搜索
308Chapter9:FundamentalsofHypothesisTesting:One-SampleTestsCHAPTER99.1H0isusedtodenotethenullhypothesis.9.2H1isusedtodenotethealternativehypothesis.9.3αisusedtodenotethesignificancelevel,orthechanceofcommittingaTypeIerror.9.4βisusedtodenotetheriskorthechanceofcommittingaTypeIIerror.9.51–βrepresentsthepowerofastatisticaltest–thatis,theprobabilityofcorrectlyrejectingthenullhypothesiswheninrealitythenullhypothesisisfalseandshouldberejected.9.6αistheprobabilityofmakingaTypeIerror–thatis,theprobabilityofincorrectlyrejectingthenullhypothesiswheninrealitythenullhypothesisistrueandshouldnotberejected.9.7βistheprobabilityofmakingaTypeIIerror–thatis,theprobabilityofincorrectlyfailingtorejectthenullhypothesiswhenitisfalse.9.8Thepowerofatestisthecomplement,whichis(1-β),oftheprobabilityβofmakingaTypeIIerror.9.9Itispossibletoincorrectlyrejectatruenullhypothesisbecausethemeanofasinglesamplecanfallintherejectionregioneventhoughthehypothesizedpopulationmeanistrue.9.10Itispossibletonotrejectafalsenullhypothesisbecausethemeanofasinglesamplecanfallinthenon-rejectionregioneventhoughthehypothesizedpopulationmeanisfalse.9.11βwillincrease.9.12Otherthingsbeingequal,thecloserthehypothesizedmeanistotheactualmean,thelargertheriskofcommittingaTypeIIerrorwillbe.9.13αistheprobabilityofincorrectlyconvictingthedefendantwhenheisinnocent.βistheprobabilityofincorrectlyfailingtoconvictthedefendantwhenheisguilty.9.14UndertheFrenchjudicialsystem,unlikeoursintheUnitedStates,thenullhypothesisassumesthedefendantisguilty,thealternativehypothesisassumesthedefendantisinnocent.ATypeIerrorwouldbenotconvictingaguiltypersonandaTypeIIerrorwouldbeconvictinganinnocentperson.SolutionstoEnd-of-SectionandChapterReviewProblems3099.15(a)ATypeIerroristhemistakeofapprovinganunsafedrug.ATypeIIerrorisnotapprovingasafedrug.(b)TheconsumergroupsaretryingtoavoidaTypeIerror.Tothem,theriskofmistakenlyapprovinganunsafedrugismostimportantand,thereforetheywanttotrytoavoidaTypeIerror.(c)TheindustrylobbyistsaretryingtoavoidaTypeIIerror.Tothem,theriskofnotapprovingasafedrugismostimportantand,therefore,theywanttotrytoavoidaTypeIIerror.(d)TolowerbothTypeIandTypeIIerrors,theFDAcanrequiremoreinformationandevidenceintheformofmorerigoroustesting.Thiscaneasilytranslateintolongertimetoapproveanewdrug.9.16H0:µ=20minutes.20minutesisadequatetraveltimebetweenclasses.H1:µ≠20minutes.20minutesisnotadequatetraveltimebetweenclasses.9.17H0:µ=70pounds.Theclothhasameanbreakingstrengthof70pounds.H1:µ≠70pounds.Theclothhasameanbreakingstrengththatdiffersfrom70pounds.9.18H0:µ=1.00.Themeanamountofpaintpercanisonegallon.H1:µ≠1.00.Themeanamountofpaintpercandiffersfromonegallon.9.19H0:µ=375hours.Themeanlifeofthemanufacturer’slightbulbsisequalto375hours.H1:µ≠375hours.Themeanlifeofthemanufacturer’slightbulbsdiffersfrom375hours.9.20Decisionrule:RejectH0ifZ–1.96orZ+1.96.Decision:SinceZcalc=+2.21isgreaterthanZcrit=+1.96,rejectH0.9.21Decisionrule:RejectH0ifZ–1.645orZ+1.645.9.22Decisionrule:RejectH0ifZ–2.58orZ+2.58.9.23Decision:SinceZcalc=–2.61islessthanZcrit=–2.58,rejectH0.9.24p-value=2(1−.9772)=0.04569.25Sincethep-valueof0.0456islessthanthe0.10levelofsignificance,thestatisticaldecisionistorejectthenullhypothesis.9.26p-value=0.16769.27At0.01levelofsignificance,thestatisticaldecisionistonotrejectthenullhypothesis.310Chapter9:FundamentalsofHypothesisTesting:One-SampleTests9.28(a)H0:µ=70pounds.Theclothhasameanbreakingstrengthof70pounds.H1:µ≠70pounds.Theclothhasameanbreakingstrengththatdiffersfrom70pounds.Decisionrule:RejectH0ifZ–1.96orZ+1.96.Teststatistic:Z=X–µσn=69.1–703.549=–1.80Decision:SinceZcalc=–1.80isbetweenthecriticalboundsof±1.96,donotrejectH0.Thereisnotenoughevidencetoconcludethattheclothhasameanbreakingstrengththatdiffersfrom70pounds.(b)p-value=2(0.0359)=0.0718Interpretation:Theprobabilityofgettingasampleof49piecesthatyieldameanstrengththatisfartherawayfromthehypothesizedpopulationmeanthanthissampleis0.0718or7.18%.(c)Decisionrule:RejectH0ifZ–1.96orZ+1.96.Teststatistic:Z=X–µσn=69.1–701.7549=–3.60Decision:SinceZcalc=–3.60islessthanthelowercriticalboundof–1.96,rejectH0.Thereisenoughevidencetoconcludethattheclothhasameanbreakingstrengththatdiffersfrom70pounds.(d)Decisionrule:RejectH0ifZ–1.96orZ+1.96.Teststatistic:Z=X–µσn=69–703.549=–2.00Decision:SinceZcalc=–2.00islessthanthelowercriticalboundof–1.96,rejectH0.Thereisenoughevidencetoconcludethattheclothhasameanbreakingstrengththatdiffersfrom70pounds.9.29(a)H0:µ=1.00.Themeanamountofpaintpercanisonegallon.H1:µ≠1.00.Themeanamountofpaintpercandiffersfromonegallon.Decisionrule:RejectH0ifZ–2.58orZ+2.58.Teststatistic:Z=X–µσn=0.995–1.000.0250=–1.77Decision:SinceZcalc=–1.77isbetweenthecriticalboundsof±2.58,donotrejectH0.Thereisnotenoughevidencetoconcludethatthemeanamountofpaintperone-galloncandiffersfromonegallon.(b)p-value=2(0.0384)=0.0768Interpretation:Theprobabilityofgettingasampleof50cansthatwillyieldameanamountthatisfartherawayfromthehypothesizedpopulationmeanthanthissampleis0.0768.(c)X±Z⋅σn=0.995±2.58⋅0.02500.9877≤µ≤1.0023(d)Samedecision.Theconfiden