商务与经济统计 第8版 英文版 (戴维 R. 安德森 著) 机械工业出版社 课后答案 461页

'课后答案网-中国第一答案下载门户答案分类爱校园社区淘答案(专业级搜索引擎为你提供服务)大学课后答案高中课后答案公共基础课|通信/电子/电气|计算机/软件/网络/高一课后答案|高二课后答案|高三课后答案信息/数学|物理/光学/声学/热学/力学|经济学/管理初中课后答案学/法学|化学/环境/生物/医学/制药|土建/机械/材初一课后答案|初二课后答案|初三课后答案料/制造|哲学/心理学/政治学|文学/史学/外语/教考试课后答案育|其它类别等级考试类答案|公务员考试答案热门答案最新求助最新答案新视野大学英语读写教高鸿业版西方经济学习概率论与数理统计教程高等数学（第五版）含C程序设计第三版(谭程答案（全）【k题答案(微观.宏观(茆诗松著)高上下册(同济大学浩强著)清华大新视野英语听力原文及理论力学第六版(哈尔线性代数(同济大学应21世纪大学英语第3册复变函数与积分变换答案课后答案【滨工业大学理论用数学系著)高（1-4）答案【khd第四版(张元林西概率与数理统计第二,C语言程序设计教程第西方经济学(微观部分)C语言程序设计教程第复变函数全解及导学[西三版(浙江大学三版(谭浩强张(高鸿业著)中二版(谭浩强张安交大第四版]社区服务社区热点进入社区http://www.khdaw.com/2009-10-15 课后答案网www.khdaw.comSolutionsManualtoAccompanyStatisticsSStatisticstatisticsforffororBusinessBBusinessusinessandananddEconomicsEEconomicsconomicsEighthEditionwww.khdaw.comDavidDDavidavidR.RR..AndersonAAndersonndersonUniversityofCincinnatiDennisDDennisennisJ.SweeneySSweeneyweeneyUniversityofCincinnatiThomasTThomashomasA.AA..WilliamsWiWilliamslliamsRochesterInstituteofTechnology©2002bySouth-Western/ThomsonLearning™Cincinnati,Ohiowww.khdaw.com 课后答案网www.khdaw.comContentsChapterCChapterhapter1.DataandStatistics2.DescriptiveStatistics:TabularandGraphicalApproaches3.DescriptiveStatistics:NumericalMethods4.IntroductiontoProbabilitywww.khdaw.com5.DiscreteProbabilityDistributions6.ContinuousProbabilityDistributions7.SamplingandSamplingDistributions8.IntervalEstimation9.HypothesisTesting10.StatisticalInferenceaboutMeansandProportionsWithTwoPopulations11.InferencesaboutPopulationVariances12.TestsofGoodnessofFitandIndependence13.AnalysisofVarianceandExperimentalDesign14.SimpleLinearRegression15.MultipleRegression16.RegressionAnalysis:ModelBuilding17.IndexNumbers18.Forecasting19.NonparametricMethods20.StatisticalMethodsforQualityControl21.SampleSurveywww.khdaw.com 课后答案网www.khdaw.comPrefaceThepurposeofStatisticsforBusinessandEconomicsistoprovidestudents,primarilyinthefieldsofbusinessadministrationandeconomics,withasoundconceptualintroductiontothefieldofstatisticsanditsmanyapplications.Thetextisapplications-orientedandhasbeenwrittenwiththeneedsofthenonmathematicianinmind.Thesolutionsmanualfurnishesassistancebyidentifyinglearningobjectivesandprovidingwww.khdaw.comdetailedsolutionsforallexercisesinthetext.AcknowledgementsAAcknowledgementscknowledgementsWewouldliketoprovidespecialrecognitiontoCatherineJ.Williamsforhereffortsinpreparingthesolutionsmanual.DavidR.AndersonDennisJ.SweeneyThomasA.Williamswww.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter1DataDDataataandananddStatisticsSStatisticstatisticsLearningLLearningearningObjectivesOObjectivesbjectives1.Obtainanappreciationforthebreadthofstatisticalapplicationsinbusinessandeconomics.2.Understandthemeaningofthetermselements,variables,andobservationsastheyareusedinstatistics.www.khdaw.com3.Obtainanunderstandingofthedifferencebetweenqualitative,quantitative,crossectionalandtimeseriesdata.4.Learnaboutthesourcesofdataforstatisticalanalysisbothinternalandexternaltothefirm.5.Beawareofhowerrorscanariseindata.6.Knowthemeaningofdescriptivestatisticsandstatisticalinference.7.Beabletodistinguishbetweenapopulationandasample.8.Understandtheroleasampleplaysinmakingstatisticalinferencesaboutthepopulation.1-1www.khdaw.com 课后答案网www.khdaw.comChapter1Solutions:SSolutions:olutions:1.Statisticscanbereferredtoasnumericalfacts.Inabroadersense,statisticsisthefieldofstudydealingwiththecollection,analysis,presentationandinterpretationofdata.2.a.9b.4c.Countryandroomratearequalitativevariables;numberofroomsandtheoverallscorearequantitativevariables.d.Countryisnominal;roomrateisordinal;numberofroomsandoverallscoreareratio.3.a.Averagenumberofrooms=808/9=89.78orapproximately90roomswww.khdaw.comb.2of9arelocatedinEngland;approximately22%c.4of9havearoomrateof$$;approximately44%4.a.10b.Fortune500largestU.S.industrialcorporationsc.Averagerevenue=$142,275.9/10=$14,227.59milliond.Usingthesampleaverage,statisticalinferencewouldletusestimatetheaveragerevenueforthepopulationof500corporationsas$14,227.59million.5.a.3b.Industrycodeisqualitative;revenuesandprofitarequantitative.c.Averageprofit=10,652.1/10=$1065.21milliond.8of10hadaprofitover$100million;80%e.1of10hadanindustrycodeof3;10%6.Questionsa,c,anddarequantitative.Questionsbandearequalitative.7.a.Thedataarenumericandthevariableisqualitative.b.Nominal8.a.2,013b.Qualitativec.Percentagessincewehavequalitativedatad.(0.28)(2013)=563.64Musthavebeen563or564.1-2www.khdaw.com 课后答案网www.khdaw.comDataandStatistics9.a.Qualitativeb.30of71;42.3%10.a.Quantitative;ratiob.Qualitative;nominalc.Qualitative(Note:Rankisanumericlabelthatidentifiesthepositionofastudentintheclass.Rankdoesnotindicatehowmuchorhowmanyandisnotquantitative.);ordinald.Qualitative;nominale.Quantitative;ratiowww.khdaw.com11.a.Quantitative;ratiob.Qualitative;ordinalc.Qualitative;ordinal(assumingemployeescanberankedbyclassification)d.Quantitative;ratioe.Qualitative;nominal12.a.ThepopulationisallvisitorscomingtothestateofHawaii.b.Sinceairlineflightscarrythevastmajorityofvisitorstothestate,theuseofquestionnairesforpassengersduringincomingflightsisagoodwaytoreachthispopulation.Thequestionnaireactuallyappearsonthebackofamandatoryplantsandanimalsdeclarationformthatpassengersmustcompleteduringtheincomingflight.Alargepercentageofpassengerscompletethevisitorinformationquestionnaire.c.Questions1and4providequantitativedataindicatingthenumberofvisitsandthenumberofdaysinHawaii.Questions2and3providequalitativedataindicatingthecategoriesofreasonforthetripandwherethevisitorplanstostay.13.a.Quantitativeb.Timeserieswith7observationsc.Numberofriverboatcasinos.d.Timeseriesshowsarapidincrease;anincreasewouldbeexpectedin1998,butitappearsthattherateofincreaseisslowing.14.a.4b.Allfourvariablesarequantitative.c.Timeseriesdatafor1993to1996.15.Crossectionaldata.Itisbasedonthe1996performancedatathatwasavailableApril1997.1-3www.khdaw.com 课后答案网www.khdaw.comChapter116.a.Wewouldliketoseedatafromproducttastetestsandtestmarketingtheproduct.b.Suchdatawouldbeobtainedfromspeciallydesignedstatisticalstudies.17.Internaldataonsalariesofotheremployeescanbeobtainedfromthepersonneldepartment.ExternaldatamightbeobtainedfromtheDepartmentofLabororindustryassociations.18.a.(48/120)100%=40%inthesamplediedfromsomeformofheartdisease.Thiscanbeusedasanestimateofthepercentageofallmales60orolderwhodieofheartdisease.b.Thedataoncauseofdeathisqualitative.19.a.AllsubscribersofBusinessWeekatthetimethe1996surveywasconducted.b.Quantitativewww.khdaw.comc.Qualitative(yesorno)d.Crossectional-1996wasthetimeofthesurvey.e.Usingthesampleresults,wecouldinferorestimate59%ofthepopulationofsubscribershaveanannualincomeof$75,000ormoreand50%ofthepopulationofsubscribershaveanAmericanExpresscreditcard.20.a.56%ofmarketbelongedtoA.C.Nielsen$387,325istheaverageamountspentpercategoryb.3.73c.$387,32521.a.ThetwopopulationsarethepopulationofwomenwhosemotherstookthedrugDESduringpregnancyandthepopulationofwomenwhosemothersdidnottakethedrugDESduringpregnancy.b.Itwasasurvey.c.63/3.980=15.8womenoutofeach1000developedtissueabnormalities.d.Thearticlereported“twice”asmanyabnormalitiesinthewomenwhosemothershadtakenDESduringpregnancy.Thus,aroughestimatewouldbe15.8/2=7.9abnormalitiesper1000womenwhosemothershadnottakenDESduringpregnancy.e.Inmanysituations,diseaseoccurrencesarerareandaffectonlyasmallportionofthepopulation.Largesamplesareneededtocollectdataonareasonablenumberofcaseswherethediseaseexists.22.a.AlladultviewersreachedbytheDenver,Coloradotelevisionstation.b.Theviewerscontactedinthetelephonesurvey.c.Asample.Itwouldclearlybetoocostlyandtimeconsumingtotrytocontactallviewers.23.a.Percentoftelevisionsetsthatweretunedtoaparticulartelevisionshowand/ortotalviewingaudience.1-4www.khdaw.com 课后答案网www.khdaw.comDataandStatisticsb.AlltelevisionsetsintheUnitedStateswhichareavailablefortheviewingaudience.Notethiswouldnotincludetelevisionsetsinstoredisplays.c.Aportionofthesetelevisionsets.Generally,individualhouseholdswouldbecontactedtodeterminewhichprogramswerebeingviewed.d.Thecancellationofprograms,theschedulingofprograms,andadvertisingcostrates.24.a.Thisisastatisticallycorrectdescriptivestatisticforthesample.b.Anincorrectgeneralizationsincethedatawasnotcollectedfortheentirepopulation.c.Anacceptablestatisticalinferencebasedontheuseoftheword“estimate.”d.Whilethisstatementistrueforthesample,itisnotajustifiableconclusionfortheentirepopulation.e.Thisstatementisnotstatisticallysupportable.Whileitistruefortheparticularsampleobserved,itwww.khdaw.comisentirelypossibleandevenverylikelythatatleastsomestudentswillbeoutsidethe65to90rangeofgrades.1-5www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter2DescriptiveDDescriptiveescriptiveStatistics:SStatistics:tatistics:TabularTTabularabularandananddGraphicalGGraphicalraphicalMethodsMMethodsethodsLearningLLearningearningObjectivesOObjectivesbjectives1.Learnhowtoconstructandinterpretsummarizationproceduresforqualitativedatasuchas:frequencyandrelativefrequencydistributions,bargraphsandpiecharts.2.Learnhowtoconstructandinterprettabularsummarizationproceduresforquantitativedatasuchas:www.khdaw.comfrequencyandrelativefrequencydistributions,cumulativefrequencyandcumulativerelativefrequencydistributions.3.Learnhowtoconstructadotplot,ahistogram,andanogiveasgraphicalsummariesofquantitativedata.4.Beabletouseandinterprettheexploratorydataanalysistechniqueofastem-and-leafdisplay.5.Learnhowtoconstructandinterpretcrosstabulationsandscatterdiagramsofbivariatedata.2-1www.khdaw.com 课后答案网www.khdaw.comChapter2Solutions:SSolutions:olutions:1.ClassFrequencyRelativeFrequencyA6060/120=0.50B2424/120=0.20C3636/120=0.301201.002.a.1-(.22+.18+.40)=.20b..20(200)=40c/dClassFrequencyPercentFrequencyA.22(200)=4422www.khdaw.comB.18(200)=3618C.40(200)=8040D.20(200)=4020Total2001003.a.360°x58/120=174°b.360°x42/120=126°c.NoOpinion16.7%Yes48.3%No35%2-2www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsd.7060504030Frequency20www.khdaw.com100YesNoNoOpinionResponse4.a.Thedataarequalitative.b.PercentTVShowFrequencyFrequencyMillionaire2448Frasier1530ChicagoHope714Charmed48Total:501002-3www.khdaw.com 课后答案网www.khdaw.comChapter2c.30252015Frequency10www.khdaw.com50MillionaireFrasierChicagoCharmedTVShowCharmed8%Chicago14%Millionaire48%Frasier30%d.Millionairehasthelargestmarketshare.Frasierissecond.5.a.MajorRelativeFrequencyPercentFrequencyManagement55/216=0.2525Accounting51/216=0.2424Finance28/216=0.1313Marketing82/216=0.3838Total1.001002-4www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsb.908070605040Frequency30www.khdaw.com20100ManagementAccountingFinanceMarketingMajorc.PieChartManagementAccounting25%24%Finance13%Marketing38%6.a.BookFrequencyPercentFrequency7Habits1016.66Millionaire1626.67Motley915.00Dad1321.67WSJGuide610.00Other610.00Total:60100.002-5www.khdaw.com 课后答案网www.khdaw.comChapter2TheErnst&YoungTaxGuide2000withafrequencyof3,InvestingforDummieswithafrequencyof2,andWhatColorisYourParachute?2000withafrequencyof1aregroupedinthe"Other"category.b.Therankorderfromfirsttofifthis:Millionaire,Dad,7Habits,Motley,andWSJGuide.c.ThepercentofsalesrepresentedbyTheMillionaireNextDoorandRichDad,PoorDadis48.33%.7.RatingFrequencyRelativeFrequencyOutstanding190.38VeryGood130.26Good100.20Average60.12Poor20.04www.khdaw.com501.00Managementshouldbepleasedwiththeseresults.64%oftheratingsareverygoodtooutstanding.84%oftheratingsaregoodorbetter.Comparingtheseratingswithpreviousresultswillshowwhetherornottherestaurantismakingimprovementsinitsratingsoffoodquality.8.a.PositionFrequencyRelativeFrequencyPitcher170.309Catcher40.0731stBase50.0912ndBase40.0733rdBase20.036Shortstop50.091LeftField60.109CenterField50.091RightField70.127551.000b.Pitchers(Almost31%)c.3rdBase(3-4%)d.RightField(Almost13%)e.Infielders(16or29.1%)toOutfielders(18or32.7%)9.a/b.StartingTimeFrequencyPercentFrequency7:003157:304208:004208:307359:00210201002-6www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsc.BarGraph876543Frequencywww.khdaw.com2107:007:308:008:309:00StartingTimed.9:007:0010%15%7:3020%8:3035%8:0020%e.Themostpreferredstartingtimeis8:30a.m..Startingtimesof7:30and8:00a.m.arenext.10.a.Thedatarefertoqualitylevelsofpoor,fair,good,verygoodandexcellent.b.RatingFrequencyRelativeFrequencyPoor20.03Fair40.07Good120.20VeryGood240.40Excellent180.30601.002-7www.khdaw.com 课后答案网www.khdaw.comChapter2c.BarGraph30252015Frequency105www.khdaw.com0PoorFairGoodVeryGoodExcellentRatingPieChartGoodFair20%7%Poor3%ExcellentVeryGood30%40%d.Thecourseevaluationdataindicateahighqualitycourse.Themostcommonratingisverygoodwiththesecondmostcommonbeingexcellent.11.ClassFrequencyRelativeFrequencyPercentFrequency12-1420.0505.015-1780.20020.018-20110.27527.521-23100.25025.524-2690.22522.5Total401.000100.02-8www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods12.ClassCumulativeFrequencyCumulativeRelativeFrequencylessthanorequalto1910.20lessthanorequalto2924.48lessthanorequalto3941.82lessthanorequalto4948.96lessthanorequalto59501.0013.1816www.khdaw.com1412108Frequency642010-1920-2930-3940-4950-591.0.8.6.4.201020304050602-9www.khdaw.com 课后答案网www.khdaw.comChapter214.a.6.08.010.012.014.016.0b/c.ClassFrequencyPercentFrequency6.0-7.94208.0-9.921010.0-11.984012.0-13.931514.0-15.931520100www.khdaw.com15.a/b.WaitingTimeFrequencyRelativeFrequency0-440.205-980.4010-1450.2515-1920.1020-2410.05Totals201.00c/d.WaitingTimeCumulativeFrequencyCumulativeRelativeFrequencyLessthanorequalto440.20Lessthanorequalto9120.60Lessthanorequalto14170.85Lessthanorequalto19190.95Lessthanorequalto24201.00e.12/20=0.6016.a.RelativePercentStockPrice($)FrequencyFrequencyFrequency10.00-19.99100.404020.00-29.9940.161630.00-39.9960.242440.00-49.9920.08850.00-59.9910.04460.00-69.9920.088Total251.001002-10www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods121086Frequency42www.khdaw.com010.00-20.00-30.00-40.00-50.00-60.00-19.9929.9939.9949.9959.9969.99StockPriceManyofthesearelowpricedstockswiththegreatestfrequencyinthe$10.00to$19.99range.b.EarningsperRelativePercentShare($)FrequencyFrequencyFrequency-3.00to-2.0120.088-2.00to-1.0100.000-1.00to-0.0120.0880.00to0.9990.36361.00to1.9990.36362.00to2.9930.1212Total251.001002-11www.khdaw.com 课后答案网www.khdaw.comChapter210987654Frequency3210-3.00to-2.00to-1.00to0.00to1.00to2.00towww.khdaw.com-2.01-1.01-0.010.991.992.99EarningsperShareThemajorityofcompanieshadearningsinthe$0.00to$2.00range.Fourofthecompanieslostmoney.17.CallDurationFrequencyRelativeFrequency2-3.950.254-5.990.456-7.940.208-9.900.0010-11.920.10Totals201.00Histogram10987654Frequency32102.0-3.94.0-5.96.0-7.98.0-9.910.0-11.9CallDuration2-12www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods18.a.Lowestsalary:$93,000Highestsalary:$178,000b.Salary($1000s)RelativePercentFrequencyFrequencyFrequency91-10540.088106-12050.1010121-135110.2222136-150180.3636151-16590.1818166-18030.066Total501.00100c.Proportion$135,000orless:20/50.www.khdaw.comd.Percentagemorethan$150,000:24%e.2018161412108Frequency642091-105106-120121-135136-150151-165166-180Salary($1000s)19.a/b.NumberFrequencyRelativeFrequency140-14920.10150-15970.35160-16930.15170-17960.30180-18910.05190-19910.05Totals201.002-13www.khdaw.com 课后答案网www.khdaw.comChapter2c/d..NumberCumulativeFrequencyCumulativeRelativeFrequencyLessthanorequalto14920.10Lessthanorequalto15990.45Lessthanorequalto169120.60Lessthanorequalto179180.90Lessthanorequalto189190.95Lessthanorequalto199201.00e.20www.khdaw.com15Frequency10514016018020020.a.Thepercentageofpeople34orlessis20.0+5.7+9.6+13.6=48.9.b.Thepercentageofthepopulationthatisbetween25and54yearsoldinclusivelyis13.6+16.3+13.5=43.4c.Thepercentageofthepopulationover34yearsoldis16.3+13.5+8.7+12.6=51.1d.Thepercentagelessthan25yearsoldis20.0+5.7+9.6=35.3.Sothereare(.353)(275)=97.075millionpeoplelessthan25yearsold.e.Anestimateofthenumberofretiredpeopleis(.5)(.087)(275)+(.126)(275)=46.6125million.21.a/b.ComputerRelativeUsage(Hours)FrequencyFrequency0.0-2.950.103.0-5.9280.566.0-8.980.169.0-11.960.1212.0-14.930.06Total501.002-14www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsc.30252015Frequency105www.khdaw.com00.0-2.93.0-5.96.0-8.99.0-11.912.0-14.9ComputerUsage(Hours)d.60504030Frequency201003691215ComputerUsage(Hours)e.Themajorityofthecomputerusersareinthe3to6hourrange.Usageissomewhatskewedtowardtherightwith3usersinthe12to15hourrange.2-15www.khdaw.com. 课后答案网www.khdaw.comChapter222.5786458702255688023523.LeafUnit=0.1637557www.khdaw.com813489361004511324.LeafUnit=101161202130671422715516028170232-16www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods25.989102466114578891224571312144www.khdaw.com15126.LeafUnit=0.104789911292001355683494856712-17www.khdaw.com 课后答案网www.khdaw.comChapter227.41366750038960114457799700013445566678880113445778990227orwww.khdaw.com413466750035896011446577997000134475566678880113448577899022972-18www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods28.a.05811133441567899223335526833677940www.khdaw.com4785560b.2000P/EPercentForecastFrequencyFrequency5-926.710-14620.015-19620.020-24620.025-2926.730-3400.035-39413.340-4413.345-4926.750-5400.055-5900.060-6413.3Total30100.029.a.y12TotalA505xB11213C21012Total1812302-19www.khdaw.com 课后答案网www.khdaw.comChapter2b.y12TotalA100.00.0100.0xB84.615.4100.0C16.783.3100.0c.www.khdaw.comy12A27.80.0xB61.116.7C11.183.3Total100.0100.0d.CategoryAvaluesforxarealwaysassociatedwithcategory1valuesfory.CategoryBvaluesforxareusuallyassociatedwithcategory1valuesfory.CategoryCvaluesforxareusuallyassociatedwithcategory2valuesfory.30.a.2-20www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods5640248y-8-24-40www.khdaw.com-40-30-20-10010203040xb.Thereisanegativerelationshipbetweenxandy;ydecreasesasxincreases.31.MealPrice($)QualityRating10-1920-2930-3940-49Good53.833.92.70.0VeryGood43.654.260.521.4Excellent2.611.936.878.6Total100.0100.0100.0100.0Asthemealpricegoesup,thepercentageofhighqualityratingsgoesup.Apositiverelationshipbetweenmealpriceandqualityisobserved.32.a.EPSRatingSales/Margins/ROE0-1920-3940-5960-7980-100TotalA189B145212C11237D3115E213Total44691336b.EPSRatingSales/Margins/ROE0-1920-3940-5960-7980-100TotalA11.1188.89100B8.3333.3341.6716.67100C14.2914.2928.5742.86100D60.0020.0020.00100E66.6733.33100HigherEPSratingsseemtobeassociatedwithhigherratingsonSales/Margins/ROE.Ofthosecompanieswithan"A"ratingonSales/Margins/ROE,88.89%ofthemhadanEPSRatingof80or2-21www.khdaw.com 课后答案网www.khdaw.comChapter2higher.Ofthe8companieswitha"D"or"E"ratingonSales/Margins/ROE,only1hadanEPSratingabove60.33.a.IndustryGroupRelativeStrengthSales/Margins/ROEABCDETotalA12249B1523112C13217D11125E123Total411710436b/c.ThefrequencydistributionsfortheSales/Margins/ROEdataisintherightmostcolumnofthecrosstabulation.ThefrequencydistributionfortheIndustryGroupRelativeStrengthdataisinthewww.khdaw.combottomrowofthecrosstabulation.d.Oncethecrosstabulationiscomplete,theindividualfrequencydistributionsareavailableinthemargins.34.a.80706050403020RelativePriceStrength100020406080100120EPSRatingb.OnemightexpectstockswithhigherEPSratingstoshowgreaterrelativepricestrength.However,thescatterdiagramusingthisdatadoesnotsupportsucharelationship.Thescatterdiagramappearssimilartotheoneshowing"NoApparentRelationship"inFigure2.19.35.a.2-22www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods900.0800.0700.0600.0500.0400.0300.0GamingRevenue200.0www.khdaw.com100.00.00.0100.0200.0300.0400.0500.0600.0700.0800.0HotelRevenueb.Thereappearstobeapositiverelationshipbetweenhotelrevenueandgamingrevenue.Highervaluesofhotelrevenueareassociatedwithhighervaluesofgamingrevenue.36.a.VehicleFrequencyPercentFrequencyF-Series1734Silverado1224Taurus816Camry714Accord612Total50100b.ThetwotopsellingvehiclesaretheFordF-SeriesPickupandtheChevroletSilverado.Accord12%F-SeriesCamry34%14%Taurus16%Silverado2-2324%www.khdaw.com 课后答案网www.khdaw.comChapter2c.37.a/b.IndustryFrequencyPercentFrequencyBeverage210Chemicals315Electronics630Food735Aerospace210Totals:20100www.khdaw.com876543Frequency210BeverageChemicalsElectronicsFoodAerospaceIndustryc.38.a.MovieFrequencyPercentFrequencyBlairWitchProject15936.0PhantomMenace8920.2Beloved8519.3PrimaryColors5712.9TrumanShow5111.62-24www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsTotal441100.0b.Truman12%ColorsWitch13%36%www.khdaw.comBeloved19%Phantom20%c.Thepercentofmailpertainingto1999coverstoriesis36.0+20.2=56.2%39.a-d.CumulativeRelativeCumulativeRelativeSalesFrequencyFrequencyFrequencyFrequency0-499130.65130.65500-99930.15160.801000-149900.00160.801500-199930.15190.952000-249910.05201.00Total201.00e.2-25www.khdaw.com 课后答案网www.khdaw.comChapter214121086Frequency4www.khdaw.com200-499500-9991000-14991500-19992000-2499Sales40.a.ClosingPriceFrequencyRelativeFrequency0-97/890.22510-197/8100.25020-297/850.12530-397/8110.27540-497/820.05050-597/820.05060-697/800.00070-797/810.025Totals401.000b.ClosingPriceCumulativeFrequencyCumulativeRelativeFrequencyLessthanorequalto97/890.225Lessthanorequalto197/8190.475Lessthanorequalto297/8240.600Lessthanorequalto397/8350.875Lessthanorequalto497/8370.925Lessthanorequalto597/8390.975Lessthanorequalto697/8390.975Lessthanorequalto797/8401.000c.2-26www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods121086Frequency42www.khdaw.com0515253545556575ClosingPriced.Over87%ofcommonstockstradeforlessthan$40ashareand60%tradeforlessthan$30pershare.41.a.RelativeExchangeFrequencyFrequencyAmerican30.15NewYork20.10OvertheCounter150.75201.00b.EarningsPerRelativeShareFrequencyFrequency0.00-0.1970.350.20-0.3970.350.40-0.5910.050.60-0.7930.150.80-0.9920.10201.00Seventypercentoftheshadowstockshaveearningspersharelessthan$0.40.ItlookslikelowEPSshouldbeexpectedforshadowstocks.Price-EarningRelativeRatioFrequencyFrequency0.00-9.930.1510.0-19.970.3520.0-29.940.2030.0-39.930.1540.0-49.920.1050.0-59.910.05201.002-27www.khdaw.com 课后答案网www.khdaw.comChapter2P-ERatiosvaryconsiderably,butthereisasignificantclusterinthe10-19.9range.42.RelativeIncome($)FrequencyFrequency18,000-21,999130.25522,000-25,999200.39226,000-29,999120.23530,000-33,99940.07834,000-37,99920.039Total511.00025www.khdaw.com201510Frequency5018,000-21,99922,000-25,99926,000-29,99930,000-33,99934,000-37,999PerCapitaIncome43.a.2-28www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethods0891022234441556666778888999201222344425683013b/c/d.NumberAnsweredRelativeCumulativeCorrectlyFrequencyFrequencyFrequencywww.khdaw.com5-920.050210-1480.2001015-19150.3752520-2490.2253425-2930.0753730-3430.07540Totals401.000e.Relativelyfewofthestudents(25%)wereabletoanswer1/2ormoreofthequestionscorrectly.ThedataseemtosupporttheJointCouncilonEconomicEducation’sclaim.However,thedegreeofdifficultyofthequestionsneedstobetakenintoaccountbeforereachingafinalaconclusion.44.a/b.HighTemperatureLowTemperature339443685750002445579614444686187357972455801146890239c.Itisclearthattherangeoflowtemperaturesisbelowtherangeofhightemperatures.Lookingatthestem-and-leafdisplayssidebyside,itappearsthattherangeoflowtemperaturesisabout20degreesbelowtherangeofhightemperatures.d.Therearetwostemsshowinghightemperaturesof80degreesorhigher.Theyshow8citieswithhightemperaturesof80degreesorhigher.e.Frequency2-29www.khdaw.com 课后答案网www.khdaw.comChapter2TemperatureHighTemp.Low.Temp.30-390140-490350-5911060-697270-794480-895090-9930Total202045.a.8075www.khdaw.com706560555045LowTemperature403530405060708090100HighTemperatureb.Thereisclearlyapositiverelationshipbetweenhighandlowtemperatureforcities.Asonegoesupsodoestheother.46.a.SatisfactionScoreOccupation30-3940-4950-5960-6970-7980-89TotalCabinetmaker243110Lawyer1521110PhysicalTherapist521210SystemsAnalyst214310Total1710118340b.SatisfactionScoreOccupation30-3940-4950-5960-6970-7980-89TotalCabinetmaker20403010100Lawyer1050201010100PhysicalTherapist50201020100SystemsAnalyst201040301002-30www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsc.Eachrowofthepercentcrosstabulationshowsapercentfrequencydistributionforanoccupation.Cabinetmakersseemtohavethehigherjobsatisfactionscoreswhilelawyersseemtohavethelowest.Fiftypercentofthephysicaltherapistshavemediocrescoresbuttherestareratherhigh.47.a.40,00035,00030,00025,00020,000www.khdaw.com15,000Revenue$mil10,0005,0000010,00020,00030,00040,00050,00060,00070,00080,00090,000100,000Employeesb.Thereappearstobeapositiverelationshipbetweennumberofemployeesandrevenue.Asthenumberofemployeesincreases,annualrevenueincreases.48.a.FuelTypeYearConstructedElecNat.GasOilPropaneOtherTotal1973orbefore4018312572471974-19792426220541980-19863738106821987-19914870201121Total14931717714504b.YearConstructedFrequencyFuelTypeFrequency1973orbefore247Electricity1491974-197954Nat.Gas3171980-198682Oil171987-1991121Propane7Total504Other14Total5042-31www.khdaw.com 课后答案网www.khdaw.comChapter2c.CrosstabulationofColumnPercentagesFuelTypeYearConstructedElecNat.GasOilPropaneOther1973orbefore26.957.770.571.450.01974-197916.18.211.828.60.01980-198624.812.05.90.042.91987-199132.222.111.80.07.1Total100.0100.0100.0100.0100.0d.Crosstabulationofrowpercentages.FuelTypeYearConstructedElecNat.GasOilPropaneOtherTotal1973orbefore16.274.14.92.02.8100.01974-197944.548.13.73.70.0100.0www.khdaw.com1980-198645.146.41.20.07.3100.01987-199139.757.81.70.00.8100.0e.ObservationsfromthecolumnpercentagescrosstabulationForthosebuildingsusingelectricity,thepercentagehasnotchangedgreatlyovertheyears.Forthebuildingsusingnaturalgas,themajoritywereconstructedin1973orbefore;thesecondlargestpercentagewasconstructedin1987-1991.Mostofthebuildingsusingoilwereconstructedin1973orbefore.Allofthebuildingsusingpropaneareolder.ObservationsfromtherowpercentagescrosstabulationMostofthebuildingsintheCG&Eserviceareauseelectricityornaturalgas.Intheperiod1973orbeforemostusednaturalgas.From1974-1986,itisfairlyevenlydividedbetweenelectricityandnaturalgas.Since1987almostallnewbuildingsareusingelectricityornaturalgaswithnaturalgasbeingtheclearleader.49.a.Crosstabulationforstockholder"sequityandprofit.Profits($000)Stockholders"Equity($000)0-200200-400400-600600-800800-10001000-1200Total0-12001011121200-24004102162400-3600433111133600-48001234800-60002316Total1816624450b.CrosstabulationofRowPercentages.Profits($000)Stockholders"Equity($1000s)0-200200-400400-600600-800800-10001000-1200Total0-120083.338.330.000.000.008.331001200-240025.0062.500.000.0012.500.001002400-360030.7723.0823.087.697.697.691003600-48000.000.000.0033.3366.671004800-60000.0033.3350.0016.670.000.001002-32www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:TabularandGraphicalMethodsc.Stockholder"sequityandprofitseemtoberelated.Asprofitgoesup,stockholder"sequitygoesup.Therelationship,however,isnotverystrong.50.a.Crosstabulationofmarketvalueandprofit.Profit($1000s)MarketValue($1000s)0-300300-600600-900900-1200Total0-8000234278000-1600044221216000-24000211424000-32000121432000-40000213Total27136450b.CrosstabulationofRowPercentages.Profit($1000s)www.khdaw.comMarketValue($1000s)0-300300-600600-900900-1200Total0-800085.1914.810.000.001008000-1600033.3333.3316.6716.6710016000-240000.0050.0025.0025.0010024000-320000.0025.0050.0025.0010032000-400000.0066.6733.330.00100c.ThereappearstobeapositiverelationshipbetweenProfitandMarketValue.Asprofitgoesup,MarketValuegoesup.51.a.ScatterdiagramofProfitvs.Stockholder"sEquity.1400.01200.01000.0800.0600.0Profit($1000s)400.0200.00.00.01000.02000.03000.04000.05000.06000.07000.0Stockholder"sEquity($1000s)b.ProfitandStockholder"sEquityappeartobepositivelyrelated.2-33www.khdaw.com 课后答案网www.khdaw.comChapter252.a.ScatterdiagramofMarketValueandStockholder"sEquity.45000.040000.035000.030000.025000.020000.0www.khdaw.com15000.010000.0MarketValue($1000s)5000.00.00.01000.02000.03000.04000.05000.06000.07000.0Stockholder"sEquity($1000s)b.ThereisapositiverelationshipbetweenMarketValueandStockholder"sEquity.2-34www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter3DescriptiveDDescriptiveescriptiveStatistics:SStatistics:tatistics:NumericalNNumericalumericalMethodsMMethodsethodsLearningLLearningearningObjectivesOObjectivesbjectives1.Understandthepurposeofmeasuresoflocation.2.Beabletocomputethemean,median,mode,quartiles,andvariouspercentiles.www.khdaw.com3.Understandthepurposeofmeasuresofvariability.4.Beabletocomputetherange,interquartilerange,variance,standarddeviation,andcoefficientofvariation.5.Understandhowzscoresarecomputedandhowtheyareusedasameasureofrelativelocationofadatavalue.6.KnowhowChebyshev’stheoremandtheempiricalrulecanbeusedtodeterminethepercentageofthedatawithinaspecifiednumberofstandarddeviationsfromthemean.7.Learnhowtoconstructa5-numbersummaryandaboxplot.8.Beabletocomputeandinterpretcovarianceandcorrelationasmeasuresofassociationbetweentwovariables.9.Beabletocomputeaweightedmean.3-1www.khdaw.com 课后答案网www.khdaw.comChapter3Solutions:SSolutions:olutions:Σx75i1.x===15n510,12,16,17,20Median=16(middlevalue)Σx96i2.x===16n610,12,16,17,20,211617+Median==16.5www.khdaw.com23.15,20,25,25,27,28,30,3220i=(8)1.6=2ndposition=20100252025+i=(8)=2=22.5100265i=(8)=5.26thposition=28100752830+i=(8)=6=291002Σx657i4.Mean===59.727n11Median=576thitemMode=53Itappears3timesΣx1106.4i5.a.x===36.88n30b.Thereareanevennumberofitems.Thus,themedianistheaverageofthe15thand16thitemsafterthedatahavebeenplacedinrankorder.36.636.7+Median==36.652c.Mode=36.4Thisvalueappears4times3-2www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodsF25Id.FirstQuartilei=HGKJ30=7.5100Roundingup,weseethatQ1isatthe8thposition.Q1=36.2F75Ie.ThirdQuartilei=HGKJ30=22.5100Roundingup,weseethatQ3isatthe23rdposition.Q3=37.9Σx1845i6.a.x===92.25www.khdaw.comn20Medianisaverageof10thand11thvaluesafterarranginginascendingorder.6695+Median==80.52DataaremultimodalΣx1334ib.x===66.7n206670+Median==682Mode=70(4brokerscharge$70)c.Comparingallthreemeasuresofcentrallocation(mean,medianandmode),weconcludethatitcostsmore,onaverage,totrade500sharesat$50pershare.d.Yes,trading500sharesat$50pershareisatransactionvalueof$25,000whereastrading1000sharesat$5pershareisatransactionvalueof$5000.Σx1380i7.a.x===46n30b.Yes,themeanhereis46minutes.Thenewspaperreportedonaverageof45minutes.45+52.9c.Median==48.952d.Q1=7(valueof8thiteminrankedorder)Q3=70.4(valueof23rditeminrankedlist)⎛40⎞e.Findpositioni=⎜⎟3012;=40thpercentileisaverageofvaluesin12thand13thpositions.⎝100⎠3-3www.khdaw.com 课后答案网www.khdaw.comChapter340thpercentile=28.8+29.1=28.9528.a.Σx=775iΣx775ix===38.75n20Themodalageis29;itappears3times.b.Medianisaverageof10thand11thitems.37+40Median==38.52www.khdaw.comDatasuggestat-homeworkersareslightlyyounger.c.ForQ1,⎛25⎞i=⎜⎟20=5⎝100⎠Sinceiisinteger,2930+Q==29.512ForQ3,⎛75⎞i=⎜⎟2015=⎝100⎠Sinceiisinteger,4649+Q==47.532⎛32⎞d.i=⎜⎟20=6.4⎝100⎠Sinceiisnotaninteger,werounduptothe7thposition.32ndpercentile=31Σx270,377i9.a.x===10,815.08Median(Position13)=8296n25b.Medianwouldbebetterbecauseoflargedatavalues.c.i=(25/100)25=6.253-4www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodsQ1(Position7)=5984i=(75/100)25=18.75Q3(Position19)=14,330d.i=(85/100)25=21.2585thpercentile(position22)=15,593.Approximately85%ofthewebsiteshavelessthan15,593uniquevisitors.10.a.Σxi=435Σx435ix===48.33www.khdaw.comn9Datainascendingorder:284245484950555860Median=49Donotreportamode;eachdatavalueoccursonce.Theindexcouldbeconsideredgoodsinceboththemeanandmedianarelessthan50.⎛25⎞b.i=⎜⎟9=2.25⎝100⎠Q1(3rdposition)=45⎛75⎞i=⎜⎟9=6.75⎝100⎠Q3(7thposition)=5552611.x==26.32015161819202122222424�����26262727303133333458Median=25Donotreportamodesincefivevaluesappeartwice.ForQ1,⎛25⎞i=⎜⎟20=5⎝100⎠3-5www.khdaw.com 课后答案网www.khdaw.comChapter32021+Q==20.512ForQ3,⎛75⎞i=⎜⎟2015=⎝100⎠3031+Q==30.53212.Usingthemeanwegetx=15.58,x=18.92citycountrywww.khdaw.comForthesamplesweseethatthemeanmileageisbetterinthecountrythaninthecity.City13.214.415.215.315.315.315.91616.116.216.216.716.8↑MedianMode:15.3Country17.217.418.318.518.618.618.719.019.219.419.420.621.1↑MedianMode:18.6,19.4Themedianandmodalmileagesarealsobetterinthecountrythaninthecity.13.a.Mean=261/15=17.4141515151616171818181819202121↑MedianModeis18(occurs4times)Interpretation:theaveragenumberofcredithourstakenwas17.4.Atleast50%ofthestudentstook18ormorehours;atleast50%ofthestudentstook18orfewerhours.Themostfrequentlyoccurringnumberofcredithourstakenwas18.b.ForQ1,⎛25⎞i=⎜⎟15=3.75⎝100⎠Q1(4thposition)=153-6www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodsForQ3,⎛75⎞i=⎜⎟1511.25=⎝100⎠Q3(12thposition)=19c.Forthe70thpercentile,⎛70⎞i=⎜⎟1510.5=www.khdaw.com⎝100⎠Roundingupweseethe70thpercentileisinposition11.70thpercentile=18Σx12,780i14.a.x===$639n20Σx1976ib.x===98.8picturesn20Σx2204ic.x===110.2minutesn20d.Thisisnotaneasychoicebecauseitisamulticriteriaproblem.Ifpricewastheonlycriterion,thelowestpricecamera(FujifilmDX-10)wouldbepreferred.Ifmaximumpicturecapacitywastheonlycriterion,themaximumpicturecapacitycamera(KodakDC280Zoom)wouldbepreferred.But,ifbatterylifewastheonlycriterion,themaximumbatterylifecamera(FujifilmDX10)wouldbepreferred.Therearemanyapproachesusedtoselectthebestchoiceinamulticriteriasituation.Theseapproachesarediscussedinmorespecializedbooksondecisionanalysis.15.Range20-10=1010,12,16,17,2025i=(5)1.25=100Q1(2ndposition)=1275i=(5)=3.75100Q3(4thposition)=17IQR=Q3-Q1=17-12=53-7www.khdaw.com 课后答案网www.khdaw.comChapter3Σx75i16.x===15n52Σ(x−x)642is===16n−14s=16=417.15,20,25,25,27,28,30,34Range=34-15=19252025+i=(8)=2Q==22.51www.khdaw.com1002752830+i=(8)=6Q==2911002IQR=Q3-Q1=29-22.5=6.5Σx204ix===25.5n82Σ(x−x)2422is===34.57n−17s=34.57=5.8818.a.Range=190-168=222b.Σ(x−x)=376i2s=376=75.25c.s=75.2=8.67⎛8.67⎞d.CoefficientofVariation=⎜⎟100=4.87⎝178⎠19.Range=92-67=25IQR=Q3-Q1=80-77=3x=78.46672∑(xi−x)=411.733322∑(xi−x)411.7333s===29.4095n−1143-8www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodss=29.4095=5.423120.a.Range=60-28=32IQR=Q3-Q1=55-45=10435b.x==48.3392Σ(x−x)=742i22Σ(xi−x)742s===92.75www.khdaw.comn−18s=92.75=9.63c.Theaverageairqualityisaboutthesame.But,thevariabilityisgreaterinAnaheim.200021.x==4005xxx−x2ii(xi−x)4104001010042040020400390400-1010040040000380400-204002000100022Σ(xi−x)1000s===250n−14s=250=15.8122.DawsonSupply:Range=11-9=24.1s==0.679J.C.Clark:Range=15-7=860.1s==2.58923.a.WinterRange=21-12=9IQR=Q3-Q1=20-16=4SummerRange=38-18=203-9www.khdaw.com 课后答案网www.khdaw.comChapter3IQR=Q3-Q1=29-18=11b.VarianceStandardDeviationWinter8.23332.8694Summer44.48896.6700c.Winter⎛s⎞⎛2.8694⎞CoefficientofVariation=⎜⎟100=⎜⎟10016.21=⎝x⎠⎝17.7⎠Summerwww.khdaw.com⎛s⎞⎛6.6700⎞CoefficientofVariation=⎜⎟100=⎜⎟100=26.05⎝x⎠⎝25.6⎠d.Morevariabilityinthesummermonths.24.a.500Sharesat$50MinValue=34MaxValue=195Range=195-34=1614550+140140+Q==47.5Q==1401322Interquartilerange=140-47.5=92.51000Sharesat$5MinValue=34MaxValue=90Range=90-34=566060.5+79.580+Q==60.25Q==79.751322Interquartilerange=79.75-60.25=19.5b.500Sharesat$5022Σ(xi−x)51,402.25s===2705.3816n−119s=2705.3816=52.011000Sharesat$53-10www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods22Σ(xi−x)5526.2s===290.8526n−119s=290.8526=17.05c.500Sharesat$50s52.01CoefficientofVariation=(100)=(100)=56.38x92.251000Sharesat$5s17.05CoefficientofVariation=(100)=(100)=25.56x66.70d.Thevariabilityisgreaterforthetradeof500sharesat$50pershare.Thisistruewhetherweusewww.khdaw.comthestandarddeviationorthecoefficientofvariationasameasure.25.s2=0.0021Productionshouldnotbeshutdownsincethevarianceislessthan.005.26.Quartermilerss=0.0564CoefficientofVariation=(s/x)100=(0.0564/0.966)100=5.8Milerss=0.1295CoefficientofVariation=(s/x)100=(0.1295/4.534)100=2.9Yes;thecoefficientofvariationshowsthatasapercentageofthemeanthequartermilers’timesshowmorevariability.4030−⎛1⎞27.a.z==21⎜−⎟=0.75Atleast75%25⎝2⎠4530−⎛1⎞b.z==31⎜−⎟=0.89Atleast89%25⎝3⎠3830−⎛1⎞c.z==1.61⎜−⎟=0.61Atleast61%25⎝1.6⎠4230−⎛1⎞d.z==2.41⎜−⎟=0.83Atleast83%25⎝2.4⎠4830−⎛1⎞e.z==3.61⎜−⎟=0.92Atleast92%25⎝3.6⎠28.a.Approximately95%3-11www.khdaw.com 课后答案网www.khdaw.comChapter3b.Almostallc.Approximately68%Σx75i29.x===15n52Σ(x−x)642is===4n−141015−10z==−1.2542015−20z==+1.254www.khdaw.com1215−12z==−0.7541715−17z==+.5041615−16z==+.254520500−30.z==+.20100650500−z==+1.50100500500−z==0.00100450500−z==−0.50100280500−z==−2.2010031.a.Thisisfrom2standarddeviationsbelowthemeanto2standarddeviationsabovethemean.Withz=2,Chebyshev’stheoremgives:11131−=1−=1−=22z244Therefore,atleast75%ofadultssleepbetween4.5and9.3hoursperday.b.Thisisfrom2.5standarddeviationsbelowthemeanto2.5standarddeviationsabovethemean.Withz=2.5,Chebyshev’stheoremgives:3-12www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods1111−=−1=−1=.8422z2.56.25Therefore,atleast84%ofadultssleepbetween3.9and9.9hoursperday.c.Withz=2,theempiricalrulesuggeststhat95%ofadultssleepbetween4.5and9.3hoursperday.TheprobabilityobtainedusingtheempiricalruleisgreaterthantheprobabilityobtainedusingChebyshev’stheorem.32.a.2hoursis1standarddeviationbelowthemean.Thus,theempiricalrulesuggeststhat68%ofthekidswatchtelevisionbetween2and4hoursperday.Sinceabell-shapeddistributionissymmetric,approximately,34%ofthekidswatchtelevisionbetween2and3hoursperday.b.1houris2standarddeviationsbelowthemean.Thus,theempiricalrulesuggeststhat95%ofthekidswatchtelevisionbetween1and5hoursperday.Sinceabell-shapeddistributionissymmetric,approximately,47.5%ofthekidswatchtelevisionbetween1and3hoursperday.Inpart(a)wewww.khdaw.comconcludedthatapproximately34%ofthekidswatchtelevisionbetween2and3hoursperday;thus,approximately34%ofthekidswatchtelevisionbetween3and4hoursperday.Hence,approximately47.5%+34%=81.5%ofkidswatchtelevisionbetween1and4hoursperday.c.Since34%ofthekidswatchtelevisionbetween3and4hoursperday,50%-34%=16%ofthekidswatchtelevisionmorethan4hoursperday.33.a.Approximately68%ofscoresarewithin1standarddeviationfromthemean.b.Approximately95%ofscoresarewithin2standarddeviationsfromthemean.c.Approximately(100%-95%)/2=2.5%ofscoresareover130.d.Yes,almostallIQscoresarelessthan145.71.0090.06−34.a.z==−0.952016890.06−b.z==3.9020c.Thez-scoreinpartaindicatesthatthevalueis0.95standarddeviationsbelowthemean.Thez-scoreinpartbindicatesthatthevalueis3.90standarddeviationsabovethemean.Thelaborcostinpartbisanoutlierandshouldbereviewedforaccuracy.35.a.xisapproximately63or$63,000,andsis4or$4000b.Thisisfrom2standarddeviationsbelowthemeanto2standarddeviationsabovethemean.Withz=2,Chebyshev’stheoremgives:11131−=1−=1−=22z244Therefore,atleast75%ofbenefitsmanagershaveanannualsalarybetween$55,000and$71,000.3-13www.khdaw.com 课后答案网www.khdaw.comChapter3c.Thehistogramofthesalarydataisshownbelow:987654www.khdaw.comFrequency321056-5858-6060-6262-6464-6666-6868-7070-7272-74SalaryAlthoughthedistributionisnotperfectlybellshaped,itdoesappearreasonabletoassumethatthedistributionofannualsalarycanbeapproximatedbyabell-shapeddistribution.d.Withz=2,theempiricalrulesuggeststhat95%ofbenefitsmanagershaveanannualsalarybetween$55,000and$71,000.TheprobabilityismuchhigherthanobtainedusingChebyshev’stheorem,butrequirestheassumptionthatthedistributionofannualsalaryisbellshaped.e.Therearenooutliersbecausealltheobservationsarewithin3standarddeviationsofthemean.36.a.xis100andsis13.88orapproximately14b.Ifthedistributionisbellshapedwithameanof100points,thepercentageofNBAgamesinwhichthewinningteamscoresmorethan100pointsis50%.Ascoreof114pointsisz=1standarddeviationabovethemean.Thus,theempiricalrulesuggeststhat68%ofthewinningteamswillscorebetween86and114points.Inotherwords,32%ofthewinningteamswillscorelessthan86pointsormorethan114points.Becauseabell-shapeddistributionissymmetric,approximately16%ofthewinningteamswillscoremorethan114points.c.Forthewinningmargin,xis11.1andsis10.77.Toseeifthereareanyoutliers,wewillfirstcomputethez-scoreforthewinningmarginthatisfarthestfromthesamplemeanof11.1,awinningmarginof32points.x−x3211.1−z===1.94s10.77Thus,awinningmarginof32pointsisnotanoutlier(z=1.94<3).Becauseawinningmarginof32pointsisfarthestfromthemean,noneoftheotherdatavaluescanhaveaz-scorethatislessthan3orgreaterthan3andhenceweconcludethattherearenooutliers3-14www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodsΣx79.86i37.a.x===3.99n204.174.20+Median==4.185(averageof10thand11thvalues)2b.Q1=4.00(averageof5thand6thvalues)Q3=4.50(averageof15thand16thvalues)2Σ(x−x)12.5080ic.s===0.8114n−1194.12−3.99d.AllisonOne:z=≈0.16www.khdaw.com0.81142.32−3.99OmniAudioSA12.3:z=≈−2.060.8114e.ThelowestratingisfortheBose501Series.It’sz-scoreis:2.14−3.99z=≈−2.280.8114Thisisnotanoutliersotherearenooutliers.38.15,20,25,25,27,28,30,34Smallest=152520+25i=(8)=2Q==22.51100225+27Median==2627528+30i=(8)=8Q==2931002Largest=3439.152025303540.5,6,8,10,10,12,15,16,18Smallest=53-15www.khdaw.com 课后答案网www.khdaw.comChapter325i=(9)=2.25Q1=8(3rdposition)100Median=1075i=(9)=6.75Q3=15(7thposition)100Largest=18www.khdaw.com510152041.IQR=50-42=8LowerLimit:Q1-1.5IQR=42-12=30UpperLimit:Q3+1.5IQR=50+12=6268isanoutlier42.a.Fivenumbersummary:59.614.519.252.7b.IQR=Q3-Q1=19.2-9.6=9.6LowerLimit:Q1-1.5(IQR)=9.6-1.5(9.6)=-4.8UpperLimit:Q3+1.5(IQR)=19.2+1.5(9.6)=33.6c.Thedatavalue41.6isanoutlier(largerthantheupperlimit)andsoisthedatavalue52.7.Thefinancialanalystshouldfirstverifythatthesevaluesarecorrect.Perhapsatypingerrorhascaused25.7tobetypedas52.7(or14.6tobetypedas41.6).Iftheoutliersarecorrect,theanalystmightconsiderthesecompanieswithanunusuallylargereturnonequityasgoodinvestmentcandidates.d.**-1052035506543.a.Median(11thposition)401925i=(21)=5.25100Q1(6thposition)=187275i=(21)15.75=1003-16www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodsQ3(16thposition)=8305608,1872,4019,8305,14138b.Limits:IQR=Q3-Q1=8305-1872=6433LowerLimit:Q1-1.5(IQR)=-7777UpperLimit:Q3+1.5(IQR)=17955c.Therearenooutliers,alldataarewithinthelimits.d.Yes,ifthefirsttwodigitsinJohnsonandJohnson"ssalesweretransposedto41,138,saleswouldwww.khdaw.comhaveshownupasanoutlier.Areviewofthedatawouldhaveenabledthecorrectionofthedata.e.03,0006,0009,00012,00015,00044.a.Mean=105.7933Median=52.7b.Q1=15.7Q3=78.3c.IQR=Q3-Q1=78.3-15.7=62.6Lowerlimitforboxplot=Q1-1.5(IQR)=15.7-1.5(62.6)=-78.2Upperlimitforboxplot=Q3+1.5(IQR)=78.3+1.5(62.6)=172.2Note:Becausethenumberofsharescoveredbyoptionsgrantscannotbenegative,thelowerlimitfortheboxplotissetat0.This,outliersarevalueinthedatasetgreaterthan172.2.Outliers:SiliconGraphics(188.8)andToysRUs(247.6)d.Meanpercentage=26.73.Thecurrentpercentageismuchgreater.45.a.FiveNumberSummary(Midsize)5171.581.596.5128FiveNumberSummary(Small)73101108.5121140b.BoxPlotsMidsize3-17www.khdaw.com 课后答案网www.khdaw.comChapter35060708090100110120130SmallSize5060708090100110120130140150c.Themidsizecarsappeartobesaferthanthesmallcars.www.khdaw.com46.a.x=37.48Median=23.67b.Q1=7.91Q3=51.92c.IQR=51.92-7.91=44.01LowerLimit:Q1-1.5(IQR)=7.91-1.5(44.01)=-58.11UpperLimit:Q3+1.5(IQR)=51.92+1.5(44.01)=117.94Russia,withapercentchangeof125.89,isanoutlier.Turkey,withapercentchangeof254.45isanotheroutlier.d.Withapercentchangeof22.64,theUnitedStatesisjustbelowthe50thpercentile-themedian.47.a.70605040y302010005101520xb.Negativerelationship3-18www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods40230c/d.Σx=40x==8Σy=230y==46ii5522Σ(x−x)(y−y)=−240Σ(x−x)=118Σ(y−y)=520iiiiΣ(x−xy)(−y)−240iis===−60xyn−151−2Σ(x−x)118is===5.4314xn−151−2Σ(y−y)520is===11.4018ywww.khdaw.comn−151−sxy−60r===−0.969xyss(5.4314)(11.4018)xyThereisastrongnegativelinearrelationship.48.a.1816141210y86420051015202530xb.Positiverelationship8050c/d.Σx=80x==16Σy=50y==10ii5522Σ(x−x)(y−y)=106Σ(x−x)=272Σ(y−y)=86iiiiΣ(x−xy)(−y)106iis===26.5xyn−151−3-19www.khdaw.com 课后答案网www.khdaw.comChapter32Σ(x−x)272is===8.2462xn−151−2Σ(y−y)86is===4.6368yn−151−sxy26.5r===0.693xyss(8.2462)(4.6368)xyApositivelinearrelationship49.a.750www.khdaw.com700650600550y=SAT5004504002.62.833.23.43.63.8x=GPAb.Positiverelationship19.83540c/d.Σx=19.8x==3.3Σy=3540y==590ii6622Σ(x−x)(y−y)=143Σ(x−x)=0.74Σ(y−y)=36,400iiiiΣ(x−xy)(−y)143iis===28.6xyn−161−2Σ(x−x)0.74is===0.3847xn−161−2Σ(y−y)36,400is===85.3229yn−161−sxy28.6r===0.8713xyss(0.3847)(85.3229)xyApositivelinearrelationship3-20www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods50.Letx=drivingspeedandy=mileage420270Σx=420x==42Σy=270y==27ii101022Σ(x−xy)(−y)=−475(Σx−x)=1660(Σy−y)=164iiiiΣ(x−xy)(−y)−475iis===−52.7778xyn−1101−2Σ(x−x)1660is===13.5810xn−1101−2Σ(y−y)164www.khdaw.comis===4.2687yn−1101−sxy−52.7778r===−.91xyss(13.5810)(4.2687)xyAstrongnegativelinearrelationship51.a.Thesamplecorrelationcoefficientis.78.b.Thereisapositivelinearrelationshipbetweentheperformancescoreandtheoverallrating.52.a.Thesamplecorrelationcoefficientis.92.b.Thereisastrongpositivelinearrelationshipbetweenthetwovariables.53.Thesamplecorrelationcoefficientis.88.Thisindicatesastrongpositivelinearrelationshipbetweenthedailyhighandlowtemperatures.Σwx6(3.2)+3(2)+2(2.5)+8(5)70.2ii54.a.x====3.69Σw6+3+2+819i3.2+2+2.5+512.7b.==3.1754455.fiMifiMi45207107091513552010025325ΣfM325iix===13n253-21www.khdaw.com 课后答案网www.khdaw.comChapter3fiMi(M−x)2fM(−x)2Mi−xiii45-864256710-3963915+2436520+74924560022ΣfMi(i−x)600s===25n−124s=25=556.a.GradexiWeightWi4(A)93(B)15www.khdaw.com2(C)331(D)30(F)060CreditHoursΣwx9(4)+15(3)+33(2)+3(1)150iix====2.50Σw9+15+33+360ib.Yes;satisfiesthe2.5gradepointaveragerequirement57.Weusetheweightedmeanformulawiththeweightsbeingtheamountsinvested.Σwx=37,830(0.00)+27,667(2.98)+31,037(2.77)+27,336(2.65)+37,553(1.58)ii+17,812(0.57)+32,660(2.00)+17,775(0.00)=375,667.1Σw=37,830+27,667+···+17,775i=229,670Σwx375,667.1iix===1.64Σw229,670i58.MififiMiMi−x(M−x)2fM(−x)2iii742148-8.74264776.4338775,656.106919271,344-3.74264714.0074072,689.4221280123,3601.2573531.580937442.6622105171,7856.25735339.1544674,111.2190232250611.257353126.7280002,914.743962716216.257353264.3015301,585.80926807,30517,399.9630Estimateoftotalgallonssold:(10.74)(120)=1288.87305x==10.746803-22www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods217,399.9630s==25.63679s=5.0659.a.ClassfiMifiMi015001101102402803853255435041400www.khdaw.comTotals5001745ΣfM1745iix===3.49n500b.22Mi−x(Mi−x)fi(Mi−x)-3.4912.18182.70-2.496.2062.00-1.492.2288.80-0.490.2420.41+0.510.2691.04Total444.952Σ(M−x)f444.952iis===0.8917s=0.8917=0.9443n−1499Σx3463i60.a.x===138.52n25Median=129(13thvalue)Mode=0(2times)b.ItappearsthatthisgroupofyoungadultseatsoutmuchmorethantheaverageAmerican.Themeanandmedianaremuchhigherthantheaverageof$65.88reportedinthenewspaper.c.Q1=95(7thvalue)Q3=169(19thvalue)d.Min=0Max=467Range=467-0=467IQR=Q3-Q1=169-95=74e.s2=9271.01s=96.293-23www.khdaw.com 课后答案网www.khdaw.comChapter3f.Thez-scoreforthelargestvalueis:467−138.52z==3.4196.29Itistheonlyoutlierandshouldbecheckedforaccuracy.61.a.Σxi=760Σx760ix===38n20Medianisaverageof10thand11thitems.www.khdaw.com3636+Median==362Themodalcashretaineris40;itappears4times.b.ForQ1,c.⎛25⎞i=⎜⎟20=5⎝100⎠Sinceiisinteger,2830+Q==2912ForQ3,⎛75⎞i=⎜⎟2015=⎝100⎠Sinceiisinteger,4050+Q==4532cRange=64–15=49Interquartilerange=45–29=1622∑(xi−x)3318d.s===174.6316n−1201−2s=s=174.6316=13.21483-24www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods⎛s⎞⎛13.2148⎞e.Coefficientofvariation=⎜⎟100=⎜⎟100=34.8⎝x⎠⎝38⎠Σx260i62.a.x===18.57n14Median=16.5(Averageof7thand8thvalues)b.s2=53.49s=7.31c.Quantexhasthebestrecord:11Days27−18.57d.z==1.157.31www.khdaw.comPackard-Bellis1.15standarddeviationsslowerthanthemean.12−18.57e.z==−0.907.31IBMis0.9standarddeviationsfasterthanthemean.f.CheckToshiba:37−18.57z==2.527.31Onthebasisofz-scores,Toshibaisnotanoutlier,butitis2.52standarddeviationsslowerthanthemean.63.x=1890.2/30=63Median(15thand16thpositions)is(63+63.5)/2=63.25Mode:60.5and63.5bothoccurtwiceb.i=(25/100)30=7.5(8thposition)Q1=55.9i=(75/100)30=22.5(23rdposition)Q3=69.064.Samplemean=7195.5Median=7019(averageofpositions5and6)Samplevariance=7,165,941Samplestandarddeviation=2676.9365.a.Thesamplemeanis83.135andthesamplestandarddeviationis16.173.3-25www.khdaw.com 课后答案网www.khdaw.comChapter3b.Withz=2,Chebyshev’stheoremgives:11131−=1−=1−=22z244Therefore,atleast75%ofhouseholdincomesarewithin2standarddeviationsofthemean.Usingthesamplemeanandsamplestandarddeviationcomputedinpart(a),therangewithin75%ofhouseholdincomesmustfallis83.135±2(16.173)=83.135±32.346;thus,75%ofhouseholdincomesmustfallbetween50.789and115.481,or$50,789to$115,481.c.Withz=2,theempiricalrulesuggeststhat95%ofhouseholdincomesmustfallbetween$50,789to$115,481.Forthesamerange,theprobabilityobtainedusingtheempiricalruleisgreaterthantheprobabilityobtainedusingChebyshev’stheorem.d.Thez-scoreforDanbury,CTis3.04;thus,theDanbury,CTobservationisanoutlier.32066.a.PublicTransportation:x==32www.khdaw.com10320Automobile:x==3210b.PublicTransportation:s=4.64Automobile:s=1.83c.Prefertheautomobile.Themeantimesarethesame,buttheautohaslessvariability.d.Datainascendingorder:Public:25282929323233343741Auto:29303131323233333435FivenumberSummariesPublic:2529323441Auto:2931323335BoxPlots:Public:2428323640Auto:3-26www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods2428323640Theboxplotsdoshowlowervariabilitywithautomobiletransportationandsupporttheconclusioninpartc.67.Datainascendingorder:4244535658616262757677787982848485888989www.khdaw.com9395969798a.FiveNumberSummary4262798998b.BoxPlot40506070809010068.Datainascendingorder:40045151157659665271174480982085290794197197510231112117412511278i=(25/100)20=5i=(75/100)20=15i=(50/100)20=10596652+Q==624129751023+Q==99932820852+Median==8362a.FiveNumberSummary3-27www.khdaw.com 课后答案网www.khdaw.comChapter34006248369991278b.4005006007008009001000110012001300c.Therearenovaluesoutsidethelimits.Thusnooutliersareidentified.Lowerlimit=624-1.5(999-624)=61.5www.khdaw.comUpperlimit=999+1.5(999-624)=1561.569.a.Thesamplecovarianceis477.5365.Becausethesamplecovarianceispositive,thereisapositivelinearrelationshipbetweenincomeandhomeprice.b.Thesamplecorrelationcoefficientis.933;thisindicatesastronglinearrelationshipbetweenincomeandhomeprice.70.a.Thescatterdiagramindicatesapositiverelationshipb.Σx=798Σy=11,688Σxy=1,058,019iiii22Σx=71,306Σy=16,058,736iiΣxyii−ΣΣ(xyii)/n1,058,019(798)(11,688)/9−r===.9856xy222222Σxi−Σ(xi)/nΣyi−Σ(yi)/n71,306(798)/916,058,736(11,688)/9−−Strongpositiverelationship71.Letxi=commissionon500sharestradeforbrokeriyi=commissionon1000sharestradeforbrokeri18291326Σx=1829x==91.45Σy=1326y==66.3ii202022Σ(x−xy)(−y)11,853.3(=Σx−x)=48,370.95(Σy−y)=8506.2iiiiΣ(x−xy)(−y)11,853.3iis===623.8579xyn−119Thecovarianceshowsthereisapositiverelationship.2Σ(x−x)48,370.95is===50.4563xn−1193-28www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethods2Σ(y−y)8506.2is===21.1588yn−119sxy623.8579r===0.5844xyss(50.4563)(21.1588)xyThecorrelationcoefficientshowsthatwhiletherelationshipispositive,itisnotrealstrong.NotethatMaxUlechargesmorethanSchwabforthe500sharetrade($195vs.$155)butlessforthe1000sharetrade($70vs.$90).72.a.Thescatterdiagramisshownbelow:www.khdaw.com3.532.521.5Earnings10.50051015202530BookValueb.Thesamplecorrelationcoefficientis.75;thisindicatesalinearrelationshipbetweenbookvalueandearnings.Σwx20(20)30(12)10(7)15(5)10(6)++++965ii73.x====11.4daysΣw2030101510++++85i74.a.(800+750+900)/3=817b.MonthJanuaryFebruaryMarchWeight123Σwx1(800)+2(750)+3(900)5000iix====833Σw1+2+36i75.fMfM22iiiiMi−x(M−x)f(M−x)iii45.522.0-6.846.24184.963-29www.khdaw.com 课后答案网www.khdaw.comChapter359.547.5-2.87.8439.20713.594.51.21.4410.08217.535.05.227.0454.08121.521.59.284.6484.64125.525.513.2174.24174.2420246.0547.20246x==12.3202547.20s==28.819s=5.3776.fiMifiMiMi−x(M−x)2f(M−x)2www.khdaw.comiii229.559.0-22484968639.5237.0-12144864449.5198.0-2416459.5238.0864256269.5139.018324648279.5159.0287841568201,030.043201030x==51.5204320s==227.3719s=15.0877.fiMifiMiMi−x(M−x)2f(M−x)2iii1047470-13.68187.14241871.4240522080-8.6875.34243013.70150578550-3.6813..54242031.361756210850+1.321.7424304.9275675025+6.3239.94242995.6815721080+11.32128.14241922.141077770+16.32266.34242663.4247528,82514,802.6428,825a.x==60.68475214,802.64b.s==31.23474s=31.23=5.593-30www.khdaw.com 课后答案网www.khdaw.comDescriptiveStatistics:NumericalMethodswww.khdaw.com3-31www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter4IntroductionIIntroductionntroductiontottooProbabilityPProbabilityrobabilityLearningLLearningearningObjectivesOObjectivesbjectives1.Obtainanappreciationoftheroleprobabilityinformationplaysinthedecisionmakingprocess.2.Understandprobabilityasanumericalmeasureofthelikelihoodofoccurrence.3.Knowthethreemethodscommonlyusedforassigningprobabilitiesandunderstandwhentheywww.khdaw.comshouldbeused.4.Knowhowtousethelawsthatareavailableforcomputingtheprobabilitiesofevents.5.Understandhownewinformationcanbeusedtoreviseinitial(prior)probabilityestimatesusingBayes’theorem.4-1www.khdaw.com 课后答案网www.khdaw.comChapter4Solutions:SSolutions:olutions:1.NumberofexperimentalOutcomes=(3)(2)(4)=24F6I6!6⋅5⋅4⋅3⋅2⋅12.===20HG3KJ3!3!(3⋅2⋅1)(3⋅2⋅1)ABCACEBCDBEFABDACFBCECDEABEADEBCFCDFABFADFBDECEFACDAEFBDFDEF66!3.P==(6)(5)(4)=1203www.khdaw.com(6−3)!BDFBFDDBFDFBFBDFDB4.a.1stToss2ndToss3rdTossH(H,H,H)TH(H,H,T)TH(H,T,H)HT(H,T,T)TH(T,H,H)TH(T,H,T)TH(T,T,H)T(T,T,T)b.Let:HbeheadandTbetail(H,H,H)(T,H,H)(H,H,T)(T,H,T)(H,T,H)(T,T,H)(H,T,T)(T,T,T)c.Theoutcomesareequallylikely,sotheprobabilityofeachoutcomesis1/8.5.P(Ei)=1/5fori=1,2,3,4,5P(Ei)≥0fori=1,2,3,4,5P(E1)+P(E2)+P(E3)+P(E4)+P(E5)=1/5+1/5+1/5+1/5+1/5=1Theclassicalmethodwasused.4-2www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbability6.P(E1)=.40,P(E2)=.26,P(E3)=.34Therelativefrequencymethodwasused.7.No.Requirement(4.3)isnotsatisfied;theprobabilitiesdonotsumto1.P(E1)+P(E2)+P(E3)+P(E4)=.10+.15+.40+.20=.858.a.Therearefouroutcomespossibleforthis2-stepexperiment;planningcommissionpositive-councilapproves;planningcommissionpositive-councildisapproves;planningcommissionnegative-councilapproves;planningcommissionnegative-councildisapproves.b.Letp=positive,n=negative,a=approves,andd=disapprovesPlanningCommissionCouncil(p,a)www.khdaw.comadp(p,d).n(n,a)ad(n,d)F50I50!50⋅49⋅48⋅479.===230,300HG4KJ4!46!4⋅3⋅2⋅110.a.Usetherelativefrequencyapproach:P(California)=1,434/2,374=.60b.Numbernotfrom4states=2,374-1,434-390-217-112=221P(Notfrom4States)=221/2,374=.09c.P(NotinEarlyStages)=1-.22=.78d.EstimateofnumberofMassachusettscompaniesinearlystageofdevelopment-(.22)390≈864-3www.khdaw.com 课后答案网www.khdaw.comChapter4e.Ifweassumethesizeoftheawardsdidnotdifferbystates,wecanmultiplytheprobabilityanawardwenttoColoradobythetotalventurefundsdisbursedtogetanestimate.EstimateofColoradofunds=(112/2374)($32.4)=$1.53billionAuthors"Note:TheactualamountgoingtoColoradowas$1.74billion.11.a.No,theprobabilitiesdonotsumtoone.Theysumto.85.b.Ownermustrevisetheprobabilitiessotheysumto1.00.12.a.Usethecountingruleforcombinations:F49I49!(49)(48)(47)(46)(45)===1,906,884HG5KJ5!44!(5)(4)(3)(2)(1)www.khdaw.comb.Verysmall:1/1,906,884=0.0000005c.Multiplytheanswertopart(a)by42togetthenumberofchoicesforthesixnumbers.No.ofChoices=(1,906,884)(42)=80,089,128ProbabilityofWinning=1/80,089,128=0.000000012513.Initiallyaprobabilityof.20wouldbeassignedifselectionisequallylikely.Datadoesnotappeartoconfirmthebeliefofequalconsumerpreference.Forexampleusingtherelativefrequencymethodwewouldassignaprobabilityof5/100=.05tothedesign1outcome,.15todesign2,.30todesign3,.40todesign4,and.10todesign5.14.a.P(E2)=1/4b.P(any2outcomes)=1/4+1/4=1/2c.P(any3outcomes)=1/4+1/4+1/4=3/415.a.S={aceofclubs,aceofdiamonds,aceofhearts,aceofspades}b.S={2ofclubs,3ofclubs,...,10ofclubs,Jofclubs,Qofclubs,Kofclubs,Aofclubs}c.Thereare12;jack,queen,orkingineachofthefoursuits.d.Fora:4/52=1/13=.08Forb:13/52=1/4=.25Forc:12/52=.234-4www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbability16.a.(6)(6)=36samplepointsb.Die212345612345672345678TotalforBoth.3456789Die1www.khdaw.com456789105678910116789101112c.6/36=1/6d.10/36=5/18e.No.P(odd)=18/36=P(even)=18/36or1/2forboth.f.Classical.Aprobabilityof1/36isassignedtoeachexperimentaloutcome.17.a.(4,6),(4,7),(4,8)b..05+.10+.15=.30c.(2,8),(3,8),(4,8)d..05+.05+.15=.25e..1518.a.0;probabilityis.05b.4,5;probabilityis.10+.10=.20c.0,1,2;probabilityis.05+.15+.35=.5519.a.Yes,theprobabilitiesareallgreaterthanorequaltozeroandtheysumtoone.b.P(A)=P(0)+P(1)+P(2)=.08+.18+.32=.584-5www.khdaw.com 课后答案网www.khdaw.comChapter4c.P(B)=P(4)=.1220.a.P(N)=56/500=.112b.P(T)=43/500=.086c.Totalin6states=56+53+43+37+28+28=245P(B)=245/500=.49AlmosthalftheFortune500companiesareheadquarteredinthesestates.21.a.P(A)=P(1)+P(2)+P(3)+P(4)+P(5)2012631=++++www.khdaw.com5050505050=.40+.24+.12+.06+.02=.84b.P(B)=P(3)+P(4)+P(5)=.12+.06+.02=.20c.P(2)=12/50=.2422.a.P(A)=.40,P(B)=.40,P(C)=.60b.P(A∪B)=P(E1,E2,E3,E4)=.80.YesP(A∪B)=P(A)+P(B).c.Ac={E3,E4,E5}Cc={E1,E4}P(Ac)=.60P(Cc)=.40d.A∪Bc={E1,E2,E5}P(A∪Bc)=.60e.P(B∪C)=P(E2,E3,E4,E5)=.8023.a.P(A)=P(E1)+P(E4)+P(E6)=.05+.25+.10=.40P(B)=P(E2)+P(E4)+P(E7)=.20+.25+.05=.50P(C)=P(E2)+P(E3)+P(E5)+P(E7)=.20+.20+.15+.05=.60b.A∪B={E1,E2,E4,E6,E7}P(A∪B)=P(E1)+P(E2)+P(E4)+P(E6)+P(E7)=.05+.20+.25+.10+.05=.65c.A∩B={E4}P(A∩B)=P(E4)=.25d.Yes,theyaremutuallyexclusive.4-6www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbabilitye.Bc={E1,E3,E5,E6};P(Bc)=P(E1)+P(E3)+P(E5)+P(E6)=.05+.20+.15+.10=.5024.P(CrashNotLikely)=1-.14-.43=.4325.LetY=highone-yearreturnM=highfive-yearreturna.P(Y)=15/30=.50P(M)=12/30=.40P(Y∩M)=6/30=.20b.P(Y∪M)=P(Y)+P(M)-P(Y∩M)www.khdaw.com=.50+.40-.20=.70c.1-P(Y∪M)=1-.70=.3026.LetY=highone-yearreturnM=highfive-yearreturna.P(Y)=9/30=.30P(M)=7/30=.23b.P(Y∩M)=5/30=.17c.P(Y∪M)=.30+.23-.17=.36P(Neither)=1-.36=.6427.Let:D=consumesorservesdomesticwineI=consumesorservesimportedwineWearegivenP(D)=0.57,P(I)=0.33,P(D∪I)=0.63P(D∩I)=P(D)+P(I)-P(D∪I)=0.57+0.33-0.63=0.2728.Let:B=rentedacarforbusinessreasonsP=rentedacarforpersonalreasonsa.P(B∪P)=P(B)+P(P)-P(B∩P)=.54+.458-.30=.698b.P(Neither)=1-.698=.302725,79029.a.P(H)=≈0.2992,425,0004-7www.khdaw.com 课后答案网www.khdaw.comChapter4537,390P(C)=≈0.2222,425,000159,877P(S)=≈0.0662,425,000b.Apersoncanhaveonlyoneprimarycauseofdeathlistedonadeathcertificate.So,theyaremutuallyexclusive.c.P(H∪C)=0.299+0.222=0.521d.P(C∪S)=0.222+0.066=0.288e.1-0.299-0.222-0.066=0.413P(A∩B).4030.a.P(AB)===.6667www.khdaw.comP(B).60P(A∩B).40b.P(BA)===.80P(A).50c.NobecauseP(A|B)≠P(A)31.a.P(A∩B)=0P(A∩B)0b.P(AB)===0P(B).4c.No.P(A|B)≠P(A);∴theevents,althoughmutuallyexclusive,arenotindependent.d.Mutuallyexclusiveeventsaredependent.32.a.SingleMarriedTotalUnder30.55.10.6530orover.20.15.35Total.75.251.00b.65%ofthecustomersareunder30.c.Themajorityofcustomersaresingle:P(single)=.75.d..554-8www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbabilitye.Let:A=eventunder30B=eventsingleP(A∩B).55P(BA)===.8462P(A).65f.P(A∩B)=.55P(A)P(B)=(.65)(.75)=.49SinceP(A∩B)≠P(A)P(B),theycannotbeindependentevents;or,sinceP(A|B)≠P(B),theycannotbeindependent.33.a.www.khdaw.comReasonforApplyingRReasonforApplyingeasonforApplyingQualityCost/ConvenienceOtherTotalFullTime.218.204.039.461PartTime.208.307.024.539.426.511.0631.00b.Itismostlikelyastudentwillcitecostorconvenienceasthefirstreason-probability=.511.Schoolqualityisthefirstreasoncitedbythesecondlargestnumberofstudents-probability=.426.c.P(Quality|fulltime)=.218/.461=.473d.P(Quality|parttime)=.208/.539=.386e.Forindependence,wemusthaveP(A)P(B)=P(A∩B).Fromthetable,P(A∩B)=.218,P(A)=.461,P(B)=.426P(A)P(B)=(.461)(.426)=.196SinceP(A)P(B)≠P(A∩B),theeventsarenotindependent.34.a.P(O)=0.38+0.06=0.44b.P(Rh-)=0.06+0.02+0.01+0.06=0.15c.P(bothRh-)=P(Rh-)P(Rh-)=(0.15)(0.15)=0.0225d.P(bothAB)=P(AB)P(AB)=(0.05)(0.05)=0.00254-9www.khdaw.com 课后答案网www.khdaw.comChapter4P(Rh-∩O).06e.P(Rh-O)===.136P(O).44f.P(Rh+)=1-P(Rh-)=1-0.15=0.85P(B∩Rh+).09P(BRh+)===.106P(Rh+).8535.a.P(UpforJanuary)=31/48=0.646b.P(UpforYear)=36/48=0.75c.P(UpforYear∩UpforJanuary)=29/48=0.604www.khdaw.comP(UpforYear|UpforJanuary)=0.604/0.646=0.935d.TheyarenotindependentsinceP(UpforYear)≠P(UpforYear|UpforJanuary)0.75≠0.93536.a.SatisfactionScoreOccupationUnder5050-5960-6970-7980-89TotalCabinetmaker.000.050.100.075.025.250Lawyer.150.050.025.025.000.250PhysicalTherapist.000.125.050.025.050.250SystemsAnalyst.050.025.100.075.000.250Total.200.250.275.200.0751.000b.P(80s)=.075(amarginalprobability)c.P(80s|PT)=.050/.250=.20(aconditionalprobability)d.P(L)=.250(amarginalprobability)e.P(L∩Under50)=.150(ajointprobability)f.P(Under50|L)=.150/.250=.60(aconditionalprobability)g.P(70orhigher)=.275(Sumofmarginalprobabilities)37.a.P(A∩B)=P(A)P(B)=(.55)(.35)=.19b.P(A∪B)=P(A)+P(B)-P(A∩B)=.55+.35-.19=.71c.P(shutdown)=1-P(A∪B)=1-.71=.295238.a.P(Telephone)=≈0.2737190b.Thisisanintersectionoftwoevents.Itseemsreasonabletoassumethenexttwomessageswillbeindependent;weusethemultiplicationruleforindependentevents.4-10www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbability⎛30⎞⎛15⎞P(E-mail∩Fax)=P(E-mail)P(Fax)=⎜⎟⎜⎟≈0.0125⎝190⎠⎝190⎠c.Thisisaunionoftwomutuallyexclusiveevents.P(Telephone∪InterofficeMail)=P(Telephone)+P(InterofficeMail)521870=+=≈0.736819019019039.a.Yes,sinceP(A1∩A2)=0b.P(A1∩B)=P(A1)P(B|A1)=.40(.20)=.08www.khdaw.comP(A2∩B)=P(A2)P(B|A2)=.60(.05)=.03c.P(B)=P(A1∩B)+P(A2∩B)=.08+.03=.11.08d.P(AB)==.72731.11.03P(AB)==.27272.1140.a.P(B∩A1)=P(A1)P(B|A1)=(.20)(.50)=.10P(B∩A2)=P(A2)P(B|A2)=(.50)(.40)=.20P(B∩A3)=P(A3)P(B|A3)=(.30)(.30)=.09.20b.P(AB)==.512.10.20.09++c.EventsP(Ai)P(B|Ai)P(Ai∩B)P(Ai|B)A1.20.50.10.26A2.50.40.20.51A3.30.30.09.231.00.391.0041.S1=successful,S2=notsuccessfulandB=requestreceivedforadditionalinformation.a.P(S1)=.50b.P(B|S1)=.75(.50)(.75).375c.P(SB)===.651(.50)(.75)(.50)(.40)+.57542.M=missedpaymentD1=customerdefaultsD2=customerdoesnotdefault4-11www.khdaw.com 课后答案网www.khdaw.comChapter4P(D1)=.05P(D2)=.95P(M|D2)=.2P(M|D1)=1P(D)P(MD)(.05)(1).0511a.P(DM)====.211P(D)P(MD)P(D)P(MD)+(.05)(1)(.95)(.2)+.241122b.Yes,theprobabilityofdefaultisgreaterthan.20.43.Let:S=smallcarSc=othertypeofvehicleF=accidentleadstofatalityforvehicleoccupantWehaveP(S)=.18,soP(Sc)=.82.AlsoP(F|S)=.128andP(F|Sc)=.05.UsingthetabularformofBayesTheoremprovides:PriorConditionalJointPosteriorwww.khdaw.comEventsProbabilitiesProbabilitiesProbabilitiesProbabilitiesS.18.128.023.36Sc.82.050.041.641.00.0641.00Fromtheposteriorprobabilitycolumn,wehaveP(S|F)=.36.So,ifanaccidentleadstoafatality,theprobabilityasmallcarwasinvolvedis.36.44.LetA1=StoryaboutBasketballTeamA2=StoryaboutHockeyTeamW="WeWin"headlineP(A1)=.60P(W|A1)=.641P(A2)=.40P(W|A2)=.462AiP(Ai)P(W|A1)P(W∩Ai)P(Ai|M)A1.60.641.3846.3846/.5694=.6754A2.40.462.1848.1848/.5694=.3246.56941.0000Theprobabilitythestoryisaboutthebasketballteamis.6754.45.a.EventsP(Di)P(S1|Di)P(Di∩S1)P(Di|S1)D1.60.15.090.2195D2.40.80.320.78051.00P(S1)=.4101.000P(D1|S1)=.2195P(D2|S1)=.7805b.EventsP(Di)P(S2|Di)P(Di∩S2)P(Di|S2)4-12www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbabilityD1.60.10.060.500D2.40.15.060.5001.00P(S2)=.1201.000P(D1|S2)=.50P(D2|S2)=.50c.EventsP(Di)P(S3|Di)P(Di∩S3)P(Di|S3)D1.60.15.090.8824D2.40.03.012.11761.00P(S3)=.1021.0000P(D1|S3)=.8824P(D2|S3)=.1176www.khdaw.comd.Usetheposteriorprobabilitiesfrompart(a)asthepriorprobabilitieshere.EventsP(Di)P(S2|Di)P(Di∩S2)P(Di|S2)D1.2195.10.0220.1582D2.7805.15.1171.84181.0000.13911.0000P(D1|S1andS2)=.1582P(D2|S1andS2)=.841846.a.P(Excellent)=.18P(PrettyGood)=.50P(PrettyGood∪Excellent)=.18+.50=.68Note:Eventsaremutuallyexclusivesinceapersonmayonlychooseonerating.b.1035(.05)=51.75Weestimate52respondentsratedUScompaniespoor.c.1035(.01)=10.35Weestimate10respondentsdidnotknowordidnotanswer.47.a.(2)(2)=4b.Lets=successfulu=unsuccessful4-13www.khdaw.com 课后答案网www.khdaw.comChapter4OilBondsE1susE2uE3www.khdaw.comsuE4c.O={E1,E2}M={E1,E3}d.O∪M={E1,E2,E3}e.O∩M={E1}f.No;sinceO∩Mhasasamplepoint.48.a.P(satisfied)=0.61b.The18-34yearoldgroup(64%satisfied)andthe65andovergroup(70%satisfied).c.P(notsatisfied)=0.26+0.04=0.3049.LetI=treatment-causedinjuryD=deathfrominjuryN=injurycausedbynegligenceM=malpracticeclaimfiled$=paymentmadeinclaimWearegivenP(I)=0.04,P(N|I)=0.25,P(D|I)=1/7,P(M|N)=1/7.5=0.1333,andP($|M)=0.50a.P(N)=P(N|I)P(I)+P(N|Ic)P(Ic)=(0.25)(0.04)+(0)(0.96)=0.014-14www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbabilityb.P(D)=P(D|I)P(I)+P(D|Ic)P(Ic)=(1/7)(0.04)+(0)(0.96)=0.006c.P(M)=P(M|N)P(N)+P(M|Nc)P(Nc)=(0.1333)(0.01)+(0)(0.99)=0.001333P($)=P($|M)P(M)+P($|Mc)P(Mc)=(0.5)(0.001333)+(0)(0.9987)=0.0006750.a.Probabilityoftheevent=P(average)+P(aboveaverage)+P(excellent)111413=++www.khdaw.com505050=.22+.28+.26=.76b.Probabilityoftheevent=P(poor)+P(belowaverage)48=+=.24505051.a.P(leases1)=168/932=0.18b.P(2orfewer)=401/932+242/932+65/932=708/932=0.76c.P(3ormore)=186/932+112/932=298/932=0.32d.P(nocars)=19/932=0.0252.a.YesNoTotal23andUnder.1026.0996.202224-26.1482.1878.336027-30.0917.1328.224531-35.0327.0956.128336andOver.0253.0837.1090Total.4005.59951.0000.4-15www.khdaw.com 课后答案网www.khdaw.comChapter4b..2022c..2245+.1283+.1090=.4618d..400553.a.P(24to26|Yes)=.1482/.4005=.3700b.P(Yes|36andover)=.0253/.1090=.2321c..1026+.1482+.1878+.0917+.0327+.0253=.5883d.P(31ormore|No)=(.0956+.0837)/.5995=.2991e.No,becausetheconditionalprobabilitiesdonotallequalthemarginalprobabilities.Forinstance,www.khdaw.comP(24to26|Yes)=.3700≠P(24to26)=.336054.LetI=importantorveryimportantM=maleF=femalea.P(I)=.49(amarginalprobability)b.P(I|M)=.22/.50=.44(aconditionalprobability)c.P(I|F)=.27/.50=.54(aconditionalprobability)d.ItisnotindependentP(I)=.49≠P(I|M)=.44andP(I)=.49≠P(I|F)=.54e.Sincelevelofimportanceisdependentongender,weconcludethatmaleandfemalerespondentshavedifferentattitudestowardrisk.P(B∩S).1255.a.P(BS)===.30P(S).40WehaveP(B|S)>P(B).Yes,continuetheadsinceitincreasestheprobabilityofapurchase.b.Estimatethecompany’smarketshareat20%.ContinuingtheadvertisementshouldincreasethemarketsharesinceP(B|S)=.30.P(B∩S).10c.P(BS)===.333P(S).30Thesecondadhasabiggereffect.56.a.P(A)=200/800=.254-16www.khdaw.com 课后答案网www.khdaw.comIntroductiontoProbabilityb.P(B)=100/800=.125c.P(A∩B)=10/800=.0125d.P(A|B)=P(A∩B)/P(B)=.0125/.125=.10e.No,P(A|B)≠P(A)=.2557.LetA=losttimeaccidentincurrentyearB=losttimeaccidentpreviousyearGiven:P(B)=.06,P(A)=.05,P(A|B)=.15a.P(A∩B)=P(A|B)P(B)=.15(.06)=.009b.P(A∪B)=P(A)+P(B)-P(A∩B)www.khdaw.com=.06+.05-.009=.101or10.1%58.Let:A=returnisfraudulentB=exceedsIRSstandardfordeductionsGiven:P(A|B)=.20,P(A|Bc)=.02,P(B)=.08,findP(A)=.3.NoteP(Bc)=1-P(B)=.92P(A)=P(A∩B)+P(A∩Bc)=P(B)P(A|B)+P(Bc)P(A|Bc)=(.08)(.20)+(.92)(.02)=.0344Weestimate3.44%willbefraudulent.59.a.P(Oil)=.50+.20=.70b.LetS=SoiltestresultsEventsP(Ai)P(S|Ai)P(Ai∩S)P(Ai|S)HighQuality(A1).50.20.10.31MediumQuality(A2).20.80.16.50NoOil(A3).30.20.06.191.00P(S)=.321.00P(Oil)=.81whichisgood;however,probabilitiesnowfavormediumqualityratherthanhighqualityoil.60.a.A1=fieldwillproduceoilA2=fieldwillnotproduceoilW=wellproducesoilEventsP(Ai)P(Wc|Ai)P(Wc∩Ai)P(Ai|Wc)OilinField.25.20.05.0625NoOilinField.751.00.75.93751.00.801.0000Theprobabilitythefieldwillproduceoilgivenawellcomesupdryis.0625.4-17www.khdaw.com 课后答案网www.khdaw.comChapter4b.EventsP(Ai)P(Wc|Ai)P(Wc∩Ai)P(Ai|Wc)OilinField.0625.20.0125.0132NoOilinField.93751.00.9375.98681.0000.95001.0000Theprobabilitythewellwillproduceoildropsfurtherto.0132.c.Supposeathirdwellcomesupdry.Theprobabilitiesarerevisedasfollows:EventsP(Ai)P(Wc|Ai)P(Wc∩Ai)P(Ai|Wc)OilinField.0132.20.0026.0026IncorrectAdjustment.98681.00.9868.99741.0000.98941.0000www.khdaw.comStopdrillingandabandonfieldifthreeconsecutivewellscomeupdry.4-18www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter5DiscreteDDiscreteiscreteProbabilityPProbabilityrobabilityDistributionsDDistributionsistributionsLearningLLearningearningObjectivesOObjectivesbjectives1.Understandtheconceptsofarandomvariableandaprobabilitydistribution.2.Beabletodistinguishbetweendiscreteandcontinuousrandomvariables.3.Beabletocomputeandinterprettheexpectedvalue,variance,andstandarddeviationforadiscretewww.khdaw.comrandomvariable.4.Beabletocomputeandworkwithprobabilitiesinvolvingabinomialprobabilitydistribution.5.BeabletocomputeandworkwithprobabilitiesinvolvingaPoissonprobabilitydistribution.6.Knowwhenandhowtousethehypergeometricprobabilitydistribution.5-1www.khdaw.com 课后答案网www.khdaw.comChapter5Solutions:SSolutions:olutions:1.a.Head,Head(H,H)Head,Tail(H,T)Tail,Head(T,H)Tail,Tail(T,T)b.x=numberofheadsontwocointossesc.OutcomeValuesofx(H,H)2(H,T)1(T,H)1www.khdaw.com(T,T)0d.Discrete.Itmayassume3values:0,1,and2.2.a.Letx=time(inminutes)toassembletheproduct.b.Itmayassumeanypositivevalue:x>0.c.Continuous3.LetY=positionisofferedN=positionisnotoffereda.S={(Y,Y,Y),(Y,Y,N),(Y,N,Y),(Y,N,N),(N,Y,Y),(N,Y,N),(N,N,Y),(N,N,N)}b.LetN=numberofoffersmade;Nisadiscreterandomvariable.c.ExperimentalOutcome(Y,Y,Y)(Y,Y,N)(Y,N,Y)(Y,N,N)(N,Y,Y)(N,Y,N)(N,N,Y)(N,N,N)ValueofN322121104.x=0,1,2,...,12.5.a.S={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}b.ExperimentalOutcome(1,1)(1,2)(1,3)(2,1)(2,2)(2,3)NumberofStepsRequired2343456.a.values:0,1,2,...,20discreteb.values:0,1,2,...discretec.values:0,1,2,...,50discreted.values:0≤x≤8continuous5-2www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionse.values:x>0continuous7.a.f(x)≥0forallvaluesofx.Σf(x)=1Therefore,itisaproperprobabilitydistribution.b.Probabilityx=30isf(30)=.25c.Probabilityx≤25isf(20)+f(25)=.20+.15=.35d.Probabilityx>30isf(35)=.408.a.xf(x)www.khdaw.com13/20=.1525/20=.2538/20=.4044/20=.20Total1.00b.f(x).4.3.2.1x1234c.f(x)≥0forx=1,2,3,4.Σf(x)=19.a.xf(x)115/462=0.032232/462=0.069384/462=0.1824300/462=0.650531/462=0.0675-3www.khdaw.com 课后答案网www.khdaw.comChapter5b.f(x)0.600.45www.khdaw.com0.300.15x012345c.Allf(x)≥0Σf(x)=0.032+0.069+0.182+0.650+0.067=1.00010.a.xf(x)10.0520.0930.0340.4250.411.00b.xf(x)10.0420.1030.1240.4650.281.00c.P(4or5)=f(4)+f(5)=0.42+0.41=0.83d.Probabilityofverysatisfied:0.28e.Seniorexecutivesappeartobemoresatisfiedthanmiddlemanagers.83%ofseniorexecutiveshaveascoreof4or5with41%reportinga5.Only28%ofmiddlemanagersreportbeingverysatisfied.5-4www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributions11.a.DurationofCallxf(x)10.2520.2530.2540.251.00b.f(x)0.30www.khdaw.com0.200.10x01234c.f(x)≥0andf(1)+f(2)+f(3)+f(4)=0.25+0.25+0.25+0.25=1.00d.f(3)=0.25e.P(overtime)=f(3)+f(4)=0.25+0.25=0.5012.a.Yes;f(x)≥0forallxandΣf(x)=.15+.20+.30+.25+.10=1b.P(1200orless)=f(1000)+f(1100)+f(1200)=.15+.20+.30=.6513.a.Yes,sincef(x)≥0forx=1,2,3andΣf(x)=f(1)+f(2)+f(3)=1/6+2/6+3/6=1b.f(2)=2/6=.333c.f(2)+f(3)=2/6+3/6=.83314.a.f(200)=1-f(-100)-f(0)-f(50)-f(100)-f(150)=1-.95=.05ThisistheprobabilityMRAwillhavea$200,000profit.b.P(Profit)=f(50)+f(100)+f(150)+f(200)=.30+.25+.10+.05=.70c.P(atleast100)=f(100)+f(150)+f(200)=.25+.10+.05=.4015.a.5-5www.khdaw.com 课后答案网www.khdaw.comChapter5xf(x)xf(x)3.25.756.503.009.252.251.006.00E(x)=µ=6.00b.xx-µ(x-µ)2f(x)(x-µ)2f(x)3-39.252.25600.500.00939.252.254.50www.khdaw.comVar(x)=σ2=4.50c.σ=4.50=2.1216.a.yf(y)yf(y)2.20.404.301.207.402.808.10.801.005.20E(y)=µ=5.20b.yy-µ(y-µ)2f(y)(y-µ)2f(y)2-3.2010.24.202.0484-1.201.44.30.43271.803.24.401.29682.807.84.10.7844.560Var(y)=4.56σ=4.56=2.1417.a/b.xf(x)xf(x)x-µ(x-µ)2(x-µ)2f(x)0.10.00-2.456.0025.6002501.15.15-1.452.1025.3153752.30.60-.45.2025.0607503.20.60.55.3025.0605004.15.601.552.4025.3603755.10.502.556.5025.6502502.452.047500E(x)=µ=2.45σ2=2.0475σ=1.43095-6www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributions18.a/b.xf(x)xf(x)x-µ(x-µ)2(x-µ)2f(x)0.010-2.35.290.05291.23.23-1.31.690.38872.41.82-0.30.090.03693.20.600.70.490.0984.10.401.72.890.2895.05.252.77.290.36452.31.23E(x)=2.3Var(x)=1.23σ=1.23=1.11Theexpectedvalue,E(x)=2.3,oftheprobabilitydistributionisthesameasthatreportedinthe1997StatisticalAbstractoftheUnitedStates.www.khdaw.com19.a.E(x)=Σxf(x)=0(.50)+2(.50)=1.00b.E(x)=Σxf(x)=0(.61)+3(.39)=1.17c.Theexpectedvalueofa3-pointshotishigher.So,iftheseprobabilitiesholdup,theteamwillmakemorepointsinthelongrunwiththe3-pointshot.20.a.xf(x)xf(x)0.900.00400.0416.001000.0330.002000.0120.004000.0140.006000.0160.001.00166.00E(x)=166.Ifthecompanychargedapremiumof$166.00theywouldbreakeven.b.GaintoPolicyHolderf(Gain)(Gain)f(Gain)-260.00.90-234.00140.00.045.60740.00.0322.201,740.00.0117.403,740.00.0137.405,740.00.0157.40-94.00E(gain)=-94.00.Thepolicyholderismoreconcernedthatthebigaccidentwillbreakhimthanwiththeexpectedannuallossof$94.00.21.a.E(x)=Σxf(x)=0.05(1)+0.09(2)+0.03(3)+0.42(4)+0.41(5)=4.05b.E(x)=Σxf(x)=0.04(1)+0.10(2)+0.12(3)+0.46(4)+0.28(5)=3.845-7www.khdaw.com 课后答案网www.khdaw.comChapter5c.Executives:σ2=Σ(x-µ)2f(x)=1.2475MiddleManagers:σ2=Σ(x-µ)2f(x)=1.1344d.Executives:σ=1.1169MiddleManagers:σ=1.0651e.Theseniorexecutiveshaveahigheraveragescore:4.05vs.3.84forthemiddlemanagers.Theexecutivesalsohaveaslightlyhigherstandarddeviation.22.a.E(x)=Σxf(x)=300(.20)+400(.30)+500(.35)+600(.15)=445www.khdaw.comThemonthlyorderquantityshouldbe445units.b.Cost:445@$50=$22,250Revenue:300@$70=21,000$1,250Loss23.a.Laptop:E(x)=.47(0)+.45(1)+.06(2)+.02(3)=.63Desktop:E(x)=.06(0)+.56(1)+.28(2)+.10(3)=1.42b.Laptop:Var(x)=.47(-.63)2+.45(.37)2+.06(1.37)2+.02(2.37)2=.4731Desktop:Var(x)=.06(-1.42)2+.56(-.42)2+.28(.58)2+.10(1.58)2=.5636c.Fromtheexpectedvaluesinpart(a),itisclearthatthetypicalsubscriberhasmoredesktopcomputersthanlaptops.Thereisnotmuchdifferenceinthevariancesforthetwotypesofcomputers.24.a.MediumE(x)=Σxf(x)=50(.20)+150(.50)+200(.30)=145Large:E(x)=Σxf(x)=0(.20)+100(.50)+300(.30)=140Mediumpreferred.b.Mediumxf(x)x-µ(x-µ)2(x-µ)2f(x)50.20-9590251805.0150.5052512.5200.30553025907.5σ2=2725.0Largeyf(y)y-µ(y-µ)2(y-µ)2f(y)5-8www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributions0.20-140196003920100.50-401600800300.30160256007680σ2=12,400Mediumpreferredduetolessvariance.25.a.www.khdaw.comSSFFSF⎛⎞22!11b.f(1)=⎜⎟(.4)(.6)=(.4)(.6)=.48⎝⎠11!1!⎛⎞22!02c.f(0)=⎜⎟(.4)(.6)=(1)(.36)=.36⎝⎠00!2!⎛⎞22!20d.f(2)=⎜⎟(.4)(.6)=(.16)(1)=.16⎝⎠22!0!e.P(x≥1)=f(1)+f(2)=.48+.16=.64f.E(x)=np=2(.4)=.8Var(x)=np(1-p)=2(.4)(.6)=.48σ=.48=.692826.a.f(0)=.3487b.f(2)=.1937c.P(x≤2)=f(0)+f(1)+f(2)=.3487+.3874+.1937=.9298d.P(x≥1)=1-f(0)=1-.3487=.6513e.E(x)=np=10(.1)=15-9www.khdaw.com 课后答案网www.khdaw.comChapter5f.Var(x)=np(1-p)=10(.1)(.9)=.9σ=.9=.948727.a.f(12)=.1144b.f(16)=.1304c.P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)=.1304+.0716+.0278+.0068+.0008=.2374d.P(x≤15)=1-P(x≥16)=1-.2374=.7626e.E(x)=np=20(.7)=14www.khdaw.comf.Var(x)=np(1-p)=20(.7)(.3)=4.2σ=4.2=2.0494⎛⎞62428.a.f(2)=⎜⎟(.33)(.67)=.3292⎝⎠2b.P(atleast2)=1-f(0)-f(1)⎛⎞6⎛⎞60615=1−⎜⎟(.33)(.67)−⎜⎟(.33)(.67)⎝⎠0⎝⎠1=1-.0905-.2673=.6422⎛10⎞010c.f(10)=⎜⎟(.33)(.67)=.0182⎝0⎠29.P(AtLeast5)=1-f(0)-f(1)-f(2)-f(3)-f(4)=1-.0000-.0005-.0031-.0123-.0350=.949130.a.Probabilityofadefectivepartbeingproducedmustbe.03foreachtrial;trialsmustbeindependent.b.Let:D=defectiveG=notdefective5-10www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionsExperimentalNumber1stpart2ndpartOutcomeDefectiveD(D,D)2DG(D,G)1.GD(G,D)1G(G,G)0c.2outcomesresultinexactlyonedefect.www.khdaw.comd.P(nodefects)=(.97)(.97)=.9409P(1defect)=2(.03)(.97)=.0582P(2defects)=(.03)(.03)=.000931.Binomialn=10andp=.0510!x10−xf(x)=(.05)(.95)x!(10−x)!a.Yes.Sincetheyareselectedrandomly,pisthesamefromtrialtotrialandthetrialsareindependent.b.f(2)=.0746c.f(0)=.5987d.P(Atleast1)=1-f(0)=1-.5987=.401332.a..90b.P(atleast1)=f(1)+f(2)2!11f(1)=(.9)(.1)1!1!=2(.9)(.1)=.182!20f(2)=(.9)(.1)2!0!=1(.81)(1)=.81∴P(atleast1)=.18+.81=.99Alternatively5-11www.khdaw.com 课后答案网www.khdaw.comChapter5P(atleast1)=1-f(0)2!02f(0)=(.9)(.1)=.010!2!Therefore,P(atleast1)=1-.01=.99c.P(atleast1)=1-f(0)3!03f(0)=(.9)(.1)=.0010!3!Therefore,P(atleast1)=1-.001=.999d.Yes;P(atleast1)becomesverycloseto1withmultiplesystemsandtheinabilitytodetectanattackwouldbecatastrophic.www.khdaw.com33.a.Usingthebinomialformulaorthetableofbinomialprobabilitieswithp=.5andn=20,wefind:xf(x)120.1201130.0739140.0370150.0148160.0046170.0011180.0002190.0000200.00000.2517Theprobability12ormorewillsendrepresentativesis0.2517.b.Usingthebinomialformulaorthetables,wefind:xf(x)00.000010.000020.000230.001140.004650.01480.0207c.E(x)=np=20(0.5)=10d.σ2=np(1-p)=20(0.5)(0.5)=5σ=5=2.236134.a.f(3)=.0634(fromtables)b.Theanswerhereisthesameaspart(a).Theprobabilityof12failureswithp=.60isthesameastheprobabilityof3successeswithp=.40.c.f(3)+f(4)+···+f(15)=1-f(0)-f(1)-f(2)5-12www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributions=1-.0005-.0047-.0219=.972935.a.f(0)+f(1)+f(2)=.0115+.0576+.1369=.2060b.f(4)=.2182c.1-[f(0)+f(1)+f(2)+f(3)]=1-.2060-.2054=.5886d.µ=np=20(.20)=436.xf(x)x-µ(x-µ)2(x-µ)2f(x)0.343-.9.81.277831.441.1.01.004412.1891.11.21.228693.0272.14.41.11907www.khdaw.com1.000σ2=.6300037.E(x)=np=30(0.29)=8.7σ2=np(1-p)=30(0.29)(0.71)=6.177σ=6.177=2.485x−33e38.a.fx()=x!2−33e9(.0498)b.f(2)===.22412!21−33ec.f(1)==3(.0498)=.14941!d.P(x≥2)=1-f(0)-f(1)=1-.0498-.1494=.8008x−22e39.a.fx()=x!b.µ=6for3timeperiodsx−66ec.fx()=x!2−22e4(.1353)d.f(2)===.27062!26−66ee.f(6)==.16066!5-13www.khdaw.com 课后答案网www.khdaw.comChapter55−44ef.f(5)==.15635!40.a.µ=48(5/60)=43-44e(64)(.0183)f(3)===.19523!6b.µ=48(15/60)=1210-1212ef(10)==.104810!c.µ=48(5/60)=4Iexpect4callerstobewaitingafter5minutes.0-4www.khdaw.comf(0)=4e=.01830!Theprobabilitynonewillbewaitingafter5minutesis.0183.d.µ=48(3/60)=2.40-2.42.4ef(0)==.09070!Theprobabilityofnointerruptionsin3minutesis.0907.41.a.30perhourb.µ=1(5/2)=5/23−(5/2)(5/2)ef(3)==.21383!0−(5/2)(5/2)e−(5/2)c.f(0)==e=.08210!χ−µµe42.a.fx()=x!2−44e16(0.0183)f(2)===8(0.0183)=0.14652!2b.Fora3-monthperiod:µ=1c.Fora6-monthperiod:µ=20−22e−2f(0)==e=0.13530!5-14www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionsTheprobabilityof1ormoreflights=1-f(0)=1-0.1353=0.86470−1010e−1043.a.f(0)==e=.0000450!b.f(0)+f(1)+f(2)+f(3)f(0)=.000045(parta)1−1010ef(1)==.000451!Similarly,f(2)=.00225,f(3)=.0075www.khdaw.comandf(0)+f(1)+f(2)+f(3)=.010245c.2.5arrivals/15sec.periodUseµ=2.50−2.52.5ef(0)==.08210!d.1-f(0)=1-.0821=.917944.Poissondistributionappliesa.µ=1.25permonth0−1.251.25eb.f(0)==0.28650!1−1.251.25ec.f(1)==0.35811!d.P(Morethan1)=1-f(0)-f(1)=1-0.2865-0.3581=0.355445.a.For1week,µ=450/52=8.651−8.658.65e−8.65b.f(0)==e=0.00020!c.Fora1-dayperiod:µ=450/365=1.230–1.231.23e–1.23f(0)==e=0.29230!1–1.231.23ef(1)==1.23(0.2923)=0.35951Probabilityof2ormoredeaths=1-f(0)-f(1)=1-0.2923-0.3595=0.34825-15www.khdaw.com 课后答案网www.khdaw.comChapter5⎛⎞⎛3103−⎞⎛3!⎞⎛7!⎞⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝141−⎠⎝1!2!⎠⎝3!4!⎠(3)(35)46.a.f(1)====.50⎛10⎞10!210⎜4⎟4!6!⎝⎠⎛⎞⎛3103−⎞⎜⎟⎜⎟⎝⎠⎝222−⎠(3)(1)b.f(2)===.067⎛10⎞45⎜⎟⎝2⎠⎛⎞⎛3103−⎞⎜⎟⎜⎟⎝⎠⎝020−⎠(1)(21)c.f(0)===.4667⎛10⎞45⎜⎟www.khdaw.com⎝2⎠⎛⎞⎛3103−⎞⎜⎟⎜⎟⎝⎠⎝242−⎠(3)(21)d.f(2)===.30⎛10⎞210⎜⎟⎝4⎠⎛⎞⎛4154−⎞⎜⎟⎜⎟⎝⎠⎝3103−⎠(4)(330)47.f(3)===.4396⎛15⎞3003⎜⎟⎝10⎠48.HypergeometricwithN=10andr=6⎛⎞⎛⎞64⎜⎟⎜⎟⎝⎠⎝⎠21(15)(4)a.f(2)===.50⎛10⎞120⎜⎟⎝3⎠b.Mustbe0or1preferCokeClassic.⎛⎞⎛⎞64⎜⎟⎜⎟⎝⎠⎝⎠12(6)(6)f(1)===.30⎛10⎞120⎜⎟⎝3⎠⎛⎞⎛⎞64⎜⎟⎜⎟⎝⎠⎝⎠03(1)(4)f(0)===.0333⎛10⎞120⎜⎟⎝3⎠P(MajorityPepsi)=f(1)+f(0)=.33335-16www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributions49.Partsa,b&cinvolvethehypergeometricdistributionwithN=52andn=2a.r=20,x=2⎛20⎞⎛32⎞⎜⎟⎜⎟⎝2⎠⎝0⎠(190)(1)f(2)===.1433⎛52⎞1326⎜⎟⎝2⎠b.r=4,x=2⎛⎞⎛448⎞⎜⎟⎜⎟⎝⎠⎝20⎠(6)(1)f(2)===.0045⎛52⎞1326⎜⎟www.khdaw.com⎝2⎠c.r=16,x=2⎛16⎞⎛36⎞⎜⎟⎜⎟⎝2⎠⎝0⎠(120)(1)f(2)===.0905⎛52⎞1326⎜⎟⎝2⎠d.Part(a)providestheprobabilityofblackjackplustheprobabilityof2acesplustheprobabilityoftwo10s.Tofindtheprobabilityofblackjackwesubtracttheprobabilitiesin(b)and(c)fromtheprobabilityin(a).P(blackjack)=.1433-.0045-.0905=.048350.N=60n=10a.r=20x=0F20IF40IF40!Ib1gHG0KJHG10KJHG10!30!KJF40!IF10!50!If(0)===F60I60!HG10!30!KJHG60!KJHG10KJ10!50!40⋅39⋅38⋅37⋅36⋅35⋅34⋅33⋅32⋅31=60⋅59⋅58⋅57⋅56⋅55⋅54⋅53⋅52⋅51≈.01b.r=20x=1F20IF40IHG1KJHG9KJF40!IF10!50!If(0)==20HGKJHGKJF60I9!31!60!HG10KJ5-17www.khdaw.com 课后答案网www.khdaw.comChapter5≈.07c.1-f(0)-f(1)=1-.08=.92d.SameastheprobabilityonewillbefromHawaii.Inpartbthatwasfoundtoequalapproximately.07.⎛1114⎞⎛⎞⎜⎟⎜⎟⎝2⎠⎝3⎠(55)(364)51.a.f(2)===.3768⎛25⎞53,130⎜⎟⎝5⎠⎛14⎞⎛11⎞⎜⎟⎜⎟⎝2⎠⎝3⎠(91)(165)b.f(2)===.2826www.khdaw.com⎛25⎞53,130⎜⎟⎝5⎠⎛14⎞⎛11⎞⎜⎟⎜⎟⎝5⎠⎝0⎠(2002)(1)c.f(5)===.0377⎛25⎞53,130⎜⎟⎝5⎠⎛14⎞⎛11⎞⎜⎟⎜⎟⎝0⎠⎝5⎠(1)(462)d.f(0)===.0087⎛25⎞53,130⎜⎟⎝5⎠52.HypergeometricwithN=10andr=2.Focusontheprobabilityof0defectives,thentheprobabilityofrejectingtheshipmentis1-f(0).a.n=3,x=0⎛⎞⎛⎞28⎜⎟⎜⎟⎝⎠⎝⎠0356f(0)===.4667⎛10⎞120⎜⎟⎝3⎠P(Reject)=1-.4667=.5333b.n=4,x=0⎛⎞⎛⎞28⎜⎟⎜⎟⎝⎠⎝⎠0470f(0)===.3333⎛10⎞210⎜⎟⎝4⎠5-18www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionsP(Reject)=1-.3333=.6667c.n=5,x=0⎛⎞⎛⎞28⎜⎟⎜⎟⎝⎠⎝⎠0556f(0)===.2222⎛10⎞252⎜⎟⎝5⎠P(Reject)=1-.2222=.7778d.Continuetheprocess.n=7wouldberequiredwiththeprobabilityofrejecting=.933353.a.,b.andc.www.khdaw.comxf(x)xf(x)x-µ(x-µ)2(x-µ)2f(x)10.180.18-2.305.290.952220.180.36-1.301.690.608430.030.09-0.300.090.008140.381.520.700.490.744850.231.151.702.893.32351.003.305.6370E(x)=µ=3.30σ2=5.6370σ=5.6370=2.374254.a.andb.xf(x)xf(x)x-µ(x-µ)2(x-µ)2f(x)10.020.02-2.646.96960.13939220.060.12-1.642.68960.16137630.280.84-0.640.40960.11468840.542.160.360.12960.06998450.100.501.361.84960.1849601.003.640.670400f(x)≥0andΣf(x)=1E(x)=µ=3.64Var(x)=σ2=0.6704c.Peopledoappeartobelievethestockmarketisovervalued.Theaverageresponseisslightlyoverhalfwaybetween“fairlyvalued”and“somewhatovervalued.”55.a.xf(x)9.3010.205-19www.khdaw.com 课后答案网www.khdaw.comChapter511.2512.0513.20b.E(x)=Σxf(x)=9(.30)+10(.20)+11(.25)+12(.05)+13(.20)=10.65Expectedvalueofexpenses:$10.65millionc.Var(x)=Σ(x-µ)2f(x)=(9-10.65)2(.30)+(10-10.65)2(.20)+(11-10.65)2(.25)+(12-10.65)2(.05)+(13-10.65)2(.20)=2.1275www.khdaw.comd.LooksGood:E(Profit)=12-10.65=1.35millionHowever,thereisa.20probabilitythatexpenseswillequal$13millionandthecollegewillrunadeficit.56.a.n=20andx=3⎛20⎞317f(3)=⎜⎟(0.04)(0.04)=0.0364⎝3⎠b.n=20andx=0⎛20⎞020f(0)=⎜⎟(0.04)(0.96)=0.4420⎝0⎠c.E(x)=np=1200(0.04)=48Theexpectednumberofappealsis48.d.σ2=np(1-p)=1200(0.04)(0.96)=46.08σ=46.08=6.788257.a.WemusthaveE(x)=np≥10Withp=.4,thisleadsto:n(.4)≥10n≥25b.Withp=.12,thisleadsto:n(.12)≥10n≥83.33So,wemustcontact84peopleinthisagegrouptohaveanexpectednumberofinternetusersofatleast10.5-20www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionsc.σ=25(.4)(.6)=2.45d.σ=25(.12)(.88)=1.6258.Sincetheshipmentislargewecanassumethattheprobabilitiesdonotchangefromtrialtotrialandusethebinomialprobabilitydistribution.a.n=5⎛⎞505f(0)=⎜⎟(0.01)(0.99)=0.9510⎝⎠0⎛⎞514b.f(1)=⎜⎟(0.01)(0.99)=0.0480www.khdaw.com⎝⎠1c.1-f(0)=1-.9510=.0490d.No,theprobabilityoffindingoneormoreitemsinthesampledefectivewhenonly1%oftheitemsinthepopulationaredefectiveissmall(only.0490).Iwouldconsideritlikelythatmorethan1%oftheitemsaredefective.59.a.E(x)=np=100(.041)=4.1b.Var(x)=np(1-p)=100(.041)(.959)=3.9319σ=3.9319=1.982960.a.E(x)=800(.41)=328b.σ=np(1−p)=800(.41)(.59)=13.91c.Forthisonep=.59and(1-p)=.41,buttheansweristhesameasinpart(b).Forabinomialprobabilitydistribution,thevarianceforthenumberofsuccessesisthesameasthevarianceforthenumberoffailures.Ofcourse,thisalsoholdstrueforthestandarddeviation.61.µ=15probof20ormorearrivals=f(20)+f(21)+···=.0418+.0299+.0204+.0133+.0083+.0050+.0029+.0016+.0009+.0004+.0002+.0001+.0001=.124962.µ=1.5probof3ormorebreakdownsis1-[f(0)+f(1)+f(2)].1-[f(0)+f(1)+f(2)]=1-[.2231+.3347+.2510]=1-.80885-21www.khdaw.com 课后答案网www.khdaw.comChapter5=.191263.µ=10f(4)=.01893−33e64.a.f(3)==0.22403!b.f(3)+f(4)+···=1-[f(0)+f(1)+f(2)]0-33e-3f(0)==e=.04980!Similarly,f(1)=.1494,f(2)=.2240∴1-[.0498+.1494+.2241]=.5767www.khdaw.com65.HypergeometricN=52,n=5andr=4.F4IF48IHG2KJHG3KJ6(17296)a.==.0399F52I2,598,960HG5KJF4IF48IHG1KJHG4KJ4(194580)b.==.2995F52I2,598,960HG5KJF4IF48IHG0KJHG5KJ1,712,304c.==.6588F52I2,598,960HG5KJd.1-f(0)=1-.6588=.341266.UsetheHypergeometricprobabilitydistributionwithN=10,n=2,andr=4.F4IF6IHG1KJHG1KJ(4)(6)a.f(1)==.5333F10I45HG2KJF4IF6IHG2KJHG0KJ(6)(1)b.f(2)==.1333F10I45HG2KJ5-22www.khdaw.com 课后答案网www.khdaw.comDiscreteProbabilityDistributionsF4IF6IHG0KJHG2KJ(1)(15)c.f(0)==.3333F10I45HG2KJwww.khdaw.com5-23www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter6ContinuousCContinuousontinuousProbabilityPProbabilityrobabilityDistributionsDDistributionsistributionsLearningLLearningearningObjectivesOObjectivesbjectives1.Understandthedifferencebetweenhowprobabilitiesarecomputedfordiscreteandcontinuousrandomvariables.2.Knowhowtocomputeprobabilityvaluesforacontinuousuniformprobabilitydistributionandbewww.khdaw.comabletocomputetheexpectedvalueandvarianceforsuchadistribution.3.Beabletocomputeprobabilitiesusinganormalprobabilitydistribution.Understandtheroleofthestandardnormaldistributioninthisprocess.4.Beabletocomputeprobabilitiesusinganexponentialprobabilitydistribution.5.UnderstandtherelationshipbetweenthePoissonandexponentialprobabilitydistributions.6-1www.khdaw.com 课后答案网www.khdaw.comChapter6Solutions:SSolutions:olutions:1.a.f(x)321x.501.01.52.0b.P(x=1.25)=0.Theprobabilityofanysinglepointiszerosincetheareaunderthecurvewww.khdaw.comaboveanysinglepointiszero.c.P(1.0≤x≤1.25)=2(.25)=.50d.P(1.20135)=(1/20)(140-135)=0.25120140+d.Ex()==130minutes24.a.f(x)1.51.0.5x0123b.P(.25.60)=1(.40)=.405.a.LengthofInterval=261.2-238.9=22.3⎧1⎪for238.9≤x≤261.2fx()=⎨22.3⎪0elsewhere⎩b.Note:1/22.3=0.045P(x<250)=(0.045)(250-238.9)=0.4995Almosthalfdrivetheballlessthan250yards.6-3www.khdaw.com 课后答案网www.khdaw.comChapter6c.P(x≥255)=(0.045)(261.2-255)=0.279d.P(245≤x≤260)=(0.045)(260-245)=0.675e.P(x≥250)=1-P(x<250)=1-0.4995=0.5005Theprobabilityofanyonedrivingit250yardsormoreis0.5005.With60players,theexpectednumberdrivingit250yardsormoreis(60)(0.5005)=30.03.Rounding,Iwouldexpect30ofthesewomentodrivetheball250yardsormore.6.a.P(12≤x≤12.05)=.05(8)=.40b.P(x≥12.02)=.08(8)=.64c.Px(<11.98)+Px(>12.02)��������������.005(8)=.04.64=.08(8)www.khdaw.comTherefore,theprobabilityis.04+.64=.687.a.P(10,000≤x<12,000)=2000(1/5000)=.40Theprobabilityyourcompetitorwillbidlowerthanyou,andyougetthebid,is.40.b.P(10,000≤x<14,000)=4000(1/5000)=.80c.Abidof$15,000givesaprobabilityof1ofgettingtheproperty.d.Yes,thebidthatmaximizesexpectedprofitis$13,000.Theprobabilityofgettingthepropertywithabidof$13,000isP(10,000≤x<13,000)=3000(1/5000)=.60.Theprobabilityofnotgettingthepropertywithabidof$13,000is.40.Theprofityouwillmakeifyougetthepropertywithabidof$13,000is$3000=$16,000-13,000.Soyourexpectedprofitwithabidof$13,000isEP($13,000)=.6($3000)+.4(0)=$1800.Ifyoubid$15,000theprobabilityofgettingthebidis1,buttheprofitifyoudogetthebidisonly$1000=$16,000-15,000.Soyourexpectedprofitwithabidof$15,000isEP($15,000)=1($1000)+0(0)=$1,000.6-4www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributions8.σ=107080901001101201309.a.www.khdaw.comσ=535404550556065b..6826since45and55arewithinplusorminus1standarddeviationfromthemeanof50.c..9544since40and60arewithinplusorminus2standarddeviationsfromthemeanof50.10.-3-2-10+1+2+3a..3413b..4332c..4772d..49386-5www.khdaw.com 课后答案网www.khdaw.comChapter611.a..3413Theseprobabilityvaluesarereaddirectlyfromthetableofareasforthestandardb..4332normalprobabilitydistribution.SeeTable1inAppendixB.c..4772d..4938e..498612.a..2967b..4418c..5000-.1700=.3300www.khdaw.comd..0910+.5000=.5910e..3849+.5000=.8849f..5000-.2612=.238813.a..6879-.0239=.6640b..8888-.6985=.1903c..9599-.8508=.109114.a.Usingthetableofareasforthestandardnormalprobabilitydistribution,theareaof.4750correspondstoz=1.96.b.Usingthetable,theareaof.2291correspondstoz=.61.c.Lookinthetableforanareaof.5000-.1314=.3686.Thisprovidesz=1.12.d.Lookinthetableforanareaof.6700-.5000=.1700.Thisprovidesz=.44.15.a.Lookinthetableforanareaof.5000-.2119=.2881.Sincethevalueweareseekingisbelowthemean,thezvaluemustbenegative.Thus,foranareaof.2881,z=-.80.b.Lookinthetableforanareaof.9030/2=.4515;z=1.66.c.Lookinthetableforanareaof.2052/2=.1026;z=.26.d.Lookinthetableforanareaof.4948;z=2.56.e.Lookinthetableforanareaof.1915.Sincethevalueweareseekingisbelowthemean,thezvaluemustbenegative.Thus,z=-.50.16.a.Lookinthetableforanareaof.5000-.0100=.4900.Theareavalueinthetableclosestto.4900providesthevaluez=2.33.b.Lookinthetableforanareaof.5000-.0250=.4750.Thiscorrespondstoz=1.96.6-6www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributionsc.Lookinthetableforanareaof.5000-.0500=.4500.Since.4500isexactlyhalfwaybetween.4495(z=1.64)and.4505(z=1.65),weselectz=1.645.However,z=1.64orz=1.65arealsoacceptableanswers.d.Lookinthetableforanareaof.5000-.1000=.4000.Theareavalueinthetableclosestto.4000providesthevaluez=1.28.17.Convertmeantoinches:µ=69a.Atx=72z=72-69=13P(x≤72)=0.5000+0.3413=0.8413P(x>72)=1-0.8413=0.1587www.khdaw.comb.Atx=60z=60-69=–33P(x≥60)=0.5000+0.4986=0.9986P(x<60)=1-0.9986=0.0014c.Atx=70z=70-69=0.333P(x≤70)=0.5000+0.1293=0.6293Atx=66z=66-69=–13P(x≤66)=0.5000-0.3413=0.1587P(66≤x≤70)=P(x≤70)-P(x≤66)=0.6293-0.1587=0.4706d.P(x≤72)=1-P(x>72)=1-0.1587=0.841318.a.FindP(x≥60)Atx=60z=60-49=0.6916P(x<60)=0.5000+0.2549=0.7549P(x≥60)=1-P(x<60)=0.2451b.FindP(x≤30)Atx=30z=30-49=–1.1916P(x≤30)=0.5000-0.3830=0.1170c.Findz-scoresothatP(z≥z-score)=0.10z-score=1.28cutsoff10%inuppertailNow,solveforcorrespondingvalueofx.6-7www.khdaw.com 课后答案网www.khdaw.comChapter6x−491.28=16x=49+(16)(1.28)=69.48So,10%ofsubscribersspend69.48minutesormorereadingTheWallStreetJournal.19.Wehaveµ=3.5andσ=.8.5.03.5−a.z=≈1.88.8P(x>5.0)=P(z>1.88)=1-P(z<1.88)=1-.9699=.0301Therainfallexceeds5inchesin3.01%oftheAprils.www.khdaw.com33.5−b.z=≈−.63.8P(x<3.0)=P(z<-.63)=P(z>.63)=1-P(z<.63)=1-.7357=.2643Therainfallislessthan3inchesin26.43%oftheAprils.c.z=1.28cutsoffapproximately.10intheuppertailofanormaldistribution.x=3.5+1.28(.8)=4.524Ifitrains4.524inchesormore,Aprilwillbeclassifiedasextremelywet.20.Weuseµ=27andσ=81127−a.z==−28P(x≤11)=P(z≤-2)=.5000-.4772=.0228Theprobabilityarandomlyselectedsubscriberspendslessthan11hoursonthecomputeris.025.4027−b.z=≈1.638P(x>40)=P(z>1.63)=1-P(z≤1.63)=1-.9484=.05165.16%ofsubscribersspendover40hoursperweekusingthecomputer.c.Az-valueof.84cutsoffanareaof.20intheuppertail.x=27+.84(8)=33.72Asubscriberwhousesthecomputer33.72hoursormorewouldbeclassifiedasaheavyuser.6-8www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributions21.Fromthenormalprobabilitytables,az-valueof2.05cutsoffanareaofapproximately.02intheuppertailofthedistribution.x=µ+zσ=100+2.05(15)=130.75Ascoreof131orbettershouldqualifyapersonformembershipinMensa.22.Useµ=441.84andσ=90a.At400400441.84−z=≈−.4690At500500441.84−z=≈.6590www.khdaw.comP(0≤z<.65)=.2422P(-.46≤z<0)=.1772P(400≤z≤500)=.1772+.2422=.4194Theprobabilityaworkerearnsbetween$400and$500is.4194.b.Mustfindthez-valuethatcutsoffanareaof.20intheuppertail.Usingthenormaltables,wefindz=.84cutsoffapproximately.20intheuppertail.So,x=µ+zσ=441.84+.84(90)=517.44Weeklyearningsof$517.44orabovewillputaproductionworkerinthetop20%.250441.84−c.At250,z=≈−2.1390P(x≤250)=P(z≤-2.13)=.5000-.4834=.0166Theprobabilityarandomlyselectedproductionworkerearnslessthan$250perweekis.0166.6080−23.a.z==−2Areatoleftis.5000-.4772=.022810b.Atx=606080−z==−2Areatoleftis.022810Atx=757580−z==−.5Areatoleftis.308510P(60≤x≤75)=.3085-.0228=.28579080−c.z==1Area=.5000-.3413=.1587106-9www.khdaw.com 课后答案网www.khdaw.comChapter6Therefore15.87%ofstudentswillnotcompleteontime.(60)(.1587)=9.522Wewouldexpect9.522studentstobeunabletocompletetheexamintime.xi24.a.x=∑=902.75n2∑(x−x)is==114.185n−1Wewillusexasanestimateofµandsasanestimateofσinparts(b)-(d)below.www.khdaw.comb.Rememberthedataareinthousandsofshares.At800800902.75−z=≈−.90114.185P(x≤800)=P(z≤-.90)=1-P(z≤.90)=1-.8159=.1841Theprobabilitytradingvolumewillbelessthan800millionsharesis.1841c.At10001000902.75−z=≈.85114.185P(x≥1000)=P(z≥.85)=1-P(z≤.85)=1-.8023=.1977Theprobabilitytradingvolumewillexceed1billionsharesis.1977d.Az-valueof1.645cutsoffanareaof.05intheuppertailx=µ+zσ=902.75+1.645(114.185)=1,090.584Theyshouldissueapressreleaseanytimesharevolumeexceeds1,091million.25.a.FindP(x>100)Atx=100z=100-110=–0.520P(x>100)=P(z≤.5)=0.6915b.FindP(x≤90)Atx=90z=90-110=–120P(x≤90)=.5000-.3413=0.15876-10www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributionsc.FindP(80≤x≤130)Atx=130z=130-110=120P(x≤130)=0.8413Atx=8080110−z==−1.5Areatoleftis.066820P(80≤x≤130)=.8413-.0668=.774526.a.P(x≤6)=1-e-6/8=1-.4724=.5276www.khdaw.comb.P(x≤4)=1-e-4/8=1-.6065=.3935c.P(x≥6)=1-P(x≤6)=1-.5276=.4724d.P(4≤x≤6)=P(x≤6)-P(x≤4)=.5276-.3935=.1341−x0/327.a.Px(≤x)1=−e0b.P(x≤2)=1-e-2/3=1-.5134=.4866c.P(x≥3)=1-P(x≤3)=1-(1-e−3/3)=e-1=.3679d.P(x≤5)=1-e-5/3=1-.1889=.8111e.P(2≤x≤5)=P(x≤5)-P(x≤2)=.8111-.4866=.324528.a.P(x<10)=1-e-10/20=.3935b.P(x>30)=1-P(x≤30)=1-(1-e-30/20)=e-30/20=.2231c.P(10≤x≤30)=P(x≤30)-P(x≤10)=(1-e-30/20)-(1-e-10/20)=e-10/20-e-30/20=.6065-.2231=.38346-11www.khdaw.com 课后答案网www.khdaw.comChapter629.a.f(x).09.08.07.06.05.04.03.02www.khdaw.com.01x6121824b.P(x≤12)=1-e-12/12=1-.3679=.6321c.P(x≤6)=1-e-6/12=1-.6065=.3935d.P(x≥30)=1-P(x<30)=1-(1-e-30/12)=.082130.a.50hoursb.P(x≤25)=1-e-25/50=1-.6065=.3935c.P(x≥100)=1-(1-e-100/50)=.135331.a.P(x<2)=1-e-2/2.78=.5130b.P(x>5)=1-P(x≤5)=1-(1-e-5/2.78)=e-5/2.78=.1655c.P(x>2.78)=1-P(x≤2.78)=1-(1-e-2.78/2.78)=e-1=.3679Thismayseemsurprisingsincethemeanis2.78minutes.But,fortheexponentialdistribution,theprobabilityofavaluegreaterthanthemeanissignificantlylessthantheprobabilityofavaluelessthanthemean.32.a.IftheaveragenumberoftransactionsperyearfollowsthePoissondistribution,thetimebetweentransactionsfollowstheexponentialdistribution.So,1µ=ofayear306-12www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributions11and==30µ1/30thenf(x)=30e-30xb.Amonthis1/12ofayearso,⎛1⎞⎛1⎞−30/12−30/12Px⎜>⎟=−1Px⎜≤⎟=−1(1−e)=e=.0821⎝12⎠⎝12⎠TheprobabilityofnotransactionduringJanuaryisthesameastheprobabilityofnotransactionduringanymonth:.0821c.Since1/2monthis1/24ofayear,wecompute,⎛1⎞−30/24www.khdaw.comPx⎜≤⎟=−1e=−1.2865=.7135⎝24⎠33.a.Letx=salesprice($1000s)⎧1⎪for200≤x≤225fx()=⎨25⎪0elsewhere⎩b.P(x≥215)=(1/25)(225-215)=0.40c.P(x<210)=(1/25)(210-200)=0.40d.E(x)=(200+225)/2=212,500Ifshewaits,herexpectedsalepricewillbe$2,500higherthanifshesellsitbacktohercompanynow.However,thereisa0.40probabilitythatshewillgetless.It’saclosecall.But,theexpectedvalueapproachtodecisionmakingwouldsuggestsheshouldwait.34.a.Foranormaldistribution,themeanandthemedianareequal.µ=63,000b.Findthez-scorethatcutsoff10%inthelowertail.z-score=-1.28Solvingforx,–1.28=x–63,00015,000x=63,000-1.28(15000)=43,800Thelower10%ofmortgagedebtis$43,800orless.6-13www.khdaw.com 课后答案网www.khdaw.comChapter6c.FindP(x>80,000)Atx=80,000z=80,000–63,000=1.1315,000P(x>80,000)=1.0000-.8708=0.1292d.Findthez-scorethatcutsoff5%intheuppertail.z-score=1.645.Solveforx.1.645=x–63,00015,000x=63,000+1.645(15,000)www.khdaw.com=87,675Theupper5%ofmortgagedebtisinexcessof$87,675.35.a.P(defect)=1-P(9.85≤x≤10.15)=1-P(-1≤z≤1)=1-.6826=.3174Expectednumberofdefects=1000(.3174)=317.4b.P(defect)=1-P(9.85≤x≤10.15)=1-P(-3≤z≤3)=1-.9972=.0028Expectednumberofdefects=1000(.0028)=2.8c.Reducingtheprocessstandarddeviationcausesasubstantialreductioninthenumberofdefects.36.a.At11%,z=-1.23–1.23=x–µ1800–2071=σσ1800–2071Therefore,σ==$220.33–1.2320002071−b.z==−.32Areatoleftis.5000-.3255=.3745220.336-14www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributions25002071−z==1.95Areatoleftis.9744220.33P(2000≤x≤2500)=.9744-.3745=.5999c.z=-1.88x=2071-1.88(220.33)=$1656.7837.µ=10,000σ=1500a.Atx=12,00012,00010,000−z==1.33Areatoleftis.90821500www.khdaw.comP(x>12,000)=1.0000-.9082=.0918b.At.95x-10,000z=1.645=1500Therefore,x=10,000+1.645(1500)=12,468.95%0.0510,00012,46812,468tubesshouldbeproduced.38.a.Atx=200200150−z==2Area=.477225P(x>200)=.5-.4772=.0228b.ExpectedProfit=ExpectedRevenue-ExpectedCost=200-150=$5039.a.FindP(80,000≤x≤150,000)Atx=150,000z=150,000–126,681=0.7830,000P(x≤150,000)=0.78236-15www.khdaw.com 课后答案网www.khdaw.comChapter6Atx=80,000z=80,000–126,681=–1.5630,000P(x≤80,000)=.5000-.4406=0.0594P(80,000≤x≤150,000)=0.7823-0.0594=0.7229b.FindP(x<50,000)Atx=50,000z=50,000–126,681=–2.5630,000P(x<50,000)=.5000-.4948=0.0052www.khdaw.comc.Findthez-scorecuttingoff95%inthelefttail.z-score=1.645.Solveforx.1.645=x–126,68130,000x=126,681+1.645(30,000)=176,031Theprobabilityis0.95thatthenumberoflostjobswillnotexceed176,031.40.a.At400,400450−z==−.500100Areatoleftis.3085At500,500450−z==+.500100Areatoleftis.6915P(400≤x≤500)=.6915-.3085=.383038.3%willscorebetween400and500.b.At630,630450−z==1.8010096.41%doworseand3.59%dobetter.c.At480,480450−z==.30100Areatoleftis.617938.21%areacceptable.6-16www.khdaw.com 课后答案网www.khdaw.comContinuousProbabilityDistributions41.a.At75,00075,00067,000−z=≈1.147,000P(x>75,000)=P(z>1.14)=1-P(z≤1.14)=1-.8729=.1271Theprobabilityofawomanreceivingasalaryinexcessof$75,000is.1271b.At75,00075,00065,500−z=≈1.367,000P(x>75,000)=P(z>1.36)=1-P(z≤1.36)=1-.9131=.0869Theprobabilityofamanreceivingasalaryinexcessof$75,000is.0869www.khdaw.comc.Atx=50,00050,00067,000−z=≈−2.437,000P(x<50,000)=P(z<-2.43)=1-P(z<2.43)=1-.9925=.0075Theprobabilityofawomanreceivingasalarybelow$50,000isverysmall:.0075d.Theanswertothisisthemalecopywritersalarythatcutsoffanareaof.01intheuppertailofthedistributionformalecopywriters.Usez=2.33x=65,500+2.33(7,000)=81,810Awomanwhomakes$81,810ormorewillearnmorethan99%ofhermalecounterparts.42.σ=.6At2%z=-2.05x=18x−µ18−µz=∴−2.05=σ.6µ=18+2.05(.6)=19.23oz.6-17www.khdaw.com 课后答案网www.khdaw.comChapter60.0218µ=19.23Themeanfillingweightmustbe19.23oz.43.a.P(x≤15)=1-e-15/36=1-.6592=.3408www.khdaw.comb.P(x≤45)=1-e-45/36=1-.2865=.7135ThereforeP(15≤x≤45)=.7135-.3408=.3727c.P(x≥60)=1-P(x<60)=1-(1-e-60/36)=.188944.a.4hoursb.f(x)=(1/4)e-x/4forx≥0c.P(x≥1)=1-P(x<1)=1-(1-e-1/4)=.7788d.P(x>8)=1-P(x≤8)=e-8/4=.13531−x/1.245.a.fx()=eforx≥01.2b.P(.5≤x≤1.0)=P(x≤1.0)-P(x≤.5)=(1-e-1/1.2)-(1-e-.5/1.2)=.5654-.3408=.2246c.P(x>1)=1-P(x≤1)=1-.5654=.4346146.a.=0.5thereforeµ=2minutes=meantimebetweentelephonecallsµb.Note:30seconds=.5minutesP(x≤.5)=1-e-.5/2=1-.7788=.2212c.P(x≤1)=1-e-1/2=1-.6065=.3935d.P(x≥5)=1-P(x<5)=1-(1-e-5/2)=.08216-18www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter7SamplingSSamplingamplingandananddSamplingSSamplingamplingDistributionsDDistributionsistributionsLearningLLearningearningObjectivesOObjectivesbjectives1.Understandtheimportanceofsamplingandhowresultsfromsamplescanbeusedtoprovideestimatesofpopulationcharacteristicssuchasthepopulationmean,thepopulationstandarddeviationand/orthepopulationproportion.www.khdaw.com2.Knowwhatsimplerandomsamplingisandhowsimplerandomsamplesareselected.3.Understandtheconceptofasamplingdistribution.4.Knowthecentrallimittheoremandtheimportantroleitplaysinsampling.5.Specificallyknowthecharacteristicsofthesamplingdistributionofthesamplemean(x)andthesamplingdistributionofthesampleproportion(p).6.Becomeawareofthepropertiesofpointestimatorsincludingunbiasedness,consistency,andefficiency.7.Learnaboutavarietyofsamplingmethodsincludingstratifiedrandomsampling,clustersampling,systematicsampling,conveniencesamplingandjudgmentsampling.8.Knowthedefinitionofthefollowingterms:simplerandomsamplingfinitepopulationcorrectionfactorsamplingwithreplacementstandarderrorsamplingwithoutreplacementunbiasednesssamplingdistributionconsistencypointestimatorefficiency7-1www.khdaw.com 课后答案网www.khdaw.comChapter7Solutions:SSolutions:olutions:1.a.AB,AC,AD,AE,BC,BD,BE,CD,CE,DEb.With10samples,eachhasa1/10probability.c.EandCbecause8and0donotapply.;5identifiesE;7doesnotapply;5isskippedsinceEisalreadyinthesample;3identifiesC;2isnotneededsincethesampleofsize2iscomplete.2.Usingthelast3-digitsofeach5-digitgroupingprovidestherandomnumbers:601,022,448,147,229,553,147,289,209Numbersgreaterthan350donotapplyandthe147canonlybeusedonce.Thus,thesimplerandomsampleoffourincludes22,147,229,and289.www.khdaw.com3.459,147,385,113,340,401,215,2,33,3484.a.6,8,5,4,1IBM,Microsoft,Intel,GeneralElectric,AT&TN!10!3,628,500b.===252nN!(−n)!5!(105)!−(120)(120)5.283,610,39,254,568,353,602,421,638,1646.2782,493,825,1807,2897.108,290,201,292,322,9,244,249,226,125,(continuingatthetopofcolumn9)147,and113.8.13,8,23,25,18,5Thesecondoccurrencesofrandomnumbers13and25areignored.Washington,Clemson,Oklahoma,Colorado,USCandWisconsin9.511,791,99,671,152,584,45,783,301,568,754,75010.finite,infinite,infinite,infinite,finite5411.a.x=Σx/n==9i62Σ(x−x)ib.s=n−1Σ(x−x)2=(-4)2+(-1)2+12(-2)2+12+52=48i48s==3.16−17-2www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributions12.a.p=75/150=.50b.p=55/150=.366746513.a.x=Σx/n==93i5b.2xi(xi−x)(xi−x)94+11100+74985-86494+1192-11www.khdaw.comTotals46501162Σ(x−x)116is===5.39n−1414.a.149/784=.19b.251/784=.32c.Totalreceivingcash=149+219+251=619619/784=.7970015.a.x=Σxn/==7yearsi102Σ(x−x)20.2ib.s===1.5yearsn−1101−16.p=1117/1400=.8017.a.595/1008=.59b.332/1008=.33c.81/1008=.0818.a.E(x)=µ=200b.σ=σ/n=50/100=5xc.NormalwithE(x)=200andσ=5xd.Itshowstheprobabilitydistributionofallpossiblesamplemeansthatcanbeobservedwithrandomsamplesofsize100.Thisdistributioncanbeusedtocomputetheprobabilitythatxiswithinaspecified±fromµ.7-3www.khdaw.com 课后答案网www.khdaw.comChapter719.a.ThesamplingdistributionisnormalwithE(x)=µ=200σ=σ/n=50/100=5xFor±5,(x-µ)=5x−µ5z===1σ5xArea=.3413x2.6826b.For±10,(x-µ)=10x−µ10www.khdaw.comz===2σ5xArea=.4772x2.954420.σ=σ/nxσ=25/50=3.54xσ=25/100=2.50xσ=25/150=2.04xσ=25/200=1.77xThestandarderrorofthemeandecreasesasthesamplesizeincreases.21.a.σ=σ/n=10/50=1.41xb.n/N=50/50,000=.001Useσ=σ/n=10/50=1.41xc.n/N=50/5000=.01Useσ=σ/n=10/50=1.41xd.n/N=50/500=.10N−nσ500−5010Useσ===1.34UsexN−1n500−150Note:Onlycase(d)wheren/N=.10requirestheuseofthefinitepopulationcorrectionfactor.Notethatσisapproximatelythesameeventhoughthepopulationsizevariesfrominfiniteto500.x7-4www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributions22.a.Usingthecentrallimittheorem,wecanapproximatethesamplingdistributionofxwithanormalprobabilitydistributionprovidedn≥30.b.n=30σ=σ/n=50/30=9.13xx400www.khdaw.comn=40σ=σ/n=50/40=7.91x40023.a.σ=σ/n=16/50=2.26xFor±2,(x−µ)=2x−µ2z===0.88σ2.26xArea=.3106xx2.621216b.σ==1.60x100x−µ2z===1.25σ1.60xArea=.3944x2.788816c.σ==1.13x200x−µ2z===1.77σ1.13xArea=.4616x2.92327-5www.khdaw.com 课后答案网www.khdaw.comChapter716d.σ==0.80x400x−µ2z===2.50σ0.80xArea=.4938x2.9876e.Thelargersampleprovidesahigherprobabilitythatthesamplemeanwillbewithin±2ofµ.24.a.www.khdaw.comσ=σ/n=4000/60=516.40xx51,800E(x)ThenormaldistributionisbasedontheCentralLimitTheorem.b.Forn=120,E(x)remains$51,800andthesamplingdistributionofxcanstillbeapproximatedbyanormaldistribution.However,σisreducedto4000/120=365.15.xc.Asthesamplesizeisincreased,thestandarderrorofthemean,σ,isreduced.Thisappearsxlogicalfromthepointofviewthatlargersamplesshouldtendtoprovidesamplemeansthatareclosertothepopulationmean.Thus,thevariabilityinthesamplemean,measuredintermsofσ,xshoulddecreaseasthesamplesizeisincreased.σ=σ/n=4000/60=516.40x51,30051,80052,30025.a.52,300-51,800z==+.97516.407-6www.khdaw.comx 课后答案网www.khdaw.comSamplingandSamplingDistributionsArea=.3340x2.6680b.σ=σ/n=4000/120=365.15x52,300-51,800z==+1.37365.15Area.4147x2.829426.a.AnormaldistributionE(x)=1.20σ=σ/n=0.10/50=0.014xwww.khdaw.com1.22−1.20b.z==1.41Area=0.42070.0141.18−1.20z==−1.41Area=0.42070.014probability=0.4207+0.4207=0.84141.21−1.20c.z==+0.71Area=0.26120.0141.19−1.20z==−0.71Area=0.26120.014probability=0.2612+0.2612=0.522427.a.E(x)=1017σ=σ/n=100/75=11.55x10271017−b.z==0.87Area=0.307811.5510071017−z==−0.87Area=0.307811.55probability=0.3078+0.3078=0.615610371017−c.z==1.73Area=0.458211.559971017−z==−1.73Area=0.458211.55probability=0.4582+0.4582=0.91647-7www.khdaw.com 课后答案网www.khdaw.comChapter7x−34,00028.a.z=σ/nError=x-34,000=250250n=30z==.68.2518x2=.50362000/30250n=50z==.88.3106x2=.62122000/50250n=100z==1.25.3944x2=.78882000/100250n=200z==1.77.4616x2=.92322000/200250www.khdaw.comn=400z==2.50.4938x2=.98762000/400b.Alargersampleincreasestheprobabilitythatthesamplemeanwillbewithinaspecifieddistancefromthepopulationmean.Inthesalaryexample,theprobabilityofbeingwithin±250ofµrangesfrom.5036forasampleofsize30to.9876forasampleofsize400.29.a.E(x)=982σ=σ/n=210/40=33.2xx−µ100z===3.01σ/n210/40.4987x2=.9974x−µ25b.z===.75σ/n210/40.2734x2=.5468c.Thesamplewithn=40hasaveryhighprobability(.9974)ofprovidingasamplemeanwithin±$100.However,thesamplewithn=40onlyhasa.5468ofprovidingasamplemeanwithin±$25.Alargersamplesizeisdesirableifthe±$25isneeded.30.a.Normaldistribution,E(x)=166,500σ=σ/n=42,000/100=4200xx−µ10,000b.z===2.38(.4913x2)=.9826σ/n4,200c.$5000z=5000/4200=1.19(.3830x2)=.7660$2500z=2500/4200=.60(.2257x2)=.45147-8www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributions$1000z=1000/4200=.24(.0948x2)=.1896d.Increasesamplesizetoimproveprecisionoftheestimate.Samplesizeof100onlyhasa.4514probabilityofbeingwithin±$2,500.31.a.σ=σ/n=5200/30=949.39x1000z==1.05Area=0.3531949.39Probability=0.3531x2=0.7062b.σ=σ/n=5200/50=735.39x1000z==1.36Area=0.4131www.khdaw.com735.39Probability=0.4131x2=0.8262c.σ=σ/n=5200/100=520x1000z==1.92Area=0.4726520Probability=0.4726x2=0.9452d.Recommendn=10032.a.n/N=40/4000=.01<.05;therefore,thefinitepopulationcorrectionfactorisnotnecessary.b.WiththefinitepopulationcorrectionfactorN−nσ4000−408.2σ===1.29xN−1n4000−140Withoutthefinitepopulationcorrectionfactorσ=σ/n=1.30xIncludingthefinitepopulationcorrectionfactorprovidesonlyaslightlydifferentvalueforσthanxwhenthecorrectionfactorisnotused.c.x−µ2z===1.541.301.30Area.4382x2.876433.a.E(p)=p=.407-9www.khdaw.com 课后答案网www.khdaw.comChapter7p(1−p)0.40(0.60)b.σ===0.0490pn100c.NormaldistributionwithE(p)=.40andσ=.0490pd.Itshowstheprobabilitydistributionforthesampleproportionp.34.a.E(p)=.40p(1−p)0.40(0.60)σ===0.0346pn200p−p0.03z===0.87σ0.0346pwww.khdaw.comArea.3078x2.6156b.p−p0.05z===1.45σ0.0346pArea.4265x2.8530p(1−p)35.σ=pn(0.55)(0.45)σ==0.0497p100(0.55)(0.45)σ==0.0352p200(0.55)(0.45)σ==0.0222p500(0.55)(0.45)σ==0.0157p1000σdecreasesasnincreasesp(0.30)(0.70)36.a.σ==0.0458p100p−p0.04z===0.87σ0.0458pArea=0.3078x2=0.61567-10www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributions(0.30)(0.70)b.σ==0.0324p200p−p0.04z===1.23σ0.0324pArea=0.3907x2=0.7814(0.30)(0.70)c.σ==0.0205pwww.khdaw.com500p−p0.04z===1.95σ0.0205pArea=0.4744x2=0.9488(0.30)(0.70)d.σ==0.0145p1000p−p0.04z===2.76σ0.0145pArea=0.4971x2=0.9942e.Withalargersample,thereisahigherprobabilitypwillbewithin±.04ofthepopulationproportionp.37.a.p(1−p)0.30(0.70)σ===0.0458pn100p0.30Thenormaldistributionisappropriatebecausenp=100(.30)=30andn(1-p)=100(.70)=70arebothgreaterthan5.b.P(.20≤p≤.40)=?7-11www.khdaw.com 课后答案网www.khdaw.comChapter7.40-.30z==2.18.0458Area.4854x2.9708c.P(.25≤p≤.35)=?.35-.30z==1.09.0458Area.3621x2.724238.a.E(p)=.76p(1−p)0.76(1−0.76)σ===0.0214www.khdaw.compn4000.79−0.76b.z==1.40Area=0.41920.02140.73−0.76z==−1.40Area=0.41920.0214probability=0.4192+0.4192=0.8384p(1−p)0.76(1−0.76)c.σ===0.0156pn7500.79−0.76z==1.92Area=0.47260.01560.73−0.76z==−1.92Area=0.47260.0156probability=0.4726+0.4726=0.945239.a.NormaldistributionE(p)=.50p(1−p)(.50)(1.50)−σ===.0206pn589p−p.04b.z===1.94σ.0206p.4738x2=.94767-12www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributionsp−p.03c.z===1.46σ.0206p.4279x2=.8558p−p.02d.z===.97σ.0206p.3340x2=.668040.a.NormaldistributionE(p)=0.25www.khdaw.comp(1−p)(0.25)(0.75)σ===0.0306pn2000.03b.z==0.98Area=0.33650.0306probability=0.3365x2=0.67300.05c.z==1.63Area=0.44840.0306probability=0.4484x2=0.896841.a.E(p)=0.37p(1−p)(0.37)(1−0.37)σ===0.0153pn10000.40−0.37b.z==1.96Area=0.47500.01530.34−0.37z==−1.96Area=0.47500.0153probability=0.4750+0.4750=0.9500p(1−p)(0.37)(1−0.37)c.σ===0.0216pn5000.40−0.37z==1.39Area=0.41770.02160.34−0.37z==−1.39Area=0.41770.0216probability=0.4177+0.4177=0.83547-13www.khdaw.com 课后答案网www.khdaw.comChapter742.a.p(1−p)0.15(0.85)σ===0.0505pn50www.khdaw.comp0.15b.P(.12≤p≤.18)=?.18-.15z==.59.0505Area.2224x2.4448c.P(p≥.10)=?.10-.15z==-.99.0505Area.3389+.5000.838943.a.E(p)=0.17p(1−p)(0.17)(1−0.17)σ===0.01328pn8000.19−0.17b.z==1.51Area=0.43450.013280.34−0.37z==−1.51Area=0.43450.01328probability=0.4345+0.4345=0.8690p(1−p)(0.17)(1−0.17)c.σ===0.0094pn16000.19−0.17z==2.13Area=0.48340.00947-14www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributions0.15−0.17z==−2.13Area=0.48340.0094probability=0.4834+0.4834=0.966844.112,145,73,324,293,875,318,61845.a.NormaldistributionE(x)=3σ1.2σ===.17xn50www.khdaw.comx−µ.25b.z===1.47σ/n1.2/50.4292x2=.858446.a.NormaldistributionE(x)=31.5σ12σ===1.70xn501b.z==0.59Area=0.22241.70probability=0.2224x2=0.44483c.z==1.77Area=0.46161.70probability=0.4616x2=0.923247.a.E(x)=$24.07σ4.80σ===0.44xn1200.50z==1.14Area=0.37290.44probability=0.3729x2=0.74581.00b.z==2.28Area=0.48870.44probability=0.4887x2=0.97747-15www.khdaw.com 课后答案网www.khdaw.comChapter7σ6048.a.σ===8.49xn50b.z=(115-115)/8.49=0Area=.5000c.z=5/8.49=.59Area=.2224z=-5/8.49=-.59Area=.2224probability=.2224+.2224=.4448σ60d.σ===6xn100www.khdaw.comz=5/6=.83Area=.2967z=-5/6=-.83Area=.2967probability=.2967+.2967=.5934N−nσ49.a.σ=xN−1nN=20002000−50144σ==20.11x2000−150N=50005000−50144σ==20.26x5000−150N=10,00010,000−50144σ==20.31x10,000−150Note:Withn/N≤.05forallthreecases,commonstatisticalpracticewouldbetoignore144thefinitepopulationcorrectionfactoranduseσ==20.36foreachcase.x50b.N=200025z==1.2420.11Area.3925x2.7850N=500025z==1.2320.26Area.39077-16www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributionsx2.7814N=10,00025z==1.2320.31Area.3907x2.7814www.khdaw.comAllprobabilitiesareapproximately.78σ50050.a.σ===20xnnn500/20=25andn=(25)2=625b.For±25,25z==1.2520Area.3944x2.788851.Samplingdistributionofxσσσ==xn300.050.05x1.9µ2.11.9+2.1µ==22Theareabetweenµ=2and2.1mustbe.45.Anareaof.45inthestandardnormaltableshowsz=1.645.Thus,2.1+2.0µ==1.645σ/30Solveforσ.7-17www.khdaw.com 课后答案网www.khdaw.comChapter7(0.1)30σ==0.331.64552.a.E(p)=0.74p(1−p)(0.74)(1−0.74)σ===0.031pn200b.z=.04/.031=1.29Area=.4015z=-.04/.031=-1.29Area=.4015probability=.4015+.4015=.8030www.khdaw.comc.z=.02/.031=.64Area=.2389z=-.02/.031=-.64Area=.2389probability=.2389+.2389=.4778p(1−p)(0.40)(0.60)53.σ===0.0245pn400P(p≥.375)=?.375-.40z==-1.02.0245Area.3461P(p≥.375)=.3461+.5000=.8461p(1−p)(.71)(1.71)−54.a.σ===.0243pn350p−p.05z===2.06σ.0243p.4803x2=.9606p−p.75.71−b.z===1.65σ.0243pArea=.4505P(p≥.75)=.5000-.4505=.049555.a.NormaldistributionwithE(p)=.15andp(1−p)(0.15)(0.85)σ===0.0292pn1507-18www.khdaw.com 课后答案网www.khdaw.comSamplingandSamplingDistributionsb.P(.12≤p≤.18)=?.18-.15z==1.03.0292Area.3485x2.6970p(1−p).25(.75)56.a.σ===.0625pnnSolveforn.25(.75)n==482www.khdaw.com(.0625)b.NormaldistributionwithE(p)=.25andσ=.0625xc.P(p≥.30)=?.30.25−z==.80.0625Area.2881ThusP(.25≤p≤.30)=.2881andP(p≥.30)=.5000-.2881=.21197-19www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter8IntervalIIntervalntervalEstimationEEstimationstimationLearningLLearningearningObjectivesOObjectivesbjectives1.Knowhowtoconstructandinterpretanintervalestimateofapopulationmeanand/orapopulationproportion.2.Understandtheconceptofasamplingerror.www.khdaw.com3.Beabletouseknowledgeofasamplingdistributiontomakeprobabilitystatementsaboutthesamplingerror.4.Understandandbeabletocomputethemarginoferror.5.Learnaboutthetdistributionanditsuseinconstructinganintervalestimateforapopulationmean.6.Beabletodeterminethesizeofasimplerandomsamplenecessarytoestimateapopulationmeanand/orapopulationproportionwithaspecifiedlevelofprecision.7.Knowthedefinitionofthefollowingterms:confidenceintervalprecisionconfidencecoefficientsamplingerrorconfidencelevelmarginoferrordegreesoffreedom8-1www.khdaw.com 课后答案网www.khdaw.comChapter8Solutions:SSolutions:olutions:1.a.σ=σ/n=5/40=0.79xb.At95%,zσ/n=1.96(5/40)=1.552.a.32±1.645(6/50)32±1.4(30.6to33.4)b.32±1.96(6/50)32±1.66(30.34to33.66)www.khdaw.comc.32±2.576(6/50)32±2.19(29.81to34.19)3.a.80±1.96(15/60)80±3.8(76.2to83.8)b.80±1.96(15/120)80±2.68(77.32to82.68)c.Largersampleprovidesasmallermarginoferror.4.126±1.96(s/n)16.071.96=4n1.96(16.07)n==7.8744n=625.a.1.96/σn=1.96(5.00/49)1.40=b.24.80±1.40or(23.40to26.20)6.x±1.96(s/n)369±1.96(50/250)369±6.20(362.80to375.20)8-2www.khdaw.com 课后答案网www.khdaw.comIntervalEstimation7.x±z(/σn).0253.37±1.96(.28/120)3.37±.05(3.32to3.42)σ8.a.x±zα/2n12,000±1.645(2,200/245)12,000±231(11,769to12,231)www.khdaw.comb.12,000±1.96(2,200/245)12,000±275(11,725to12,275)c.12,000±2.576(2,200/245)12,000±362(11,638to12,362)d.Intervalwidthmustincreasesincewewanttomakeastatementaboutµwithgreaterconfidence.9.a.Usingacomputer,x=$12.41b.Usingacomputer,s=3.64c.x±1.96(s/n)12.41±1.96(3.64/60)12.41±0.92(11.49to13.33)s10.x±z.025n3.457.751.96±1807.75±.50(7.25to8.25)11.UsingMinitabweobtainedasamplestandarddeviationof2.163.Theconfidenceintervaloutputisshownbelow:THEASSUMEDSIGMA=2.16NMEANSTDEVSEMEAN95.0PERCENTC.I.Miami506.3402.1630.306(5.740,6.940)The95%confidenceintervalestimateis5.74to6.94.8-3www.khdaw.com 课后答案网www.khdaw.comChapter8Σx114i12.a.x===3.8minutesn302Σ(x−x)ib.s==2.26minutesn−1s2.26MarginofError=z=1.96=.81minutes.025n30sc.x±z.025n3.8±.81(2.99to4.61)www.khdaw.com13.a..95b..90c..01d..05e..95f..8514.a.1.734b.-1.321c.3.365d.-1.761and+1.761e.-2.048and+2.0488015.a.x=Σx/n==10i82Σ(x−x)84ib.s===3.464n−18−1c.With7degreesoffreedom,t.025=2.365x±t.025(s/n)10±2.365(3.464/8)10±2.90(7.10to12.90)8-4www.khdaw.com 课后答案网www.khdaw.comIntervalEstimation16.a.17.25±1.729(3.3/20)17.25±1.28(15.97to18.53)b.17.25±2.09(3.3/20)17.25±1.54(15.71to18.79)c.17.25±2.861(3.3/20)17.25±2.11(15.14to19.36)17.At90%,80±t.05(s/n)withdf=17t.05=1.740www.khdaw.com80±1.740(10/18)80±4.10(75.90to84.10)At95%,80±2.11(10/18)withdf=17t.05=2.11080±4.97(75.03to84.97)Σx18.96i18.a.x===$1.58n122Σ(x−x).239ib.s===.1474n−1121−c.t.025=2.201x±t.025(s/n)1.58±2.201(.1474/12)1.58±.09(1.49to1.67)19.x=Σx/n=6.53minutesi2Σ(x−x)is==0.54minutesn−1x±t.025(s/n)6.53±2.093(0.54/20)6.53±.25(6.28to6.78)8-5www.khdaw.com 课后答案网www.khdaw.comChapter820.a.22.4±1.96(5/61)22.4±1.25(21.15to23.65)b.Withdf=60,t.025=2.00022.4±2(5/61)22.4±1.28(21.12to23.68)c.x±t.025(s/n)Confidenceintervalsareessentiallythesameregardlessofwhetherzortisused.Σx864i21.x===$108www.khdaw.comn82Σ(x−x)654is===9.6658n−181−t.025=2.365x±t.025(s/n)108±2.365(9.6658/8)108±8.08(99.92to116.08)22.a.Usingacomputer,x=6.86s=0.78b.x±t.025(s/n)t.025=2.064df=246.86±2.064(0.78/25)6.86±0.32(6.54to7.18)2222zσ(1.96)(25).02523.n===96.04Usen=9722E524.a.Planningvalueofσ=Range/4=36/4=92222zσ(1.96)(9).025b.n===34.57Usen=3522E322(1.96)(9)c.n==77.79Usen=78228-6www.khdaw.com 课后答案网www.khdaw.comIntervalEstimation22(1.96)(6.82)25.n==79.41Usen=802(1.5)22(1.645)(6.82)n==31.47Usen=32222222zσ(1.96)(9400)26.a.n===339.44Use34022E(1000)22(1.96)(9400)b.n==1357.78Use13582(500)22(1.96)(9400)c.n==8486.09Use8487www.khdaw.com20022(1.96)(2,000)27.a.n==61.47Usen=622(500)22(1.96)(2,000)b.n==384.16Usen=3852(200)22(1.96)(2,000)c.n==1536.64Usen=15372(100)2222zσ(1.645)(220)28.a.n===52.39Use5322E(50)22(1.96)(220)b.n==74.37Use752(50)22(2.576)(220)c.n==128.47Use1292(50)d.Mustincreasesamplesizetoincreaseconfidence.22(1.96)(6.25)29.a.n==37.52Usen=382222(1.96)(6.25)b.n==150.06Usen=1512122(1.96)(7.8)30.n==58.43Usen=59228-7www.khdaw.com 课后答案网www.khdaw.comChapter831.a.p=100/400=0.25p(1−p)0.25(0.75)b.==0.0217n400p(1−p)c.p±z.025n.25±1.96(.0217).25±.0424(.2076to.2924)0.70(0.30)32.a..70±1.645www.khdaw.com800.70±.0267(.6733to.7267)0.70(0.30)b..70±1.96800.70±.0318(.6682to.7318)22zp(1−p)(1.96)(0.35)(0.65).02533.n===349.59Usen=35022E(0.05)34.Useplanningvaluep=.502(1.96)(0.50)(0.50)n==1067.11Usen=10682(0.03)35.a.p=562/814=0.6904p(1−p)0.6904(1−0.6904)b.1.645=1.645=0.0267n814c.0.6904±0.0267(0.6637to0.7171)36.a.p=152/346=.4393p(1−p).4393(1.4393)−b.σ===.0267pn346p±zσ.025p.4393±1.96(.0267).4393±.0523(.3870to.4916)8-8www.khdaw.com 课后答案网www.khdaw.comIntervalEstimationp(1−p)37.p±1.96n182p==.28650(0.28)(0.72).28±1.966500.28±0.0345(0.2455to0.3145)p(1−p)(0.26)(0.74)38.a.1.96=1.96=0.0430n400www.khdaw.comb.0.26±0.0430(0.2170to0.3030)21.96(0.26)(0.74)c.n==821.25Usen=8222(0.03)22zp(1−p)(1.96)(.33)(1.33)−.02539.a.n===943.75Use94422E(.03)22zp(1−p)(2.576)(.33)(1.33)−.005b.n===1630.19Use163122E(.03)40.a.p=255/1018=0.2505(0.2505)(1−0.2505)b.1.96=0.02661018p(1−p).16(1.16)−41.σ===.0102pn1285MarginofError=1.96σ=1.96(.0102)=.02p.16±1.96σp.16±.02(.14to.18)p(1−p).50(1.50)−42.a.σ===.0226pn491zσ=1.96(.0226)=.0442.025p2zp(1−p).025b.n=2E8-9www.khdaw.com 课后答案网www.khdaw.comChapter821.96(.50)(1.50)−Septembern==600.25Use6012.0421.96(.50)(1.50)−Octobern==1067.11Use10682.0321.96(.50)(1.50)−Novembern==24012.0221.96(.50)(1.50)−Pre-Electionn==96042.0121.96(0.5)(1−0.5)43.a.n==600.25Usen=6012www.khdaw.com(0.04)b.p=445/601=0.7404(0.7404)(0.2596)c.0.7404±1.966010.7404±0.0350(0.7054to0.7755)s20,50044.a.z=1.96=2009.025n400b.x±z.025(s/n)50,000±2009(47,991to52,009)45.a.x±z.025(s/n)252.45±1.96(74.50/64)252.45±18.25or$234.20to$270.70b.Yes.thelowerlimitforthepopulationmeanatNiagaraFallsis$234.20whichisgreaterthan$215.60.46.a.Usingacomputer,x=49.8minutesb.Usingacomputer,s=15.99minutesc.x±1.96(s/n)49.8±1.96(15.99/200)49.8±2.22(47.58to52.02)8-10www.khdaw.com 课后答案网www.khdaw.comIntervalEstimation47.a.Usingacomputer,x=16.8ands=4.25With19degreesoffreedom,t.025=2.093x±2.093(s/n)16.8±2.093(4.25/20)16.8±1.99(14.81to18.79)b.Usingacomputer,x=24.1ands=6.2124.1±2.093(6.21/20)www.khdaw.com24.1±2.90(21.2to27.0)c.16.8/24.1=0.697or69.7%orapproximately70%13248.a.x=Σx/n==13.2i102Σ(x−x)547.6ib.s===7.8n−19c.Withdf=9,t.025=2.262x±t.025(s/n)13.2±2.262(7.8/10)13.2±5.58(7.62to18.78)d.The±5.58showspoorprecision.Alargersamplesizeisdesired.21.96(0.45)49.n==77.79Usen=7821022(2.33)(2.6)50.n==36.7Usen=372122(1.96)(8)51.n==61.47Usen=622222(2.576)(8)n==106.17Usen=1072222(1.96)(675)52.n==175.03Usen=17621008-11www.khdaw.com 课后答案网www.khdaw.comChapter8p(1−p)53.a.p±1.96n(0.47)(0.53)0.47±1.964500.47±0.0461(0.4239to0.5161)(0.47)(0.53)b.0.47±2.5764500.47±0.0606(0.4094to0.5306)www.khdaw.comc.Themarginoferrorbecomeslarger.54.a.p=200/369=0.5420p(1−p)(0.5420)(0.4580)b.1.96=1.96=0.0508n369c.0.5420±0.0508(0.4912to0.5928)55.a.p=504/1400=.36(0.36)(0.64)b.1.96=0.025114002(2.33)(0.70)(0.30)56.a.n==1266.74Usen=12672(0.03)2(2.33)(0.50)(0.50)b.n==1508.03Usen=15092(0.03)57.a.p=110/200=0.55(0.55)(0.45)0.55±1.96200.55±.0689(.4811to.6189)2(1.96)(0.55)(0.45)b.n==380.32Usen=3812(0.05)58.a.p=340/500=.68p(1−p).68(1.68)−b.σ===.0209pn5008-12www.khdaw.com 课后答案网www.khdaw.comIntervalEstimationp±zσ.025p.68±1.96(.0209).68±.0409(.6391to.7209)2(1.96)(0.3)(0.7)59.a.n==2016.84Usen=20172(0.02)b.p=520/2017=0.2578p(1−p)c.p±1.96www.khdaw.comn(0.2578)(0.7422)0.2578±1.9620170.2578±0.0191(0.2387to0.2769)60.a.p=618/1993=.3101p(1−p)b.p±1.961993(0.3101)(0.6899)0.3101±1.961993.3101±.0203(.2898to.3304)2zp(1−p)c.n=2E2(1.96)(0.3101)(0.6899)z==8218.64Usen=82192(0.01)No;thesampleappearsunnecessarilylarge.The.02marginoferrorreportedinpart(b)shouldprovideadequateprecision.8-13www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter9HypothesisHHypothesisypothesisTestingTTestingestingLearningLLearningearningObjectivesOObjectivesbjectives1.Learnhowtoformulateandtesthypothesesaboutapopulationmeanand/orapopulationproportion.2.Understandthetypesoferrorspossiblewhenconductingahypothesistest.www.khdaw.com3.Beabletodeterminetheprobabilityofmakingvariouserrorsinhypothesistests.4.Knowhowtocomputeandinterpretp-values.5.Beabletodeterminethesizeofasimplerandomsamplenecessarytokeeptheprobabilityofhypothesistestingerrorswithinacceptablelimits.6.Knowthedefinitionofthefollowingterms:nullhypothesisone-tailedtestalternativehypothesistwo-tailedtesttypeIerrorp-valuetypeIIerroroperatingcharacteristiccurvecriticalvaluepowercurvelevelofsignificance9-1www.khdaw.com 课后答案网www.khdaw.comChapter9Solutions:SSolutions:olutions:1.a.H0:µ≤600Manager’sclaim.Ha:µ>600b.Wearenotabletoconcludethatthemanager’sclaimiswrong.c.Themanager’sclaimcanberejected.Wecanconcludethatµ>400.2.a.H0:µ≤14Ha:µ>14Researchhypothesiswww.khdaw.comb.Thereisnostatisticalevidencethatthenewbonusplanincreasessalesvolume.c.Theresearchhypothesisthatµ>14issupported.Wecanconcludethatthenewbonusplanincreasesthemeansalesvolume.3.a.H0:µ=32SpecifiedfillingweightHa:µ≠32Overfillingorunderfillingexistsb.Thereisnoevidencethattheproductionlineisnotoperatingproperly.Allowtheproductionprocesstocontinue.c.Concludeµ≠32andthatoverfillingorunderfillingexists.Shutdownandadjusttheproductionline.4.a.H0:µ≥220Ha:µ<220Researchhypothesistoseeifmeancostislessthan$220.b.Weareunabletoconcludethatthenewmethodreducescosts.c.Concludeµ<220.Considerimplementingthenewmethodbasedontheconclusionthatitlowersthemeancostperhour.5.a.TheTypeIerrorisrejectingH0whenitistrue.Inthiscase,thiserroroccursiftheresearcherconcludesthatthemeannewspaper-readingtimeforindividualsinmanagementpositionsisgreaterthanthenationalaverageof8.6minuteswheninfactitisnot.b.TheTypeIIerrorisacceptingH0whenitisfalse.Inthiscase,thiserroroccursiftheresearcherconcludesthatthemeannewspaper-readingtimeforindividualsinmanagementpositionsislessthanorequaltothenationalaverageof8.6minuteswheninfactitisgreaterthan8.6minutes.6.a.H0:µ≤1Thelabelclaimorassumption.Ha:µ>1b.Claimingµ>1whenitisnot.Thisistheerrorofrejectingtheproduct’sclaimwhentheclaimistrue.9-2www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingc.Concludingµ≤1whenitisnot.Inthiscase,wemissthefactthattheproductisnotmeetingitslabelspecification.7.a.H0:µ≤8000Ha:µ>8000Researchhypothesistoseeiftheplanincreasesaveragesales.b.Claimingµ>8000whentheplandoesnotincreasesales.Amistakecouldbeimplementingtheplanwhenitdoesnothelp.c.Concludingµ≤8000whentheplanreallywouldincreasesales.Thiscouldleadtonotimplementingaplanthatwouldincreasesales.8.a.H0:µ≥220Ha:µ<220www.khdaw.comb.Claimingµ<220whenthenewmethoddoesnotlowercosts.Amistakecouldbeimplementingthemethodwhenitdoesnothelp.c.Concludingµ≥220whenthemethodreallywouldlowercosts.Thiscouldleadtonotimplementingamethodthatwouldlowercosts.9.a.z=-1.645RejectH0ifz<-1.645x−µ9.4610−b.z===−1.91s/n2/50RejectH0;concludeHaistrue.10.a.z=2.05RejectH0ifz>2.05x−µ16.515−b.z===1.36s/n7/40c.Areafromz=0toz=1.36is.4131p-value=.5000-.4131=.0869d.DonotrejectH011.RejectH0ifz<-1.645x−µ2225−a.z===−2.50RejectH0σ/n12/1002425−b.z==−.83DonotrejectH012/10023.525−c.z==−1.25DonotrejectH012/1009-3www.khdaw.com 课后答案网www.khdaw.comChapter922.825−d.z==−1.83RejectH012/10012.a.p-value=.5000-.4656=.0344RejectH0b.p-value=.5000-.1736=.3264DonotrejectH0c.p-value=.5000-.4332=.0668DonotrejectH0d.z=3.09isthelargesttablevaluewith.5000-.4990=.001areaintail.Forz=3.30,thep-valueislessthan.001orapproximately0.RejectH0.e.Sincezistotheleftofthemeanandtherejectionregionisintheuppertail,p-value=.5000+.3413=.8413.DonotrejectH0.www.khdaw.com13.a.H0:µ≥1056Ha:µ<1056b.RejectH0ifz<-1.645x−µ9101056−0c.z===−1.83s/n1600/400d.RejectH0andconcludethatthemeanrefundof“lastminute”filersislessthan$1056.e.p-value=.5000-.4664=.033614.a.z.01=2.33RejectH0ifz>2.33x−µ7.256.70−b.z===3.11s/n2.5/200c.RejectH0;concludethemeantelevisionviewingtimeperdayisgreaterthan6.70.15.a.z.05=1.645RejectH0ifz<-1.645x−µ930010,192−b.z===−1.98s/n4500/100c.RejectH0;concludethatthemeansalespriceofusedcarsislessthanthenationalaverage.16.a.H0:µ≥13Ha:µ<13b.z.01=2.339-4www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingRejectH0ifz<-2.33x−µ10.813−c.z===−2.88s/n9.2/145d.RejectH0;concludeCanadianmeaninternetusageislessthan13hourspermonth.Note:p-value=.00217.a.H0:µ≤15Ha:µ>15x−µ1715−b.z===2.96s/n4/35www.khdaw.comc.p-value=.5000-.4985=.0015d.RejectH0;thepremiumrateshouldbecharged.18.a.H0:µ≤5.72Ha:µ>5.72x−µ5.985.72−b.z===2.12s/n1.24/102c.p-value=.5000-.4830=.0170d.p-value<α;rejectH0.ConcludeteensinChicagohaveameanexpendituregreaterthan5.72.19.a.H0:µ≥181,900Ha:µ<181,900x−µ166,400181,900−b.z===−2.93s/n33,500/40c.p-value=.5000-.4983=.0017d.p-value<α;rejectH0.ConcludemeansellingpriceinSouthislessthanthenationalmeansellingprice.20.a.H0:µ≤37,000Ha:µ>37,000x−µ38,10037,000−b.z===1.47s/n5200/48c.p-value=.5000-.4292=.07089-5www.khdaw.com 课后答案网www.khdaw.comChapter9d.p-value>α;donotrejectH0.CannotconcludepopulationmeansalaryhasincreasedinJune2001.21.a.RejectH0ifz<-1.96orz>1.96x−µ10.8−10b.z===2.40s/n2.5/36RejectH0;concludeHaistrue.22.a.RejectH0ifz<-2.33orz>2.33x−µ14.2−15b.z===−1.13s/n5/50www.khdaw.comc.p-value=(2)(.5000-.3708)=.2584d.DonotrejectH023.RejectH0ifz<-1.96orz>1.962225−a.z==−2.68RejectH010/802725−b.z==1.79DonotrejectH010/8023.525−c.z==−1.34DonotrejectH010/802825−d.z==2.68RejectH010/8024.a.p-value=2(.5000-.4641)=.0718DonotrejectH0b.p-value=2(.5000-.1736)=.6528DonotrejectH0c.p-value=2(.5000-.4798)=.0404RejectH0d.approximately0RejectH0e.p-value=2(.5000-.3413)=.3174DonotrejectH025.a.z.025=1.96RejectH0ifz<-1.96orz>1.96x−µ38.539.2−b.z===−1.54s/n4.8/112c.DonotrejectH0.Cannotconcludeachangeinthepopulationmeanhasoccurred.d.p-value=2(.5000-.4382)=.123626.a.H0:µ=89-6www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingHa:µ≠8RejectH0ifz<-1.96orifz>1.96x−µ7.5−80b.z===−1.71s/n3.2/120c.DonotrejectH0;cannotconcludethemeanwaitingtimediffersfromeightminutes.27.a.H0:µ=16ContinueproductionHa:µ≠16ShutdownRejectH0ifz<-1.96orifz>1.96x−µ16.3216−0b.z===2.19www.khdaw.comσ/n.8/30RejectH0andshutdownforadjustment.x−µ15.8216−0c.z===−1.23σ/n.8/30DonotrejectH0;continuetorun.d.Forx=16.32,p-value=2(.5000-.4857)=.0286Forx=15.82,p-value=2(.5000-.3907)=.218628.a.H0:µ=1075Ha:µ≠1075x−µ11601075−b.z===1.43s/n840/200c.p-value=2(.5000-.4236)=.1528d.DonotrejectH0.Cannotconcludeachangeinmeanamountofcharitablegiving.29.a.H0:µ=15.20Ha:µ≠15.20x−µ14.30−15.200z===−1.06s/n5/35b.p-value=2(.5000-.3554)=.2892c.DonotrejectH0;thesampledoesnotprovideevidencetoconcludethattherehasbeenachange.30.a.H0:µ=26,1339-7www.khdaw.com 课后答案网www.khdaw.comChapter9Ha:µ≠26,133x−µ25,45726,133−b.z===−2.09s/n7600/410c.p-value=2(.5000-.4817)=.0366d.p-value<α;rejectH0.ConcludepopulationmeanwageinCollierCountydiffersfromthestatemeanwage.σ31.a.x±z.025n⎛180⎞9351.96±⎜⎟⎝200⎠www.khdaw.com935±25or910to960Since900isnotintheinterval,rejectH0andconcludeµ≠900.b.RejectH0ifz<-1.96orifz>1.96x−µ935900−0z===2.75σ/n180/200RejectH0c.p-value=2(.5000-.4970)=.006032.a.Theupper95%confidencelimitiscomputedasfollows:⎛s⎞x+z⎜⎟.05⎝n⎠⎛.60⎞14.501.645+⎜⎟=14.66⎝36⎠Thus,weare95%confidentthatµis$14.66perhourorless.b.Since$15.00isnotintheinterval$14.66perhourorless,werejectH0.Concludethatthemeanwagerateislessthan$15.00.33.a.With15degreesoffreedom,t.05=1.753RejectH0ift>1.753x−µ11−100b.t===1.33DonotrejectH0s/n3/1634.a.x=∑x/n=108/6=18i9-8www.khdaw.com 课后答案网www.khdaw.comHypothesisTesting∑(x−x)10ib.s===1.414n−161−c.RejectH0ift<-2.571ort>2.571x−µ18−200d.t===−3.46s/n1.414/6e.RejectH0;concludeHaistrue.35.RejectH0ift<-1.72113−15a.t==−1.17DonotrejectH08/2211.5−15b.t==−2.05RejectH0www.khdaw.com8/2215−15c.t==0DonotrejectH08/2219−15d.t==2.35DonotrejectH08/2236.Usethetdistributionwith15degreesoffreedoma.p-value=.01RejectH0b.p-value=.10DonotrejectH0c.p-valueisbetween.025and.05RejectH0d.p-valueisgreaterthan.10DonotrejectH0e.p-valueisapproximately0RejectH037.a.H0:µ=3.00Ha:µ≠3.00b.t.025=2.262RejectH0ift<-2.262orift>2.262Σx28ic.x===2.80n102Σ(x−x).44id.s===.70n−1101−x−µ2.803.00−e.t===−.90s/n.70/109-9www.khdaw.com 课后答案网www.khdaw.comChapter9f.DonotrejectH0;cannotconcludethepopulationmeansearningpersharehaschanged.g.t.10=1.383p-valueisgreaterthan.10x2=.20Actualp-value=.391638.a.t.025=2.06424degreesoffreedomRejectH0ift<-2.064orift>2.064x−µ84.5090−b.t===−1.90www.khdaw.coms/n14.50/25c.DonotrejectH0;cannotconcludethemeanexpenditureinCorningdiffersfromtheU.S.meanexpenditure.39.a.t.05=1.8957degreesoffreedomΣx475ib.x===59.375n82Σ(x−x)123.87ic.s===4.21n−181−x−µ59.3855−d.t===2.94s/n4.21/8c.RejectH0;concludethatthemeannumberofhoursworkedexceeds55.40.a.H0:µ=4000Ha:µ≠4000b.t.05=2.16013degreesoffreedomRejectH0ift<-2.160orift>2.160x−µ41204000−c.t===+1.63s/n275/14d.DonotrejectH0;CannotconcludethatthemeancostinNewCitydiffersfrom$4000.e.With13degreesoffreedomt.05=1.771t.10=1.3509-10www.khdaw.com 课后答案网www.khdaw.comHypothesisTesting1.63isbetween1.350and1.771.Thereforethep-valueisbetween.10and.20.41.a.H0:µ≤280Ha:µ>280b.286.9-280=6.9yardsc.t.05=1.860with8degreesoffreedomx−µ286.9280−d.t===2.07s/n10/9www.khdaw.come.RejectH0;ThepopulationmeandistanceofthenewdriverisgreaterthantheUSGAapproveddriver..f.t.05=1.860t.025=2.306p-valueisbetween.025and.05Actualp-value=.036142.a.H0:µ≤2Ha:µ>2b.With9degreesoffreedom,rejectH0ift>1.833c.x=∑x/n=24/10=2.4i2∑(x−x)2.40id.s===.516n−19x−µ2.42−0e.t===2.45s/n.516/10f.RejectH0andclaimµisgreaterthan2hours.Forcostestimatingpurposes,considerusingmorethan2hoursoflabortime.g.t.025=2.262,t.01=2.821p-valueisbetween.025and.01.43.a.RejectH0ifz>1.645.50(.50)b.σ==.0354p2009-11www.khdaw.com 课后答案网www.khdaw.comChapter9p−p.57−.50z===1.98RejectH0σ.0354p44.a.RejectH0ifz<-1.96orz>1.96.20(.80)b.σ==.02p400p−p.175−.20z===−1.25σ.02pc.p-value=2(.5000-.3944)=.2122www.khdaw.comd.DonotrejectH0.45.RejectH0ifz<-1.645.75(.25)a.σ==.0250p300p−p.68−.75z===−2.80σ.025pp-value=.5000-.4974=.0026RejectH0.p−p.72−.75b.z===−1.20σ.025pp-value=.5000-.3849=.1151DonotrejectH0.p−p.70−.75c.z===−2.00σ.025pp-value=.5000-.4772=.0228RejectH0.p−p.77−.75d.z===.80σ.025pp-value=.5000+.2881=.7881DonotrejectH0.46.a.H0:p≤.409-12www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingHa:p>.40b.RejectH0ifz>1.645c.p=188/420=.4476p(1−p).40(1.40)−σ===.0239pn420p−p.4476.40−z===1.99σ.0239pd.RejectH0.Concludethattherehasbeenanincreaseintheproportionofusersreceivingmorethantene-mailsperday.www.khdaw.com47.a.z.05=1.645RejectH0ifz<-1.645b.p=52/100=.52p(1−p).64(1.64)−σ===.0480pn100p−p.52.64−z===−2.50σ.0480pc.RejectH0.Concludelessthan64%ofshoppersbelievesupermarketketchupisasgoodasthenationalbrand.d.p-value=.5000-.4938=.006248.a.p=285/500=.57p(1−p).50(1.50)−b.σ===.0224pn500p−p.57.50−z===3.13σ.0224pc.z=3.13isnotinthetable.Closestvalueisz=3.09.Thus,p-valueisapproximately.5000-.4990=.001d.p-value<.01,RejectH0.Over50%preferBurgerKingfries.e.Yes;thestatisticalevidenceshowsBurgerKingfriesarepreferred.Thegive-awaywasagoodwaytogetpotentialcustomerstotrythenewfries.49.a.H0:p=.48Ha:p≠.489-13www.khdaw.com 课后答案网www.khdaw.comChapter9b.z.025=1.96RejectH0ifz<-1.96orifz>1.96c.p=368/800=.45p(1−p).48(1.48)−d.σ===.0177pn800p−p.45.48−z===−1.70σ.0177pd.DonotrejectH0.Cannotconcludetheproportionofdriverswhodonotstophaschanged.www.khdaw.com50.a.p=67/105=.6381(about64%)p(1−p).50(1.50)−b.σ===.0488pn105p−p.6381.50−z===2.83σ.0488pc.p-value=2(.5000-.4977)=.0046d.p-value<.01,rejectH0.Concludepreferenceisforthefourten-hourdayschedule.51.a.H0:p=.44Ha:p≠.44b.p=205/500=.41p(1−p).44(1.44)−σ===.0222pn500p−p.41.44−z===−1.35σ.0222pp-value=2(.5-.4115)=.1770DonotrejectH0.Cannotconcludethattherehasbeenachangeintheproportionofrepeatcustomers.c.p=245/500=.49p−p.49.44−z===2.25σ.0222pp-value=2(.5-.4878)=.02449-14www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingRejectH0.concludethattheproportionofrepeatcustomershaschanged.Thepointestimateofthepercentageofrepeatcustomersisnow49%.p(1−p).75(1.75)−52.a.σ===.025pn300p−p.72.75−z===−1.20σ.025pb.p-value=.5000-.3849=.1151c.DonotrejectH0.Cannotconcludethemanager"sclaimiswrongbasedonthissampleevidence.www.khdaw.com53.H0:p≤.15Ha:p>.15RejectH0ifz>2.33p(1−p).15(.85)σ===.0160pn500p=88/500=.176p−p.176.15−0z===1.63σ.0160pDonotrejectH0;p≤.15cannotberejected.Thusthespecialoffershouldnotbeinitiated.p-value=.5000-.4484=.051654.a.H0:p≥.047Ha:p<.047b.p=35/1182=.0296.047(1.047)−c.σ==.0062p1182p−p.0296.047−z===−2.82σ.0062pd.p-value=.5000-.4976=.0024e.p-value<α,rejectH0.TheerrorrateforBrooksRobinsonislessthantheoverallerrorrate.55.H0:p≥.209-15www.khdaw.com 课后答案网www.khdaw.comChapter9Ha:p<.20RejectH0ifz<-1.645p=83/596=.1393p(1−p).20(1.20)−σ===.0164pn596p−p.1393.20−z===−3.71σ.0164pp-value≈0RejectH0;concludethatlessthan20%ofworkerswouldworkforlesspayinordertohavemorewww.khdaw.compersonalandleisuretime.σ556.σ===.46xn120cHa:µ<10H0:µ≥10.05x10c=10-1.645(5/120)=9.25RejectH0ifx<9.25a.Whenµ=9,9.259−z==.555/120Prob(H0)=(.5000-.2088)=.2912b.TypeIIerrorc.Whenµ=8,9-16www.khdaw.com 课后答案网www.khdaw.comHypothesisTesting9.258−z==2.745/120β=(.5000-.4969)=.003157.RejectH0ifz<-1.96orifz>1.96σ10σ===.71xn200www.khdaw.comHa:µ≠20H0:µ=20Ha:µ≠20.025.025xxc120c2c1=20-1.96(10/200)=18.61c2=20+1.96(10/200)=21.39a.µ=1818.6118−z==.8610/200β=.5000-.3051=.1949b.µ=22.521.3922.5−z==−1.5710/200β=.5000-.4418=.0582c.µ=2121.3921−z==.5510/2009-17www.khdaw.com 课后答案网www.khdaw.comChapter9β=.5000+.2088=.708858.a.H0:µ≤15Ha:µ>15Concludingµ≤15whenthisisnottrue.Fowlewouldnotchargethepremiumrateeventhoughtherateshouldbecharged.b.RejectH0ifz>2.33x−µx−150z===2.33σ/n4/35Solveforx=16.58www.khdaw.comDecisionRule:AcceptH0ifx≤16.58RejectH0ifx>16.58Forµ=17,16.5817−z==−.624/35β=.5000-.2324=.2676c.Forµ=18,16.5818−z==−2.104/35β=.5000-.4821=.017959.a.H0:µ≥25Ha:µ<25RejectH0ifz<-2.05x−µx−250z===−2.05σ/n3/30Solveforx=23.88DecisionRule:AcceptH0ifx≥23.88RejectH0ifx<23.889-18www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingb.Forµ=23,23.8823−z==1.613/30β=.5000-.4463=.0537c.Forµ=24,23.8824−z==−.22www.khdaw.com3/30β=.5000+.0871=.5871d.TheTypeIIerrorcannotbemadeinthiscase.Notethatwhenµ=25.5,H0istrue.TheTypeIIerrorcanonlybemadewhenH0isfalse.60.a.AcceptingH0andconcludingthemeanaverageagewas28yearswhenitwasnot.b.RejectH0ifz<-1.96orifz>1.96x−µx−280z==σ/n6/100Solvingforx,wefindatz=-1.96,x=26.82atz=+1.96,x=29.18DecisionRule:AcceptH0if26.82≤x≤29.18RejectH0ifx<26.82orifx>29.18Atµ=26,26.8226−z==1.376/100β=.5000+.4147=.0853Atµ=27,26.8227−z==−.306/1009-19www.khdaw.com 课后答案网www.khdaw.comChapter9β=.5000+.1179=.6179Atµ=29,29.1829−z==.306/100β=.5000+.1179=.6179Atµ=30,www.khdaw.com29.1830−z==−1.376/100β=.5000-.4147=.0853c.Power=1-βatµ=26,Power=1-.0853=.9147Whenµ=26,thereisa.9147probabilitythatthetestwillcorrectlyrejectthenullhypothesisthatµ=28.61.a.AcceptingH0andlettingtheprocesscontinuetorunwhenactuallyover-fillingorunder-fillingexists.b.DecisionRule:RejectH0ifz<-1.96orifz>1.96indicatesAcceptH0if15.71≤x≤16.29RejectH0ifx<15.71orifx>16.29Forµ=16.516.2916.5−z==−1.44.8/30β=.5000-.4251=.07499-20www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingcβxwww.khdaw.com16.2916.5c.Power=1-.0749=.9251d.ThepowercurveshowstheprobabilityofrejectingH0forvariouspossiblevaluesofµ.Inparticular,itshowstheprobabilityofstoppingandadjustingthemachineunderavarietyofunderfillingandoverfillingsituations.Thegeneralshapeofthepowercurveforthiscaseis1.00.75Power.50.25.0015.615.816.016.216.4PossibleValuesofuσ462.c=µ+z=152.33+=16.320.01n5016.3217−Atµ=17z==−1.204/50β=.5000-.3849=.115116.3218−Atµ=18z==−2.974/50β=.5000-.4985=.0015IncreasingthesamplesizereducestheprobabilityofmakingaTypeIIerror.9-21www.khdaw.com 课后答案网www.khdaw.comChapter963.a.Acceptµ≤100whenitisfalse.b.Criticalvaluefortest:σ75c=µ+z=1001.645+=119.510.05n40119.51120−Atµ=120z==−.0475/40β=.5000-.0160=.4840119.51130−c.Atµ=13z==−.8875/40www.khdaw.comβ=.5000-.3106=.1894d.Criticalvaluefortest:σ75c=µ+z=1001.645+=113.790.05n80113.79120−Atµ=120z==−.7475/80β=.5000-.2704=.2296113.79130−Atµ=130z==−1.9375/80β=.5000-.4732=.0268Increasingthesamplesizefrom40to80reducestheprobabilityofmakingaTypeIIerror.2222(z+z)σ(1.6451.28)(5)+αβ64.n===21422(µ−µ)(109)−0a2222(z+z)σ(1.961.645)(10)+αβ65.n===32522(µ−µ)(2022)−0a66.Atµ0=3,α=.01.z.01=2.33Atµa=2.9375,β=.10.z.10=1.28σ=.182222(z+z)σ(2.331.28)(.18)+αβn===108.09Use10922(µ−µ)(32.9375)−0a9-22www.khdaw.com 课后答案网www.khdaw.comHypothesisTesting67.Atµ0=400,α=.02.z.02=2.05Atµa=385,β=.10.z.10=1.28σ=302222(z+z)σ(2.051.28)(30)+αβn===44.4Use4522(µ−µ)(400385)−0a68.Atµ0=28,α=.05.Notehoweverforthistwo-tailedtest,zα/2=z.025=1.96Atµa=29,β=.15.z.15=1.04σ=62222(z+z)σ(1.961.04)(6)+α/2βwww.khdaw.comn===32422(µ−µ)(2829)−0a69.Atµ0=25,α=.02.z.02=2.05Atµa=24,β=.20.z.20=.84σ=32222(z+z)σ(2.05.84)(3)+αβn===75.2Use7622(µ−µ)(2524)−0a70.a.H0:µ≤45,250Ha:µ>45,250x−µ47,00045,250−b.z===2.71s/n6300/95c.p-value=.5000-.4966=.0034d.p-value<α;rejectH0.NewYorkCityschoolteachersmusthaveahighermeanannualsalary.71.H0:µ≥30Ha:µ<30RejectH0ifz<–2.33x−µ29.530−0z===−1.96s/n1.8/50p-value=.5000-.4750=.0250DonotrejectH0;thesampleevidencedoesnotsupporttheconclusionthattheFordTaurusprovideslessthan30milespergallon.9-23www.khdaw.com 课后答案网www.khdaw.comChapter972.H0:µ≤25,000Ha:µ>25,000RejectH0ifz>1.645x−µ26,00025,000−0z===2.26s/n2,500/32p-value=.5000-.4881=.0119RejectH0;theclaimshouldberejected.Themeancostisgreaterthan$25,000.73.H0:µ=120www.khdaw.comHa:µ≠120Withn=10,useatdistributionwith9degreesoffreedom.RejectH0ift<-2.262oroft>2.262Σxix==118.9n2Σ(x−x)is==4.93n−1x−µ118.9120−0t===−.71s/n4.93/10DonotrejectH0;theresultsdonotpermitrejectionoftheassumptionthatµ=120.74.a.H0:µ=550Ha:µ≠550RejectH0ifz<-1.96orifz>1.96x−µ562−5500z===1.80s/n40/36DonotrejectH0;theclaimof$550permonthcannotberejected.b.p-value=2(.5000-.4641)=.071875.a.H0:µ≤75Ha:µ>75RejectH0ifz>1.6459-24www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingx−µ82.5075.00−0b.z===1.58s/n30/40DonotrejectH0;thereisnoevidencetoconcludeanincreaseinmaintenancecostexists.c.p-value=.5000-.4429=.0571Since.0571>.05,donotrejectH0.76.a.H0:µ≤72Ha:µ>72x−7280−72z===2.19s/n20/30www.khdaw.comp-value=.5000-.4857=.0143b.Sincep-value<.05,rejectH0;themeanidletimeexceeds72minutesperday.77.a.H0:p≤.60Ha:p>.60RejectH0ifz>1.645p(1−p).60(.40)σ===.0775pn40p=27/40=.675p−p.675.60−z===.97σ.0775pDonotrejectH0;thesampleresultsdonotjustifytheconclusionthatp>.60forMidwesterners.b.p-value=.5000-.3340=.166078.a.p=355/546=.6502p(1−p).67(1.67)−b.σ===.0201pn546p−p.6502.67−z===−.98σ.0201pc.p-value=2(.5000-.3365)=.3270d.p-value≥α,donotrejectH0.Theassumptionoftwo-thirdscannotberejected.9-25www.khdaw.com 课后答案网www.khdaw.comChapter979.H0:p≥.79Ha:p<.79RejectH0ifz<-1.645p=360/500=.72p−p.72−.790z===−3.84σp(.79)(.21)500RejectH0;concludethattheproportionislessthan.79in1995.80.a.Theresearchisattemptingtoseeifitcanbeconcludedthatlessthan50%oftheworkingpopulationwww.khdaw.comholdjobsthattheyplannedtohold..50(.50)b.σ==.0136p1350.41−.50z==−6.62.0136p-value≈0RejectH0ifz<-2.33RejectH0;itcanbeconcludedthatlessthan50%oftheworkingpopulationholdjobsthattheyplannedtohold.Themajorityholdjobsduetochance,lackofchoice,orsomeotherunplannedreason..75(.25)81.σ==.0229p356p=313/356=.88.88−.75z==5.68.0229p-value≈0RejectH0;concludep≠.75.Datasuggestthat88%ofwomenwearshoesthatareatleastonesizetoosmall.82.a.p=330/400=.825p(1−p).78(1.78)−b.σ===.0207pn4009-26www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingp−p.825.78−z===2.17σ.0207pc.p-value=2(.5000-.4850)=.03d.p-value<α,rejectH0.Arrivalratehaschangedfrom78%.Serviceappearstobeimproving.83.H0:p≥.90Ha:p<.90RejectH0ifz<-1.645.90(.10)σ==.0394pwww.khdaw.com58p=49/58=.845p−p.845.90−z===−1.40σ.0394pp-value=.5000-.4192=.0808DonotrejectH0;thestation’sclaimcannotberejected84.a.p=44/125=.352p(1−p).47(1.47)−b.σ===.0446pn125p−p.352.47−z===−2.64σ.0446pc.p-value=.5000-.4959=.0041d.RejectH0;concludethattheproportionoffoodsamplecontainingpesticideresidueshasbeenreduced.85.a.H0:µ≤72Ha:µ>72RejectH0ifz>1.645x−µx−720z===1.645σ/n20/30Solveforx=78DecisionRule:9-27www.khdaw.com 课后答案网www.khdaw.comChapter9AcceptH0ifx≤78RejectH0ifx>78Forµ=807880−z==−.5520/30β=.5000-.2088=.2912b.Forµ=75,7875−z==.8220/30www.khdaw.comβ=.5000+.2939=.7939c.Forµ=70,H0istrue.InthiscasetheTypeIIerrorcannotbemade.d.Power=1-β1.0.8Po.6wer.4.272747678808284PossibleValuesofµHoFalse86.H0:µ≥15,000Ha:µ<15,000Atµ0=15,000,α=.02.z.02=2.05Atµa=14,000,β=.05.z.10=1.6452222(z+z)σ(2.051.645)(4,000)+αβn===218.5Use21922(µ−µ)(15,00014,000)−0a87.H0:µ=120Ha:µ≠1209-28www.khdaw.com 课后答案网www.khdaw.comHypothesisTestingAtµ0=120,α=.05.Withatwo-tailedtest,zα/2=z.025=1.96Atµa=117,β=.02.z.02=2.052222(z+z)σ(1.962.05)(5)+α/2βn===44.7Use4522(µ−µ)(120117)−0ab.Examplecalculationforµ=118.RejectH0ifz<-1.96orifz>1.96x−µx−1200z==σ/n5/45Solveforx.Atz=-1.96,x=118.54www.khdaw.comAtz=+1.96,x=121.46DecisionRule:AcceptH0if118.54≤x≤121.46RejectH0ifx<118.54orifx>121.46Forµ=118,118.54118−z==.725/45β=.5000+.2642=.2358OtherResults:Ifµiszβ1172.07.0192118.72.2358119-.62.7291121+.62.7291122+.72.2358123-2.07.01929-29www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter10StatisticalSStatisticaltatisticalInferenceIInferencenferenceaboutababoutoutMeansMMeanseansandananddProportionsPProportionsroportionswithwwithithTwoTTwowoPopulationsPPopulationsopulationsLearningLLearningearningObjectivesOObjectivesbjectives1.Beabletodevelopintervalestimatesandconducthypothesistestsaboutthedifferencebetweenthemeansoftwopopulations.www.khdaw.com2.Knowthepropertiesofthesamplingdistributionofthedifferencebetweentwomeans(x1−x2).3.Beabletousethetdistributiontoconductstatisticalinferencesaboutthedifferencebetweenthemeansoftwonormalpopulationswithequalvariances.4.Understandtheconceptanduseofapooledvarianceestimate.5.Learnhowtoanalyzethedifferencebetweenthemeansoftwopopulationswhenthesamplesareindependentandwhenthesamplesarematched.6.Beabletodevelopintervalestimatesandconducthypothesistestsaboutthedifferencebetweentheproportionsoftwopopulations.7.Knowthepropertiesofthesamplingdistributionofthedifferencebetweentwoproportions(p−p).1210-1www.khdaw.com 课后答案网www.khdaw.comChapter10Solutions:SSolutions:olutions:1.a.x−x=13.6-11.6=2122222ss(2.2)(3)12b.s=+=+=0.595x1−x2nn5035122±1.645(.595)2±.98(1.02to2.98)c.2±1.96(.595)2±1.17(0.83to3.17)2.a.x−x=22.5-20.1=2.4www.khdaw.com1222222(n1−1)s1+(n2−1)s29(2.5)+7(2)b.s===5.27n+n−210+8−2122F11IF11Ic.s=s+=5.27+=1.09x1−x2HGnnKJHG108KJ1216degreesoffreedom,t.025=2.122.4±2.12(1.09)2.4±2.31(.09to4.71)3.a.x=∑x/n=54/6=91ix=∑x/n=42/6=72i2∑(x−x)18i1b.s===1.901n−161−12∑(x−x)16i2s===1.792n−161−2c.x−x=9-7=21222222(n1−1)s1+(n2−1)s25(1.90)+5(1.79)d.s===3.41n+n−26+6−212e.With10degreesoffreedom,t.025=2.2282F11IF11Is=s+=3.41+=1.07x1−x2HGKJHGn1n2KJ6610-2www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulations2±2.228(1.07)2±2.37(-0.37to4.37)4.a.x−x=1.58-0.98=$0.60122222ss.12.0812b.s=+=+=.021x1−x2nn504212x1−x2±z.025sx1−x2.60±1.96(.021)www.khdaw.com.60±.04(.56to.64)5.a.22.5-18.6=3.9milesperdayb.x1−x2±zα/2sx1−x22222ss(8.4)(7.4)12s=+=+=1.58x1−x2nn50501222.5-18.6±1.96(1.58)3.9±3.1or0.6to7.06.LAMiamix6.726.34s2.3742.163x−x±zs12α/2x1−x22222ss(2.374)(2.163)12s=+=+=0.454x1−x2nn5050126.72-6.34±1.96(.454).38±.89or-.51to1.277.a.x−x=14.9-10.3=4.6years122222ss5.23.812b.s=+=+=.66x1−x2nn1008512z.025sx−x=1.96(.66)=1.31210-3www.khdaw.com 课后答案网www.khdaw.comChapter10c.x1−x2±z.025sx1−x24.6±1.3(3.3to5.9)8.a.x−x=15,700-14,500=1,20012b.Pooledvariance2227(700)+11(850)s==632,08318⎛11⎞s=632,083⎜+⎟=362.88x1−x2www.khdaw.com⎝812⎠With18degreesoffreedomt.025=2.1011200±2.101(362.88)1200±762(438to1962)c.Populationsarenormallydistributedwithequalvariances.9.a.n1=10n2=8x=21.2x=22.812s1=2.70s2=3.55x−x=21.2-22.8=-1.612Kitchensarelessexpensiveby$1,600.b.x1−x2±zα/2sx1−x2Degreesoffreedom=n1+n2-2=16t.05=1.7462229(2.70)+7(3.55)s==9.631082+−⎛11⎞s=9.63⎜+⎟=1.47x1−x2⎝108⎠-1.6±1.746(1.47)-1.6±2.57(-4.17to+.97)10.a.x=17.54x=15.361210-4www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulationsx−x=17.54-15.36=$2.18perhourgreaterforunionworkers.1222222(n1−1)s1+(n2−1)s214(2.24)+19(1.99)b.s===4.41n+n−215+20−212c.x−x±ts12α/2x1−x2⎛11⎞s=4.41⎜+⎟=0.72x1−x2⎝1520⎠17.5415.36−±t(.72)α/22.18±t(.72)α/2Note:Valuesfort.025arenotlistedfor33degreesoffreedom;for30d.f.t.025=2.042andfor40www.khdaw.comd.f.t.025=2.021.Wewillusethemoreconservativevalueof2.042asanapproximation.2.18±2.042(.72)2.18±1.47or0.71to3.652222ss(5.2)(6)1211.a.s=+=+=1.18x1−x2nn405012(25.222.8)−z==2.031.18RejectH0ifz>1.645RejectH0;concludeHaistrueandµ1−µ2>0.b.p-value=.5000-.4788=.02122222ss(8.4)(7.6)1212.a.s=+=+=1.31x1−x2nn807012(x−x)−(µ−µ)(104−106)−01212z===−1.53s1.31x1−x2RejectH0ifz<-1.96orz>1.96DonotrejectH0b.p-value=2(.5000-.4370)=.126013.a.x−x=1.4–1.0=0.41222222(n1−1)s1+(n2−1)s27(.4)+6(.6)s===0.2523n+n−28+7−21210-5www.khdaw.com 课后答案网www.khdaw.comChapter10⎛11⎞s=0.2523⎜+⎟=0.26x1−x2⎝87⎠With13degreesoffreedom.t.025=2.16RejectH0ift<-2.16ort>2.16(x−x)−(µ−µ)0.41212t===1.54s0.26x1−x2DonotrejectH014.a.H0:µ1-µ2=0www.khdaw.comHa:µ1−µ2≠0b.RejectH0ifz<-1.96orifz>1.962222ss16.815.212c.s=+=+=1.79x1−x2nn15017512(x1−x2)−0(39.335.4−)−0z===2.18s1.79x1−x2d.RejectH0;concludethepopulationmeansdiffer.e.p-value=2(.5000-.4854)=.029215.H0:µ1-µ2=0Ha:µ1−µ2≠0RejectH0ifz<-1.96orifz>1.96(x−x)0−(4035)−12z===2.412222σσ(9)(10)12++nn364912RejectH0;customersatthetwostoresdifferintermsofmeanages.p-value=2(.5000-.4920)=.016016.H0:µ1−µ2≤0Ha:µ1−µ2>0RejectH0ifz>2.0510-6www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulations(x1−x2)−(µ1−µ2)(547525)0−−z===4.992222σσ837812++nn56285212RejectH0;concludethatthefemaleshaveahighermeanverbalscore.p-value≈017.Population1issupplierA.Population2issupplierB.H0:µ1−µ2≤0StaywithsupplierAwww.khdaw.comHa:µ1−µ2>0ChangetosupplierBRejectH0ifz>1.645(x−x)(−µ−µ)(1412.5)0−−1212z===2.682222σσ(3)(2)12++nn503012p-value=.5000-.4963=.0037RejectH0;changetosupplierB.18.a.H0:µ1−µ2=0Ha:µ1−µ2≠02222σσ2.52.512σ=+=+=.36x1−x2nn1128412(x1−x2)−069.9569.56−z===1.08σ.36x1−x2b.p-value=2(.5000-.3599)=.2802c.DonorejectH0.Cannotconcludethatthereisadifferencebetweenthepopulationmeanscoresforthetwogolfers.19.a.H0:µ1−µ2=0Ha:µ1−µ2≠0b.t.025=2.021df=n1+n2-2=22+20-2=40RejectH0ift<-2.021orift>2.02110-7www.khdaw.com 课后答案网www.khdaw.comChapter1022222(n1−1)s1+(n2−1)s2(221)(.8)−+(201)(1.1)−c.s===.9108n+n−222202+−122⎛11⎞⎛11⎞s=s⎜+⎟=.9108⎜+⎟=.2948x1−x2⎝n1n2⎠⎝2220⎠(x1−x2)−02.52.1−t===1.36s.2948x1−x2d.DonotrejectH0.Cannotconcludethatadifferencebetweenpopulationmeanexists.e.Withdf=40,t.05=1.684andt.10=1.303Withtwotails,p-valueisbetween.10and.20.www.khdaw.com20.a.H0:µ1−µ2≤0Ha:µ1−µ2>0b.t.05=1.711df=n1+n2-2=16+10-2=24RejectH0ift>1.71122222(n1−1)s1+(n2−1)s2(161)(.64)−+(101)(.75)−c.s===.4669n+n−216102++122⎛11⎞⎛11⎞s=s⎜+⎟=.4669⎜+⎟=.2755x1−x2⎝n1n2⎠⎝1610⎠(x1−x2)−06.826.25−t===2.07s.2755x1−x2d.RejectH0.Concludethattheconsultantwiththemoreexperiencehasthehigherpopulationmeanrating.e.With24df,t.025=2.064p-valueisapproximately.02521.a.1,2,0,0,2b.d=∑d/n=5/5=1i2∑(d−d)4ic.s===1dn−15−1d.With4degreesoffreedom,t.05=2.132RejectH0ift>2.13210-8www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulationsd−µ1−0dt===2.24s/n1/5dp-valueisbetween.025and.05RejectH0;concludeµd>0.22.a.3,-1,3,5,3,0,1b.d=∑d/n=14/7=2i2∑(d−d)26ic.s===2.082dn−17−1www.khdaw.comd.d=2e.With6degreesoffreedomt.025=2.44722.4472.082/7±()2±1.93(.07to3.93)23.Difference=ratingafter-ratingbeforeH0:µd≤0Ha:µd>0With7degreesoffreedom,rejectH0ift>1.895d=.625ands=1.3025dd−µ.625−0dt===1.36s/n1.3025/8dp-valueisgreaterthan.10DonotrejectH0;wecannotconcludethatseeingthecommercialimprovesthemeanpotentialtopurchase.24.Differences:.20,.29,.39,.02,.24,.20,.20,.52,.29,.20d=∑dn/=2.55/10=.255i2∑(d−d)is==.1327dn−1Withdf=9,t.025=2.262sdd±t.025n10-9www.khdaw.com 课后答案网www.khdaw.comChapter10⎛.1327⎞.255±2.262⎜⎟⎝10⎠.255±.095(.16to.35)25.Differences:8,9.5,6,10.5,15,9,11,7.5,12,5d=93.5/10=9.35ands=2.954dt.025=2.2629.35±2.262e2.954/10j=9.35±2.11www.khdaw.comIntervalestimateis7.24to11.4626.H0:µd=0Ha:µd≠0RejectH0ift<-2.365orift>2.365df=7Differences-.01,.03,-.06,.16,.21,.17,-.09,.11d=∑dn/=.52/8=.065i2∑(d−d)is==.1131dn−1d−0.065t===1.63s.1131dn8DonotrejectH0.Cannotconcludethatthepopulationmeansdiffer.27.Usingmatchedsamples,thedifferencesareasfollows:4,-2,8,8,5,6,-4,-2,-3,0,11,-5,5,9,5H0:µd≤0Ha:µd>0d=3ands=5.21dd−µ3−0dt===2.23s/n5.21/15dp-valueisbetween.01and.025With14degreesoffreedom,rejectH0ift>1.76110-10www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulationsRejectH0.Concludethatthepopulationofreadersspendsmoretime,onaverage,watchingtelevisionthanreading.28.a.H0:µ1-µ2=0Ha:µ1-µ2≠0Withdf=11,t.025=2.201RejectH0ift<-2.201orift>2.201Calculatethedifference,di,foreachstock.d=∑d/n=85/12=7.08i2∑(d−d)is==3.34dwww.khdaw.comn−1x−µt==7.34s/ndp-value≈0RejectH0;adecreaseinP/Eratiosisbeingprojectedfor1998.sdb.d±t.025n⎛3.34⎞7.082.201±⎜⎟⎝12⎠7.08±2.12(4.96to9.21)29.a.Difference=Pricedeluxe-PriceStandardH0:µd=10Ha:µd≠10With6degreesoffreedom,rejectH0ift<-2.447orift>2.447d=8.86ands=2.61dd−µ8.86−10dt===−1.16s/n2.61/7dp-valueisgreaterthan.20DonotrejectH0;wecannotrejectthehypothesisthata$10pricedifferentialexists.10-11www.khdaw.com 课后答案网www.khdaw.comChapter10sdb.d±tα/2n⎛2.61⎞8.862.447±⎜⎟⎝7⎠8.86±2.41(6.45to11.27)30.a.(p−p)=.48-.36=.1212p(1−p)p(1−p)0.48(0.52)0.36(0.64)1122b.s=+=+=0.0373p1−p2nn400300www.khdaw.com120.12±1.645(0.0373)0.12±0.0614(0.0586to0.1814)c.0.12±1.96(0.0373)0.12±0.0731(0.0469to0.1931)np+np200(0.22)+300(0.16)112231.a.p===0.184n+n200+30012F11Is=(0.184)(0.816)+=0.0354p1−p2HGKJ200300RejectH0ifz>1.645(.22.16)0−−z==1.69.0354RejectH0b.p-value=(.5000-.4545)=.045532.p=220/400=0.551p=192/400=0.4820.55(0.45)0.48(0.52)s=+=0.0353p1−p2400400p−p±1.96s12p1−p20.55-0.48±1.96(0.0353)10-12www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulations0.07±0.0691(0.0009to0.1391)7%moreexecutivesarepredictinganincreaseinfull-timejobs.Theconfidenceintervalshowsthedifferencemaybefrom0%to14%.33.p−p±zs12α/2p1−p2p(1−p)p(1−p)(0.25)(0.75)(0.16)(0.84)1122s=+=+=0.025p1−p2nn496505120.25-0.16±1.96(0.25)0.09±0.05or0.04to0.1434.a.p=682/1082=.6303(63%)www.khdaw.com1p=413/1008=.4097(41%)2p−p=.6303-.4097=.2206(22%)12p(1−p)p(1−p).6303(1.6303)−.4097(1.4097)−1122b.σ=+=+=.0213p1−p2nn1082100812p−p±1.96σ12p1−p2.2206±1.96(.0213).2206±.0418(.1788to.2624)35.a.p=279/300=0.931p=255/300=0.852b.H0:p1-p2=0Ha:p1-p2≠0RejectH0ifz<-1.96orifz>1.96279+255p==0.89300+300F11Is=(0.89)(0.11)+=0.0255p1−p2HGKJ300300p−p−00.93−0.8512z===3.13s0.0255p1−p2p-valueislessthan.001RejectH0;womenandmendifferonthisquestion.10-13www.khdaw.com 课后答案网www.khdaw.comChapter10c.p−p±1.96s12p1−p2(0.93)(0.07)(0.85)(0.15)s=+=0.0253p1−p23003000.93-0.85±1.96(0.0253)0.08±0.05(0.03to0.13)95%confident,3%to13%morewomenthanmenagreewiththisstatement.36.H0:p1≤p2www.khdaw.comHa:p1>p2(p1−p2)−bp1−p2gz=sp1−p2np+np1545(0.675)+1691(0.608)1122p===0.64n+n1545+169112F11IF11Is=p(1−p)+=(0.64)(0.36)+=0.017p1−p2HGKJHGn1n2KJ15451691(0.675−0.608)−0z==3.940.017Since3.94>z.05=1.645,werejectH0p-value≈0Conclusion:Theproportionofmenthatfeelthatthedivisionofhouseworkisfairisgreaterthantheproportionofwomenthatfeelthatthedivisionofhouseworkisfair.37.H0:p1-p2=0Ha:p1-p2≠0RejectH0ifz<-1.96orifz>1.9663+60p==0.3514150+200F11Is=(0.3514)(0.6486)+=0.0516p1−p2HGKJ150200p=63/150=0.42p=60/200=0.3012(p1−p2)−bp1−p2g(0.42−0.30)−0z===2.33s0.0516p1−p210-14www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulationsp-value=2(.5000-.4901)=.0198RejectH0;thereisadifferencebetweentherecallratesforthetwocommercials.0.42(58)0.30(0.70)b.(0.42−0.30)±1.96+150200.12±.10(.02to.22)np+np232(.815)210(.724)+112238.p===.7718n+n232210+12⎛11⎞⎛11⎞s=p(1−p)⎜+⎟=(.7718)(17718)−⎜+⎟=.04p1−p2⎝n1n2⎠⎝232210⎠www.khdaw.comz=(p1−p2)−0=.815.724−=2.28s.04p1−p2p-value=2(.5-.4887)=.0226p-value<.05,rejectH0.Thepopulationproportionsdiffer.NYSEisshowingagreaterproportionofstocksbelowtheir1997highs.39.H0:p1-p2≤0Ha:p1-p2>0np+np240(.40)250(.32)+1122p===.3592n+n240250+12⎛11⎞⎛11⎞s=p(1−p)⎜+⎟=(.3592)(1.3592)−⎜+⎟=.0434p1−p2⎝n1n2⎠⎝240250⎠(p1−p2)−0.40.32−z===1.85s.0434p1−p2p-value=.5000-.4678=.0322p-value<.05,rejectH0.TheproportionofusersatworkisgreaterinWashingtonD.C.22ss1240.x−x±z+12.05nn1222(2500)(2000)40,00035,0001.645−±+60805000±646(4354to5646)10-15www.khdaw.com 课后答案网www.khdaw.comChapter1041.H0:µ1-µ2=0Ha:µ1-µ2≠0RejectH0ifz<-1.96orifz>1.96(x−x)(−µ−µ)(4.13.3)0−−1212z===3.192222σσ(2.2)(1.5)12++nn12010012RejectH0;adifferenceexistswithsystemBhavingthelowermeancheckouttime.www.khdaw.com42.a.H0:µ1-µ2≤0Ha:µ1-µ2>0RejectH0ifz>1.645b.Usingthecomputer,n1=30n2=30x=16.23x=15.7012s1=3.52s2=3.3122(3.52)(3.31)s=+=0.88x1−x23030(x−x)−0(16.23−15.70)12z===0.59s0.88x1−x2DonotrejectH0;cannotconcludethatthemutualfundswithaloadhaveagreatermeanrateofreturn.Loadfunds16.23%;noloadfunds15.7%c.Atz=0.59,Area=0.2224p-value=0.5000-0.2224=0.277643.H0:µ1-µ2=0Ha:µ1-µ2≠0Use25degreesoffreedom.RejectH0ift<-2.06orift>2.0622211(8)+14(10)s==84.162510-16www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulations(x1−x2)−(µ1−µ2)(7278−)−0t===−1.69⎛11⎞⎛11⎞2+84.16⎜+⎟s⎜⎟⎝nn⎠⎝1215⎠12p-valueisbetween.10and.20DonotrejectH0;cannotconcludeadifferenceexists.44.Difference=before-afterH0:µd≤0Ha:µd>0www.khdaw.comWith5degreesoffreedom,rejectH0ift>2.015d=6.167ands=6.585dd−µ6.167−0dt===2.29s/n6.585/6dp-valueisbetween.05and.10RejectH0;concludethattheprogramprovidesweightloss.45.a.Population1-1996Population2-1997H0:µ1-µ2≤0Ha:µ1-µ2>0b.d=∑d/n=1.74/14=0.12i2∑(d−d)is==0.33dn−1Degreesoffreedom=13;t.05=1.771RejectH0ift>1.771d−00.12t===1.42s/n0.33/14dp-valueisbetween.05and.10DonotrejectH0.Thesampleof14companiesshowsearningsaredowninthefourthquarterbyameanof0.12pershare.However,datadoesnotsupporttheconclusionthatmeanearningsforallcompaniesaredownin1997.10-17www.khdaw.com 课后答案网www.khdaw.comChapter1046.a.H0:p1-p2≤0Ha:p1-p2>0b.p=704/1035=.6802(68%)1p=582/1004=.5797(58%)2p−p=.6802-.5797=.100512np+np1035(0.6802)1004(0.5797)+1122p===.6307n+n10351004+12⎛11⎞⎛11⎞s=p(1−p)⎜+⎟=(.6307)(1.6307)−⎜+⎟=.0214p1−p2www.khdaw.com⎝n1n2⎠⎝10351004⎠(p−p)0−.6802.5797−12z===4.70s.0214p1−p2p-value≈0c.RejectH0;proportionindicatinggood/excellentincreased.47.a.H0:p1-p2=0Ha:p1-p2≠0RejectH0ifz<-1.96orifz>1.9676+90p==0.1277400+900F11Is=(0.1277)(0.8723)+=0.02p1−p2HGKJ400900p=76/400=0.19p=90/900=0.1012(p−p)−(p−p)(0.19−0.10)−01212z===4.50s0.02p1−p2p-value≈0RejectH0;thereisadifferencebetweenclaimrates.0.19(0.81)0.10(0.90)b.0.09±1.96+400900.09±.0432(.0468to.1332)10-18www.khdaw.com 课后答案网www.khdaw.comStatisticalInferenceaboutMeansandProportionswithTwoPopulations9+51448.p===0.0341142+268410F11Is=(0.0341)(0.9659)+=0.0188p1−p2HGKJ142268p=9/142=0.0634p=5/268=0.018712p−p=0.0634−0.0187=0.0447120.0447−0z==2.380.0188p-value=2(0.5000-0.4913)=0.0174RejectH0;Thereisasignificantdifferenceindrugresistancebetweenthetwostates.NewJerseywww.khdaw.comhasthehigherdrugresistancerate.10-19www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter11InferencesIInferencesnferencesAboutAAboutboutPopulationPPopulationopulationVariancesVVariancesariancesLearningLLearningearningObjectivesOObjectivesbjectives1.Understandtheimportanceofvarianceinadecision-makingsituation.2Understandtheroleofstatisticalinferenceindevelopingconclusionsaboutthevarianceofasinglepopulation.www.khdaw.com3.Knowthesamplingdistributionof(n-1)s2/σ2hasachi-squaredistributionandbeabletousethisresulttodevelopaconfidenceintervalestimateofσ2.4.Knowhowtotesthypothesesinvolvingσ2.5.Understandtheroleofstatisticalinferenceindevelopingconclusionsaboutthevariancesoftwopopulations.226.Knowthatthesamplingdistributionofs/shasanFdistributionandbeabletousethisresultto12testhypothesesinvolvingthevariancesoftwopopulations.11-1www.khdaw.com 课后答案网www.khdaw.comChapter11Solutions:SSolutions:olutions:1.a.11.0705b.27.4884c.9.59083d.23.2093e.9.390462.s2=2522a.With19degreesoffreedomχ=30.1435andχ=10.1170.05.9519(25)219(25)www.khdaw.com≤σ≤30.143510.117015.76≤σ2≤46.9522b.With19degreesoffreedomχ=32.8523andχ=8.90655.025.97519(25)219(25)≤σ≤32.85238.9065514.46≤σ2≤53.33c.3.8≤σ2≤7.323.With15degreesoffreedomχ=24.9958.052RejectH0ifχ>24.9958222(n−1)s(161)(8)−χ===19.22σ50DonotrejectH04.a.n=18s2=.3622χ=27.5871χ=8.67176(17degreesoffreedom).05.9517(.36)217(.36)≤σ≤27.58718.67176.22≤σ2≤.71b..47≤σ≤.8411-2www.khdaw.com 课后答案网www.khdaw.comInferencesAboutPopulationVariances22Σ(x2−x)5.a.s==31.07n−1s=31.07=5.5722b.χ=16.0128χ=1.68987.025.975(81)(31.07)−2(81)(31.07)−≤σ≤16.01281.6898713.58≤σ2≤128.71c.3.69≤σ≤11.3422Σ(xi−x)www.khdaw.com6.a.s==176.96n−1s=176.96=13.3022b.χ=11.1433χ=0.484419.025.975(51)(176.96)−2(51)(176.96)−≤σ≤11.14330.48441963.52≤σ2≤1461.217.97≤σ≤38.2322Σ(xi−x)7.a.s==2.62n−1s=2.62=1.6222b.χ=16.0128χ=1.68987.025.095(81)(2.62)−2(81)(2.62)−≤σ≤16.01281.689871.14≤σ2≤10.85c.1.07≤σ≤3.2922Σ(xi−x).09298.a.s===.00845n−1121−b.s=.00845=.091911-www.khdaw.com 课后答案网www.khdaw.comChapter11c.11degreesoffreedom22χ=21.92χ=3.81575.025.97522(n−1)s2(n−1)s≤σ≤22χχ.025.975(121).00845−2(121).00845−≤σ≤21.923.81575.0042≤σ2≤.0244d..0651≤σ≤.1561www.khdaw.com9.H0:σ2≤.0004Ha:σ2>.0004n=302χ=42.5569(29degreesoffreedom).052(29)(.0005)χ==36.25.0004DonotrejectH0;theproductspecificationdoesnotappeartobeviolated.10.H0:σ2≤.75Ha:σ2>.752χ=42.5569(29degreesoffreedom).05222(n−1)s(29)(2)χ===206.2222σ(.75)02Sinceχ=206.22>42.5569,rejectH0ThestandarddeviationfortelevisionsetsisgreaterthanthestandarddeviationforVCR’s.11.19degreesoffreedom22χ=8.90655χ=32.8523.975.02522RejectH0ifχ<8.90655orifχ>32.8523222(n−1)s(201)(.114)−χ===26.792σ.009216DonotrejectH0.Cannotconcludethevarianceininterestrateshaschanged.11-4www.khdaw.com 课后答案网www.khdaw.comInferencesAboutPopulationVariances22Σ(xi−x)12.s==.8106n−1H0:σ2=.94Ha:σ2≠.9422(n−1)s(11)(.8106)χ===9.492σ.9402222With11degreesoffreedom,rejectifχ<χ=3.81575orχ>χ=21.92..975.0252Sinceχ=9.49isnotintherejectionregion,wecannotrejectH0.www.khdaw.com13.a.F.05=2.91b.F.025=2.76c.F.01=4.5011d.F===.29.975F3.42.025,20,10RemembertoreversethedegreesoffreedomintheF.025above.14.F.05,15,19=2.23RejectH0ifF>2.232s5.81F===2.422s2.4222RejectH0:concludeσ1>σ215.Werecommendplacingthelargersamplevarianceinthenumerator.Withα=.05,F.025,20,24=2.33.RejectifF>2.33.F=8.2/4.0=2.05DonotrejectH0OrifwehadthelowertailFvalue,11F===.41.025,20,24F2.41.025,24,20F=4.0/8.2=.49∴F>.41DonotrejectH02216.H:σ≤σ01222H:σ>σa1211-www.khdaw.com 课后答案网www.khdaw.comChapter11F.01,24,29=2.49RejectH0ifF>2.4922s941F===2.6322s582RejectH0;Concludeadultshaveagreatervarianceinonlinetimesthanteens.217.a.Letσ=varianceinrepaircosts(4yearoldautomobiles)12σ=varianceinrepaircosts(2yearoldautomobiles)222H:σ≤σ01222H:σ>σa12www.khdaw.comb.s2=(170)2=28,9001s2=(100)2=10,00022s28,9001F===2.892s10,0002F.01,24,24=2.66RejectH;concludethat4yearoldautomobileshavealargervarianceinannualrepaircosts0comparedto2yearoldautomobiles.Thisisexpectedduetothefactthatolderautomobilesaremorelikelytohavesomeveryexpensiverepairswhichleadtogreatervarianceintheannualrepaircosts.2218.H:σ=σ01222H:σ≠σa12Fα/2=F.025,9,6=5.5222s4.271F===3.5422s2.272DonotrejectH;Cannotconcludeanydifferencebetweenvariancesofthetwoindustries.02219.H:σ=σ01222H:σ≠σa12F.025=2.37(Degreesoffreedomare24numerator,21denominator)11-6www.khdaw.com 课后答案网www.khdaw.comInferencesAboutPopulationVariancesUsingMinitab,Machine1:n1=25s1=.2211x1=3.328Machine1:n1=22s1=.0768x1=3.27822s(.2211)1F===8.2922s(.0768)2RejectH;theprocessvariancesaresignificantlydifferent.Machine1offersthebestopportunity0forprocessqualityimprovements.Notethatthesamplemeansaresimilarwiththemeanbagweightsofapproximately3.3grams.www.khdaw.comHowever,theprocessvariancesaresignificantlydifferent.2220.H:σ=σ01222H:σ≠σa12F.025=2.37(Degreesoffreedomare24numerator,24denominator)With11.1thelargersamplevariance,wehaveF=11.1/2.1=5.29RejectH;thevariancesarenotequalforseniorsandmanagers.022Σ(xi−x)21.a.s=n−122s=9663.57s=19,237.73NovDec22b.H:σ=σ0NovDec22H:σ≠σaNovDec2s19,237.73DecF===1.992s9663.57NovF.05,9,9=3.18SinceF=1.99<3.18,donotrejectH0Thereisnoevidencethatthepopulationvariancesdiffer.2222.H:σ≤σ0wetdry22H:σ>σawetdry11-www.khdaw.com 课后答案网www.khdaw.comChapter112222s=32=1024s=16=256wetdryF.05=2.402s1024wetF===42s256drySinceF=4>2.40,rejectHandconcludethatthereisgreatervariabilityinstoppingdistanceson0wetpavement.b.Drivecarefullyonwetpavementbecauseoftheuncertaintyinstoppingdistances.23.a.s2=(30)2=90022b.χ=30.1435andχ=10.1170(19degreesoffreedom)www.khdaw.com.05.95(19)(900)2(19)(900)≤σ≤30.143510.1170567.29≤σ2≤1690.22c.23.82≤σ≤41.1124.With12degreesoffreedom,22χ=23.3367χ=4.40379.025.97522(12)(14.95)2(12)(14.95)≤σ≤23.33674.40379114.93≤σ2≤609.0310.72≤σ≤24.68Σxi25.a.x==$260.16n22Σ(xi−x)b.s==4996.79n−1s=4996.79=70.6922c.χ=32.8523χ=8.90655.025.975(201)(4996.78)−2(201)(4996.78)−≤σ≤32.85238.906552889.87≤σ2≤10,659.4553.76≤σ≤103.2411-8www.khdaw.com 课后答案网www.khdaw.comInferencesAboutPopulationVariances26.a.H0:σ2≤.0001Ha:σ2>.00012χ=21.0642(14degreesoffreedom).1022(14)(.014)χ==27.44.0001RejectH;σ2exceedsmaximumvariancerequirement.022b.χ=23.6848andχ=6.57063(14degreesoffreedom)www.khdaw.com.05.9522(14)(.014)2(14)(.014)≤σ≤23.68486.57063.00012≤σ2≤.0004227.H0:σ2≤.02Ha:σ2>.022χ=55.7585(40degreesoffreedom).0522(40)(.16)χ==51.2.02DonotrejectH;thevariancedoesnotappeartobeexceedingthestandard.028.n=22s2=1.5H0:σ2≤1Ha:σ2>12χ=29.6151(21degreesoffreedom).102(21)(1.5)χ==31.51RejectH;concludethatσ2>1.022Σ(xi−x)101.5629.s===12.69n−191−H:σ2=10011-www.khdaw.com 课后答案网www.khdaw.comChapter11Ha:σ2≠1022(n−1)s(8)(12.69)χ===10.162σ100With8degreesoffreedom,rejectif2222χ<χ=2.73264orχ>χ=15.5073.95.052Sinceχ=10.16isnotintherejectionregion,wecannotrejectH.030.a.Tryn=1522χ=26.1190χ=5.62872(14degreesoffreedom).025.975www.khdaw.com(14)(64)2(14)(64)≤σ≤26.11905.6287234.30≤σ2≤159.185.86≤σ≤12.62∴Asamplesizeof15wasused.b.n=25;expectedthewidthoftheintervaltobesmaller.22χ=39.3641χ=12.4011(24degreesoffreedom).05.97522(24)(8)2(24)(8)≤σ≤39.364112.401139.02≤σ2≤126.866.25≤σ≤11.132231.H:σ=σ01222H:σ≠σa12Fα/2=F.05,9,9=3.1822s15.81F===422s7.92RejectH.ConcludethevariancesdifferwithNASDAQstocksshowingthegreatervariance.02232.H:σ=σ01222H:σ≠σa1211-10www.khdaw.com 课后答案网www.khdaw.comInferencesAboutPopulationVariancesF.025=1.46622s.9401F===1.3922s.7972DonotrejectH;Wearenotabletoconcludestudentswhocompletethecourseandstudentswho0dropouthavedifferentvariancesofgradepointaverages.233.n=16s=5.4112n=16s=2.32222H:σ=σ01222H:σ≠σwww.khdaw.coma12F.05=2.40(Degreesoffreedomare15numerator,15denominator)2s5.41F===2.352s2.32DonotrejectH;datadoesnotindicateadifferencebetweenthepopulationvariances.02234.H:σ=σ01222H:σ≠σa12F.05=1.94(30numeratorand24denominatordegreesoffreedom)2s251F===2.082s122RejectH;concludethatthevariancesofassemblytimesarenotequal.011-www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter12TestsTTestsestsofGoodnessGGoodnessoodnessofFitFFititandananddIndependenceIIndependencendependenceLearningLLearningearningObjectivesOObjectivesbjectives1.Knowhowtoconductagoodnessoffittest.www.khdaw.com2.Knowhowtousesampledatatotestforindependenceoftwovariables.3.Understandtheroleofthechi-squaredistributioninconductingtestsofgoodnessoffitandindependence.4.Beabletoconductagoodnessoffittestforcaseswherethepopulationishypothesizedtohaveeitheramultinomial,aPoisson,oranormalprobabilitydistribution.5.Foratestofindependence,beabletosetupacontingencytable,determinetheobservedandexpectedfrequencies,anddetermineifthetwovariablesareindependent.12-1www.khdaw.com 课后答案网www.khdaw.comChapter12Solutions:SSolutions:olutions:1.Expectedfrequencies:e1=200(.40)=80,e2=200(.40)=80e3=200(.20)=40Actualfrequencies:f1=60,f2=120,f3=20(60-80)2(120-80)2(20-40)22χ=++8080404001600400=++808040=5+20+10=352www.khdaw.comχ.01=9.21034withk-1=3-1=2degreesoffreedom2Sinceχ=35>9.21034rejectthenullhypothesis.Thepopulationproportionsarenotasstatedinthenullhypothesis.2.Expectedfrequencies:e1=300(.25)=75,e2=300(.25)=75e3=300(.25)=75,e4=300(.25)=75Actualfrequencies:f1=85,f2=95,f3=50,f4=70(85-75)2(95-75)2(50-75)2(70-75)22χ=+++7575757510040062525=+++757575751150=75=15.332χ=7.81473withk-1=4-1=3degreesoffreedom.05Sinceχ2=15.33>7.81473rejectH0Weconcludethattheproportionsarenotallequal.3.H0=pABC=.29,pCBS=.28,pNBC=.25,pIND=.18Ha=TheproportionsarenotpABC=.29,pCBS=.28,pNBC=.25,pIND=.1812-2www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependenceExpectedfrequencies:300(.29)=87,300(.28)=84300(.25)=75,300(.18)=54e1=87,e2=84,e3=75,e4=54Actualfrequencies:f1=95,f2=70,f3=89,f4=462χ=7.81(3degreesoffreedom).0522222(95-87)(70-84)(89-75)(46-54)χ=+++87847554=6.87www.khdaw.comDonotrejectH0;thereisnosignificantchangeintheviewingaudienceproportions.4.ObservedExpectedHypothesizedFrequencyFrequencyCategoryProportion(fi)(ei)(fi-ei)2/eiBrown0.30177151.84.18Yellow0.20135101.211.29Red0.2079101.24.87Orange0.104150.61.82Green0.103650.64.21Blue0.103850.63.14Totals:50629.512χ=11.07(5degreesoffreedom).05Since29.51>11.07,weconcludethatthepercentagefiguresreportedbythecompanyhavechanged.5.ObservedExpectedHypothesizedFrequencyFrequencyCategoryProportion(fi)(ei)(fi-ei)2/eiFullService1/3264249.330.86Discount1/3255249.330.13Both1/3229249.331.66Totals:7482.652χ=4.61(2degreesoffreedom).10Since2.65<4.61,thereisnosignificantdifferenceinpreferenceamongthethreeservicechoices.6.12-3www.khdaw.com 课后答案网www.khdaw.comChapter12ObservedExpectedHypothesizedFrequencyFrequencyCategoryProportion(fi)(ei)(fi-ei)2/eiNewsandOpinion1/62019.17.04GeneralEditorial1/61519.17.91FamilyOriented1/63019.176.12Business/Financial1/62219.17.42FemaleOriented1/61619.17.52African-American1/61219.172.68Totals:11510.692χ=9.24(5degreesoffreedom).10Since10.69>9.24,weconcludethatthereisadifferenceintheproportionofadswithguiltappealsamongthesixtypesofmagazines.www.khdaw.com7.Expectedfrequencies:ei=(1/3)(135)=452222(43-45)(53-45)(39-45)χ=++=2.314545452With2degreesoffreedom,χ=5.99.05DonotrejectH0;thereisnojustificationforconcludingadifferenceinpreferenceexists.8.H0:p1=.03,p2=.28,p3=.45,p4=.242df=3χ=11.34.01RejectH0ifχ2>11.34RatingObservedExpected(fi-ei)2/eiExcellent24.03(400)=1212.00Good124.28(400)=1121.29Fair172.45(400)=180.36Poor80.24(400)=962.67400400χ2=16.31RejectH0;concludethattheratingsdiffer.Acomparisonofobservedandexpectedfrequenciesshowtelephoneserviceisslightlybetterwithmoreexcellentandgoodratings.9.H0=ThecolumnvariableisindependentoftherowvariableHa=ThecolumnvariableisnotindependentoftherowvariableExpectedFrequencies:ABCP28.539.945.6Q21.530.134.412-4www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependence2222222(20-28.5)(44-39.9)(50-45.6)(30-21.5)(26-30.1)(30-34.4)χ=+++++28.539.945.621.530.134.4=7.862χ=7.37776with(2-1)(3-1)=2degreesoffreedom.025Sinceχ2=7.86>7.37776RejectH0Concludethatthecolumnvariableisnotindependentoftherowvariable.10.H0=ThecolumnvariableisindependentoftherowvariableHa=Thecolumnvariableisnotindependentoftherowvariablewww.khdaw.comExpectedFrequencies:ABCP17.500030.625021.8750Q28.750050.312535.9375R13.750024.062517.18752222(20-17.5000)(30-30.6250)(30-17.1875)χ=++⋅⋅⋅+17.500030.625017.1875=19.782χ=9.48773with(3-1)(3-1)=4degreesoffreedom.05Sinceχ2=19.78>9.48773RejectH0Concludethatthecolumnvariableisnotindependentofftherowvariable.11.H0:TypeofticketpurchasedisindependentofthetypeofflightHa:Typeofticketpurchasedisnotindependentofthetypeofflight.ExpectedFrequencies:e11=35.59e12=15.41e21=150.73e22=65.27e31=455.68e32=197.32ObservedExpectedFrequencyFrequencyTicketFlight(fi)(ei)(fi-ei)2/eiFirstDomestic2935.591.22FirstInternational2215.412.82BusinessDomestic95150.7320.61BusinessInternational12165.2747.59FullFareDomestic518455.688.52FullFareInternational135197.3219.68Totals:920100.4312-5www.khdaw.com 课后答案网www.khdaw.comChapter122χ=5.99with(3-1)(2-1)=2degreesoffreedom.05Since100.43>5.99,weconcludethatthetypeofticketpurchasedisnotindependentofthetypeofflight.12.a.ObservedFrequency(fij)DomesticEuropeanAsianTotalSame1255568248Different140105107352Total265160175600ExpectedFrequency(eij)DomesticEuropeanAsianTotalSame109.5366.1372.33248Different155.4793.87102.67352www.khdaw.comTotal265160175600ChiSquare(fij-eij)2/eijDomesticEuropeanAsianTotalSame2.181.870.264.32Different1.541.320.183.04χ2=7.362Degreesoffreedom=2χ=5.99.05RejectH0;concludebrandloyaltyisnotindependentofmanufacturer.b.BrandLoyaltyDomestic125/265=.472(47.2%)←HighestEuropean55/160=.344(34.4%)Asian68/175=.389(38.9%)13.IndustryMajorOilChemicalElectricalComputerBusiness3022.517.530Engineering3022.517.530Note:Valuesshownabovearetheexpectedfrequencies.2χ=11.3449(3degreesoffreedom:1x3=3).01χ2=12.39RejectH0;concludethatmajorandindustrynotindependent.12-6www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependence14.ExpectedFrequencies:e11=31.0e12=31.0e21=29.5e22=29.5e31=13.0e32=13.0e41=5.5e42=5.5e51=7.0e52=7.0e61=14.0e62=14.0ObservedExpectedFrequencyFrequencyMostDifficultGender(fi)(ei)(fi-ei)2/eiSpouseMen3731.01.16SpouseWomen2531.01.16ParentsMen2829.50.08www.khdaw.comParentsWomen3129.50.08ChildrenMen713.02.77ChildrenWomen1913.02.77SiblingsMen85.51.14SiblingsWomen35.51.14In-LawsMen47.01.29In-LawsWomen107.01.29OtherRelativesMen1614.00.29OtherRelativesWomen1214.00.29Totals:20013.432χ=11.0705with(6-1)(2-1)=5degreesoffreedom.05Since13.43>11.0705.weconcludethatgenderisnotindependentofthemostdifficultpersontobuyfor.15.ExpectedFrequencies:e11=17.16e12=12.84e21=14.88e22=11.12e31=28.03e32=20.97e41=22.31e42=16.69e51=17.16e52=12.84e61=15.45e62=11.55ObservedExpectedFrequencyFrequencyMagazineAppeal(fi)(ei)(fi-ei)2/eiNewsGuilt2017.160.47NewsFear1012.840.63GeneralGuilt1514.880.00GeneralFear1111.120.00FamilyGuilt3028.030.14FamilyFear1920.970.18BusinessGuilt2222.310.00BusinessFear1716.690.01FemaleGuilt1617.160.08FemaleFear1412.840.1112-7www.khdaw.com 课后答案网www.khdaw.comChapter12African-AmericanGuilt1215.450.77African-AmericanFear1511.551.03Totals:2013.412χ=15.09with(6-1)(2-1)=5degreesoffreedom.01Since3.41<15.09,thehypothesisofindependencecannotberejected.34.a.ObservedFrequency(fij)PharmConsumerComputerTelecomTotalCorrect207136151178672Incorrect3491228Total210140160190700ExpectedFrequency(eij)www.khdaw.comPharmConsumerComputerTelecomTotalCorrect201.6134.4153.6182.4672Incorrect8.45.66.47.628Total210140160190700ChiSquare(fij-eij)2/eijPharmConsumerComputerTelecomTotalCorrect.14.02.04.11.31Incorrect3.47.461.062.557.53χ2=7.852Degreesoffreedom=3χ=7.81473.05DonotrejectH0;concludeorderfulfillmentisnotindependentofindustry.b.Thepharmaceuticalindustryisdoingthebestwith207of210(98.6%)correctlyfilledorders.17.ExpectedFrequencies:PartQualitySupplierGoodMinorDefectMajorDefectA88.766.075.14B173.0911.8310.08C133.159.107.75χ2=7.962χ=9.48773(4degreesoffreedom:2x2=4).05DonotrejectH0;concludethattheassumptionofindependencecannotberejected18.ExpectedFrequencies:PartyAffiliationEducationLevelDemocraticRepublicanIndependentDidnotcompletehighschool282814Highschooldegree323216Collegedegree40402012-8www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependenceχ2=13.422χ=13.2767(4degreesoffreedom:2x2=4).01RejectH0;concludethatpartyaffiliationisnotindependentofeducationlevel.19.ExpectedFrequencies:e11=11.81e12=8.44e13=24.75e21=8.40e22=6.00e23=17.60e31=21.79e32=15.56e33=45.65ObservedExpectedFrequencyFrequencySiskelEbert(fi)(ei)(fi-ei)2/eiConCon2411.8112.57ConMixed88.440.02www.khdaw.comConPro1324.755.58MixedCon88.400.02MixedMixed136.008.17MixedPro1117.602.48ProCon1021.796.38ProMixed915.562.77ProPro6445.657.38Totals:16045.362χ=13.28with(3-1)(3-1)=4degreesoffreedom.01Since45.36>13.28,weconcludethattheratingsarenotindependent.20.Firstestimateµfromthesampledata.Samplesize=120.0(39)1(30)2(30)3(18)4(3)++++156µ===1.3120120Therefore,weusePoissonprobabilitieswithµ=1.3tocomputeexpectedfrequencies.ObservedPoissonExpectedDifferencexFrequencyProbabilityFrequency(fi-ei)039.272532.7006.300130.354342.516-12.516230.230327.6362.364318.099811.9766.0244ormore3.04305.160-2.160222222(6.300)(-12.516)(2.364)(6.024)(-2.160)χ=++++32.70042.51627.63611.9765.160=9.03482χ=7.81473with5-1-1=3degreesoffreedom.05Sinceχ2=9.0348>7.81473RejectH012-9www.khdaw.com 课后答案网www.khdaw.comChapter12ConcludethatthedatadonotfollowaPoissonprobabilitydistribution.21.Withn=30wewillusesixclasseswith162/3%oftheprobabilityassociatedwitheachclass.x=22.80s=6.2665Thezvaluesthatcreate6intervals,eachwithprobability.1667are-.98,-.43,0,.43,.98zCutoffvalueofx-.9822.8-.98(6.2665)=16.66-.4322.8-.43(6.2665)=20.11022.8+0(6.2665)=22.80.4322.8+.43(6.2665)=25.49www.khdaw.com.9822.8+.98(6.2665)=28.94ObservedExpectedIntervalFrequencyFrequencyDifferencelessthan16.6635-216.66-20.1175220.11-22.8055022.80-25.4975225.49-28.9435-228.94andup5502222222(2)−(2)(0)(2)(2)−(0)16χ=+++++==3.2055555552χ=9.34840with6-2-1=3degreesoffreedom.025Sinceχ2=3.20≤9.34840DonotrejectH0Theclaimthatthedatacomesfromanormaldistributioncannotberejected.0(34)1(25)2(11)3(7)4(3)++++22.µ==180UsePoissonprobabilitieswithµ=1.12-10www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependencePoissonxObservedProbabilitiesExpected034.367929.432125.367929.432211.183914.71237.06134.904combineinto143.01531.224categoryof3or}moretomake5ormore-.0037.296ei≥5.www.khdaw.comχ2=4.302χ=5.99147(2degreesoffreedom).05DonotrejectH0;theassumptionofaPoissondistributioncannotberejected.0(15)1(31)2(20)3(15)4(13)5(4)6(2)++++++23.µ==2100PoissonxObservedProbabilitiesExpected015.135313.53131.270727.07220.270727.07315.180418.04413.09029.025ormore6.05275.27χ2=4.982χ=7.77944(4degreesoffreedom).10DonotrejectH0;theassumptionofaPoissondistributioncannotberejected.24.x=24.5s=3n=30Use6classesObservedExpectedIntervalFrequencyFrequencylessthan21.565521.56-23.214523.21-24.503524.50-25.797525.79-27.447527.41up45χ2=2.812-11www.khdaw.com 课后答案网www.khdaw.comChapter122χ=6.25139(3degreesoffreedom:6-2-1=3).10DonotrejectH0;theassumptionofanormaldistributioncannotberejected.25.x=71s=17n=25Use5classesObservedExpectedIntervalFrequencyFrequencylessthan56.77556.7-66.57566.5-74.61574.6-84.51584.5up95χ2=11.22χ=9.21034(2degreesoffreedom)www.khdaw.com.01RejectH0;concludethedistributionisnotanormaldistribution.26.Observed60455936Expected50505050χ2=8.042χ=7.81473(3degreesoffreedom).05RejectH0;concludethattheorderpotentialsarenotthesameineachsalesterritory.27.Observed48323791663Expected37.03306.82126.9621.1637.032222(48–37.03)(323–306.82)(63–37.03)χ=++•••+37.03306.8237.03=41.692χ=13.2767(4degreesoffreedom).01Since41.69>13.2767,rejectH0.Mutualfundinvestors"attitudestowardcorporatebondsdifferfromtheirattitudestowardcorporatestock.28.Observed20204060Expected3535353512-12www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependence22222(20–35)(20–35)(40–35)(60–35)χ=+++35353535=31.432χ=7.81473(3degreesoffreedom).05Since31.43>7.81473,rejectH0.Theparkmanagershouldnotplanonthesamenumberattendingeachday.PlanonalargerstaffforSundaysandholidays.29.Observed1316281716Expected1818181818www.khdaw.comχ2=7.442χ=9.48773.05DonotrejectH0;theassumptionthatthenumberofridersisuniformlydistributedcannotberejected.30.ObservedExpectedHypothesizedFrequencyFrequencyCategoryProportion(fi)(ei)(fi-ei)2/eiVerySatisfied0.281051408.75SomewhatSatisfied0.462352300.11Neither0.1255600.42SomewhatDissatisfied0.10905032.00VeryDissatisfied0.0415201.25Totals:50042.532χ=9.49(4degreesoffreedom).05Since42.53>9.49,weconcludethatthejobsatisfactionforcomputerprogrammersisdifferentthanthejobsatisfactionforISmanagers.31.ExpectedFrequencies:QualityShiftGoodDefective1st368.4431.562nd276.3323.673rd184.2215.78χ2=8.112χ=5.99147(2degreesoffreedom).05RejectH0;concludethatshiftandqualityarenotindependent.32.ExpectedFrequencies:e11=1046.19e12=632.81e21=28.66e22=17.3412-13www.khdaw.com 课后答案网www.khdaw.comChapter12e31=258.59e32=156.41e41=516.55e42=312.45ObservedExpectedFrequencyFrequencyEmploymentRegion(fi)(ei)(fi-ei)2/eiFull-TimeEastern11051046.193.31Full-timeWestern574632.815.46Part-TimeEastern3128.660.19Part-TimeWestern1517.340.32Self-EmployedEastern229258.593.39Self-EmployedWestern186156.415.60NotEmployedEastern485516.551.93NotEmployedWestern344312.453.19Totals:296923.372χ=7.81with(4-1)(2-1)=3degreesoffreedomwww.khdaw.com.05Since23.37>7.81,weconcludethatemploymentstatusisnotindependentofregion.33.Expectedfrequencies:LoanApprovalDecisionLoanOfficesApprovedRejectedMiller24.8615.14McMahon18.6411.36Games31.0718.93Runk12.437.57χ2=2.212χ=7.81473(3degreesoffreedom).05DonotrejectH0;theloandecisiondoesnotappeartobedependentontheofficer.34.a.ObservedFrequency(fij)NeverMarriedMarriedDivorcedTotalMen23410610350Women21616816400Total45027426750ExpectedFrequency(eij)NeverMarriedMarriedDivorcedTotalMen210127.8712.13350Women240146.1313.87400Total45027426750ChiSquare(fij-eij)2/eijNeverMarriedMarriedDivorcedTotalMen2.743.74.386.86Women2.403.27.336.0012-14www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependenceχ2=12.862Degreesoffreedom=2χ=9.21.01RejectH0;concludemartialstatusisnotindependentofgender.b.MartialStatusNeverMarriedMarriedDivorcedMen66.9%30.3%2.9%Women54.0%42.0%4.0%Men100-66.9=33.1%havebeenmarriedWomen100-54.0=46.0%havebeenmarriedwww.khdaw.com35.ExpectedFrequencies:(50)(18)(50)(24)(50)(12)e==9,e==12,,⋅⋅⋅e==61112251001001002222(49)−(1012)−(46)−χ=++⋅⋅⋅+=9.7691262χ=9.48773(4degreesoffreedom).05Since9.76<9.48773,rejectH0.BankingtendstohavelowerP/Eratios.WecanconcludethatindustrytypeandP/Eratioarerelated.36.ExpectedFrequencies:DaysoftheWeekCountySunMonTuesWedThurFriSatTotalUrban56.747.655.156.760.172.644.2393Rural11.39.410.911.311.914.48.878Total68576668728753471χ2=6.202χ=12.5916(6degreesoffreedom).05DonotrejectH0;theassumptionofindependencecannotberejected.37.x=76.83s=12.43ObservedExpectedIntervalFrequencyFrequencylessthan62.545562.54-68.503568.50-72.856572.85-76.835576.83-80.815580.81-85.167585.16-91.124512-15www.khdaw.com 课后答案网www.khdaw.comChapter1291.12up55χ2=22χ=11.0705(5degreesoffreedom).05DonotrejectH0;theassumptionofanormaldistributioncannotberejected.38.ExpectedFrequencies:LosAngelesSanDiegoSanFranciscoSanJoseTotalOccupied165.7124.3186.4165.7642Vacant34.325.738.634.3133Total200.0150.0225.0200.0775www.khdaw.com2222(160-165.7)(116-124.3)(26-34.3)χ=++⋅⋅⋅+165.7124.334.3=7.782χ=7.81473with3degreesoffreedom.05Sinceχ2=7.78≤7.81473DonotrejectH0.Wecannotconcludethatofficevacanciesaredependentonmetropolitanarea,butitisclose:thep-valueisslightlylargerthan.05.39.a.ObservedBinomialProb.ExpectedxFrequenciesn=4,p=.30Frequencies030.240124.01132.411641.16225.264626.46310.07567.5643.0081.81100100.00Theexpectedfrequencyofx=4is.81.Combinex=3andx=4intoonecategorysothatallexpectedfrequenciesare5ormore.ObservedExpectedxFrequenciesFrequencies03024.0113241.1622526.463or4138.37100100.00b.With3degreesoffreedom,χ2=7.81473.RejectHifχ2>7.81473..05022(fi−ei)χ=Σ=6.17ei12-16www.khdaw.com 课后答案网www.khdaw.comTestsofGoodnessofFitandIndependenceDonotrejectH0;concludethattheassumptionofabinomialdistributioncannotberejected.www.khdaw.com12-17www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter13AnalysisAAnalysisnalysisofVarianceVVariancearianceandananddExperimentalEExperimentalxperimentalDesignDDesignesignLearningLLearningearningObjectivesOObjectivesbjectives1.Understandhowtheanalysisofvarianceprocedurecanbeusedtodetermineifthemeansofmorethantwopopulationsareequal.2.Knowtheassumptionsnecessarytousetheanalysisofvarianceprocedure.www.khdaw.com3.UnderstandtheuseoftheFdistributioninperformingtheanalysisofvarianceprocedure.4.KnowhowtosetupanANOVAtableandinterprettheentriesinthetable.5.Beabletouseoutputfromcomputersoftwarepackagestosolveanalysisofvarianceproblems.6.KnowhowtouseFisher’sleastsignificantdifference(LSD)procedureandFisher’sLSDwiththeBonferroniadjustmenttoconductstatisticalcomparisonsbetweenpairsofpopulationsmeans.7.Understandthedifferencebetweenacompletelyrandomizeddesign,arandomizedblockdesign,andfactorialexperiments.8.Knowthedefinitionofthefollowingterms:comparisonwiseTypeIerrorratepartitioningexperimentwiseTypeIerrorrateblockingfactormaineffectlevelinteractiontreatmentreplication13-1www.khdaw.com 课后答案网www.khdaw.comChapter13Solutions:SSolutions:olutions:1.a.x=(30+45+36)/3=37k2SSTR=∑n(x−x)=5(30-37)2+5(45-37)2+5(36-37)2=570jjj=1MSTR=SSTR/(k-1)=570/2=285k2b.SSE=∑(nj−1)sj=4(6)+4(4)+4(6.5)=66j=1MSE=SSE/(nT-k)=66/(15-3)=5.5www.khdaw.comc.F=MSTR/MSE=285/5.5=51.82F.05=3.89(2degreesoffreedomnumeratorand12denominator)SinceF=51.82>F.05=3.89,werejectthenullhypothesisthatthemeansofthethreepopulationsareequal.d.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments570228551.82Error66125.5Total636142.a.x=(153+169+158)/3=160k2SSTR=∑n(x−x)=4(153-160)2+4(169-160)2+4(158-160)2=536jjj=1MSTR=SSTR/(k-1)=536/2=268k2b.SSE=∑(nj−1)sj=3(96.67)+3(97.33)+3(82.00)=828.00j=1MSE=SSE/(nT-k)=828.00/(12-3)=92.00c.F=MSTR/MSE=268/92=2.91F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=2.91F.05=3.89werejectthenullhypothesisthatthemeansofthethreepopulationsareequal.d.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1020251013.36Error4581238.17Total1478144.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1200340080Error300605Total150063b.F.05=2.76(3degreesoffreedomnumeratorand60denominator)SinceF=80>F.05=2.76werejectthenullhypothesisthatthemeansofthe4populationsareequal.5.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments12026020Error216723Total33674b.F.05=3.15(2numeratordegreesoffreedomand60denominator)F.05=3.07(2numeratordegreesoffreedomand120denominator)Thecriticalvalueisbetween3.07and3.15SinceF=20mustexceedthecriticalvalue,nomatterwhatitsactualvalue,werejectthenullhypothesisthatthe3populationmeansareequal.13-www.khdaw.com 课后答案网www.khdaw.comChapter136.Manufacturer1Manufacturer2Manufacturer3SampleMean232821SampleVariance6.674.673.33x=(23+28+21)/3=24k2SSTR=∑n(x−x)=4(23-24)2+4(28-24)2+4(21-24)2=104jjj=1MSTR=SSTR/(k-1)=104/2=52k2SSE=∑(nj−1)sj=3(6.67)+3(4.67)+3(3.33)=44.01j=1www.khdaw.comMSE=SSE/(nT-k)=44.01/(12-3)=4.89F=MSTR/MSE=52/4.89=10.63F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=10.63>F.05=4.26werejectthenullhypothesisthatthemeantimeneededtomixabatchofmaterialisthesameforeachmanufacturer.7.SuperiorPeerSubordinateSampleMean5.755.55.25SampleVariance1.642.001.93x=(5.75+5.5+5.25)/3=5.5k2SSTR=∑n(x−x)=8(5.75-5.5)2+8(5.5-5.5)2+8(5.25-5.5)2=1jjj=1MSTR=SSTR/(k-1)=1/2=.5k2SSE=∑(nj−1)sj=7(1.64)+7(2.00)+7(1.93)=38.99j=1MSE=SSE/(nT-k)=38.99/21=1.86F=MSTR/MSE=0.5/1.86=0.27F.05=3.47(2degreesoffreedomnumeratorand21denominator)SinceF=0.27F.05=3.68,werejectthenullhypothesisthatthemeanperceptionscoreisthesameforthethreegroupsofspecialists.9.RealEstateAgentArchitectStockbrokerSampleMean67.7361.1365.80SampleVariance117.72180.10137.12x=(67.73+61.13+65.80)/3=64.89k2SSTR=∑n(x−x)=15(67.73-64.89)2+15(61.13-64.89)2+15(65.80-64.89)2=345.47jjj=1MSTR=SSTR/(k-1)=345.47/2=172.74k2SSE=∑(nj−1)sj=14(117.72)+14(180.10)+14(137.12)=6089.16j=1MSE=SSE/(nT-k)=6089.16/(45-3)=144.98F=MSTR/MSE=172.74/144.98=1.19F.05=3.22(2degreesoffreedomnumeratorand42denominator)Note:Table4doesnotshowavaluefor2degreesoffreedomnumeratorand42denominator.However,thevalueof3.23correspondingto2degreesoffreedomnumeratorand40denominatorcanbeusedasanapproximation.13-www.khdaw.com 课后答案网www.khdaw.comChapter13SinceF=1.19α=0.05,wecannotrejectthenullhypothesisthatthatthemeanprice/earningsratioisthesameforthesethreegroupsoffirms.⎛11⎞⎛11⎞11.aLSD=tMSE⎜+⎟=t5.5⎜+⎟=2.1792.2=3.23α/2⎜⎟.025⎝ninj⎠⎝55⎠x−x=3045−=15>LSD;significantdifference12x−x=3036−=6>LSD;significantdifference13x−x=4536−=9>LSD;significantdifference23⎛11⎞b.x−x±tMSE⎜+⎟12α/2nn⎝12⎠⎛11⎞(3045)2.1795.5−±⎜+⎟nn⎝12⎠-15±3.23=-18.23to-11.7712.a.Sample1Sample2Sample3SampleMean517758SampleVariance96.6797.3481.99x=(51+77+58)/3=62k2SSTR=∑n(x−x)=4(51-62)2+4(77-62)2+4(58-62)2=1,448jjj=113-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignMSTR=SSTR/(k-1)=1,448/2=724k2SSE=∑(nj−1)sj=3(96.67)+3(97.34)+3(81.99)=828j=1MSE=SSE/(nT-k)=828/(12-3)=92F=MSTR/MSE=724/92=7.87F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=7.87>F.05=4.26,werejectthenullhypothesisthatthemeansofthethreepopulationsareequal.⎛11⎞⎛11⎞b.LSD=tMSE⎜+⎟=t92⎜+⎟=2.26246=15.34α/2⎜⎟.025www.khdaw.com⎝ninj⎠⎝44⎠x−x=5177−=26>LSD;significantdifference12x−x=5158−=7LSD;significantdifference23⎛11⎞⎛11⎞13.LSD=tMSE⎜+⎟=t4.89⎜+⎟=2.2622.45=3.54α/2.025⎝n1n3⎠⎝44⎠Sincex−x=2321−=2<3.54,theredoesnotappeartobeanysignificantdifferencebetween13themeansofpopulation1andpopulation3.14.x−x±LSD1223-28±3.54-5±3.54=-8.54to-1.4615.Sincethereareonly3possiblepairwisecomparisonswewillusetheBonferroniadjustment.α=.05/3=.017t.017/2=t.0085whichisapproximatelyt.01=2.602⎛11⎞⎛11⎞BSD=2.602MSE⎜+⎟=2.602.5⎜+⎟=1.06⎜⎝ninj⎟⎠⎝66⎠x−x=54.5−=.5<1.06;nosignificantdifference12x−x=56−=<11.06;nosignificantdifference1313-www.khdaw.com 课后答案网www.khdaw.comChapter13x−x=4.56−=1.5>1.06;significantdifference2316.a.Machine1Machine2Machine3Machine4SampleMean7.19.19.911.4SampleVariance1.21.93.701.02x=(7.1+9.1+9.9+11.4)/4=9.38k2SSTR=∑n(x−x)=6(7.1-9.38)2+6(9.1-9.38)2+6(9.9-9.38)2+6(11.4-9.38)2=57.77jjj=1MSTR=SSTR/(k-1)=57.77/3=19.26k2www.khdaw.comSSE=∑(nj−1)sj=5(1.21)+5(.93)+5(.70)+5(1.02)=19.30j=1MSE=SSE/(nT-k)=19.30/(24-4)=.97F=MSTR/MSE=19.26/.97=19.86F.05=3.10(3degreesoffreedomnumeratorand20denominator)SinceF=19.86>F.05=3.10,werejectthenullhypothesisthatthemeantimebetweenbreakdownsisthesameforthefourmachines.b.Note:tα/2isbasedupon20degreesoffreedom⎛11⎞⎛11⎞LSD=tMSE⎜+⎟=t0.97⎜+⎟=2.086.3233=1.19α/2⎜⎟.025⎝ninj⎠⎝66⎠x−x=9.111.4−=2.3>LSD;significantdifference2417.C=6[(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]α=.05/6=.008andα/2=.004Sincethesmallestvalueforα/2inthettableis.005,wewilluset.005=2.845asanapproximationfort.004(20degreesoffreedom)⎛11⎞BSD=2.8450.97⎜+⎟=1.62⎝66⎠Thus,iftheabsolutevalueofthedifferencebetweenanytwosamplemeansexceeds1.62,thereissufficientevidencetorejectthehypothesisthatthecorrespondingpopulationmeansareequal.Means(1,2)(1,3)(1,4)(2,3)(2,4)(3,4)|Difference|22.84.30.82.31.5Significant?YesYesYesNoYesNo13-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesign18.n1=12n2=8n3=10tα/2isbasedupon27degreesoffreedomComparing1and2⎛11⎞LSD=t13⎜+⎟=2.0522.7083=3.38.025⎝128⎠9.9514.75−=4.8>LSD;significantdifferenceComparing1and3⎛11⎞LSD=2.05213⎜+⎟=2.0522.3833=3.17www.khdaw.com⎝1210⎠|9.95-13.5|=3.55>LSD;significantdifferenceComparing2and3⎛11⎞LSD=2.05213⎜+⎟=2.0522.9250=3.51⎝810⎠|14.75-13.5|=1.25F.05=3.68,werejectthehypothesisthatthemeansforthethreetreatmentsareequal.20.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments148827445.50www.khdaw.comError203015135.3Total351817⎛11⎞⎛11⎞b.LSD=tMSE⎜+⎟=2.131135.3⎜+⎟=14.31α/2⎜⎟⎝ninj⎠⎝66⎠|156-142|=14<14.31;nosignificantdifference|156-134|=22>14.31;significantdifference|142-134|=8<14.31;nosignificantdifference21.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments30047514.07Error160305.33Total4603422.a.H0:u1=u2=u3=u4=u5Ha:Notallthepopulationmeansareequalb.F.05=2.69(4degreesoffreedomnumeratorand30denominator)SinceF=14.07>2.69werejectH023.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1502754.80Error2501615.63Total40018F.05=3.63(2degreesoffreedomnumeratorand16denominator)SinceF=4.80>F.05=3.63,werejectthenullhypothesisthatthemeansofthethreetreatmentsareequal.13-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesign24.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments1200260043.99Error6004413.64Total180046F.05=3.23(2degreesoffreedomnumeratorand40denominator)F.05=3.15(2degreesoffreedomnumeratorand60denominator)ThecriticalFvalueisbetween3.15and3.23.SinceF=43.99exceedsthecriticalvalue,werejectthehypothesisthatthetreatmentmeansareequal.25.www.khdaw.comABCSampleMean119107100SampleVariance146.8996.43173.788(119)10(107)10(100)++x==107.9328k2SSTR=∑n(x−x)=8(119-107.93)2+10(107-107.93)2+10(100-107.93)2=1617.9jjj=1MSTR=SSTR/(k-1)=1617.9/2=809.95k2SSE=∑(nj−1)sj=7(146.86)+9(96.44)+9(173.78)=3,460j=1MSE=SSE/(nT-k)=3,460/(28-3)=138.4F=MSTR/MSE=809.95/138.4=5.85F.05=3.39(2degreesoffreedomnumeratorand25denominator)SinceF=5.85>F.05=3.39,werejectthenullhypothesisthatthemeansofthethreetreatmentsareequal.26.a.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments4560222809.87Error624027231.11Total1080029b.F.05=3.35(2degreesoffreedomnumeratorand27denominator)SinceF=9.87>F.05=3.35,werejectthenullhypothesisthatthemeansofthethreeassemblymethodsareequal.13-www.khdaw.com 课后答案网www.khdaw.comChapter1327.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFBetween61.64320.5517.56Error23.41201.17Total85.0523F.05=3.10(3degreesoffreedomnumeratorand20denominator)SinceF=17.56>F.05=3.10,werejectthenullhypothesisthatthemeanbreakingstrengthofthefourcablesisthesame.28.50°60°70°SampleMean332928SampleVariance3217.59.5www.khdaw.comx=(33+29+28)/3=30k2SSTR=∑n(x−x)=5(33-30)2+5(29-30)2+5(28-30)2=70jjj=1MSTR=SSTR/(k-1)=70/2=35k2SSE=∑(nj−1)sj=4(32)+4(17.5)+4(9.5)=236j=1MSE=SSE/(nT-k)=236/(15-3)=19.67F=MSTR/MSE=35/19.67=1.78F.05=3.89(2degreesoffreedomnumeratorand12denominator)SinceF=1.78F.05=3.55,werejectthenullhypothesisthatthemeansforthethreegroupsareequal.30.Paint1Paint2Paint3Paint4SampleMean13.3139136144www.khdaw.comSampleVariance47.5.502154.5x=(133+139+136+144)/3=138k2SSTR=∑n(x−x)=5(133-138)2+5(139-138)2+5(136-138)2+5(144-138)2=330jjj=1MSTR=SSTR/(k-1)=330/3=110k2SSE=∑(nj−1)sj=4(47.5)+4(50)+4(21)+4(54.5)=692j=1MSE=SSE/(nT-k)=692/(20-4)=43.25F=MSTR/MSE=110/43.25=2.54F.05=3.24(3degreesoffreedomnumeratorand16denominator)SinceF=2.54F.05=3.89,werejectthenullhypothesisthatthemeanmilespergallonratingsarethesameforthethreeautomobiles.32.Note:degreesoffreedomfortα/2are18⎛11⎞⎛11⎞LSD=tMSE⎜+⎟=t5.09⎜+⎟=2.1011.4543=2.53α/2⎜⎟.025www.khdaw.com⎝ninj⎠⎝77⎠x−x=17.020.4−=3.4>2.53;significantdifference12x−x=17.025.0−=>82.53;significantdifference13x−x=20.425−=4.6>2.53;significantdifference2333.Note:degreesoffreedomfortα/2are12⎛11⎞⎛11⎞LSD=tMSE⎜+⎟=t2⎜+⎟=2.179.8=1.95α/2⎜⎟.025⎝ninj⎠⎝55⎠x−x=2021−=<11.95;nosignificantdifference12x−x=2025−=>51.95;significantdifference13x−x=2125−=4>1.95;significantdifference2334.TreatmentMeans:x=13.6x=11.0x=10.6i1i2i3BlockMeans:x=9x=7.67x=15.67x=18.67x=7.671i2i3i4i5iOverallMean:x=176/15=11.7313-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignStep12SST=∑∑(x−x)=(10-11.73)2+(9-11.73)2+···+(8-11.73)2=354.93ijijStep22SSTR=b∑(xi−x)=5[(13.6-11.73)2+(11.0-11.73)2+(10.6-11.73)2]=26.53jjStep32SSBL=k∑(xi−x)=3[(9-11.73)2+(7.67-11.73)2+(15.67-11.73)2+iiwww.khdaw.com(18.67-11.73)2+(7.67-11.73)2]=312.32Step4SSE=SST-SSTR-SSBL=354.93-26.53-312.32=16.08SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments26.53213.276.60Blocks312.32478.08Error16.0882.01Total354.9314F.05=4.46(2degreesoffreedomnumeratorand8denominator)SinceF=6.60>F.05=4.46,werejectthenullhypothesisthatthemeansofthethreetreatmentsareequal.35.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments310477.517.69Blocks85242.5Error3584.38Total43014F.05=3.84(4degreesoffreedomnumeratorand8denominator)SinceF=17.69>F.05=3.84,werejectthenullhypothesisthatthemeansofthetreatmentsareequal.36.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments900330012.60Blocks400757.14Error5002123.81Total180031F.05=3.07(3degreesoffreedomnumeratorand21denominator)SinceF=12.60>F.05=3.07,werejectthenullhypothesisthatthemeansofthetreatmentsareequal.13-www.khdaw.com 课后答案网www.khdaw.comChapter1337.TreatmentMeans:x=56x=44i1i2BlockMeans:x=46x=49.5x=54.51i2i3iOverallMean:x=300/6=50Step12www.khdaw.comSST=∑∑(x−x)=(50-50)2+(42-50)2+···+(46-50)2=310ijijStep22SSTR=b∑(xi−x)=3[(56-50)2+(44-50)2]=216jjStep32SSBL=k∑(xi−x)=2[(46-50)2+(49.5-50)2+(54.5-50)2]=73iiStep4SSE=SST-SSTR-SSBL=310-216-73=21SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments216121620.57Blocks73236.5Error21210.5Total3105F.05=18.51(1degreeoffreedomnumeratorand2denominator)SinceF=20.57>F.05=18.51,werejectthenullhypothesisthatthemeantuneuptimesarethesameforbothanalyzers.38.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments45411.257.12Blocks36312Error19121.58Total10019F.05=3.26(4degreesoffreedomnumeratorand12denominator)13-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignSinceF=7.12>F.05=3.26,werejectthenullhypothesisthatthemeantotalaudittimesforthefiveauditingproceduresareequal.39.TreatmentMeans:x=16x=15x=21i1i2i3BlockMeans:x=18.67x=19.33x=15.33x=14.33x=191i2i3i4i5iOverallMean:x=260/15=17.33www.khdaw.comStep12SST=∑∑(x−x)=(16-17.33)2+(16-17.33)2+···+(22-17.33)2=175.33ijijStep22SSTR=b∑(xi−x)=5[(16-17.33)2+(15-17.33)2+(21-17.33)2]=103.33jjStep32SSBL=k∑(xi−x)=3[(18.67-17.33)2+(19.33-17.33)2+···+(19-17.33)2]=64.75iiStep4SSE=SST-SSTR-SSBL=175.33-103.33-64.75=7.25SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatments100.33251.6756.78Blocks64.75416.19Error7.258.91Total175.3314F.05=4.46(2degreesoffreedomnumeratorand8denominator)SinceF=56.78>F.05=4.46,werejectthenullhypothesisthatthemeantimesforthethreesystemsareequal.40.TheMinitaboutputforthesedataisshownbelow:13-www.khdaw.com 课后答案网www.khdaw.comChapter13ANALYSISOFVARIANCEBPMSOURCEDFSSMSBlock92796311Treat3198056602ERROR277949294TOTAL3930550Individual95%CITreatMean----+---------+---------+---------+-------1178.0(-----*-----)2171.0(-----*----)3175.0(-----*----)4123.6(-----*----)----+---------+---------+---------+-------www.khdaw.com120.0140.0160.0180.0F.05=2.96(3degreesoffreedomnumeratorand27denominator)SinceF=6602/294=22.46>2.96,werejectthenullhypothesesthatthemeanheartrateforthefourmethodsareequal.41.FactorBFactorALevel1Level2Level3MeansLevel1=150=78=84=104FactorALevel2=110=116=128=118FactorBMeans=130=97=106=111Step12SST=∑∑∑(x−x)=(135-111)2+(165-111)2+···+(136-111)2=9,028ijkijkStep22SSA=br∑(xi−x)=3(2)[(104-111)2+(118-111)2]=588jiStep313-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesign2SSB=ar∑(xi−x)=2(2)[(130-111)2+(97-111)2+(106-111)2]=2,328jjStep42SSAB=r∑∑(x−xi−xi+x)=2[(150-104-130+111)2+(78-104-97+111)2+ijijij···+(128-118-106+111)2]=4,392Step5SSE=SST-SSA-SSB-SSAB=9,028-588-2,328-4,392=1,720SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFFactorA58815882.05www.khdaw.comFactorB2328211644.06Interaction4392221967.66Error17206286.67Total902811F.05=5.99(1degreeoffreedomnumeratorand6denominator)F.05=5.14(2degreesoffreedomnumeratorand6denominator)SinceF=2.05F.05=5.14,Interactionissignificant.42.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFFactorA2638.673.72FactorB23211.504.94Interaction175629.1712.52Error56242.33Total28035F.05=3.01(3degreesoffreedomnumeratorand24denominator)SinceF=3.72>F.05=3.01,FactorAissignificant.F.05=3.40(2degreesoffreedomnumeratorand24denominator)SinceF=4.94>F.05=3.40,FactorBissignificant.F.05=2.51(6degreesoffreedomnumeratorand24denominator)SinceF=12.52>F.05=2.51,Interactionissignificant43.13-www.khdaw.com 课后答案网www.khdaw.comChapter13FactorBFactorBSmallLargeMeansA=10=10=10FactorAB=18=28=23C=14=16=15FactorBMeans=14=18=16www.khdaw.comStep12SST=∑∑∑(x−x)=(8-16)2+(12-16)2+(12-16)2+···+(14-16)2=544ijkijkStep22SSA=br∑(xi−x)=2(2)[(10-16)2+(23-16)2+(15-16)2]=344iiStep32SSB=ar∑(xi−x)=3(2)[(14-16)2+(18-16)2]=48jjStep42SSAB=r∑∑(x−xi−xi+x)=2[(10-10-14+16)2+···+(16-15-18+16)2]=56ijijijStep5SSE=SST-SSA-SSB-SSAB=544-344-48-56=96SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFFactorA3442172172/16=10.75FactorB4814848/16=3.00Interaction5622828/16=1.75Error96616Total54411F.05=5.14(2degreesoffreedomnumeratorand6denominator)SinceF=10.75>F.05=5.14,FactorAissignificant,thereisadifferenceduetothetypeofadvertisementdesign.13-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignF.05=5.99(1degreeoffreedomnumeratorand6denominator)SinceF=3F.05=4.26,FactorA(gender)isnotsignificant.SinceF=13.25>F.05=3.40,FactorB(occupation)issignificant.SinceF=5.53>F.05=3.40,Interactionissignificant.46.x=(1.13+1.56+2.00)/3=1.5631ix=(0.48+1.68+2.86)/3=1.6732ix=(1.13+0.48)/2=0.805i1x=(1.56+1.68)/2=1.620i2x=(2.00+2.86)/2=2.43i3x=(1.13+1.56+2.00+0.48+1.68+2.86)/6=1.618Step1SST=327.50(giveninproblemstatement)Step22SSA=br∑(xi−x)=3(25)[(1.563-1.618)2+(1.673-1.618)2]=0.4538iiStep32SSB=ar∑(xi−x)=2(25)[(0.805-1.618)2+(1.62-1.618)2+(2.43-1.618)2]=66.0159jj13-www.khdaw.com 课后答案网www.khdaw.comChapter13Step42SSAB=r∑∑(x−xi−xi+x)=25[(1.13-1.563-0.805+1.618)2+(1.56-1.563-1.62ijijij+1.618)2+···+(2.86-1.673-2.43+1.618)2]=14.2525Step5SSE=SST-SSA-SSB-SSAB=327.50-0.4538-66.0159-14.2525SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFFactorA0.453810.45380.2648FactorB66.1059233.008019.2608Interaction14.252527.12634.1583Error246.77781441.7137www.khdaw.comTotal327.5000149F.05for1degreeoffreedomnumeratorand144degreesoffreedomdenominatorisbetween3.92and3.84.F.05for2degreesoffreedomnumeratorand144denominatorisbetween3.07and3.00.Since0.2648F.05=3.07,FactorBissignificantSince4.1583>F.05=3.07,Interactionissignificant47.a.Area1Area2SampleMean9694SampleVariance504022s+s5040+12pooledestimate===4522⎛11⎞estimateofstandarddeviationofx−x=45⎜+⎟=4.7412⎝44⎠x−x9694−12t===.424.744.74t.025=2.447(6degreesoffreedom)Sincet=.42F.05=4.26werejectthenullhypothesisthatthemeanaskingpricesforallthreeareasareequal.48.TheMinitaboutputforthesedataisshownbelow:AnalysisofVarianceSourceDFSSMSFPFactor2753.3376.618.590.000Error27546.920.3Total291300.2Individual95%CIsForMeanBasedonPooledStDevLevelNMeanStDev---------+---------+---------+-------SUV1058.6004.575(-----*-----)Small1048.8004.211(-----*----)FullSize1060.1004.701(-----*-----)---------+---------+---------+-------PooledStDev=4.50150.055.060.013-www.khdaw.com 课后答案网www.khdaw.comChapter13Becausethep-value=.000<α=.05,wecanrejectthenullhypothesisthatthemeanresalevalueisthesame.Itappearsthatthemeanresalevalueforsmallpickuptrucksismuchsmallerthanthemeanresalevalueforsportutilityvehiclesorfull-sizepickuptrucks.49.FoodPersonalCareRetailSampleMean52.2562.2555.75SampleVariance22.2515.584.92x=(52.25+62.25+55.75)/3=56.75k2SSTR=∑n(x−x)=4(52.25-56.75)2+4(62.25-56.75)2+4(55.75-56.75)2=206jjj=1www.khdaw.comMSTR=SSTR/(k-1)=206/2=103k2SSE=∑(nj−1)sj=3(22.25)+3(15.58)+3(4.92)=128.25j=1MSE=SSE/(nT-k)=128.25/(12-3)=14.25F=MSTR/MSE=103/14.25=7.23F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=7.23exceedsthecriticalFvalue,werejectthenullhypothesisthatthemeanageofexecutivesisthesameinthethreecategoriesofcompanies.50.PhysicalCabinetSystemsLawyerTherapistMakerAnalystSampleMean50.063.769.161.2SampleVariance124.22164.68105.88136.6250.063.769.161.2+++x==614k2SSTR=∑n(x−x)=10(50.0-61)2+10(63.7-61)2+10(69.1-61)2+10(61.2-61)2=1939.4jjj=1MSTR=SSTR/(k-1)=1939.4/3=646.47k2SSE=∑(nj−1)sj=9(124.22)+9(164.68)+9(105.88)+9(136.62)=4,782.60j=1MSE=SSE/(nT-k)=4782.6/(40-4)=132.85F=MSTR/MSE=646.47/132.85=4.87F.05=2.84(3degreesofnumeratorand40denominator)13-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignF.05=2.76(3degreesoffreedomnumeratorand60denominator)Thus,thecriticalFvalueisbetween2.76and2.84.SinceF=4.87exceedsthecriticalFvalue,werejectthenullhypothesisthatthemeanjobsatisfactionratingisthesameforthefourprofessions.51.TheMinitaboutputforthesedataisshownbelow:AnalysisofVarianceSourceDFSSMSFPFactor2433921693.660.039Error2715991592Total2920330Individual95%CIsForMeanBasedonPooledStDevwww.khdaw.comLevelNMeanStDev---+---------+---------+---------+---West10108.0023.78(-------*-------)South1091.7019.62(-------*-------)NE10121.1028.75(-------*------)---+---------+---------+---------+---PooledStDev=24.3480100120140Becausethep-value=.039<α=.05,wecanrejectthenullhypothesisthatthemeanrateforthethreeregionsisthesame.52.TheMintaboutputisshownbelow:ANALYSISOFVARIANCESOURCEDFSSMSFpFACTOR31271.0423.78.740.000ERROR361744.248.4TOTAL393015.2INDIVIDUAL95PCTCI"SFORMEANBASEDONPOOLEDSTDEVLEVELNMEANSTDEV--+---------+---------+---------+----West1060.0007.218(------*-----)South1045.4007.610(------*-----)N.Cent1047.3006.778(------*-----)N.East1052.1006.152(-----*------)--+---------+---------+---------+----POOLEDSTDEV=6.96142.049.056.063.0Sincethep-value=0.000<α=0.05,wecanrejectthenullhypothesisthatthatthemeanbasesalaryforartdirectorsisthesameforeachofthefourregions.13-www.khdaw.com 课后答案网www.khdaw.comChapter1353.TheMinitaboutputforthesedataisshownbelow:AnalysisofVarianceSourceDFSSMSFPFactor212.4026.2019.330.001Error3724.5960.665Total3936.998Individual95%CIsForMeanBasedonPooledStDevLevelNMeanStDev------+---------+---------+---------+Receiver157.41330.8855(-------*------)Guard136.10770.7399(-------*------)Tackle127.05830.8005(-------*-------)------+---------+---------+---------+www.khdaw.comPooledStDev=0.81536.006.607.207.80Becausethep-value=.001<α=.05,wecanrejectthenullhypothesisthatthemeanratingforthethreepositionsisthesame.Itappearsthatwidereceiversandtackleshaveahighermeanratingthanguards.54.XYZSampleMean929784SampleVariance30635.33x=(92+97+44)/3=91k2SSTR=∑n(x−x)=4(92-91)2+4(97-91)2+4(84-91)2=344jjj=1MSTR=SSTR/(k-1)=344/2=172k2SSE=∑(nj−1)sj=3(30)+3(6)+3(35.33)=213.99j=1MSE=SSE/(nT-k)=213.99/(12-3)=23.78F=MSTR/MSE=172/23.78=7.23F.05=4.26(2degreesoffreedomnumeratorand9denominator)SinceF=7.23>F.05=4.26,werejectthenullhypothesisthatthemeanabsorbencyratingsforthethreebrandsareequal.55.FirstYearSecondYearThirdYearFourthYearSampleMean1.03-0.9915.249.81SampleVariance416.93343.04159.3155.43x=(1.03-.99+15.24+9.81)/4=6.2713-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignk2SSTR=∑n(x−x)=7(1.03-6.27)2+7(-.99-6.27)2+7(15.24-6.27)2+(9.81-6.27)2jjj=1=1,212.10MSTR=SSTR/(k-1)=1,212.10/3=404.03k2SSE=∑(nj−1)sj=6(416.93)+6(343.04)+6(159.31)+6(55.43)=5,848.26j=1MSE=SSE/(nT-k)=5,848.26/(28-4)=243.68F=MSTR/MSE=404.03/243.68=1.66www.khdaw.comF.05=3.01(3degreesoffreedomnumeratorand24denominator)SinceF=1.66F.05=4.26,werejectthenullhypothesisthatthemeanlifetimeinhoursisthesameforthethreedesigns.59.a.NonbrowserLightBrowserHeavyBrowserSampleMean4.255.255.75SampleVariance1.071.071.36x=(4.25+5.25+5.75)/3=5.0813-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignk2SSTR=∑n(x−x)=8(4.25-5.08)2+8(5.25-5.08)2+8(5.75-5.08)2=9.33jjj=1MSB=SSB/(k-1)=9.33/2=4.67k2SSW=∑(nj−1)sj=7(1.07)+7(1.07)+7(1.36)=24.5j=1MSW=SSW/(nT-k)=24.5/(24-3)=1.17F=MSB/MSW=4.67/1.17=3.99F.05=3.47(2degreesoffreedomnumeratorand21denominator)SinceF=3.99>F.05=3.47,werejectthenullhypothesisthatthemeancomfortscoresarethesamewww.khdaw.comforthethreegroups.⎛11⎞⎛11⎞b.LSD=tMSW⎜+⎟=2.0801.17⎜+⎟=1.12α/2⎜⎟⎝ninj⎠⎝88⎠Sincetheabsolutevalueofthedifferencebetweenthesamplemeansfornonbrowsersandlightbrowsersis4.255.25−=1,wecannotrejectthenullhypothesisthatthetwopopulationmeansareequal.60.TreatmentMeans:x=22.8x=24.8x=25.80⋅1⋅2⋅3BlockMeans:x=19.67x=25.67x=31x=23.67x=22.331⋅2⋅3⋅4⋅5⋅OverallMean:x=367/15=24.47Step12SST=∑∑(x−x)=(18-24.47)2+(21-24.47)2+···+(24-24.47)2=253.73ijijStep22SSTR=b∑(xi−x)=5[(22.8-24.47)2+(24.8-24.47)2+(25.8-24.47)2]=23.33jj13-www.khdaw.com 课后答案网www.khdaw.comChapter13Step32SSBL=k∑(xi−x)=3[(19.67-24.47)2+(25.67-24.47)2+···+(22.33-24.47)2]=217.02iiStep4SSE=SST-SSTR-SSBL=253.73-23.33-217.02=13.38SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFTreatment23.33211.676.99Blocks217.02454.2632.49Error13.3881.67Total253.7314www.khdaw.comF.05=4.46(2degreesoffreedomnumeratorand8denominator)SinceF=6.99>F.05=4.46werejectthenullhypothesisthatthemeanmilespergallonratingsforthethreebrandsofgasolineareequal.61.IIIIIISampleMean22.824.825.8SampleVariance21.29.227.2x=(22.8+24.8+25.8)/3=24.47k2SSTR=∑n(x−x)=5(22.8-24.47)2+5(24.8-24.47)2+5(25.8-24.47)2=23.33jjj=1MSTR=SSTR/(k-1)=23.33/2=11.67k2SSE=∑(nj−1)sj=4(21.2)+4(9.2)+4(27.2)=230.4j=1MSE=SSE/(nT-k)=230.4/(15-3)=19.2F=MSTR/MSE=11.67/19.2=.61F.05=3.89(2degreesoffreedomnumeratorand12denominator)SinceF=.61F.05=5.14,FactorBissignificant;thatis,thereisasignificantdifferenceduetothelanguagetranslated.TypeofsystemandinteractionarenotsignificantsincebothFvaluesarelessthanthecriticalvalue.64.FactorBFactorBManualAutomaticMeansMachine1=32=28=36FactorAMachine2=21=26=23.5FactorBMeans=26.5=27=26.7513-www.khdaw.com 课后答案网www.khdaw.comAnalysisofVarianceandExperimentalDesignStep12SST=∑∑∑(x−x)=(30-26.75)2+(34-26.75)2+···+(28-26.75)2=151.5ijkijkStep22SSA=br∑(xi−x)=2(2)[(30-26.75)2+(23.5-26.75)2]=84.5iiStep32SSB=ar∑(xi−x)=2(2)[(26.5-26.75)2+(27-26.75)2]=0.5jwww.khdaw.comjStep42SSAB=r∑∑(x−xi−xi+x)=2[(30-30-26.5+26.75)2+···+(28-23.5-27+26.75)2]ijijij=40.5Step5SSE=SST-SSA-SSB-SSAB=151.5-84.5-0.5-40.5=26SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFFactorA84.5184.513FactorB.51.5.08Interaction40.5140.56.23Error2646.5Total151.57F.05=7.71(1degreeoffreedomnumeratorand4denominator)SinceF=13>F.05=7.71,FactorA(TypeofMachine)issignificant.TypeofLoadingSystemandInteractionarenotsignificantsincebothFvaluesarelessthanthecriticalvalue.13-www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter14SimpleSSimpleimpleLinearLLinearinearRegressionRRegressionegressionLearningLLearningearningObjectivesOObjectivesbjectives1.Understandhowregressionanalysiscanbeusedtodevelopanequationthatestimatesmathematicallyhowtwovariablesarerelated.2.Understandthedifferencesbetweentheregressionmodel,theregressionequation,andtheestimatedregressionequation.www.khdaw.com3.Knowhowtofitanestimatedregressionequationtoasetofsampledatabasedupontheleast-squaresmethod.4.Beabletodeterminehowgoodafitisprovidedbytheestimatedregressionequationandcomputethesamplecorrelationcoefficientfromtheregressionanalysisoutput.5.Understandtheassumptionsnecessaryforstatisticalinferenceandbeabletotestforasignificantrelationship.6.Learnhowtousearesidualplottomakeajudgementastothevalidityoftheregressionassumptions,recognizeoutliers,andidentifyinfluentialobservations.7.Knowhowtodevelopconfidenceintervalestimatesofygivenaspecificvalueofxinboththecaseofameanvalueofyandanindividualvalueofy.8.Beabletocomputethesamplecorrelationcoefficientfromtheregressionanalysisoutput.9.Knowthedefinitionofthefollowingterms:independentanddependentvariablesimplelinearregressionregressionmodelregressionequationandestimatedregressionequationscatterdiagramcoefficientofdeterminationstandarderroroftheestimateconfidenceintervalpredictionintervalresidualplotstandardizedresidualplotoutlierinfluentialobservationleverage14-1www.khdaw.com 课后答案网www.khdaw.comChapter14Solutions:SSolutions:olutions:1a.161412108y6420www.khdaw.com0123456xb.Thereappearstobealinearrelationshipbetweenxandy.c.Manydifferentstraightlinescanbedrawntoprovidealinearapproximationoftherelationshipbetweenxandy;inpartdwewilldeterminetheequationofastraightlinethat“best”representstherelationshipaccordingtotheleastsquarescriterion.d.Summationsneededtocomputetheslopeandy-interceptare:2Σx=15Σy=40(Σx−xy)(−y)=26(Σx−x)=10iiiiiΣ(x−xy)(−y)26iib===2.612Σ(x−x)10ib=y−bx=8−(2.6)(3)=0.201ŷ=0.2−2.6xe.ŷ=0.2−2.6(4)=10.614-2www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression2.a.35302520y1510500123456789www.khdaw.comxb.Thereappearstobealinearrelationshipbetweenxandy.c.Manydifferentstraightlinescanbedrawntoprovidealinearapproximationoftherelationshipbetweenxandy;inpartdwewilldeterminetheequationofastraightlinethat“best”representstherelationshipaccordingtotheleastsquarescriterion.d.Summationsneededtocomputetheslopeandy-interceptare:2Σx=19Σy=116(Σx−xy)(−y)=−57.8(Σx−x)=30.8iiiiiΣ(x−xy)(−y)−57.8iib===−1.876612Σ(x−x)30.8ib=y−bx=23.2−(−1.8766)(3.8)=30.331101ŷ=30.33−1.88xe.ŷ=30.33−1.88(6)=19.0514-3www.khdaw.com 课后答案网www.khdaw.comChapter143.a.7654y32100123456789www.khdaw.comxb.Summationsneededtocomputetheslopeandy-interceptare:2Σx=26Σy=17(Σx−xy)(−y)11.6(=Σx−x)=22.8iiiiiΣ(x−xy)(−y)11.6iib===0.508812Σ(x−x)22.8ib=y−bx=3.4−(0.5088)(5.2)=0.754201ŷ=0.75+0.51xc.ŷ=0.75+0.51(4)=2.7914-4www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression4.a.135130125120y115110105100616263646566676869www.khdaw.comxb.Thereappearstobealinearrelationshipbetweenxandy.c.Manydifferentstraightlinescanbedrawntoprovidealinearapproximationoftherelationshipbetweenxandy;inpartdwewilldeterminetheequationofastraightlinethat“best”representstherelationshipaccordingtotheleastsquarescriterion.d.Summationsneededtocomputetheslopeandy-interceptare:2Σx=325Σy=585(Σx−xy)(−y)110(=Σx−x)=20iiiiiΣ(x−xy)(−y)110iib===5.512Σ(x−x)20ib=y−bx=117−(5.5)(65)=−240.501ŷ=−240.5+5.5xe.ŷ=−240.5+5.5x=−240.5+5.5(63)=106pounds14-5www.khdaw.com 课后答案网www.khdaw.comChapter145.a.210019001700150013001100y900700500300www.khdaw.com100020406080100120140xb.Thereappearstobealinearrelationshipbetweenxandy.c.Manydifferentstraightlinescanbedrawntoprovidealinearapproximationoftherelationshipbetweenxandy;inpartdwewilldeterminetheequationofastraightlinethat“best”representstherelationshipaccordingtotheleastsquarescriterion.Summationsneededtocomputetheslopeandy-interceptare:2Σx=420.6Σy=5958.7(Σx−xy)(−y)142,040.3443=(Σx−x)=9847.6486iiiiiΣ(x−xy)(−y)142,040.3443iib===14.423812Σ(x−x)9847.6486ib=y−bx=851.2429−(14.4238)(60.0857)=−15.4201ŷ=−15.42+14.42xd.Aonemilliondollarincreaseinmediaexpenditureswillincreasecasesalesbyapproximately14.42million.e.ŷ=−15.42+14.42x=−15.42+14.42(70)=993.9814-6www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression6.a.1.41.210.8y0.60.40.2www.khdaw.com066687072747678808284xb.Thereappearstobealinearrelationshipbetweenxandy.c.Summationsneededtocomputetheslopeandy-interceptare:2Σx=667.2Σy=7.18(Σx−xy)(−y)=−9.0623(Σx−x)=128.7iiiiiΣ(x−xy)(−y)−9.0623iib===−0.070412Σ(x−x)128.7ib=y−bx=0.7978−(−0.0704)(74.1333)=6.0201ŷ=6.02−0.07xd.Aonepercentincreaseinthepercentageofflightsarrivingontimewilldecreasethenumberofcomplaintsper100,000passengersby0.07.eŷ=6.02−0.07x=6.02−0.07(80)=0.4214-7www.khdaw.com 课后答案网www.khdaw.comChapter14155015001450S&P14001350www.khdaw.com13009600980010000102001040010600108001100011200DJIA7.a.b.Letx=DJIAandy=S&P.Summationsneededtocomputetheslopeandy-interceptare:2Σx=104,850Σy=14,233(Σx−xy)(−y)=268,921(Σx−x)=1,806,384iiiiiΣ(x−xy)(−y)268,921iib===0.1488712Σ(x−x)1,806,384ib=ybx−=1423.3(.14887)(10,485)−=−137.62901yˆ=−137.630.1489+xc.yˆ=−137.630.1489(11,000)1500.27+=orapproximately15008.a.Summationsneededtocomputetheslopeandy-interceptare:2Σx=121Σy=1120.9(Σx−xy)(−y)=544.0429(Σx−x)=177.4286iiiiiΣ(x−xy)(−y)544.0429iib===3.066312Σ(x−x)177.4286ib=y−bx=160.1286−(3.0663)(17.2857)=107.1301ŷ=107.13+3.07xb.Increasingthenumberoftimesanadisairedbyonewillincreasethenumberofhouseholdexposuresbyapproximately3.07million.c.ŷ=107.13+3.07x=107.13+3.07(15)=153.214-8www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression9.a.150140130120110100y90www.khdaw.com8070605002468101214xb.Summationsneededtocomputetheslopeandy-interceptare:2Σx=70Σy=1080(Σx−xy)(−y)=568(Σx−x)=142iiiiiΣ(x−xy)(−y)568iib===412Σ(x−x)142ib=y−bx=108−(4)(7)=8001ŷ=80+4xc.ŷ=80+4x=80+4(9)=11614-9www.khdaw.com 课后答案网www.khdaw.comChapter149590858075OverallRating706560www.khdaw.com100120140160180200220240260PerformanceScore10.a.b.Letx=performancescoreandy=overallrating.Summationsneededtocomputetheslopeandy-interceptare:2Σx=2752Σy=1177(Σx−xy)(−y)1723.73(=Σx−x)=11,867.73iiiiiΣ(x−xy)(−y)1723.73iib===0.145212Σ(x−x)11,867.73ib=ybx−=78.4667(.1452)(183.4667)−=51.8201yˆ=51.820.145+xc.yˆ=51.820.145(225)+=84.4orapproximately8414-10www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression11.a.900.0800.0700.0600.0500.0y400.0300.0200.0www.khdaw.com100.00.00.0100.0200.0300.0400.0500.0600.0700.0800.0xb.Thereappearstobealinearrelationshipbetweenthevariables.c.Thesummationsneededtocomputetheslopeandthey-interceptare:2Σx=2973.3Σy=3925.6(Σx−xy)(−y)=453,345.042(Σx−x)=483,507.581iiiiiΣ(x−xy)(−y)453,345.042iib===0.938512Σ(x−x)483,507.581ib=y−bx=392.56−(0.9385)(297.33)=113.5201ŷ=113.52+0.94xd.ŷ=113.52+0.94x=113.52+0.94(500)=583.514-11www.khdaw.com 课后答案网www.khdaw.comChapter1412.a.4000035000300002500020000Revenue1500010000www.khdaw.com50000020000400006000080000100000NumberofEmployeesb.Thereappearstobeapositivelinearrelationshipbetweenthenumberofemployeesandtherevenue.c.Letx=numberofemployeesandy=revenue.Summationsneededtocomputetheslopeandy-interceptare:2Σx=4200Σy=1669(Σx−xy)(−y)=4,658,594,168(Σx−x)=14,718,343,803iiiiiΣ(x−xy)(−y)4,658,594,168iib===0.31651612Σ(x−x)14,718,343,803ib=ybx−=14,048(.316516)(40,299)1293−=01yˆ=12930.3165+xd.yˆ=1293.3165(75,000)+=25,03114-12www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression13.a.30.025.020.015.0y10.0www.khdaw.com5.00.00.020.040.060.080.0100.0120.0140.0xb.Thesummationsneededtocomputetheslopeandthey-interceptare:2Σx=399Σy=97.1(Σx−xy)(−y)1233.7(=Σx−x)=7648iiiiiΣ(x−xy)(−y)1233.7iib===0.1613112Σ(x−x)7648ib=y−bx=13.87143−(0.16131)(57)=4.6767501ŷ=4.68+0.16xc.ŷ=4.68+0.16x=4.68+0.16(52.5)=13.08orapproximately$13,080.Theagent"srequestforanauditappearstobejustified.14-13www.khdaw.com 课后答案网www.khdaw.comChapter1414.a.858075y70www.khdaw.com6560607080x90100110b.Thesummationsneededtocomputetheslopeandthey-interceptare:2Σx=1677.25Σy=1404.3(Σx−xy)(−y)=897.9493Σ(x−x)=3657.4568iiiiiΣ(x−xy)(−y)897.9493iib===0.245512Σ(x−x)3657.4568ib=y−bx=70.215−(0.2455)(83.8625)=49.6301ŷ=49.63+.2455xc.ŷ=49.63+.2455x=49.63+.2455(80)=69.3%15.a.Theestimatedregressionequationandthemeanforthedependentvariableare:ŷ=0.2+2.6xy=8iiThesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−ŷ)=12.40SST=∑(y−y)=80iiiThus,SSR=SST-SSE=80-12.4=67.6b.r2=SSR/SST=67.6/80=.845Theleastsquareslineprovidedaverygoodfit;84.5%ofthevariabilityinyhasbeenexplainedbytheleastsquaresline.c.r=.845=+.919214-14www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression16.a.Theestimatedregressionequationandthemeanforthedependentvariableare:yˆ=30.331.88−xy=23.2iThesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−yˆ)=6.33SST=∑(y−y)=114.80iiiThus,SSR=SST-SSE=114.80-6.33=108.47b.r2=SSR/SST=108.47/114.80=.945Theleastsquareslineprovidedanexcellentfit;94.5%ofthevariabilityinyhasbeenexplainedbytheestimatedregressionequation.www.khdaw.comc.r=.945=−.9721Note:thesignforrisnegativebecausetheslopeoftheestimatedregressionequationisnegative.(b1=-1.88)17.Theestimatedregressionequationandthemeanforthedependentvariableare:yˆ=.75.51+xy=3.4iThesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−yˆ)=5.3SST=∑(y−y)=11.2iiiThus,SSR=SST-SSE=11.2-5.3=5.9r2=SSR/SST=5.9/11.2=.527Weseethat52.7%ofthevariabilityinyhasbeenexplainedbytheleastsquaresline.r=.527=+.725918.a.Theestimatedregressionequationandthemeanforthedependentvariableare:yˆ=1790.5581.1+xy=3650Thesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−yˆ)=85,135.14SST=∑(y−y)=335,000iiiThus,SSR=SST-SSE=335,000-85,135.14=249,864.86b.r2=SSR/SST=249,864.86/335,000=.746Weseethat74.6%ofthevariabilityinyhasbeenexplainedbytheleastsquaresline.c.r=.746=+.863714-15www.khdaw.com 课后答案网www.khdaw.comChapter1419.a.Theestimatedregressionequationandthemeanforthedependentvariableare:yˆ=−137.63.1489+xy=1423.3Thesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−yˆ)=7547.14SST=∑(y−y)=47,582.10iiiThus,SSR=SST-SSE=47,582.10-7547.14=40,034.96b.r2=SSR/SST=40,034.96/47,582.10=.84Weseethat84%ofthevariabilityinyhasbeenexplainedbytheleastsquaresline.c.r=.84=+.92www.khdaw.com20.a.Letx=incomeandy=homeprice.Summationsneededtocomputetheslopeandy-interceptare:2Σx=1424Σy=2455.5(Σx−xy)(−y)=4011(Σx−x)=1719.618iiiiiΣ(x−xy)(−y)4011iib===2.332512Σ(x−x)1719.618ib=ybx−=136.4167(2.3325)(79.1111)−=−48.1101yˆ=−48.112.3325+xb.Thesumofsquaresduetoerrorandthetotalsumofsquaresare22SSE=∑(y−yˆ)=2017.37SST=∑(y−y)=11,373.09iiiThus,SSR=SST-SSE=11,373.09–2017.37=9355.72r2=SSR/SST=9355.72/11,373.09=.82Weseethat82%ofthevariabilityinyhasbeenexplainedbytheleastsquaresline.r=.82=+.91c.yˆ=−48.112.3325(95)173.5+=orapproximately$173,50021.a.Thesummationsneededinthisproblemare:2Σx=3450Σy=33,700(Σx−xy)(−y)=712,500(Σx−x)=93,750iiiiiΣ(x−xy)(−y)712,500iib===7.612Σ(x−x)93,750ib=ybx−=5616.67(7.6)(575)1246.67−=01ŷ=1246.67+7.6x14-16www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionb.$7.60c.Thesumofsquaresduetoerrorandthetotalsumofsquaresare:22SSE=∑(y−yˆ)=233,333.33SST=∑(y−y)=5,648,333.33iiiThus,SSR=SST-SSE=5,648,333.33-233,333.33=5,415,000r2=SSR/SST=5,415,000/5,648,333.33=.9587Weseethat95.87%ofthevariabilityinyhasbeenexplainedbytheestimatedregressionequation.d.ŷ=1246.67+7.6x=1246.67+7.6(500)=$5046.6722.a.Thesummationsneededinthisproblemare:2Σx=613.1Σy=70(Σx−xy)(−y)=5766.7(Σx−x)=45,833.9286www.khdaw.comiiiiiΣ(x−xy)(−y)5766.7iib===0.125812Σ(x−x)45,833.9286ib=ybx−=10(0.1258)(87.5857)−=−1.018301yˆ=−1.01830.1258+xb.Thesumofsquaresduetoerrorandthetotalsumofsquaresare:22SSE=∑(y−yˆ)=1272.4495SST=∑(y−y)=1998iiiThus,SSR=SST-SSE=1998-1272.4495=725.5505r2=SSR/SST=725.5505/1998=0.3631Approximately37%ofthevariabilityinchangeinexecutivecompensationisexplainedbythetwo-yearchangeinthereturnonequity.c.r=0.3631=+0.6026Itreflectsalinearrelationshipthatisbetweenweakandstrong.23.a.s2=MSE=SSE/(n-2)=12.4/3=4.133b.s=MSE=4.133=2.0332c.Σ(x−x)=10is2.033s===0.643b1()210Σx−xib2.61d.t===4.04s.643b114-17www.khdaw.com 课后答案网www.khdaw.comChapter14t.025=3.182(3degreesoffreedom)Sincet=4.04>t.05=3.182werejectH0:β1=0e.MSR=SSR/1=67.6F=MSR/MSE=67.6/4.133=16.36F.05=10.13(1degreeoffreedomnumeratorand3denominator)SinceF=16.36>F.05=10.13werejectH0:β1=0SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFRegression67.6167.616.36Error12.434.133Total80.04www.khdaw.com24.a.s2=MSE=SSE/(n-2)=6.33/3=2.11b.s=MSE=2.11=1.4532c.Σ(x−x)=30.8is1.453s===0.262b1()230.8Σx−xib−1.881d.t===−7.18s.262b1t.025=3.182(3degreesoffreedom)Sincet=-7.18<-t.025=-3.182werejectH0:β1=0e.MSR=SSR/1=8.47F=MSR/MSE=108.47/2.11=51.41F.05=10.13(1degreeoffreedomnumeratorand3denominator)SinceF=51.41>F.05=10.13werejectH0:β1=0SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFRegression108.471108.4751.41Error6.3332.11Total114.80425.a.s2=MSE=SSE/(n-2)=5.30/3=1.77s=MSE=1.77=1.332b.Σ(x−x)=22.8i14-18www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressions1.33s===0.28b1()222.8Σx−xib.511t===1.82s.28b1t.025=3.182(3degreesoffreedom)Sincet=1.82t.025=2.776werejectH0:β1=0b.MSR=SSR/1=249,864.86/1=249.864.86F=MSR/MSE=249,864.86/21,283.79=11.74F.05=7.71(1degreeoffreedomnumeratorand4denominator)SinceF=11.74>F.05=7.71werejectH0:β1=0c.SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFRegression249864.86111.74249864.86Error85135.14421283.79Total335000527.Thesumofsquaresduetoerrorandthetotalsumofsquaresare:14-19www.khdaw.com 课后答案网www.khdaw.comChapter142SSE=Σ(y−yˆ)=170SST=2442iiThus,SSR=SST-SSE=2442-170=2272MSR=SSR/1=2272SSE=SST-SSR=2442-2272=170MSE=SSE/(n-2)=170/8=21.25F=MSR/MSE=2272/21.25=106.92F.05=5.32(1degreeoffreedomnumeratorand8denominator)SinceF=106.92>F.05=5.32werejectH0:β1=0.www.khdaw.comYearsofexperienceandsalesarerelated.28.SST=411.73SSE=161.37SSR=250.36MSR=SSR/1=250.36MSE=SSE/(n-2)=161.37/13=12.413F=MSR/MSE=250.36/12.413=20.17F.05=4.67(1degreeoffreedomnumeratorand13denominator)SinceF=20.17>F.05=4.67werejectH0:β1=0.29.SSE=233,333.33SST=5,648,333.33SSR=5,415,000MSE=SSE/(n-2)=233,333.33/(6-2)=58,333.33MSR=SSR/1=5,415,000F=MSR/MSE=5,415,000/58,333.25=92.83SourceofVariationSumofSquaresDegreesofFreedomMeanSquareFRegression5,415,000.0015,415,00092.83Error4233,333.3358,333.33Total5,648,333.335F.05=7.71(1degreeoffreedomnumeratorand4denominator)SinceF=92.83>7.71werejectH0:β1=0.Productionvolumeandtotalcostarerelated.30.UsingthecomputationsfromExercise22,SSE=1272.4495SST=1998SSR=725.5505s=254.4899=15.952∑(x−x)=45,833.9286i14-20www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressions15.95s===0.0745b1()245,833.9286Σx−xib0.12581t===1.69s0.0745b1t.025=2.571Sincet=1.69<2.571,wecannotrejectH0:β1=0Thereisnoevidenceofasignificantrelationshipbetweenxandy.31.SST=11,373.09SSE=2017.37SSR=9355.72www.khdaw.comMSR=SSR/1=9355.72MSE=SSE/(n-2)=2017.37/16=126.0856F=MSR/MSE=9355.72/126.0856=74.20F.01=8.53(1degreeoffreedomnumeratorand16denominator)SinceF=74.20>F.01=8.53werejectH0:β1=0.32.a.s=2.0332x=3(Σx−x)=10i221(x−x)1(43)−psyˆ=s+2=2.033+=1.11pnΣ(x−x)510ib.ŷ=0.2+2.6x=0.2+2.6(4)=10.6ŷ±tspα/2ŷp10.6±3.182(1.11)=10.6±3.53or7.07to14.13221(x−x)1(43)−pc.s=s1++=2.0331++=2.32ind2nΣ(x−x)510id.ŷ±tspα/2ind10.6±3.182(2.32)=10.6±7.38or3.22to17.9814-21www.khdaw.com 课后答案网www.khdaw.comChapter1433.a.s=1.4532b.x=3.8(Σx−x)=30.8i221(x−x)1(33.8)−psyˆ=s+2=1.453+=.068pnΣ(x−x)530.8iŷ=30.33−1.88x=30.33−1.88(3)=24.69ŷ±tspα/2ŷpwww.khdaw.com24.69±3.182(.68)=24.69±2.16or22.53to26.85221(x−x)1(33.8)−pc.s=s1++=1.4531++=1.61ind2nΣ(x−x)530.8id.ŷ±tspα/2ind24.69±3.182(1.61)=24.69±5.12or19.57to29.8134.s=1.332x=5.2(Σx−x)=22.8i221(x−x)1(35.2)−psyˆ=s+2=1.33+=0.85pnΣ(x−x)522.8iŷ=0.75+0.51x=0.75+0.51(3)=2.28ŷ±tspα/2ŷp2.28±3.182(.85)=2.28±2.70or-.40to4.98221(x−x)1(35.2)−ps=s1++=1.331++=1.58ind2nΣ(x−x)522.8iŷ±tspα/2ind2.28±3.182(1.58)=2.28±5.0314-22www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionor-2.27to7.3135.a.s=145.892x=3.2(Σx−x)=0.74i221(x−x)1(33.2)−psyˆ=s+2=145.89+=68.54pnΣ(x−x)60.74iŷ=290.54+581.08x=290.54+581.08(3)=2033.78ŷ±tspα/2ŷpwww.khdaw.com2,033.78±2.776(68.54)=2,033.78±190.27or$1,843.51to$2,224.05221(x−x)1(33.2)−pb.s=s1++=145.891++=161.19ind2nΣ(x−x)60.74iŷ±tspα/2ind2,033.78±2.776(161.19)=2,033.78±447.46or$1,586.32to$2,481.2436.a.yˆ=51.819.1452+x=51.819.1452(200)+=80.859b.s=3.52322x=183.4667(Σx−x)=11,867.73i221(x−x)1(200183.4667)−psyˆ=s+2=3.5232+=1.055pnΣ(x−x)1511,867.73iŷ±tspα/2ŷp80.859±2.160(1.055)=80.859±2.279or78.58to83.14221(x−x)1(200183.4667)−pc.s=s1++=3.52321++=3.678ind2nΣ(x−x)1511,867.73iŷ±tspα/2ind80.859±2.160(3.678)=80.859±7.94414-23www.khdaw.com 课后答案网www.khdaw.comChapter14or72.92to88.80237.a.x=57(Σx−x)=7648is2=1.88s=1.37221(x−x)1(52.557)−psyˆ=s+2=1.37+=0.52pnΣ(x−x)77648iŷ±tspα/2ŷp13.08±2.571(.52)=13.08±1.34www.khdaw.comor11.74to14.42or$11,740to$14,420b.sind=1.4713.08±2.571(1.47)=13.08±3.78or9.30to16.86or$9,300to$16,860c.Yes,$20,400ismuchlargerthananticipated.d.Anydeductionsexceedingthe$16,860upperlimitcouldsuggestanaudit.38.a.ŷ=1246.67+7.6(500)=$5046.672b.x=575(Σx−x)=93,750is2=MSE=58,333.33s=241.52221(x−x)1(500575)−ps=s1++=241.521++=267.50ind2nΣ(x−x)693,750iŷ±tspα/2ind5046.67±4.604(267.50)=5046.67±1231.57or$3815.10to$6278.24c.Basedononemonth,$6000isnotoutoflinesince$3815.10to$6278.24isthepredictioninterval.However,asequenceoffivetosevenmonthswithconsistentlyhighcostsshouldcauseconcern.39.a.Summationsneededtocomputetheslopeandy-interceptare:2Σx=227Σy=2281.7(Σx−xy)(−y)=6003.41(Σx−x)=1032.1iiiiiΣ(x−xy)(−y)6003.41iib===5.81669412Σ(x−x)1032.1i14-24www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionb=ybx−=228.17(5.816694)(27.7)−=67.04757601ŷ=67.0476+5.8167xb.SST=39,065.14SSE=4145.141SSR=34,920.000r2=SSR/SST=34,920.000/39,065.141=0.894Theestimatedregressionequationexplained89.4%ofthevariabilityiny;averygoodfit.c.s2=MSE=4145.141/8=518.143s=518.143=22.76221(x−x)1(3527.7)−pwww.khdaw.comsyˆ=s+2=22.76+=8.86pnΣ(x−x)101032.1iŷ=67.0476+5.8167x=67.0476+5.8167(35)=270.63ŷ±tspα/2ŷp270.63±2.262(8.86)=270.63±20.04or250.59to290.67221(x−x)1(3527.7)−pd.s=s1++=22.761++=24.42ind2nΣ(x−x)101032.1iŷ±tspα/2ind270.63±2.262(24.42)=270.63±55.24or215.39to325.8740.a.9b.yˆ=20.0+7.21xc.1.3626d.SSE=SST-SSR=51,984.1-41,587.3=10,396.8MSE=10,396.8/7=1,485.3F=MSR/MSE=41,587.3/1,485.3=28.00F.05=5.59(1degreeoffreedomnumeratorand7denominator)SinceF=28>F.05=5.59werejectH0:B1=0.14-25www.khdaw.com 课后答案网www.khdaw.comChapter14e.yˆ=20.0+7.21(50)=380.5or$380,50041.a.yˆ=6.1092+.8951xb−B.89510−11b.t===6.01s.149b1t.025=2.306(1degreeoffreedomnumeratorand8denominator)Sincet=6.01>t.025=2.306werejectH0:B1=0c.yˆ=6.1092+.8951(25)=28.49or$28.49permonthwww.khdaw.com42a.yˆ=80.0+50.0xb.30c.F=MSR/MSE=6828.6/82.1=83.17F.05=4.20(1degreeoffreedomnumeratorand28denominator)SinceF=83.17>F.05=4.20werejectH0:B1=0.Branchofficesalesarerelatedtothesalespersons.d.yˆ=80+50(12)=680or$680,00043.a.TheMinitaboutputisshownbelow:TheregressionequationisPrice=-11.8+2.18IncomePredictorCoefSECoefTPConstant-11.8012.84-0.920.380Income2.18430.27807.860.000S=6.634R-Sq=86.1%R-Sq(adj)=84.7%AnalysisofVarianceSourceDFSSMSFPRegression12717.92717.961.750.000ResidualError10440.144.0Total113158.0PredictedValuesforNewObservationsNewObsFitSEFit95.0%CI95.0%PI175.792.47(70.29,81.28)(60.02,91.56)b.r2=.861.Theleastsquareslineprovidedaverygoodfit.c.The95%confidenceintervalis70.29to81.28or$70,290to$81,280.14-26www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressiond.The95%predictionintervalis60.02to91.56or$60,020to$91,560.44.a/b.Thescatterdiagramshowsalinearrelationshipbetweenthetwovariables.c.TheMinitaboutputisshownbelow:TheregressionequationisRental$=37.1-0.779Vacancy%PredictorCoefSECoefTPConstant37.0663.53010.500.000Vacancy%-0.77910.2226-3.500.003S=4.889R-Sq=43.4%R-Sq(adj)=39.8%www.khdaw.comAnalysisofVarianceSourceDFSSMSFPRegression1292.89292.8912.260.003ResidualError16382.3723.90Total17675.26PredictedValuesforNewObservationsNewObsFitSEFit95.0%CI95.0%PI117.592.51(12.27,22.90)(5.94,29.23)228.261.42(25.26,31.26)(17.47,39.05)ValuesofPredictorsforNewObservationsNewObsVacancy%125.0211.3d.Sincethep-value=0.003islessthanα=.05,therelationshipissignificant.e.r2=.434.Theleastsquareslinedoesnotprovideaverygoodfit.f.The95%confidenceintervalis12.27to22.90or$12.27to$22.90.g.The95%predictionintervalis17.47to39.05or$17.47to$39.05.245.a.Σx=14Σy=76(Σx−xy)(−y)=200(Σx−x)=126iiiiiΣ(x−xy)(−y)200iib===1.587312Σ(x−x)126ib=ybx−=15.2(1.5873)(14)−=−7.022201yˆ=−7.021.59+xb.Theresidualsare3.48,-2.47,-4.83,-1.6,and5.2214-27www.khdaw.com 课后答案网www.khdaw.comChapter14c.6420-2Residuals-4www.khdaw.com-60510152025xWithonly5observationsitisdifficulttodetermineiftheassumptionsaresatisfied.However,theplotdoessuggestcurvatureintheresidualsthatwouldindicatethattheerrortermassumptionsarenotsatisfied.Thescatterdiagramforthesedataalsoindicatesthattheunderlyingrelationshipbetweenxandymaybecurvilinear.2d.s=23.78221(x−x)1(x−14)iih=+=+i2nΣ(x−x)5126iThestandardizedresidualsare1.32,-.59,-1.11,-.40,1.49.e.Thestandardizedresidualplothasthesameshapeastheoriginalresidualplot.Thecurvatureobservedindicatesthattheassumptionsregardingtheerrortermmaynotbesatisfied.46.a.yˆ=2.32.64+xb.14-28www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression43210-1Residuals-2-3-40246810xwww.khdaw.comTheassumptionthatthevarianceisthesameforallvaluesofxisquestionable.Thevarianceappearstoincreaseforlargervaluesofx.47.a.Letx=advertisingexpendituresandy=revenueyˆ=29.41.55+xb.SST=1002SSE=310.28SSR=691.72MSR=SSR/1=691.72MSE=SSE/(n-2)=310.28/5=62.0554F=MSR/MSE=691.72/62.0554=11.15F.05=6.61(1degreeoffreedomnumeratorand5denominator)SinceF=11.15>F.05=6.61weconcludethatthetwovariablesarerelated.c.1050-5Residuals-10-15253035404550556065PredictedValues14-29www.khdaw.com 课后答案网www.khdaw.comChapter14d.Theresidualplotleadsustoquestiontheassumptionofalinearrelationshipbetweenxandy.Eventhoughtherelationshipissignificantatthe.05levelofsignificance,itwouldbeextremelydangeroustoextrapolatebeyondtherangeofthedata.www.khdaw.com48.a.yˆ=804+x86420Residuals-2-4-6-802468101214xb.Theassumptionsconcerningtheerrortermappearreasonable.49.a.Letx=returnoninvestment(ROE)andy=price/earnings(P/E)ratio.yˆ=−32.133.22+xb.14-30www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegression21.510.50-0.5-1StandardizedResiduals-1.50102030405060xwww.khdaw.comc.Thereisanunusualtrendintheresiduals.Theassumptionsconcerningtheerrortermappearquestionable.50.a.TheMINITABoutputisshownbelow:TheregressionequationisY=66.1+0.402XPredictorCoefStdevt-ratiopConstant66.1032.062.060.094X0.40230.22761.770.137s=12.62R-sq=38.5%R-sq(adj)=26.1%AnalysisofVarianceSOURCEDFSSMSFpRegression1497.2497.23.120.137Error5795.7159.1Total61292.9UnusualObservationsObs.XYFitStdev.FitResidualSt.Resid1135145.00120.424.8724.582.11RRdenotesanobs.withalargest.resid.Thestandardizedresidualsare:2.11,-1.08,.14,-.38,-.78,-.04,-.41Thefirstobservationappearstobeanoutliersinceithasalargestandardizedresidual.b.14-31www.khdaw.com 课后答案网www.khdaw.comChapter142.4+-*STDRESID---1.2+----*0.0+*--**-*--1.2+*www.khdaw.com---+---------+---------+---------+---------+---------+----YHAT110.0115.0120.0125.0130.0135.0Thestandardizedresidualplotindicatesthattheobservationx=135,y=145maybeanoutlier;notethatthisobservationhasastandardizedresidualof2.11.c.Thescatterdiagramisshownbelow-Y-*--135+--**--120+**---*-105+--*----+---------+---------+---------+---------+---------+--X105120135150165180Thescatterdiagramalsoindicatesthattheobservationx=135,y=145maybeanoutlier;theimplicationisthatforsimplelinearregressionanoutliercanbeidentifiedbylookingatthescatterdiagram.51.a.TheMinitaboutputisshownbelow:TheregressionequationisY=13.0+0.425X14-32www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionPredictorCoefStdevt-ratiopConstant13.0022.3965.430.002X0.42480.21162.010.091s=3.181R-sq=40.2%R-sq(adj)=30.2%AnalysisofVarianceSOURCEDFSSMSFpRegression140.7840.784.030.091Error660.7210.12Total7101.50UnusualObservationsObs.XYFitStdev.FitResidualSt.Resid712.024.0018.101.205.902.00Rwww.khdaw.com822.019.0022.352.78-3.35-2.16RXRdenotesanobs.withalargest.resid.Xdenotesanobs.whoseXvaluegivesitlargeinfluence.Thestandardizedresidualsare:-1.00,-.41,.01,-.48,.25,.65,-2.00,-2.16Thelasttwoobservationsinthedatasetappeartobeoutlierssincethestandardizedresidualsfortheseobservationsare2.00and-2.16,respectively.b.UsingMINITAB,weobtainedthefollowingleveragevalues:.28,.24,.16,.14,.13,.14,.14,.76MINITABidentifiesanobservationashavinghighleverageifhi>6/n;forthesedata,6/n=6/8=.75.Sincetheleveragefortheobservationx=22,y=19is.76,MINITABwouldidentifyobservation8asahighleveragepoint.Thus,weconcludethatobservation8isaninfluentialobservation.c.24.0+*-Y---20.0+*-*-*--16.0+*-*--*-12.0+*-+---------+---------+---------+---------+---------+------X14-33www.khdaw.com 课后答案网www.khdaw.comChapter140.05.010.015.020.025.0Thescatterdiagramindicatesthattheobservationx=22,y=19isaninfluentialobservation.52.a.TheMinitaboutputisshownbelow:TheregressionequationisAmount=4.09+0.196MediaExpPredictorCoefSECoefTPConstant4.0892.1681.890.096MediaExp0.195520.036355.380.001S=5.044R-Sq=78.3%R-Sq(adj)=75.6%AnalysisofVariancewww.khdaw.comSourceDFSSMSFPRegression1735.84735.8428.930.001ResidualError8203.5125.44Total9939.35UnusualObservationsObsMediaExpAmountFitSEFitResidualStResid112036.3027.553.308.752.30RRdenotesanobservationwithalargestandardizedresidualb.Minitabidentifiesobservation1ashavingalargestandardizedresidual;thus,wewouldconsiderobservation1tobeanoutlier.53.a.TheMinitaboutputisshownbelow:TheregressionequationisExposure=-8.6+7.71AiredPredictorCoefSECoefTPConstant-8.5521.65-0.390.703Aired7.71490.511915.070.000S=34.88R-Sq=96.6%R-Sq(adj)=96.2%AnalysisofVarianceSourceDFSSMSFPRegression1276434276434227.170.000ResidualError897351217Total9286169UnusualObservationsObsAiredExposureFitSEFitResidualStResid195.0758.8724.432.034.42.46RXRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.14-34www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionb.Minitabidentifiesobservation1ashavingalargestandardizedresidual;thus,wewouldconsiderobservation1tobeanoutlier.Minitabalsoidentifiesobservation1asaninfluentialobservation.54.a.TheMinitaboutputisshownbelow:TheregressionequationisSalary=707+0.00482MktCapPredictorCoefSECoefTPConstant707.0118.05.990.000MktCap0.00481540.00080765.960.000S=379.8R-Sq=66.4%R-Sq(adj)=64.5%AnalysisofVarianceSourceDFSSMSFPwww.khdaw.comRegression15129071512907135.550.000ResidualError182596647144258Total197725718UnusualObservationsObsMktCapSalaryFitSEFitResidualStResid65072173325.03149.5338.6175.51.02X17120967116.21289.586.4-1173.3-3.17RRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.b.Minitabidentifiesobservation6ashavingalargestandardizedresidualandobservation17asanobservationwhosexvaluegivesitlargeinfluence.Astandardizedresidualplotagainstthepredictedvaluesisshownbelow:3210-1-2-3StandardizedResiduals-4500100015002000250030003500PredictedValues55.No.Regressionorcorrelationanalysiscanneverprovethattwovariablesarecasuallyrelated.56.Theestimateofameanvalueisanestimateoftheaverageofallyvaluesassociatedwiththesamex.Theestimateofanindividualyvalueisanestimateofonlyoneoftheyvaluesassociatedwithaparticularx.14-35www.khdaw.com 课后答案网www.khdaw.comChapter1457.Todeterminewhetherornotthereisasignificantrelationshipbetweenxandy.However,ifwerejectB1=0,itdoesnotimplyagoodfit.58.a.TheMinitaboutputisshownbelow:TheregressionequationisPrice=9.26+0.711SharesPredictorCoefSECoefTPConstant9.2651.0998.430.000Shares0.71050.14744.820.001S=1.419R-Sq=74.4%R-Sq(adj)=71.2%AnalysisofVarianceSourceDFSSMSFPwww.khdaw.comRegression146.78446.78423.220.001ResidualError816.1162.015Total962.900b.Sincethep-valuecorrespondingtoF=23.22=.001<α=.05,therelationshipissignificant.2c.r=.744;agoodfit.Theleastsquareslineexplained74.4%ofthevariabilityinPrice.d.yˆ=9.26.711(6)13.53+=59.a.TheMinitaboutputisshownbelow:TheregressionequationisOptions=-3.83+0.296CommonPredictorCoefSECoefTPConstant-3.8345.903-0.650.529Common0.295670.0264811.170.000S=11.04R-Sq=91.9%R-Sq(adj)=91.2%AnalysisofVarianceSourceDFSSMSFPRegression11520815208124.720.000ResidualError111341122Total1216550b.yˆ=−3.83.296(150)+=40.57;approximately40.6millionsharesofoptionsgrantsoutstanding.2c.r=.919;averygoodfit.Theleastsquareslineexplained91.9%ofthevariabilityinOptions.60.a.TheMinitaboutputisshownbelow:Theregressionequationis14-36www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionIBM=0.275+0.950S&P500PredictorCoefStDevTPConstant0.27470.90040.310.768S&P5000.94980.35692.660.029S=2.664R-Sq=47.0%R-Sq(adj)=40.3%AnalysisofVarianceSourceDFSSMSFPRegression150.25550.2557.080.029Error856.7817.098Total9107.036b.Sincethep-value=0.029islessthanα=.05,therelationshipissignificant.www.khdaw.comc.r2=.470.Theleastsquareslinedoesnotprovideaverygoodfit.d.Woolworthhashigherriskwithamarketbetaof1.25.61.a.10090807060HighTemperature5040354555657585LowTemperatureb.Itappearsthatthereisapositivelinearrelationshipbetweenthetwovariables.c.TheMinitaboutputisshownbelow:Theregressionequationis14-37www.khdaw.com 课后答案网www.khdaw.comChapter14High=23.9+0.898LowPredictorCoefSECoefTPConstant23.8996.4813.690.002Low0.89800.11218.010.000S=5.285R-Sq=78.1%R-Sq(adj)=76.9%AnalysisofVarianceSourceDFSSMSFPRegression11792.31792.364.180.000ResidualError18502.727.9Total192294.9d.Sincethep-valuecorrespondingtoF=64.18=.000<α=.05,therelationshipissignificant.2www.khdaw.come.r=.781;agoodfit.Theleastsquareslineexplained78.1%ofthevariabilityinhightemperature.f.r=.781=+.8862.TheMINITABoutputisshownbelow:TheregressionequationisY=10.5+0.953XPredictorCoefStdevt-ratiopConstant10.5283.7452.810.023X0.95340.13826.900.000s=4.250R-sq=85.6%R-sq(adj)=83.8%AnalysisofVarianceSOURCEDFSSMSFpRegression1860.05860.0547.620.000Error8144.4718.06Total91004.53FitStdev.Fit95%C.I.95%P.I.39.131.49(35.69,42.57)(28.74,49.52)a.yˆ=10.5+.953xb.Sincethep-valuecorrespondingtoF=47.62=.000<α=.05,werejectH0:β1=0.c.The95%predictionintervalis28.74to49.52or$2874to$4952d.Yes,sincetheexpectedexpenseis$3913.63.a.TheMinitaboutputisshownbelow:TheregressionequationisDefects=22.2-0.148SpeedPredictorCoefSECoefTP14-38www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionConstant22.1741.65313.420.000Speed-0.147830.04391-3.370.028S=1.489R-Sq=73.9%R-Sq(adj)=67.4%AnalysisofVarianceSourceDFSSMSFPRegression125.13025.13011.330.028ResidualError48.8702.217Total534.000PredictedValuesforNewObservationsNewObsFitSEFit95.0%CI95.0%PI114.7830.896(12.294,17.271)(9.957,19.608)www.khdaw.comb.Sincethep-valuecorrespondingtoF=11.33=.028<α=.05,therelationshipissignificant.2c.r=.739;agoodfit.Theleastsquareslineexplained73.9%ofthevariabilityinthenumberofdefects.d.UsingtheMinitaboutputinpart(a),the95%confidenceintervalis12.294to17.271.64.a.Thereappearstobeanegativelinearrelationshipbetweendistancetoworkandnumberofdaysabsent.b.TheMINITABoutputisshownbelow:TheregressionequationisY=8.10-0.344XPredictorCoefStdevt-ratiopConstant8.09780.808810.010.000X-0.344200.07761-4.430.002s=1.289R-sq=71.1%R-sq(adj)=67.5%AnalysisofVarianceSOURCEDFSSMSFpRegression132.69932.69919.670.002Error813.3011.663Total946.000FitStdev.Fit95%C.I.95%P.I.6.3770.512(5.195,7.559)(3.176,9.577)c.Sincethep-valuecorrespondingtoF=419.67is.002<α=.05.WerejectH0:β1=0.14-39www.khdaw.com 课后答案网www.khdaw.comChapter14d.r2=.711.Theestimatedregressionequationexplained71.1%ofthevariabilityiny;thisisareasonablygoodfit.e.The95%confidenceintervalis5.195to7.559orapproximately5.2to7.6days.65.a.LetX=theageofabusandY=theannualmaintenancecost.TheMINITABoutputisshownbelow:TheregressionequationisY=220+132XPredictorCoefStdevt-ratiopConstant220.0058.483.760.006X131.6717.807.400.000www.khdaw.coms=75.50R-sq=87.3%R-sq(adj)=85.7%AnalysisofVarianceSOURCEDFSSMSFpRegression131205031205054.750.000Error8456005700Total9357650FitStdev.Fit95%C.I.95%P.I.746.729.8(678.0,815.4)(559.5,933.9)b.Sincethep-valuecorrespondingtoF=54.75is.000<α=.05,werejectH0:β1=0.c.r2=.873.Theleastsquareslineprovidedaverygoodfit.d.The95%predictionintervalis559.5to933.9or$559.50to$933.9066.a.LetX=hoursspentstudyingandY=totalpointsearnedTheMINITABoutputisshownbelow:TheregressionequationisY=5.85+0.830XPredictorCoefStdevt-ratiopConstant5.8477.9720.730.484X0.82950.10957.580.000s=7.523R-sq=87.8%R-sq(adj)=86.2%AnalysisofVarianceSOURCEDFSSMSFpRegression13249.73249.757.420.000Error8452.856.6Total93702.5FitStdev.Fit95%C.I.95%P.I.84.653.67(76.19,93.11)(65.35,103.96)14-40www.khdaw.com 课后答案网www.khdaw.comSimpleLinearRegressionb.Sincethep-valuecorrespondingtoF=57.42is.000<α=.05,werejectH0:β1=0.c.84.65pointsd.The95%predictionintervalis65.35to103.9667.a.TheMinitaboutputisshownbelow:TheregressionequationisAudit%=-0.471+0.000039IncomePredictorCoefSECoefTPConstant-0.47100.5842-0.810.431Income0.000038680.000017312.230.038www.khdaw.comS=0.2088R-Sq=21.7%R-Sq(adj)=17.4%AnalysisofVarianceSourceDFSSMSFPRegression10.217490.217494.990.038ResidualError180.784510.04358Total191.00200PredictedValuesforNewObservationsNewObsFitSEFit95.0%CI95.0%PI10.88280.0523(0.7729,0.9927)(0.4306,1.3349)b.Sincethep-value=0.038islessthanα=.05,therelationshipissignificant.c.r2=.217.Theleastsquareslinedoesnotprovideaverygoodfit.d.The95%confidenceintervalis.7729to.9927.14-41www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter15MultipleMMultipleultipleRegressionRRegressionegressionLearningLLearningearningObjectivesOObjectivesbjectives1.Understandhowmultipleregressionanalysiscanbeusedtodeveloprelationshipsinvolvingonedependentvariableandseveralindependentvariables.2.Beabletointerpretthecoefficientsinamultipleregressionanalysis.www.khdaw.com3.Knowtheassumptionsnecessarytoconductstatisticaltestsinvolvingthehypothesizedregressionmodel.4.Understandtheroleofcomputerpackagesinperformingmultipleregressionanalysis.5.Beabletointerpretandusecomputeroutputtodeveloptheestimatedregressionequation.6.Beabletodeterminehowgoodafitisprovidedbytheestimatedregressionequation.7.Beabletotestforthesignificanceoftheregressionequation.8.Understandhowmulticollinearityaffectsmultipleregressionanalysis.9.Knowhowresidualanalysiscanbeusedtomakeajudgementastotheappropriatenessofthemodel,identifyoutliers,anddeterminewhichobservationsareinfluential.15-1www.khdaw.com 课后答案网www.khdaw.comChapter15Solutions:SSolutions:olutions:1.a.b1=.5906isanestimateofthechangeinycorrespondingtoa1unitchangeinx1whenx2isheldconstant.b2=.4980isanestimateofthechangeinycorrespondingtoa1unitchangeinx2whenx1isheldconstant.2.a.Theestimatedregressionequationisyˆ=45.06+1.94x1Anestimateofywhenx1=45iswww.khdaw.comyˆ=45.06+1.94(45)=132.36b.Theestimatedregressionequationisyˆ=85.22+4.32x2Anestimateofywhenx2=15isyˆ=85.22+4.32(15)=150.02c.Theestimatedregressionequationisyˆ=-18.37+2.01x1+4.74x2Anestimateofywhenx1=45andx2=15isyˆ=-18.37+2.01(45)+4.74(15)=143.183.a.b1=3.8isanestimateofthechangeinycorrespondingtoa1unitchangeinx1whenx2,x3,andx4areheldconstant.b2=-2.3isanestimateofthechangeinycorrespondingtoa1unitchangeinx2whenx1,x3,andx4areheldconstant.b3=7.6isanestimateofthechangeinycorrespondingtoa1unitchangeinx3whenx1,x2,andx4areheldconstant.b4=2.7isanestimateofthechangeinycorrespondingtoa1unitchangeinx4whenx1,x2,andx3areheldconstant.4.a.yˆ=235+10(15)+8(10)=255;salesestimate:$255,000b.Salescanbeexpectedtoincreaseby$10foreverydollarincreaseininventoryinvestmentwhenadvertisingexpenditureisheldconstant.Salescanbeexpectedtoincreaseby$8foreverydollarincreaseinadvertisingexpenditurewheninventoryinvestmentisheldconstant.15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression5.a.TheMinitaboutputisshownbelow:TheregressionequationisRevenue=88.6+1.60TVAdvPredictorCoefSECoefTPConstant88.6381.58256.020.000TVAdv1.60390.47783.360.015S=1.215R-Sq=65.3%R-Sq(adj)=59.5%AnalysisofVarianceSourceDFSSMSFPRegression116.64016.64011.270.015ResidualError68.8601.477www.khdaw.comTotal725.500b.TheMinitaboutputisshownbelow:TheregressionequationisRevenue=83.2+2.29TVAdv+1.30NewsAdvPredictorCoefSECoefTPConstant83.2301.57452.880.000TVAdv2.29020.30417.530.001NewsAdv1.30100.32074.060.010S=0.6426R-Sq=91.9%R-Sq(adj)=88.7%AnalysisofVarianceSourceDFSSMSFPRegression223.43511.71828.380.002ResidualError52.0650.413Total725.500SourceDFSeqSSTVAdv116.640NewsAdv16.795c.No,itis1.60inpart2(a)and2.99above.Inthisexerciseitrepresentsthemarginalchangeinrevenueduetoanincreaseintelevisionadvertisingwithnewspaperadvertisingheldconstant.d.Revenue=83.2+2.29(3.5)+1.30(1.8)=$93.56or$93,5606.a.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=49.8+0.0151WeightPredictorCoefSECoefTPConstant49.7819.112.610.021Weight0.0151040.0060052.520.02515-www.khdaw.com 课后答案网www.khdaw.comChapter15S=7.000R-Sq=31.1%R-Sq(adj)=26.2%AnalysisofVarianceSourceDFSSMSFPRegression1309.95309.956.330.025Error14686.0049.00Total15995.95b.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=80.5-0.00312Weight+0.105HorsepwrPredictorCoefSECoefTPwww.khdaw.comConstant80.4879.1398.810.000Weight-0.0031220.003481-0.900.386Horsepwr0.104710.013317.860.000S=3.027R-Sq=88.0%R-Sq(adj)=86.2%AnalysisofVarianceSourceDFSSMSFPRegression2876.80438.4047.830.000ResidualError13119.159.17Total15995.957.a.TheMinitaboutputisshownbelow:TheregressionequationisSales=66.5+0.414Compet$-0.270Heller$PredictorCoefSECoefTPConstant66.5241.881.590.156Compet$0.41390.26041.590.156Heller$-0.269780.08091-3.330.013S=18.74R-Sq=65.3%R-Sq(adj)=55.4%AnalysisofVarianceSourceDFSSMSFPRegression24618.82309.46.580.025ResidualError72457.3351.0Total97076.1b.b1=.414isanestimateofthechangeinthequantitysold(1000s)oftheHellermowerwithrespecttoa$1changeinpriceincompetitor’smowerwiththepriceoftheHellermowerheldconstant.b2=-.270isanestimateofthechangeinthequantitysold(1000s)oftheHellermowerwithrespecttoa$1changeinitspricewiththepriceofthecompetitor’smowerheldconstant.c.yˆ=66.5+0.414(170)-0.270(160)=93.68or93,680units15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression8.a.TheMinitaboutputisshownbelow:TheregressionequationisReturn=247-32.8Safety+34.6ExpRatioPredictorCoefSECoefTPConstant247.4110.42.240.039Safety-32.8413.95-2.350.031ExpRatio34.5914.132.450.026S=16.98R-Sq=58.2%R-Sq(adj)=53.3%AnalysisofVarianceSourceDFSSMSFPRegression26823.23411.611.840.001www.khdaw.comResidualError174899.7288.2Total1911723.0b.yˆ=24732.8(7.5)34.6(2)−+=70.29.a.TheMinitaboutputisshownbelow:Theregressionequationis%College=26.7-1.43Size+0.0757SatScorePredictorCoefSECoefTPConstant26.7151.670.520.613Size-1.42980.9931-1.440.170SatScore0.075740.039061.940.072S=12.42R-Sq=38.2%R-Sq(adj)=30.0%AnalysisofVarianceSourceDFSSMSFPRegression21430.4715.24.640.027ResidualError152312.7154.2Total173743.1b.yˆ=26.7-1.43(20)+0.0757(1000)=73.8Estimateis73.8%10.a.TheMinitaboutputisshownbelow:TheregressionequationisRevenue=33.3+7.98CarsPredictorCoefSECoefTPConstant33.3483.080.400.695Cars7.98400.632312.630.000S=226.7R-Sq=92.5%R-Sq(adj)=91.9%15-www.khdaw.com 课后答案网www.khdaw.comChapter15AnalysisofVarianceSourceDFSSMSFPRegression181920678192067159.440.000Error1366793651380Total148860003b.Anincreaseof1000carsinservicewillresultinanincreaseinrevenueof$7.98million.c.TheMinitaboutputisshownbelow:TheregressionequationisRevenue=106+8.94Cars-0.191LocationPredictorCoefSECoefTPwww.khdaw.comConstant105.9785.521.240.239Cars8.94270.774611.550.000Location-0.19140.1026-1.870.087S=207.7R-Sq=94.2%R-Sq(adj)=93.2%AnalysisofVarianceSourceDFSSMSFPRegression28342186417109396.660.000Error1251781743151Total14886000311.a.SSE=SST-SSR=6,724.125-6,216.375=507.752SSR6,216.375b.R===.924SST6,724.12522n−1101−c.R=−1(1−R)=−1(1.924)−=.902an−p−11021−−d.Theestimatedregressionequationprovidedanexcellentfit.2SSR14,052.212.a.R===.926SST15,182.922n−1101−b.R=−1(1−R)=−1(1.926)−=.905an−p−11021−−c.Yes;afteradjustingforthenumberofindependentvariablesinthemodel,weseethat90.5%ofthevariabilityinyhasbeenaccountedfor.2SSR176013.a.R===.975SST180515-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression22n−1301−b.R=−1(1−R)=−1(1.975)−=.971an−p−13041−−c.Theestimatedregressionequationprovidedanexcellentfit.2SSR12,00014.a.R===.75SST16,00022n−19b.R=−1(1−R)=−1.25=.68an−p−17c.Theadjustedcoefficientofdeterminationshowsthat68%ofthevariabilityhasbeenexplainedbythetwoindependentvariables;thus,weconcludethatthemodeldoesnotexplainalargeamountofwww.khdaw.comvariability.2SSR23.43515.a.R===.919SST25.522n−181−R=−1(1−R)=−1(1.919)−=.887an−p−1821−−b.MultipleregressionanalysisispreferredsincebothR2andR2showanincreasedpercentageoftheavariabilityofyexplainedwhenbothindependentvariablesareused.16.Note:theMinitaboutputisshownwiththesolutiontoExercise6.a.No;R-Sq=31.1%b.MultipleregressionanalysisispreferredsincebothR-SqandR-Sq(adj)showanincreasedpercentageofthevariabilityofyexplainedwhenbothindependentvariablesareused.2SSR1430.417.a.R===.382SST3743.122n−1181−R=−1(1−R)=−1(1.382)−=.30an−p−11821−−b.Thefitisnotverygood18.Note:TheMinitaboutputisshownwiththesolutiontoExercise10.a.R-Sq=94.2%R-Sq(adj)=93.2%b.Thefitisverygood.19.a.MSR=SSR/p=6,216.375/2=3,108.188SSE507.75MSE===72.536n−p−11021−−15-www.khdaw.com 课后答案网www.khdaw.comChapter15b.F=MSR/MSE=3,108.188/72.536=42.85F.05=4.74(2degreesoffreedomnumeratorand7denominator)SinceF=42.85>F.05=4.74theoverallmodelissignificant.c.t=.5906/.0813=7.26t.025=2.365(7degreesoffreedom)Sincet=2.365>t.025=2.365,β1issignificant.d.t=.4980/.0567=8.78Sincet=8.78>t.025=2.365,β2issignificant.www.khdaw.com20.AportionoftheMinitaboutputisshownbelow.TheregressionequationisY=-18.4+2.01X1+4.74X2PredictorCoefSECoefTPConstant-18.3717.97-1.020.341X12.01020.24718.130.000X24.73780.94845.000.002S=12.71R-Sq=92.6%R-Sq(adj)=90.4%AnalysisofVarianceSourceDFSSMSFPRegression214052.27026.143.500.000ResidualError71130.7161.5Total915182.9a.Sincethep-valuecorrespondingtoF=43.50is.000<α=.05,werejectH0:β1=β2=0;thereisasignificantrelationship.b.Sincethep-valuecorrespondingtot=8.13is.000<α=.05,werejectH0:β1=0;β1issignificant.c.Sincethep-valuecorrespondingtot=5.00is.002<α=.05,werejectH0:β2=0;β2issignificant.21.a.Inthetwoindependentvariablecasethecoefficientofx1representstheexpectedchangeinycorrespondingtoaoneunitincreaseinx1whenx2isheldconstant.Inthesingleindependentvariablecasethecoefficientofx1representstheexpectedchangeinycorrespondingtoaoneunitincreaseinx1.b.Yes.Ifx1andx2arecorrelatedonewouldexpectachangeinx1tobeaccompaniedbyachangeinx2.15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression22.a.SSE=SST-SSR=16000-12000=40002SSE4000s===571.43np--17SSR12000MSR===6000p2b.F=MSR/MSE=6000/571.43=10.50F.05=4.74(2degreesoffreedomnumeratorand7denominator)SinceF=10.50>F.05=4.74,werejectH0.Thereisasignificantrelationshipamongthevariables.www.khdaw.com23.a.F=28.38F.01=13.27(2degreesoffreedom,numeratorand1denominator)SinceF>F.01=13.27,rejectH0.Alternatively,thep-valueof.002leadstothesameconclusion.b.t=7.53t.025=2.571Sincet>t.025=2.571,β1issignificantandx1shouldnotbedroppedfromthemodel.c.t=4.06t.025=2.571Sincet>t.025=2.571,β2issignificantandx2shouldnotbedroppedfromthemodel.24.Note:TheMinitaboutputisshowninpart(b)ofExercise6a.F=47.83F.05=3.81(2degreesoffreedomnumeratorand13denominator)SinceF=47.83>F.05=3.81,werejectH0:β1=β2=0.Alternatively,sincethep-value=.000<α=.05wecanrejectH0.b.ForWeight:H0:β1=0Ha:β1≠0Sincethep-value=0.386>α=0.05,wecannotrejectH015-www.khdaw.com 课后答案网www.khdaw.comChapter15ForHorsepower:H0:β2=0Ha:β2≠0Sincethep-value=0.000<α=0.05,wecanrejectH025.a.TheMinitaboutputisshownbelow:TheregressionequationisP/E=6.04+0.692Profit%+0.265Sales%PredictorCoefSECoefTPConstant6.0384.5891.320.211Profit%0.69160.21333.240.006Sales%0.26480.18711.420.180www.khdaw.comS=5.456R-Sq=47.2%R-Sq(adj)=39.0%AnalysisofVarianceSourceDFSSMSFPRegression2345.28172.645.800.016ResidualError13387.0029.77Total15732.28b.Sincethep-value=0.016<α=0.05,thereisasignificantrelationshipamongthevariables.c.ForProfit%:Sincethep-value=0.006<α=0.05,Profit%issignificant.ForSales%:Sincethep-value=0.180>α=0.05,Sales%isnotsignificant.26.Note:TheMinitaboutputisshownwiththesolutiontoExercise10.a.Sincethep-valuecorrespondingtoF=96.66is0.000<α=.05,thereisasignificantrelationshipamongthevariables.b.ForCars:Sincethep-value=0.000<α=0.05,Carsissignificantc.ForLocation:Sincethep-value=0.087>α=0.05,Locationisnotsignificant27.a.yˆ=29.1270+.5906(180)+.4980(310)=289.8150b.Thepointestimateforanindividualvalueisyˆ=289.8150,thesameasthepointestimateofthemeanvalue.28.a.UsingMinitab,the95%confidenceintervalis132.16to154.16.b.UsingMinitab,the95%predictionintervalis111.13to175.18.15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression29.a.yˆ=83.2+2.29(3.5)+1.30(1.8)=93.555or$93,555Note:InExercise5b,theMinitaboutputalsoshowsthatb0=83.230,b1=2.2902,andb2=1.3010;hence,yˆ=83.230+2.2902x1+1.3010x2.Usingthisestimatedregressionequation,weobtainyˆ=83.230+2.2902(3.5)+1.3010(1.8)=93.588or$93,588Thedifference($93,588-$93,555=$33)issimplyduetothefactthatadditionalsignificantdigitsareusedinthecomputations.Fromapracticalpointofview,however,thedifferenceisnotenoughtobeconcernedabout.Inpractice,acomputersoftwarepackageisalwaysusedtoperformthecomputationsandthiswillnotbeanissue.TheMinitaboutputisshownbelow:www.khdaw.comFitStdev.Fit95%C.I.95%P.I.93.5880.291(92.840,94.335)(91.774,95.401)NotethatthevalueofFIT(yˆ)is93.588.b.Confidenceintervalestimate:92.840to94.335or$92,840to$94,335c.Predictionintervalestimate:91.774to95.401or$91,774to$95,40130.a.Sinceweightisnotstatisticallysignificant(seeExercise24),wewilluseanestimatedregressionequationwhichusesonlyHorsepowertopredictthespeedat1/4mile.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=72.6+0.0968HorsepwrPredictorCoefSECoefTPConstant72.6502.65527.360.000Horsepwr0.0967560.0098659.810.000S=3.006R-Sq=87.3%R-Sq(adj)=86.4%AnalysisofVarianceSourceDFSSMSFPRegression1869.43869.4396.210.000ResidualError14126.529.04Total15995.95UnusualObservationsObsHorsepwrSpeedFitSEFitResidualStResid2290108.000100.7090.8147.2912.52R6450116.200116.1902.0360.0100.00XRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.Theoutputshowsthatthepointestimateisaspeedof101.290milesperhour.15-www.khdaw.com 课后答案网www.khdaw.comChapter15b.The95%confidenceintervalis99.490to103.089milesperhour.c.The95%predictionintervalis94.596to107.984milesperhour.31.a.UsingMinitabthe95%confidenceintervalis58.37%to75.03%.b.UsingMinitabthe95%predictionintervalis35.24%to90.59%.32.a.E(y)=β0+β1x1+β2x2wherex2=0iflevel1and1iflevel2b.E(y)=β0+β1x1+β2(0)=β0+β1x1www.khdaw.comc.E(y)=β0+β1x1+β2(1)=β0+β1x1+β2d.β2=E(y|level2)-E(y|level1)β1isthechangeinE(y)fora1unitchangeinx1holdingx2constant.33.a.twob.E(y)=β0+β1x1+β2x2+β3x3wherex2x3Level001102013c.E(y|level1)=β0+β1x1+β2(0)+β3(0)=β0+β1x1E(y|level2)=β0+β1x1+β2(1)+β3(0)=β0+β1x1+β2E(y|level3)=β0+β1x1+β2(0)+β3(0)=β0+β1x1+β3β2=E(y|level2)-E(y|level1)β3=E(y|level3)-E(y|level1)β1isthechangeinE(y)fora1unitchangeinx1holdingx2andx3constant.34.a.$15,300b.Estimateofsales=10.1-4.2(2)+6.8(8)+15.3(0)=56.1or$56,100c.Estimateofsales=10.1-4.2(1)+6.8(3)+15.3(1)=41.6or$41,60035.a.LetType=0ifamechanicalrepairType=1ifanelectricalrepairTheMinitaboutputisshownbelow:15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionTheregressionequationisTime=3.45+0.617TypePredictorCoefSECoefTPConstant3.45000.54676.310.000Type0.61670.70580.870.408S=1.093R-Sq=8.7%R-Sq(adj)=0.0%AnalysisofVarianceSourceDFSSMSFPRegression10.9130.9130.760.408ResidualError89.5631.195Total910.476www.khdaw.comb.Theestimatedregressionequationdidnotprovideagoodfit.Infact,thep-valueof.408showsthattherelationshipisnotsignificantforanyreasonablevalueofα.c.Person=0ifBobJonesperformedtheserviceandPerson=1ifDaveNewtonperformedtheservice.TheMinitaboutputisshownbelow:TheregressionequationisTime=4.62-1.60PersonPredictorCoefSECoefTPConstant4.62000.319214.470.000Person-1.60000.4514-3.540.008S=0.7138R-Sq=61.1%R-Sq(adj)=56.2%AnalysisofVarianceSourceDFSSMSFPRegression16.40006.400012.560.008ResidualError84.07600.5095Total910.4760d.Weseethat61.1%ofthevariabilityinrepairtimehasbeenexplainedbytherepairpersonthatperformedtheservice;anacceptable,butnotgood,fit.36.a.TheMinitaboutputisshownbelow:TheregressionequationisTime=1.86+0.291Months+1.10Type-0.609PersonPredictorCoefSECoefTPConstant1.86020.72862.550.043Months0.291440.083603.490.013Type1.10240.30333.630.011Person-0.60910.3879-1.570.167S=0.4174R-Sq=90.0%R-Sq(adj)=85.0%15-www.khdaw.com 课后答案网www.khdaw.comChapter15AnalysisofVarianceSourceDFSSMSFPRegression39.43053.143518.040.002ResidualError61.04550.1743Total910.4760b.Sincethep-valuecorrespondingtoF=18.04is.002<α=.05,theoverallmodelisstatisticallysignificant.c.Thep-valuecorrespondingtot=-1.57is.167>α=.05;thus,theadditionofPersonisnotstatisticallysignificant.PersonishighlycorrelatedwithMonths(thesamplecorrelationcoefficientis-.691);thus,oncetheeffectofMonthshasbeenaccountedfor,Personwillnotaddmuchtothemodel.www.khdaw.com37.a.LetPosition=0ifaguardPosition=1ifanoffensivetackle.b.TheMinitaboutputisshownbelow:TheregressionequationisRating=11.2+0.732Position+0.0222Weight-2.28SpeedPredictorCoefSECoefTPConstant11.2234.5232.480.022Position0.73240.28932.530.019Weight0.022190.010392.140.045Speed-2.27750.9290-2.450.023S=0.6936R-Sq=47.5%R-Sq(adj)=40.1%AnalysisofVarianceSourceDFSSMSFPRegression39.15623.05216.350.003ResidualError2110.10140.4810Total2419.2576c.Sincethep-valuecorrespondingtoF=6.35is.003<α=.05,thereisasignificantrelationshipbetweenratingandtheindependentvariables.d.ThevalueofR-Sq(adj)is40.1%;theestimatedregressionequationdidnotprovideaverygoodfit.e.Sincethep-valueforPositionist=2.53<α=.05,positionisasignificantfactorintheplayer’srating.f.yˆ=11.2.732(1).0222(300)2.28(5.1)++−=6.9615-www.khdaw.com 课后答案网www.khdaw.comMultipleRegression38.a.TheMinitaboutputisshownbelow:TheregressionequationisRisk=-91.8+1.08Age+0.252Pressure+8.74SmokerPredictorCoefSECoefTPConstant-91.7615.22-6.030.000Age1.07670.16606.490.000Pressure0.251810.045235.570.000Smoker8.7403.0012.910.010S=5.757R-Sq=87.3%R-Sq(adj)=85.0%AnalysisofVarianceSourceDFSSMSFPwww.khdaw.comRegression33660.71220.236.820.000ResidualError16530.233.1Total194190.9b.Sincethep-valuecorrespondingtot=2.91is.010<α=.05,smokingisasignificantfactor.c.UsingMinitab,thepointestimateis34.27;the95%predictionintervalis21.35to47.18.Thus,theprobabilityofastroke(.2135to.4718atthe95%confidencelevel)appearstobequitehigh.ThephysicianwouldprobablyrecommendthatArtquitsmokingandbeginsometypeoftreatmentdesignedtoreducehisbloodpressure.39.a.TheMinitaboutputisshownbelow:TheregressionequationisY=0.20+2.60XPredictorCoefSECoefTPConstant0.2002.1320.090.931X2.60000.64294.040.027S=2.033R-Sq=84.5%R-Sq(adj)=79.3%AnalysisofVarianceSourceDFSSMSFPRegression167.60067.60016.350.027ResidualError312.4004.133Total480.000b.UsingMinitabweobtainedthefollowingvalues:StandardizedxiyiyˆiResidual132.8.16275.4.94358.0-1.6541110.6.2451413.2.6215-www.khdaw.com 课后答案网www.khdaw.comChapter15Thepoint(3,5)doesnotappeartofollowthetrendofremainingdata;however,thevalueofthestandardizedresidualforthispoint,-1.65,isnotlargeenoughforustoconcludethat(3,5)isanoutlier.c.UsingMinitab,weobtainedthefollowingvalues:StudentizedxiyiDeletedResidual13.1327.9135-4.42411.19514.54www.khdaw.comt.025=4.303(n-p-2=5-1-2=2degreesoffreedom)Sincethestudentizeddeletedresidualfor(3,5)is-4.42<-4.303,weconcludethatthe3rdobservationisanoutlier.40.a.TheMinitaboutputisshownbelow:TheregressionequationisY=-53.3+3.11XPredicatorCoefStdevt-ratiopConstant-53.2805.786-9.210.003X3.11000.201615.430.001s=2.851R-sq=98.8%R-sq(adj)=98.3%AnalysisofVarianceSOURCEDFSSMSFpRegression11934.41934.4238.030.001Error324.48.1Total41598.8b.UsingtheMinitabweobtainedthefollowingvalues:StudentizedxiyiDeletedResidual2212-1.942421-.1226311.792835.404070-1.90t.025=4.303(n-p-2=5-1-2=2degreesoffreedom)Sincenoneofthestudentizeddeletedresidualsarelessthan-4.303orgreaterthan4.303,noneoftheobservationscanbeclassifiedasanoutlier.15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionc.UsingMinitabweobtainedthefollowingvalues:xiyihi2212.382421.282631.222835.204070.92Thecriticalvalueis3(p+1)3(11)+==1.2n5www.khdaw.comSincenoneofthevaluesexceed1.2,weconcludethattherearenoinfluentialobservationsinthedata.d.UsingMinitabweobtainedthefollowingvalues:xiyiDi2212.602421.002631.262835.03407011.09SinceD5=11.09>1(ruleofthumbcriticalvalue),weconcludethatthefifthobservationisinfluential.41.a.TheMinitaboutputappearsinthesolutiontopart(b)ofExercise5;theestimatedregressionequationis:Revenue=83.2+2.29TVAdv+1.30NewsAdvb.UsingMinitabweobtainedthefollowingvalues:StandardizedyˆResiduali96.63-1.6290.41-1.0894.341.2292.21-.3794.391.1094.24-.4094.42-1.1293.351.08Withtherelativelyfewobservations,itisdifficulttodetermineifanyoftheassumptionsregardingtheerrortermhavebeenviolated.Forinstance,anargumentcouldbemadethattheredoesnotappeartobeanypatternintheplot;alternativelyanargumentcouldbemadethatthereisacurvilinearpatternintheplot.15-www.khdaw.com 课后答案网www.khdaw.comChapter15c.Thevaluesofthestandardizedresidualsaregreaterthan-2andlessthan+2;thus,usingtest,therearenooutliers.Asafurthercheckforoutliers,weusedMinitabtocomputethefollowingstudentizeddeletedresiduals:StudentizedObservationDeletedResidual1-2.112-1.1031.314-.3351.136-.367-1.1681.10www.khdaw.comt.025=2.776(n-p-2=8-2-2=4degreesoffreedom)Sincenoneofthestudentizeddeletedresidualsislesstan-2.776orgreaterthan2.776,weconcludethattherearenooutliersinthedata.d.UsingMinitabweobtainedthefollowingvalues:ObservationhiDi1.631.522.65.703.30.224.23.015.26.146.14.017.66.818.13.06Thecriticalaveragevalueis3(p+1)3(21)+==1.125n8Sincenoneofthevaluesexceed1.125,weconcludethattherearenoinfluentialobservations.However,usingCook’sdistancemeasure,weseethatD1>1(ruleofthumbcriticalvalue);thus,weconcludethefirstobservationisinfluential.FinalConclusion:observations1isaninfluentialobservation.42.a.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=71.3+0.107Price+0.0845HorsepwrPredictorCoefSECoefTPConstant71.3282.24831.730.000Price0.107190.039182.740.017Horsepwr0.0844960.0093069.080.000S=2.485R-Sq=91.9%R-Sq(adj)=90.7%15-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionAnalysisofVarianceSourceDFSSMSFPRegression2915.66457.8374.120.000ResidualError1380.306.18Total15995.95SourceDFSeqSSPrice1406.39Horsepwr1509.27UnusualObservationsObsPriceSpeedFitSEFitResidualStResid293.8108.000105.8822.0072.1181.45Xwww.khdaw.comXdenotesanobservationwhoseXvaluegivesitlargeinfluence.b.Thestandardizedresidualplotisshownbelow.Thereappearstobeaveryunusualtrendinthestandardizedresiduals.--xxx1.2+-xSRES1-x-x-x0.0+xx-x-xxx---1.2+x--x-x---------+---------+---------+---------+---------+--------FITS190.096.0102.0108.0114.0c.TheMinitaboutputshowninpart(a)didnotidentifyanyobservationswithalargestandardizedresidual;thus,theredoesnotappeartobeanyoutliersinthedata.d.TheMinitaboutputshowninpart(a)identifiesobservation2asaninfluentialobservation.43.a.TheMinitaboutputisshownbelow:Theregressionequationis%College=-26.6+0.0970SatScorePredictorCoefSECoefTPConstant-26.6137.22-0.720.485SatScore0.097030.037342.600.01915-www.khdaw.com 课后答案网www.khdaw.comChapter15S=12.83R-Sq=29.7%R-Sq(adj)=25.3%AnalysisofVarianceSourceDFSSMSFPRegression11110.81110.86.750.019ResidualError162632.3164.5Total173743.1UnusualObservationsObsSatScore%CollegeFitSEFitResidualStResid371640.0042.8610.79-2.86-0.41XXdenotesanobservationwhoseXvaluegivesitlargeinfluence.b.TheMinitaboutputshowninpartaidentifiesobservation3asaninfluentialobservation.c.TheMinitaboutputappearsinthesolutiontoExercise9;theestimatesregressionequationiswww.khdaw.com%College=26.7-1.43Size+0.0757SATScored.ThefollowingMinitaboutputwasalsoprovidedaspartoftheregressionoutputforpartc.UnusualObservationsObs.Size%CollegeFitStdev.FitResidualSt.Resid330.040.038.0410.971.960.34XXdenotesanobs.whoseXvaluegivesitlargeinfluence.Observation3isstillidentifiedasaninfluentialobservation.44.a.Theexpectedincreaseinfinalcollegegradepointaveragecorrespondingtoaonepointincreaseinhighschoolgradepointaverageis.0235whenSATmathematicsscoredoesnotchange.Similarly,theexpectedincreaseinfinalcollegegradepointaveragecorrespondingtoaonepointincreaseintheSATmathematicsscoreis.00486whenthehighschoolgradepointaveragedoesnotchange.b.yˆ=-1.41+.0235(84)+.00486(540)=3.1945.a.Jobsatisfactioncanbeexpectedtodecreaseby8.69unitswithaoneunitincreaseinlengthofserviceifthewageratedoesnotchange.Adollarincreaseinthewagerateisassociatedwitha13.5pointincreaseinthejobsatisfactionscorewhenthelengthofservicedoesnotchange.b.yˆ=14.4-8.69(4)+13.5(6.5)=67.3946.a.Thecomputeroutputwiththemissingvaluesfilledinisasfollows:TheregressionequationisY=8.103+7.602X1+3.111X2PredicatorCoefStdevt-ratioConstant8.1032.6673.04X17.6022.1053.61X23.1110.6135.0815-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressions=3.35R-sq=92.3%R-sq(adj)=91.0%AnalysisofVarianceSOURCEDFSSMSFRegression2161280671.82Error12134,6711.2225Total141746.67b.t.025=2.179(12DF)forβ1:3.61>2.179;rejectH0:β1=0forβ2:5.08>2.179;rejectH0:β2=0c.Seecomputeroutput.www.khdaw.com214d.R=−1(1.923)−=.91a1247.a.TheregressionequationisY=-1.41+.0235X1+.00486X2PredictorCoefStdevt-ratioConstant-1.40530.4848-2.90X10.0234670.0086662.71X2.004860.0010774.51s=0.1298R-sq=93.7%R-sq(adj)=91.9%AnalysisofVarianceSOURCEDFSSMSFRegression21.76209.88152.44Error7.1179.0168Total91.88000b.F.05=4.74(2DFnumerator,7DFdenominator)F=52.44>F.05;significantrelationship.2SSRc.R==.937SST29R=−1(1.937)−=.919a7goodfitd.t.025=2.365(7DF)forB1:t=2.71>2.365;rejectH0:B1=015-www.khdaw.com 课后答案网www.khdaw.comChapter15forB2:t=4.51>2.365;rejectH0:B2=048.a.TheregressionequationisY=14.4-8.69X1+13.52X2PredictorCoefStdevt-ratioConstant14.4488.1911.76X1-8.691.555-5.59X213.5172.0856.48s=3.773R-sq=90.1%R-sq(adj)=86.1%AnalysisofVarianceSOURCEDFSSMSFRegression2648.83324.41522.79www.khdaw.comError571.1714.234Total7720.00b.F.05=5.79(5DF)F=22.79>F.05;significantrelationship.2SSRc.R==.901SST27R=−1(1.901)−=.861a5goodfitd.t.025=2.571(5DF)forβ1:t=-5.59<-2.571;rejectH0:β1=0forβ2:t=6.48>2.571;rejectH0:β2=049.a.TheMinitaboutputisshownbelow:TheregressionequationisPrice=12.8+2.26BookValPredictorCoefSECoefTPConstant12.7936.6241.930.064BookVal2.26490.66313.420.002S=19.50R-Sq=29.4%R-Sq(adj)=26.9%AnalysisofVarianceSourceDFSSMSFPRegression14433.94433.911.670.002Error2810642.3380.115-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionTotal2915076.1b.ThevalueofR-sqis29.4%;theestimatedregressionequationdoesnotprovideagoodfit.c.TheMinitaboutputisshownbelow:TheregressionequationisPrice=5.88+2.54BookVal+0.484ReturnEqPredictorCoefSECoefTPConstant5.8775.5451.060.299BookVal2.53560.53314.760.000ReturnEq0.48410.11744.120.000S=15.55R-Sq=56.7%R-Sq(adj)=53.5%www.khdaw.comAnalysisofVarianceSourceDFSSMSFPRegression28544.24272.117.660.000Error276531.9241.9Total2915076.1Sincethep-valuecorrespondingtotheFtestis0.000,therelationshipissignificant.50.a.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=97.6+0.0693Price-0.00082Weight+0.0590Horsepwr-2.48Zero60PredictorCoefSECoefTPConstant97.5711.798.270.000Price0.069280.038051.820.096Weight-0.0008160.002593-0.310.759Horsepwr0.059010.015433.820.003Zero60-2.48360.9601-2.590.025S=2.127R-Sq=95.0%R-Sq(adj)=93.2%AnalysisofVarianceSourceDFSSMSFPRegression4946.18236.5552.280.000ResidualError1149.774.52Total15995.95b.Sincethep-valuecorrespondingtotheFtestis0.000,therelationshipissignificant.c.Sincethep-valuescorrespondingtothettestforbothHorsepwr(p-value=.003)andZero60(p-value=.025)arelessthan.05,bothoftheseindependentvariablesaresignificant.15-www.khdaw.com 课后答案网www.khdaw.comChapter15d.TheMinitaboutputisshownbelow:TheregressionequationisSpeed=103+0.0558Horsepwr-3.19Zero60PredictorCoefSECoefTPConstant103.1039.44810.910.000Horsepwr0.055820.014523.840.002Zero60-3.18760.9658-3.300.006S=2.301R-Sq=93.1%R-Sq(adj)=92.0%AnalysisofVarianceSourceDFSSMSFPwww.khdaw.comRegression2927.12463.5687.540.000ResidualError1368.845.30Total15995.95SourceDFSeqSSHorsepwr1869.43Zero60157.68UnusualObservationsObsHorsepwrSpeedFitSEFitResidualStResid2290108.000103.3521.0154.6482.25R1215584.60082.7471.7731.8531.26XRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.e.Thestandardizedresidualplotisshownbelow:-SRES-x--1.5+-xx---2xx0.0+xx2-x--xx--1.5+-xx-----+---------+---------+---------+---------+---------+--FIT84.090.096.0102.0108.0114.015-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionThereisanunusualtrendintheplotandoneobservationappearstobeanoutlier.f.TheMinitaboutputindicatesthatobservation2isanoutlierg.TheMinitaboutputindicatesthatobservation12isaninfluentialobservation.51.a.TheMinitaboutputisshownbelow:640+-xExposure---480+-x-www.khdaw.com--x320+----x160+x3x-x------+---------+---------+---------+---------+---------+TimesAir153045607590b.TheMinitaboutputisshownbelow:TheregressionequationisExposure=53.2+6.74TimesAirPredictorCoefSECoefTPConstant53.2416.533.220.012TimesAir6.74270.447215.080.000S=31.70R-Sq=96.6%R-Sq(adj)=96.2%AnalysisofVarianceSourceDFSSMSFPRegression1228520228520227.360.000Error880411005Total9236561Sincethep-valueis0.000,therelationshipissignificant.c.TheMinitaboutputisshownbelow:TheregressionequationisExposure=73.1+5.04TimesAir+101BigAds15-www.khdaw.com 课后答案网www.khdaw.comChapter15PredictorCoefSECoefTPConstant73.0637.5079.730.000TimesAir5.03680.326815.410.000BigAds101.1115.996.320.000S=13.08R-Sq=99.5%R-Sq(adj)=99.3%AnalysisofVarianceSourceDFSSMSFPRegression2235363117682687.840.000Error71198171Total9236561www.khdaw.comd.Thep-valuecorrespondingtothettestforBigAdsis0.000;thus,thedummyvariableissignificant.e.Thedummyvariableenablesustofittwodifferentlinestothedata;thisapproachisreferredtoaspiecewiselinearapproximation.52.a.TheMinitaboutputisshownbelow:Resale%=38.8+0.000766PricePredictorCoefSECoefTPConstant38.7724.3488.920.000Price0.00076560.00019004.030.000S=5.421R-Sq=36.7%R-Sq(adj)=34.4%AnalysisofVarianceSourceDFSSMSFPRegression1477.25477.2516.240.000ResidualError28822.9229.39Total291300.17Sincethep-valuecorrespondingtoF=16.24is.000<α=.05,thereisasignificantrelationshipbetweenResale%andPrice.b.R-Sq=36.7%;notaverygoodfit.c.LetType1=0andType2=0ifasmallpickup;Type1=1andType2=0ifafull-sizepickup;andType1=0andType2=1ifasportutility.TheMinitaboutputusingType1,Type2,andPriceisshownbelow:TheregressionequationisResale%=42.6+9.09Type1+7.92Type2+0.000341PricePredictorCoefSECoefTPConstant42.5543.56211.950.000Type19.0902.2484.040.000Type27.9172.1633.660.001Price0.00034150.00018001.900.06915-www.khdaw.com 课后答案网www.khdaw.comMultipleRegressionS=4.298R-Sq=63.1%R-Sq(adj)=58.8%AnalysisofVarianceSourceDFSSMSFPRegression3819.77273.2614.790.000ResidualError26480.4018.48Total291300.17d.Sincethep-valuecorrespondingtoF=14.79is.000<α=.05,thereisasignificantrelationshipbetweenResale%andtheindependentvariables.Notethatindividually,Priceisnotsignificantatthe.05levelofsignificance.IfwereruntheregressionusingjustType1andType2thevalueofR-Sq(adj)decreasesto54.4%,adropofonly4%.Thus,itappearsthatforthesedata,thetypeofvehicleisthestrongestpredictoroftheresalevalue.www.khdaw.com15-www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter16RegressionRRegressionegressionAnalysis:AAnalysis:nalysis:ModelMModelodelBuildingBBuildinguildingLearningLLearningearningObjectivesOObjectivesbjectives1.Learnhowthegenerallinearmodelcanbeusedtomodelproblemsinvolvingcurvilinearrelationships.2.Understandtheconceptofinteractionandhowitcanbeaccountedforinthegenerallinearmodel.3.UnderstandhowanFtestcanbeusedtodeterminewhentoaddordeleteoneormorevariables.www.khdaw.com4.Developanappreciationforthecomplexitiesinvolvedinsolvinglargerregressionanalysisproblems.5.Understandhowvariableselectionprocedurescanbeusedtochooseasetofindependentvariablesforanestimatedregressionequation.6.KnowhowtheDurban-Watsontestcanbeusedtotestforautocorrelation.7.Learnhowanalysisofvarianceandexperimentaldesignproblemscanbeanalyzedusingaregressionmodel.Solutions:SSolutions:olutions:16-1www.khdaw.com 课后答案网www.khdaw.comChapter161.a.TheMinitaboutputisshownbelow:TheregressionequationisY=-6.8+1.23XPredictorCoefStdevt-ratiopConstant-6.7714.17-0.480.658X1.22960.46972.620.059s=7.269R-sq=63.1%R-sq(adj)=53.9%AnalysisofVarianceSOURCEDFSSMSFpRegression1362.13362.136.850.059Error4211.3752.84www.khdaw.comTotal5573.50b.Sincethep-valuecorrespondingtoF=6.85is0.59>α=.05,therelationshipisnotsignificant.c.-40+*-Y-**-*-30+----*20+----*10+------+---------+---------+---------+---------+---------+X20.025.030.035.040.045.0Thescatterdiagramsuggeststhatacurvilinearrelationshipmaybeappropriate.d.TheMinitaboutputisshownbelow:TheregressionequationisY=-169+12.2X-0.177XSQPredictorCoefStdevt-ratiopConstant-168.8839.79-4.240.024X12.1872.6634.580.020XSQ-0.177040.04290-4.130.026s=3.248R-sq=94.5%R-sq(adj)=90.8%16-2www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingAnalysisofVarianceSOURCEDFSSMSFpRegression2541.85270.9225.680.013Error331.6510.55Total5573.50e.Sincethep-valuecorrespondingtoF=25.68is.013<α=.05,therelationshipissignificant.f.yˆ=-168.88+12.187(25)-0.17704(25)2=25.1452.a.TheMinitaboutputisshownbelow:TheregressionequationisY=9.32+0.424XPredictorCoefStdevt-ratiopwww.khdaw.comConstant9.3154.1962.220.113X0.42420.19442.180.117s=3.531R-sq=61.4%R-sq(adj)=48.5%AnalysisofVarianceSOURCEDFSSMSFpRegression159.3959.394.760.117Error337.4112.47Total496.80Thehighp-value(.117)indicatesaweakrelationship;notethat61.4%ofthevariabilityinyhasbeenexplainedbyx.b.TheMinitaboutputisshownbelow:TheregressionequationisY=-8.10+2.41X-0.0480XSQPredictorCoefStdevt-ratiopConstant-8.1014.104-1.970.187X2.41270.44095.470.032XSQ-0.047970.01050-4.570.045s=1.279R-sq=96.6%R-sq(adj)=93.2%AnalysisofVarianceSOURCEDFSSMSFpRegression293.52946.76528.600.034Error23.2711.635Total496.800Atthe.05levelofsignificance,therelationshipissignificant;thefitisexcellent.c.yˆ=-8.101+2.4127(20)-0.04797(20)2=20.96516-3www.khdaw.com 课后答案网www.khdaw.comChapter163.a.Thescatterdiagramshowssomeevidenceofapossiblelinearrelationship.b.TheMinitaboutputisshownbelow:TheregressionequationisY=2.32+0.637XPredictorCoefStdevt-ratiopConstant2.3221.8871.230.258X0.63660.30442.090.075s=2.054R-sq=38.5%R-sq(adj)=29.7%AnalysisofVarianceSOURCEDFSSMSFpRegression118.46118.4614.370.075www.khdaw.comError729.5394.220Total848.000c.Thefollowingstandardizedresidualplotindicatesthattheconstantvarianceassumptionisnotsatisfied.--*-1.2+*---*-**0.0+--**---1.2+-**--+---------+---------+---------+---------+---------+------YHAT3.04.05.06.07.08.0d.Thelogarithmictransformationdoesnotappeartoeliminatethewedged-shapedpatternintheaboveresidualplot.Thereciprocaltransformationdoes,however,removethewedge-shapedpattern.Neithertransformationprovidesagoodfit.TheMinitaboutputforthereciprocaltransformationandthecorrespondingstandardizedresidualpotareshownbelow.Theregressionequationis1/Y=0.275-0.0152XPredictorCoefStdevt-ratiopConstant0.274980.046015.980.000X-0.0151820.007421-2.050.080s=0.05009R-sq=37.4%R-sq(adj)=28.5%16-4www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingAnalysisofVarianceSOURCEDFSSMSFpRegression10.0105010.0105014.190.080Error70.0175630.002509Total80.028064-*---1.0+*-*---*www.khdaw.com0.0+*---**--1.0+-**---+---------+---------+---------+---------+---------+----YHAT0.1400.1600.1800.2000.2200.2404.a.TheMinitaboutputisshownbelow:TheregressionequationisY=943+8.71XPredictorCoefStdevt-ratiopConstant943.0559.3815.880.000X8.7141.5445.640.005s=32.29R-sq=88.8%R-sq(adj)=86.1%AnalysisofVarianceSOURCEDFSSMSFpRegression1332233322331.860.005Error441721043Total537395b.Thep-valueof.005<α=.01;rejectH016-5www.khdaw.com 课后答案网www.khdaw.comChapter165.TheMinitaboutputisshownbelow:TheregressionequationisY=433+37.4X-0.3831/YPredictorCoefStdevt-ratiopConstant432.6141.23.060.055X37.4297.8074.790.0171/Y-0.38290.1036-3.700.034s=15.83R-sq=98.0%R-sq(adj)=96.7%AnalysisofVarianceSOURCEDFSSMSFpwww.khdaw.comRegression2366431832273.150.003Error3751250Total537395b.Sincethelinearrelationshipwassignificant(Exercise4),thisrelationshipmustbesignificant.Notealsothatsincethep-valueof.005<α=.05,wecanrejectH0.c.Thefittedvalueis1302.01,withastandarddeviationof9.93.The95%confidenceintervalis1270.41to1333.61;the95%predictionintervalis1242.55to1361.47.6.a.Thescatterdiagramisshownbelow:-*1.60+-DISTANCE---1.20+-*---0.80+*--*--**0.40++---------+---------+---------+---------+---------+------NUMBER8.012.016.020.024.028.0b.No;therelationshipappearstobecurvilinear.c.Severalpossiblemodelscanbefittedtothesedata,asshownbelow:yˆ=2.90-0.185x+.00351x2R2=.91a16-6www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuilding⎛1⎞2yˆ=−0.046814.4+⎜⎟R=.91a⎝x⎠7.a.TheMinitaboutputisshownbelow:-36+x-Shipment---24+-x--x-xwww.khdaw.com12+--xx-2xx-0++---------+---------+---------+---------+---------+-----Media$0255075100125b.TheMinitaboutputisshownbelow:TheregressionequationisShipment=4.09+0.196Media$PredictorCoefSECoefTPConstant4.0892.1681.890.096Media$0.195520.036355.380.000S=5.044R-Sq=78.3%R-Sq(adj)=75.6%AnalysisofVarianceSourceDFSSMSFPRegression1735.84735.8428.930.000Error8203.5125.44Total9939.35UnusualObservationsObsMedia$ShipmentFitStDevFitResidualStResid112036.3027.553.308.752.30RRdenotesanobservationwithalargestandardizedresidualSimplelinearregressionappearstodogoodjobinexplainingthevariabilityinshipments.However,thescatterdiagraminpart(a)indicatesthatacurvilinearrelationshipmaybemoreappropriate.16-7www.khdaw.com 课后答案网www.khdaw.comChapter16c.TheMinitaboutputisshownbelow:TheregressionequationisShipment=5.51+0.00182Media$SqPredictorCoefSECoefTPConstant5.5061.6863.270.011Media$Sq0.00182250.00027926.530.000S=4.308R-Sq=84.2%R-Sq(adj)=82.2%AnalysisofVarianceSourceDFSSMSFPRegression1790.88790.8842.620.000Error8148.4718.56www.khdaw.comTotal9939.35UnusualObservationsObsMedia$SqShipmentFitStDevFitResidualStResid31002015.9023.772.26-7.87-2.15RRdenotesanobservationwithalargestandardizedresidual8.a.Thescatterdiagramisshownbelow:807060504030202005ProjectedUsers(%)100010203040501999InternetUsers(%)Itappearsthatasimplelinearregressionmodelisnotappropriatebecausethereiscurvatureintheplot.16-8www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingb.TheMinitaboutputisshownbelow:Theregressionequationis2005%=17.1+3.151999%-0.04451999%SqPredictorCoefSECoefTPConstant17.0994.6393.690.0031999%3.14620.49716.330.0001999%Sq-0.044540.01018-4.370.001S=5.667R-Sq=89.7%R-Sq(adj)=88.1%AnalysisofVarianceSourceDFSSMSFPRegression23646.31823.256.780.000www.khdaw.comResidualError13417.432.1Total154063.8c.TheMinitaboutputisshownbelow:TheregressionequationisLog2000%=1.17+0.449Log1999%PredictorCoefSECoefTPConstant1.174200.0746815.720.000Log1999%0.448950.059787.510.000S=0.08011R-Sq=80.1%R-Sq(adj)=78.7%AnalysisofVarianceSourceDFSSMSFPRegression10.361990.3619956.400.000ResidualError140.089850.00642Total150.45184d.Theestimatedregressioninpart(b)ispreferredbecauseitexplainsahigherpercentageofthevariabilityinthedependentvariable.9.a.TheMinitaboutputisshownbelow:TheregressionequationisDistance=268+1.52Wind-0.0177WindSqPredictorCoefSECoefTPConstant267.8412.358113.600.000Wind1.521430.0771819.710.000WindSq-0.0176980.004456-3.970.001S=7.073R-Sq=95.7%R-Sq(adj)=95.3%16-9www.khdaw.com 课后答案网www.khdaw.comChapter16AnalysisofVarianceSourceDFSSMSFPRegression22023310117202.200.000Error1890150Total2021134b.Theestimateddistanceis267.841+1.52143(-15)-0.017698(15)2=241.04.c.Theestimateddistanceis267.841+1.52143(25)-0.017698(25)2=269.34.10.a.SSR=SST-SSE=1030MSR=1030MSE=520/25=20.8F=1030/20.8=49.52www.khdaw.comF.05=4.24(25DF)Since49.52>4.24werejectH0:β1=0andconcludethatx1issignificant.(520100)/2−b.F==48.3100/23F.05=3.42(2degreesoffreedomnumeratorand23denominator)Since48.3>3.42theadditionofvariablesx2andx3isstatisticallysignificant11.a.SSE=SST-SSR=1805-1760=45MSR=1760/4=440MSE=45/25=1.8F=440/1.8=244.44F.05=2.76(4degreesoffreedomnumeratorand25denominator)Since244.44>2.76,variablesx1andx4contributesignificantlytothemodelb.SSE(x1,x2,x3,x4)=45c.SSE(x2,x3)=1805-1705=100(10045)/2−d.F==15.281.8F.05=3.39(2numeratorand25denominatorDF)Since15.28>3.39weconcludethatx1andx3contributesignificantlytothemodel.12.a.TheMinitaboutputisshownbelow.TheregressionequationisPoints=170+6.61TeamIntPredictorCoefSECoefTP16-10www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingConstant170.1344.023.860.002TeamInt6.6132.2582.930.013S=43.93R-Sq=41.7%R-Sq(adj)=36.8%AnalysisofVarianceSourceDFSSMSFPRegression116546165468.570.013ResidualError12231571930Total1339703UnusualObservationsObsTeamIntPointsFitSEFitResidualStResid1333.0340.0388.434.2-48.4-1.75XXdenotesanobservationwhoseXvaluegivesitlargeinfluence.www.khdaw.comb.TheMinitaboutputisshownbelow.TheregressionequationisPoints=280+5.18TeamInt-0.0037Rushing-3.92OpponIntPredictorCoefSECoefTPConstant280.3481.423.440.006TeamInt5.1762.0732.500.032Rushing-0.003730.03336-0.110.913OpponInt-3.9181.651-2.370.039S=37.84R-Sq=63.9%R-Sq(adj)=53.1%AnalysisofVarianceSourceDFSSMSFPRegression32538684625.910.014ResidualError10143171432Total1339703SourceDFSeqSSTeamInt116546Rushing1776OpponInt18064SSE(reduced)-SSE(full)23,15714,317−#extraterms2c.F===3.09MSE(full)1432F.05=4.10(2numeratorand10denominatorDF)Since3.09<4.10theadditionofthetwoindependentvariablesisnotsignificant.Note:Supposethatweconsideredaddingonlythenumberofinterceptionsmadebytheopponents;thecorrespondingMinitaboutputisshownbelow:Points=274+5.23TeamInt-3.96OpponIntPredictorCoefSECoefTP16-11www.khdaw.com 课后答案网www.khdaw.comChapter16Constant273.7753.815.090.000TeamInt5.2271.9312.710.020OpponInt-3.9651.524-2.600.025S=36.10R-Sq=63.9%R-Sq(adj)=57.3%AnalysisofVarianceSourceDFSSMSFPRegression225368126849.730.004ResidualError11143351303Total1339703SourceDFSeqSSTeamInt116546OpponInt18822www.khdaw.comUnusualObservationsObsTeamIntPointsFitSEFitResidualStResid1017.0312.00247.6416.8564.362.02RRdenotesanobservationwithalargestandardizedresidualInthiscase,23,15714,335−1F==6.771303F.05=4.84(1numeratorand11denominatorDF)Since6.77>4.84theadditionofthenumberofinterceptionsmadebytheopponentsissignificant.13.a.TheMinitaboutputisshownbelow:Points=218+0.0252Passing+4.39TeamInt-4.38OpponIntPredictorCoefSECoefTPConstant218.3869.073.160.010Passing0.025200.020391.240.245TeamInt4.3872.0052.190.053OpponInt-4.3761.525-2.870.017S=35.26R-Sq=68.7%R-Sq(adj)=59.3%AnalysisofVarianceSourceDFSSMSFPRegression32726990907.310.007ResidualError10124351243Total1339703SourceDFSeqSSPassing13416TeamInt113617OpponInt11023516-12www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingUnusualObservationsObsPassingPointsFitSEFitResidualStResid103247312.00247.8916.4664.112.06RRdenotesanobservationwithalargestandardizedresidualb.TheMinitaboutputisshownbelow:Points=235+0.0266Passing+4.18TeamInt-4.26OpponInt-0.0115RushingPredictorCoefSECoefTPConstant235.4087.522.690.025Passing0.026630.021741.220.252www.khdaw.comTeamInt4.1852.1801.920.087OpponInt-4.2561.635-2.600.029Rushing-0.011450.03316-0.350.738S=36.93R-Sq=69.1%R-Sq(adj)=55.4%AnalysisofVarianceSourceDFSSMSFPRegression42743168585.030.021ResidualError9122721364Total1339703SourceDFSeqSSPassing13416TeamInt113617OpponInt110235Rushing1163SSE(reduced)-SSE(full)12,43512,272−#extraterms1c.F===.1195MSE(full)1364F.05=5.12(1numeratorand9denominatorDF)Since.1195<5.12theadditionofRushingisnotsignificant.Note:Sinceonly1variablewasaddedtothemodelinpart(a),thetestcanalsobeperformedusingthet-ratioforRushingintheMinitaboutput.14.a.TheMinitaboutputisshownbelow:Risk=-111+1.32Age+0.296PressurePredictorCoefSECoefTPConstant-110.9416.47-6.740.000Age1.31500.17337.590.000Pressure0.296400.051075.800.00016-13www.khdaw.com 课后答案网www.khdaw.comChapter16S=6.908R-Sq=80.6%R-Sq(adj)=78.4%AnalysisofVarianceSourceDFSSMSFPRegression23379.61689.835.410.000ResidualError17811.347.7Total194190.9SourceDFSeqSSAge11772.0Pressure11607.7UnusualObservationsObsAgeRiskFitSEFitResidualStResidwww.khdaw.com1766.08.0025.051.67-17.05-2.54RRdenotesanobservationwithalargestandardizedresidualb.TheMinitaboutputisshownbelow:Risk=-123+1.51Age+0.448Pressure+8.87Smoker-0.00276AgePressPredictorCoefSECoefTPConstant-123.1656.94-2.160.047Age1.51300.77961.940.071Pressure0.44830.34571.300.214Smoker8.8663.0742.880.011AgePress-0.0027560.004807-0.570.575S=5.881R-Sq=87.6%R-Sq(adj)=84.3%AnalysisofVarianceSourceDFSSMSFPRegression43672.11918.0326.540.000ResidualError15518.8434.59Total194190.95SourceDFSeqSSAge11771.98Pressure11607.66Smoker1281.10AgePress111.37UnusualObservationsObsAgeRiskFitSEFitResidualStResid1766.08.0020.912.01-12.91-2.34RRdenotesanobservationwithalargestandardizedresidualSSE(reduced)-SSE(full)811.3518.84−#extraterms2c.F===4.23MSE(full)34.5916-14www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingF.05=3.68(2numeratorand15denominatorDF)Since4.23>3.68theadditionofthetwotermsissignificant.15.a.LetPos1=0andPos2=0ifplayerisaguard;Pos1=1andPos2=0ifplayerisanoffensivetackle;Pos1=0andPos2=1ifplayerisawidereceiver.b.TheMinitaboutputisshownbelow:TheregressionequationisRating=12.5+0.706Pos1+1.92Pos2+0.0242Weight-2.63SpeedPredictorCoefSECoefTPConstant12.4904.2052.970.005Pos10.70550.29232.410.021Pos21.91790.92112.080.045www.khdaw.comWeight0.0241570.0076653.150.003Speed-2.63180.8600-3.060.004S=0.7111R-Sq=52.2%R-Sq(adj)=46.7%AnalysisofVarianceSourceDFSSMSFPRegression419.29994.82509.540.000ResidualError3517.69790.5057Total3936.9978c.Thep-valuecorrespondingtoF=9.54is0.000<α=.05;thus,theestimatedregressionequationissignificant.d.TheMinitaboutputusingonlyWeightandSpeedisshownbelow:TheregressionequationisRating=19.8+0.0183Weight-3.59SpeedPredictorCoefSECoefTPConstant19.8392.8007.090.000Weight0.0182970.0060153.040.004Speed-3.59430.8576-4.190.000S=0.7763R-Sq=39.7%R-Sq(adj)=36.5%AnalysisofVarianceSourceDFSSMSFPRegression214.70167.350812.200.000ResidualError3722.29610.6026Total3936.9978TheFstatisticusedtodetermineiftheadditionofPos1andPos2resultsinasignificantreductionintheerrorsumofsquaresis22.296117.6979−2F==4.55.505716-15www.khdaw.com 课后答案网www.khdaw.comChapter16TheFtablesintheappendixdonotshowavaluefortwonumeratorand35denominatordegreesoffreedom.But,theFvaluefortwonumeratorand30denominatordegreesoffreedomis3.32andtheFvaluefortwonumeratorand40denominatordegreesoffreedomis3.23.Thus,theFvaluefortwonumeratorand35denominatordegreesoffreedomisbetween3.23and3.32.BecausethecomputedFexceedsthisvalue,theadditionofPos1andPos2isstatisticallysignificant.Inotherwords,positionisasignificantfactorintheplayer’srating.16.a.TheMinitaboutputisshownbelow:Theregressionequationis%College=-26.6+0.0970SATScorePredictorCoefSECoefTPConstant-26.6137.22-0.720.485www.khdaw.comSATScore0.097030.037342.600.019S=12.83R-Sq=29.7%R-Sq(adj)=25.3%AnalysisofVarianceSourceDFSSMSFPRegression11110.81110.86.750.019ResidualError162632.3164.5Total173743.1UnusualObservationsObsSATScore%CollegeFitSEFitResidualStResid371640.0042.8610.79-2.86-0.41XXdenotesanobservationwhoseXvaluegivesitlargeinfluence.b.Alpha-to-Enter:0.15Alpha-to-Remove:0.15Responseis%Collegeon5predictors,withN=18Step12Constant-26.61-26.93SATScore0.0970.084T-Value2.602.46P-Value0.0190.026%TakeSAT0.204T-Value2.21P-Value0.043S12.811.5R-Sq29.6846.93R-Sq(adj)25.2839.86C-p6.93.816-16www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingc.Backwardeliminationprocedure:Backwardelimination.Alpha-to-Remove:0.1Responseis%Collegeon5predictors,withN=18Step1234Constant33.7117.46-32.47-26.93Size-1.56-1.39T-Value-1.43-1.42P-Value0.1790.178www.khdaw.comSpending-0.0024-0.0026-0.0019T-Value-1.47-1.75-1.31P-Value0.1680.1040.212Salary-0.00026T-Value-0.40P-Value0.693SATScore0.0770.0810.0950.084T-Value2.062.362.772.46P-Value0.0620.0340.0150.026%TakeSAT0.2850.2740.2910.204T-Value2.472.532.602.21P-Value0.0290.0250.0210.043S11.210.911.211.5R-Sq59.6559.1052.7146.93R-Sq(adj)42.8346.5142.5839.86C-p6.04.24.13.8d.Responseis%CollegeSS%pATeSTanaSkSdlceiiaoSznrrAVarsR-SqR-Sq(adj)C-pSegyeT129.725.36.912.826X125.520.88.213.203X246.939.93.811.508XX238.230.06.412.417XX352.742.64.111.244XXX349.538.75.011.618XXX459.146.54.210.852XXXX16-17www.khdaw.com 课后答案网www.khdaw.comChapter16452.838.36.011.660XXXX559.642.86.011.219XXXXX17.a.Thecorrelationcoefficientsareasfollows:WinsPointsRushingPassingTeamIntPoints-0.6640.010Rushing0.527-0.3180.0530.268Passing0.2060.2930.1330.4790.3090.651TeamInt-0.6710.646-0.2850.290www.khdaw.com0.0090.0130.3240.314OpponInt0.506-0.6310.3120.120-0.2760.0650.0150.2780.6820.340CellContents:PearsoncorrelationP-ValueThevariablemosthighlycorrelatedwithWinsisTeamInt.TheMinitaboutputforthismodelusingTeamInttopredictWinsisshownbelow:TheregressionequationisWins=14.3-0.373TeamIntPredictorCoefSECoefTPConstant14.2942.3186.170.000TeamInt-0.37300.1189-3.140.009S=2.313R-Sq=45.1%R-Sq(adj)=40.5%AnalysisofVarianceSourceDFSSMSFPRegression152.65252.6529.840.009ResidualError1264.2055.350Total13116.857UnusualObservationsObsTeamIntWinsFitSEFitResidualStResid415.04.0008.6980.765-4.698-2.15R1333.04.0001.9831.8002.0171.39XRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.16-18www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingb.Stepwiseregressionprocedure:Alpha-to-Enter:0.15Alpha-to-Remove:0.15ResponseisWinson5predictors,withN=14Step123Constant14.2947.58511.199TeamInt-0.37-0.44-0.28T-Value-3.14-4.14-2.45www.khdaw.comP-Value0.0090.0020.034Passing0.002560.00288T-Value2.273.02P-Value0.0440.013Points-0.026T-Value-2.37P-Value0.040S2.311.991.67R-Sq45.0662.6376.04R-Sq(adj)40.4855.8368.85C-p11.36.53.3c.Backwardeliminationprocedure:ResponseisWinson5predictors,withN=14Step123Constant8.0727.82711.199Points-0.024-0.023-0.026T-Value-1.54-2.08-2.37P-Value0.1630.0670.040Rushing0.00180.0018T-Value1.131.19P-Value0.2920.263Passing0.002610.002570.00288T-Value2.362.653.02P-Value0.0460.0260.013TeamInt-0.26-0.26-0.28T-Value-2.11-2.30-2.45P-Value0.0680.0470.034OpponInt-0.01T-Value-0.08P-Value0.93916-19www.khdaw.com 课后答案网www.khdaw.comChapter16S1.741.641.67R-Sq79.3379.3176.04R-Sq(adj)66.4170.1268.85C-p6.04.03.3d.ResponseisWinsORPTpPuaepossaoihsmnwww.khdaw.comniiIItnnnnVarsR-SqR-Sq(adj)C-pSsggtt145.140.511.32.3131X144.039.411.72.3344X262.655.86.51.9925XX261.654.76.82.0187XX376.068.93.31.6732XXX369.360.15.91.8932XXX479.370.14.01.6389XXXX476.065.45.31.7637XXXX579.366.46.01.7377XXXXX18.a.TheMinitaboutputisshownbelow:TheregressionequationisScoreAvg=81.4-0.147Green%PredictorCoefSECoefTPConstant81.4241.93042.200.000Green%-0.146810.02775-5.290.000S=0.4380R-Sq=60.9%R-Sq(adj)=58.7%AnalysisofVarianceSourceDFSSMSFPRegression15.37015.370127.990.000Error183.45370.1919Total198.8238UnusualObservationsObsGreen%ScoreAvgFitStDevFitResidualStResid378.670.350069.88490.27230.46511.36X1969.172.140071.27970.09840.86032.02RRdenotesanobservationwithalargestandardizedresidualXdenotesanobservationwhoseXvaluegivesitlargeinfluence.16-20www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingb.TheMinitaboutputisshownbelow:TheregressionequationisScoreAvg=58.2-0.00996Distance-0.152Green%+0.869PuttsPredictorCoefSECoefTPConstant58.1986.6448.760.000Distance-0.0099560.009111-1.090.291Green%-0.151860.02437-6.230.000Putts0.86860.22513.860.001www.khdaw.comS=0.3306R-Sq=80.2%R-Sq(adj)=76.5%AnalysisofVarianceSourceDFSSMSFPRegression37.07552.358521.580.000Error161.74830.1093Total198.8238UnusualObservationsObsDistanceScoreAvgFitStDevFitResidualStResid325670.350069.73690.21040.61312.40RRdenotesanobservationwithalargestandardizedresidualc.Theestimatedregressionequationappearstobereasonable;thatis,increasingDistanceandGreen%lowerstheaveragescore,whereasincreasingPuttsincreasestheaveragescore.d.Estimatedaveragescore=58.198-0.009956(231.6)-0.15186(65.2)+0.8686(30.69)=72.65.19.LetHealth=1ifhealth-drugsHealth=0ifenergy-internationalorotherTheregressionequationisP/E=10.8+0.430Sales%+10.6HealthPredictorCoefSECoefTPConstant10.8173.1433.440.004Sales%0.42970.18132.370.034Health10.6002.7503.850.002S=5.012R-Sq=55.4%R-Sq(adj)=48.5%AnalysisofVarianceSourceDFSSMSFPRegression2405.69202.858.070.005Error13326.5925.1216-21www.khdaw.com 课后答案网www.khdaw.comChapter16Total15732.28SourceDFSeqSSSales%132.36Health1373.3320.SeethesolutiontoExercise14inthischapter.TheMinitaboutputusingthebestsubsetsregressionprocedureisshownbelow:ResponseisRiskPArgeSesmPsorAukewww.khdaw.comgresVarsR-SqR-Sq(adj)C-pSeers163.361.328.59.2430X146.343.349.111.182X280.678.49.56.9083XX279.577.110.87.1058XX387.385.03.35.7566XXX386.283.74.76.0051XXX487.684.35.05.8813XXXXThisoutputsuggeststhatthemodelinvolvingAge,Pressure,andSmokeristhepreferredmodel;theMinitaboutputforthismodelisshownbelow:Risk=-91.8+1.08Age+0.252Pressure+8.74SmokerPredictorCoefSECoefTPConstant-91.7615.22-6.030.000Age1.07670.16606.490.000Pressure0.251810.045235.570.000Smoker8.7403.0012.910.010S=5.757R-Sq=87.3%R-Sq(adj)=85.0%AnalysisofVarianceSourceDFSSMSFPRegression33660.71220.236.820.000ResidualError16530.233.1Total194190.9SourceDFSeqSSAge11772.0Pressure11607.7Smoker1281.1UnusualObservationsObsAgeRiskFitSEFitResidualStResid1766.08.0021.111.94-13.11-2.42R16-22www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuilding21.a.TheMinitaboutputisshownbelow:TheregressionequationisP/E=6.51+0.569%PROFITPredictorCoefSECoefTpConstant6.5071.5094.310.000%PROFIT0.56910.12814.440.000s=2.580R-sq=53.7%R-sq(adj)=51.0%www.khdaw.comAnalysisofVarianceSOURCEDFSSMSFpRegression1131.40131.4019.740.000Error17113.146.66Total18244.54b.Theresidualplotasafunctionoftheorderinwhichthedataarepresentedisshownbelow:RESIDUAL---33.50+6-8-71-13-420.00+09-27-5-5-69-3.50+8--4-+---------+---------+---------+---------+05101520Theredoesnotappeartobeanypatternindicativeofpositiveautocorrelation.c.TheDurban-Watsonstatistic(obtainedfromMinitab)isd=2.34.Atthe.05levelofsignificance,dL=1.18anddU=1.39.Sinced>dUthereisnosignificantpositiveautocorrelation.22.FromMinitab,d=1.60.Atthe.05levelofsignificance,dL=1.04anddU=1.77.SincedL≤d,thetestisinconclusive.23.Thedummyvariablesaredefinedasfollows:16-23www.khdaw.com 课后答案网www.khdaw.comChapter16x1x2x3Treatment000A100B010C001DE(y)=β0+β1x1+β2x2+β3x324.Thedummyvariablesaredefinedasfollows:x1x2Treatment001102013www.khdaw.comx3=0ifblock1and1ifblock2E(y)=β0+β1x1+β2x2+β3x325.FactorAx1=0iflevel1and1iflevel2FactorBx2x3Level001102013E(y)=β0+β1x1+β2x2+β3x1x2+β4x1x326.a.Thedummyvariablesaredefinedasfollows:D1D2Mfg.001102013E(y)=β0+β1D1+β2D2b.TheMinitaboutputisshownbelow:TheregressionequationisTIME=23.0+5.00D1-2.00D2PredictorCoefSECoefTpConstant23.0001.10620.800.000D15.0001.5633.200.011D2-2.0001.563-1.280.233s=2.211R-sq=70.3%R-sq(adj)=63.7%16-24www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingAnalysisofVarianceSOURCEDFSSMSFpRegression2104.00052.00010.640.004Error944.0004.889Total11148.000c.H0:β1=β2=0d.Thep-valueof.004islessthanα=.05;therefore,wecanrejectH0andconcludethatthemeantimetomixabatchofmaterialismostthesameforeachmanufacturer.27.a.Thedummyvariablesaredefinedasfollows:D1D2D3Paint000110020103www.khdaw.com0014TheMinitaboutputisshownbelow:TheregressionequationisTIME=133+6.00D1+3.00D2+11.0D3PredictorCoefSECoefTpConstant133.0002.94145.220.000D16.0004.1591.440.168D23.0004.1590.720.481D311.0004.1592.640.018s=6.576R-sq=32.3%R-sq(adj)=19.6%AnalysisofVarianceSOURCEDFSSMSFpRegression3330.00110.002.540.093Error16692.0043.25Total191022.00Theappropriatehypothesistestis:H0:β1=β2=β3=0Thep-valueof.093isgreaterthanα=.05;therefore,atthe5%levelofsignificancewecannotrejectH0.b.Note:Estimatingthemeandryingforpaint2usingtheestimatedregressionequationsdevelopedinpart(a)maynotbethebestapproachbecauseatthe5%levelofsignificance,wecannotrejectH0.But,ifwewanttousetheoutput,wewouldprocedeasfollows.D1=1D2=0D3=0TIME=133+6(1)+3(0)+11(0)=13928.X1=0ifcomputerizedanalyzer,1ifelectronicanalyzer16-25www.khdaw.com 课后答案网www.khdaw.comChapter16X2andX3aredefinedasfollows:X2X3Car001102013ThecompletedatasetandtheMinitaboutputareshownbelow:YX1X2X35000055010630014210044110www.khdaw.com46101TheregressionequationisY=52.0-12.0X1+3.50X2+8.50X3PredictorCoefStdevt-ratiopConstant52.0002.64619.650.003X1-12.0002.646-4.540.045X23.5003.2401.080.393X38.5003.2402.620.120s=3.240R-sq=93.2%R-sq(adj)=83.1%AnalysisofVarianceSOURCEDFSSMSFpRegression3289.0096.339.170.100Error221.0010.50Total5310.00TotestforanysignificantdifferencebetweenthetwoanalyzerswemusttestH0:β1=0.Sincethep-valuecorrespondingtot=-4.54is.045<α=.05,werejectH0:β0=0;thetimetodoatuneupisnotthesameforthetwoanalyzers.29.X1=0ifasmalladvertisementand1ifalargeadvertisementX2andX3aredefinedasfollows:X2X3Design00A10B01C16-26www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingThecompletedatasetandtheMinitaboutputareshownbelow:YX1X2X3X1X2X1X3800000120000012100008100002201000www.khdaw.com1401000261101030110101000100180010018101011410101TheregressionequationisY=10.0+0.00X1+8.00X2+4.00X3+10.0X1X2+2.00X1X3PredictorCoefSECoefTpConstant10.0002.8283.540.012X10.0004.0000.001.000X28.0004.0002.000.092X34.0004.0001.000.356X1X210.0005.6571.770.128X1X32.0005.6570.350.736s=4.000R-sq=82.4%R-sq(adj)=67.6%AnalysisofVarianceSOURCEDFSSMSFpRegression5448.0089.605.600.029Error696.0016.00Total11544.0030.a.LetExSdenotetheinteractionbetweentheexpenseratio(%)andthesafetyrating.TheregressionequationisPerform%=23.3+222Expense%-28.9ExSPredictorCoefSECoefTPConstant23.3219.821.180.256Expense%222.4341.935.300.000ExS-28.8696.636-4.350.000S=13.01R-Sq=69.0%R-Sq(adj)=65.3%16-27www.khdaw.com 课后答案网www.khdaw.comChapter16AnalysisofVarianceSourceDFSSMSFPRegression26396.43198.218.900.000ResidualError172876.6169.2ResidualError3517.69790.5057Total3936.9978Thetypeoffund(loadornoload)doesnotappeartobeasignificantfactorinpredictingtheone-yearperformance.Theinteractionbetweentheexpenseratioandthesafetyratingissignificant,andisaccountedforbytheExStermintheestimatedregressionequation.b.Thefitprovidedbytheestimatedregressionequationshowninpart(a)isnotbad;R-Sq(adj)=65.3%yˆ=23.2222(1.12)28.9[(1.12)(7.6)]+−=25.8;thus,theestimatedone-yearperformanceforAcornwww.khdaw.comInternationalisapproximately26%.31.a.TheMinitaboutputisshownbelow:TheregressionequationisAUDELAY=80.4+11.9INDUS-4.82PUBLIC-2.62ICQUAL-4.07INTFINPredictorCoefStdevt-ratiopConstant80.4295.91613.600.000INDUS11.9443.7983.150.003PUBLIC-4.8164.229-1.140.263ICQUAL-2.6241.184-2.220.033INTFIN-4.0731.851-2.200.035s=10.92R-sq=38.3%R-sq(adj)=31.2%AnalysisofVarianceSOURCEDFSSMSFpRegression42587.7646.95.420.002Error354176.3119.3Total396764.0b.Thelowvalueoftheadjustedcoefficientofdetermination(31.2%)doesnotindicateagoodfit.c.Thescatterdiagramisshownbelow:96+-**AUDELAY-3-**-*80+*2*-**-3*-3*2-2*64+*-****-*16-28www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuilding-3-*48+*-2--+---------+---------+---------+---------+---------+----INTFIN0.01.02.03.04.05.0Thescatterdiagramsuggestsacurvilinearrelationshipbetweenthesetwovariables.d.Theoutputfromthestepwiseprocedureisshownbelow,whereINTFINSQisthesquareofINTFIN.ResponseisAUDELAYon5predictors,withN=40Step12Constant112.4112.8www.khdaw.comINDUS11.511.6T-Value3.673.80P-Value0.0010.001PUBLIC-1.0T-Value-0.29P-Value0.775ICQUAL-2.45-2.49T-Value-2.51-2.60P-Value0.0170.014INTFIN-36.0-36.6T-Value-4.61-4.91P-Value0.0000.000INTFINSQ6.56.6T-Value4.174.44P-Value0.0000.000S9.018.90R-Sq59.1559.05R-Sq(adj)53.1454.37C-p6.04.132.Thecomputeroutputisshownbelow:TheregressionequationisAUDELAY=63.0+11.1INDUSPredictorCoefSECoefTpConstant63.0003.39318.570.000INDUS11.0744.1302.680.011s=12.23R-sq=15.9%R-sq(adj)=13.7%AnalysisofVarianceSOURCEDFSSMSFpRegression11076.11076.17.190.01116-29www.khdaw.com 课后答案网www.khdaw.comChapter16Error385687.9149.7Total396764.0UnusualObservationsObs.INDUSAUDELAYFitStdev.FitResidualSt.Resid50.0091.0063.003.3928.002.38R381.0046.0074.072.35-28.07-2.34RDurban-Watsonstatistic=1.55Atthe.05levelofsignificance,dL=1.44anddU=1.54.Sinced=1.55>dU,thereisnosignificantpositiveautocorrelation.33.a.TheMinitaboutputisshownbelow:Theregressionequationiswww.khdaw.comAUDELAY=70.6+12.7INDUS-2.92ICQUALPredictorCoefSECoefTpConstant70.6344.55815.500.000INDUS12.7373.9663.210.003ICQUAL-2.9191.238-2.360.024s=11.56R-sq=26.9%R-sq(adj)=22.9%AnalysisofVarianceSOURCEDFSSMSFpRegression21818.6909.36.800.003Error374945.4133.7Total396764.0SOURCEDFSEQSSINDUS11076.1ICQUAL1742.4UnusualObservationsObs.INDUSAUDELAYFitStdev.FitResidualSt.Resid50.0091.0067.713.7823.292.13R381.0046.0071.702.44-25.70-2.27RRdenotesanobs.withalargest.resid.Durban-Watsonstatistic=1.4316-30www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingb.Theresidualplotasafunctionoftheorderinwhichthedataarepresentedisshownbelow:RESID--5-617.5+0-8989-647www.khdaw.com-1-4670190.0+172480235-05-372-33-1946-17.5+2-5-8-+---------+---------+---------+---------+010203040Thereisnoobviouspatterninthedataindicativeofpositiveautocorrelation.c.Atthe.05levelofsignificance,dL=1.44anddU=1.54.Sinced=1.43>dU,thereisnosignificantpositiveautocorrelation.34.Thedummyvariablesaredefinedasfollows:D1D2Type00Non10Light01HeavyTheMinitaboutputisshownbelow:TheregressionequationisScore=4.25+1.00D1+1.50D2PredictorCoefSECoefTpConstant4.25000.381911.130.000D11.00000.54011.850.078D21.50000.54012.780.011s=1.080R-sq=27.6%R-sq(adj)=20.7%AnalysisofVariance16-31www.khdaw.com 课后答案网www.khdaw.comChapter16SOURCEDFSSMSFpRegression29.3334.6674.000.034Error2124.5001.167Total2333.833Sincethep-value=.034islessthanα=.05,therearesignificantdifferencesbetweencomfortlevelsforthethreetypesofbrowsers.35.First,wewillusesimplelinearregressiontoestimatethechangeintheDowJonesIndustrialAverageusingjustparty.InthefollowingMinitaboutputthedummyvariablePartyiscodedasfollows:Party=0ifDemocrat,1ifRepublican.TheregressionequationisChangeDJ=7.13+0.92PartyPredictorCoefSECoefTPwww.khdaw.comConstant7.1254.7681.490.146Party0.9206.0310.150.880S=16.52R-Sq=0.1%R-Sq(adj)=0.0%AnalysisofVarianceSourceDFSSMSFPRegression16.36.30.020.880Error308184.3272.8Total318190.6UnusualObservationsObsPartyChangeDJFitStDevFitResidualStResid101.00-27.608.053.69-35.65-2.21RRdenotesanobservationwithalargestandardizedresidualTherelationshipbetweenthesetwovariablesisnotstatisticallysignificant.Tomodeltheeffectoftheyearinthepresidentialterm,threedummyvariableswereused:Year1,Year2,andYear3.Thesevariableswerecodedasfollows:Year1=1iftheobservationcorrespondstothefirstyearinthetermofoffice,0otherwise;Year2=1iftheobservationcorrespondstothesecondyearinthetermofoffice,0otherwise;andYear3=1iftheobservationcorrespondstothethirdyearinthetermofoffice,0otherwise.UsingParty,Year1,Year2,andYear3asindependentvariables,theonlyvariablethatprovedtobesignificantatthe.05levelofsignificanceisYear3.TheMinitaboutputisshownbelow:TheregressionequationisChangeDJ=4.42+13.1Year3PredictorCoefSECoefTPConstant4.4253.1541.400.171Year313.1006.3072.080.046S=15.45R-Sq=12.6%R-Sq(adj)=9.7%AnalysisofVarianceSourceDFSSMSFP16-32www.khdaw.com 课后答案网www.khdaw.comRegressionAnalysis:ModelBuildingRegression11029.71029.74.310.046Error307161.0238.7Total318190.6UnusualObservationsObsYear3ChangeDJFitStDevFitResidualStResid100.00-27.604.423.15-32.02-2.12RRdenotesanobservationwithalargestandardizedresidualwww.khdaw.com16-33www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter17IndexIIndexndexNumbersNNumbersumbersLearningLLearningearningObjectivesOObjectivesbjectives1.Knowhowtocomputepricerelativesandunderstandhowtheyrepresentpricechangesovertime.2.Knowhowtocomputeaggregatepriceindexesandunderstandhowthechoiceofabaseperiodaffectstheindex.www.khdaw.com3.BecomefamiliarwiththeConsumerPriceIndex,theProducerPriceIndexandtheDowJonesaverages.4.Learnhowtodeflateatimeseriestomeasurechangesovertimeinconstantdollars.5.Learnhowtocomputeanaggregatequantityindexandhowtointerpretit.17-1www.khdaw.com 课后答案网www.khdaw.comChapter17Solutions:SSolutions:olutions:1.a.ItemPriceRelativeA103=(7.75/7.50)B238=(1500/630)7.751500.00+1507.75b.I=(100)=(100)=23720017.50630.00+637.507.75(1500)1500.00(2)+14,625.00c.I=(100)=(100)117=20017.50(1500)630.00(2)+12,510.007.75(1800)1500.00(1)+15,450.00d.I=(100)=(100)109=2001www.khdaw.com7.50(1800)630.00(1)+14,130.002.a.Fromthepricerelativeweseethepercentageincreasewas32%.b.Dividethecurrentcostbythepricerelativeandmultiplyby100.$10.751990cost=(100)=$8.141323.a.PriceRelativesA=(6.00/5.45)100=110B=(5.95/5.60)100=106C=(6.20/5.50)100=1136.005.956.20++b.I=(100)110=20015.455.605.50++6.00(150)5.95(200)6.20(120)++c.I=(100)109=20015.45(150)5.60(200)5.50(120)++9%increaseoverthetwoyearperiod.16.25(35,000)64.00(5,000)10.00(60,000)++4.I=(100)105=200115.00(35,000)60.00(5,000)9.80(60,000)++.19(500)1.80(50)4.20(100)13.20(40)+++5.I=(100)104=.15(500)1.60(50)4.50(100)12.00(40)+++Paascheindex17-2www.khdaw.com 课后答案网www.khdaw.comIndexNumbers6.PriceBasePeriodBasePeriodWeightedPriceItemRelativePriceUsageWeightRelativesA15022.002044066,000B905.005025022,500C12014.004056067,2001250155,700155,700I==12512507.a.PriceRelativesA=(3.95/2.50)100=158B=(9.90/8.75)100=113www.khdaw.comC=(.95/.99)100=96b.PriceBaseWeightWeightedPriceItemRelativesPriceQuantityPi0QiRelativesA1582.502562.59875B1138.7515131.314837C96.996059.45702253.23041430414I==120253.2Costofrawmaterialsisup20%forthechemical.8.PriceBaseWeightedPriceStockRelativesPriceQuantityWeightRelativesHoliday11015.505007750852500NYElectric10918.502003700403300KYGas9726.75500133751297375PQSoaps10842.253001267513689003750039220753922075I==10537500Portfolioup5%9.PriceBaseWeightedPriceItemRelativesPriceQuantityWeightRelativesBeer10815.0035,000525,00056,700,000Wine10760.005,000300,00032,100,000SoftDrink1029.8060,000588,00059,976,0001,413,000148,776,000148,776,000I==1051,413,00017-3www.khdaw.com 课后答案网www.khdaw.comChapter17$7.2710.a.Deflated1980wages:(100)=$8.8282.4$14.36Deflated2000wages:(100)=$8.32172.614.36b.(100)197.5=Thepercentageincreaseinactualwagesis97.5%.7.278.32c.(100)=94.3Thechangeinreadwagesisadecreaseof5.7%.8.8211.199611.76(100/156.9)=7.49199712.23(100/160.5)=7.62www.khdaw.com199812.84(100/163.0)=7.88199913.35(100/166.6)=8.01200013.82(100/172.6)=8.018.01=1.02theincreaseinrealwagesandsalariesfrom1998to2000is2%.7.8812.a.19973929(100/160.5)=244819984052(100/163.0)=248619994260(100/166.6)=2557Manufacturers"shipmentsareincreasingslightlyinconstantdollarswhendeflatedusingtheCPI.b.19973929(100/131.8)=298119984052(100/130.7)=310019994260(100/133.0)=3203c.ThePPIisabetterdeflatorsincemanufacturingshipmentsreflectpricespaidbymanufacturers.13.DeflatedYearRetailSales($)CPIRetailSales($)1982380,00096.5393,7821987520,000113.6457,7461992700,000140.3498,9311997870,000160.5542,0562000940,000172.6544,612Intermsofconstantdollars,thefirm"ssalesareincreasingmoderately.300(18.00)400(4.90)850(15.00)++20,11014.I=(100)=(100)110=350(18.00)220(4.90)730(15.00)++18,32895(1200)75(1800)50(2000)70(1500)+++15.I=(100)=99120(1200)86(1800)35(2000)60(1500)+++Quantitiesaredownslightly.17-4www.khdaw.com 课后答案网www.khdaw.comIndexNumbers16.QuantityBaseWeightedQuantityModelRelativesQuantityPrice($)WeightRelativesSedan8520015,2003,040,000258,400,000Sport8010017,0001,700,000136,000,000Wagon807516,8001,260,000100,800,0006,000,000495,200,000495,200,000I==836,000,00017.a/b.PriceIndexYear1996Base1997Base1996100.095.9www.khdaw.com1997104.3100.01998108.9104.51999114.3109.618.a.PriceRelativesA=(15.90/10.50)(100)=151B=(32.00/16.25)(100)=197C=(17.40/12.20)(100)=143D=(35.50/20.00)(100)=17815.90(2000)32.00(5000)17.40(6500)35.50(2+++500)b.I=(100)170=10.50(2000)16.25(5000)12.20(6500)20.00(2+++500)15.90(4000)32.00(3000)17.40(7500)35.50(3+++000)19.I=(100)164=10.50(4000)16.25(3000)12.20(7500)20.00(3+++000)32.75(100)59(150)42(75)16.5(50)+++20.I=(100)=96Jan31.50(100)65(150)40(75)18(50)+++32.50(100)57.5(150)39.5(75)13.75(50)+++I=(100)=92Mar31.50(100)65(150)40(75)18(50)+++Marketisdowncomparedto1999.21.PriceRelatives:JanMarOil(32.75/31.50)(100)=104103Computer(59.00/65.00)(100)=9188Steel(42.00/40.00)(100)=10599RealEstate(16.5/18.00)(100)=9276IJan=96IMar=9217-5www.khdaw.com 课后答案网www.khdaw.comChapter1722.BaseWeightedPriceProductRelativesPriceQuantityWeightRelativesCorn1132.3014273282370,866Soybeans1235.513501929237,2675211608,133608,133I==117521123.a.FruitPriceRelativesBananas(.51/.41)(100)=124.4Apples(.85/.71)(100)=119.7www.khdaw.comOranges(.61/.56)(100)=108.9Pears(.98/.64)(100)=153.1b.Weights(PioQio)PriceRelativeProduct9.963124.41239.397214.129119.71691.24137.784108.9847.67762.048153.1313.5488Totals33.9244091.86494091.8649I==120.633.924Fruitpriceshaveincreasedby20.6%overthe10-yearperiodaccordingtotheindex.24.Salariesinconstant(1982-84)dollarsarecomputedasfollows:1970$14,000(100/38.8)=$36,0821975$17,500(100/53.8)=$32,5281980$23,000(100/82.4)=$27,9131985$37,000(100/107.6)=$34,3871990$53,000(100/130.7)=$40,5511995$65,000(100/152.4)=$42,6512000$80,000(100/172.6)=$46,350Inconstantdollarterms,realstartingsalarieshaveincreasedabout28%overthisperiod.25.Thestockmarketpricesinconstant(1982-84)dollarsarecomputedasfollows:1996$51.00(100/156.9)=$32.501997$54.00(100/160.5)=$33.641998$58.00(100/163.0)=$35.581999$59.50(100/166.6)=$35.712000$59.00(100/172.6)=$34.18Thevalueofthestock,inrealdollars,isonlyslightlymorein2000thanitisin1996.Ofcourse,ifthestockpaidahighdividenditmaystillhavebeenagoodinvestmentoverthisperiod.17-6www.khdaw.com 课后答案网www.khdaw.comIndexNumbers1200(30)500(20)500(25)++26.I=(100)143=800(30)600(20)200(25)++Quantityisup43%.www.khdaw.com17-7www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter18ForecastingFForecastingorecastingLearningLLearningearningObjectivesOObjectivesbjectives1.Understandthatthelong-runsuccessofanorganizationisoftencloselyrelatedtohowwellmanagementisabletopredictfutureaspectsoftheoperation.2.Knowthevariouscomponentsofatimeseries.www.khdaw.com3.Beabletousesmoothingtechniquessuchasmovingaveragesandexponentialsmoothing.4.Beabletousetheleastsquaresmethodtoidentifythetrendcomponentofatimeseries.5.Understandhowtheclassicaltimeseriesmodelcanbeusedtoexplainthepatternorbehaviorofthedatainatimeseriesandtodevelopaforecastforthetimeseries.6.Beabletodetermineanduseseasonalindexesforatimeseries.7.Knowhowregressionmodelscanbeusedinforecasting.8.Knowthedefinitionofthefollowingterms:timeseriesmeansquarederrorforecastmovingaveragestrendcomponentweightedmovingaveragescyclicalcomponentsmoothingconstantseasonalcomponentseasonalconstantirregularcomponent18-1www.khdaw.com 课后答案网www.khdaw.comChapter18Solutions:SSolutions:olutions:1.a.Time-SeriesForecastWeekValueForecastError(Error)2182133154171252551615116916-74975Forecastforweek7is(17+16+9)/3=14www.khdaw.comb.MSE=75/3=25c.Smoothingconstant=.3.Time-SeriesValueForecastErrorSquaredErrorWeektYtForecastFtYt-Ft(Yt-Ft)2182138.005.0025.003159.006.0036.0041710.206.8046.2451611.564.4419.716912.45-3.4511.90138.85138.85Forecastforweek7is.2(9)+.8(12.45)=11.76d.Fortheα=.2exponentialsmoothingforecastMSE=138.85/5=27.77.Sincethethree-weekmovingaveragehasasmallerMSE,itappearstoprovidethebetterforecasts.e.Smoothingconstant=.4.Time-SeriesValueForecastErrorSquaredErrorWeektYtForecastFtYt-Ft(Yt-Ft)2182138.05.025.0031510.05.025.0041712.05.025.0051614.02.04.006914.8-5.833.64112.64MSE=112.64/5=22.53.Asmoothingconstantof.4appearstoprovidebetterforecasts.Forecastforweek7is.4(9)+.6(14.8)=12.4818-2www.khdaw.com 课后答案网www.khdaw.comForecasting2.a.4-WeekMoving5-WeekMovingTime-SeriesAverageAverageWeekValueForecast(Error)2Forecast(Error)211722131942351820.004.0061620.2518.0619.6012.9672019.001.0019.400.3681819.251.5619.201.4492218.0016.0019.009.00102019.001.0018.801.44111520.0025.0019.2017.64122218.7510.5619.009.00www.khdaw.com77.1851.84b.MSE(4-Week)=77.18/8=9.65MSE(5-Week)=51.84/7=7.41c.Forthelimiteddataprovided,the5-weekmovingaverageprovidesthesmallestMSE.3.a.Time-SeriesWeightedMovingForecastWeekValueAverageForecastError(Error)211722131942319.333.6713.4751821.33-3.3311.0961619.83-3.8314.6772017.832.174.7181818.33-0.330.1192218.333.6713.47102020.33-0.330.11111520.33-5.3328.41122217.834.1717.39103.43b.MSE=103.43/9=11.49Prefertheunweightedmovingaveragehere.c.Youcouldalwaysfindaweightedmovingaverageatleastasgoodastheunweightedone.Actuallytheunweightedmovingaverageisaspecialcaseoftheweightedoneswheretheweightsareequal.18-3www.khdaw.com 课后答案网www.khdaw.comChapter184.WeekTime-SeriesValueForecastError(Error)211722117.004.0016.0031917.401.602.5642317.565.4429.5951818.10-0.100.0161618.09-2.094.3772017.882.124.4981818.10-0.100.0192218.093.9115.29102018.481.522.31111518.63-3.6313.18122218.273.7313.91101.72101.72www.khdaw.comMSE=101.72/11=9.25α=.2providedalowerMSE;thereforeα=.2isbetterthanα=.15.a.F13=.2Y12+.16Y11+.64(.2Y10+.8F10)=.2Y12+.16Y11+.128Y10+.512F10F13=.2Y12+.16Y11+.128Y10+.512(.2Y9+.8F9)=.2Y12+.16Y11+.128Y10+.1024Y9+.4096F9F13=.2Y12+.16Y11+.128Y10+.1024Y9+.4096(.2Y8+.8F8)=.2Y12+.16Y11+.128Y10+.1024Y9+.08192Y8+.32768F8b.Themorerecentdatareceivesthegreaterweightorimportanceindeterminingtheforecast.Themovingaveragesmethodweightsthelastndatavaluesequallyindeterminingtheforecast.6.a.3-MonthMovingα=2MonthYtAveragesForecast(Error)2Forecast(Error)218028280.004.0038480.4012.9648382.001.0081.123.5358383.000.0081.502.2568483.330.4581.804.8478583.332.7982.247.6288484.000.0082.791.4698284.335.4383.031.06108383.670.4582.830.03118483.001.0082.861.30128383.000.0083.090.0111.1239.06MSE(3-Month)=11.12/9=1.24MSE(α=.2)=39.06/11=3.55Use3-monthmovingaverages.b.(83+84+83)/3=83.318-4www.khdaw.com 课后答案网www.khdaw.comForecasting7.a.Time-Series3-MonthMoving4-MonthMovingMonthValueAverageForecast(Error)2AverageForecast(Error)219.529.339.449.69.400.0459.89.430.149.450.1269.79.600.019.530.0379.89.700.019.630.03810.59.770.539.730.5999.910.000.019.950.00109.710.070.149.980.08119.610.030.189.970.14129.69.730.029.920.101.081.09www.khdaw.comMSE(3-Month)=1.08/9=.12MSE(4-Month)=1.09/8=.14Use3-Monthmovingaverages.b.Forecast=(9.7+9.6+9.6)/3=9.63c.Forthelimiteddataprovided,the5-weekmovingaverageprovidesthesmallestMSE.8.a.Time-Series3-MonthMovingα=.2MonthValueAverageForecast(Error)2Forecast(Error)212402350240.0012100.003230262.001024.004260273.33177.69255.6019.365280280.000.00256.48553.196320256.674010.69261.183459.797220286.674444.89272.952803.708310273.331344.69262.362269.579240283.331877.49271.891016.9710310256.672844.09265.511979.3611240286.672178.09274.411184.0512230263.331110.89267.531408.5017,988.5227,818.49MSE(3-Month)=17,988.52/9=1998.72MSE(α=.2)=27,818.49/11=2528.95BasedontheaboveMSEvalues,the3-monthmovingaveragesappearsbetter.However,exponentialsmoothingwaspenalizedbyincludingmonth2whichwasdifficultforanymethodtoforecast.Usingonlytheerrorsformonths4to12,theMSEforexponentialsmoothingisrevisedtoMSE(α=.2)=14,694.49/9=1632.72Thus,exponentialsmoothingwasbetterconsideringmonths4to12.18-5www.khdaw.com 课后答案网www.khdaw.comChapter18b.Usingexponentialsmoothing,F13=αY12+(1-α)F12=.20(230)+.80(267.53)=2609.a.Smoothingconstant=.3.Time-SeriesValueForecastErrorSquaredErrorMonthtYtForecastFtYt-Ft(Yt-Ft)21105.02135.0105.0030.00900.003120.0114.006.0036.004105.0115.80-10.80116.64590.0112.56-22.56508.956120.0105.7914.21201.927145.0110.0534.951221.508140.0120.5419.46378.699100.0126.38-26.38695.90www.khdaw.com1080.0118.46-38.461479.1711100.0106.92-6.9247.8912110.0104.855.1526.52Total5613.18MSE=5613.18/11=510.29Forecastformonth13:F13=.3(110)+.7(104.85)=106.4b.Smoothingconstant=.5Time-SeriesValueForecastErrorMonthtYtForecastFtYt-FtSquaredError(Yt-Ft)21105213510530.00900.003120.5(135)+.5(105)=1200.000.004105.5(120)+.5(120)=120-15.00225.00590.5(105)+.5(120)=112.50-22.50506.256120.5(90)+.5(112.5)=101.2518.75351.567145.5(120)+.5(101.25)=110.6334.371181.308140.5(145)+.5(110.63)=127.8112.19148.609100.5(140)+.5(127.81)=133.91-33.911149.891080.5(100)+.5(133.91)=116.95-36.951365.3011100.5(80)+.5(116.95)=98.481.522.3112110.5(100)+.5(98.48)=99.2410.76115.785945.99MSE=5945.99/11=540.55Forecastformonth13:F13=.5(110)+.5(99.24)=104.62Conclusion:asmoothingconstantof.3isbetterthanasmoothingconstantof.5sincetheMSEislessfor0.3.18-6www.khdaw.com 课后答案网www.khdaw.comForecasting10.a/b.Time-Seriesα=.2α=.3WeekValueForecast(Error)2Forecast(Error)217.3527.407.35.00257.35.002537.557.36.03617.36.036147.567.40.02567.42.019657.607.43.02897.46.019667.527.46.00367.50.000477.527.48.00167.51.000187.707.48.04847.51.036197.627.53.00817.57.0025107.557.55.00007.58.0009.1548.1178www.khdaw.comc.MSE(α=.2)=.1548/9=.0172MSE(α=.3)=.1178/9=.0131Useα=.3.F11=.3Y10+.7F10=.3(7.55)+.7(7.58)=7.5711.a.MethodForecastMSE3-Quarter80.732.534-Quarter80.552.81The3-quartermovingaverageforecastisbetterbecauseithasthesmallestMSE.b.MethodForecastMSEα=.480.402.40α=.580.572.01Theα=.5smoothingconstantisbetterbecauseithasthesmallestMSE.c.Theα=.5isbetterbecauseithasthesmallestMSE.12.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t15∑t=15∑Y=55∑tY=186tt∑tY−∑∑(tY)/n186(15)(55)/5−ttb===2.1122∑t2−∑(t)/n55(15)/5−b=Y−bt=112.1(3)−=4.701Tt=4.7+2.1tForecast:T6=4.7+2.1(6)=17.318-7www.khdaw.com 课后答案网www.khdaw.comChapter1813.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t21∑t=91∑Y=1171∑tY=4037ttComputationofslope:∑tY−∑∑(tY)/n4037(21)(1171)/6−ttb===−3.5143122∑t2−∑(t)/n91(21)/6−Computationofintercept:b=Y−bt=195.1667(3.5143)(3.5)−−=207.466801www.khdaw.comEquationforlineartrend:Tt=207.467-3.514tForecast:T6=207.467-3.514(7)=182.8714.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t21∑t=91∑Y=117.1∑tY=403.7ttComputationofslope:∑tY−∑∑(tY)/n403.7(21)(117.1)/6−ttb===−0.3514122∑t2−∑(t)/n91(21)/6−Computationofintercept:b=Y−bt=19.5167(0.3514)(3.5)−−=20.746601Equationforlineartrend:Tt=20.7466-0.3514tConclusion:enrollmentappearstobedecreasingbyanaverageofapproximately351studentsperyear.15.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t28∑t=140∑Y=213,400∑tY=865,400ttComputationofslope:∑tY−∑∑(tY)/n865,400(28)(213,400)/7−ttb===421.429122∑t2−∑(t)/n140(28)/7−Computationofintercept:b=Y−bt=30,485.714421.429(4)−=28,80001Equationforlineartrend:Tt=28,800+421.429t18-8www.khdaw.com 课后答案网www.khdaw.comForecasting16.Alineartrendmodelisnotappropriate.Anonlinearmodelwouldprovideabetterapproximation.17.a.Alineartrendappearstobereasonable.b.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t36∑t=204∑Y=223.8∑tY=1081.6ttComputationofslope:∑tY−∑∑(tY)/n1081.6(36)(223.8)/8−ttb===1.7738122∑t2−∑(t)/n204(36)/8−www.khdaw.comComputationofintercept:b=Y−bt=27.9751.7738(4.5)19.993−=01Equationforlineartrend:Tt=19.993+1.774tConclusion:Thefirmhasbeenrealizinganaveragecostincreaseof$1.77perunitperyear.18.a.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t55∑t=385∑Y=14.26∑tY=94.34ttComputationofslope:∑tY−∑∑(tY)/n94.35(55)(14.26)/10−ttb===.19297122∑t2−∑(t)/n385(55)/10−Computationofintercept:b=Y−bt=1.426.19297(5.5)−=.36501Equationforlineartrend:Tt=.365+.193tForecast:Tt=.365+.193(11)=$2.49b.Overthepasttenyearstheearningspersharehavebeenincreasingattheaveragerateof$.193peryear.AlthoughthisisapositiveindicatorofWalgreen’sperformance.Moreinformationwouldbenecessarytoconclude“goodinvestment.”19.a.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t21∑t=91∑Y=45.5∑tY=160.15ttComputationofslope:∑tY−∑∑(tY)/n160.15(21)(45.5)/6−ttb===0.0514122∑t2−∑(t)/n91(21)/6−18-9www.khdaw.com 课后答案网www.khdaw.comChapter18Computationofintercept:b=Y−bt=7.5833-0.0514(3.5)=7.403301Equationforlineartrend:Tt=7.4033+0.0514tThenumberofapplicationsisincreasingbyapproximately1630peryear.b.1996:Tt=7.4033+0.0514(7)=7.7633orabout7.76%1997:Tt=7.4033+0.0514(8)=7.8148orabout7.81%20.a.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t55∑t=385∑Y=41841∑tY=262,923www.khdaw.comttComputationofslope:∑tY−∑∑(tY)/n262,923(55)(41,841)/10−ttb===397.545122∑t2−∑(t)/n385(55)/10−Computationofintercept:b=Y−bt=4184.1-397.545(5.5)=1997.601Equationforlineartrend:Tt=1997.6+397.545tb.T11=1997.6+397.545(11)=6371T12=1997.6+397.545(12)=676821.a.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t21∑t=91∑Y=118.2∑tY=549.7ttComputationofslope:∑tY−∑∑(tY)/n549.7(21)(118.2)/6−ttb===7.7714122∑t2−∑(t)/n91(21)/6−Computationofintercept:b=Y−bt=(118.2/6)-7.7714(21/6)=-7.501Equationforlineartrend:Tt=-7.5+7.7714tb.7.7714($M)peryearc.1998forecast:T8=-7.5+7.7714(7)=46.918-10www.khdaw.com 课后答案网www.khdaw.comForecasting22.a.Four-QuarterCenteredYearQuarterYtMovingAverageMovingAverage114223.50333.7504.00454.1254.252164.5004.75235.0005.25www.khdaw.com355.3755.50475.8756.253176.3756.50266.6256.753648b.CenteredSeasonal-IrregularYearQuarterYtMovingAverageComponent11422333.7500.8000454.1251.21212164.5001.3333235.0000.6000355.3750.9302475.8751.19153176.3751.0980266.6250.9057364818-11www.khdaw.com 课后答案网www.khdaw.comChapter18Seasonal-IrregularAdjustedSeasonalQuarterComponentValuesSeasonalIndexIndex11.3333,1.09801.21571.20502.60000,.90570.75290.74633.80000,.90320.86510.867541.2121,1.19151.20181.19124.0355Note:Adjustmentforseasonalindex=4.000/4.0355=0.991223.a.Fourquartermovingaveragesbeginningwith(1690+940+2625+2500)/4=1938.75www.khdaw.comOthermovingaveragesare1966.252002.501956.252052.502025.002060.001990.002123.75b.Seasonal-IrregularAdjustedSeasonalQuarterComponentValuesSeasonalIndexIndex10.9040.9000.90200.90020.4480.5260.49700.48631.3441.4531.39851.39641.2751.1641.21951.2174.0070Note:Adjustmentforseasonalindex=4.000/4.007=0.9983c.Thelargestseasonaleffectisinthethirdquarterwhichcorrespondstotheback-to-schooldemandduringJuly,August,andSeptemberofeachyear.24.Seasonal-IrregularMonthComponentValuesSeasonalIndexAdjustedSeasonalIndex10.720.700.710.70720.800.750.780.77730.830.820.830.82740.940.990.970.96651.011.021.021.01661.251.361.311.30571.491.511.501.49481.191.261.231.22590.980.970.980.976100.981.000.990.986110.930.940.940.936120.780.800.790.78712.0518-12www.khdaw.com 课后答案网www.khdaw.comForecastingNotes:1.Adjustmentforseasonalindex=12/12.05=0.9962.Theadjustmentisreallynotnecessaryinthisproblemsinceitimpliesmoreaccuracythaniswarranted.Thatis,theseasonalcomponentvaluesandtheseasonalindexwereroundedtotwodecimalplaces.25.a.Useatwelveperiodmovingaverages.Aftercenteringthemovingaverages,youshouldobtainthefollowingseasonalindexes:HourSeasonalIndexHourSeasonalIndex10.77171.20720.86480.99430.95490.85041.392100.64751.571110.579www.khdaw.com61.667120.504b.ThehoursofJuly18arenumber37to48inthetimeseries.Thusthetrendcomponentfor7:00a.m.onJuly18(period37)wouldbeT37=32.983+.3922(37)=47.49AsummaryofthetrendcomponentsforthetwelvehoursonJuly18isasfollows:HourTrendComponentHourTrendComponent147.49749.85247.89850.24348.28950.63448.671051.02549.061151.42649.461251.81c.MultiplythetrendcomponentinpartbbytheseasonalindexesinpartatoobtainthetwelvehourlyforecastsforJuly18.Forexample,47.49x(.771)=36.6orroundedto37,wouldbetheforecastfor7:00a.m.onJuly18th.TheseasonallyadjustedhourlyforecastsforJuly18areasfollows:HourForecastHourForecast13776024185034694346810335771130682122626.a.Yes,thereisaseasonaleffectoverthe24hourperiod.TimePeriodSeasonalIndex12-4a.m.1.6964-8a.m.1.4588-120.71112-4p.m.0.3264-8p.m.0.44818-13www.khdaw.com 课后答案网www.khdaw.comChapter188-121.362b.TimePeriodForecast12-4p.m.166,761.134-8p.m.146,052.9927.a.3-MonthMovingForecastMonthTime-SeriesValueAverageForecastError(Error)2134.8750235.6250334.6875433.562535.0625-1.5002.2500532.625034.6250-2.0004.0000634.000033.62500.37500.1406733.625033.39580.22920.0525www.khdaw.com835.062533.41671.64582.7088934.062534.2292-0.16670.02781034.125034.2500-0.12500.01561133.250034.4167-1.16671.36111232.062533.8125-1.75003.0625Note:MSE=13.6189/9=1.51ForecastforDecemberis(34.1250+33.2500+32.0625)/3=33.1458b.Theweightedmovingaverageforecastsformonths4-12are35.1000,34.4250,33.4125,33.3625,33.5750,34.2750,34.3750,34.2875and33.7625.Note:MSE=12.3047/9=1.37ForecastforDecemberis0.2(34.125)+0.4(33.2500)+0.4(32.0625)=32.9500c.Theexponentialsmoothingforecastsformonths2-12are34.8750,35.1375,34.9800,34.4839,33.8333,33.8916,33.7983,34.2408,34.1784,34.1597and33.8413.Note:MSE=11.1881=1.24ForecastforDecemberis0.35(32.0625)+0.65(33.8413)=33.2187d.MethodMSEMovingAverage1.51WeightedMovingAverage1.37ExponentialSmoothing1.24ExponentialSmoothingisthebestofthethreeapproachesbecauseithasthesmallestMSE.18-14www.khdaw.com 课后答案网www.khdaw.comForecasting28.a.Time-Seriesα=0.1α=0.1α=0.2α=0.2YearValueForecast(Error)2Forecast(Error)219725519743855.0289.055.0289.019765453.30.551.65.819783753.4268.052.1227.419805351.71.649.115.519824051.9140.749.997.019845350.75.447.926.219863650.9222.248.9166.519885049.40.346.313.519903749.5155.647.1101.219925548.245.945.099.119943948.998.147.064.6www.khdaw.com19964947.91.245.412.7Totals:1228.41118.5SmoothingConstantMSE0.11228.4/12=102.40.21118.5/12=93.2Asmoothingconstantof0.2isbetter.b.Usingα=0.2Forecastfor1998=0.2(49)+0.8(45.4)=46.129.a.TimeSeriesα=.2α=.3α=.4PeriodValueForecastsForecastsForecasts128.9231.029.8029.8029.80329.930.0430.1630.28430.130.0130.0830.13532.230.0330.0930.12631.530.4630.7230.95732.030.6730.9531.17831.930.9431.2731.50930.031.1331.4631.66MSE(α=.2)=1.40MSE(α=.3)=1.27MSE(α=.4)=1.23α=.4providesthebestforecastb.Usingα=.4,F10=.4(.30)+.6(31.66)=31.0018-15www.khdaw.com 课后答案网www.khdaw.comChapter1830.Time-SeriesValueForecastForecastErrorSquaredErrorWeektYtFtYt-Ft(Yt-Ft)212221822.00-4.0016.0032321.201.803.2442121.56-0.560.3151721.45-4.4519.8062420.563.4411.8372021.25-1.251.5681921.00-2.004.0091820.60-2.606.76102120.080.920.8564.35www.khdaw.comTotalMSE=64.35/9=7.15Forecastforweek11:F11=0.2(21)+0.8(20.08)=20.2631.tYtFtYt-Ft(Yt-Ft)212,75023,1002,750.00350.00122,500.0033,2502,890.00360.00129,600.0042,8003,034.00-234.0054,756.0052,9002,940.40-40.401,632.1663,0502,924.24125.7615,815.5873,3002,974.54325.46105,924.2183,1003,104.73-4.7322.3792,9503,102.84-152.8423,260.07103,0003,041.70-41.701,738.89113,2003,025.02174.9830,618.00123,1503,095.0154.993,023.90Total:488,991.18MSE=488,991.18/11=44,453.74Forecastforweek13:F13=0.4(3,150)+0.6(3,095.01)=3,117.0132.a.SmoothingConstantMSEα=.34,492.37α=.42,964.67α=.52,160.31Theα=.5smoothingconstantisbetterbecauseithasthesmallestMSE.b.Tt=244.778+22.088t18-16www.khdaw.com 课后答案网www.khdaw.comForecastingMSE=357.81c.TrendprojectionprovidesmuchbetterforecastsbecauseithasthesmallestMSE.ThereasonMSEissmallerfortrendprojectionisthatsalesareincreasingovertime;asaresult,exponentialsmoothingcontinuouslyunderestimatesthevalueofsales.Ifyoulookattheforecasterrorsforexponentialsmoothingyouwillseethattheforecasterrorsarepositiveforperiods2through18.33.a.ForecastforJulyis236.97ForecastforAugust,usingforecastforJulyastheactualsalesinJuly,is236.97.Exponentialsmoothingprovidesthesameforecastforeveryperiodinthefuture.Thisiswhyitisnotusuallyrecommendedforlong-termforecasting.www.khdaw.comb.Tt=149.719+18.451tForecastforJulyis278.88ForecastforAugustis297.33c.Theproposedsettlementisnotfairsinceitdoesnotaccountfortheupwardtrendinsales.Basedupontrendprojection,thesettlementshouldbebasedonforecastedlostsalesof$278,880inJulyand$297,330inAugust.34.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t28∑t=140∑Y=1575∑tY=6491ttComputationofslope:∑tY−∑∑(tY)/n6491(28)(1575)/7−ttb===6.8214122∑t2−∑(t)/n140(28)/7−Computationofintercept:b=Y−bt=225-6.8214(4)=197.71401Equationforlineartrend:Tt=197.714+6.821tForecast:T8=197.714+6.821(8)=252.28T9=65.025+4.735(9)=259.1035.Thefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t120∑t=1240∑Y=578,400∑tY=5,495,900ttComputationofslope:∑tY−∑∑(tY)/n5,495,900(120)(578,400)/15−ttb===3102.5122∑t2−∑(t)/n1240(120)/15−18-17www.khdaw.com 课后答案网www.khdaw.comChapter18Computationofintercept:b=Y−bt=(578,400/15)-3102.5(120/15)=13,74001Equationforlineartrend:Tt=13,740+3102.5tb.1995forecast:Tt=13,740+3102.5(16)=60,277.51996forecast:Tt=13,740+3102.5(17)=66,482.536.a.Agraphofthesedatashowsalineartrend.bThefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t15∑t=55∑Y=200∑tY=750ttwww.khdaw.comComputationofslope:∑tY−∑∑(tY)/n750(15)(200)/5−ttb===15122∑t2−∑(t)/n55(15)/5−Computationofintercept:b=Y−bt=40-15(3)=-501Equationforlineartrend:Tt=-5+15tConclusion:averageincreaseinsalesis15unitsperyear37.a.Yes,alineartrendappearstoexist.bThefollowingvaluesareneededtocomputetheslopeandintercept:2∑=t28∑t=140∑Y=595∑tY=2815ttComputationofslope:∑tY−∑∑(tY)/n2815(28)(595)/7−ttb===15.5357122∑t2−∑(t)/n140(28)/7−Computationofintercept:b=Y−bt=85-15.5357(4)=22.85701Equationforlineartrend:Tt=22.857+15.536tc.Forecast:T8=22.857+15.536(8)=147.1538.a.Alineartrendappearstobeappropriate.b.T2=12,899.98+2092.066t18-18www.khdaw.com 课后答案网www.khdaw.comForecastingc.$2092.066or$2,092,066d.1997:T13=12,899.98+2092.066(13)=40,096.838or$40,096,8381998:T14=12,899.98+2092.066(14)=42,188.904or$42,188,90439.Alineartrenddoesnotseemappropriate.Theplotindicatessometypeofcurvilinearrelationshipovertimesuchas2T=b+bt+btt01t2t16000Xwww.khdaw.comNumberX8000XXXXXX01.63.24.86.48.0Year40.a.CenteredSeasonal-IrregulartSalesMovingAverageComponent162153109.2501.0814410.1250.39551011.1250.89961812.1251.48571513.0001.1548714.5000.48391416.5000.848102618.1251.434112319.3751.187121220.2500.593131920.7500.916142821.7501.287152522.8751.093161824.0000.750172225.1250.876183425.8751.314192826.5001.057202127.0000.778212427.5000.873223627.6251.303233028.0001.071242029.0000.69018-19www.khdaw.com 课后答案网www.khdaw.comChapter18252830.1250.929264031.6251.26527352827b.Seasonal-IrregularSeasonalQuarterComponentValuesIndex10.899,0.848,0.916,0.876,0.873,0.9290.89021.485,1.434,1.287,1.314,1.303,1.2651.34831.081,1.154,1.187,1.093,1.057,1.0711.10740.395,0.483,0.593,0.750,0.778,0.6900.615Total3.960www.khdaw.comQuarterAdjustedSeasonalIndex10.89921.36231.11840.621Note:Adjustmentforseasonalindex=4.00/3.96=1.0101c.HudsonMarineexperiencesthelargestseasonalincreaseinquarter2.Sincethisquarteroccurspriortothepeaksummerboatingseason,thisresultseemsreasonable.41.a.CenteredSeasonal-IrregulartSalesMovingAverageComponent1422313.2500.308453.7501.333564.3751.371645.8750.681747.5000.5338147.8751.7789107.8751.2701038.2500.3641158.7500.57112169.7501.641131210.7501.11614911.7500.76615713.2500.528162214.1251.558171815.0001.200181017.3750.57619132035Seasonal-IrregularSeasonalQuarterComponentValuesIndex18-20www.khdaw.com 课后答案网www.khdaw.comForecasting11.371,1.270,1.116,1.2001.23920.681,0.364,0.776,0.5760.59730.308,0.533,0.571,0.5280.48541.333,1.778,1.641,1.5581.578Total3.899QuarterAdjustedSeasonalIndex11.27120.61330.49841.619Note:Adjustmentforseasonalindex=4/3.899=1.026b.Thelargesteffectisinquarter4;thisseemsreasonablesinceretailsalesaregenerallyhigherduringwww.khdaw.comOctober,November,andDecember.42.a.Note:Tosimplifythecalculationstheseasonalindexescalculatedinproblem40havebeenroundedtotwodecimalplaces.SeasonalFactorDeseasonalizedSalesYearQuarterSalesYtStYt/St=TtIt1160.906.672151.3611.033101.128.93440.626.4521100.9011.112181.3613.243151.1213.39470.6211.2931140.9015.562261.3619.123231.1220.544120.6219.3541190.9021.112281.3620.593251.1222.324180.6229.0351220.9024.442341.3625.003281.1225.004210.6233.8761240.9026.672361.3626.473301.1226.794200.6232.2671280.9031.112401.3629.413351.1231.254270.6243.5518-21www.khdaw.com 课后答案网www.khdaw.comChapter18tYttYtt2(deseasonalized)16.676.671211.0322.06438.9326.79946.4525.8016511.1155.5525613.2479.4436713.3993.7349811.2990.3264www.khdaw.com915.56140.04811019.12191.201001120.54225.941211219.35232.201441321.11274.431691420.59288.261961522.32334.802251629.03464.482561724.44415.482891825.00450.003241925.00475.003612033.87677.404002126.67560.074412226.47582.344842326.79616.175292432.26774.245762531.11777.756252629.41764.666762731.25843.757292843.551,219.40784406605.5510,707.347,714t=14.5=21.627Yb=1.055b=6.329T=6.3291.055+t10tb/c.tTrendForecast2936.923037.983139.033240.09YearQuarterTrendForecastSeasonalIndexQuarterlyForecast8136.920.9033.23237.981.3651.65329.031.1243.71440.090.6224.8618-22www.khdaw.com 课后答案网www.khdaw.comForecasting43.aNote:Tosimplifythecalculationstheseasonalindexesinproblem40havebeenroundedtotwodecimalplaces.SeasonalFactorDeseasonalizedSalesYearQuarterSalesYtStYt/St=TtIt1141.273.15220.613.28310.502.00451.623.092161.274.72www.khdaw.com240.616.56340.508.004141.628.6431101.277.87230.614.92350.5010.004161.629.8841121.279.45290.6114.75370.5014.004221.6213.5851181.2714.172100.6116.393130.5026.004351.6221.60Ytt(deseasonalized)tYtt213.153.15123.286.56432.006.00943.0912.361654.7223.602566.5639.363678.0056.004988.6469.126497.8770.8381104.9249.201001110.00110.00121129.88118.56144139.45122.851691414.75206.501961514.00210.002251613.58217.282561714.17240.892891816.39295.023241926.00494.003612021.60432.00400210202.052783.28287018-23www.khdaw.com 课后答案网www.khdaw.comChapter18t=10.5Y=10.1025b=.995b=−.345T=−.345.995+t10tb.yTrendForecast2120.552221.552322.542423.54c.www.khdaw.comTrendSeasonalQuarterlyYearQuarterForecastIndexForecast6120.551.2726.10221.550.6113.15322.540.5011.27423.541.6238.1318-24www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter19NonparametricNNonparametriconparametricMethodsMMethodsethodsLearningLLearningearningObjectivesOObjectivesbjectives1.Learnthedifferencebetweenparametricandnonparametricmethods.2.Knowtheparticularadvantagesofnonparametricmethodsandwhentheyareandwhentheyarenotapplicable.www.khdaw.com3.Learnhowtousethesigntestfortheanalysisofpairedcomparisons.4.Beabletousethesigntesttoconducthypothesistestsaboutamedian.5.BeabletousetheWilcoxonsigned-ranktestandtheMann-Whitney-Wilcoxontesttodeterminewhetherornottwopopulationshavethesamedistribution.6.BeabletousetheKruskal-Wallistestsforthecomparisonofkpopulations.7.BeabletocomputetheSpearmanrankcorrelationcoefficientandtestforasignificantcorrelationbetweentwosetsofrankings.19-1www.khdaw.com 课后答案网www.khdaw.comChapter19Solutions:SSolutions:olutions:1.BinomialProbabilitiesforn=10,p=.50.xProbabilityxProbability0.00106.20511.00987.11722.04398.04393.11729.00984.205110.00105.2461P(0)+P(1)=.0108;addingP(2),exceeds.025requiredinthetail.Therefore,rejectH0ifthenumberofplussignsislessthan2orgreaterthan8.Numberofplussignsis7.www.khdaw.comDonotrejectH0;concludethatthereisnoindicationthatadifferenceexists.2.Therearen=27casesinwhichavaluedifferentfrom150isobtained.Usethenormalapproximationwithµ=np=.5(27)=13.5andσ=.25n=.25(27)=2.6Usex=22asthenumberofplussignsandobtainthefollowingteststatistic:x−µ2213.5−z===3.27σ2.6Withα=.01,werejectifz>2.33;sincez=3.27>2.33werejectH0.Conclusion:themedianisgreaterthan150.3.a.Letp=probabilitythesharesheldwillbeworthmoreafterthesplitH0:p≤.50Ha:p>.50IfH0cannotberejected,thereisnoevidencetoconcludestocksplitscontinuetoaddvaluetostockholdings.b.Letxbethenumberofplussigns(increasesinvalue).Usethebinomialprobabilitytableswithn=18(therewere2tiesinthe20cases)P(x>12)=.0482RejectH0ifthenumberof+signsisgreaterthan12.c.Withx=14,werejectH0.Theresultssupporttheconclusionthatstocksplitsarebeneficialforshareholders.19-2www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods4.Weneedtodeterminethenumberwhosaidbetterandthenumberwhosaidworse.Thesumofthetwoisthesamplesizeusedforthestudy.n=.34(1253)+.29(1253)=789.4Usethelargesampletestusingthenormaldistribution.Thismeansthevalueofn(n=789.4above)neednotbeinteger.Hence,µ=.5n=.5(789.4)=394.7σ=.25n=.25(789.4)=14.05Letp=proportionofadultswhofeelchildrenwillhaveabetterfuture.H0:p≤.50www.khdaw.comHa:p>.50Withx=.34(1253)=426x−µ426394.7−z===2.23σ14.05Withα=.05,werejectH0ifz>1.645Sincez=2.23>1.645,werejectH0Conclusion:morethanhalfoftheadultsfeeltheirchildrenwillhaveabetterfuture.5.n=185+165=350µ=0.5n=0.5(350)=175σ=.25n=.25(350)=9.35RejectH0ifz<-1.96orifz>1.96185175−z==1.079.35DonotrejectH0;cannotconcludethereisadifferenceinpreferenceforthetwoshows.6.n=202+158=360µ=0.5n=0.5(360)=180σ=.25n=.25(360)=9.49RejectH0ifz<-1.96orifz>1.96202180−z==2.329.4919-3www.khdaw.com 课后答案网www.khdaw.comChapter19RejectH0;concludePackardBellandCompaqhavedifferentmarketshares.7.µ=0.5n=0.5(300)=150σ=.25n=.25(300)=8.66165150−z==1.738.66p-value=2(.5000-.4582)=.0836DonotrejectH0;weareunabletoconcludethatthemedianannualincomediffers.8.µ=.5n=.5(150)=75www.khdaw.comσ=.25n=.25(150)=6.12Onetailedtest:rejectH0ifz>1.645For98+signs9875−z==3.766.12RejectH0;concludethatahometeamadvantageexists.9.H0:Median≤15Ha:Median>15Usebinomialprobabilitieswithn=8andp=.50:Onetailtestwithα=.05,P(8+’s)=.0039P(7+’s)=.0312.0351P(6+’s)=.1094RejectH0if7or8+’s.With7+’sinthesample,rejectH0.Datadoesenableustoconcludethattherehasbeenanincreaseinthemediannumberofpart-timeemployees.10.H0:Median=152Ha:Median≠152µ=.5n=.5(225)=112.5σ=.25n=.25(225)=7.5RejectH0ifz<-1.96orifz>1.9619-4www.khdaw.com 课后答案网www.khdaw.comNonparametricMethodsFor122cases122112.5−z==1.277.5DonotrejectH0;weareunabletoconcludethatthemedianannualincomeneededdiffersfromthatreportedinthesurvey.11.n=50µ=0.5n=0.5(50)=25σ=.25n=.25(50)=3.54RejectH0ifz>1.645www.khdaw.com33hadwagesgreaterthan$5853325−z==2.263.54RejectH0;concludethatthemedianweeklywageisgreaterthan$585.12.H0:ThepopulationsareidenticalHa:ThepopulationsarenotidenticalAdditive1Additive2DifferenceAbsoluteValueRankSignedRank20.1218.052.072.079+923.5621.771.791.797+722.0322.57-.54.543-319.1517.062.092.0910+1021.2321.22.01.011+124.7723.80.97.974+416.1617.20-1.041.045-518.5514.983.573.5712+1221.8720.031.841.848+824.2321.153.083.0811+1123.2122.78.43.432+225.0223.701.321.326+6Total62µT=0nn(+1)(2n+1)12(13)(25)σ===25.5T66T−µ620−Tz===2.43σ25.5TTwo-tailedtest.RejectH0ifz<-1.96orifz>1.9619-5www.khdaw.com 课后答案网www.khdaw.comChapter19Sincez=2.43>1.96werejectH0.Conclusion:thereisasignificantdifferenceintheadditives.13.WithoutWithRankofAbsoluteRelaxantRelaxantDifferenceDifferenceSignedRank151059912102332212101010811-36.5-6.510911175233810-23-310736.56.5141136.56.59636.56.5www.khdaw.comTotal36µT=0nn(+1)(2n+1)10(11)(21)σ===19.62T66T−µ36Tz===1.83σ19.62TOne-tailedtest.RejectH0ifz>1.645Sincez=1.83>1.645werejectH0.Conclusion:thereisasignificantdifferenceinfavoroftherelaxant.14.AbsoluteAirportDifferenceDifferenceSignedRankBostonLogan0.190.1910ChicagoMidway-0.020.02-3.5ChicagoO"Hare0.050.056Denver0.040.045FortLauderdale-0.010.01-1.5LosAngeles0.060.067Miami0.020.023.5NewYork(JFK)0.090.098OrangeCounty(CA)0.160.169Washington(Dulles)0.010.011.5T=45nn(+1)(2n+1)10(11)(21)σ===19.62T66T−µ450−Tz===2.29σ19.62T19-6www.khdaw.com 课后答案网www.khdaw.comNonparametricMethodsRejectH0ifz<-1.96orifz>1.96.RejectH0;concludeadifferenceexistswithAvishigher.15.RankofAbsoluteService#1Service#2DifferenceDifferenceSignedRank24.528.0-3.57.5-7.526.025.50.51.51.528.032.0-4.09.5-9.521.020.01.044.018.019.5-1.56-6.036.028.08.01111.025.029.0-4.09.5-9.521.022.0-1.04-4.024.023.50.51.51.5www.khdaw.com26.029.5-3.57.5-7.531.030.01.044.0T=-22.0µT=0nn(+1)(2n+1)11(12)(23)σ===22.49T66T−µ−22Tz===−.98σ22.49TRejectH0ifz<-1.96orifz>1.96.Sincez=-.98,donotrejectH0;thereisnosignificantdifference.16.1997P/ERatioEst.1998P/ERatioDifferenceRankSignedRank4032899242222.52.52123-22.5-2.53023788251966.56.519190002017344291910101035201511111718-11-1332766.56.52016455T=59n=11(discardingthe0)µT=0nn(+1)(2n+1)11(12)(23)σ===22.49T6619-7www.khdaw.com 课后答案网www.khdaw.comChapter19RejectH0ifz<-1.96orifz>1.96590−z==2.6222.49RejectH0;concludeadifferenceexistsbetweenthe1997P/Eratiosandtheestimated1998P/Eratios.17.RankofAbsolutePrecampaignPostcampaignDifferenceDifferenceSignedRank130160-3010-10100105-52.5-2.5120140-209-9959052.52.5140130104.54.58082-21-1www.khdaw.com6555104.54.590105-157.5-7.5140152-126-6125140-157.5-7.5T=-32µT=0nn(+1)(2n+1)10(11)(21)σ===19.62T66T−µ−32Tz===−1.63σ19.62TRejectH0ifz<-1.63DonotrejectH0;thedifferenceisnotsignificantattheα=.05level.18.Rankthecombinedsamplesandfindtheranksumforeachsample.Thisisasmallsampletestsincen1=7andn2=9Additive1Additive2MPGRankMPGRank17.3218.78.518.4617.8419.11021.31516.7121.01418.2522.11618.6718.78.517.5319.8113420.71320.212102T=34Withα=.05,n1=7andn2=919-8www.khdaw.com 课后答案网www.khdaw.comNonparametricMethodsTL=41andTU=7(7+9+1)-41=78SinceT=34<41,werejectH0Conclusion:thereisasignificantdifferenceingasolinemileage19.a.PublicFinancialAccountantRankPlannerRank25.2524.0233.81924.2331.31628.11033.21830.91529.21326.98.530.01428.61125.9624.7434.52028.912www.khdaw.com31.71726.8726.98.523.91136.573.511µ=nn(+n+1)=10(10101)105++=T11222T=136.511σ=nnn(+n+1)=(10)(10)(21)=13.23T12121212RejectH0ifz<-1.96orifz>1.96136.5105−z==2.3813.23RejectH0;salariesdiffersignificantlyforthetwoprofessions.b.PublicAccountant:$30,200FinancialPlanner:$26,700Conclusion:thereisasignificantdifferenceinstartingsalaries20.a.Median→4thsalaryforeachMen49.9Women35.4b.MenRankWomenRank30.6444.5875.51435.4545.2927.9362.21340.5738.2625.8249.91147.51055.31224.8119-9www.khdaw.com 课后答案网www.khdaw.comChapter19T=36FromTablesTL=37T1.645Sincez=2.05>1.645werejectH0Conclusion:thereisasignificantdifferencebetweenthepopulations.22.H0:ThereisnodifferenceinthedistributionsofP/EratiosHa:ThereisadifferencebetweenthedistributionsofP/EratiosWewillrejectH0ifz<-2.33orz>2.33JapanUnitedStatesCompanyP/ERatioRankCompanyP/ERatioRankSumitomoCorp.15320Gannet196Kinden218Motorola2411.5Heiwa185Schlumberger2411.5NCRJapan12519OracleSystems4316SuzukiMotor3113Gap2210FujiBank21321Winn-dixie142SumitomoChemical6417Ingersoll-Rand218SeibuRailway66622Am.Elec.Power142Shiseido3314Hercules218TohoGas6818TimesMirror3815Total157WellPointHealth154No.StatesPower142Total9611µ=nn(+n+1)=10(10121)115++=T1122219-10www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods11σ=nnn(+n+1)=(10)(12)(10121)++=15.17T12121212T=157157115−z==2.7715.17Sincez=2.77>2.33,rejectH0.WeconcludethatthereisasignificantdifferenceinP/Eratiosforthetwocountries.23.Sumofranks(Winter)=71.5Sumofranks(Summer)=138.5www.khdaw.comUseT=71.511µ=nn(+n+1)=10(21)105=T1122211σ=nnn(+n+1)=(10)(10)(21)=13.23T12121212T−µ71.5105−Tz===−2.53σ13.23TRejectH0ifz<-1.96orifz>1.96RejectH0;thereisasignificantdifference24.Sumofranks(Dallas)=116Sumofranks(SanAntonio)=160UseT=11611µ=nn(+n+1)=10(24)120=T1122211σ=nnn(+n+1)=(10)(13)(24)=16.12T12121212T−µ116120−Tz===−.25σ16.12TRejectH0ifz<-1.96orifz>1.96DonotrejectH0;thereisnotsignificantevidencetoconcludethatthereisadifference.19-11www.khdaw.com 课后答案网www.khdaw.comChapter1925.KitchenRankMasterRankBedroom25,2001618,000417,400222,9001122,8001026,4001721,900924,8001519,7005.526,9001823,0001217,800319,7005.524,6001416,900121,000721,800889www.khdaw.com23,6001382FromAppendixB,TL=73TU=n1(n1+n2+1)-TL=10(10+8+1)-73=117RejectH0ifT<73orifT>117SinceT=82,donotrejectThereisnosignificantdifferencebetweenthecosts.26.ABC41178142101513126913534652122212⎡(34)(65)(21)⎤W=⎢++⎥−3(16)10.22=(15)(16)⎣555⎦2χ=5.99147(2degreesoffreedom).05RejectH0;concludethattheratingsfortheproductsdiffer.27.ABC11.55.017.02.511.520.08.02.515.010.04.08.019-12www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods8.06.016.018.01.019.013.014.058.043.0109.022212⎡(58)(43)(109)⎤W=⎢++⎥−3(21)=9.06(20)(21)⎣677⎦2χ=9.21034(2degreesoffreedom).01DonotrejectH0;wecannotconcludethatthereisasignificantdifferenceintestpreparationprograms.28.SwimmingRankTennisRankCyclingRank408841593855www.khdaw.com380448514250142511450132953400642010402742712530152682Sum41611822212⎡(41)(61)(18)⎤W=⎢++⎥−3(151)+=9.26(15)(151)+⎣555⎦2χ=5.99147.05Since9.26>5.99147,rejectH0;concludethatthereisasignificantdifferenceincaloriesamongthethreeactivities.29.ABC221274.5144.5910.52713710.51522.53364.522212⎡(22.5)(33)(64.5)⎤W=⎢++⎥−3(16)=9.555(15)(16)⎣555⎦2χ=5.99147(2degreesoffreedom).05Since9.555>5.99147werejectH0andconcludethatthereisasignificantdifferenceingasmileageamongthethreeautomobiles.30.Course1Course2Course3Course432192014716419-13www.khdaw.com 课后答案网www.khdaw.comChapter19101915125186131117852267953222212⎡(52)(26)(79)(53)⎤W=⎢+++⎥−3(21)=8.03(20)(21)⎣5555⎦2χ=7.81473(3degreesoffreedom).05Since8.03>7.81473,werejectH0andconcludethatthereisasignificantdifferenceinthequalityofcoursesofferedbythefourmanagementdevelopmentcenters.31.M&MsKitKatMilkyWayIIwww.khdaw.com10.593756131441512210.58156481622212⎡(56)(48)(16)⎤W=⎢++⎥−3(16)=8.96(15)(16)⎣555⎦2χ=5.99147(2degreesoffreedom).05Since8.96>5.99147werejectH0Therearesignificantdifferencesincaloriecontentamongthethreecandies.232.a.Σd=52i26Σd6(52)ir=−1=−1=.68s2nn(−1)10(99)11b.σ===.33rsn−19r−0.68sz===2.06σ.33rsRejectifz<-1.96orifz>1.96Sincez=2.06>1.96,werejectH0.Concludethatsignificantrankcorrelationexists.33.Case1:19-14www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods2Σd=0i26Σd6(0)ir=−1=−1=1s2nn(−1)6(361)−Case2:2Σd=70i26Σd6(70)ir=−1=−1=−1s2nn(−1)6(361)−Withperfectagreement,rs=1.Withexactoppositeranking,rs=-1.www.khdaw.com34.Σd2=250i26Σd6(250)ir=−1=−1=−.136s2nn(−1)11(120)11σ===.32rsn−110r−0−.136sz===−.425σ.32rsRejectifz<-1.96orifz>1.96Sincez=-.425,wecannotrejectH0.Concludethatthereisnotasignificantrelationshipbetweentherankings.235.a.Σd=54i26Σd6(54)ir=−1=−1=.6722nn(−1)10(10−1)b.H0:pr≤0Ha:pr>011σ===.3333rsn−1101−rs−µr.670−z=s==2.02σ.3333rsp-value=.5000-.4783=.021719-15www.khdaw.com 课后答案网www.khdaw.comChapter19c.RejectH0:Concludeasignificantpositiverankcorrelation.36.DrivingDistancePuttingd2idi15-41656-11410-6369274967-1110374928-63639-636www.khdaw.com7439817492Σd=282i26Σd6(282)ir=−1=−1=−.709s2nn(−1)10(1001)−µ=0rs11σ===.333rsn−19RejectH0ifz<-1.645orifz>1.645−.7090−z==−2.13.333RejectH0;thereisasignificantnegativerankcorrelationbetweendrivingdistanceandputting.237.Σd=38i26Σd6(38)ir=−1=−1=.77s2nn(−1)10(99)µ=0rs11σ===.3333rsn−19RejectH0ifz<-1.645orifz>1.645r−0.77sz===2.31σ.3333rsRejectH0;thereisasignificantrankcorrelationbetweencurrentstudentsandrecentgraduates.19-16www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods38.n=905+1045=1950µ=.5n=.5(1950)=975σ=.25n=.25(1950)=22.01RejectH0ifz<-1.96orifz>1.96905975−z==−3.1722.01RejectH0;thedifferenceinthefavor-opposeopinionissignificant.39.a.n=11+32=43www.khdaw.comH0:Median≥$118,000Ha:Median<$118,000µ=.5n=.5(43)=21.5σ=.25n=.25(43)=3.2787x−µ1121.5−z===−3.20σ3.2787Sincez=-3.20<-1.645,rejectH0.WeconcludethatthemedianresalepriceforhomesinHouston,Texasisbelowthenationalmedian.b.n=27+13=40H0:Median≤$118,000Ha:Median>$118,000µ=.5n=.5(40)=20σ=.25n=.25(40)=3.1623x−µ2720−z===2.21σ3.1623Sincez=2.21>1.645,rejectH0.WeconcludethatthemedianresalepriceforhomesinBoston,Massachusettsisabovethenationalmedian.19-17www.khdaw.com 课后答案网www.khdaw.comChapter1940.UsetheWilcoxonSignedRankTestHomemakerDifferenceSignedRank1-250-11240235034-150-65-330-126-180-77-190-8.5www.khdaw.com8-230-109-100-510-190-8.511-90-412201T=-66µ=0Tnn(+1)(2n+1)12(13)(25)σ===25.5T66RejectH0ifz<-1.96orifz>1.96T−µ−66Tz===−2.59σ25.5TRejectH0;concludethatthemodelsdifferintermsofsellingprices.41.RankofDifferenceAbsoluteDifferenceSignedRank1.51010.01.299.0-.22.5-2.50⎯⎯.544.0.766.0.877.01.088.00⎯⎯.655.0.22.52.5-.011-1.0T=4819-18www.khdaw.com 课后答案网www.khdaw.comNonparametricMethodsnn(+1)(2n+1)10(11)(21)σ===19.62T66RejectH0ifz>1.645T−µ48Tz===2.45σ19.62TRejectH0;concludethatthereisasignificantweightgain.42.UsetheMWWtest.Sumofranks(line1)=70www.khdaw.comSumofranks(line2)=183T=7011µ=nn(+n+1)=10(23)115=T1122211σ=nnn(+n+1)=(10)(12)(23)=15.17T12121212RejectH0ifz<-1.645orifz>1.645T−µ70115−Tz===−2.97σ15.17TRejectH0;concludethattheweightsdifferforthetwoproductionlines.43.Method1Method2Method38.54.52.015.014.07.06.016.010.017.08.51.018.012.53.012.511.04.577.066.527.522212⎡(77)(66.5)(27.5)⎤W=⎢++⎥−3(19)=7.956(18)(19)⎣666⎦2χ=5.99147(3degreesoffreedom).05Since7.956>5.99147,werejectH0andconcludethatthereisasignificantdifferenceamongthemethods.19-19www.khdaw.com 课后答案网www.khdaw.comChapter1944.NoProgramCompanyProgramOffSiteProgram1612792011017415192116313188145www.khdaw.com741063022212⎡(74)(106)(30)⎤W=⎢++⎥−3(21)12.61=(20)(21)⎣677⎦2χ=7.37776(2degreesoffreedom).05Since12.61>7.37776,werejectH0;thereisasignificantdifferenceamongtheprograms.45.BlackJenningsSwansonWilson22.520.522.59.59.527.06.017.58.07.02.51.02.517.512.55.026.028.517.524.04.028.512.520.517.515.025.014.011.072.5171.5113.577.5222212⎡(72.5)(171.5)(113.5)(77.5)⎤W=⎢+++⎥−3(30)=6.344(29)(30)⎣6896⎦2χ=7.81473(3degreesoffreedom).05Since6.344<7.81473wecannotrejectH0.Wecannotconcludethatthereisasignificantdifferenceamongthecourseevaluationratingsforthe4instructors.246.Σd=136i26Σd6(136)ir=−1=−1=.76s2nn(−1)15(224)19-20www.khdaw.com 课后答案网www.khdaw.comNonparametricMethods11σ===.2673rsn−114RejectH0ifz<-1.645orifz>1.645rs−µr.76z=s==2.84σ.2673rsRejectH0;concludethatthereisasignificantrankcorrelationbetweenthetwoexams.www.khdaw.com19-21www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter20StatisticalSStatisticaltatisticalMethodsMMethodsethodsforffororQualityQQualityualityControlCControlontrolLearningLLearningearningObjectivesOObjectivesbjectives1.Learnabouttheimportanceofqualitycontrolandhowstatisticalmethodscanassistinthequalitycontrolprocess.2.Learnaboutacceptancesamplingprocedures.www.khdaw.com3.Knowthedifferencebetweenconsumer’sriskandproducer’srisk.4.Beabletousethebinomialprobabilitydistributiontodevelopacceptancesamplingplans.5.Knowwhatismeantbymultiplesamplingplans.6.Beabletoconstructqualitycontrolchartsandunderstandhowtheyareusedforstatisticalprocesscontrol.7.Knowthedefinitionsofthefollowingterms:producer"sriskassignablecausesconsumer"sriskcommoncausesacceptancesamplingcontrolchartsacceptablecriterionuppercontrollimitoperatingcharacteristiccurvelowercontrollimit20-1www.khdaw.com 课后答案网www.khdaw.comChapter20Solutions:SSolutions:olutions:1.a.Forn=4UCL=µ+3(σ/n)=12.5+3(.8/4)=13.7LCL=µ-3(σ/n)=12.5-3(.8/4)=11.3b.Forn=8UCL=µ+3(.8/8)=13.35LCL=µ-3(.8/8)=11.65Forn=16UCL=µ+3(.8/16)=13.10www.khdaw.comLCL=µ-3(.8/16)=11.90c.UCLandLCLbecomeclosertogetherasnincreases.Iftheprocessisincontrol,thelargersamplesshouldhavelessvarianceandshouldfallcloserto12.5.677.52.a.µ==5.4225(5)b.UCL=µ+3(σ/n)=5.42+3(.5/5)=6.09LCL=µ-3(σ/n)=5.42-3(.5/5)=4.751353.a.p==0.054025(100)p(1−p)0.0540(0.9460)b.σ===0.0226pn100c.UCL=p+3σ=0.0540+3(0.0226)=0.1218pLCL=p-3σ=0.0540-3(0.0226)=-0.0138pUseLCL=04.RChart:UCL=RD=1.6(1.864)=2.984LCL=RD=1.6(0.136)=0.223xChart:UCL=x+AR=28.5+0.373(1.6)=29.102LCL=x−AR=28.5-0.373(1.6)=27.9025.a.UCL=µ+3(σ/n)=128.5+3(.4/6)=128.99LCL=µ-3(σ/n)=128.5-3(.4/6)=128.0120-2www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControl772.4b.x=Σx/n==128.73incontroli6774.3c.x=Σx/n==129.05outofcontroli620.12+19.906.ProcessMean==20.012UCL=µ+3(σ/n)=20.01+3(σ/5)=20.12Solveforσ:(20.12−20.01)5σ==0.082www.khdaw.com37.SampleNumberObservationsxRii131422833.6714226183526.3317325303429.679417252121.008538293534.009641423639.676721172922.3312832262828.676941343336.0081029173025.33131126314032.33141223192522.3361317243224.33151443351731.67261518252924.00111630423134.33121728363232.0081840293133.33111918292825.00112022342627.3312R=11.4andx=29.17RChart:UCL=RD=11.4(2.575)=29.354LCL=RD=11.4(0)=03xChart:UCL=x+AR=29.17+1.023(11.4)=40.82LCL=x−AR=29.17-1.023(11.4)=17.5220-3www.khdaw.com 课后答案网www.khdaw.comChapter20RChart:30UCL=29.320R=11.4www.khdaw.com100LCL=01234567891011121314151617181920SampleNumberxChart:UCL=40.840=30x=29.1720LCL=17.51234567891011121314151617181920SampleNumber1418.a.p==0.047020(150)p(1−p)0.0470(0.9530)b.σ===0.0173pn150UCL=p+3σ=0.0470+3(0.0173)=0.0989pLCL=p-3σ=0.0470-3(0.0173)=-0.0049p20-4www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControlUseLCL=012c.p==0.08150Processshouldbeconsideredincontrol.d.p=.047,n=150UCL=np+3np(1−p)=150(0.047)+3150(0.047)(0.953)=14.826LCL=np-3np(1−p)=150(0.047)-3150(0.047)(0.953)=-0.726Thus,theprocessisoutofcontrolifmorethan14defectivepackagesarefoundinasampleof150.www.khdaw.come.Processshouldbeconsideredtobeincontrolsince12defectivepackageswerefound.f.Thenpchartmaybepreferredbecauseadecisioncanbemadebysimplycountingthenumberofdefectivepackages.9.a.Totaldefectives:165165p==0.041320(200)p(1−p)0.0413(0.9587)b.σ===0.0141pn200UCL=p+3σ=0.0413+3(0.0141)=0.0836pLCL=p-3σ=0.0413+3(0.0141)=-0.0010pUseLCL=020c.p==0.10Outofcontrol200d.p=.0413,n=200UCL=np+3np(1−p)=200(0.0413)+3200(0.0413)(0.9587)=16.702LCL=np-3np(1−p)=200(0.0413)-3200(0.0413)(0.9587)=0.1821e.Theprocessisoutofcontrolsince20defectivepistonswerefound.n!xn−x10.f(x)=p(1−p)x!(n−x)!Whenp=.02,theprobabilityofacceptingthelotis25!025f(0)=(0.02)(1−0.02)=0.60350!(25−0)!20-5www.khdaw.com 课后答案网www.khdaw.comChapter20Whenp=.06,theprobabilityofacceptingthelotis25!025f(0)=(0.06)(1−0.06)=0.21290!(25−0)!11.a.Usingbinomialprobabilitieswithn=20andp0=.02.P(Acceptlot)=f(0)=.6676Producer’srisk:α=1-.6676=.3324b.P(Acceptlot)=f(0)=.2901www.khdaw.comProducer’srisk:α=1-.2901=.709912.Atp0=.02,then=20andc=1planprovidesP(Acceptlot)=f(0)+f(1)=.6676+.2725=.9401Producer’srisk:α=1-.9401=.0599Atp0=.06,then=20andc=1planprovidesP(Acceptlot)=f(0)+f(1)=.2901+.3703=.6604Producer’srisk:α=1-.6604=.3396Foragivensamplesize,theproducer’sriskdecreasesastheacceptancenumbercisincreased.13.a.Usingbinomialprobabilitieswithn=20andp0=.03.P(Acceptlot)=f(0)+f(1)=.5438+.3364=.8802Producer’srisk:α=1-.8802=.1198b.Withn=20andp1=.15.P(Acceptlot)=f(0)+f(1)=.0388+.1368=.1756Consumer’srisk:β=.1756c.Theconsumer’sriskisacceptable;however,theproducer’sriskassociatedwiththen=20,c=1planisalittlelargerthandesired.20-6www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControl14.P(Accept)Producer’sP(accept)Consumer’scp0=.05Riskαp1=.30Riskβ(n=10)0.5987.4013.0282.02821.9138.0862.1493.14932.9884.0116.3828.3828(n=15)0.4633.5367.0047.00471.8291.1709.0352.03522.9639.0361.1268.12683.9946.0054.2968.2968www.khdaw.com(n=20)0.3585.6415.0008.00081.7359.2641.0076.00762.9246.0754.0354.03543.9842.0158.1070.1070Theplanwithn=15,c=2isclosewithα=.0361andβ=.1268.However,theplanwithn=20,c=3isnecessarytomeetbothrequirements.15.a.P(Accept)shownforpvaluesbelow:cp=.01p=.05p=.08pp=.15=.100.8179.3585.1887.1216.03881.9831.7359.5169.3918.17562.9990.9246.7880.6770.4049TheoperatingcharacteristiccurveswouldshowtheP(Accept)versuspforeachvalueofc.b.P(Accept)cAtp0Producer’sRiskAtp1=.08Consumer’sRisk=.010.8179.1821.1887.18871.9831.0169.5169.51692.9990.0010.7880.7880Σx190816.a.µ===95.42020b.UCL=µ+3(σ/n)=95.4+3(.50/5)=96.07LCL=µ-3(σ/n)=95.4-3(.50/5)=94.73c.No;allwereincontrol17.a.Forn=1020-7www.khdaw.com 课后答案网www.khdaw.comChapter20UCL=µ+3(σ/n)=350+3(15/10)=364.23LCL=µ-3(σ/n)=350-3(15/10)=335.77Forn=20UCL=350+3(15/20)=360.06LCL=350-3(15/20)=339.94Forn=30UCL=350+3(15/30)=358.22LCL=350-3(15/30)=343.78www.khdaw.comb.Bothcontrollimitscomeclosertotheprocessmeanasthesamplesizeisincreased.c.Theprocesswillbedeclaredoutofcontrolandadjustedwhentheprocessisincontrol.d.Theprocesswillbejudgedincontrolandallowedtocontinuewhentheprocessisoutofcontrol.e.Allhavez=3wherearea=.4986P(TypeI)=1-2(.4986)=.002818.RChart:UCL=RD=2(2.115)=4.234LCL=RD=2(0)=03xChart:UCL=x+AR=5.42+0.577(2)=6.572LCL=x−AR=5.42-0.577(2)=4.272EstimateofStandardDeviation:R2σ̂===0.86d2.326219.R=0.665x=95.398xChart:UCL=x+AR=95.398+0.577(0.665)=95.7822LCL=x−AR=95.398-0.577(0.665)=95.0142RChart:UCL=RD=0.665(2.115)=1.4064LCL=RD=0.665(0)=0320-8www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControlTheRchartindicatedtheprocessvariabilityisincontrol.Allsamplerangesarewithinthecontrollimits.However,theprocessmeanisoutofcontrol.Sample11(x=95.80)andSample17(x=94.82)falloutsidethecontrollimits.20.R=.053x=3.082xChart:UCL=x+AR=3.082+0.577(0.053)=3.1122LCL=x−AR=3.082-0.577(0.053)=3.0512RChart:UCL=RD=0.053(2.115)=0.1121www.khdaw.com4LCL=RD=0.053(0)=03Alldatapointsarewithinthecontrollimitsforbothcharts.21.a..08UCL.06.04.02LCL0Warning:Processshouldbechecked.Allpointsarewithincontrollimits;however,allpointsarealsogreaterthantheprocessproportiondefective.20-9www.khdaw.com 课后答案网www.khdaw.comChapter20b.25UCL24www.khdaw.com23LCL22Warning:Processshouldbechecked.AllpointsarewithincontrollimitsyetthetrendinpointsshowamovementorshifttowardUCLout-of-controlpoint.22.a.p=.04p(1−p)0.04(0.96)σ===0.0139pn200UCL=p+3σ=0.04+3(0.0139)=0.0817pLCL=p-3σ=0.04-3(0.0139)=-0.0017pUseLCL=0b.20-10www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControloutofcontrolUCL(.082).04LCL(0)Formonth1p=10/200=0.05.Othermonthlyvaluesare.075,.03,.065,.04,and.085.Onlythelastmonthwithp=0.085isanout-of-controlsituation.www.khdaw.com23.a.Usebinomialprobabilitieswithn=10.Atp0=.05,P(Acceptlot)=f(0)+f(1)+f(2)=.5987+.3151+.0746=.9884Producer’sRisk:α=1-.9884=.0116Atp1=.20,P(Acceptlot)=f(0)+f(1)+f(2)=.1074+.2684+.3020=.6778Consumer’srisk:β=.6778b.Theconsumer’sriskisunacceptablyhigh.Toomanybadlotswouldbeaccepted.c.Reducingcwouldhelp,butincreasingthesamplesizeappearstobethebestsolution.24.a.P(Accept)areshownbelow:(Usingn=15)p=.01p=.02p=.03p=.04p=.05f(0).8601.7386.6333.5421.4633f(1).1303.2261.2938.3388.3658.9904.9647.9271.8809.8291α=1-P(Accept).0096.0353.0729.1191.1709Usingp0=.03sinceαiscloseto.075.Thus,.03isthefractiondefectivewheretheproducerwilltoleratea.075probabilityofrejectingagoodlot(only.03defective).b.p=.25f(0).0134f(1).066820-11www.khdaw.com 课后答案网www.khdaw.comChapter20β=.080225.a.P(Accept)whenn=25andc=0.Usethebinomialprobabilityfunctionwithn!xn−xf(x)=p(1−p)x!(n−x)!or25!02525f(0)=p(1−p)=(1−p)0!25!Iff(0)p=.01.7778p=.03.4670p=.10.0718p=.20.0038www.khdaw.comb.1.0.8.6P(Accept).4.2.00.02.04.06.08.10.12.14.16.18.20PercentDefectivec.1-f(0)=1-.778=.22226.a.µ=np=250(.02)=5σ=np(1−p)=250(0.02)(0.98)=2.21P(Accept)=P(x≤10.5)10.5−5z==2.492.21P(Accept)=.5000+.4936=.9936Producer’sRisk:α=1-.9936=.006420-12www.khdaw.com 课后答案网www.khdaw.comStatisticalMethodsforQualityControlb.µ=np=250(.08)=20σ=np(1−p)=250(0.08)(0.92)=4.29P(Accept)=P(x≤10.5)10.5−5z==−2.214.29P(Accept)=1-.4864=.0136Consumer’sRisk:β=.0136c.Theadvantageistheexcellentcontrolovertheproducer’sandtheconsumer’srisk.Thedisadvantageiswww.khdaw.comthecostoftakingalargesample.20-13www.khdaw.com 课后答案网www.khdaw.comChapterCChapterhapter21SampleSSampleampleSurveySSurveyurveyLearningLLearningearningObjectivesOObjectivesbjectives1.Learnwhatasamplesurveyisandhowitdiffersfromanexperimentasamethodofcollectingdata.2.Knowaboutthemethodsofdatacollectionforasurvey.www.khdaw.com3.Knowthedifferencebetweensamplingandnonsamplingerror.4.Learnaboutfoursampledesigns:(1)simplerandomsampling,(2)stratifiedsimplerandomsampling,(3)clustersampling,and(4)systematicsampling.5.Leanhowtoestimateapopulationmean,apopulationtotal,andapopulationproportionusingtheabovesampledesigns.6.Understandtherelationshipbetweensamplesizeandprecision.7.Learnhowtochoosetheappropriatesamplesizeusingstratifiedandsimplerandomsampling.8.Learnhowtoallocatethetotalsampletothevariousstratausingstratifiedsimplerandomsampling.21-1www.khdaw.com 课后答案网www.khdaw.comChapter21Solutions:SSolutions:olutions:1.a.x=215isanestimateofthepopulationmean.2080050−b.s==2.7386x50800c.215±2(2.7386)or209.5228to220.47722.a.Estimateofpopulationtotal=Nx=400(75)=30,000b.EstimateofStandardError=Nsx⎛8⎞40080−Ns=400⎜⎟=320xwww.khdaw.com⎝80⎠400c.30,000±2(320)or29,360to30,6403.a.p=.30isanestimateofthepopulationproportion⎛1000100−⎞⎛(.3)(.7)⎞b.s=⎜⎟⎜⎟=.0437p⎝1000⎠⎝99⎠c..30±2(.0437)or.2126to.38744.B=152(70)4900n===72.983022(15)(70)67.1389+4450Asamplesizeof73willprovideanapproximate95%confidenceintervalofwidth30.5.a.x=149,670ands=73,4207715073,420−⎛⎞s=⎜⎟=10,040.83x771⎝50⎠approximate95%confidenceinterval149,670±2(10,040.83)or$129,588.34to$169,751.66b.�X=Nx=771(149,670)=115,395,570sx̂=Nsx=771(10,040.83)=7,741,479.9321-2www.khdaw.com 课后答案网www.khdaw.comSampleSurveyapproximate95%confidenceinterval115,395,770±2(7,741,479.93)or$99,912,810.14to$130,878,729.86⎛77150−⎞⎛(.36)(.64)⎞c.p=18/50=0.36ands=⎜⎟⎜⎟=.0663p⎝771⎠⎝49⎠approximate95%confidenceinterval0.36±2(0.0663)or0.2274to0.4926www.khdaw.comThisisaratherlargeinterval;samplesizesmustberatherlargetoobtaintightconfidenceintervalsonapopulationproportion.6.B=5000/2=2500Usethevalueofsforthepreviousyearintheformulatodeterminethenecessarysamplesize.2(31.3)979.69n===336.005122(2.5)(31.3)2.9157+4724Asamplesizeof337willprovideanapproximate95%confidenceintervalofwidthnolargerthan$5000.7.a.Stratum1:=138Stratum2:x=1032Stratum3:x=2103b.Stratum1x=1381⎛30⎞20020−s=⎜⎟=6.3640x1⎝20⎠200138±2(6.3640)or125.272to150.728Stratum2x=103221-www.khdaw.com 课后答案网www.khdaw.comChapter21⎛25⎞25030−s=⎜⎟=4.2817x2⎝30⎠250103±2(4.2817)or94.4366to111.5634Stratum3x=2103⎛50⎞10025−s=⎜⎟=8.6603x3⎝25⎠100210±2(8.6603)www.khdaw.comor192.6794to227.3206⎛200⎞⎛250⎞⎛100⎞c.x=⎜⎟138+⎜⎟103+⎜⎟210st⎝550⎠⎝550⎠⎝550⎠=50.1818+46.8182+38.1818=135.1818222⎛1⎞⎛(30)(25)(50)⎞s=⎜⎟⎜200(180)+250(220)+100(75)⎟xst2⎝(550)⎠⎝203025⎠⎛1⎞=⎜⎟3,515,833.3=3.40922⎝(550)⎠approximate95%confidenceinterval135.1818±2(3.4092)or128.3634to142.00028.a.Stratum1:Nx=200(138)=27,60011Stratum2:Nx=250(103)=25,75022Stratum3:Nx=100(210)=21,00033b.Nx=27,600+25,750+21,000=74,350stNote:thesumoftheestimateforeachstratumtotalequalsNxstc.Nx=550(3.4092)=1875.06(see7c)stapproximate95%confidenceinterval21-4www.khdaw.com 课后答案网www.khdaw.comSampleSurvey74,350±2(1875.06)or70,599.88to78,100.129.a.Stratum1p=.501⎛20020−⎞⎛(.50)(.50)⎞s=⎜⎟⎜⎟=.1088p1⎝200⎠⎝19⎠50±2(.1088)or.2824to.7176www.khdaw.comStratum2p=.782⎛25030−⎞⎛(.78)(.22)⎞s=⎜⎟⎜⎟=.0722p2⎝250⎠⎝29⎠.78±2(.0722)or.6356to.9244Stratum3p=.213⎛10025−⎞⎛(.21)(.79)⎞s=⎜⎟⎜⎟=.0720p3⎝100⎠⎝24⎠.21±2(.0720)or.066to.354200250100b.p=(.50)+(.78)+(.21)=.5745st550550550⎛1⎞⎛(.5)(.5)(.78)(.22)(.21)(.79)⎞c.s=⎜⎟⎜200(180)+250(220)+100(75)⎟pst2⎝(550)⎠⎝192924⎠⎛1⎞=⎜⎟(473.6842325.448351.8438)++=.05302⎝(550)⎠d.approximate95%confidenceinterval.5745±2(.0530)21-www.khdaw.com 课后答案网www.khdaw.comChapter21or.4685to.68052[300(150)600(75)500(100)++](140,000)210.a.n===92.83592⎛(20)⎞196,000,00015,125,000+222(1400)⎜⎟+⎡300(150)600(75)++500(100)⎤2⎣⎦⎝⎠Roundingupwechooseatotalsampleof93.⎛300(150)⎞n=93⎜⎟=301⎝140,000⎠⎛600(75)⎞n=93⎜⎟=302www.khdaw.com⎝140,000⎠⎛500(100)⎞n=93⎜⎟=333⎝140,000⎠b.WithB=10,thefirstterminthedenominatorintheformulafornchanges.22(140,000)(140,000)n===305.65302⎛(10)⎞49,000,00015,125,000+2(1400)⎜⎟+15,125,000⎝4⎠Roundingup,weseethatasamplesizeof306isneededtoprovidethislevelofprecision.⎛300(150)⎞n=306⎜⎟=981⎝140,000⎠⎛600(75)⎞n=306⎜⎟=982⎝140,000⎠⎛500(100)⎞n=306⎜⎟=1093⎝140,000⎠Duetorounding,thetotaloftheallocationstoeachstrataonlyaddto305.Notethateventhoughthesamplesizeislarger,theproportionallocatedtoeachstratumhasnotchanged.22(140,000)(140,000)n===274.60602(15,000)56,250,00015,125,000++15,125,0004Roundingup,weseethatasamplesizeof275willprovidethedesiredlevelofprecision.Theallocationstothestrataareinthesameproportionasforpartsaandb.21-6www.khdaw.com 课后答案网www.khdaw.comSampleSurvey⎛300(150)⎞n=275⎜⎟=981⎝140,000⎠⎛600(75)⎞n=275⎜⎟=882⎝140,000⎠⎛500(100)⎞n=275⎜⎟=983⎝140,000⎠Again,duetorounding,thestratumallocationsdonotaddtothetotalsamplesize.Anotheritemcouldbesampledfrom,say,stratum3ifdesired.11.a.x=29.5333x=64.77512x=45.2125x=53.0300www.khdaw.com34b.Indianapolis⎛13.3603⎞386−29.5332±⎜⎟⎝6⎠3829.533±10.9086(.9177)or19.5222to39.5438Louisville⎛25.0666⎞458−64.7752±⎜⎟⎝8⎠4564.775±17.7248(.9068)or48.7022to80.8478St.Louis⎛19.4084⎞808−45.21252±⎜⎟⎝8⎠8045.2125±(13.7238)(.9487)or32.1927to58.2323Memphis⎛29.6810⎞7010−53.03002±⎜⎟⎝10⎠7053.0300±18.7719(.9258)or35.6510to70.409021-www.khdaw.com 课后答案网www.khdaw.comChapter21⎛38⎞⎛⎞⎛145⎞⎛⎞⎛580⎞⎛⎞⎛370⎞⎛5⎞c.p=⎜⎟⎜⎟⎜+⎟⎜⎟⎜+⎟⎜⎟⎜+⎟⎜⎟=.4269st⎝233⎠⎝⎠⎝6233⎠⎝⎠⎝8233⎠⎝⎠⎝8233⎠⎝10⎠⎛⎞⎛⎞15⎜⎟⎜⎟p(1−p)⎝⎠⎝⎠6611d.NN(−n)=38(32)=33.7778111n−151⎛⎞⎛⎞53⎜⎟⎜⎟p(1−p)⎝⎠⎝⎠8822NN(−n)=45(37)=55.7478222n−172⎛⎞⎛⎞35⎜⎟⎜⎟p(1−p)⎝⎠⎝⎠88www.khdaw.com33NN(−n)=80(72)=192.8571333n−173⎛5⎞⎛5⎞⎜⎟⎜⎟p(1−p)⎝10⎠⎝10⎠44NN(−n)=70(60)=116.6667444n−194⎛1⎞1spst=⎜2⎟[33.777855.7478192.8571116.6667+++]=2(399.0494)=.0857⎝(233)⎠(233)approximate95%confidenceinterval.4269±2(.0857)or.2555to.598312.a.St.Louistotal=Nx=80(45.2125)=361711Indollars:$3,617,000b.Indianapolistotal=Nx=38(29.5333)=1122.265411Indollars:$1,122,265⎛38⎞⎛45⎞⎛80⎞⎛70⎞c.x=⎜⎟29.5333+⎜⎟64.775+⎜⎟45.2125+⎜⎟53.0300=48.7821st⎝233⎠⎝233⎠⎝233⎠⎝233⎠22s(13.3603)1NN(−n)=38(32)=36,175.517111n6122s(25.0666)2NN(−n)=45(37)=130,772.1222n8221-8www.khdaw.com 课后答案网www.khdaw.comSampleSurvey22s(19.4084)3NN(−n)=80(72)=271,213.91333n8322s(29.6810)4NN(−n)=70(60)=370,003.94444n104⎛1⎞sxst=⎜2⎟[36,175.517130,772.1271,213.91370,003.94+++]⎝(233)⎠1=(808,165.47)=3.85832(233)approximate95%confidenceintervalx±2swww.khdaw.comstxst48.7821±2(3.8583)or41.0655to56.4987Indollars:$41,066to$56,499d.approximate95%confidenceintervalNx±2Nsstxst233(48.7821)±2(233)(3.8583)11,366.229±1797.9678or9,568.2612to13,164.197Indollars:$9,568,261to$13,164,1972[50(80)38(150)35(45)++](11,275)213.n===27.3394⎛(30)2⎞3,404,0251,245,875+2222(123)⎜⎟+⎡50(80)+38(150)+35(45)⎤4⎣⎦⎝⎠Roundingupweseethatasamplesizeof28isnecessarytoobtainthedesiredprecision.⎛50(80)⎞n=28⎜⎟=101⎝11,275⎠⎛38(150)⎞n=28⎜⎟=142⎝11,275⎠⎛35(45)⎞n=28⎜⎟=43⎝11,275⎠21-www.khdaw.com 课后答案网www.khdaw.comChapter2122[50(100)38(100)35(100)++][123(100)]b.n===3322⎛(30)⎞3,404,025123(100)+2222(123)⎜⎟+⎡50(100)+38(100)+35(100)⎤4⎣⎦⎝⎠⎛50(100)⎞n=33⎜⎟=131⎝12,300⎠⎛38(100)⎞n=33⎜⎟=102⎝12,300⎠⎛35(100)⎞n=33⎜⎟=93⎝12,300⎠Thisisthesameasproportionalallocation.Notethatforeachstratumwww.khdaw.com⎛Nh⎞n=n⎜⎟h⎝N⎠∑x750i14.a.x===15c∑M50i�X=Mx=300(15)=4500c∑a15ip===.30c∑M50i2b.∑(x−xM)=[95-15(7)]2+[325-15(18)]2+[190-15(15)]2+[140-15(10)]2ici=(-10)2+(55)2+(-35)2+(-10)2=4450⎛254−⎞⎛4450⎞s=⎜⎟⎜⎟=1.4708xc2⎝(25)(4)(12)⎠⎝3⎠sX�=Msx=300(1.4708)=441.24c2∑(a−pM)=[1-.3(7)]2+[6-.3(18)]2+[6-.3(15)]2+[2-.3(10)]2ici=(-1.1)2+(.6)2+(1.5)2+(-1)2=4.82⎛254−⎞⎛4.82⎞s=⎜⎟⎜⎟=.0484pc2⎝(25)(4)(12)⎠⎝3⎠c.approximate95%confidenceIntervalforPopulationMean:15±2(1.4708)or12.0584to17.941621-10www.khdaw.com 课后答案网www.khdaw.comSampleSurveyd.approximate95%confidenceIntervalforPopulationTotal:4500±2(441.24)or3617.52to5382.48e.approximate95%confidenceIntervalforPopulationProportion:.30±2(.0484)or.2032to.396810,40015.a.x==80cwww.khdaw.com130�X=Mx=600(80)=48,000c13p==.10c1302b.∑(x−xM)=[3500-80(35)]2+[965-80(15)]2+[960-80(12)]2ici+[2070-80(23)]2+[1100-80(20)]2+[1805-80(25)]2=(700)2+(-235)2+(0)2+(230)2+(-500)2+(-195)2=886,150⎛306−⎞⎛886,150⎞s=⎜⎟⎜⎟=7.6861xc2⎝(30)(6)(20)⎠⎝5⎠approximate95%confidenceIntervalforPopulationMean:80±2(7.6861)or64.6278to95.3722c.s�X=600(7.6861)=4611.66approximate95%confidenceIntervalforPopulationTotal:48,000±2(4611.66)or38,776.68to57,223.322∑(a−pM)=[3-.1(35)]2+[0-.1(15)]2+[1-.1(12)]2+[4-.1(23)]2ici+[3-.1(20)]2+[2-.1(25)]2=(-.5)2+(-1.5)2+(-.2)2+(1.7)2+(1)2+(-.5)2=6.6821-www.khdaw.com 课后答案网www.khdaw.comChapter21⎛306−⎞⎛6.68⎞s=⎜⎟⎜⎟=.0211pc2⎝(30)(6)(20)⎠⎝5⎠approximate95%confidenceIntervalforPopulationProportion:.10±2(.0211)or.0578to.1422200016.a.x==40cwww.khdaw.com50Estimateofmeanageofmechanicalengineers:40years35b.p==.70c50Estimateofproportionattendinglocaluniversity:.702c.∑(x−xM)=[520-40(12)]2+···+[462-40(13)]2ici=(40)2+(-7)2+(-10)2+(-11)2+(30)2+(9)2+(22)2+(8)2+(-23)2+(-58)2=7292⎛12010−⎞⎛7292⎞s=⎜⎟⎜⎟=2.0683xc2⎝(120)(10)(50/12)⎠⎝9⎠approximate95%confidenceIntervalforMeanage:40±2(2.0683)or35.8634to44.13662d.∑(a−pM)=[8-.7(12)]2+···+[12-.7(13)]2ici=(-.4)2+(-.7)2+(-.4)2+(.3)2+(-1.2)2+(-.1)2+(-1.4)2+(.3)2+(.7)2+(2.9)2=13.3⎛12010−⎞⎛13.3⎞s=⎜⎟⎜⎟=.0883pc2⎝(120)(10)(50/12)⎠⎝9⎠approximate95%confidenceIntervalforProportionAttendingLocalUniversity:21-12www.khdaw.com 课后答案网www.khdaw.comSampleSurvey.70±2(.0883)or.5234to.876617(37)35(32)++⋅⋅⋅+57(44)11,24017.a.x===36.9737c1735++⋅⋅⋅+57304Estimateofmeanage:36.9737yearsb.ProportionofCollegeGraduates:128/304=.4211ProportionofMales:112/304=.36842c.∑(x−xM)=[17(37)-(36.9737)(17)]2+···+[57(44)-(36.9737)(44)]2ici=(.4471)2+(-174.0795)2+(-25.3162)2+(-460.2642)2+(173.1309)2www.khdaw.com+(180.3156)2+(-94.7376)2+(400.4991)2=474,650.68⎛1508−⎞⎛474,650.68⎞s=⎜⎟⎜⎟=2.2394xc2⎝(150)(8)(40)⎠⎝7⎠approximate95%confidenceIntervalforMeanAgeofAgents:36.9737±2(2.2394)or32.4949to41.45252d.∑(a−pM)=[3-.4211(17)]2+···+[25-.4211(57)]2ici=(-4.1587)2+(-.7385)2+(-2.9486)2+(10.2074)2+(-.1073)2+(-3.0532)2+(-.2128)2+(.9973)2=141.0989⎛1508−⎞⎛141.0989⎞s=⎜⎟⎜⎟=.0386pc2⎝(150)(8)(40)⎠⎝7⎠approximate95%confidenceIntervalforProportionofAgentsthatareCollegeGraduates:.4211±2(.0386)or.3439to.49832e.∑(a−pM)=[4-.3684(17)]2+···+[26-.3684(57)]2ici=(-2.2628)2+(-.8940)2+(-2.5784)2+(3.6856)2+(-3.8412)2+(1.5792)2+(-.6832)2+(5.0012)2=68.878721-www.khdaw.com 课后答案网www.khdaw.comChapter21⎛1508−⎞⎛68.8787⎞s=⎜⎟⎜⎟=.0270pc2⎝(150)(8)(40)⎠⎝7⎠approximate95%confidenceIntervalforProportionofAgentsthatareMale:.3684±2(.0270)or.3144to.422418.a.p=0.19(0.19)(0.81)s==0.0206p363www.khdaw.comApproximate95%ConfidenceInterval:0.19±2(0.0206)or0.1488to0.2312b.p=0.31(0.31)(0.69)s==0.0243p363Approximate95%ConfidenceInterval:0.31±2(0.0243)or0.2615to0.3585c.p=0.17(0.17)(0.83)s==0.0197p373Approximate95%ConfidenceInterval:0.17±2(0.0197)or0.1306to0.2094d.Thelargeststandarderroriswhenp=.50.Atp=.50,weget(0.5)(0.5)s==0.0262p36321-14www.khdaw.com 课后答案网www.khdaw.comSampleSurveyMultiplyingby2,wegetaboundofB=2(.0262)=0.0525Forasampleof363,then,theyknowthatintheworstcase(p=0.50),theboundwillbeapproximately5%.e.Ifthepollwasconductedbycallingpeopleathomeduringthedaythesampleresultswouldonlyberepresentativeofadultsnotworkingoutsidethehome.ItislikelythattheLouisHarrisorganizationtookprecautionsagainstthisandotherpossiblesourcesofbias.19.a.Assume(N-n)/N≈1p=.55www.khdaw.com(0.55)(0.45)s==0.0222p504b.p=.31(0.31)(0.69)s==0.0206p504c.Theestimateofthestandarderrorinpart(a)islargerbecausepiscloserto.50.d.Approximate95%Confidenceinterval:.55±2(.0222)or.5056to.5944e.Approximate95%Confidenceinterval:.31±2(.0206).2688to.351230002003000−20.a.s==204.9390x3000200Approximate95%ConfidenceIntervalforMeanAnnualSalary:23,200±2(204.9390)or$22,790to$23,610b.Nx=3000(23,200)=69,600,000ŝ=3000(204.9390)=614,817xApproximate95%ConfidenceIntervalforPopulationTotalSalary:21-www.khdaw.com 课后答案网www.khdaw.comChapter2169,600,000±2(614,817)or$68,370,366to$70,829,634c.p=.73⎛3000200−⎞⎛(.73)(.27)⎞s=⎜⎟⎜⎟=.0304p⎝3000⎠⎝199⎠Approximate95%ConfidenceIntervalforProportionthatareGenerallySatisfied:.73±2(.0304)orwww.khdaw.com.6692to.7908d.Ifmanagementadministeredthequestionnaireandanonymitywasnotguaranteedwewouldexpectadefiniteupwardbiasinthepercentreportingtheywere“generallysatisfied”withtheirjob.Aprocedureforguaranteeinganonymityshouldreducethebias.21.a.p=1/3⎛38030−⎞⎛(1/3)(2/3)⎞s=⎜⎟⎜⎟=.0840p⎝380⎠⎝29⎠Approximate95%ConfidenceInterval:.3333±2(.0840)or.1653to.5013b.�X2=760(19/45)=320.8889c.p=19/45=.4222⎛76045−⎞⎛(19/45)(26/45)⎞s=⎜⎟⎜⎟=.0722p⎝760⎠⎝44⎠Approximate95%ConfidenceInterval:.4222±2(.0722)or.2778to.5666⎛380⎞⎛10⎞⎛760⎞⎛19⎞⎛260⎞⎛7⎞d.p=⎜⎟⎜⎟⎜+⎟⎜⎟⎜+⎟⎜⎟=.3717st⎝1400⎠⎝30⎠⎝1400⎠⎝45⎠⎝1400⎠⎝25⎠⎡p(1−p)⎤(1/3)(2/3)hh∑NN(−n)⎢⎥=380(350)hhh⎣n−1⎦29h21-16www.khdaw.com 课后答案网www.khdaw.comSampleSurvey(19/45)(26/45)(7/25)(18/25)+760(715)+260(235)4424=1019.1571+3012.7901+513.2400=4545.1892⎛1⎞s=⎜⎟4545.1892=.0482pst2⎝(1400)⎠Approximate95%ConfidenceInterval:.3717±2(.0482)orwww.khdaw.com.2753to.468122.a.�X=380(9/30)+760(12/45)+260(11/25)=431.0667Estimateapproximately431deathsduetobeating.⎛380⎞⎛9⎞⎛760⎞⎛12⎞⎛260⎞⎛11⎞b.p=⎜⎟⎜⎟⎜+⎟⎜⎟⎜+⎟⎜⎟=.3079st⎝1400⎠⎝30⎠⎝1400⎠⎝45⎠⎝1400⎠⎝25⎠[ph(1−ph)]∑NN(−n)hhhn−1h=(380)(380-30)(9/30)(21/30)/29+(760)(760-45)(12/45)(33/45)/44+(260)(260-25)(11/25)(14/25)/24=4005.5079⎛1⎞s=⎜⎟4005.5079=.0452pst2⎝(1400)⎠Approximate95%ConfidenceInterval:.3079±2(.0452)or.2175to.3983⎛380⎞⎛21⎞⎛760⎞⎛34⎞⎛260⎞⎛15⎞c.p=⎜⎟⎜⎟⎜+⎟⎜⎟⎜+⎟⎜⎟=.7116st⎝1400⎠⎝30⎠⎝1400⎠⎝45⎠⎝1400⎠⎝25⎠[ph(1−ph)]∑NN(−n)hhhn−1h=(380)(380-30)(21/30)(9/30)/29+(760)(760-45)(34/45)(11/45)/44+(260)(260-25)(15/25)(10/25)/2421-www.khdaw.com 课后答案网www.khdaw.comChapter21=3855.0417⎛1⎞s=⎜⎟3855.0417=.0443pst2⎝(1400)⎠Approximate95%ConfidenceInterval:.7116±2(.0443)or.6230to.8002d.�X=1400(.7116)=996.24Estimateoftotalnumberofblackvictims≈9962www.khdaw.com[3000(80)600(150)250(220)100(700)50(3000)++++]23.a.n=22⎛(20)⎞22222(4000)⎜⎟+3000(80)+600(150)+250(220)+100(700)+50(3000)⎝4⎠366,025,000,000==170.73651,600,000,000543,800,000+Roundingup,weneedasamplesizeof171forthedesiredprecision.⎛3000(80)⎞n=171⎜⎟=681⎝605,000⎠⎛600(150)⎞n=171⎜⎟=252⎝605,000⎠⎛250(220)⎞n=171⎜⎟=163⎝605,000⎠⎛100(700)⎞n=171⎜⎟=204⎝605,000⎠⎛50(3000)⎞n=171⎜⎟=425⎝605,000⎠14(61)7(74)96(78)23(69)71(73)29(84)+++++18,06624.a.x===75.275c14796237129+++++240Estimateofmeanageisapproximately75yearsold.1223081022+++++84b.p===.35c14796237129+++++2402∑(a−pM)=[12-.35(14)]2+[2-.35(7)]2+[30-.35(96)]2ici21-18www.khdaw.com 课后答案网www.khdaw.comSampleSurvey+[8-.35(23)]2+[10-.35(71)]2+[22-.35(29)]2=(7.1)2+(-.45)2+(-3.6)2+(-.05)2+(-14.85)2+(11.85)2=424.52⎛1006−⎞⎛424.52⎞s=⎜⎟⎜⎟=.0760pc2⎝(100)(6)(48)⎠⎝5⎠Approximate95%ConfidenceInterval:.35±2(.0760)orwww.khdaw.com.198to.502�X=4800(.35)=1680EstimateoftotalnumberofDisabledPersonsis1680.21-www.khdaw.com'