计量经济学 斯托克 沃森 第三版 答案

ForStudentsSolutionstoOdd-NumberedEnd-of-ChapterExercises©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter2ReviewofProbability2.1.(a)ProbabilitydistributionfunctionforYOutcome(numberofheads)Y0Y1Y2Probability0.250.500.25(b)CumulativeprobabilitydistributionfunctionforYOutcome(numberofheads)Y00Y11Y2Y2Probability00.250.751.0(c)=()(00.25)(10.50)(20.25)1.00YEY.,.dFFqUsingKeyConcept2.3:22var()()[()],YEYEYand(|)iiuXsothat222var()()[()]1.50(1.00)0.50.YEYEY2.3.Forthetwonewrandomvariables36WXand207,VYwehave:(a)()(207)207()2070781454,()(36)36()3607072EVEYEYEWEXEX(b)222222var(36)636021756,var(207)(7)490171684084WXVYXY(c)cov(36,207)6(7)cov(,)4200843528WVXYXY3528corr(,)0442575684084WVWVWVSolutionstoOdd-NumberedEnd-of-ChapterExercises3©2011PearsonEducation,Inc.PublishingasAddisonWesley2.5.LetXdenotetemperatureinFandYdenotetemperatureinC.RecallthatY0whenX32andY100whenX212;thisimplies(100/180)(32)or17.78(5/9).YXYXUsingKeyConcept2.3,X70oFimpliesthat17.78(5/9)7021.11C,YandX7oFimplies(5/9)73.89C.Y2.7.Usingobviousnotation,;CMFthusCMFand2222cov(,).CMFMFThisimplies(a)4045$85,000Cperyear.(b)cov(,)corr(,)MFMFMF,sothatcov(,)corr(,).MFMFMFThuscov(,)MF12180.80172.80,wheretheunitsaresquaredthousandsofdollarsperyear.(c)2222cov(,),CMFMFsothat22212182172.80813.60,Cand813.6028.524Cthousanddollarsperyear.(d)FirstyouneedtolookupthecurrentEuro/dollarexchangerateintheWallStreetJournal,theFederalReservewebpage,orotherfinancialdataoutlet.Supposethatthisexchangerateise(saye0.80Eurosperdollar);each1dollaristhereforewitheEuros.ThemeanisthereforeeC(inunitsofthousandsofEurosperyear),andthestandarddeviationiseC(inunitsofthousandsofEurosperyear).Thecorrelationisunit-free,andisunchanged.ValueofY2.9.1422304065ProbabilityDistributionofX10.020.050.100.030.010.2150.170.150.050.020.010.40ValueofX80.020.030.150.100.090.39ProbabilitydistributionofY0.210.230.300.150.111.00(a)Theprobabilitydistributionisgiveninthetableabove.22222222()140.21220.23300.30400.15650.1130.15()140.21220.23300.30400.15650.111127.23var()()[()]218.2114.77YEYEYYEYEY4Stock/Watson•IntroductiontoEconometrics,ThirdEdition©2011PearsonEducation,Inc.PublishingasAddisonWesley(b)TheconditionalprobabilityofY|X8isgiveninthetablebelowValueofY14223040650.02/0.390.03/0.390.15/0.390.10/0.390.09/0.39222222(|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)39.21(|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)1778.7EYXEYX28var()1778.739.21241.6515.54YXY(c)()(1140.02)(122:0.05)(8650.09)171.7EXYcov(,)()()()171.75.3330.1511.0XYEXYEXEYcorr(,)cov(,)/()11.0/(2.6014.77)0.286XYXYXY2.11.(a)0.90(b)0.05(c)0.05(d)When210~,Ythen10,/10~.YF(e)2,YZwhere~(0,1),ZNthusPr(1)Pr(11)0.32.YZ2.13.(a)2222()Var()101;()Var()1000100.YWEYYEWW(b)YandWaresymmetricaround0,thusskewnessisequalto0;becausetheirmeaniszero,thismeansthatthethirdmomentiszero.(c)Thekurtosisofthenormalis3,so443()/YYEY;solvingyields4E()3;YasimilarcalculationyieldstheresultsforW.(d)First,conditionon0,Xsothat:SW2342.(|0)0;(|0)100,(|0)0,(|0)3100ESXESXESXESXSimilarly,234(|1)0;(|1)1,(|1)0,(|1)3.ESXESXESXESXFromthelawofiteratedexpectations2223334442()(|0)Pr(X0)(|1)Pr(1)0()(|0)Pr(X0)(|1)Pr(1)1000.0110.991.99()(|0)Pr(X0)(|1)Pr(1)0()(|0)Pr(X0)(|1)Pr(1)31000.013ESESXESXXESESXESXXESESXESXXESESXESXX10.99302.97SolutionstoOdd-NumberedEnd-of-ChapterExercises5©2011PearsonEducation,Inc.PublishingasAddisonWesley(e)()0,SESthus33()()0SESESfrompart(d).Thusskewness0.Similarly,222()()1.99,SSESESand44()()302.97.SESESThus,2kurtosis302.97/(1.99)76.52.15.(a)9.6101010.410Pr(9.610.4)Pr4/4/4/9.61010.410Pr4/4/YYnnnZnnwhereZ~N(0,1).Thus,(i)n20;9.61010.410PrPr(0.890.89)0.634/4/ZZnn(ii)n100;9.61010.410PrPr(2.002.00)0.9544/4/ZZnn(iii)n1000;9.61010.410PrPr(6.326.32)1.0004/4/ZZnn(b)10Pr(1010)Pr4/4/4/Pr.4/4/cYccYcnnnccZnnAsngetlarge4/cngetslarge,andtheprobabilityconvergesto1.(c)Thisfollowsfrom(b)andthedefinitionofconvergenceinprobabilitygiveninKeyConcept2.6.2.17.Y=0.4and20.40.60.24Y(a)(i)P(Y0.43)0.40.430.40.4PrPr0.61240.270.24/0.24/0.24/YYnnn(ii)P(Y0.37)0.40.370.40.4PrPr1.220.110.24/0.24/0.24/YYnnn(b)WeknowPr(1.96Z1.96)0.95,thuswewantntosatisfy0.410.400.411.9624/nand0.390.401.96.24/nSolvingtheseinequalitiesyieldsn9220.6Stock/Watson•IntroductiontoEconometrics,ThirdEdition©2011PearsonEducation,Inc.PublishingasAddisonWesley2.19.(a)11Pr()Pr(,)Pr(|)Pr()ljijiljiiiYyXxYyYyXxXx(b)111111()Pr()Pr(|)Pr()Pr(|)Pr()(|)Pr()iiikkljjjjiijjikljjijiliEYyYyyYyXxXxyYyXxXxEYXxXx(c)WhenXandYareindependent,Pr(,)Pr()Pr()ijijXxYyXxYyso1111[()()]()()Pr(,)()()Pr()Pr()XYXYlkiXjYijijlkiXjYijijEXYxyXxYyxyXxYy11()Pr()()Pr(()()000,0corr(,)0lkiXijYjijXYXYXYXYxXxyYyEXEYXY2.21.(a)3232222332233223323()[()()][22]()3()3()()3()()3()[()][()]()3()()2()EXEXXEXXXXXEXEXEXEXEXEXEXEXEXEXEXEXEX(b)43223432233223443223443224()[(33)()][3333]()4()()6()()4()()()()4[()][()]6[()][()]3[()]EXEXXXXEXXXXXXXEXEXEXEXEXEXEXEXEXEXEXEXEXEXSolutionstoOdd-NumberedEnd-of-ChapterExercises7©2011PearsonEducation,Inc.PublishingasAddisonWesley2.23.XandZaretwoindependentlydistributedstandardnormalrandomvariables,so220,1,0.XZXZXZ(a)BecauseoftheindependencebetweenXand,ZPr(|)Pr(),ZzXxZzand(|)()0.EZXEZThus2222(|)(|)(|)(|)0EYXEXZXEXXEZXXX(b)222()1,XXEXand22()()101YZEXZEX(c)33()()()().EXYEXZXEXEZXUsingthefactthattheoddmomentsofastandardnormalrandomvariableareallzero,wehave3()0.EXUsingtheindependencebetweenXand,Zwehave()0.ZXEZXThus3()()()0.EXYEXEZX(d)cov()[()()][(0)(1)]()()()0000corr(,)0XYXYXYXYXYEXYEXYEXYXEXYEXXY2.25.(a)12312311()()nninniiiaxaxaxaxaxaxxxxax(b)11221121211()()()()niinninnnniiiixyxyxyxyxxxyyyxy(c)1()niaaaaana(d)222222112222211111()(222)222nniiiiiiiiiinnnnniiiiiiiiiiiabxcyabxcyabxacybcxynabxcyabxacybcxy2.27(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,thedistributionofthesampleaverage(Y)isapproximately2,YYNwith22.YYnGivenapopulation100,Y2430,Ywehave(a)100,n2243043,100YYnand100101100Pr(101)Pr(1.525)09364043043YY(b)64,n224306719,64YYnand101100100103100Pr(101103)Pr067190671906719(36599)(12200)099990888801111YY(c)165,n224302606,165YYnand10098100Pr(98)1Pr(98)1Pr02606026061(39178)(39178)10000(roundedtofourdecimalplaces)YYY3.3.Denoteeachvoter’spreferenceby.Y1Yifthevoterpreferstheincumbentand0Yifthevoterprefersthechallenger.YisaBernoullirandomvariablewithprobabilityPr(1)YpandPr(0)1.YpFromthesolutiontoExercise3.2,Yhasmeanpandvariance(1).pp(a)215ˆ05375.400p(b)Theestimatedvarianceofpˆis4ˆˆ(1)0.5375(10.5375)ˆvar()6214810.400pppnThestandarderrorisSE12ˆˆ()(var())00249.ppSolutionstoOdd-NumberedEnd-of-ChapterExercises9©2011PearsonEducation,Inc.PublishingasAddisonWesley(c)Thecomputedt-statisticis0ˆ05375051506ˆSE()00249pactptpBecauseofthelargesamplesize(400),nwecanuseEquation(3.14)inthetexttogetthep-valueforthetest005Hpvs.105:Hp-value2(||)2(1506)200660132actpt(d)UsingEquation(3.17)inthetext,thep-valueforthetest005Hpvs.105Hpis-value1()1(1506)109340066actpt(e)Part(c)isatwo-sidedtestandthep-valueistheareainthetailsofthestandardnormaldistributionoutside(calculatedt-statistic).Part(d)isaone-sidedtestandthep-valueistheareaunderthestandardnormaldistributiontotherightofthecalculatedt-statistic.(f)Forthetest005Hpvs.105,Hpwecannotrejectthenullhypothesisatthe5%significancelevel.Thep-value0.066islargerthan0.05.Equivalentlythecalculatedt-statistic1506islessthanthecriticalvalue1.64foraone-sidedtestwitha5%significancelevel.Thetestsuggeststhatthesurveydidnotcontainstatisticallysignificantevidencethattheincumbentwasaheadofthechallengeratthetimeofthesurvey.3.5.(a)(i)ThesizeisgivenbyˆPr(|0.5|.02),pwheretheprobabilityiscomputedassumingthat0.5.pˆˆPr(|0.5|0.02)1Pr(0.020.5.02)ˆ0.020.50.021Pr0.50.5/10550.50.5/10550.50.5/1055ˆ0.51Pr1.301.300.50.5/10550.19ppppwherethefinalequalityusingthecentrallimittheoremapproximation.(ii)ThepowerisgivenbyˆPr(|0.5|0.02),pwheretheprobabilityiscomputedassumingthatp0.53.ˆˆPr(|0.5|0.02)1Pr(0.020.5.02)ˆ0.020.50.021Pr0.530.47/10550.530.47/10550.530.47/1055ˆ0.050.530.011Pr0.530.47/10550.530.47/10550.530.47/1055ˆ0.531Pr3.25ppppp0.65.530.47/10550.74wherethefinalequalityusingthecentrallimittheoremapproximation.10Stock/Watson•IntroductiontoEconometrics,ThirdEdition©2011PearsonEducation,Inc.PublishingasAddisonWesley(b)(i)0.540.502.61,andPr(||2.61)0.01,(0.540.46)/1055ttsothatthenullisrejectedatthe5%level.(ii)Pr(2.61).004,tsothatthenullisrejectedatthe5%level.(iii)0.541.96(0.540.46)/10550.540.03,or0.51to0.57.(iv)0.542.58(0.540.46)/10550.540.04,or0.50to0.58.(v)0.540.67(0.540.46)/10550.540.01,or0.53to0.55.(c)(i)Theprobabilityis0.95isanysinglesurvey,thereare20independentsurveys,sotheprobabilityif200.950.36(ii)95%ofthe20confidenceintervalsor19.(d)Therelevantequationisˆ1.96SE().01or1.96(1)/.01.pppnThusnmustbechosensothat221.96(1),0.01ppnsothattheanswerdependsonthevalueofp.Notethatthelargestvaluethatp(1−p)cantakeonis0.25(thatis,p0.5makesp(1p)aslargeaspossible).Thusif221.960.259604,0.01nthenthemarginoferrorislessthan0.01forallvaluesofp.3.7.Thenullhypothesisisthatthesurveyisarandomdrawfromapopulationwithp=0.11.Thet-statisticisˆ0.11,ˆSE()ptpwhereˆˆˆSE()(1)/.pppn(AnalternativeformulaforSE(pˆ)is0.11(10.11)/,nwhichisvalidunderthenullhypothesisthat0.11).pThevalueofthet-statisticis2.71,whichhasap-valueofthatislessthan0.01.Thusthenullhypothesis0.11p(thesurveyisunbiased)canberejectedatthe1%level.3.9.Denotethelifeofalightbulbfromthenewprocessby.YThemeanofYisandthestandarddeviationofYis200Yhours.Yisthesamplemeanwithasamplesize100.nThestandarddeviationofthesamplingdistributionofYis20020100YYnhours.Thehypothesistestis0:2000Hvs.12000.HThemanagerwillacceptthealternativehypothesisif2100Yhours.(a)Thesizeofatestistheprobabilityoferroneouslyrejectinganullhypothesiswhenitisvalid.Thesizeofthemanager’stestis7sizePr(2100|2000)1Pr(2100|2000)2000210020001Pr|200020201(5)1099999971328710,YYYSolutionstoOdd-NumberedEnd-of-ChapterExercises11©2011PearsonEducation,Inc.PublishingasAddisonWesleywherePr(2100|2000)Ymeanstheprobabilitythatthesamplemeanisgreaterthan2100hourswhenthenewprocesshasameanof2000hours.(b)Thepowerofatestistheprobabilityofcorrectlyrejectinganullhypothesiswhenitisinvalid.Wecalculatefirsttheprobabilityofthemanagererroneouslyacceptingthenullhypothesiswhenitisinvalid:215021002150Pr(2100|2150)Pr|21502020(25)1(25)10993800062YYThepowerofthemanager’stestingis110006209938.(c)Foratestwith5%,therejectionregionforthenullhypothesiscontainsthosevaluesofthet-statisticexceeding1.645.2000164520001645202032920actactactYtYThemanagershouldbelievetheinventor’sclaimifthesamplemeanlifeofthenewproductisgreaterthan2032.9hoursifshewantsthesizeofthetesttobe5%.3.11.Assumethatnisanevennumber.ThenYisconstructedbyapplyingaweightof1/2tothen/2“odd”observationsandaweightof3/2totheremainingn/2observations.1211212222211313()()()()()2222113222211919var()var()var()var()var()44441191254242nnYYYnnYYYEYEYEYEYEYnnnnYYYYYnnnnn3.13(a)Samplesize420,nsampleaverageY646.2samplestandarddeviation195.YsThestandarderrorofYisSE19.5()09515.420YsYnThe95%confidenceintervalforthemeantestscoreinthepopulationis196SE()646219609515(6443464806)YY(b)Thedataare:samplesizeforsmallclasses1238,nsampleaverage16574,Ysamplestandarddeviation1194;ssamplesizeforlargeclasses2182,nsampleaverage26500,Ysamplestandarddeviation2179.sThestandarderrorof12YYis222212121219.417.9()18281.238182ssSEYYnnThehypothesistestsforhigheraveragescoresinsmallerclassesis12Stock/Watson•IntroductiontoEconometrics,ThirdEdition©2011PearsonEducation,Inc.PublishingasAddisonWesley0121120vs0HHThet-statisticis12126574650040479SE()18281actYYtYYThep-valuefortheone-sidedtestis:5-value1()1(40479)109999741472585310actptWiththesmallp-value,thenullhypothesiscanberejectedwithahighdegreeofconfidence.Thereisstatisticallysignificantevidencethatthedistrictswithsmallerclasseshavehigheraveragetestscores.3.15.Fromthetextbookequation(2.46),weknowthatE(Y)Yandfrom(2.47)weknowthatvar(Y)2Yn.Inthisproblem,becauseYaandYbareBernoullirandomvariables,ˆapaY,ˆbpbY,2Yapa(1–pa)and2Ybpb(1–pb).Theanswersto(a)followfromthis.Forpart(b),notethatvar(ˆap–ˆbp)var(ˆap)var(ˆbp)–2cov(ˆap,ˆbp).But,theyareindependent(andthushaveˆˆcov(,)abpp0becauseˆapandˆbpareindependent(theydependondatachosenfromindependentsamples).Thusvar(ˆap–ˆbp)var(ˆap)var(ˆbp).Forpart(c),useequation3.21fromthetext(replacingYwithpˆandusingtheresultin(b)tocomputetheSE).For(d),applytheformulain(c)toobtain95%CIis(.859–.374)±1.960.859(10.859)0.374(10.374)58014249or0.485±0.017.3.17.(a)The95%confidenceintervalis,2008,1992,2008,19921.96SE()mmmmYYYYwhere2222,2008,1992,2008,1992,2008,199211.7810.17SE()0.37;18381594mmmmmmssYYnnthe95%confidenceintervalis(24.9823.27)±0.73or1.71±0.73.(b)The95%confidenceintervalis,2008,1992,2008,19921.96SE()wwwwYYYYwhere2222,2008,1992,2008,1992,2008,19929.667.78SE()0.31;18711368wwwwwwssYYnnthe95%confidenceintervalis(20.8720.05)0.60or0.820.60.(c)The95%confidenceintervalis,2004,1992,2004,1992,2008,1992()()1.96SE[()mmwwmmYYYYYY,2008,1992()],wwYYwhere,2008,1992,2008,1992SE[()()]mmwwYYYY2222,2008,1992,2008,1992,2008,1993,2008,1992mmwwmmwwssssnnnn222211.7810.179.667.7818381594187113680.48.The95%confidenceintervalis(24.98-23.27)−(20.87−20.05)±1.960.48or0.89±0.95.3.19.(a)No.2222()and()for.iYYijYEYEYYijThusSolutionstoOdd-NumberedEnd-of-ChapterExercises13©2011PearsonEducation,Inc.PublishingasAddisonWesley22222221111111()()()nnniiijYYiiijiEYEYEYEYYnnnn(b)Yes.IfYgetsarbitrarilyclosetoYwithprobabilityapproaching1asngetslarge,then2Ygetsarbitrarilycloseto2Ywithprobabilityapproaching1asngetslarge.(Asitturnsout,thisisanexampleofthe“continuousmappingtheorem”discussedinChapter17.)3.21.Setnmnwn,anduseequation(3.19)writethesquaredSEofmwYYas22112221111()()(1)(1)[()]()().(1)nnimimiwiwmwnnimimiwiwYYYYnnSEYYnnYYYYnnSimilarly,usingequation(3.23)22112221111()()2(1)(1)[()]2()().(1)nnimimiwiwpooledmwnnimimiwiwYYYYnnSEYYnYYYYnn©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter4LinearRegressionwithOneRegressor4.1.(a)Thepredictedaveragetestscoreis52045822239236TestScore(b)Thepredictedchangeintheclassroomaveragetestscoreis(58219)(58223)2328TestScore(c)Usingtheformulafor0ˆinEquation(4.8),weknowthesampleaverageofthetestscoresacrossthe100classroomsis01ˆˆ520458221439585TestScoreCS(d)Usetheformulaforthestandarderroroftheregression(SER)inEquation(4.19)togetthesumofsquaredresiduals:22(2)(1002)11512961SSRnSERUsetheformulafor2RinEquation(4.16)togetthetotalsumofsquares:22129611304411008SSRTSSRThesamplevarianceis2YsTSS130441991318.nThus,standarddeviationis2115.YYss4.3.(a)Thecoefficient9.6showsthemarginaleffectofAgeonAWE;thatis,AWEisexpectedtoincreaseby$9.6foreachadditionalyearofage.696.7istheinterceptoftheregressionline.Itdeterminestheoverallleveloftheline.(b)SERisinthesameunitsasthedependentvariable(Y,orAWEinthisexample).ThusSERismeasuredindollarsperweek.(c)R2isunitfree.(d)(i)696.79.625$936.7;(ii)696.79.645$1,128.7(e)No.Theoldestworkerinthesampleis65yearsold.99yearsisfaroutsidetherangeofthesampledata.(f)No.Thedistributionofearningispositivelyskewedandhaskurtosislargerthanthenormal.(g)01ˆˆ,YXsothat01ˆˆ.YXThusthesamplemeanofAWEis696.79.641.6$1,096.06.SolutionstoOdd-NumberedEnd-of-ChapterExercises15©2011PearsonEducation,Inc.PublishingasAddisonWesley4.5.(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)490.249070.6;490.2412077.8;490.2415085.0(ii)0.24102.4.4.7.Theexpectationof0ˆisobtainedbytakingexpectationsofbothsidesofEquation(4.8):010111011101ˆˆˆ()()1ˆ()()niiniiEEYXEXuXnEXEunwherethethirdequalityintheaboveequationhasusedthefactsthatE(ui)0andE[(1ˆ−1)X]E[(E(1ˆ−1)|X)X]because11ˆ[()|]0EX(seetextequation(4.31).)4.9.(a)With10ˆˆ0,,Yand0ˆˆ.iYYThusESS0andR20.(b)IfR20,thenESS0,sothatiˆYYforalli.But01ˆˆˆ,iiYXsothatiˆYYforalli,whichimpliesthat1ˆ0,orthatXiisconstantforalli.IfXiisconstantforalli,then21()0niiXXand1ˆisundefined(seeequation(4.7)).4.11.(a)Theleastsquaresobjectivefunctionis211().niiiYbXDifferentiatingwithrespecttob1yields211111()2().niiniiiiiYbXXYbXbSettingthiszero,andsolvingfortheleastsquaresestimatoryields1121ˆ.niiiniiXYX(b)Followingthesamestepsin(a)yields1121(4)ˆ.niiiniiXYX16Stock/Watson•IntroductiontoEconometrics,ThirdEdition©2011PearsonEducation,Inc.PublishingasAddisonWesley4.13.TheanswerfollowsthederivationsinAppendix4.3in“Large-SampleNormalDistributionoftheOLSEstimator.”Inparticular,theexpressionforiisnowi(XiX)ui,sothatvar(i)3var[(XiX)ui],andtheterm2carrythroughtherestofthecalculations.©2011PearsonEducation,Inc.PublishingasAddisonWesleyChapter5RegressionwithaSingleRegressor:HypothesisTestsandConfidenceIntervals5.1(a)The95%confidenceintervalfor1is{582196221},thatis11015214884.(b)Calculatethet-statistic:11ˆ058226335ˆSE()221acttThep-valueforthetest010Hvs.110His-value2(||)2(26335)20004200084actptThep-valueislessthan0.01,sowecanrejectthenullhypothesisatthe5%significancelevel,andalsoatthe1%significancelevel.(c)Thet-statisticis11ˆ(5.6)0