2025年CFA二级数量分析题库

2025年CFA二级数量分析题库考试时间：______分钟总分：______分姓名：______1.Aportfolioconsistsoftwoassets,AssetAandAssetB.AssetAhasanexpectedreturnof12%andastandarddeviationof15%.AssetBhasanexpectedreturnof8%andastandarddeviationof10%.ThecorrelationcoefficientbetweenthereturnsofAssetAandAssetBis0.4.Whatistheexpectedreturnofaportfolioinvested60%inAssetAand40%inAssetB?2.Thedailyreturnsofastockarenormallydistributedwithameanof0.1%andastandarddeviationof1.0%.Whatistheprobabilitythatthestock'sreturnwillbelessthan-1.5%inagivenday?3.Ananalystgathersthefollowingdatafortwostocks,StockXandStockY:StockX:Meanreturn=10%,Standarddeviation=20%StockY:Meanreturn=15%,Standarddeviation=25%ThecorrelationcoefficientbetweenStockXandStockYis0.6.Whichofthefollowingstatementsismostaccurate?A.StockYhashighervolatilitythanStockX.B.ThereturnsofStockXandStockYtendtomoveinthesamedirection.C.Aportfolioconsistingof50%StockXand50%StockYwillhaveastandarddeviationgreaterthanthestandarddeviationofStockY.4.Asimplelinearregressionanalysisyieldsthefollowingoutput:Y=5+2XR-squared=0.81Standarderroroftheestimate=3Samplesize(n)=30Whatisthe95%confidenceintervalfortheexpectedvalueofYwhenX=4?5.Consideraportfolioofthreeassetswiththefollowingweightsandexpectedreturns:Asset1:Weight=30%,ExpectedReturn=14%Asset2:Weight=50%,ExpectedReturn=12%Asset3:Weight=20%,ExpectedReturn=16%Thestandarddeviationsoftheassetsare15%,10%,and12%,respectively.Thecorrelationcoefficientsbetweentheassetsareasfollows:Correlation(1,2)=0.5,Correlation(1,3)=0.2,Correlation(2,3)=0.7Whatistheexpectedreturnoftheportfolio?6.Youaretestingthenullhypothesisthatthemeanreturnofastockis0%.Thealternativehypothesisisthatthemeanreturnisgreaterthan0%.Youcollectasampleof50daysandcalculatethesamplemeanreturntobe0.5%withasamplestandarddeviationof2.0%.Assumingthereturnsarenormallydistributed,whatisthep-valueforthistestatthe5%significancelevel?7.Asampleof100observationsisusedtoestimatethemeanofapopulation.Thesamplemeanis50,andthesamplestandarddeviationis5.Whatisthe95%confidenceintervalforthepopulationmean?8.Afinancialanalystusesasimplelinearregressionmodeltopredictthepriceofastockbasedonitsearningspershare(EPS).Themodelisfoundtobestatisticallysignificant(p-valuefortheinterceptis0.03,p-valuefortheslopeis0.01).Whichofthefollowingstatementsiscorrect?A.Themodelexplains100%ofthevariabilityinthestockprice.B.Earningspersharehaveastatisticallysignificantpositiverelationshipwiththestockprice.C.Theinterceptoftheregressionlineisnotstatisticallydifferentfromzero.9.Astockhasanexpectedreturnof18%andavolatility(standarddeviation)of30%.Therisk-freerateis5%.AccordingtotheCapitalAssetPricingModel(CAPM),whatisthestock'sbetaifthemarketexpectedreturnis12%?10.Aportfoliomanagerwantstoconstructaportfoliousingtwoassets,AssetPandAssetF.AssetPhasanexpectedreturnof10%andastandarddeviationof15%.AssetFhasanexpectedreturnof6%andastandarddeviationof10%.ThecorrelationcoefficientbetweenthereturnsofAssetPandAssetFis-0.2.WhatistheminimumpossiblestandarddeviationofaportfolioconsistingofAssetPandAssetF?11.Thefollowingarethereturnsoftwoassetsoverthepast5years:Year1:AssetX=10%,AssetY=5%Year2:AssetX=-5%,AssetY=-10%Year3:AssetX=15%,AssetY=10%Year4:AssetX=0%,AssetY=-2%Year5:AssetX=20%,AssetY=15%WhatisthecovariancebetweenthereturnsofAssetXandAssetY?12.Asampleof20observationsisusedtoestimatethepopulationvariance.Thesamplevarianceis25.Whatisthe90%confidenceintervalforthepopulationvariance?13.Youareconductingahypothesistestforthepopulationmean.ThenullhypothesisisH0:μ=100.Youtakeasampleof25observationsandcalculatethesamplemeantobe105withasamplestandarddeviationof15.Ifyousetthesignificancelevel(α)at0.05,whatisthecriticalvalueforatwo-tailedtest?14.Astockhasabetaof1.2.Themarketriskpremium(E(Rm)-Rf)is5%.Therisk-freerateis3%.AccordingtotheCapitalAssetPricingModel(CAPM),whatistheexpectedreturnofthestock?15.Aportfolioconsistsof100%investedinarisk-freeassetwithareturnof4%and100%investedinariskyassetwithanexpectedreturnof15%andastandarddeviationof20%.Whatistheexpectedreturnandstandarddeviationoftheportfolio?16.AsimplelinearregressionwasperformedtopredictYusingX.TheestimatedregressionequationisY=5+3X.AnewobservationhasX=10.TheanalystpredictsthevalueofYtobe35.TheactualvalueofYforthisobservationis40.Whatisthepredictionerrorforthisobservation?17.ThereturnsofAssetAarenormallydistributedwithameanof12%andastandarddeviationof18%.ThereturnsofAssetBarenormallydistributedwithameanof8%andastandarddeviationof14%.ThecorrelationcoefficientbetweenAssetAandAssetBreturnsis0.3.Whatisthevarianceofaportfolioinvested40%inAssetAand60%inAssetB?18.Aninvestorisconsideringaddinganewassettotheirportfolio.Theassethasanexpectedreturnof14%andastandarddeviationof25%.Thecorrelationcoefficientbetweentheasset'sreturnsandtheportfolio'sreturnsis0.6.Thecurrentportfoliohasanexpectedreturnof10%andastandarddeviationof15%.Whatistheexpectedreturnandstandarddeviationofthenewportfolioiftheassetisaddedwithaweightof20%?19.Aresearcheristestingwhetherthereisasignificantdifferenceinthemeanheightsoftwopopulations,Population1andPopulation2.Theytakeindependentsamplesfromeachpopulation:Sample1(Population1):n1=30,mean=170cm,standarddeviation=10cmSample2(Population2):n2=35,mean=168cm,standarddeviation=12cmAssumingequalpopulationvariances,whatistheteststatisticforthetwo-samplet-testatthe5%significancelevel?20.Astock'sreturnsarenormallydistributedwithameanof10%andastandarddeviationof20%.Whatistheprobabilitythatthestock'sreturnwillbebetween-10%and30%inagivenyear?试卷答案1.9.8%解析思路：E(Rp)=wA*E(RA)+wB*E(RB)=0.6*12%+0.4*8%=7.2%+3.2%=9.8%2.0.0455or4.55%解析思路：Z=(X-μ)/σ=(-1.5%-0.1%)/1.0%=-1.6.查标准正态分布表，P(Z<-1.6)=0.0548.由于正态分布对称，P(Z>1.6)=0.0548.P(X<-1.5%)=P(Z<-1.6)=0.0548.题目问的是"lessthan",所以概率为5.48%或0.0548.3.B解析思路：A.标准差衡量波动性，20%<25%，所以StockYhashighervolatility.该选项错误。B.相关系数ρ=0.6>0，说明两股票回报同向变动趋势。该选项正确。C.联合方差σp^2=wA^2*σA^2+wB^2*σB^2+2*wA*wB*ρ*σA*σB=0.5^2*0.2^2+0.5^2*0.25^2+2*0.5*0.5*0.6*0.2*0.25=0.01+0.0156+0.015=0.0406.σp=√0.0406≈0.2015.σY=0.25.0.2015<0.25,所以thestandarddeviationoftheportfoliowillbelessthanthestandarddeviationofStockY.该选项错误。4.[3.12,6.88]解析思路：预测值Y_pred=5+2*4=13.标准误差oftheestimate(sε)=3.样本量(n)=30.自由度(df)=n-2=28.查t分布表，t_(0.025,28)≈2.048.置信区间=Y_pred±t*(sε/√n)=13±2.048*(3/√30)≈13±2.048*0.5477≈13±1.12.置信区间约为[11.88,14.12].考虑到标准误差乘以t值后可能有小数，题目中给出的选项[3.12,6.88]并不符合计算结果。按照标准计算，结果应为[11.88,14.12].题目可能存在选项错误或计算设定偏差。若必须选择，需确认题目或选项来源。按标准计算，答案为[11.88,14.12].5.12.2%解析思路：E(Rp)=w1*E(R1)+w2*E(R2)+w3*E(R3)=0.3*14%+0.5*12%+0.2*16%=4.2%+6%+3.2%=13.4%.(注：此题未使用协方差或相关系数计算风险，仅计算预期收益。)6.0.0013or0.13%解析思路：H0:μ=0%,H1:μ>0%.样本量n=50.样本均值X̄=0.5%.样本标准差s=2.0%.计算检验统计量t=(X̄-μ0)/(s/√n)=(0.5%-0%)/(2.0%/√50)=0.005/(0.02/7.071)=0.005/0.002828≈1.77.自由度df=n-1=49.查t分布表，双边检验α=0.05，临界值为t_(0.025,49)≈2.0096.由于是单边检验（大于），p-value=P(T>1.77)<P(T>2.0096)=0.025.使用计算器或软件计算精确p-value，t=1.77,df=49,one-tailedp-value≈0.0384.题目要求的α=0.05，p-value(0.0384)<α(0.05)，拒绝H0.p-value约为0.0384或3.84%。若题目选项与此值接近，可能存在四舍五入或选项设置问题。按标准计算，p-value≈0.0384.7.[48.84,51.16]解析思路：样本量n=100.样本均值X̄=50.样本标准差s=5.估计总体标准误SE=s/√n=5/√100=0.5.查Z分布表，双边检验α=0.05，临界值Z_(0.025)=1.96.置信区间=X̄±Z*SE=50±1.96*0.5=50±0.98.置信区间为[49.02,50.98].题目中给出的选项[48.84,51.16]并不符合标准计算结果。按标准计算，答案为[49.02,50.98].8.B解析思路：A.R-squared表示模型解释的变异比例，题中未给出R-squared值，无法判断为100%。该选项错误。B.检验斜率系数的p-value(0.01)<α(通常为0.05)，拒绝原假设（斜率=0），说明存在显著关系。斜率系数为3，表示X每变动1单位，Y预计变动3单位。正斜率表示正向关系。该选项正确。C.检验截距系数的p-value(0.03)<α(0.05)，拒绝原假设（截距=0），说明截距在统计上显著不为零。该选项错误。9.1.2解析思路：根据CAPM公式:E(Ri)=Rf+βi*[E(Rm)-Rf].18%=5%+βi*(12%-5%).13%=βi*7%.βi=13%/7%=1.857≈1.86.10.0.1354or13.54%解析思路：已知组合权重wP=0.6,wF=0.4.计算组合预期收益E(Rp)=wP*E(RP)+wF*E(RF)=0.6*10%+0.4*6%=6%+2.4%=8.4%.计算组合方差σp^2=wP^2*σP^2+wF^2*σF^2+2*wP*wF*ρ*σP*σF=(0.6)^2*(15%)^2+(0.4)^2*(10%)^2+2*(0.6)*(0.4)*(-0.2)*(15%)*(10%)=0.36*0.0225+0.16*0.01+2*0.6*0.4*(-0.2)*0.15*0.1=0.0081+0.0016-0.0036=0.0061.组合标准差σp=√0.0061≈0.0781or7.81%.最小标准差通常发生在相关系数为-1时。计算相关系数为-1时的方差：σp_min^2=wP^2*σP^2+wF^2*σF^2-2*wP*wF*σP*σF=(0.6)^2*(15%)^2+(0.4)^2*(10%)^2-2*(0.6)*(0.4)*(15%)*(10%)=0.0081+0.0016-0.0096=0.0001.σp_min=√0.0001=0.01or1.0%.因此，最小标准差为1.0%.11.0.0184or1.84%解析思路：计算每对观测值的离差乘积之和：Σ[(X_i-X̄)(Y_i-Ȳ)]/(n-1)其中X̄=(10-5+15+0+20)/5=10,Ȳ=(5-10+10-2+15)/5=6.Σ[(X_i-10)(Y_i-6)]=(10-10)(5-6)+(-5-10)(-10-6)+(15-10)(10-6)+(0-10)(-2-6)+(20-10)(15-6)=0+15*16+5*4+(-10)*(-8)+10*9=0+240+20+80+90=430.Cov(X,Y)=430/(5-1)=430/4=107.5.标准答案0.0184=107.5/(20*14)=107.5/280=0.3839/20.5≈0.0186.计算结果与选项略有差异，可能源于原始数据或计算过程。按标准公式计算结果为107.5.12.[15.40,44.90]解析思路：样本量n=20.样本方差s^2=25.估计总体方差的标准误SE=s/√(n-1)=√25/√19=5/√19≈5/4.359=1.147.自由度df=n-1=19.查χ^2分布表：α/2=0.05/2=0.025,df=19,χ^2_(0.025,19)≈32.852.1-α/2=0.975,df=19,χ^2_(0.975,19)≈8.907.置信区间为[1/χ^2_(0.975,19),1/χ^2_(0.025,19)]=[1/8.907,1/32.852]≈[0.112,0.0304].将结果乘以样本方差s^2=25:置信区间=[25*0.112,25*0.0304]=[2.8,0.76].题目中给出的选项[15.40,44.90]并不符合标准计算结果。按标准计算，答案为[2.8,7.6].13.2.064解析思路：检验统计量t=(X̄-μ0)/(s/√n)=(105-100)/(15/√25)=5/(15/5)=5/3≈1.667.自由度df=n-1=25-1=24.查t分布表，双边检验α=0.05，df=24，临界值t_(0.025,24)≈2.064.(单边检验α=0.05，df=24，临界值t_(0.05,24)≈1.711).题目为双边检验，α=0.05，临界值约为±2.064.14.13.0%解析思路：根据CAPM公式:E(Ri)=Rf+βi*[E(Rm)-Rf].E(Ri)=3%+1.2*5%=3%+6%=9%.15.E(Rp)=8.0%,σp=4.0%解析思路：A.E(Rp)=wRf*Rf+wRisky*E(Risky)=1.0*4%+1.0*15%=4%+15%=19%.σp=0*0+1*20%=20%.(注：此题权重为100%在风险资产，100%在无风险资产，构成一个纯粹的杠杆投资组合，预期收益为风险资产收益，波动性为风险资产波动性，但题目可能意图是不同权重组合或简化场景。若按标准公式计算100%投资于风险资产)：E(Rp)=15%,σp=20%.若题目意图是不同权重，例如w=0.5,1-w=0.5,则E(Rp)=0.5*4%+0.5*15%=9.5%,σp=0.5*0*20%+0.5*20%=10%.题目未明确权重，无法确定唯一答案。按最直接理解，若权重为1:1（50%-50%），则E(Rp)=9.5%,σp=10%.若理解为题目描述有误，应为不同权重组合。假设题目本意是标准组合，则预期收益为19%，标准差为20%.但若必须从给定选项中选择，且选项给出了具体数值，需确认题目来源。按最常见组合理解，若权重非0/1，则E(Rp)=9.5%,σp=10%.若必须选择，需题目明确。此处按标准公式计算100%投资于风险资产，E(Rp)=15%,σp=20%.但题目选项给出E(Rp)=8.0%,σp=4.0%，与计算结果均不符。若题目描述权重为0.5,则E(Rp)=9.5%,σp=10%.16.5解析思路：预测值Y_pred=5+3*10=35.实际值Y_actual=40.预测误差=Y_actual-Y_pred=40-35=5.17.0.0269or2.69%解析思路：计算组合方差σp^2=wA^2*σA^2+wB^2*σB^2+2*wA*wB*ρ*σA*σB=(0.4)^2*(0.18)^2+(0.6)^2*(0.14)^2+2*(0.4)*(0.6)*(0.3)*(0.18)*(0.14)=0.16*0.0324+0.36*0.0196+2*0.4*0.6*0.3*0.0252=0.005184+0.007056+0.0017648=0.0140028.方差σp^2≈0.0140028.标准差σp=√0.0140028≈0.1183or11.83%.方差约为0.0140或1.40%.(注意：计算中可能存在微小差异来源，但标准结果接近0.014).题目选项2.69%对应的方差约为(0.0269)^2≈0.000723.标准计算结果为0.0140.若题目选项或计算过程有特定设定，需确认。按标准计算，方差约为0.0140.18.E(RnewPortfolio)=10.8%,σnewPortfolio≈15.81%解析思路：A.E(RnewPortfolio)=wcurrP*E(RcurrP)+wnew*A*E(new*A)=0.8*10%+0.2*14%=8%+2.8%=10.8%.B.计算新组合的方差：σnew^2=wcurrP^2*σcurrP^2+wnew*A^2*σnew*A^2+2*wcurrP*wnew*A*ρcurrA_newA*σcurrP*σnew*A.σcurrP^2=(15%)^2=0.0225.σnew*A^2=(25%)^2=0.0625.ρcurrA_newA=0.6.σcurrP*σnew*A=15%*25%=0.375.σnew^2=(0.8)^2*0.0225+(0.2)^2*0.0625+2*(0.8)*(0.2)*(0.6)*(0.375)=0.0144+0.0025+2*0.8*0.2*0.6*0.375=0.0144+0.0025+0.036=0.0534.σnew=√0.0534≈0.2312or23.12%.σnewPortfolio≈23.12%.19.1.747解析思路：计算两样本均值差的标准误：SE(μ1-μ2)=√[(s1^2/n1)+(s2^2/n2)]=√[(10^2/30)+(12^2/35)]=√[(1