统计建模与R软件第七章习题答案.pdf

统计建模与R软件第七章习题答案(方差分析)Ex7.1(1)>lamp<-data.frame(X=c(115,116,98,83,103,107,118,116,73,89,85,97),A=factor(rep(1:3,c(4,4,4))))>lamp.aov<-aov(X~A,data=lamp);summary(lamp.aov)DfSumSqMeanSqFvaluePr(>F)A21304652.04.9230.0359*Residuals91192132.4P值小于0.05，有显著差异。(2)对甲的区间估计：>a<-c(115,116,98,83)>t.test(a)OneSamplet-testdata:at=13.1341,df=3,p-value=0.0009534alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:78.04264127.95736sampleestimates:meanofx103或者用这个命令更简单：>attach(lamp)>t.test(X[A==1])乙的均值估计为111，95%置信区间为99.59932,122.40068。丙的均值估计为86，95%置信区间为70.08777,101.91223。(3)多重检验：>attach(lamp)P值不做调整：>pairwise.t.test(X,A,p.adjust.method="none")PairwisecomparisonsusingttestswithpooledSDdata:XandA1220.351-30.0660.013P值进行Holm调整：Pvalueadjustmentmethod:none>pairwise.t.test(X,A,p.adjust.method="holm",data)PairwisecomparisonsusingttestswithpooledSDdata:XandA1220.35-30.130.04Pvalueadjustmentmethod:holm不论采取哪种方法，都可看出乙和丙有显著差异。Ex7.2(1)>lamp<-data.frame(X=c(20,18,18,17,15,16,13,18,22,17,26,19,26,28,23,25,24,25,18,22,27,24,12,14),A=factor(rep(1:4,c(10,6,6,2))))>lamp.aov<-aov(X~A,data=lamp);summary(lamp.aov)DfSumSqMeanSqFvaluePr(>F)A3351.7117.2415.112.28e-05***Residuals20155.27.76P值小于0.05，可认为四个厂生产的产品的变化率有显著差异。(2)>attach(lamp)P值不做调整：>pairwise.t.test(X,A,p.adjust.method="none")PairwisecomparisonsusingttestswithpooledSDdata:XandA12328.0e-05--30.000530.47666-40.054906.1e-050.00020Pvalueadjustmentmethod:noneP值进行Holm调整：>pairwise.t.test(X,A,p.adjust.method="holm")PairwisecomparisonsusingttestswithpooledSDdata:XandA12320.00040--30.001580.47666-40.109790.000360.00079Pvalueadjustmentmethod:holm由此可得，除了A1和A4，A2和A3这两组的差异不显著外，其他组合的差异都很显著。Ex7.3>lamp1<-data.frame(X=c(30,27,35,35,29,33,32,36,26,41,33,31,43,45,53,44,51,53,54,37,47,57,48,42,82,66,66,86,56,52,76,83,72,73,59,53),A=factor(rep(1:3,c(12,12,12))))>attach(lamp1)正态性检验：>shapiro.test(X[A==1])Shapiro-Wilknormalitytestdata:X[A==1]W=0.9731,p-value=0.9407>shapiro.test(X[A==2])Shapiro-Wilknormalitytestdata:X[A==2]W=0.9708,p-value=0.9193>shapiro.test(X[A