CFA一级百题进阶题：数量

2.Quantitative

2.1.进阶题

Q-1.

Thetablebelowshowsthreemutuallyexclusive$2,000,000mortgagechoices.Eachof

thethreechoicesiscompoundedmonthly.

Quotedannualinterestrateat

Mortgagetype

initiation

32-yearfixedrate

24-yearfixedrate

6.5%

6.0%

4.5%

32-yearadjustablerate

Theadjustable-ratemortgagewillresetitsinterestrateto6.2%attheendoftheyear4.

Afterresettingtheinterestrateattheendofyear4,whichmortgagewillhavethe

largestmonthlypayment?

A.32-yearfixedratemortgage.

B.24-yearfixed-ratemortgage.

C.32-yearadjustable-ratemortgage.

Solution:B.

Afteryear4,the24-yearfixed-ratemortgagehasthelargestpayment.

Theloanpaymentsaresummarizedinthetablebelow.

Mortgagetype

32-yearfixed

InitialPayment($)

12,389.92

Paymentafteradjustment($)

12,389.92

24-yearfixed

13,119.56

13,119.56

32-yearadjustable

9,836.93

11,785.90

Paymentonthe32-yearfixediscalculatedas:

N=12×32=384,I/Y=6.5/12,PV=-2,000,000,FV=0;CPTPMT=12,389.92

Paymentonthe24-yearfixediscalculatedas:

N=12×24=288,I/Y=6/12,PV=-2,000,000,FV=0;CPTPMT=13,119.56

Paymentonthe32-yearadjustableiscalculatedas:

Initialpayment

N=12×32=384,I/Y=4.5/12,PV=-2,000,000;FV=0;CPTPMT=9,836.93

Balanceatendofyear4:

N=12×28=336,I/Y=4.5/12,FV=0,PMT=9,836.93;CPTPV=-1,877,349.82

Paymentaftertheendofyear4:

N=336,I/Y=6.2/12,PV=-1,877,349.82;FV=0;CPTPMT=11,785.90

1-16

Q-2.

Whenrollingtwosix-sideddiceandsummingtheiroutcomes,whichofthefollowing

sumsismostlikelytooccur?

A.Nine

B.Six

C.Five

Solution:B.

Thisscenarioprovidesanexampleofadiscreterandomvariable.Thepairedoutcomesforthe

diceareindicatedinthefollowingtable.Theoutcomeofthedicesummingtosixisthemost

likelytooccurofthethreechoicesbecauseitcanoccurinfivedifferentways,whereasthe

summationtofiveandninecanoccurinonlyfourdifferentways.

SummedOutcomePairedOutcomes(Die1,Die2)PossibleCombinations

5(1,4),(2,3),(3,2),and(4,1)4

6(1,5),(2,4),(3,3),(4,2),and(5,1)5

9(3,6),(4,5),(5,4),and(6,3)4

2-16

Q-3.

Independentsamplesdrawnfromnormallydistributedpopulationsexhibitthe

followingcharacteristics:

Sample

Size

28

SampleMean

SampleStandardDeviation

A

B

210

195

50

65

21

Assumingthatthevariancesoftheunderlyingpopulationsareequal,thepooled

estimateofthecommonvarianceis3,377.13.Thet-teststatisticappropriatetotestthe

hypothesisthatthetwopopulationmeansareequalisclosestto:

A.1.80.

B.0.31.

C.0.89.

Solution:C.

Thet-statisticforthegiveninformation(normallydistributedpopulations,populationvariances

assumedequal)iscalculatedas:

(210−195)−0

t=

=0.89

3377.133377.13

(

+

)0.5

28

21

3-16

Q-4.

Twodistributionshavethesamemean.Oneisnegativelyskew,theotherispositive

skew.Whichonehasthelargermedian?

A.Distributionwithnegativeskew

B.Distributionwithpositiveskew

C.Thesame

Solution:A.

Asshowninthefollowingfigure,themedianissmallerthanthemeanforthepositiveskew.In

contrast,themedianislargerthanthemeanforthenegativeskew.

Mode<Median<Mean

Positive(right)skew

Mean<Median<Mode

Negative(left)skew

Therefore,ifthetwomeansequal,themedianofthenegativeskewislargerthanthatofpositive

skew.

4-16

Q-5.

Population

1

2

Samplesize

n=6

1

n=6

2

2

2

Samplevariance

S=5

S=30

1

2

Thesamplesaredrawnindependently,andbothpopulations

areassumedtobenormallydistributed

Usingtheabovedata,ananalystistryingtotestthenullhypothesisthatthepopulation

variancesareequal(H:흈ퟐ=흈ퟐ)againstthealternativehypothesisthatthevariances

0

ퟏ

ퟐ

arenotequal(H:흈ퟐ≠흈ퟐ)atthe5%levelofsignificance.Thetableofthe

a

ퟏ

ퟐ

F-Distributionisprovidedbelow.

TableoftheF-Distribution

PanelA:Criticalvaluesforright-handtailareasequalto0.05

df1(readacross)

1

2

3

4

5

df2

(readdown)

1

2

3

4

5

161

200

216

225

230

18.5

10.1

7.71

6.61

19.0

9.55

6.94

5.79

19.2

9.28

9.59

5.41

19.2

9.12

6.39

5.19

19.3

9.01

6.26

5.05

PanelB:Criticalvaluesforright-handtailareasequalto0.025

df1(readacross)

1

2

3

4

5

df2

(readdown)

1

2

3

4

5

648

799

864

900

922

38.51

17.44

12.22

10.01

39.00

16.04

10.65

8.43

39.17

15.44

9.98

7.76

39.25

15.10

9.60

7.39

39.30

14.88

9.36

7.15

Whichofthefollowingstatementsismostappropriate?Thecriticalvalueis:

A.9.36andrejectthenull.

B.9.60anddonotrejectthenull.

C.7.15anddonotrejectthenull.

Solution:C.

Identifytheappropriateteststatisticandinterprettheresultsforahypothesistestconcerning1)

thevarianceofanormallydistributedpopulation,and2)theequalityofthevariancesoftwo

normallydistributedpopulationsbasedontwoindependentrandomsamples.

5-16

numerator.Here,theteststatisticis30÷5=6.Thedegreesoffreedomare5by5.Becauseitisa

two-tailedtest,thecorrectcriticalvalueatα=5%is7.15.Andbecausetheteststatisticisless

thanthecriticalvalue,wecannotrejectthenullhypothesis.

6-16

Q-6.

Usingthefollowingsampleresultsdrawnas25pairedobservationsfromtheir

underlyingdistributions,testwhetherthemeanreturnsofthetwoportfoliosdiffer

fromeachotheratthe1%levelofstatisticalsignificance.Assumetheunderlying

distributionsofreturnsforeachportfolioarenormalandthattheirpopulation

variancesarenotknown.

Portfolio1

15.00

15.50

Portfolio2

20.25

Difference

5.25

Meanreturn

Standarddeviation

15.75

6.25

t-statisticfor24degreesoffreedomandatthe1%levelofstatisticalsignificance=

1.711

Nullhypothesis(H):Meandifferenceofreturns=0

0

Basedonthepairedcomparisonstestofthetwoportfolios,themostappropriate

conclusionisthatH0shouldbe:

A.acceptedbecausethecomputedteststatisticexceeds1.711.

B.rejectedbecausethecomputedteststatisticexceeds1.711.

C.acceptedbecausethecomputedteststatisticislessthan1.711.

Solution:B.

푑−휇̅푑0

Theteststatisticis:

wheredisthemeandifference,휇̅isthehypothesizeddifference

푑

0

푆푑/√푛

inthemeans,sisthesamplestandarddeviationofdifferences,andnisthesamplesize.Inthis

d

(5.25−0)

case,theteststatisticequals:

=4.20.Because4.20>1.711,thenull

⁄(6.25⁄√25)

hypothesisthatthemeandifferenceiszeroisrejected.

7-16

Q-7.

Ifthepopulationdistributionisunknown,themethodthatwillleadtotheleastreliable

estimationofaparameteristo:

A.usepointestimatesinsteadofconfidenceintervalestimates.

B.uset-distributioninsteadofstandardnormaldistributiontoestablishconfidenceintervals

C.drawmoresamples

Solution:A.

Pointestimatesarelessreliablethanconfidenceintervalestimates.

Usingthet-distributionratherthanthenormaldistributionisamoreconservativeapproachto

constructconfidenceintervals,andthusincreasethereliabilityoftheconfidenceinterval.

Increasingthesamplesizecanalsoincreasethereliabilityoftheconfidenceinterval.

8-16

Q-8.

Thetablebelowreportstheannualreturnsfortwoactiveportfoliosinthesame

industry,namely,theirreturnsaredependentwitheachother.

Year

2013

2014

2015

2016

2017

2018

PortfolioA(%)

PortfolioB(%)

11

-10

1

9

4

-3

12

23

-4

8

21

2

Ifwewanttotestwhetherthetwoportfolioshavethesamemeanreturnata5%

significancelevel,theteststatisticsweshalluseisclosestto:

A.1.96.

B.1.66.

C.0.45.

Solution:C.

First,calculatethereturndifferenceeachyear:

Year

2013

2014

2015

2016

2017

2018

PortfolioA(%)

PortfolioB(%)

Differences(%)

11

-10

1

9

4

-2

14

-4

4

-3

12

23

-4

8

21

2

2

-6

1

Andcalculatethemeandifferenceofreturnsusingafinancialcalculator:d

d1.33%

i

n

Then,calculatethesamplestandarddeviationandthestandarderrorofthemeandifference

usingafinancialcalculator:

(dd)

2

Sd

i

=7.23%

n1

Sd

n

7.23%

6

S

2.95%

d

d0

Finally,calculatethet-statistic:t

0.45

Sd

9-16

Q-9.

Inaheadandshoulderspattern,ifthenecklineisat$23,theshouldersat$28,andthe

headat$33.Thepricetargetisclosesttowhichofthefollowing:

A.$13.

B.$19.

C.$40.

Solution:A.

Headandshoulderspattern:Pricetarget=neckline–(head–neckline)=23–(33–23)=13.

10-16

Q-10.

Ananalysthasestablishedthefollowingpriorprobabilitiesregardingacompany'snext

quarter'searningspershare(EPS)exceeding,equaling,orbeingbelowtheconsensus

estimate.

PriorPrababilities

EPSexceedconsensus

EPSequalconsensus

23%

56%

21%

EPSarelessthanconsensus

Severaldaysbeforereleasingitsearningsstatement,thecompanyannouncesacutin

itsdividend.Giventhisnewinformation,theanalystreviseshisopinionregardingthe

likelihoodthatthecompanywillhaveEPSbelowtheconsensusestimate.Heestimates

thelikelihoodthecompanywillcutthedividend,giventhatEPSexceeds/meets/falls

belowconsensus,asreportedbelow.

ProbabilitiestheCompanyCutsDividends,Conditional

onEPSExceeding/Equaling/Fallingbelowconsensus

P(Cutdiv/EPSexceed)

P(Cutdiv/EPSequal)

P(Cutdiv/EPSbelow)

3%

11%

86%

UsingBayes'formula,theupdated(posterior)probabilitythatthecompany'sEPSare

belowtheconsensusisclosestto:

A.73%.

B.84%.

C.22%.

Solution:A.

Bayes'formula:P(A/B)=[P(B/A)P(A)]/P(B)

Updatedprobabilityofeventgiventhenewinformation:

where

Updatedprobabilityofeventgiventhenewinformation:P(EPSbelow│Cutdiv);

Probabilityofthenewinformationgivenevent:P(Cutdiv│EPSbelow)=86%;

Unconditionalprobabilityofthenewinformation:P(Cutdiv)=P(Cutdiv/EPSexceed)P(EPS

exceed)+P(Cutdiv/EPSequal)P(EPSequal)+P(Cutdiv/EPSbelow)P(EPS

below)=23%*3%+56%*11%+21%*86%=0.69%+6.16%+18.06%=24.91%;

Priorprobabilityofevent:P(EPSbelow)=21%.

Therefore,theprobabilityofEPSfallingbelowtheconsensusisupdatedas:

P(EPSbelow│Cutdiv)=[P(Cutdiv│EPSbelow)/P(Cutdiv)]×P(EPSbelow)

=(0.86/0.2491)×0.21≈73%

11-16

Q-11.

Samplesofsize(n,n)aredrawnrespectivelyfromtwopopulations(X,X)with

1212

associatedsamplemeansandstandarddeviationsof(퐗̅,퐗̅)and(S,S)and

ퟏ

ퟐ

1

2

associatedpopulationmeansandstandarddeviationsof(μ,μ)and(σ,σ)where

1

2

1

2

(σ≠σ).Inaddition,퐝̅isthesamplemeanof퐗̅,퐗̅withastandarderrorof퐒퐝̅

1

2

ퟏ

ퟐ

andapopulationmeanof훍풅and퐒ퟐisapooledestimatorofthecommonvariance.

퐩

Themostappropriateteststatistictodeterminetheequalityofthetwopopulation

meansassumingXandXareindependentandnormallydistributedis:

1

2

푑̅−μ푑0

A.t=

B.t=

푆푑̅

(푋̅−푋̅)−(μ−μ)

1

2

1

2

2

2

푆

푝

푆

푝

(

+

)0.5

푛1푛2

(푋̅−푋̅)−(μ−μ)

C.t=

1

2

1

2

2

2

푆

2

푆

1

)0.5

(

+

푛1푛2

Solution:C.

Themostappropriateteststatisticforthedifferencebetweentwopopulationmeans(unequal

(푋̅−푋̅)−(μ−μ)

andunknownpopulationvariances)ist=

1

2

1

2

.

2

2

푆

2

푆

1

)0.5

(

+

푛1푛2

12-16

Q-12.

MonteCarlosimulationisbestdescribedas:

A.arestrictiveformofscenarioanalysis.

B.providingadistributionofpossiblesolutionstocomplexfunctions.

C.anapproachtobacktestdata.

Solution:B.

MonteCarlosimulationprovidesadistributionofpossiblesolutionstocomplexfunctions.The

centraltendencyandthevarianceofthedistributionofsolutionsgiveimportantcluestodecision

makersregardingexpectedresultsandrisk.

13-16

Q-13.

Whichofthefollowingmostaccuratelydescribe