ProbabilityandStatisticsSolutions9GuanghuaSchoolofManagementPekingUniversityDecember7,20121/9Question1Thedataisfromanormaldistributionandthevarianceisunknown,letα=1−γ,whereγistheconfidencecoefficient,fromthefactthatU=√n(¯Xn−µ)S∼tn−1,where¯Xn=1nn�i=1Xi,S2=1n−1n�i=1(Xi−¯Xn)2,n=8.Thentheconfidenceintervalwillbe[¯Xn−tn−1,α/2S/√n,¯Xn+tn−1,α/2S/√n],2/9Question1wheretn−1,α/2istheupperα/2quantileofthetdistributionwithdegreeoffreedomof(n−1).Whenγ=0.9,α=0.1,t7,0.05=1.8946.Sotheconfidenceintervalwithconfidencecoefficient0.9willbe[2.7191,3.4058].Similarly,wecancalculatetheconfidenceintervalforconfidencecoefficients0.95and0.99willbe[2.634,3.491]and[2.4284,3.6966].3/9Question2Thedataisfromthenormaldistributionwithknownvarianceσ2,thenfromZ=√n(¯Xn−µ)σ∼N(0,1),Sotheconfidenceintervalwithconfidencecoefficient0.95is[¯Xn−zα/2σ/√n,¯Xn+zα/2σ/√n],(1)whereα=0.05andzα/2istheupperα/2quantileofstandardnormaldistribution.4/9Question2Thelengthofconfidenceintervalis2zα/2σ√n=2×1.96σ√n≤0.1σ.Thenn≥1537.5/9Question3LetCdenotethecriticalregionofthetest,for∀β∈Ω,thenthepowerfunctionisπ(β|δ)=Pr(X∈C|β)=Pr(X≥1|β)=�∞11βe−βxdx=e−β.Thesizeofthetestwillbeα(δ)=supθ|δπ(β|δ)=supβ≥1e−β=e−1.6/9Question4WefirstlycalculatethedistributionofYn.Ify≤0,thenFYn(y)=0.Ify∈[0,θ],FYn(y)=P(Yn≤y)=n�i=1Pr(Xi≤y)=(yθ)n.Ify≥θ,thenFYn(y)=1.Thenthepowerfunctionofthetestisπ(θ|δ)=Pr(Yn≤1.5|θ)=FYn,θ(1.5)=���1θ≤1.5(1.5θ)nθ>1.57/9Question4Thesizeofthetestisα(δ)=supθ∈Ω0π(θ|δ)=supθ≥2π(θ|δ)=(34)n.8/9Thepowerofthetestisπ(θ|δ)=Pr(Y≤1orY≥7|p)=1−Pr(2≤Y≤6|p)=1−6�k=2Ck20pk(1−p)(20−k).Sowhenpequals0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,thepoweris1,0.3941,0.1558,0.3996,0.7505,0.9423,0.9935,0.9997,0.9999,0.9999,1.Thesizeofthetestisα=Pr(Y≤1orY≥7|p=0.2)=0.1558.9/9
