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1、Chapter14:AnalysisofCategoricalData14.1a.H0:p1=.41,p2=.10,p3=.04,p4=.45vs.Ha:notH0.Theobservedandexpectedcountsare:ABABOobserved89181281expected200(.41)=82200(.10)=20200(.04)=8200(.45)=9022222(89?82)(18?20)(12?8)(81?90)Thechi–squarestatisticisX=+++=3.696with4–1=382208902degree
2、soffreedom.Sinceχ=7.81473,wefailtorejectH0;thereisnotenough.05evidencetoconcludetheproportionsdiffer.2b.UsingtheApplet,p–value=P(χ>3.696)=.29622.14.2a.H0:p1=.60,p2=.05,p3=.35vs.Ha:notH0.Theobservedandexpectedcountsare:admittedunconditionallyadmittedconditionallyrefusedobserved
3、32943128expected500(.60)=300500(.05)=25500(.35)=1752222(329?300)(43?25)(128?175)Thechi–squareteststatisticisX=++=28.386with3–1=2300251752degreesoffreedom.Sinceχ=7.37776,wecanrejectH0andconcludethatthecurrent.05admissionratesdifferfromthepreviousrecords.2b.UsingtheApplet,p–valu
4、e=P(χ>28.386)=.00010.14.3ThenullhypothesisisH10:p1=p2=p3=p4=vs.Ha:notH0.Theobservedand4expectedcountsare:lane1234observed294276238192expected25025025025022222(294?250)+(276?250)+(238?250)+(192?250)Thechi–squarestatisticisX==24.48with4–1=32502degreesoffreedom.Sinceχ=7.81473,wer
5、ejectH0andconcludethatthelanesare.05notpreferredequally.FromTable6,p–value<.005.NotethatRcanbeusedby:>lanes<-c(294,276,238,192)>chisq.test(lanes,p=c(.25,.25,.25,.25))#pisnotnecessaryhereChi-squaredtestforgivenprobabilitiesdata:lanesX-squared=24.48,df=3,p-value=1.983e-05287288C
6、hapter14:AnalysisofCategoricalDataInstructor’sSolutionsManual14.4ThenullhypothesisisH10:p1=p2=…=p7=vs.Ha:notH0.Theobservedand7expectedcountsare:SUMTWRFSAobserved24362726322629expected28.57128.57128.57128.57128.57128.57128.5712222(24?28.571)+(36?28.571)+…+(29?28.571)Thechi–squa
7、restatisticisX==24.48with7–1=628.5712degreesoffreedom.Sinceχ=12.5916,wecanrejectthenullhypothesisandconclude.05thatthereisevidenceofadifferenceinpercentagesofheartattacksforthedaysoftheweek14.5a.Letp=proportionofheartattacksonMondays.Then,H110:p=vs.Ha:p>.Then,77p?=36/200=.18an
8、dfromSection8.3,theteststatisticis.18?1/7z==1.50.(1/7)(6/7)20
