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1、2.Quantitative2.1.进阶题Q-l.Thetablebelowshowsthreemutuallyexclusive$2,000,000mortgagechoices.Eachofthethreechoicesiscompoundedmonthly.MortgagetypeQuotedannualinterestrateatinitiation32-yearfixedrate6.5%24-yearfixedrate6.0%32-yearadjustablerate4.5%Theadjustable-ratemortgagewillresetitsinterestrateto6.2
2、%attheendoftheyear4.Afterresettingtheinterestrateattheendofyear4,whichmortgagewillhavethelargestmonthlypayment?A.32-yearfixedratemortgage.B.24-yearfixed-ratemortgage.C.32-yearadjustable-ratemortgage.Solution:B.Afteryear4,the24-yearfixed-ratemortgagehasthelargestpayment.Theloanpaymentsaresummarizedin
3、thetablebelow.MortgagetypeInitialPayment($)Paymentafteradjustment($)32-yearfixed12z3S9.9212,389.9224-yearfixed13,119.5613,119.5632-yearadjustable9,836.9311,785.90Paymentonthe32-yearfixediscalculatedas:N=1232=384zl/Y=6.5/12,PV=-2,000,000,FV=0;CPTPMT=12,389.92Paymentonthe24-yearfixediscalculatedas:N=1
4、224=288,l/Y=6/12,PV=-2,0f000zFV=O;CPTPMT=13,119.56Paymentonthe32-yearadjustableiscalculatedas:InitialpaymentN=1232=384zl/Y=4.5/12,PV=2,000,000;FV=0;CPTPMT=9,836.93Balanceatendofyear4:N=1228=336,l/Y=4.5/12,FV=0,PMT=9,83693;CPTPV=1,877,349.82Paymentaftertheendofyear4:N=336,l/Y=6.2/12,PV=-1,877,349.82;
5、FV=0;CPTPMT=llz785.90Q-2.Whenrollingtwosix-sideddiceandsummingtheiroutcomes,whichofthefollowingsumsismostlikelytooccur?A.NineB.SixC.FiveSolution:B.Thisscenarioprovidesanexampleofadiscreterandomvariable.Thepairedoutcomesforthediceareindicatedinthefollowingtable.Theoutcomeofthedicesummingtosixisthemos
6、tlikelytooccurofthethreechoicesbecauseitcanoccurinfivedifferentways,whereasthesummationtofiveandninecanoccurinonlyfourdifferentways.SummedOutcomePairedOutcomes(Die1,Die2)PossibleCombinations5(1,4)/2,3),E2),and1)46(1,5),(2,4),(3,3),(4,2),and(5,1)59(3,6),(4,5),(5,4),and(6,3)4Q-3.Independentsamplesdraw
7、nfromnormallydistributedpopulationsexhibitthefollowingcharacteristics:SampleSizeSampleMeanSampleStandardDeviationA2821050B2119565Assumingthatthevariancesoftheunderlyingpopulationsareequal,thepooledestimateofthecommonvarianceis3,377.13.Thet-teststatisticappropriatetotestthehypothesisthatthetwopopulat
8、ionmeansareequalisclosestto:A.1.80.B.0.31.C0.89.Solution:C.Thet-statisticforthegiveninformation(normallydistributedpopulations,populationvariances0.89assumedequal)iscalculatedas:(210-195)-0t=3377.133377.13)0.5二Q-4.Twodistributionshavethesamemean.Oneisnegativelyskew,theotherispositiveskew.Whichonehas
9、thelargermedian?A.DistributionwithnegativeskewB.DistributionwithpositiveskewC.ThesameSolution:A.Asshowninthefollowingfigure,themedianissmallerthanthemeanforthepositiveskew.Incontrastthemedianislargerthanthemeanforthenegativeskew.ModeMedianMeanMeanMedianModePositive(right)skewNegative(left)skewTheref
10、ore,ifthetwomeansequal,themedianofthenegativeskewislargerthanthatofpositiveskew.Population12Samplesizen二6n三6SamplevarianceSi2=5&=30Q-5.Thesamplesaredrawnindependently,andbothpopulationsareassumedtobenormallydistributedUsingtheabovedata,ananalystistryingtotestthenullhypothesisthatthepopulationvarianc
11、esareequal(Ho：妾=底)againstthealternativehypothesisthatthevariancesarenotequal(Ha:娈娈)atthe5%levelofsignificance.ThetableoftheF-Distributionisprovidedbelow.TableoftheF-DistributionPanelA:Criticalvaluesforright-handtailareasequalto0.05dfl(readacross)df2(readdown)123451161200216225230218.519.019.219.219.
12、3310.19.559.289.129.0147.716.949.596,396,2656.615795,415195+05PanelB:Criticalvaluesforright-handtailareasequalto0.025dfl(readacross)df2(readdown)123451648799864900922238.5139.0039.1739.2539.30317.4416,0415.4415.1014.88412.2210,659.989.60936510.018.437.767.397.15Whichofthefollowingstatementsismostapp
13、ropriate?Thecriticalvalueis:A.B.C.9.36andrejectthenull.9.60anddonotrejectthenull.7.15anddonotrejectthenull.Solution:C.Identifytheappropriateteststatisticandinterprettheresultsforahypothesistestconcerning1)thevarianceofanormallydistributedpopulation,and2)theequalityofthevariancesoftwonormallydistribu
14、tedpopulationsbasedontwoindependentrandomsamples.numerator.Here,theteststatisticis305=6.Thedegreesoffreedomare5by5.Becauseitisatwo-tailedtest,thecorrectcriticalvalueat=5%is7.15.Andbecausetheteststatisticislessthanthecriticalvalue,wecannotrejectthenullhypothesis.Q-6.Usingthefollowingsampleresultsdraw
15、nas25pairedobservationsfromtheirunderlyingdistributions,testwhetherthemeanreturnsofthetwoportfoliosdifferfromeachotheratthe1%levelofstatisticalsignificance.Assumetheunderlyingdistributionsofreturnsforeachportfolioarenormalandthattheirpopulationvariancesarenotknown.Portfolio1Portfolio2DifferenceMeanreturn15.0020.255.25Standarddeviation15.5015.75625t-statisticfor24degre
