%

ofintervalsconstructedcontainμ;(

)100%donot.SamplingDistributionoftheMeanxx1x2DCOVAExampleAsampleof11circuitsfromalargenormalpopulationhasameanresistanceof2.20ohms.Weknowfrompasttestingthatthepopulationstandarddeviationis0.35ohms.Determinea95%confidenceintervalforthetruemeanresistanceofthepopulation.DCOVAExampleAsampleof11circuitsfromalargenormalpopulationhasameanresistanceof2.20ohms.Weknowfrompasttestingthatthepopulationstandarddeviationis0.35ohms.

Solution:(continued)DCOVAInterpretationWeare95%confidentthatthetruemeanresistanceisbetween1.9932and2.4068ohms

Althoughthetruemeanmayormaynotbeinthisinterval,95%ofintervalsformedinthismannerwillcontainthetruemeanDCOVAConfidenceIntervalsPopulationMean

σ

UnknownConfidenceIntervalsPopulationProportion

σKnownDCOVADoYouEverTrulyKnowσ?Probablynot!Invirtuallyallrealworldbusinesssituations,σisnotknown.Ifthereisasituationwhereσisknownthenμisalsoknown(sincetocalculateσyouneedtoknowμ.)Ifyoutrulyknowμtherewouldbenoneedtogatherasampletoestimateit.Ifthepopulationstandarddeviationσisunknown,wecansubstitutethesamplestandarddeviation,S

Thisintroducesextrauncertainty,sinceSisvariablefromsampletosampleSoweusethetdistributioninsteadofthenormaldistributionConfidenceIntervalforμ

(σUnknown)DCOVAAssumptionsPopulationstandarddeviationisunknownPopulationisnormallydistributedIfpopulationisnotnormal,uselargesample(n>30)UseStudent’stDistributionConfidenceIntervalEstimate:(wheretα/2isthecriticalvalueofthetdistributionwithn-1degreesoffreedomandanareaofα/2ineachtail)ConfidenceIntervalforμ

(σUnknown)(continued)DCOVAStudent’stDistributionThetisafamilyofdistributionsThetα/2valuedependsondegreesoffreedom(d.f.)Numberofobservationsthatarefreetovaryaftersamplemeanhasbeencalculated

d.f.=n-1DCOVAIfthemeanofthesethreevaluesis8.0,thenX3

mustbe9

(i.e.,X3isnotfreetovary)DegreesofFreedom(df)Here,n=3,sodegreesoffreedom=n–

1=3–1=2(2valuescanbeanynumbers,butthethirdisnotfreetovaryforagivenmean)Idea:Numberofobservationsthatarefreetovary aftersamplemeanhasbeencalculatedExample:

Supposethemeanof3numbersis8.0LetX1=7 LetX2=8 Whatis

X3?DCOVAStudent’stDistributiont0t(df=5)t(df=13)t-distributionsarebell-shapedandsymmetric,buthave‘fatter’tailsthanthenormalStandardNormal(twithdf=∞)Note:tZasnincreasesDCOVAStudent’stTableDCOVAUpperTailAreadf.10.05.02513.0786.31412.70621.88631.6382.3533.182t02.920Thebodyofthetablecontainstvalues,notprobabilitiesLet:n=3

df=n-1=2

=0.10

/2=0.05

/2=0.054.3032.920SelectedtdistributionvaluesWithcomparisontotheZvalueConfidencetttZ

Level

(10d.f.)

(20d.f.)

(30d.f.)

(∞d.f.)0.80 1.3721.3251.3101.280.901.8121.7251.6971.6450.952.2282.0862.0421.960.993.1692.8452.7502.58Note:tZasnincreasesDCOVAExampleoftdistributionconfidenceintervalArandomsampleofn=25hasX=50and S=8.Forma95%confidenceintervalforμd.f.=n–1=24,soTheconfidenceintervalis

46.698≤μ

≤53.302DCOVAExampleoftdistributionconfidenceintervalInterpretingthisintervalrequirestheassumptionthatthepopulationyouaresamplingfromisapproximatelyanormaldistribution(especiallysincenisonly25).Thisconditioncanbecheckedbycreatinga:NormalprobabilityplotorBoxplot(continued)DCOVAConfidenceIntervalsPopulationMean

σ

UnknownConfidenceIntervalsPopulationProportion

σ

KnownDCOVAConfidenceIntervalsforthe

PopulationProportion,πAnintervalestimateforthepopulationproportion(π)canbecalculatedbyaddinganallowanceforuncertaintytothesampleproportion(p)

DCOVAConfidenceIntervalsforthe

PopulationProportion,πRecallthatthedistributionofthesampleproportionisapproximatelynormalifthesamplesizeislarge,withstandarddeviationWewillestimatethiswithsampledata:(continued)DCOVAConfidenceIntervalEndpointsUpperandlowerconfidencelimitsforthepopulationproportionarecalculatedwiththeformulawhereZα/2isthestandardnormalvalueforthelevelofconfidencedesiredpisthesampleproportionnisthesamplesizeNote:musthavenp>5andn(1-p)>5DCOVAExampleArandomsampleof100peopleshowsthat25areleft-handed.Forma95%confidenceintervalforthetrueproportionofleft-handersDCOVAExampleArandomsampleof100peopleshowsthat25areleft-handed.Forma95%confidenceintervalforthetrueproportionofleft-handers.(continued)DCOVAInterpretationWeare95%confidentthatthetruepercentageofleft-handersinthepopulationisbetween16.51%and33.49%.Althoughtheintervalfrom0.1651to0.3349mayormaynotcontainthetrueproportion,95%ofintervalsformedfromsamplesofsize100inthismannerwillcontainthetrueproportion.DCOVADeterminingSampleSizeFortheMeanDeterminingSampleSizeFortheProportionDCOVASamplingErrorTherequiredsamplesizecanbefoundtoreachadesiredmarginoferror(e)withaspecifiedlevelofconfidence(1-

)Themarginoferrorisalsocalledsamplingerrortheamountofimprecisionintheestimateofthepopulationparametertheamountaddedandsubtractedtothepointestimatetoformtheconfidenceinterval

DCOVADeterminingSampleSizeFortheMeanDeterminingSampleSizeSamplingerror(marginoferror)DCOVADeterminingSampleSizeFortheMeanDeterminingSampleSize(continued)NowsolveforntogetDCOVADeterminingSampleSizeTodeterminetherequiredsamplesizeforthemean,youmustknow:Thedesiredlevelofconfidence(1-

),whichdeterminesthecriticalvalue,Zα/2Theacceptablesamplingerror,eThestandarddeviation,σ(continued)DCOVARequiredSampleSizeExampleIf=45,whatsamplesizeisneededtoestimatethemeanwithin±5with90%confidence?(Alwaysroundup)Sotherequiredsamplesizeisn=220DCOVAIfσisunknownIfunknown,σcanbeestimatedwhenusingtherequiredsamplesizeformulaUseavalueforσthatisexpectedtobeatleastaslargeasthetrueσSelectapilotsampleandestimateσwiththesamplestandarddeviation,S

DCOVADeterminingSampleSizeDeterminingSampleSizeFortheProportionNowsolveforntoget(continued)DCOVADeterminingSampleSizeTodeterminetherequiredsamplesizefortheproportion,youmustknow:Thedesiredlevelofconfidence(1-

),whichdeterminesthecriticalvalue,Zα/2Theacceptablesamplingerror,eThetrueproportionofeventsofinterest,ππcanbeestimatedwithapilotsampleifnecessary(orconservativelyuse0.5asanestimateofπ)(continued)DCOVARequiredSampleSizeExampleHowlargeasamplewouldbenecessarytoestimatethetrueproportionofdefectivesinalargepopulationwithin±3%,

with95%confidence?(Assumeapilotsampleyieldsp=0.12)DCOVARequiredSampleSizeExampleSolution:For95%confidence,useZα/2=1.96e=0.03p=0.12,sousethistoestimateπSousen=451(continued)DCOVAEthicalIssuesAconfidenceintervalestimate(reflectingsamplingerror)shouldalwaysbeincludedwhenreportingapointestimateThelevelofconfidenceshouldalwaysbereportedThesamplesizeshouldbereportedAninterpretationoftheconfidenceintervalestimateshouldalsobeprovidedChapterSummaryInthischapterwediscussed:

TheconstructionandinterpretationofconfidenceintervalestimatesforthepopulationmeanandthepopulationproportionThedeterminationofthesamplesizenecessarytodevelopaconfidenceintervalforthepopulationmeanorpopulationproportionOnLineTopic:BootstrappingChapter8BootstrappingIsAMethodToUseWhenPopulationIsNotNormal Toestimateapopulationmeanusingbootstrapping,youwould:SelectarandomsampleofsizenwithoutreplacementfromapopulationofsizeN.Resampletheinitialsamplebyselectingnvalueswithreplacementfromtheinitialsample.ComputeXfromthisresample.Repeatsteps2&3mdifferenttimes.ConstructtheresamplingdistributionofX.ConstructanorderedarrayoftheentiresetofresampledX’s.Inthisorderedarrayfindthevaluethatcutsoffthesmallestα/2(100%)andthevaluethatcutsoffthelargestα/2(100%).Thesevaluesprovidethelowerandupperlimitsofthebootstrapconfidenceintervalestimateofμ.DCOVABootstrappingRequiresTheUseofSoftwareAsMinitaborJMPTypicallyaverylargenumber(thousands)ofresamplesareused.Softwareisneededto:Automatetheresamplingproces