斯托克、沃森《计量经济学（第三版）》课后习题答案

ForStudentsSolutionstoOdd-NumberedEnd-of-ChapterExercisesChapter2ReviewofProbability2.1.(a)ProbabilitydistributionfunctionforYOutcome(numberofheads)Y0Y1Y2Probability0.250.500.25(b)CumulativeprobabilitydistributionfunctionforYOutcome(numberofheads)Y00Y11Y2Y2Probability00.250.751.0d(c)Y=EY()(00.25)(10.50)(20.25)1.00.FFq,.UsingKeyConcept2.3:var(YEYEY)(22)[()],and(|uXii)sothatvar(YEYEY)(22)[()]1.50(1.00)20.50.2.3.ForthetwonewrandomvariablesWX36andVY207,wehave:(a)EV()E(207Y)207EY()2070781454,EW()E(36)36()3607072XEX222(b)WXvar(36X)636021756,222VYvar(207Y)(7)490171684084(c)WVcov(36XY,207)6(7)cov(XY,)42008435283528corr(WV,)WV04425WV75684084©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises32.5.LetXdenotetemperatureinFandYdenotetemperatureinC.RecallthatY0whenX32andY100whenX212;thisimpliesYXYX(100/180)(32)or17.78(5/9).UsingKeyooConcept2.3,X70FimpliesthatY17.78(5/9)7021.11C,andX7FimpliesY(5/9)73.89C.2222.7.Usingobviousnotation,CMF;thusCMFandCMF2cov(M,F).Thisimplies(a)C4045$85,000peryear.cov(M,F)(b)corr(MF,),sothatcov(M,FMF)MFcorr(,).Thuscov(MF,)MF12180.80172.80,wheretheunitsaresquaredthousandsofdollarsperyear.222222(c)CMF2cov(M,F),sothatC12182172.80813.60,andC813.6028.524thousanddollarsperyear.(d)FirstyouneedtolookupthecurrentEuro/dollarexchangerateintheWallStreetJournal,theFederalReservewebpage,orotherfinancialdataoutlet.Supposethatthisexchangerateise(saye0.80Eurosperdollar);each1dollaristhereforewitheEuros.ThemeanisthereforeeC(inunitsofthousandsofEurosperyear),andthestandarddeviationiseC(inunitsofthousandsofEurosperyear).Thecorrelationisunit-free,andisunchanged.2.9.ValueofYProbabilityDistributionof1422304065XValueofX10.020.050.100.030.010.2150.170.150.050.020.010.4080.020.030.150.100.090.39ProbabilitydistributionofY0.210.230.300.150.111.00(a)Theprobabilitydistributionisgiveninthetableabove.EY()140.21220.23300.30400.15650.1130.15EY(22)140.212220.233020.304020.156520.111127.23var(YEYEY)(22)[()]218.21Y14.77©2011PearsonEducation,Inc.PublishingasAddisonWesley4Stock/Watson•IntroductiontoEconometrics,ThirdEdition(b)TheconditionalprobabilityofY|X8isgiveninthetablebelowValueofY14223040650.02/0.390.03/0.390.15/0.390.10/0.390.09/0.39E(YX|8)14(0.02/0.39)22(0.03/0.39)30(0.15/0.39)40(0.10/0.39)65(0.09/0.39)39.21EY(22|X8)14(0.02/0.39)222(0.03/0.39)302(0.15/0.39)4022(0.10/0.39)65(0.09/0.39)1778.7var(Y)1778.739.212241.65YX815.54(c)EXY()(1140.02)(122:0.05)(8650.09)171.7cov(XY,)EXY()EXEY()()171.75.3330.1511.0corr(XY,)cov(XY,)/(XY)11.0/(2.6014.77)0.2862.11.(a)0.90(b)0.05(c)0.052(d)WhenY~,10thenYF/10~10,.(e)YZ2,whereZN~(0,1),thusPr(YZ1)Pr(11)0.32.2.13.(a)EY(22)Var(Y)101;EW(2)Var(W)21000100.YW(b)YandWaresymmetricaround0,thusskewnessisequalto0;becausetheirmeaniszero,thismeansthatthethirdmomentiszero.444(c)Thekurtosisofthenormalis3,so3(EYYY)/;solvingyieldsE(Y)3;asimilarcalculationyieldstheresultsforW.(d)First,conditiononX0,sothatSW:ESX(|0)0;ES(2342|X0)100,ES(|X0)0,ES(|X0)3100.Similarly,234ESX(|1)0;ES(|X1)1,ES(|X1)0,ES(|X1)3.FromthelawofiteratedexpectationsES()ESX(|0)Pr(X0)ESX(|1)Pr(X1)0ES(22)ES(|X0)Pr(X0)ES(2|X1)Pr(X1)1000.0110.991.99ES()33ES(|X0)Pr(X0)(|ES3X1)Pr(1)0XES(44)ES(|X0)Pr(X0)ES(4|X1)Pr(X1)231000.01310.99302.97©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises533(e)SES()0,thusES()()0SESfrompart(d).Thusskewness0.Similarly,22244SSES()ES()1.99,andES(S)ES()302.97.Thus,kurtosis302.97/(1.992)76.59.610Y1010.4102.15.(a)Pr(9.6Y10.4)Pr4/nn4/4/n9.61010.410PrZ4/nn4/whereZ~N(0,1).Thus,9.61010.410(i)n20;PrZZPr(0.890.89)0.634/nn4/9.61010.410(ii)n100;PrZZPr(2.002.00)0.9544/nn4/9.61010.410(iii)n1000;PrZZPr(6.326.32)1.0004/nn4/cY10c(b)Pr(10cY10c)Pr4/nnn4/4/ccPrZ.4/nn4/cAsngetlargegetslarge,andtheprobabilityconvergesto1.4/n(c)Thisfollowsfrom(b)andthedefinitionofconvergenceinprobabilitygiveninKeyConcept2.6.2.17.=0.4and20.40.60.24YYYY0.40.430.40.4(a)(i)P(Y0.43)PrPr0.61240.270.24/nn0.24/0.24/nYY0.40.370.40.4(ii)P(Y0.37)PrPr1.220.110.24/nn0.24/0.24/n0.410.40(b)WeknowPr(1.96Z1.96)0.95,thuswewantntosatisfy0.411.9624/n0.390.40and1.96.Solvingtheseinequalitiesyieldsn9220.24/n©2011PearsonEducation,Inc.PublishingasAddisonWesley6Stock/Watson•IntroductiontoEconometrics,ThirdEditionl2.19.(a)Pr(Yyjij)Pr(XxYy,)i1lPr(YyXx|)Pr(Xx)jiii1kkl(b)E(YyYyy)jPr(jj)Pr(YyXxXxji|)Pr(i)jji111lkyYyXxXxPr(|)Pr()jjiii1j1lEYXx(|)Pr(Xx)iii1(c)WhenXandYareindependent,Pr(XxYij,y)Pr(Xxi)Pr(Yyj)soXYEX[(X)(YY)]lk()()Pr(,)xyXxYyiXjYijij11lk(x)(yXxYy)Pr()Pr()iXjYijij11lk(xiX)Pr(Xxi)(yjY)Pr(Yyjij11EX()()000,XYEY0corr(XY,)XY0XYXY2.21.(a)EX()32E[(X)(X)]EX[3222232XXX2X]EX()3()32EX3()EX2332EX()3()()EXEX3(EX)[(EX)][(23EX)]EX()3()()2()32EXEXEX3(b)EX()43223E[(X3X3X)(X)]EXXXXXXX[34332233322343]E(XEXEXEXEXEXEXEX43)4()()6(22)()4()()34()EX(43224)4[EX()][EX()]6[EX()][EX()]3[EX()]©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises72.23.XandZaretwoindependentlydistributedstandardnormalrandomvariables,so22XZ0,XZ1,XZ0.(a)BecauseoftheindependencebetweenXandZ,Pr(ZzX|x)Pr(Zz),andEZX(|)EZ()0.ThusE(|YX)EX(22ZX|)EX(|X)EZX(|)X220X22222(b)EX()XX1,andYZEX()()101ZEX(c)E()XYE(X33ZX)E()XE().ZXUsingthefactthattheoddmomentsofastandardnormalrandomvariableareallzero,wehaveEX()0.3Usingtheindependencebetween3XandZ,wehaveEZX()ZX0.ThusEXY()EX()EZX()0.(d)cov(XY)E[(XXY)(Y)]E[(X0)(Y1)]EXY()()()XEXYEX0000corr(XY,)XY0XYXYnn2.25.(a)axinni()()ax123axaxaxax123xxxaxi1i1n(b)()(xyiixyxy1122xnny)i1()()x12xxyyynn12nnxyiiii11n(c)aaaa()anai1nn(d)222222()(abxcyiiabxcyii222)abxacybcxyiiiiii11nnnnn22222nabxiicy222abxiacyibcxiiyii11i1i1i12.27(a)E(W)E[E(W|Z)]E[E(XX)|Z]E[E(X|Z)E(X|Z)]0.(b)E(WZ)E[E(WZ|Z)]E[ZE(W)|Z]E[Z0]0(c)Usingthehint:VWh(Z),sothatE(V2)E(W2)E[h(Z)2]2E[Wh(Z)].Usinganargumentlikethatin(b),E[Wh(Z)]0.Thus,E(V2)E(W2)E[h(Z)2],andtheresultfollowsbyrecognizingthatE[h(Z)2]0becauseh(z)20foranyvalueofz.©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter3ReviewofStatistics3.1.Thecentrallimittheoremsuggeststhatwhenthesamplesize(n)islarge,thedistributionofthe222Ysampleaverage(Y)isapproximatelyN,with.Givenapopulation100,YYYnY2Y430,wehave243(a)n100,2Y043,andYn100Y100101100Pr(Y101)Pr(1.525)09364043043243(b)n64,2Y06719,andYn64101100Y100103100Pr(101Y103)Pr067190671906719(36599)(12200)099990888801111243(c)n165,2Y02606,andYn165Y10098100Pr(YY98)1Pr(98)1Pr02606026061(39178)(39178)10000(roundedtofourdecimalplaces)3.3.Denoteeachvoter’spreferencebyY.Y1ifthevoterpreferstheincumbentandY0ifthevoterprefersthechallenger.YisaBernoullirandomvariablewithprobabilityPr(1)YpandPr(0)1.YpFromthesolutiontoExercise3.2,Yhasmeanpandvariancepp(1).215(a)pˆ05375.400ppˆˆ(1)0.5375(10.5375)(b)Theestimatedvarianceofpˆisvar(pˆ)62148104.Then4001standarderrorisSE(ppˆˆ)(var())200249.©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises9(c)Thecomputedt-statisticispˆ0537505tactp01506SE(pˆ)00249Becauseofthelargesamplesize(400),nwecanuseEquation(3.14)inthetexttogetthep-valueforthetestHp005vs.Hp105:pt-value2(|act|)2(1506)200660132(d)UsingEquation(3.17)inthetext,thep-valueforthetestHp005vs.Hp105ispt-value1(act)1(1506)109340066(e)Part(c)isatwo-sidedtestandthep-valueistheareainthetailsofthestandardnormaldistributionoutside(calculatedt-statistic).Part(d)isaone-sidedtestandthep-valueistheareaunderthestandardnormaldistributiontotherightofthecalculatedt-statistic.(f)ForthetestHp005vs.Hp105,wecannotrejectthenullhypothesisatthe5%significancelevel.Thep-value0.066islargerthan0.05.Equivalentlythecalculatedt-statistic1506islessthanthecriticalvalue1.64foraone-sidedtestwitha5%significancelevel.Thetestsuggeststhatthesurveydidnotcontainstatisticallysignificantevidencethattheincumbentwasaheadofthechallengeratthetimeofthesurvey.3.5.(a)(i)ThesizeisgivenbyPr(|pˆ0.5|.02),wheretheprobabilityiscomputedassumingthatp0.5.Pr(|ppˆˆ0.5|0.02)1Pr(0.020.5.02)0.02pˆ0.50.021Pr0.50.5/10550.50.5/10550.50.5/1055pˆ0.51Pr1.301.300.50.5/10550.19wherethefinalequalityusingthecentrallimittheoremapproximation.(ii)ThepowerisgivenbyPr(|pˆ0.5|0.02),wheretheprobabilityiscomputedassumingthatp0.53.Pr(|ppˆˆ0.5|0.02)1Pr(0.020.5.02)0.02pˆ0.50.021Pr0.530.47/10550.530.47/10550.530.47/10550.05pˆ0.530.011Pr0.530.47/10550.530.47/10550.530.47/1055pˆ0.531Pr3.250.65.530.47/10550.74wherethefinalequalityusingthecentrallimittheoremapproximation.©2011PearsonEducation,Inc.PublishingasAddisonWesley10Stock/Watson•IntroductiontoEconometrics,ThirdEdition0.540.50(b)(i)tt2.61,andPr(||2.61)0.01,sothatthenullisrejectedatthe(0.540.46)/10555%level.(ii)Pr(t2.61).004,sothatthenullisrejectedatthe5%level.(iii)0.541.96(0.540.46)/10550.540.03,or0.51to0.57.(iv)0.542.58(0.540.46)/10550.540.04,or0.50to0.58.(v)0.540.67(0.540.46)/10550.540.01,or0.53to0.55.(c)(i)Theprobabilityis0.95isanysinglesurvey,thereare20independentsurveys,sotheprobabilityif0.95200.36(ii)95%ofthe20confidenceintervalsor19.(d)Therelevantequationis1.96SE(pppnˆ).01or1.96(1)/.01.Thusnmustbe1.962p(1p)chosensothatn,sothattheanswerdependsonthevalueofp.Notethatthe0.012largestvaluethatp(1−p)cantakeonis0.25(thatis,p0.5makesp(1p)aslargeas1.9620.25possible).Thusifn9604,thenthemarginoferrorislessthan0.01forall0.012valuesofp.3.7.Thenullhypothesisisthatthesurveyisarandomdrawfromapopulationwithp=0.11.Thet-pˆ0.11statisticist,whereSE(ppˆˆ)(1pnˆ)/.(AnalternativeformulaforSE(pˆ)isSE(pˆ)0.11(10.11)/n,whichisvalidunderthenullhypothesisthatp0.11).Thevalueofthet-statisticis2.71,whichhasap-valueofthatislessthan0.01.Thusthenullhypothesisp0.11(thesurveyisunbiased)canberejectedatthe1%level.3.9.DenotethelifeofalightbulbfromthenewprocessbyY.ThemeanofYisandthestandarddeviationofYisY200hours.Yisthesamplemeanwithasamplesizen100.The200standarddeviationofthesamplingdistributionofYisY20hours.TheYn100hypothesistestisH0:2000vs.H12000.ThemanagerwillacceptthealternativehypothesisifY2100hours.(a)Thesizeofatestistheprobabilityoferroneouslyrejectinganullhypothesiswhenitisvalid.Thesizeofthemanager’stestissizePr(YY2100|2000)1Pr(2100|2000)Y2000210020001Pr|200020201(5)10999999713287107,©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises11wherePr(Y2100|2000)meanstheprobabilitythatthesamplemeanisgreaterthan2100hourswhenthenewprocesshasameanof2000hours.(b)Thepowerofatestistheprobabilityofcorrectlyrejectinganullhypothesiswhenitisinvalid.Wecalculatefirsttheprobabilityofthemanagererroneouslyacceptingthenullhypothesiswhenitisinvalid:Y215021002150Pr(Y2100|2150)Pr|21502020(25)1(25)10993800062Thepowerofthemanager’stestingis110006209938.(c)Foratestwith5%,therejectionregionforthenullhypothesiscontainsthosevaluesofthet-statisticexceeding1.645.actactY2000actt1645Y20001645202032920Themanagershouldbelievetheinventor’sclaimifthesamplemeanlifeofthenewproductisgreaterthan2032.9hoursifshewantsthesizeofthetesttobe5%.3.11.Assumethatnisanevennumber.ThenYisconstructedbyapplyingaweightof1/2tothen/2“odd”observationsandaweightof3/2totheremainingn/2observations.11313EY()EY()12EY()EY(nn1)EY()n222211nn3YYYn222211919var(YYYYY)2var(12)var()var(nn1)var()n4444211nn229Y2YY125nn42423.13(a)Samplesizen420,sampleaverageY646.2samplestandarddeviationsY195.Thes19.5standarderrorofYisSE(Y)Y09515.The95%confidenceintervalforthen420meantestscoreinthepopulationisYY196SE()646219609515(6443464806)(b)Thedataare:samplesizeforsmallclassesn1238,sampleaverageY16574,samplestandarddeviations194;samplesizeforlargeclassesn182,sampleaverage12Y26500,samplestandarddeviations2179.ThestandarderrorofYY12is2222ss1219.417.9SE(Y12Y)18281.Thehypothesistestsforhigheraveragenn12238182scoresinsmallerclassesis©2011PearsonEducation,Inc.PublishingasAddisonWesley12Stock/Watson•IntroductiontoEconometrics,ThirdEditionHH0120vs1120Thet-statisticis65746500tactYY1240479SE(YY12)18281Thep-valuefortheone-sidedtestis:pt-value1(act)1(40479)1099997414725853105Withthesmallp-value,thenullhypothesiscanberejectedwithahighdegreeofconfidence.Thereisstatisticallysignificantevidencethatthedistrictswithsmallerclasseshavehigheraveragetestscores.3.15.Fromthetextbookequation(2.46),weknowthatE(Y)Yandfrom(2.47)weknowthat2Yvar(Y).Inthisproblem,becauseYaandYbareBernoullirandomvariables,pˆY,pˆnaab22Yb,Yapa(1–pa)andYbpb(1–pb).Theanswersto(a)followfromthis.Forpart(b),notethatvar(pˆa–pˆb)var(pˆa)var(pˆb)–2cov(pˆa,pˆb).But,theyareindependent(andthushavecov(ppˆˆab,)0becausepˆaandpˆbareindependent(theydependondatachosenfromindependentsamples).Thusvar(pˆa–pˆb)var(pˆa)var(pˆb).Forpart(c),useequation3.21fromthetext(replacingYwithpˆandusingtheresultin(b)tocomputetheSE).For(d),applytheformulain(c)toobtain0.859(10.859)0.374(10.374)95%CIis(.859–.374)±1.96or0.485±0.017.580142493.17.(a)The95%confidenceintervalisYYmm,2008,19921.96SE(YYmm,2008,1992)where2222ssmm,2008,199211.7810.17SE(YYmm,2008,1992)0.37;the95%confidencennmm,2008,199218381594intervalis(24.9823.27)±0.73or1.71±0.73.(b)The95%confidenceintervalisYYww,2008,19921.96SE(YYww,2008,1992)where2222ssww,2008,19929.667.78SE(YYww,2008,1992)0.31;the95%confidenceintervalnnww,2008,199218711368is(20.8720.05)0.60or0.820.60.(c)The95%confidenceintervalis(YYmm,2004,1992)(YYww,2004,1992)1.96SE[(YYmm,2008,1992)()],YYww,2008,1992whereSE[(YYmm,2008,1992)(YYww,2008,1992)]ssss222211.78210.172229.667.78mmww,2008,1992,2008,19920.48.The95%nnnnmmww,2008,1993,2008,19921838159418711368confidenceintervalis(24.98-23.27)−(20.87−20.05)±1.960.48or0.89±0.95.22223.19.(a)No.E()YEYYijiYYand(ijY)for.Thus©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises13211nn1n1EY()2222EYEY()EYY()ii22ijYYnnii11niji1n(b)Yes.IfYgetsarbitrarilyclosetoYwithprobabilityapproaching1asngetslarge,then22YgetsarbitrarilyclosetoYwithprobabilityapproaching1asngetslarge.(Asitturnsout,thisisanexampleofthe“continuousmappingtheorem”discussedinChapter17.)3.21.Setnmnwn,anduseequation(3.19)writethesquaredSEofYYmwas11nn()YY22()YY(1)nnimim11(1)iwiw[(SEYY)]2mwnnnn()()YY22YYimim11iwiw.nn(1)Similarly,usingequation(3.23)11nn()YY22()YY2(nn1)imim11(1)iwiw[()]SEYY2pooledmw2nnn()()YY22YYimim11iwiw.nn(1)©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter4LinearRegressionwithOneRegressor4.1.(a)ThepredictedaveragetestscoreisTestScore52045822239236(b)ThepredictedchangeintheclassroomaveragetestscoreisTestScore(58219)(58223)2328(c)UsingtheformulaforˆinEquation(4.8),weknowthesampleaverageofthetestscores0acrossthe100classroomsisTestScoreˆˆCS52045822143958501(d)Usetheformulaforthestandarderroroftheregression(SER)inEquation(4.19)togetthesumofsquaredresiduals:SSR(n2)SER22(1002)11512961UsetheformulaforR2inEquation(4.16)togetthetotalsumofsquares:SSR12961TSS1304411008R222TSS130442ThesamplevarianceissYn1991318.Thus,standarddeviationisssYY115.4.3.(a)Thecoefficient9.6showsthemarginaleffectofAgeonAWE;thatis,AWEisexpectedtoincreaseby$9.6foreachadditionalyearofage.696.7istheinterceptoftheregressionline.Itdeterminestheoverallleveloftheline.(b)SERisinthesameunitsasthedependentvariable(Y,orAWEinthisexample).ThusSERismeasuredindollarsperweek.(c)R2isunitfree.(d)(i)696.79.625$936.7;(ii)696.79.645$1,128.7(e)No.Theoldestworkerinthesampleis65yearsold.99yearsisfaroutsidetherangeofthesampledata.(f)No.Thedistributionofearningispositivelyskewedandhaskurtosislargerthanthenormal.ˆˆˆˆ(g)01YX,sothatYX01.ThusthesamplemeanofAWEis696.79.641.6$1,096.06.©2011PearsonEducation,Inc.PublishingasAddisonWesleySolutionstoOdd-NumberedEnd-of-ChapterExercises154.5.(a)uirepresentsfactorsotherthantimethatinfluencethestudent’sperformanceontheexamincludingamountoftimestudying,aptitudeforthematerial,andsoforth.Somestudentswillhavestudiedmorethanaverage,otherless;somestudentswillhavehigherthanaverageaptitudeforthesubject,otherslower,andsoforth.(b)BecauseofrandomassignmentuiisindependentofXi.SinceuirepresentsdeviationsfromaverageE(ui)0.BecauseuandXareindependentE(ui|Xi)E(ui)0.(c)(2)issatisfiedifthisyear’sclassistypicalofotherclasses,thatis,studentsinthisyear’sclasscanbeviewedasrandomdrawsfromthepopulationofstudentsthatenrollintheclass.(3)issatisfiedbecause0Yi100andXicantakeononlytwovalues(90and120).(d)(i)490.249070.6;490.2412077.8;490.2415085.0(ii)0.24102.4.ˆ4.7.Theexpectationof0isobtainedbytakingexpectationsofbothsidesofEquation(4.8):nˆˆ1ˆE()01011EY(X)EXuiXni1nˆ1011EXEu()()ini10ˆwherethethirdequalityintheaboveequationhasusedthefactsthatE(ui)0andE[(1−1)X]ˆˆE[(E(1−1)|X)X]becauseEX[(11)|]0(seetextequation(4.31).)ˆˆˆˆ24.9.(a)With100,Y,andYYi0.ThusESS0andR0.2ˆˆˆˆˆ(b)IfR0,thenESS0,sothatYYiforalli.ButYXii01,sothatYYiforalli,ˆwhichimpliesthat10,orthatXiisconstantforalli.IfXiisconstantforalli,thenn()0XX2andˆisundefined(seeequation(4.7)).i1i1n4.11.(a)Theleastsquaresobjectivefunctionis().YbX2Differentiatingwithrespecttobi1ii11n2()YbXii1nyieldsi12(XYbX).Settingthiszero,andsolvingfortheleasti1ii1ib1nXiiYˆi1squaresestimatoryields1n.2Xii1nXYii(4)ˆi1(b)Followingthesamestepsin(a)yields1n.2Xii1©2011PearsonEducation,Inc.PublishingasAddisonWesley16Stock/Watson•IntroductiontoEconometrics,ThirdEdition4.13.TheanswerfollowsthederivationsinAppendix4.3in“Large-SampleNormalDistributionoftheOLSEstimator.”Inparticular,theexpressionforiisnowi(XiX)ui,sothatvar(i)32var[(XiX)ui],andthetermcarrythroughtherestofthecalculations.©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter5RegressionwithaSingleRegressor:HypothesisTestsandConfidenceIntervals5.1(a)The95%confidenceintervalfor1is{582196221},thatis10152114884.(b)Calculatethet-statistic:ˆ0582tact126335SE(ˆ)2211Thep-valueforthetestH010vs.H110ispt-value2(|act|)2(26335)20004200084Thep-valueislessthan0.01,sowecanrejectthenullhypothesisatthe5%significancelevel,andalsoatthe1%significancelevel.(c)Thet-statisticisˆ(5.6)022tact10.10SE(ˆ)2211Thep-valueforthetestH01:5.6vs.H11:5.6ispt-value2(|act|)2(0.10)0.92Thep-valueislargerthan0.10,sowecannotrejectthenullhypothesisatthe10%,5%or1%significancelevel.Because15.6isnotrejectedatthe5%level,thisvalueiscontainedinthe95%confidenceinterval.(d)The99%confidenceintervalfor0is{520.42.5820.4},thatis,467.70573.0.5.3.The99%confidenceintervalis1.5{3.942.580.31)or4.71lbsWeightGain7.11lbs.5.5(a)Theestimatedgainfrombeinginasmallclassis13.9points.Thisisequaltoapproximately1/5ofthestandarddeviationintestscores,amoderateincrease.13.9(b)Thet-statisticistact5.56,whichhasap-valueof0.00.Thusthenullhypothesisis2.5rejectedatthe5%(and1%)level.(c)13.92.582.513.96.45.3.2