浙大朱军生物统计作业2答案

生物统计与试验设计——作业二1.IfarandomvariableXfollowsthenormaldistributionN(2,25),letrandomvariableY=𝑋;25,𝒛=*2𝑋−𝑋+,1)what’stheexpectationandvarianceofrandomvariableY，what’sthedistributionthisrandomvariablefollows?E(Y)=E(𝑋;25)=𝐸(𝑋);25=2;25=0;Var(Y)=Var(𝑋;25)=Var(𝑋5)=125σ2（X）=2525=1；ThisrandomvariableYfollowsstandardnormaldistributionY~N(0,1)2)deduce(推断)theexpectationandvarianceofrandomvectorz，what’sdistributionitfollows？E(z)=E*2𝑋−𝑋+=[𝐸(2𝑋)𝐸(−𝑋)]=[2𝐸(𝑋)−𝐸(𝑋)]=*4−2+Var(z)=Var*2𝑋−𝑋+=Var(z)=E{[z－E(z)][z－E(z)]T}=E,[*2X−X+－*4−2+][*2X−X+－*4−2+]𝑇-=E{*2𝑋−4−𝑋+2+*2𝑋−4−𝑋+2+𝑇}=E,*2𝑋−4−𝑋+2+[2𝑋−4−𝑋+2]-=E[4(𝑋−2)2−2(𝑋−2)2−2(𝑋−2)2(𝑋−2)2]=[4𝐸(𝑋−2)2−2𝐸(𝑋−2)2−2𝐸(𝑋−2)2𝐸(𝑋−2)2]T=𝑋;𝜇𝜎~N(0,1)X=σ∙T+μ=5T+2E(T)=0𝐸(𝑋−2)2=E(5T+2−2)2=E(5T)2=25E(T2)=25∫𝑡2:∞;∞𝜑(𝑡2)𝑑𝑡=25∫𝑡2:∞;∞1√2𝜋𝑒;𝑡2/2𝑑𝑡=25∴Var(z)=*4×25−2×25−2×2525+=*100−50−5025+∴z~MVN(*4−2+,*100−50−5025+)2.Tostudyontherelationshipbetweentheleanmeat(瘦肉量y)witheyemusclearea(眼肌面积x1)、legmeat(腿肉量𝑥2)、waistmeat(腰肉量𝑥3),asampleof25pigswassampled,andtraitinvestigationwasconduced.Themodel𝑦𝑖=𝑏0+𝑏1𝑥𝑖1+𝑏2𝑥𝑖2+𝑏3𝑥𝑖3+𝑒𝑖wasanalyzedforthethemultiplevariablelinearregressionofleanmeatoneyemusclemeat,legmeat,waistmeat.SupposeSSR=23.865，R2=0.842，𝑏̂0(SE)=0.857(1.384),𝑏̂1(SE)=0.0187(0.0296),𝑏̂2(SE)=2.073(0.270)，𝑏̂3(SE)=1.938(0.513).1)PleasewriteoutSASstatementsforthestepofprocedureifthedatasethasbeenestablished;datelean;inputeyelegwaist@@;card;………………；procreg;modellean=eyelegwaist;run;2)CalculatethetotalsumofsquaresSSTO,residualsumsquaresSSEandadjusteddeterminantcoefficientforthemodel𝑅𝑎2;𝑅2=𝑆𝑆𝑅𝑆𝑆𝑇𝑂SSTO=𝑆𝑆𝑅𝑅2=23.8650.842=28.343SSTO=SSR+SSESSE=SSTO-SSR=28.343-23.865=4.478MSE=SSE/(n-p-1)=4.478/(25-3-1)=0.213MSTO=SSTO/(n-1)=28.343/24=1.181𝑅𝑎2=1−𝑀𝑆𝐸/𝑀𝑆𝑇𝑂=1-0.213/1.181=0.8203)Conductstatisticaltestonthelinearrelationshipofleanmeatoneyemusclemeat,legmeat,waistmeat,andgiveoutappropriatestatisticalinference（统计推断）;𝑦𝑖=0.857+0.0187𝑥𝑖1+2.073𝑥𝑖2+1.938𝑥𝑖3+𝑒𝑖原假设𝐻0：𝑏1=𝑏2=𝑏3=0备择假设H1：上式不成立𝐹∗=𝑀𝑆𝑅𝑀𝑆𝐸=23.8650.213=112.042F(0.05;p,n-p-1)=F(0.05;3,21)=3.07∴𝐹∗F(0.05;3,21)否定原假设，接受备择假设。方差分析表明，F检验达到极显著水平。因而瘦肉量y与眼肌面积x1、腿肉量x2、腰肉量x3存在回归关系。大约82%的瘦肉量是由这三个产量构成因素决定的。4)Conductstatisticaltestonthesignificanceofeachregressionparameter,accordingtotheresultsofstatisticaltest,isitnecessarytomodify(修改)themodel?why?上面第二题检验只是否定了所有自变量的回归参数均为零的假设𝐻0：𝑏1=𝑏2=𝑏3=0，但这并不能进一步判断认为每个自变量的回归参数均不为零。下面对每个自变量用t检验：t(α/2;n-p-1)=t(0.025;21)=2.4138𝑡1∗=𝑏̂1SE(𝑏̂1)=0.01870.0296=0.6081t(0.025;21)故𝑥1不显著𝑡2∗=𝑏̂2SE(𝑏̂2)=2.0730.270=7.6778t(0.025;21)故𝑥2显著𝑡3∗=𝑏̂3SE(𝑏̂3)=1.9380.513=3.7778t(0.025;21)故𝑥3显著所以此线性模型需要修改，修改为：𝑦𝑖=𝑏0+𝑏2𝑥𝑖2+𝑏3𝑥𝑖3+𝑒𝑖5)Iftheequationofb1=0.02,b3=2.0isrequiredtobetestedinonegeneralizedlineartest,writeoutthematrixKandvectorb.K=*01000001+b=[𝑏0𝑏1𝑏2𝑏3]3．Thefollowingdataareaportionoftheresponsescollectedduringaninter-laboratorystudy.Eachoftheseverallaboratorieswassentanumberofmaterialsthatwerecarefullychosentohavedifferentmeasurementvaluesonthecharacteristicofinterest.Thelaboratorieswererequiredtoperformthreeseparateanalysesofthetestmaterial.LaboratoryMaterialReplicate（重复）1A12.212.312.2B15.515.015.3C18.118.118.22A12.612.312.7B15.015.515.2C18.518.318.63A12.712.812.7B15.315.215.2C18.018.217.91)What’stheexperimentaldesign?Experimentaldesignisabodyofknowledgeandtechniquethatenableaninvestigatortoconductbetterexperiments,analyzedataefficiently,andmaketheconnectionbetweentheconclusionsfromtheanalysisandtheoriginalobjectiveoftheinvestigation.InthisexercisetheexperimentaldesignisTwo-wayNestedDesigns(双因素巢式设计)2)Writeoutthefactorsandtheirlevelsofthistrial;factorsMaterialLaboratoryReplicatelevels1、2、31、2、31、2、33)SetupanANOVAmodelforthistrial,defineeachofvariablesinthemodel;FactorLFactorMinFactorL𝐿1𝑀1(1)𝑀2(1)𝑀3(1)𝐿2𝑀1(2)𝑀2(2)𝑀3(2)𝐿3𝑀1(3)𝑀2(3)𝑀3(3)Y𝑖𝑗𝑘=𝜇+𝑙𝑖+𝑚𝑗(i)+𝜀𝑖𝑗𝑘i=1,2,3;j=1,2,3;k=1,2,3;μ是总体的均值𝑙𝑖是第i项laboratory的效应𝑚𝑖是第j项material的效应i表示第i个实验室j表示第j个实验材料k表示第k次重复εijk是残差效应，独立正态随机变量εijk~N(0，𝜎𝜀2)4)Ifthemodelisarandommodel(eacheffectisrandomeffectinmodel),writeoutthestatisticfortestingsignificanceofeacheffect.(不要求采用上表数据进行实际计算)Y𝑖𝑗𝑘=𝜇+𝑙𝑖+𝑚𝑗(i)+𝜀𝑖𝑗𝑘~N(𝜇,𝜎𝜀2+𝜎𝑙2+𝜎𝑚2)i=1,2,3;j=1,2,3;k=1,2,3;εijk~N(0，𝜎𝜀2)𝐿𝑖~N(0，𝜎𝑙2𝑀𝑗(i)~N(0，𝜎𝑚2)变异来源自由度MSE(MS)随即L3-1=2MSL𝜎𝜀2+3𝜎𝑙2+9𝜎𝑚2M(L)3*(3-1)=6MSM(L)𝜎𝜀2+3𝜎𝑙2Error3*3*（3-1）=18MSE𝜎𝜀2F测验:NullHypothesisH0:𝜎𝑙2=0AlternativeHypothesis𝐻1::𝜎𝑙2≠0StatisticF∗=𝑀𝑆𝐿/𝑀𝑆𝑀(𝐿)~F(2,6)NullHypothesisH0:𝜎𝑚2=0AlternativeHypothesis𝐻1::𝜎𝑚2≠0StatisticF∗=𝑀𝑆𝑀(𝐿)/𝑀𝑆𝐸~F(6,18)