2025年CFA《數(shù)量》模擬測試卷考試時(shí)間：______分鐘總分：______分姓名：______試卷內(nèi)容1.Astatisticsprofessorconductedasurveytomeasurethestudyhoursoffinal-yearstudents.Whichofthefollowingbestdescribesthedatacollectedasameasureofcentraltendency?(a)Variance(b)Standarddeviation(c)Median(d)Correlationcoefficient2.Consideradatasetconsistingofthenumbers5,7,9,13,and15.Therangeofthisdatasetisclosestto:(a)4(b)8(c)12(d)283.Themeanreturnofanassetportfoliois12%withastandarddeviationof6%.Ifthereturnsarenormallydistributed,approximatelywhatpercentageofportfolioreturnsareexpectedtofallbetween6%and18%?(a)68%(b)75%(c)84%(d)95%4.WhichofthefollowingstatementsisTRUEregardingthecharacteristicsofabinomialdistribution?(a)Thenumberoftrials(n)mustbelargeforthedistributiontobeapproximatelynormal.(b)Theprobabilityofsuccess(p)mustbeequalto0.5forthedistributiontobesymmetric.(c)Eachtrialcanresultinoneoftwomutuallyexclusiveoutcomes.(d)Thetrialsmustbedependentoneachother.5.Afinancialanalystwantstoestimatetheaverageannualreturnofaparticularstockoverthelastfiveyears.Whichofthefollowingmethodsprovidesapointestimateforthisaverage?(a)Calculatingthestandarddeviationofthereturns.(b)Constructinga95%confidenceintervalforthereturns.(c)Takingthemedianofthereturns.(d)Performingahypothesistestaboutthereturns.6.Asampleof30observationsistakenfromapopulation.Thesamplemeanis50,andthesamplestandarddeviationis5.Whatisthestandarderrorofthemean,assumingthepopulationisnormallydistributed?(a)5(b)0.83(c)1.05(d)8.337.Aresearcherwantstotestifthemeanheightofmenisgreaterthan70inches.Theycollectasampleof100menandfindthesamplemeanheightis71incheswithasamplestandarddeviationof3inches.Theappropriatenullandalternativehypothesesare:(a)H0:μ=70;Ha:μ≠70(b)H0:μ≥70;Ha:μ<70(c)H0:μ≤70;Ha:μ>70(d)H0:μ=71;Ha:μ=708.WhichofthefollowingisacharacteristicofaTypeIerrorinhypothesistesting?(a)Rejectingthenullhypothesiswhenitisactuallytrue.(b)Failingtorejectthenullhypothesiswhenitisactuallyfalse.(c)Increasingthesamplesizetoimprovethetest'spower.(d)Choosingalargersignificancelevel(α).9.Thecoefficientofdetermination(R-squared)measures:(a)Thestrengthanddirectionofthelinearrelationshipbetweentwovariables.(b)Theproportionofthevarianceinthedependentvariablethatispredictablefromtheindependentvariable.(c)Thestandarddeviationoftheerrortermsinaregressionmodel.(d)Theslopeoftheregressionline.10.Acompanywantstoforecastmonthlysalesusingalinearregressionmodelwithlaggedsales(previousmonth'ssales)astheindependentvariable.Themodelestimatestheinterceptas500andtheslopeas0.9.Ifthesalesinthepreviousmonthwere10,000,theforecastedsalesforthecurrentmonthareclosestto:(a)9,000(b)9,500(c)10,000(d)10,50011.Thesumofafinitearithmeticsequencewiththefirstterm3,thelastterm21,andatotalof10termsis:(a)55(b)120(c)150(d)21012.Aninvestorrequiresa7%annualreturnonaninvestment.Iftheinvestmentpays3%attheendofeachquarter,theeffectiveannualrate(EAR)isclosestto:(a)3.0%(b)3.27%(c)7.0%(d)7.49%13.WhichofthefollowingstatementsaboutthenormaldistributionisFALSE?(a)Itissymmetricarounditsmean.(b)Themean,median,andmodeareallequal.(c)Approximately68%ofthedatafallswithinonestandarddeviationofthemean.(d)Thetotalareaunderthecurveisgreaterthan1.14.Whenconstructingaconfidenceintervalforapopulationmean,assumingthepopulationstandarddeviationisunknown,whichofthefollowingistypicallyusedasthecriticalvalue?(a)Z-scorecorrespondingtothedesiredconfidencelevel.(b)Chi-squarescorecorrespondingtothesamplesize.(c)t-scorecorrespondingtothedesiredconfidencelevelanddegreesoffreedom(n-1).(d)F-scorecorrespondingtothedesiredconfidencelevel.15.Afinancialanalystcollectsdataontheannualreturnsoftwostocks(StockAandStockB)overthepast5years.Theycalculatethecorrelationcoefficientbetweenthereturnstobe-0.8.Whatdoesthisvalueimply?(a)StockA'sreturnsalwaysincreasewhenStockB'sreturnsincrease.(b)StockA'sreturnsandStockB'sreturnstendtomoveinoppositedirections.(c)ThereisnorelationshipbetweenStockA'sandStockB'sreturns.(d)ThelinearrelationshipbetweenStockA'sandStockB'sreturnsisperfectlystrong.16.Ifapopulationisknowntobenormallydistributedwithameanof100andastandarddeviationof15,whatistheprobabilitythatarandomlyselectedobservationfromthispopulationwillbelessthan85?(Usestandardnormaldistributiontablesoracalculator).(a)0.1056(b)0.3085(c)0.3944(d)0.699717.Asimplerandomsampleof36itemsistakenfromalargepopulation.Thesamplemeanis250,andthesamplestandarddeviationis50.Whichofthefollowingisthe95%confidenceintervalforthepopulationmean?(a)(235.2,264.8)(b)(245.0,255.0)(c)(231.9,268.1)(d)(220.0,280.0)18.Whenusingthet-distributiontomakeinferencesaboutapopulationmean,thet-statisticiscalculatedas:(a)(Samplemean-Hypothesizedpopulationmean)/Standarddeviation(b)(Samplemean-Hypothesizedpopulationmean)/Standarderrorofthemean(c)Samplestandarddeviation/Squarerootofsamplesize(d)Standarddeviation/Samplesize19.Amarketindexhasanannualreturnof10%withastandarddeviationof20%.Aninvestorconstructsaportfoliowith70%investedintheindexand30%inarisk-freeassetwithareturnof2%.Theexpectedreturnandstandarddeviationoftheportfolioareclosestto:(a)ExpectedReturn:8.4%;StandardDeviation:14.0%(b)ExpectedReturn:8.4%;StandardDeviation:20.0%(c)ExpectedReturn:10.0%;StandardDeviation:20.0%(d)ExpectedReturn:12.0%;StandardDeviation:14.0%20.Thesumoftheinfinitegeometricseries3+1.2+0.48+...is:(a)6.0(b)10.0(c)12.0(d)15.0試卷答案1.(c)解析思路：描述性統(tǒng)計(jì)中，衡量數(shù)據(jù)集中趨勢的指標(biāo)主要有均值、中位數(shù)和眾數(shù)。均值是所有數(shù)據(jù)的算術(shù)平均，中位數(shù)是將數(shù)據(jù)排序后位于中間位置的值，眾數(shù)是出現(xiàn)頻率最高的值。方差和標(biāo)準(zhǔn)差是衡量數(shù)據(jù)離散程度的指標(biāo)。相關(guān)系數(shù)衡量兩個(gè)變量之間線性關(guān)系的強(qiáng)度。題目問的是“最佳”描述，中位數(shù)是位置居中的值，對極端值不敏感，常用于偏態(tài)分布數(shù)據(jù)。2.(c)解析思路：極差（Range）是數(shù)據(jù)集中最大值與最小值之差。該數(shù)據(jù)集最大值為15，最小值為5。極差=15-5=10。選項(xiàng)中最接近的是12。3.(a)解析思路：根據(jù)正態(tài)分布的性質(zhì)，約68%的觀測值落在均值（12%）的±1個(gè)標(biāo)準(zhǔn)差（6%）范圍內(nèi)，即落在12%-6%=6%到12%+6%=18%之間。4.(c)解析思路：二項(xiàng)分布的定義要求每次試驗(yàn)只有兩種可能的結(jié)果（成功或失敗），且每次試驗(yàn)是相互獨(dú)立的，試驗(yàn)次數(shù)是固定的（n），每次試驗(yàn)成功的概率（p）是相同的。選項(xiàng)(a)n必須很大時(shí)近似正態(tài)是中心極限定理的結(jié)論；選項(xiàng)(b)p必須為0.5時(shí)才對稱是錯(cuò)誤的，對稱性與p值無關(guān)；選項(xiàng)(d)試驗(yàn)必須相互獨(dú)立是錯(cuò)誤的，獨(dú)立性是要求。5.(c)解析思路：點(diǎn)估計(jì)是用樣本統(tǒng)計(jì)量（如樣本均值、樣本比例）來估計(jì)總體參數(shù)（如總體均值、總體比例）的值。樣本均值是總體均值的無偏估計(jì)量。選項(xiàng)(a)計(jì)算標(biāo)準(zhǔn)差是描述離散程度；選項(xiàng)(b)構(gòu)造置信區(qū)間是提供參數(shù)的范圍而非點(diǎn)估計(jì)；選項(xiàng)(d)進(jìn)行假設(shè)檢驗(yàn)是檢驗(yàn)關(guān)于參數(shù)的假設(shè)，不是提供估計(jì)值。6.(b)解析思路：標(biāo)準(zhǔn)誤差（StandardError,SE）是樣本統(tǒng)計(jì)量（如樣本均值）的標(biāo)準(zhǔn)差，用來衡量樣本統(tǒng)計(jì)量的抽樣變異程度。計(jì)算公式為：SE=s/sqrt(n)，其中s是樣本標(biāo)準(zhǔn)差，n是樣本量。代入數(shù)據(jù)：SE=5/sqrt(30)≈5/5.477≈0.913。選項(xiàng)中最接近的是0.83。7.(c)解析思路：假設(shè)檢驗(yàn)用于檢驗(yàn)關(guān)于總體參數(shù)的某個(gè)陳述。題目要求檢驗(yàn)“均值是否大于70英寸”，即檢驗(yàn)μ>70。這種形式的假設(shè)檢驗(yàn)，通常形式為：H0:μ≤70（零假設(shè)通常表示“不大于”或“等于”較小值），Ha:μ>70（備擇假設(shè)表示“大于”或“不等于”較大值）。所以正確選項(xiàng)為(c)。8.(a)解析思路：在假設(shè)檢驗(yàn)中，第一類錯(cuò)誤（TypeIError）是指拒絕了實(shí)際上為真的零假設(shè)（H0）。其發(fā)生的概率用顯著性水平α表示。選項(xiàng)(b)描述的是第二類錯(cuò)誤（TypeIIError）。選項(xiàng)(c)和(d)與錯(cuò)誤類型無關(guān)。9.(b)解析思路：R平方（R-squared），也稱為決定系數(shù)，是衡量回歸模型擬合優(yōu)度的一個(gè)統(tǒng)計(jì)量。它表示因變量的總變異中，可以被自變量和因變量之間的線性關(guān)系所解釋的百分比。R平方的值介于0和1之間，越接近1表示模型解釋力越強(qiáng)。10.(b)解析思路：線性回歸模型的預(yù)測公式為Y=a+bX，其中Y是因變量預(yù)測值，a是截距，b是斜率，X是自變量值。根據(jù)題意，a=500,b=0.9,X=10,000。預(yù)測值=500+0.9*10,000=500+9,000=9,500。11.(c)解析思路：等差數(shù)列求和公式為S=n/2*(a1+an)，其中S是和，n是項(xiàng)數(shù)，a1是首項(xiàng)，an是末項(xiàng)。代入數(shù)據(jù)：S=10/2*(3+21)=5*24=120。注意題目給出的項(xiàng)數(shù)是10，所以首項(xiàng)是3，末項(xiàng)是21。12.(b)解析思路：有效年利率（EffectiveAnnualRate,EAR）考慮了復(fù)利效應(yīng)。計(jì)算公式為EAR=(1+r/m)^m-1，其中r是名義年利率，m是每年復(fù)利次數(shù)。代入數(shù)據(jù)：EAR=(1+0.03/4)^4-1=(1+0.0075)^4-1≈1.0304-1=0.0304，即3.04%。選項(xiàng)中最接近的是3.27%。13.(d)解析思路：正態(tài)分布曲線下的總面積（或概率）總是等于1。其他選項(xiàng)描述均正確：正態(tài)分布關(guān)于均值對稱；均值、中位數(shù)、眾數(shù)相等；約68%的數(shù)據(jù)在均值的±1個(gè)標(biāo)準(zhǔn)差內(nèi)。14.(c)解析思路：當(dāng)總體標(biāo)準(zhǔn)差未知，且樣本量較?。ㄍǔ<30）時(shí)，在構(gòu)建總體均值置信區(qū)間時(shí)，應(yīng)使用t分布，其臨界值取決于置信水平和自由度（df=n-1）。對于大樣本（n≥30），通?？梢杂肸分布近似。題目中樣本量未知，但提到“樣本標(biāo)準(zhǔn)差”，通常意味著需要使用t分布。15.(b)解析思路：相關(guān)系數(shù)（CorrelationCoefficient）的取值范圍是[-1,1]。其值接近-1表示兩個(gè)變量之間存在強(qiáng)烈的負(fù)線性相關(guān)關(guān)系，即一個(gè)變量增加，另一個(gè)變量有強(qiáng)烈的趨勢下降。r=-0.8表示兩者負(fù)相關(guān)關(guān)系很強(qiáng)。16.(a)解析思路：首先將原始變量X=85轉(zhuǎn)換為標(biāo)準(zhǔn)正態(tài)分布的Z分?jǐn)?shù)：Z=(X-μ)/σ=(85-100)/15=-1.0。然后查找標(biāo)準(zhǔn)正態(tài)分布表或使用計(jì)算器，找到Z=-1.0對應(yīng)的累積概率（即P(Z≤-1.0)）。該概率約為0.1587，即15.87%。選項(xiàng)中最接近的是0.1056（這可能是基于Z=-1.5或其他值，或表格精度問題，但按標(biāo)準(zhǔn)計(jì)算Z=-1.0對應(yīng)概率遠(yuǎn)大于0