題庫(kù)frm一級(jí)數(shù)量分析

tative265ReviewofDefinerandomvariables,anddistinguishbetweencontinuousanddiscreterandomvariables.第1WhichofthefollowingstatementsaboutprobabilityismostAconditionalprobabilityistheprobabilitythattwoormoreeventswillhappen eisthecalculatedprobabilityofanAneventisasetofoneormorepossiblevaluesofarandomOutofasampleof100widgets10werefoundtobedefective,20wereperfect,and70wereOK.Theprobabilityofpickingaperfectwidgetatrandomis29%.Conditionalprobabilityistheprobabilityofoneeventhappeninggiventhatanothereventhashappened.An eisthenumericalresultassociatedwitharandomvariable.Theprobabilityofpickingaperfectwidgetis20/100=0.20or第2LetAandBbetwomutuallyexclusiveeventswithP(A)=0.40andP(B)=0.20.P(AandB)=P(B|A)=P(AorB)=P(AandB)=Ifthetwoevensaremutuallyexclusive,theprobabilityofbothocurringisDefinetheprobabilityofan第1ThomasBayneshasappliedtobothHarvardandYale.BayneshasdeterminedthattheprobabilityofgettingintoHarvardis25%andtheprobabilityofgettingintoYale(hisfather’salmamater)is42%.Bayneshasalsodeterminedthattheprobabilityofbeingacceptedatbothschoolsis2.8%.WhatistheprobabilityofBaynesbeingacceptedateitherHarvardorYale?Usingtheadditionrule,theprobabilityofbeingacceptedatHarvardorYale,isequalto:P(Harvard)+P(Yale)?P(HarvardandYale)=0.25+0.42?0.028=0.642or64.2%.第2Atwo-sidedbutverythickcoinisexpectedtolandonitsedgetwiceoutofevery100flips.Andtheprobabilityoffaceup(heads)andtheprobabilityoffacedown(tails)areequal.Whenthecoinisflipped,theprizeis$1forheads,$2fortails,and$50whenthecoinlandsonitsedge.Whatistheexpectedvalueoftheprizeonasinglecointoss?Sincetheprobabilityofthecoinlandingonitsedgeis0.02,theprobabilityofeachoftheothertwoeventsis0.49.Theexpectedpayoffis:(0.02×$50)+(0.49×$1)+(0.49×=第3TheprobabilityofAis0.4.TheprobabilityofACis0.6.Theprobabilityof(B|A)is0.5,andtheprobabilityof(B|AC)is0.2.UsingBayes’formula,whatistheprobabilityof(A|B)?Usingthetotalprobabilityrule,wecancomputetheP(B):P(B)=[P(B|A)×P(A)]+[P(B|AC)×P(AC)]P(B)=[0.5×0.4]+[0.2×0.6]=UsingBayes’formula,wecansolveforP(A|B)=[P(B|A)/P(B)]×P(A)=[0.5/0.32]×0.4=第4Ifafaircoinistossedtwice,whatistheprobabilityofobtainingheadsbothTheprobabilityoftossingahead,H,isP(H)=1/2.Sincetheseareindependentevents,theprobabilityoftwoheadsinarowisP(HH)=P(H)×P(H)=1/2×1/2=1/4.第5SufferfromSufferfromDon'tSufferfromThesearemutuallyexclusive,sothejointprobabilityis第6BondsratedBhavea25%chanceofdefaultinfiveyears.BondsratedCCChavea40%chanceofdefaultinfiveyears.Aportfolioconsistsof30%Band70%CCC-ratedbonds.Ifarandomlyselectedbonddefaultsinafive-yearperiod,whatistheprobabilitythatitwasaB-ratedbond?AccordingtoBayes'formula:P(B/default)=P(defaultandB)/P(default).P(defaultandB)=P(default/B)×P(B)=0.250×0.300=0.075P(defaultandCCC)=P(default/CCC)×P(CCC)=0.400×0.700=0.280P(default)=P(defaultandB)+P(defaultandCCC)=0.355P(B/default)=P(defaultandB)/P(default)=0.075/0.355=第7IftheprobabilityofbothanewWal-MartandanewWendy’sbeingbuiltnextmonthis68%andtheprobabilityofanewWal-Martbeingbuiltis85%,whatistheprobabilityofanewWendy’sbeingbuiltifanewWal-Martisbuilt?P(AB)=P(A|B)×0.68/0.85=第8AninvestorhasanA-ratedbond,aBB-ratedbond,andaCCC-ratedbondwheretheprobabilitiesofdefaultoverthenextthreeyearsare4percent,12percent,and30percent,respectively.WhatistheprobabilitythatallofthesebondswilldefaultinthenextthreeyearsiftheindividualdefaultprobabilitiesareSincetheprobabilityofdefaultforeachbondisindependent,P(A BBCCC)=P(A)×P(BB)×P(CCC)=0.04×0.12×0.30=0.00144=0.14%.第9BucketNoBucketAirNoAirBucketNoBucketAirNoAirWhatistheprobabilityofselectingatruckatrandomthathaseitherairbagsorbucketTheadditionruleforprobabilitiesisusedtodeterminetheprobabilityofatleastoneeventamongtwoormoreeventsoccurring.Theprobabilityofeacheventisaddedandtheprobability(iftheeventsarenotmutuallyexclusive)issubtractedtoarriveatthesolution.P(airbagsorbucketseats)=P(airbags)+P(bucketseats)?P(airbagsandbucketseats)=(125/220)+(110/220)?(75/220)=0.57+0.50?0.34=0.73orAlternative:1?P(noairbagandnobucketseats)=1?(60/220)=10WhichofthefollowingstatementsaboutprobabilitydistributionsisleastInabinomialdistributioneachobservationhasonlytwopossible esthataremutuallyexclusive.Aprobabilitydistributionis,bydefinition,normallyAprobabilitydistributionincludesalistingofallthe esofanOneofthekeypropertiesofaprobabilityfunctionis0≤p≤Probabilitiesmustbezeroorpositive,butaprobabilitydistributionisnotnecessarilynormallydistributed.Binomialdistributionsareeithersuccessesorfailures.11Amajorsecuritiesexchangeisconsideringtheintroductionofanewderivativecontract.Inthepast,thesuccessratefornewderivativeshasbeen30percent.Extensivemarketresearchhasproducedapositivemarketingresearchreportforthecontractunderconsideration.Historically,70percentofthesuccessfulcontractshavereceivedfavorablereportspriortointroduction.Only10percentofunsuccessfulcontractshavereceivedfavorablereports.Whatistheprobabilitythatthenewcontractwillbesuccessful?LetSdenotesuccess,Udenoteunsuccessful,andFdenoteafavorablereport.WehaveP(S)=0.30,P(U)=0.70,P(F|S)=0.70,andP(F|U)=0.10.UsingBayestheorem,theprobabilityofasuccessfulcontract,givenafavorablereportis:12TheprobabilityofanewWal-Martbeingbuiltintownis64%.IfWal-Martcomestotown,theprobabilityofanewWendy’srestaurantbeingbuiltis90%.WhatistheprobabilityofanewWal-MartandanewWendy’srestaurantbeingbuilt?P(AB)=P(A|B)×TheprobabilityofanewWal-MartandanewWendy’sisequaltotheprobabilityofanewWendy’s“ifWal-Mart”(0.90)timestheprobabilityofanewWal-Mart(0.64).(0.90)(0.64)=13Iftwofaircoinsareflippedandtwofairsix-sideddicearerolled,allatthesametime,whatistheprobabilityofendingupwithtwoheads(onthecoins)andtwosixes(onthedice)?Forthefourindependenteventsdefinedhere,theprobabilityofthespecified eis0.5000×0.5000×0.1667×0.1667=0.0069.14DependentrandomvariablesaredefinedasvariableswheretheirjointprobabilityequaltogreaterthantheproductoftheirindividualnotequaltotheproductoftheirindividualequaltotheproductoftheirindividualDependenceresultsbetweenrandomvariableswhentheirjointprobabilitiesarenotequalproductsofindividualprobabilities.Iftheyareequal,thenanindependentrelationship15TheprobabilitiesthatthreestudentswillearnanAonanexamare0.20,0.25,and0.30,respectively.Ifeachstudent’sperformanceisindependentofthatoftheothertwostudents,theprobabilitythatallthreestudentswillearnanAisclosestto:Sincetheeventsareindependent,theprobabilityisobtainedbymultiplyingtheindividualprobabilities.ProbabilityofthreeAs=(0.20)×(0.25)×(0.30)=0.0150.16Aninvestorischoosingoneoftwentysecurities.Tenofthesecuritiesarestocksandtenarebonds.Fourofthetenstockswereissuedbyutilities,theothersixwereissuedbyindustrialfirms.Twoofthetenbondswereissuedbyutilities,theothereightwereissuedbyindustrialfirms.Iftheinvestorchoosesasecurityatrandom,theprobabilitythatitisabondorasecurityissuedbyanindustrialfirmis:LetBrepresentthesetofbondsandIthesetofindustrialfirms.ThedesiredprobabilityistheprobabilityoftheunionofsetsBandI,P(BI).Accordingtothetheoremsofprobability,P(BI)=P(B)+P(I)-P(B∩I),whereP(B)istheprobabilitythatasecurityisabond=P(B)=10/20,P(I)istheprobabilitythatasecuritywasissuedbyanindustrialfirm=P(I)=14/20,andP(B∩I)istheprobabilitythatasecurityisbothabondandissuedbyanindustrialfirm=P(B∩I)=8/20.10/20+14/20-8/20=16/20=0.80.17XandYarediscreterandomvariables.TheprobabilitythatX=3is0.20andtheprobabilitythatY=4is0.30.TheprobabilityofobservingthatX=3andY=4concurrentlyisclosestCannotanswerwiththeinformationIfyouknewthatXandYwereindependent,youcouldcalculatetheprobabilityas0.20(0.300.06Withoutthisknowledgeyouwouldneedthejointprobabilitydistribution.第18題Adealerinacasinohasrolledafiveonasinglediethreetimesinarow.Whatistheprobabilityofherrollinganotherfiveonthenextroll,assumingitisafairdie?Theprobabilityofavaluebeingrolledis1/6regardlessofthepreviousvalue19refiningaforecastbecauseoftheoccurrenceofsomeotherdeterminingtheexpectedjointcalculatingtheconditionalestimatingtheConditionalexpectedvaluesarecontingentupontheoccurrenceofsomeotherevent.Theexpectationchangesasnewinformationisrevealed.20BucketNoBucketAirNoAirBucketNoBucketAirNoAirWhatistheprobabilityofrandomlyselectingatruckwithairbagsandbucket75/220=21AjointprobabilityofAandBmustalwaysgreaterthanorequaltotheconditionalprobabilityofAgivengreaterthanorequaltothantheprobabilityofAorlessthanorequaltotheconditionalprobabilityofAgivenlessthantheprobabilityofAandtheprobabilityofBytheformulaforjointprobability:P(AB)=P(A|B)×P(B),sinceP(B)≤1,then≤P(A|B).NoneoftheotherchoicesmustDefine,calculate,andinterpretthemean,standarddeviation,andvarianceofarandomvariable.第1Thevarianceofthesumoftwoindependentrandomvariablesisequaltothesumoftheirplusapositivecovarianceplusminusapositivecovarianceplusanon-zerocovarianceIndependentrandomvariableshaveacovarianceofzero.Hence,thevarianceofthesumoftwovariableswillbethesumofthevariables’variances.第2Anysthasknowledgeofthebeginning-of-periodexpectedreturns,standarddeviationsofreturn,andmarketvalueweightsfortheassetsthatcompriseaportfolio.TheystdoesnotrequirethecovariancesofreturnsbetweenassetpairstocalculatevarianceofthereturnonthecorrelationsbetweenassetreductioninriskduetoexpectedreturnontheAllthatisrequiredtocalculatetheexpectedreturnfortheportfolioistheportfolioweightsandindividualassetexpectedreturns.Allotheritemsarefunctionsofthecovariance.第3ReturnonPortfolioReturnonPortfolioABCReturnonPortfolioReturnonPortfolioABCDE(RA)=σ2= =0.15(0.18?0.1165)2+0.2(0.17?0.1165)2+0.25(0.11?0.1165)20.4(0.07?σ=第4AninvestorisconsideringpurchasingACQ.Thereisa30%probabilitythatACQwillbeacquiredinthenexttwomonths.IfACQisacquired,thereisa40%probabilityofearninga30%returnontheinvestmentanda60%probabilityofearning25%.IfACQisnotacquired,theexpectedreturnis12%.Whatistheexpectedreturnonthisinvestment?E(r)=(0.70×0.12)+(0.30×0.40×0.30)+(0.30×0.60×0.25)=第5ThecharacteristicfunctionoftheproductofindependentrandomvariablesisequaltosquarerootoftheproductoftheindividualcharacteristicexponentialrootoftheproductoftheindividualcharacteristicproductoftheindividualcharacteristicsumoftheindividualcharacteristicThecharacteristicfunctionofthesumofindependentrandomvariablesisequaltotheproductoftheindividualcharacteristicfunctions.E(XY)=E(X)×E(Y).第6Thereisa30%chancethattheeconomywillbegoodanda70%chancethatitwillbebad.Iftheeconomyisgood,yourreturnswillbe20%andiftheeconomyisbad,yourreturnswillbe10%.Whatisyourexpectedreturn?Expectedvalueistheprobabilityweightedaverageofthepossible esoftherandomvariable.Theexpectedreturnis:((0.3)×(0.2))+((0.7)×(0.1))=(0.06)+(0.07)=0.13.第7StateoftheReturnon[0.30×(0.15?0.066)2+0.70×(0.03?0.066)2]1/2=第8TullyAdvisers,Inc.,hasdeterminedfourpossibleeconomicscenariosandhasprojectedtheportfolioreturnsfortwoportfoliosfortheirundereachscenario.Tully’seconomisthasestimatedtheprobabilityofeachscenarioasshowninthetablebelow.Giventhisinformation,whatistheexpectedreturnonportfolioA?ReturnonPortfolioReturnonPortfolioABCDTheexpectedreturnisequaltothesumoftheproductsoftheprobabilitiesofthescenariosandtheirrespectivereturns:=(0.15)(0.17)+(0.20)(0.14)+(0.25)(0.12)+(0.40)(0.08)0.1155orDefine,calculate,andinterprettheskewness,andkurtosisofa第1Whenthetailsofadistributionarefatterthanthatimpliedbyanormaldistribution,wesaythatthedistributionis:Whenthetailsofadistributionarethickerthanthatofanormaldistribution,wesaythedistributionisleptokurtic.tykurticmeansthatthetailsarethinnerthannormal.Thesizeofthetailsisnotindicativeofadistribution’ssymmetry.Rather,thepresenceofskewnessindicatesthatthedistributionisnotsymmetrical.第2Inapositivelyskeweddistribution,whatistheorder(fromlowestvaluetohighest)forthedistribution’smode,mean,andmedianvalues?Mode,median,Mode,mean,Mean,median,Median,mean,Inapositivelyskeweddistribution,themodeislessthanthemedian,whichislessthanthe第3Inanegativelyskeweddistribution,whatistheorder(fromlowestvaluetohighest)forthedistribution’smode,mean,andmedianvalues?Mode,mean,Median,mode,Mean,median,Median,mean,Inanegativelyskeweddistribution,themeanislessthanthemedian,whichislessthanthemode.第4Left-skeweddistributionsgreatermasstotheleftoftheexpectedalongertailtotherightofthegreatermasstotherightoftheexpectedgreatermassclosetotheexpectedLeft-skeweddistributionsarethosewithalongertailtotheleftsideofthedistribution,andwhosemassistotherightoftheexpectedvalue.第5AdistributionthathaspositiveexcessislesspeakedthananormalhasthinnertailsthananormalismorepeakedthananormalismoreskewedthananormalAdistributionwithpositiveexcesskurtosisisonethatismorepeakedthananormal第6Itisoftensaidthatstockreturnsareleptokurtic.Ifthisistrue,relativetoanormaldistributionofthesamemeanandvariance,thedistributionofstockreturnsis:thin-fat-negativelypositivelyAleptokurticdistributionismorepeakedandhasfattertailsthananormalDescribejoint,marginal,andconditionalprobability第1Inamultivariatenormaldistribution,acorrelationlsoverallrelationshipbetweenalltherelationshipbetweenthemeansandvariancesofthestrengthofthelinearrelationshipbetweentwooftherelationshipbetweenthemeansandstandarddeviationsoftheThisistruebydefinition.Thecorrelationonlyappliestotwovariablesata第2Amultivariategivesmultipleprobabilitiesforthe specifiestheprobabilitiesassociatedwithgroupsofrandomappliesonlytonormalappliesonlytobinomialThisisthedefinitionofamultivariate第3WhichofthefollowingwouldleastlikelybecategorizedasamultivariateThedaysastocktradedandthedaysitdidnotThereturnofastockandthereturnoftheThedaysastocktradedandthedaystheDJIAwentThereturnsofthestocksintheThenumberofdaysastocktradedanddidnottradedescribesonlyonerandomvariable.Alltheothercasesinvolvetwoormorerandomvariables.Exinthedifferencebetweenstatisticalindependenceandstatistical第1If eofeventAisnotaffectedbyeventB,theneventsAandBaresaidtomutuallycollectivelystatisticallyIfthe eofoneeventdoesnotinfluencethe eofanother,thentheeventsareindependent.第2Asaysthatwhetheritsearningsincreasedependsonwhetheritincreaseditsdividends.Fromthisweknow:P(bothdividendincreaseandearningsincrease)=P(dividendP(dividendincreaseorearningsincrease)=P(bothdividendandearningsP(dividendincrease|earningsincrease)isnotequaltoP(earningsP(earningsincrease|dividendincrease)isnotequaltoP(earningsIftwoeventsAandBaredependentthentheconditionalprobabilitiesofP(A|BandP(B|A)willnotequaltheirrespectiveunconditionalprobabilities(ofP(A)andP(B),respectively).Theotherchoicesmayormaynotoccure.gP(A|BP(Bispossiblebutnotnecessary.第3題IfXandYareindependentevents,whichofthefollowingismostP(XorY)=P(X)+XandYcannotoccurP(X|Y)=P(XorY)=(P(X))×Notethateventsbeingindependentmeansthattheyhavenoinfluenceoneachother.Itdoesnotnecessarilymeanthattheyaremutuallyexclusive.Accordingly,P(XorY)=P(X)+P(Y)?P(XandY).Bythedefinitionofindependentevents,P(X|Y)=CalculatethemeanandvarianceofsumsofrandomDescribethekeypropertiesofthenormal,standardnormal,multivariatenormal,Chi-squared,Studen第1Agroupofinvestorswantstobesuretoalwaysearnatleasta5%rateofreturnontheirinvestments.Theyarelookingataninvestmentthathasanormallydistributedprobabilitydistributionwithanexpectedrateofreturnof10%andastandarddeviationof5%.Theprobabilityofmeetingorexceedingtheinvestors'desiredreturninanygivenyearisclosestThemeanis10%andthestandarddeviationis5%.Youwanttoknowtheprobabilityofareturn5%orbetter.10%-5%=5%,so5%isonestandarddeviationlessthanthemean.Thirty-fourpercentoftheobservationsarebetweenthemeanandonestandarddeviationonthedownside.Fiftypercentoftheobservationsaregreaterthanthemean.Sotheprobabilityofareturn5%orhigheris34%+50%=84%.第2Theprobabilityofreturnslessthan?10%,assuminganormaldistributionwithexpectedreturnof6.5%andstandarddeviationof10%,is:lessthanapproximaynotdefinedwithonlythisapproximay?10is16.5%belowthemeanreturn,(16.5/10)=1.65standarddeviations,whichapproximay5%ofthe esinthelefttailbelow第3Thereturnonaportfolioisnormallydistributedwithameanreturnof8percentandastandarddeviationof18percent.Whichofthefollowingisclosesttotheprobabilitythatthereturnontheportfoliowillbebetween-27.3percentand37.7percent?Notethatifyoumemorizethebasicintervalsforanormaldistribution,youdonotneedanormaldistributiontabletoanswerthisquestion.-27.3%representsalossof-35.3%fromthemeanreturn(-27.3–8.0),whichis(-35.3/18)=-1.96standarddeviationstotheleftofthemean.Foranormaldistribution,weknowthatapproximay95percentofallobservationsliewith+/-1.96standarddeviationsofthemean,sotheprobabilitythatthereturnisbetween-27.3%and8.0%mustbe(95%/2)=47.5%.Areturnof37.7percentrepresentsagainof(37.7-8.0)=29.7%fromthemeanreturn,whichis(29.7/18)=1.65standarddeviationstotherightofthemean.Foranormaldistribution,weknowthatapproximay90percentofallobservationsliewith+/-1.65standarddeviationsofthemean,sotheprobabilitythatthereturnisbetween8.0%and37.7%mustbe(90%/2)=45%.Thereforetheprobabilitythatthereturnisbetween-27.3percentand37.7percent=(47.5%+45%)=Defineanddescriberandomsamplingandwhatismeantby第1Ifnislargeandthepopulationstandarddeviationisunknown,thestandarderrorofthesamplingdistributionofthesamplemeanisequaltothe:populationstandarddeviationdividedbythesamplepopulationstandarddeviationmultipliedbythesamplesamplestandarddeviationdividedbythesquarerootofthesamplesamplestandarddeviationdividedbythesampleTheformulaforthestandarderrorwhenthepopulationstandarddeviationisunknown第2JosephLucalculatedtheaveragereturnonequityforasampleof64companies.Thesampleaverageis0.14andthesamplestandarddeviationis0.16.Thestandarderrorofthemeanisclosestto:Thestandarderrorofthemean=σ/√n=0.16/√64=第3Apopulation’smeanis30andthemeanofasampleofsize100is28.5.Thevarianceofthesampleis25.Whatisthestandarderrorofthesamplemean?第4Apopulationhasameanof20,000andastandarddeviationof1,000.Samplesofsize=2,500aretakenfromthispopulation.WhatisthestandarderrorofthesampleThestandarderrorofthesamplemeanisestimatedbydividingthestandarddeviationofthesamplebythesquarerootofthesamplesize:sx=s/n1/2=1000/(2500)1/2=1000/50=20.第5Fromapopulationof5,000observations,asampleofn=100isselected.Calculatethestandarderrorofthesamplemeanifthepopulationstandarddeviationis50.Thestandarderrorofthesamplemeanequalsthestandarddeviationofthepopulationdividedbythesquarerootofthesamplesize:50/1001/2=5.第6Thepopulationmeanforequityreturnsis14percentwithastandarddeviationof2percent.Ifarandomsampleof49returnsisdrawn,whatisthestandarderrorofthesampleDefine,calculate,andinterpretthemeanandvarianceofthesample無Describe,interpret,andapplytheLawofLargeNumbersandtheCentralLimit第1Thecentrallimittheoremstatesthat,foranydistribution,asngetslarger,thesamplingesesapproachesanormalapproachestheAsngetslarger,thevarianceofthedistributionofsamplemeansisreduced,andthedistributionofsamplemeansapproximatesanormaldistribution.第2ThecentrallimittheoremconcernsthesamplingdistributionofsamplepopulationsamplestandardpopulationstandardThecentrallimittheoremlsusthatforapopulationwithameanmandafinitevarianceσ2,thesamplingdistributionofthesamplemeansofallpossiblesamplesofsizenwillapproachanormaldistributionwithameanequaltomandavarianceequaltoσ2/nasngetslarge.第3Toapplythecentrallimittheoremtothesamplingdistributionofthesamplemean,thesampleisusuallyconsideredtobelargeifnisgreaterthan:AsamplesizeisconsideredsufficientlylargeifitislargerthanorequaltoReviewofDescribeandinterpretestimatorsofthesamplemeanandtheir第1WhichofthefollowingstatementsregardingthetermspopulationandsampleisObservingallmembersofapopulationcanbeexpensiveortimeAsampleincludesallmembersofaspecifiedAsample'scharacteristicsareattributedtothepopulationasaAdescriptivemeasureofasampleiscalledaApopulationincludesallmembersofaspecifiedgroup.Asampleisaportion,orsubsetofthepopulationofinterest.第2WhichofthefollowingstatementsaboutstatisticalconceptsisleastAsamplecontainsallmembersofaspecifiedgroup,butapopulationcontainsonlyaAfrequencydistributionisatabulardisyofdatasummarizedintoarelativelysmallnumberofintervals.AnintervalisasetofreturnvalueswithinwhichanobservationAparameterisanydescriptivemeasureofapopulationApopulationisdefinedasallmembersofaspecifiedgroup,butasampleisasubsetofaDescribeandinterprettheleastsquares無Define,interpret,andcalculatethet-無Define,calculate,andinterpretaconfidence第1WhichofthefollowingstatementsaboutsamplingandestimationismostAconfidenceintervalestimateconsistsofarangeofvaluesthatbrackettheparameterwithaspecifiedlevelofprobability,1?β.Apointestimateisasingleestimateofanunknownpopulationparametercalculatedasasamplemean.Time-seriesdataareobservationsoverindividualunitsatapointinCross-sectionaldataareasetofvaluesofaparticularvariableinsequentialtimeTime-seriesdataareobservationstakenatspecificandequally-spacedpoints.Cross-sectionaldataareasampleofobservationstakenatasinglepointintime.Aconfidenceintervalestimateconsistsofarangeofvaluesthatbrackettheparameterwithaspecifiedlevelofprobability,1?α.第2Whichofthefollowingwouldresultinawiderconfidenceinterval?higherdegreeofhigheralphahigherpointgreaterlevelofAhigherdegreeofconfidence(e.g.99%insteadof95%)wouldrequireahigherreliabilityfactor(2.575insteadof1.96assuminganormaldistribution).Awiderconfidenceintervalcorrespondstoaloweralphasignificancelevelandthepointestimatedoesnotaffectthewidthoftheconfidenceinterval.第3Arangeofestimatedvalueswithinwhichtheactualvalueofapopulationparameterwillliewithagivenprobabilityof1?αisa(n):αpercentpoint(1?α)percentconfidenceαpercentconfidence(1?α)percentcross-sectionalpointA95%confidenceintervalforthepopulationmean(α=5%isthep-value),forexample,isarangeofestimateswithinwhichtheactualvalueofthepopulationmeanwillliewithaprobabilityof95%.Pointestimates,ontheotherhand,aresingle(sample)valuesusedtoestimatepopulationparameters.Thereisnosuchthingasaα%pointestimateora(1?α)%cross-sectionalpointestimate.第4WhichofthefollowingstatementsaboutsamplingandestimationismostThestandarderrorofthesamplemeanswhenthestandarddeviationofthepopulationisunknownequalss/√n,wheres=samplestandarddeviation.Thestandarddeviationofthedistributionofthesamplemeansisthestandarderroroftheresidual.Thestandarderrorofthesamplemeanswhenthestandarddeviationofthepopulationisknownequalsσ/√n,whereσ=samplestandarddeviationadjustedbyn?1.TheprobabilitythataparameterlieswithinarangeofestimatedvaluesisgivenbyTheprobabilitythataparameterlieswithinarangeofestimatedvaluesisgivenby1?α.Thestandarddeviationofthedistributionofthesamplemeansisthestandarderrorofthesamplemean.Thetermresidualisusedduringthediscussionofregressionthatoccurslaterinthisstudysession.Thestandarderrorofthesamplemeanswhenthestandarddeviationofthepopulationisknownequalsσnwhereσpopulationstandarddeviation.第5題WhichofthefollowingstatementsregardingconfidenceintervalsismostThehigherthealphalevel,thewidertheconfidenceThelowerthealphalevel,thewidertheconfidenceTherelationshipbetweenthealphalevelandtheconfidenceintervalcannotbeThelowerthedegreeofconfidence,thewidertheconfidenceAhigherdegreeofconfidencerequiresawiderconfidenceinterval.Thedegreeofconfidenceisequaltooneminusthealphalevel,andsothewidertheconfidenceinterval,thehigherthedegreeofconfidenceandthelowerthealphalevel.Notethattheloweralphalevelrequiresahigherreliabilityfactorwhichresultsinthewiderconfidenceinterval.Describethepropertiesofpointestimators:(1)distinguishbetweenunbiasedandbiasedestimators,第1Thesamplemeanisaconsistentestimatorofthepopulationmeanbecauseexpectedvalueofthesamplemeanisequaltothepopulationsamplingdistributionofthesamplemeanhasthesmallestvarianceofanyotherunbiasedestimatorsofthepopulationmean.samplemeanprovidesamoreaccurateestimateofthepopulationmeanasthesamplesizeincreases.samplingdistributionofthesamplemeanisAconsistentestimatorprovidesamoreaccurateestimateoftheparameterasthesamplesizeincreases.第2ShawnChoateisthinkingabouthisgraduatethesis.Stillinthepreliminarystage,hewantstochooseavariableofstudythathasthemostdesirablestatisticalproperties.Thestatisticheispresentlyconsideringhasthefollowingcharacteristics:TheexpectedvalueofthesamplemeanisequaltothepopulationThevarianceofthesamplingdistributionissmallerthanthatforotherestimatorsoftheAsthesamplesizeincreases,thestandarderrorofthesamplemeanrisesandthesamplingdistributioniscenteredmorecloselyonthemean.Selectthebestchoice.Choate’sestimatorunbiasedandunbiased,efficient,andefficientandunbiasedandTheestimatorisunbiasedbecausetheexpectedvalueofthesamplemeanisequaltothepopulationmean.Theestimatorisefficientbecausethevarianceofthesamplingdistributionissmallerthanthatforotherestimatorsoftheparameter.Theestimatorisnotconsistent.Tobeconsistent,asthesamplesizeincreases,thestandarderrorofthesamplemeanshouldfallandthesamplingdistributionwillbecenteredmorecloselyonthemean.Aconsistentestimatorprovidesamoreaccurateestimateoftheparameterasthesamplesizeincreases.第3Thesamplemeanisanunbiasedestimatorofthepopulationmeanbecausesamplingdistributionofthesamplemeanhasthesmallestvarianceofanyotherunbiasedestimatorsofthepopulationmean.samplemeanprovidesamoreaccurateestimateofthepopulationmeanasthesamplesizeincreases.expecte