中級計量-第四講

1MultipleRegressionAnalysis:Inference

多元回歸分析：推斷

(1)y=b0+b1x1+b2x2+...bkxk+u2ChapterOutline本章提綱SamplingDistributionsoftheOLSEstimators OLS估計量的樣本分布TestingHypothesisAboutaSinglePopulationParameter:Thettest 單個總體參數(shù)的假設檢驗：t檢驗ConfidenceIntervals 置信區(qū)間TestingHypothesesAboutaSingleLinearCombinationoftheParameters 參數(shù)線性組合的假設檢驗（一維情形）TestingMultipleLinearRestrictions:TheFTest 多個線性約束的假設檢驗：F檢驗ReportingRegressionResults 報告回歸結果3LectureOutline本課提綱SamplingDistributions:Review 樣本分布：復習CLMassumptionsandSamplingDistributionsoftheOLSEstimators 經(jīng)典假設與OLS估計量的樣本分布Backgroundreviewofhypothesistesting 假設檢驗的背景知識One-sidedandtwo-sidedttests 單邊與雙邊t檢驗Calculatingthepvalues 計算p值4SamplingDistribution:Review

樣本分布：復習Simplerandomsamplingisthecasewherenobjectsareselectedatrandomfromapopulation,andeachmemberofthepopulationisequallylikelytobeincludedinthesample. 簡單隨機抽樣是指從總體中隨機取樣n次，使得總體中的每個元素在樣本中的出現(xiàn)的可能性相同。Wheny1,y2,…,ynaredrawnfromthesamedistributionandareindependentlydistributed,theyaresaidtobeindependentlyandidenticallydistributed,ori.i.d.如果y1,y2,…,yn

來自于同一分布且相互獨立，則稱這一組隨機變量獨立同分布（i.i.d.）5SamplingDistribution:Review

樣本分布：復習Samplingdistributionsplayacentralroleinthedevelopmentofstatisticalandeconometricprocedures. 樣本分布在統(tǒng)計學和計量經(jīng)濟學發(fā)展中具有核心地位Itistheprobabilitydistributionofanestimatoroverallpossiblees. 它是指一個估計量在其所有可能取值上的概率分布Therearetwoapproachescharacteringsamplingdistributions:an“exact”approachandan“approximate”approach. 刻畫樣本分布的兩種方式：“準確”方式和“近似”方式6SamplingDistribution:Review

樣本分布：復習The“exact”approachentailsderivingaformulaforthesamplingdistributionthatholdsexactlyforanyvalueofn. “準確”方式需要對任何n的取值都得到樣本分布的精確表達式。Suchdistributioniscalledtheexactdistributionorfinite-sampledistribution. 這樣的分布被稱為小樣本（有限樣本）的準確分布Forexample,ifyisnormallydistributed,andy1,y2,…,ynarei.i.d,thentheiraveragehasanexactdistributionofnormaldistribution.例如，如果y服從正態(tài)分布，且y1,y2,…,yn

獨立同分布，則其均值恰好服從正態(tài)分布7SamplingDistribution:Review

樣本分布:復習The“approximate”approachusesapproximationstothesamplingdistributionsthatrelyonthesamplesizebeinglarge. “近似”方式對樣本分布進行大樣本下的近似。Thelargesampleapproximationtothesamplingdistributionisoftencalledtheasymptoticdistribution. 對樣本分布的大樣本近似常稱為漸近分布。Theasymptoticdistributionscanbecountedontoprovidedgoodapproximationstotheexactsamplingdistribution,aslongasthesamplesizeislarge. 只要樣本量足夠大，漸近分布就是對準確分布的很好的近似。Twokeytools:thelawoflargenumbers,andthecentrallimittheorem. 兩個重要工具：大數(shù)定律，中心極限定理8TheLawofLargeNumbers

大數(shù)定律Thelawsaysthatundergeneralconditions,thesampleaveragewillbeclosetothepopulationmeanwithveryhighprobabilitywhenthesamplesizeislarge. 大數(shù)定律：在一般情形下，當樣本量充分大時，樣本均值將以很高的概率逼近總體均值。Theconditionsforthelawoflargenumberthatwewilluseinthiscoursearethatyarei.i.drandomdrawsandtheitsvarianceisfinite. 本課中，為了應用大數(shù)定律，我們假設y為獨立同分布具有有限方差的隨機取樣。9TheCentralLimitTheorem

中心極限定理Thistheoremstatesthatundergeneralconditions,thesamplingdistributionofthestandardizedsampleaverageiswellapproximatedbyastandardnormaldistributionwhenthesamplesizeislarge. 這個定理說明，在一般條件下，如果樣本足夠大，標準化的樣本均值的樣本分布可以由標準正態(tài)分布近似(scanStockandWatson,p45,47,and49forillustration) 參考：StockandWatson10SamplingDistributionofOLSEstimators

OLS估計量的樣本分布WehavediscussedtheexpectedvalueandvariancesoftheOLSestimators,buttoperformstatisticalinference,wewishtoknowthefullsamplingdistribution. 我們已經(jīng)討論了OLS估計量的期望和方差，但是為了進行統(tǒng)計推斷，我們?nèi)韵Ｍ罉颖痉植肌hesamplingdistributionsoftheOLSestimatorsdependontheunderlyingdistributionoftheerrors. OLS估計量的樣本分布依賴于對誤差項分布的假設。11AssumptionMLR.6(Normality)

假設MLR.6（正態(tài)）

Sofar,weknowthatgiventheGauss-Markovassumptions,OLSisBLUE, 我們已經(jīng)知道當Gauss－Markov假設成立時，OLS是最優(yōu)線性無偏估計。Inordertodoclassicalhypothesistesting,weneedtoaddanotherassumption(beyondtheGauss-Markovassumptions) 為了進行經(jīng)典的假設檢驗，我們要在Gauss－Markov假設之外增加另一假設。

AssumptionMLR.6(Normality):Assumethatuisindependentofx1,x2,…,xkanduisnormallydistributedwithzeromeanandvariances2:u~Normal(0,s2) 假設MLR.6（正態(tài)）：假設u與x1,x2,…,xk獨立，且u服從均值為0，方差為s2的正態(tài)分布。12CLMAssumptions

經(jīng)典線性模型假設AssumptionsMLR.1–MLR.6arecalledtheclassicallinearmodel(CLM)assumptions. 假設MLR.1-MLR.被稱為經(jīng)典線性模型假設Werefertothemodelunderthesesixassumptionsastheclassicallinearmodel. 我們將滿足這六個假設的模型稱為經(jīng)典線性模型UnderCLM,OLSisnotonlyBLUE,butalsotheminimumvarianceunbiasedestimator,thatis,amonglinearandnonlinearestimators,OLSestimatorgivesthesmallestvariance.在經(jīng)典線性模型假設下，OLS不僅是BLUE，而且是最小方差無偏估計量，即在所有線性和非線性的估計量中，OLS估計量具有最小的方差。13CLMAssumptions

經(jīng)典線性模型假設WecansummarizethepopulationassumptionsofCLMasfollows 我們對總體的經(jīng)典線性模型假設做個總結

y|x~Normal(b0+b1x1+…+bkxk,s2)Whilefornowwejustassumenormality,sometimesthisisnotthecase 盡管現(xiàn)在我們假設了正態(tài)，但有時候并不是這種情況14CLMAssumptions

經(jīng)典線性模型假設Whatshouldwedowhenthenormalityassumptionfails? 如果正態(tài)假設不成立怎么辦？Usingatransformation,especiallytakingthelog,oftenyieldsadistributionthatisclosertonormal. 通過變換，特別是通過取自然對數(shù)，往往可以得到接近于正態(tài)的分布。Largesampleswillletusdropnormality(theapproximateapproach) 大樣本允許我們放棄正態(tài)假設（近似方式）15..x1x2Thehomoskedasticnormaldistributionwithasingleexplanatoryvariable同方差正態(tài)分布——單解釋變量情形E(y|x)=b0+b1xyf(y|x)Normaldistributions16Theorem4.1NormalSamplingDistributions

定理4.1正態(tài)樣本分布17BriefProofofTheorem4.1

證明提要18BriefProofofTheorem4.1

證明提要19Theorem4.1canbeextended.Anylinearcombinationsofisalsonormallydistributed,andanysubsetofhasajointnormaldistribution.

可以擴展定理4.1。的任意線性組合服從正態(tài)分布，任意子集服從聯(lián)合正態(tài)。Wewillusethesefactsinhypothesistesting. 我們將利用這些事實來進行假設檢驗Theorem4.1NormalSamplingDistributions

定理4.1正態(tài)樣本分布20TestingHypothesesaboutaSinglePopulationParameter:thet-test

對單個總體參數(shù)的假設檢驗：t檢驗ConsiderapopulationmodelwhichsatisfiestheCLMassumptions.Wenowstudyhowtotesthypothesesaboutaparticular

考慮總體中滿足CLM的模型我們現(xiàn)在研究如何對一個特定的進行假設檢驗21BackgroundReview

背景知識回顧Thehypothesistobetestediscalledthenullhypothesis. 被檢驗的假設稱為零假設Hypothesistestingentailsusingdatatocomparethenullhypothesiswithasecondhypothesis,i.e.,thealternativehypothesis. 假設檢驗利用數(shù)據(jù)將零假設和另一個假設（替代假設）進行比較22BackgroundReview

背景知識回顧Thealternativehypothesisspecifieswhatistrueifthenullhypothesisisnot. 替代假設給出在零假設不成立時的真實情況。Ourgoal:usetheevidenceinarandomlyselectedsampleofdatatodecidewhethertoacceptthenullhypothesis. 我們的目的：利用一個隨機選取的樣本提供給我們的證據(jù)來決定是否應當接受零假設。23BackgroundReview

背景知識回顧Twokindsofmistakesarepossibleinhypothesistesting. 在假設檢驗中存在兩種可能的錯誤。TypeIerror:rejectthenullhypothesiswhenitisinfacttrue. 第一類錯誤：當零假設為真時拒絕零假設（棄真）TypeIIerror:failtorejectthenullwhenitisactuallyfalse.第二類錯誤：當零假設為假時未拒絕零假設（取偽）24BackgroundReview

背景知識回顧HypothesistestingrulesareconstructedtomaketheprobabilityofcommittingtypeIerrorfairlysmall. 我們建立一些假設檢驗的規(guī)則使發(fā)生第一類錯誤的概率非常小。Thesignificancelevel(level)ofatestistheprobabilityofaTypeIerror. 一個檢驗的顯著性水平是發(fā)生第一類錯誤的概率。Commonlyspecifiedsignificancelevels:0.1,0.05,0.01.Ifitequals0.05,itmeanstheresearcheriswillingtofalselyrejectthenullat5%ofthetime. 通常設定的限制性水平為：0.1,0.05,0.01。如果為0.05意味著研究者愿意在5％的檢驗中錯誤地拒絕零假設。25BackgroundReview

背景知識回顧Thecriticalvalueoftheteststatisticisthevalueofthestatisticforwhichthetestjustrejectthenullhypothesisatthegivensignificancelevel. 檢驗統(tǒng)計量的臨界值是使得零假設剛好在給定顯著性水平上被拒絕的統(tǒng)計量的值。Thesetofvaluesoftheteststatisticforwhichthetestrejectsthenullistherejectionregion,

andthevaluesoftheteststatisticforwhichitdoesnotrejectthenullistheacceptanceregion.假設檢驗中，使得零假設被拒絕的檢驗統(tǒng)計量的取值范圍稱為拒絕域，使得零假設不能被拒絕的檢驗統(tǒng)計量的取值范圍成為接受域。26BackgroundReview

背景知識回顧Theprobabilitythatatestactuallyincorrectlyrejectsthenullhypothesiswhenthenullistrueisthesizeofthetest. 當零假設為真而錯誤地拒絕零假設的概率稱為檢驗的容量。Theprobabilitythatatestcorrectlyrejectthenullwhenthealternativeistrueisthepowerofthetest. 當替代假設為真并正確地拒絕零假設的概率稱為檢驗的勢。27BackgroundReview

背景知識回顧Ateststatistic(T)issomefunctionoftherandomsample.Whenwecomputethestatisticforaparticularsample,weobtainaneoftheteststatistic(t). 一個檢驗統(tǒng)計量（T）是關于隨機樣本的一個函數(shù)。當我們用某一特定樣本計算此統(tǒng)計量時，我們得到這個檢驗統(tǒng)計量的一個實現(xiàn)（t）。28Theorem4.2tDistributionfortheStandardizedEstimators

定理4.2:標準化估計量的t分布29ThetTest(cont)

t檢驗

Knowingthesamplingdistributionforthestandardizedestimatorallowsustocarryouthypothesistests 知道標準化估計量的樣本分布后，便可以進行假設檢驗Startwithanullhypothesis 由零假設出發(fā)Forexample,H0:bj=0 例如，H0:bj=0Ifacceptnull,thenacceptthatxjhasnopartialeffectony,controllingforotherx’s 如果接受零假設，則認為控制x其它分量后，

xj對y沒有邊際影響。30ThetTest(cont)31ThetTest(cont)Thetstatisticmeasureshowmanyestimatedstandarddeviationsisawayfromzero. t統(tǒng)計量度量了估計值相對0偏離了多少個估計的標準差。Itssignisthesameas 它的符號與相同Noticewearetestinghypothesesaboutthepopulationparameters,nottestinghypothesesabouttheestimatesfromaparticularsample. 注意我們檢驗的是關于總體參數(shù)的假設，而不是關于來自某一特定樣本的估計值的假設。32tTest:One-SidedAlternatives

t檢驗：單邊替代假設

Besidesournull,H0,weneedanalternativehypothesis,H1,andasignificancelevel 除了零假設外，我們需要替代假設H1，并設定顯著性水平H1maybeone-sided,ortwo-sided H1可以是單邊或雙邊的H1:bj>0andH1:bj<0areone-sided H1:bj>0和H1:bj<0是單邊的H1:bj

0isatwo-sidedalternative H1:bj

0是雙邊替代假設Ifwewanttohaveonlya5%probabilityofrejectingH0ifitisreallytrue,thenwesayoursignificancelevelis5% 如果我們愿意在5％的概率上錯誤地拒絕實際上為真的零假設，則說我們的顯著水平為5％33One-SidedAlternatives(cont)

單邊替代假設

Havingpickedasignificancelevel,a,welookupthe(1–a)thpercentileinatdistributionwithn–k–1degreeoffreedomandcallthisc,thecriticalvalue. 取定顯著性水平a后，找到自由度為n–k–1的t分布的(1–a)分位數(shù)c，即臨界值34One-SidedAlternatives(cont)IfH0:bj=0versusH1:bj>0,werejectH0iftbj>c,failtoreject

H0iftbj<=c. 如果H0:bj=0對H1:bj>0，當tbj>c時我們拒絕H0，當tbj<=c，則不能拒絕H0Becausetdistributionissymmetric,ifH0:bj=0versusH1:bj<0,werejectH0iftbj<-c,failstoreject

H0iftbj>=-c. 由于t分布是對稱的，如果H0:bj=0對H1:bj<0,當tbj<-c時我們拒絕H0，當tbj>=-c

，則不能拒絕H035yi=b0+b1xi1+…

+bkxik+uiH0:bj=0H1:bj>0c0a(1-a)One-SidedAlternatives(cont)單邊替代假設Failtorejectreject36Thetdistributionversusnormaldistribution

t分布與正態(tài)分布Noticethatasthedegreeoffreedominthetdistributiongetslarge,thetdistributionapproachesthestandardnormaldistribution. 注意：當t分布的自由度增大時，t分布趨近于標準正態(tài)分布。37Example:StudentPerformanceandSchoolSize

例子：學生表現(xiàn)與學校規(guī)模Question:Doeslargerclasssizeresultsinpoorerstudentperformance? 問題：是不是較大的班級意味著較差的學生表現(xiàn)？Use408highschoolsinMichiganforyear1993,performthefollowingregression: 應用1993年408個密歇根州中學的數(shù)據(jù)，進行如下回歸^p+0.048staff–.0002enroll (6.113) (0.0001)(0.04) (0.00022) math10:percentageofstudentspassingtheMEAPstandardizedgrade10mathtest 通過MEAP標準化10年級數(shù)學測驗的學生百分比p:averageannualteacher’scompensation 平均教師年度補償 staff:thenumberofstaffperonethousandstudents 每千個學生對應的工作人員數(shù)目 enroll :studentenrollment 學生錄取 38Decidethetestinghypotheses: 確定被檢驗的假設H0:βenroll=0versusH1:βenroll<0Computethetstatistic,計算t統(tǒng)計量 t=-0.0002/0.00022=-0.91Sincen-k-1=404,weusethestandardnormalcriticalvalue.Atthe5%level,thecriticalvalueis–1.65. 由于n-k-1=404，我們使用標準正態(tài)的臨界值。在5％顯著水平下，臨界值位－1.65Because-0.91>–1.65,wefailtorejectthenull. 由于-0.91>-1.65，我們不能拒絕零假設Example:StudentPerformanceandSchoolSize

例子：學生表現(xiàn)與學校規(guī)模39Ifwearealsointerestedinaskingwhetherbetter-paidteachersleadstobetterstudentperformance,wecantest 如果我們同樣感興趣是否高收入的教師會使學生表現(xiàn)更好，我們可以檢驗：H0:βp=0versusH1:βp>0Thecalculatedtstatisticequals4.6.Since4.6>2.326,thereforerejectingtheH0at1%level. 計算得到的t統(tǒng)計量為4.6。由于4.6>2.326，故在1%顯著水平下拒絕零假設。Example:StudentPerformanceandSchoolSize

例子：學生表現(xiàn)與學校規(guī)模40TheTwo-sidedAlternatives

雙邊替代假設H1:bj

0isatwo-sidedalternative.Underthisalternative,wehavenotspecifiedthesignofthepartialeffectofxj

ony. H1:bj

0為雙邊替代假設。在此替代假設下，我們并未規(guī)定xj

對y影響的符號。Foratwo-sidedtest,wesetthecriticalvaluebasedona/2andrejectH0

iftheabsolutevalueofthetstatistic>c.cisthe97.5thpercentileinthetdistributionwithn-k-1degreesoffreedomifa=0.05. 對于雙邊檢驗，我們根據(jù)a/2計算臨界值。當t的絕對值大于臨界值c時，拒絕零假設。當a=0.05時，

c是n-k-1自由度的t分布的97.5分位數(shù)。41yi=b0+b1Xi1+…

+bkXik+uiH0:bj=0H1:bj?=0c0a/2(1-a)-ca/2Two-SidedAlternatives雙邊替代假設rejectrejectfailtoreject42Example:StudentPerformanceandSchoolSizecontinued

例子：學生表現(xiàn)與學校規(guī)模Wehaveestimated 我們已經(jīng)得到^p+0.048staff–.0002enroll (6.113) (0.0001)(0.04) (0.00022)Ifthequestionis:whetherthenumberofteachershasimpactsonstudentperformance,wecanformhypothesesof:H0:bstaff=0,H1:bstaff?=0. 如果問題是：教師數(shù)目是否對學生表現(xiàn)有影響，我們可以檢驗如下假設：H0:bstaff=0,H1:bstaff?=0.Thecalculatedtratiois1.2.The5%criticalvalueofstandardnormalis1.96.Since1.2<1.96,wefailtorejectthenull 計算得到的t值為1.2。標準正態(tài)分布的在5%的顯著水平對應的臨界值為1.96。由于1.2<1.96，我們不能拒絕零假設。

43SummaryforH0:bj=0

總結

Unlessotherwisestated,thealternativeisassumedtobetwo-sided 除非特別指出，我們總認為替代假設是雙邊的Ifwerejectthenull,wetypicallysay“xjisstatisticallysignificantatthea%level” 如果拒絕了零假設，我們通常說“xj

在a%水平下顯著”Ifwefailtorejectthenull,wetypicallysay“xjisstatisticallyinsignificantatthea%level” 如果不能拒絕零假設，我們通常說“xj

在a%水平下不顯著”44Testingotherhypotheses

其他假設檢驗AmoregeneralformofthetstatisticrecognizesthatwemaywanttotestsomethinglikeH0:bj=aj

如果我們想對形如H0:bj=aj

的假設進行檢驗，需要更一般的t統(tǒng)計量Inthiscase,theappropriatetstatisticis 此時，恰當?shù)膖統(tǒng)計量是45Example:CampusCrimeandEnrollment

例子：校園犯罪與錄取Question:Will1%increaseinenrollmentincreasecampuscrimebymorethan1%? 問題：錄取量提高1%是否會導致校園犯罪增加超過1%?Supposetotalnumberofcrimesisdeterminedby 假設犯罪總數(shù)由下式?jīng)Q定 crime=exp(b0

)enrollb1exp(u).Onecanestimate 可以估計 log(crime)=b0+b1log(enroll)+u46Example:CampusCrimeandEnrollment

例子：校園犯罪與錄取Andtest

H0:b1=1H1:b1>1.UsingdatafromtheFBI’suniformCrimereports(97observations),theestimatedequationis利用FBI犯罪報告（97個觀察值）的數(shù)據(jù)，估計得到方程 ^log(crime)=-6.63+1.27log(enroll) (1.03)(0.11) Thecorrecttratio=(1.27-1)/0.11=2.45.The1%one-sidedcriticalvalueforatdistributionwith95degreesoffreedomis2.37<2.45.Thereforerejectthenull. t值=(1.27-1)/0.11=2.45。對于95自由度的t分布，1%顯著水平下單邊檢驗的臨界值為2.37<2.45，拒絕零假設。47Computingp-valuesfortTests

計算t檢驗的p值Thestepsinclassicalhypothesistesting: 經(jīng)典假設檢驗的步驟Statethenullandthealternativehypothesis 表述零假設和替代假設Decideasignificancelevelandfindtherelatedcriticalvalue 決定顯著水平，找到臨界值Calculatethetstatisticbasedonthesampledata 根據(jù)樣本數(shù)據(jù)計算t統(tǒng)計量Comparethetstatisticwiththecriticalvaluetodecidewhethertorejectthenull比較t值與臨界值，決定是否拒絕零假設。48Computingp-valuesfortTests

計算t檢驗的p值Supposeat40degreesoffreedom,acalculatedtratiois2.423,therelated5%and1%criticalvaluesare2.021and2.704,respectively.Shouldwerejectornottorejectthenull? 假設自由度為40，算得t值為2.423，對應5%和1%的臨界值分別為2.021和2.704。我們是否應當拒絕零假設？Committingtoasignificancelevelaheadoftimecanhideusefulinformationabouttheeofahypothesistest. 提前確定顯著水平可能會隱藏關于假設檢驗的一些有用信息。“Tobe,ornottobe:thatisthequestion.”--Hamlet(3.1.64-98).49Computingp-valuesfortTests

計算t檢驗的p值Analternativetotheclassicalapproach:Ifthecalculatedtstatisticisusedascriticalvalue,whatisthesmallestsignificancelevelatwhichthenullhypothesiswouldberejected? 另一種想法：如果將算得的t統(tǒng)計量作為臨界值，那么使得零假設被拒絕的最小顯著水平是多少？Thislevelisknownasthep-value.Foratwo-sidedalternative, 這個水平稱為p值。對于雙邊檢驗

p-value=P(|T|>|t|).

50Computingp-valuesfortTests

計算t檢驗的p值C0.025C0.025C0.005C0.005C0.01C0.01pα/2pα/2Intheaboveexample,itmustbetruethat1%<p<5%.p-value=P(|T|>2.423)=2P(T>2.423)=0.02.51Usefulinformationaboutp-values

一些關于p值的信息Becauseitisaprobability,itsrangeisbetween0and1. 由于這是一個概率，其取值范圍在0，1之間Smallpvaluesareevidenceagainstthenull,largepvaluesprovidelittleevidenceagainstthenull. 小p值提供了拒絕零假設的證據(jù)，大p值不能提供證據(jù)拒絕零假設。Anotherangletounderstandpvalues:Ifthenullistrue,whatistheprobabilityofobservingatstatisticlikethecalculatedone? 理解p值的其它角度：如果零假設為真，那么有多大的概率可以觀察到算得的t值？52EconomicSignificanceversusstatisticalsignificance

經(jīng)濟重要性與統(tǒng)計顯著性Statisticalsignificanceisdeterminedtotallybyhowlargeisthetstatistic. 統(tǒng)計顯著性完全由t統(tǒng)計量的大小決定。Economicsignificancefocusonthemagnitudeoftheestimatedcoefficients. 經(jīng)濟上的重要性強調(diào)估計系數(shù)的大小。Bothneedtobeconsideredtogetabalancedviewaboutaregressor’spartialeffectony. 權衡兩者來判斷解釋變量對被解釋變量的邊際影響53MultipleRegressionAnalysis:Inference

多元回歸分析：推斷

(2)y=b0+b1x1+b2x2+...bkxk+u54ChapterOutlineSamplingDistributionsoftheOLSEstimators OLS估計量的樣本分布TestingHypothesisAboutaSinglePopulationParameter:Thettest 單總體參數(shù)的假設檢驗：t檢驗ConfidenceIntervals 置信區(qū)間TestingHypothesesAboutaSingleLinearCombinationoftheParameters 參數(shù)線性組合的假設檢驗（一維情形）TestingMultipleLinearRestrictions:TheFTest 多個線性約束的假設檢驗：F檢驗ReportingRegressionResults 報告回歸結果55LectureOutlineConfidenceIntervals:Review 置信區(qū)間：復習Howtoconstructconfidenceintervals 如何構造置信區(qū)間TestingLinearcombinations 檢驗線性組合56ConfidenceIntervals

置信區(qū)間Becauseofrandomsamplingerror,itisimpossibletolearntheexactvalueofb

usingonlytheinformationinthesample. 由于隨機取樣誤差的存在，我們不可能通過樣本知道b的準確值。Butitispossibletousedatafromarandomsampletoconstructasetofvaluesthatcontainsthetruevaluewithacertainspecifiedprobability. 但是利用來自隨機樣本的數(shù)據(jù)構造一個取值的集合，使得真值在給定概率下屬于這個集合是可能的。57ConfidenceIntervals

置信區(qū)間Suchasetiscalledaconfidenceset,andthepre-specifiedprobabilitythatthetruevalueiscontainedinthissetiscalledtheconfidencelevel. 這樣的集合稱為置信集，預先設定的真值屬于此集合的概率稱為置信水平（置信度）。Theconfidencesetturnsouttobeallthepossiblevaluesbetweenalowerandanupperlimit,sothattheconfidencesetisaninterval,i.e.,confidenceinterval. 置信集是下限和上限之間所有可能的取值，故置信集為一個區(qū)間，稱為置信區(qū)間58ConstructingConfidenceIntervals

構造置信區(qū)間Considerthefollowingsimulationstudythatthetruevalueisknown. 考慮一個模擬研究，此時真值已知Supposethesampley1,y2,…,y10arei.i.ddrawsfromN(2,1),and20ofsuchrandomsamplesaredrawn.Wethencalculatethesampleaverageandtherelatedtstatistic. 假設樣本y1,y2,…,y10

服從N(2,1),且獨立同分布，取20個樣本。計算樣本均值和對應的t值5960ConstructingConfidenceIntervals

構造置信區(qū)間In95%ofallsamples,wecorrectlyacceptthemeanμtobe2. 在95%的樣本中，我們正確地接受了μ＝2Thevaluesincolumn2canbeputintotheconfidenceset,exceptthevalueinrow19. 第二列的數(shù)值均落入置信集，除了第19行的值Ifnislarge,andmorereplicationsareperformed,theprobabilityofrejectingthenullat5%levelis5%. 如果n足夠大，而且將試驗重復更多次，在5%水平上拒絕零假設的概率為5%61ConstructingConfidenceIntervals

構造置信區(qū)間Whenμisunknown,inourminds,thefollowingstepscanbeused. 當μ未知時，我們認為可以應用下述過程Beginbypickingsomearbitraryrandomsampleandestimateμ,callthisμ1. 任意選取隨機樣本并估計μ，記為μ1TestH0:μ=μ1

,H1:μ?=μ1bycomputingthetstatistic,t1. 通過計算t統(tǒng)計量t1檢驗H0:μ=μ1

，H1:μ?=μ1

62ConstructingConfidenceIntervals

構造置信區(qū)間Ift1<1.96,wecannotrejectthenull.Putμ1intheconfidenceset. 如果t1<1.96，我們不能拒絕零假設，將μ1歸入置信集。Repeattheabovestepsforallpossiblerandomsamples. 對所有可能的隨機取樣重復上述過程Continuingthisprocessyieldsthesetofallvaluesofμthatcannotberejectedatthe5%levelofatwo-sidedhypothesistest. 這樣可以得到在5％水平的雙邊檢驗下所有不能被拒絕的μ的取值63ConstructingConfidenceIntervals

構造置信區(qū)間Inotherwords,in95%ofallsamples,theconfidencesetwillcontainthetruevalueofμ. 換言之，95％的樣本中，置信集會包含μ的真值。Thissethasaremarkableproperty:theprobabilitythatitcontainsthetruevalueofis95%. 重要性質(zhì)：此集合包含真值的概率為95%Therefore,wehaveconstructeda95%confidencesetforμ. 因此，我們構造了μ的95%置信區(qū)間。64ConstructingConfidenceIntervals

構造置信區(qū)間Inrealitydrawingallsamplestoconstructconfidenceintervalisimpractical. 現(xiàn)實中，抽取所有樣本來構造置信區(qū)間是不實際的。Butbecausethesampleaverageisnormallydistributedwithmeanμandvariance1/n,theconfidenceintervalcanbecalculatedforaspecificrandomsample. 但是由于樣本均值服從均值為μ，方差為1/n的正態(tài)分布，置信區(qū)間可以由一特定隨機樣本計算。65ConstructingConfidenceIntervals

構造置信區(qū)間66ConfidenceIntervalsforb

b的置信區(qū)間

Theaboveanalysiscanbeextendedtoconstructaconfidenceintervalforbusingthesamecriticalvalueaswasusedforatwo-sidedtest. 通過對上述分析進行擴展，我們可以利用雙邊檢驗的臨界值來構造b的置信區(qū)間。Ifhasatdistributionwithn-k-1degreesoffreedom,simplemanipulationleadstoaconfidenceintervalsfortheunknownbj 如果服從n-k-1自由度的t分布，簡單的運算可以得到關于未知的bj的置信區(qū)間67ConfidenceIntervalsforb

b的置信區(qū)間68ConfidenceIntervalsforb

b的置信區(qū)間Fordf=n-k-1=25,a95%confidenceintervalforanybjisgivenby 如果自由度為25，那么對任意bj

，95％的置信區(qū)間為Whenn-k-1>120,thet(n-k-1)distributioniscloseenoughtonormaltousethe97.5thpercentileinastandardnormaldistributionforconstructinga95%CI: 當n-k-1>120，t(n-k-1)分布與正態(tài)分布充分接近，可以用標準正態(tài)分布的97.5分位數(shù)來構造95%置信區(qū)間69ConfidenceIntervalsforb

b的置信區(qū)間Onceaconfidenceintervalisconstructed,onecancarryouttwo-tailedhypothesestests. 構造了置信區(qū)間之后，可以進行雙尾假設檢驗IfthenullhypothesisisH0:bj=aj,thenitisrejectedagainstH1:bj~=ajatthe5%significancelevelifandonlyifajisnotinthe95%confidenceinterval. 零假設為H0:bj=aj，當且僅當aj不在95％的置信區(qū)間內(nèi)時，零假設相對于H1:bj~=aj在5％的顯著水平上被拒絕。70Example:HedonicPriceModelforHouses

例子：住房的效用價格模型Log(price)=7.46+0.634log(sqrft)-0.066bdrms+0.158bthrms

(1.15)(0.184)(0.059)(0.075)n=19R-square=0.806

df=19-4=15,c=2.13at95%

The95%confidenceintervalforthecoefficientonlog(sqrft) log(sqrft)對應系數(shù)的95％置信區(qū)間 [0.634-2.131*0.184,0.634+2.131*0.184]=[0.242,1.026]71Stataandp-values,ttests,etc.

Stata，p值，t檢驗

Mostcomputerpackageswillcomputethep-valueforyou,assumingatwo-sidedtest 大部分軟件包可以在假定計算雙邊檢驗的基礎上計算p值Ifyoureallywantaone-sidedalternative,justdividethetwo-sidedp-valueby2 單邊檢驗p值是雙邊檢驗的p值的一半Stataprovidesthetstatistic,p-value,and95%confidenceintervalforH0:bj=0foryou,incolumnslabeled“t”,“P>|t|”and“[95%Conf.Interval]”,respectively Stata提供了關于零假設H0:bj=0的t值，p值和95%置信區(qū)間，分別標記為“t”,“P>|t|”，“[95%Conf.Interval]”72TestingaLinearCombination

檢驗線性組合

Supposeinsteadoftestingwhetherb1isequaltoaconstant,youwanttotestifitisequaltoanotherparameter,thatisH0:b1=b2 假設我們要檢驗是否一個參數(shù)等于另一個參數(shù)H0:b1=b2，而不是檢驗b1是否等于一個常數(shù)。Usesamebasicprocedureforformingatstatistic 應用與構造t統(tǒng)計量相同的程序73TestingLinearCombo(cont)

檢驗線性組合74TestingaLinearCombo(cont)

檢驗線性組合

So,touseformula,needs12,whichstandardoutputdoesnothave 需要s12帶入上式，標準的程序并不報告此值。Manypackageswillhaveanoptiontogetit,orwilljustperformthetestforyou 許多軟件有計算此值的選項，或是可以直接進行檢驗InStata,afterregyx1x2…xkyouwouldtypetestx1=x2

togetap-valueforthetest Stata中，在regyx1x2…xk后，可以輸入testx1=x2得到檢驗的p值Moregenerally,youcanalwaysrestatetheproblemtogetthetestyouwant 可以重復地輸入此命令來得到檢驗的結果75Example:

例子：Supposeyouareinterestedintheeffectofcampaignexpendituresones: 假設你感興趣的是競選支出對選舉結果的影響 voteA=b0+b1log(expendA)+b2log(expendB)+b3prtystrA+uH0:b1=-b2,orH0:q1=b1+b2=0b1=q1–b2,sosubstituteinandrearrange 令b1=q1–b2,帶入并移項可得 voteA=b0+q1log(expendA)+b2log(expendB-expendA)+b3prtystrA+u76Example(cont):

Thisisthesamemodelasoriginally,butnowyougetastandarderrorforb1–b2=q1directlyfromthebasicregression 這個模型與原模型相同，但是此時可以直接從回歸中得到b1–b2=q1的標準差Anylinearcombinationofparameterscouldbetestedinasimilarmanner 參數(shù)的任何線性組合都可以用類似的手段進行檢驗。Otherexamplesofhypothesesaboutasinglelinearcombinationofparameters: 關于檢驗參數(shù)的單個線性組合的其它例子 b1=1+b2;b1=5b2;b1=-1/2b2;etc77MultipleRegressionAnalysis:

Inference

多元回歸分析：推斷

(3)y=b0+b1x1+b2x2+...bkxk+u78ChapterOutline本章提綱SamplingDistributionsoftheOLSEstimators OLS估計量的樣本分布TestingHypothesisAboutaSinglePopulationParameter:Thettest 單總體參數(shù)的假設檢驗：t檢驗ConfidenceIntervals置信區(qū)間TestingHypothesesAboutaSingleLinearCombinationoftheParameters 參數(shù)線性組合的假設檢驗（一維情形）TestingMultipleLinearRestrictions:TheFTest 多個線性約束的假設檢驗：F檢驗ReportingRegressionResults 報告回歸結果79LectureOutline TheFtest F檢驗80MultipleLinearRestrictions

多線性約束

Everythingwe’vedonesofarhasinvolvedtestingasinglelinearrestriction,(e.g.b1