概率與統(tǒng)計(jì)(英文)chapter8TestsofHypothesesBasedona課件

Chapter8

TestsofHypothesesBasedonaSingleSample

8.1HypothesesandTestProcedures8.2TestsAboutaPopulationMean8.3TestsConcerningPopulationProportion8.4P-Values8.5SomeCommentsonSelectingaTestProcedureChapter8

TestsofHypot1IntroductionAparametercanbeestimatedfromsampledataeitherbyasinglenumber(apointestimated)oranentireintervalofplausiblevalues(aconfidenceinterval).Frequently,however,theobjectiveofaninvestigationisnottoestimateaparameterbuttodecidewhichoftwocontradictoryclaimsabouttheparameteriscorrect.

Methodsforaccomplishingthiscomprisethepartofstatisticalinferencecalledhypothesistesting.Inthischapter,wefirstdiscusssomeofthebasicconceptsandterminologyinhypothesistestingandthendevelopdecisionproceduresforthemostfrequentlyencounteredtestingproblemsbasedonasamplefromasinglepopulation.IntroductionAparameter28.1HypothesesandTestProceduresAstatisticalhypothesis,orhypothesis,isaclaimeitheraboutthevalueofasinglepopulationcharacteristicoraboutthevaluesofseveralpopulationcharacteristics.Oneexampleofahypothesisistheclaimμ=0.75,whereμisthetrueaverageinsidediameterofacertaintypeofPVCpipe.Anotherexampleisthestatementp<0.10,wherepistheproportionofdetectivecircuitboardsamongallcircuitboardsproducedbyacertainmanufacture.Ifanddenotethetrueaveragebreakingstrengthsoftwodifferenttypesoftwine,onehypothesisistheassertionthat-=0,andanotheristhestatement->5.8.1HypothesesandTestProced3Inanyhypothesis-testingproblem,therearetwocontradictoryhypothesesunderconsideration.Onehypothesismightbetheclaimμ=0.75andtheotherμ.75,orthetwocontradictorystatementsmightbe

p≥.10andp<.10.Theobjectiveistodecide,basedonsampleinformation,whichofthetwohypothesisiscorrect.Thereisafamiliaranalogytothisinacriminaltrial.Oneclaimistheassertionthattheaccusedindividualisinnocent.IntheU.S.judicialsystem,thisistheclaimthatisinitiallybelievedtobetrue.Onlyinthefaceofstrongevidencetothecontraryshouldthejuryrejectthisclaiminfavorofthealternativeassertionthattheaccusedisguilty.Inanyhypothesis-testi4Inthissense,theclaimofinnocenceisthefavoredorprotectedhypothesis,andtheburdenofproofisplacedonthosewhobelieveinthealternativeclaim.Similarly,intestingstatisticalhypotheses,theproblemwillbeformulatedsothatoneoftheclaimsisinitiallyfavored.Thisinitiallyfavoredclaimwillnotberejectedinfavorofthealternativeclaimunlesssampleevidencecontradictsitandprovidesstrongsupportforthealternativeassertion.Inthissense,thecla5DEFINITIONThenullhypothesis,denotedby,istheclaimaboutoneormorepopulationcharacteristicsthatisinitiallyassumedtobetrue(the“priorbelief”claim).Thealternativehypothesis,denotedby,istheassertionthatiscontradictoryto.Thenullhypothesiswillberejectedinfavorofthealternativehypothesisonlyifsampleevidencesuggeststhat

isfalse.Ifthesampledoesnotstronglycontradict,wewillcontinuetobelieveinthetruthofthenullhypothesis.Thetwopossibleconclusionsfromahypothesis-testinganalysisarethenreject

orfailtoreject

.DEFINITIONThenullhypot6Atestofhypotheses

isamethodofusingsampledatatodecidewhetherthenullhypothesisshouldberejected.Thus,wemighttest:μ=.75

againstthealternative:μ.75.Onlyifsampledatastronglysuggeststhatμissomethingotherthan.75shouldthenullhypothesisberejected.Intheabsenceofsuchevidence,

shouldnotberejected,sinceitisstillquiteplausible.Sometimesaninvestigatordoesnotwanttoacceptaparticularassertionunlessanduntildatacanprovidestrongsupportfortheassertion.Atestofhypothesesisameth7ProposeaHypothese(step1)Example1:Supposeacompanyisconsideringputtinganewtypeofcoatingonbearingsthatitproduces.Thetrueaveragewearlifewiththecurrentcoatingisknowntobe1000hours.Withμdenotingthetrueaveragelifeforthenewcoating,thecompanywouldnotwanttomakeachangeunlessevidencestronglysuggestedthatμexceeds1000.ProposeaHypothese(step1)8Example2:Anengineerhassuggestedachangeintheproductionprocessinthebeliefthatitwillresultinareduceddefectiverate.Letpdenotethetrueproportionofdetectiveboardsresultingfromthechangedprocess,andp=0.1inunchangedprocess.Pleasegivethereasonablehypothesis.ORORExample2:Anengineerha9Scientificresearchofteninvolvestryingtodecidewhetheracurrenttheoryshouldbereplacedbyamoreplausibleandsatisfactoryexplanationofthephenomenonunderinvestigation.Aconservativeapproachistoidentifythecurrenttheorywith

andtheresearcher’salternativeexplanationwith.

Scientificresearchofteninvo10SummaryofFormsforNullandAlternativeHypotheses

Letμ0denotethespecificnumericalvaluebeingconsideredinthenullandalternativehypotheses.Ingeneral,ahypothesistestaboutthevaluesofapopulationmeanμmusttakeoneofthefollowingthreeforms.Inmanysituations,thechoiceofH0andHaisnotobviousandjudgmentisnecessarytoselecttheproperform.However,astheprecedingformsshow,theequalitypartoftheexpression(either≥,≤,or=)alwaysappearsinthenullhypothesis.

SummaryofFormsforNulland11Rejectionofthecurrenttheorywillthenoccuronlywhenevidenceismuchmoreconsistentwiththenewtheory.Inmanysituations,

isreferredtoasthe“researcher’shypothesis,”sinceitistheclaimthattheresearcherwouldreallyliketovalidate.Thewordnull

means“ofnovalue,effect,orconsequence,”whichsuggeststhatshouldbeidentifywiththehypothesisofnochange(fromcurrentopinion),nodifference,noimprovement,andsoon.Suppose,forexample,that10%ofallcircuitboardsproducedbyacertainmanufacturerduringarecentperiodweredefective.Rejectionofthecurrent12Inourtreatmentofhypothesistesting,

willalwaysbestatedasanequalityclaim.Ifdenotestheparameterofinterest,thenullhypothesiswillhavetheform:,whereisaspecifiednumbercalledthenullvalueoftheparameter(valueclaimedforbythenullhypothesis).Asanexample,considerthecircuitboardsituationjustdiscussed.ThereasonablehypothesisisInourtreatmentofhypothesis13Exercise1P319Foreachofthefollowingassertions,statewhetheritisalegitimatestatisticalhypothesisandwhy.(a)(b)(c)(d)(e)(f)Whereλistheparameterofanexponentialdistributionusedtomodelcomponentlifetime.Exercise1P319Foreachofth14Exercise2:Forthefollowingpairsofassertions,indicatewhichdonotcomplywithourrulesforsettinguphypothesesandwhy.Exercise2:Forthefollowin15

Exercise3:Todeterminewhetherthepipeweldsinanuclearpowerplantmeetspecifications,arandomsampleofweldsisselected,andtestsareconducedoneachweldinthesample.Weldstrengthismeasuredastheforcerequiredtobreaktheweld.Supposethespecificationsstatethatmeanstrengthofweldsshouldexceed100lb/in2;theinspectionteamdecidestotestH0:μ=100versusHa:μ>100.ExplainwhyitmightbepreferabletousethisHaratherthan:μ<100Exercise3:Todeterminewhet16TestProceduresAtestprocedureisarule,basedonsampledata,fordecidingwhethertoreject.Atestof:p=0.10versus:p<0.10inthecircuitboardproblemmightbebasedonexaminingarandomsampleofn=200boards.LetXdenotethenumberofdefectiveboardsinthesample,abinomialrandomvariable;xrepresentstheobservedvalueofX.Ifistrue,E(X)=np=200(.10)=20,whereaswecanexpectfewerthan20detectiveboardsifistrue.

TestProceduresAtestpro17Avaluexjustabitbelow20doesnotstronglycontradict,soitisreasonabletorejectonlyifxissubstantiallylessthan20.Onesuchtestprocedureistorejectifx≤15andnotrejectotherwise.Thisprocedurehastwoconstituents:(1)ateststatisticorfunctionofthesampledatausedtomakeadecisionand(2)arejectionregionconsistingofthosexvaluesforwhichwillberejectedinfavorof.Fortherulejustsuggested,therejectionregionconsistsofx=0,1,2,….,and15.willnotberejectedifx=16,17,….,199,or200.

Avaluexjustabitbelow2018Atestprocedureisspecifiedbythefollowing:1.Ateststatistic,afunctionofthesampledataonwhichthedecision(rejectordonotreject)istobebased2.Arejectionregion,thesetofallteststatisticvaluesforwhichwillberejectedThenullhypothesiswillthenberejectedifandonlyiftheobservedorcomputedteststatisticvaluefallsintherejectionregion.Atestprocedureisspecified19StepsofHypothesisTesting

1.Determinethenullandalternativehypothesesthatareappropriatefortheapplication.2.Selecttheteststatisticthatwillbeusedtodecidewhethertorejectthenullhypothesis.3.Specifythelevelofsignificance

forthetest.4.UsethelevelofsignificancetodeveloptherejectionregionthatindicatesthevaluesoftheteststatisticthatwillleadtotherejectionofH05.Collectthesampledataandcomputethevalueoftheteststatistic.6.ComparethevalueoftheteststatistictothecriticalvaluespecifiedintherejectionregiontodeterminewhetherH0shouldberejectedStepsofHypothesisTesting20Example:SupposeacigarettemanufacturerclaimsthattheaveragenicotinecontentofbrandBis(atmost)1.5mg,itwouldbeunwisetorejectthemanufacturer’sclaimwithoutstrongcontradictoryevidence.Considerthefollowinghypothesistest.Consideradecisionrulebasedonanalyzingarandomsampleof32cigarettes.Letdenotethesampleaveragenicotinecontent.IfH0istrue,,whereasifH0isfalse,thenSo,wemightuseasateststatisticalongwiththerejectionregionExample:Supposeacigarette21Example:

Considerthefollowinghypothesistest.Asampleof40providesasamplemeanof16.5andsamplestandarddeviationof7.a.Ata=0.02,whatisthecriticalvalueforz,andwhatistherejectionregion?b.Computethevalueoftheteststatisticzc.Whatisyourconclusion?RejectH0ifz>za=2.06Don’trejectH0Example:Considerthefollow22Example:

Individualsfiling1994federalincometaxreturnspriortoMarch31,1995,hadanaveragerefundof$1056.Considerthepopulationof“l(fā)ast-minute”filerswhomailtheirreturnsduringthelastfivedaysoftheincometaxperiod.a.Aresearchsuggeststhatoneofthereasonsindividualwaituntilthelastfivedaystofiletheirreturnsisthatonaveragethoseindividualshavealowerrefundthanearlyfilers.DevelopappropriatehypothesessuchthatrejectionofH0willsupporttheresearcher’scontention.b.Forasampleof400individualswhofiledareturnbetweenApril10andApril15,thesamplemeanrefundwas$910andthesamplestandarddeviationwas$1600.Ata=0.05,whatisyourconclusion?RejectH0ifz<-za=-1.645SorejectH0Solution:Example:Individualsfiling123Example:

NewtiresmanufacturedbyacompanyinFindlay,Ohio,aredesignedtoprovideameanofatleast28000miles.Testswith30randomlyselectedtiresshowedasamplemeanof27500milesandasamplestandarddeviationof1000miles.Usinga0.05levelofsignificance,testwhetherthereissufficientevidencetorejecttheclaimofameanofatleast28000miles.

RejectH0ifz<-za=-1.645So,thereissufficientevidencetorejectH0.Solution:Example:Newtiresmanufacture24Example:

theaverageU.S.householdspends$90perday.Assumeasamplepf25householdsinCorning,NewYork,showedasamplemeandailyexpenditureof$84.50withasamplestandarddeviationof$14.50.a.TestH0:μ=90andHa:μ≠90toseewhetherthepopulationmeaninCorning,NewYork,differsfromU.S.mean.Usea0.05levelofsignificance,.Whatisyourconclusion?

RejectH0if>2.064or<-2.064Don’trejectH0Solution:Example:theaverageU.S.hou25Summary:One-TailTestaboutaPopulationMeanLarge-sample(n>30)hypothesistestaboutapopulationmeanforaone-tailedtestoftheformTeststatistic:σknownTeststatistic:σunknownRejectionruleatalevelofsignificanceofaRejectH0ifz<-zaSummary:One-TailTestabouta26Summary:One-TailTestaboutaPopulationMeanLarge-sample(n>30)hypothesistestaboutapopulationmeanforaone-tailedtestoftheformTeststatistic:σknownTeststatistic:σunknownRejectionruleatalevelofsignificanceofaRejectH0ifz>zaSummary:One-TailTestabouta27Teststatistic:σknownTeststatistic:σunknownRejectionruleatalevelofsignificanceofaRejectH0ifz<-za/2orz>za/2Summary:Two-tailedtestsaboutapopulationmeanWhenLargesampleTeststatistic:σknownTestst28ErrorsinHypothesisTesting

Thebasisforchoosingaparticularrejectionregionliesanunderstandingoftheerrorsthatonemightbefacedwithindrawingaconclusion.Considertherejectionregionx≤15(n=200)inthecircuitboardproblem.Evenwhen:

H0:p=0.10istrue,itmighthappenthatanunusualsampleresultsinx=13,sothatH0iserroneouslyrejected.Example:AtestofH0:p=0.10versusHa:p<0.10inthecircuitboardproblemmightbebasedonexaminingarandomsampleofn=200boards.Therejectionregionconsistsofx=0,1,2,….,and15ErrorsinHypothesisTesting29Ontheotherhand,evenwhenHa:p<0.10istrue,anunusualsamplemightyieldx=20,inwhichcaseH0wouldnotberejected,againanincorrectconclusion.Thus,itispossiblethatH0mayberejectedwhenitistrueorthatH0maynotberejectedwhenitisfalse.Thesepossibleerrorsarenotconsequencesofafoolishlychosenrejectionregion.Eitheroneofthesetwoerrorsmightresultwhenregionx≤14isemployed,orindeedwhenanyotherregionisused.Ontheotherhand,evenwhen30DEFINITIONAtypeIerrorconsistsofrejectingthenullhypothesiswhenitistrue.AtypeIIerrorinvolvesnotrejectingwhenisfalse.PopulationconditionConclusionH0trueHatrueAcceptH0CorrectconclusionTypeⅡerrorRejectH0TypeⅠerrorCorrectconclusionα=theprobabilityofmakingaTypeⅠerrorβ=theprobabilityofmakingaTypeⅡerrorsDEFINITIONAtypeIerrorconsi31Acertaintypeofautomobileisknowntosustainnovisibledamage25%ofthetimein10-mphcrashtests.Amodifiedbumperdesignhasbeenproposedinanefforttoincreasethispercentage.Letpdenotetheproportionofall10-mphcrasheswiththisnewbumperthatresultinnovisibledamage.Thetestwillbebasedonasampleofn=20Thehypothesesare:Example8.1

Teststatistic:X=thenumberofcrasheswithnovisibledamageRejectionregion:R8={8,9,10,…,19,20}Acertaintypeofautomobi32β(0.3)=P(typeIIerrorwhenp=0.3)=P(isnotrejectedwhenitisfalsebecausep=0.3)=P(X≤7whenX~Bin(2,0.3))=B(7;20,0.3)=0.772Whenpisactually0.3ratherthan0.25(a“small”departurefrom),roughly77%ofallexperimentsofthistypewouldresultinbeingincorrectlynotrejected!Incontrasttoα,thereisnotasingleβ.Instead,thereisadifferentβforeachdifferentpthatexceeds0.25.Thusthereisavalueofβforp=0.3(inwhichcaseX~Bin(20,.3)),anothervalueofβforp=0.5,andsoon.Forexample,β(0.3)=P(typeIIerrorwhen33Theaccompanyingtabledisplaysβforselectedvaluesofp(eachcalculatedfortherejectionregion).Clearly,βdecreaseasthevalueofpmovesfarthertotherightofthenullvalue.25.Intuitively,thegreaterthedeparturefrom,thelesslikelyitisthatsuchadeparturewillnotbedetected.pβ(p)0.7720.4160.1320.0210.0010.000Theaccompanyingtabledisplay34Example8.2Thedryingtimeofacertaintypeofpaintunderspecifiedtestconditionsisknowntobenormallydistributedwithmeanvalue75minandstandarddeviation9min.Chemistshaveproposedanewadditivedesignedtodecreaseaveragedryingtime.Letμdenotethetrueaveragedryingtimewhentheadditiveisused.Rejectregion{}α=P(typeIerror)=P(isrejectedwhenitistrue)=P(≤70.8when～normalwith,σ=1.8)=Example8.2Thedryingtimeof35

=P(typeⅡerrorwhen=72)=P(isnotrejectedwhenitisfalsebecause)=P(>70.8when~normalwith=72and

=P(typeⅡerrorwhen36Example8.3LetususethesameexperimentandteststatisticXaspreviouslydescribedintheautomobilebumperproblem,butnowconsidertherejectionregionR9={9,10,…,20}.Xstillhasabinomialdistributionwithparametersn=20andp,supposethehypothesisisstillCalculatetheprobabilityoftypeIerrorandprobabilityoftypeIIerror.Solution:ThetypeIerrorprobabilityhasbeendecreasedbyusingthenewrejection.However,apricehasbeenpaidforthisdecrease.Example8.3Letususethesam37Example8.4Thetrueofcutoffvaluec=70.8inthepaint-dryingexampleresultedinaverysmallvalueofbutratherlarge‘s.Considerthesameexperimentandteststatisticwiththenewrejectionregion.BecauseisstillnormallydistributedwithmeanvalueandDetermineprobabilityoftypeIerrorandtypeIIerrorSolution:Example8.4Thetrueofcutoff38PropositionSupposeanexperimentandasamplesizearefixed,andateststatisticischosen.Thendecreasingthesizeoftherejectionregiontoobtainasmallervalue

ofresultsinalargervalueofforanyparticularparametervalueconsistentwithHa.Thispropositionsaysthatoncetheteststatisticandnarefixed,thereisnorejectionregionthatwillsimultaneouslymakebothand‘ssmall.AregionmustbechosentoeffectacompromisebetweenandBecauseofthesuggestedguidelinesforspecifyingH0andHa,atypeIerrorisusuallymoreseriousthanatypeIIerror(thiscanalwaysbeachievedbyproperchoiceofthehypotheses).Theapproachadheredtobymoststatisticalpractitionersisthentospecifythelargestvalueofαthatcanbetoleratedandfindarejectionregionhavingthatvalueofαratherthananythingsmaller.ThismakesβassmallaspossiblesubjecttotheboundonαPropositionSupposeanexperime39Theresultingvalueofαisoftenreferredtoasthesignificancelevelofthetest.Traditionallevelsofsignificanceare0.10,0.05,and0.01,thoughthelevelinanyparticularproblemwilldependontheseriousnessofatypeIerror—themoreseriousthiserror,thesmallershouldbethesignoficancelevel.Thecorrespondingtestprocedureiscalledalevelαtest(e.g.alevel0.05testoralevel0.01test).AtestwithsignificancelevelαisoneforwhichthetypeIerrorprobabilityiscontrolledatthespecifiedlevel.Theresultingvalueofαis40ExerciseP3194Letμdenotethetrueaverageradioactivitylevel(picocuriesperliter).Thevalue5PCi/Lisconsideredthedividinglinebetweensafeandunsafewater.WouldyourecommendtestingH0:μ=5versusHa:μ>5orH0:μ=5versusHa:μ<5?ExplainyourreasoningExerciseP3194Letμdenoteth41ExerciseP3195Beforeagreeingtopurchasealargeorderofpolyethylenesheathsforaparticulartypeofhigh-pressureoil-filledsubmarinepowercable,acompanywantstoseeconclusiveevidencethatthetruestandarddeviationofsheaththicknessislessthan0.05mm.whathypothesesshouldbetested,andwhy?Inthiscontext,whatarethetypeIandtypeIIerrors?ExerciseP3195Beforeagreein42ExerciseP3196Manyolderhomeshaveelectricalsystemsthatusefusesratherthancircuitbreakers.Amanufacturerof40-ampfuseswantstomakesurethatthemeanamperageatwhichitsfusesburnoutisinfact40.customerswillcomplainbecausethefusesrequirereplacementtoooften.Ifthemeanamperageishigherthan40,themanufacturermightbeliablefordamagetoanelectricalsystemduetofusemalfunction.Toverifytheamperageofthefuses,asampleoffusesistobeselectedandinspected.Ifahypothesistestweretobeperformedontheresultingdata,whatnullandalternativehypotheseswouldbeofinteresttothemanufacturer?DescribetypeIandtypeIIerrorsinthecontextofthisproblemsituation.ExerciseP3196Manyolderhom43ExerciseP3207Watersamplesaretakenfromwaterusedfrocoolingasitisbeingdischargedfromapowerplantintoariver.Ithasbeendeterminedthataslongasthemeantemperatureofthedischargedwaterisatmost150.F,therewillbenonegativeeffectsontheriver’secosystem.Toinvestigatewhethertheplantisincompliancewithregulationsthatprohibitameandischarge-watertemperatureabove150.,50watersampleswillbetakenatrandomlyselectedtimes,andthetemperatureofeachsamplerecorded.TheresultingdatawillbeusedtotestthehypothesesH0:μ=150verseHa:μ>150.Inthecontextofthissituation,describetypeⅠandtypeⅡerrors.Whichtypeoferrorwouldyouconsidermoreserious?Explain.

ExerciseP3207Watersamples44ExerciseP3208Aregulartypeoflaminateiscurrentlybeingusedbyamanufactureofcircuitboards.Aspeciallaminatehasbeendevelopedtoreducewarpage.Theregularlaminatewillbeusedononesampleofspecimensandthespeciallaminateonanothersample,andtheamountofwarpagewillthenbedeterminedforeachspecimen.Themanufacturewillthenswitchtothespeciallaminateonlyifitcanbedemonstratedthatthetrueaverageamountofwarpageforthatlaminateislessthanfortheregularlaminate.Statetherelevanthypotheses,anddescribethetypeIandtypeIIerrorsinthecontextofthissituation.ExerciseP3208Aregulartype45ExerciseP3209Twodifferentcompanieshaveappliedtoprovidecabletelevisionserviceinacertainregion.Letpdenotetheprobabilityofallpotentialsubscriberswhofavorthefirstcompanyoverthesecond.Considertestingversusbasedonarandomsampleof25individuals.LetXdenotethenumberinthesamplewhofavorthefirstcompanyandxrepresenttheobservedvalueofX.a.Whichofthefollowingrejectionregionsismostappropriateandwhy?b.Inthecontextofthisproblemsituation,describewhattypeⅠtypeⅡerrorsare.c.WhatisprobabilitydistributionoftheteststatisticXwhenH0istrue?UseittocomputetheprobabilityofatypeⅠerror.d.ComputetheprobabilityofatypeⅡerrorfortheselectregionwhenp=0.3.e.Usingtheselectedregion,whatwouldyouconcludeif6ofthe25queriedfavoredcompany1?ExerciseP3209Twodifferent46ExerciseP32010AmixtureofpulverizedfuelashandPortlandcementtobeusedforgroutingshouldhaveacompressivestrengthofmorethan1300KN/m2,.Themixturewillnotbeusedunlessexperimentalevidenceindicatesconclusivelythatthestrengthspecificationhasbeenmet.Supposecompressivestrengthforspecimensofthismixtureisnormallydistributedwithσ=60.Letμdenotethetrueaveragecompressivestrength.Whataretheappropriatenullandalternativehypotheses?LetXdenotethesampleaveragecompressivestrengthforn=20randomlyselectedspecimens.ConsiderthetestprocedurewithteststatisticXandrejectionregionx≥1331.26.whatistheprobabilitydistributionoftheteststatisticwhenH0istrue?WhatistheprobabilityofatypeIerrorforthetestprocedure?Whatistheprobabilitydistributionoftheteststatisticwhenμ=1350?Usingthetestprocedureofpart(b),whatistheprobabilitythatthemixturewillbejudgedunsatisfactorywheninfactμ=1350(atypeIIerror)?ExerciseP32010Amixtureof47d.Howwouldyouchangethetestprocedureofpart(b)toobtainatestwithsignificanclevel0.05?Whatimpactwouldthischangehaveontheerrorprobabilityofpart??e.ConsiderthestandardizedteststatisticWhatarethevaluesofZcorrespondingtotherejectionregionofpart(b)?d.Howwouldyouchangethete48ExerciseP32011Thecalibrationofascaleistobecheckedbyweighinga10-kgtestspecimen25times.Supposethattheresultsofdifferentweightingsareindependentofoneanotherandthattheweightoneachtrialisnormallydistributedwithσ=0.200kg.Letμdenotethetrueaverageweightreadingonthescale.Whathypothesesshouldbetested?Supposethescaleistoberecalibratedifeitherx≥10.1032orx≤9.8968.whatistheprobabilitythatrecalibrationiscarriedoutwhenitisactuallyunnecessary?Whatistheprobabilitythatrecalibrationisjudgedunecessarywheninfactμ=10.1?whenμ=9.8?Let.Forwhatvaluecistherejectionregionofpart(b)equivalenttothe“two-tailed”regioneitherz≥corz≤c?ExerciseP32011Thecalibrati49e.Ifthesamplesizewereonly10ratherthan25,howshouldtheprocedureofpart(d)bealteredsothatα=0.05?f.Usingthetestofpart(e),whatwouldyouconcludefromthefollowingsampledata:9.98110.0069.85710.1079.8889.72810.43910.21410.1909.793g.Reexpressthetestprocedureofpart(b)intermsofthestandardizedteststatistice.Ifthesamplesizewereonl50c.Whatisthesignificancelevelfortheappropriateregionofpart(b)?Howwouldyouchangetheregiontoobtainatestwithα=0.001?d.Whatistheprobabilitythatthenewdesignisnotimplementedwhenitstrueaveragebrakingdistanceisactu