商务与经济统计习题答案（第8版，中文版）SBE8-SM20

20-1Chapter20StatisticalMethodsforQualityControlLearningObjectives1.Learnabouttheimportanceofqualitycontrolandhowstatisticalmethodscanassistinthequalitycontrolprocess.2.Learnaboutacceptancesamplingprocedures.3.Knowthedifferencebetweenconsumer’sriskandproducer’srisk.4.Beabletousethebinomialprobabilitydistributiontodevelopacceptancesamplingplans.5.Knowwhatismeantbymultiplesamplingplans.6.Beabletoconstructqualitycontrolchartsandunderstandhowtheyareusedforstatisticalprocesscontrol.7.Knowthedefinitionsofthefollowingterms:producer'sriskassignablecausesconsumer'sriskcommoncausesacceptancesamplingcontrolchartsacceptablecriterionuppercontrollimitoperatingcharacteristiccurvelowercontrollimitChapter2020-2Solutions:1.a.Forn=4UCL=+3(/n)=12.5+3(.8/4)=13.7LCL=-3(/n)=12.5-3(.8/4)=11.3b.Forn=8UCL=+3(.8/8)=13.35LCL=-3(.8/8)=11.65Forn=16UCL=+3(.8/16)=13.10LCL=-3(.8/16)=11.90c.UCLandLCLbecomeclosertogetherasnincreases.Iftheprocessisincontrol,thelargersamplesshouldhavelessvarianceandshouldfallcloserto12.5.2.a.6775255542.().b.UCL=+3(/n)=5.42+3(.5/5)=6.09LCL=-3(/n)=5.42-3(.5/5)=4.753.a.p1352510000540().b.pppn().(.).1005400946010000226c.UCL=p+3p=0.0540+3(0.0226)=0.1218LCL=p-3p=0.0540-3(0.0226)=-0.0138UseLCL=04.RChart:UCL=RD4=1.6(1.864)=2.98LCL=RD3=1.6(0.136)=0.22xChart:UCL=2xAR=28.5+0.373(1.6)=29.10LCL=xAR2=28.5-0.373(1.6)=27.905.a.UCL=+3(/n)=128.5+3(.4/6)=128.99LCL=-3(/n)=128.5-3(.4/6)=128.01StatisticalMethodsforQualityControl20-3b.xxni/..7724612873incontrolc.xxni/..7743612905outofcontrol6.ProcessMean=2012199022001...UCL=+3(/n)=20.01+3(/5)=20.12Solvefor:(..).201220015300827.SampleNumberObservationsxiRi131422833.6714226183526.3317325303429.679417252121.008538293534.009641423639.676721172922.3312832262828.676941343336.0081029173025.33131126314032.33141223192522.3361317243224.33151443351731.67261518252924.00111630423134.33121728363232.0081840293133.33111918292825.00112022342627.3312R=11.4andx2917.RChart:UCL=RD4=11.4(2.575)=29.35LCL=RD3=11.4(0)=0xChart:UCL=xAR2=29.17+1.023(11.4)=40.8LCL=xAR2=29.17-1.023(11.4)=17.5Chapter2020-4RChart:xChart:8.a.p1412015000470().b.pppn().(.).1004700953015000173UCL=p+3p=0.0470+3(0.0173)=0.0989LCL=p-3p=0.0470-3(0.0173)=-0.0049StatisticalMethodsforQualityControl20-5UseLCL=0c.p12150008.Processshouldbeconsideredincontrol.d.p=.047,n=150UCL=np+3npp()1=150(0.047)[phone](.)(.)=14.826LCL=np-3npp()1=150(0.047)-315000470953(.)(.)=-0.726Thus,theprocessisoutofcontrolifmorethan14defectivepackagesarefoundinasampleof150.e.Processshouldbeconsideredtobeincontrolsince12defectivepackageswerefound.f.Thenpchartmaybepreferredbecauseadecisioncanbemadebysimplycountingthenumberofdefectivepackages.9.a.Totaldefectives:165p1652020000413().b.pppn().(.).1004130958720000141UCL=p+3p=0.0413+3(0.0141)=0.0836LCL=p-3p=0.0413+3(0.0141)=-0.0010UseLCL=0c.p20200010.Outofcontrold.p=.0413,n=200UCL=np+3npp()1=200(0.0413)[phone](.)(.)=16.702LCL=np-3npp()1=200(0.0413)-32000041309587(.)(.)=0.1821e.Theprocessisoutofcontrolsince20defectivepistonswerefound.10.fxnxnxppxnx()!!()!()1Whenp=.02,theprobabilityofacceptingthelotisf()!)!(.)(.).0250!(250002100206035025Chapter2020-6Whenp=.06,theprobabilityofacceptingthelotisf()!)!(.)(.).0250!(25000610060212902511.a.Usingbinomialprobabilitieswithn=20andp0=.02.P(Acceptlot)=f(0)=.6676Producer’srisk:=1-.6676=.3324b.P(Acceptlot)=f(0)=.2901Producer’srisk:=1-.2901=.709912.Atp0=.02,then=20andc=1planprovidesP(Acceptlot)=f(0)+f(1)=.6676+.2725=.9401Producer’srisk:=1-.9401=.0599Atp0=.06,then=20andc=1planprovidesP(Acceptlot)=f(0)+f(1)=.2901+.3703=.6604Producer’srisk:=1-.6604=.3396Foragivensamplesize,theproducer’sriskdecreasesastheacceptancenumbercisincreased.13.a.Usingbinomialprobabilitieswithn=20andp0=.03.P(Acceptlot)=f(0)+f(1)=.5438+.3364=.8802Producer’srisk:=1-.8802=.1198b.Withn=20andp1=.15.P(Acceptlot)=f(0)+f(1)=.0388+.1368=.1756Consumer’srisk:=.1756c.Theconsumer’sriskisacceptable;however,theproducer’sriskassociatedwiththen=20,c=1planisalittlelargerthandesired.StatisticalMethodsforQualityControl20-714.cP(Accept)p0=.05Producer’sRiskP(accept)p1=.30Consumer’sRisk(n=10)0.5987.4013.0282.02821.9138.0862.1493.14932.9884.0116.3828.3828(n=15)0.4633.5367.0047.00471.8291.1709.0352.03522.9639.0361.1268.12683.9946.0054.2968.2968(n=20)0.3585.6415.0008.00081.7359.2641.0076.00762.9246.0754.0354.03543.9842.0158.1070.1070Theplanwithn=15,c=2isclosewith=.0361and=.1268.However,theplanwithn=20,c=3isnecessarytomeetbothrequirements.15.a.P(Accept)shownforpvaluesbelow:cp=.01p=.05p=.08p=.10p=.150.8179.3585.1887.1216.03881.9831.7359.5169.3918.17562.9990.9246.7880.6770.4049TheoperatingcharacteristiccurveswouldshowtheP(Accept)versuspforeachvalueofc.b.P(Accept)cAtp0=.01Producer’sRiskAtp1=.08Consumer’sRisk0.8179.1821.1887.18871.9831.0169.5169.51692.9990.0010.7880.788016.a.x20190820954.b.UCL=+3(/n)=95.4+3(.50/5)=96.07LCL=-3(/n)=95.4-3(.50/5)=94.73c.No;allwereincontrol17.a.Forn=10Chapter2020-8UCL=+3(/n)=350+3(15/10)=364.23LCL=-3(/n