时间序列图课件

點圖(DotDiagram)~&Two1DOEClass90a直方圖(Histogram)&Two2DOEClass90a盒形圖(BoxPlot)&Two3DOEClass90a時間序列圖(TimeSeriesPlot)&Two4DOEClass90a期望值與變異數之公式母體平均數(m

)=隨機變數之期望值

E(X)母體變異數(s

2)=隨機變數之變異數V(X)&Two5DOEClass90a期望值與變異數之公式&Two6DOEClass90aSampleandSampling&Two7DOEClass90a點估計(PointEstimation)以抽樣得來之樣本資料,依循某一公式計算出單一數值,來估計母體參數,稱為點估計.好的點估計公式之條件:不偏性最小變異常用之點估計:母體平均數(m)母體變異數(s2)&Two8DOEClass90aCentralLimitTheorem&Two9DOEClass90a假設檢定(HypothesisTesting)“Apersonisinnocentuntilprovenguiltybeyondareasonabledoubt.”

在沒有充分證據證明其犯罪之前,任何人皆是清白的.假設檢定

H0:m=50cm/s H1:m

50cm/sNullHypothesis(H0)Vs.AlternativeHypothesis(H1)One-sidedandtwo-sidedHypothesesAstatisticalhypothesisisastatementabouttheparametersofoneormorepopulations.&Two10DOEClass90aAboutTestingCriticalRegionAcceptanceRegionCriticalValues&Two11DOEClass90aErrorsinHypothesisTesting檢定結果可能為TypeIError(a):RejectH0whileH0istrue.TypeIIError(b):FailtorejectH0whileH0isfalse.&Two12DOEClass90a&Two13DOEClass90aMakingConclusionsWealwaysknowtheriskofrejectingH0,i.e.,a,thesignificantlevelortherisk.WethereforedonotknowtheprobabilityofcommittingatypeIIerror(b).Twowaysofmakingconclusion: 1.RejectH0 2.FailtorejectH0,(DonotsayacceptH0) orthereisnotenoughevidencetorejectH0.&Two14DOEClass90aSignificantLevel(a)a=P(typeIerror)=P(rejectH0whileH0istrue)n=10,s=2.5s/n=0.79&Two15DOEClass90a&Two16DOEClass90a&Two17DOEClass90a&Two18DOEClass90a&Two19DOEClass90aThePowerofaStatisticalTestPower=1-bPower=thesensitivityofastatisticaltest&Two20DOEClass90a1.Fromtheproblemcontext,identifytheparameterofinterest.2.Statethenullhypothesis,H0.3.Specifyanappropriatealternativehypothesis,H1.4.Chooseasignificancelevela.5.Stateanappropriateteststatistic.6.Statetherejectionregionforthestatistic.7.Computeanynecessarysamplequantities,substitutetheseintotheequationfortheteststatistic,andcomputethatvalue.8.DecidewhetherornotH0shouldberejectedandreportthatintheproblemcontext.GeneralProcedureforHypothesisTesting&Two21DOEClass90aInferenceontheMeanofaPopulation

-VarianceKnownH0:m=m0 H1:m

m0,wherem0isaspecifiedconstant.Samplemeanistheunbiasedpointestimatorforpopulationmean.&Two22DOEClass90aExample8-2Aircrewescapesystemsarepoweredbyasolidpropellant.Theburningrateofthispropellantisanimportantproductcharacteristic.Specificationsrequirethatthemeanburningratemustbe50cm/s.Weknowthatthestandarddeviationofburningrateis2cm/s.TheexperimenterdecidestospecifyatypeIerrorprobabilityorsignificancelevelofα=0.05.Heselectsarandomsampleofn=25andobtainsasampleaverageoftheburningrateofx=51.3cm/s.Whatconclusionsshouldbedrawn?&Two23DOEClass90aTheparameterofinterestism,themeaningburningrate.H0:m=50cm/sH1:m

50cm/sa=0.05Theteststatisticsis:RejectH0ifZ0>1.96orZ0<-1.96(becauseZa/2=Z0.025=1.96)Computations:Conclusions:SinceZ0=3.25>1.96,werejectH0:m=50atthe0.05levelofsignificance.Weconcludethatthemeanburningratediffersfrom50cm/s,basedonasampleof25measurements.Infact,thereisstringevidencethatthemeanburningrateexceeds50cm/s.&Two24DOEClass90aP-ValuesinHypothesisTestsWhereZ0istheteststatistic,and(z)isthestandardnormalcumulativefunction.&Two25DOEClass90aTheSampleSize(I)Givenvaluesofaandd,findtherequiredsamplesizentoachieveaparticularlevelofb..&Two26DOEClass90aTheOperatingCharacteristicCurves

-Normaltest(z-test)UsetoperformingsamplesizeortypeIIerrorcalculations.Theparameterdisdefinedas: sothatitcanbeusedforallproblemsregardlessofthevaluesofm0ands.課本41頁之公式為兩平均數差之假設檢定所需之樣本數公式。&Two27DOEClass90a&Two28DOEClass90aConstructionoftheC.I.FromCentralLimitTheory,UsestandardizationandthepropertiesofZ,&Two29DOEClass90aInferenceontheMeanofaPopulation

-VarianceUnknownLetX1,X2,…,Xn

bearandomsampleforanormaldistributionwithunknownmeanmandunknownvariances2.Thequantity hasatdistributionwithn-1degreesoffreedom.&Two30DOEClass90aInferenceontheMeanofaPopulation

-VarianceUnknownH0:m=m0 H1:m

m0,wherem0isaspecifiedconstant.Varianceunknown,therefore,usesinsteadofsintheteststatistic.Ifnislargeenough(30),wecanuseZ-test.However,nisusuallysmall.Inthiscase,T0willnotfollowthestandardnormaldistribution.&Two31DOEClass90aInferencefortheDifferenceinMeans

-TwoNormalDistributionsand

VarianceUnknownWhy?

&Two32DOEClass90a&Two33DOEClass90aisdistributedapproximatelyastwithdegreesoffreedomgivenby&Two34DOEClass90aC.I.ontheDifferenceinMeans&Two35DOEClass90aC.I.ontheDifferenceinMeans&Two36DOEClass90aPairedt-TestWhentheobservationsonthetwopopulationsofinterestarecollectedinpairs.Let(X11,X21),(X12,X22),…,(X1n,X2n)beasetofnpairedobservations,inwhichX1j~(m1,s12)andX2j~(m2,s22)andDj

=X1j–X2j,j=1,2,…,n.Then,totestH0:m1=m2isthesameasperformingaone-samplet-testH0:mD

=0since

mD

=E(X1-X2)=E(X1)-E(X2)=m1-m2

&Two37DOEClass90a&Two38DOEClass90aInferenceontheVarianceofaNormalPopulation(I)H0:s2=s02 H1:s2

s02,wheres02isaspecifiedconstant.Samplingfromanormaldistributionwithunknownmeanmandunknownvariances2,thequantity hasaChi-squaredistributionwithn-1degreesoffreedom.Thatis,&Two39DOEClass90aInferenceontheVarianceofaNormalPopulation(II)LetX1,X2,…,Xn

bearandomsampleforanormaldistributionwithunknownmeanmandunknownvariances2.Totestthehypothesis H0:s2=s02 H1:s2

s02,wheres02isaspecifiedconstant. WeusethestatisticIfH0istrue,thenthestatistichasachi-squaredistributionwithn-1d.f..&Two40DOEClass90a&Two41DOEClass90aTheReasoningForH0tobetrue,thevalueof02cannotbetoolargeortoosmall.Whatvaluesof02shouldwerejectH0?(basedonavalue) Whatvaluesof02shouldweconcludethatthereisnotenoughevidencetorejectH0?&Two42DOEClass90a&Two43DOEClass90aExample8-11

Anautomaticfillingmachineisusedtofillbottleswithliquiddetergent.Arandomsampleof20bottlesresultsinasamplevarianceoffillvolumeofs2=0.0153(fluidounces)2.Ifthevarianceoffillvolumeexceeds0.01(fluidounces)2,anunacceptableproportionofbottleswillbeunderfilledandoverfilled.Isthereevidenceinthesampledatatosuggestthatthemanufacturerhasaproblemwithunderfilledandoverfil