R软件计算题--统计学专业

例4.15P179（一个正态总体的区间估计）为估计一件物体的重量a，将其称了10次，得到的重量（单位：kg）为10.1,10,9.8,10.5,9.7,10.1,9.9,10.2,10.3,9.9，假设所称出物体重量服从正态分布求该物体重量a的置信系数为0.95的置信区间。x<-c(10.1,10,9.8,10.5,9.7,10.1,9.9,10.2,10.3,9.9)t.test(x)程序结果：OneSamplet-testdata:xt=131.59,df=9,p-value=4.296e-16alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:9.87722510.222775sampleestimates:meanofx10.05得到的区间估计为：[9.88,10.22]12021/3/11例4.18P185(均值差的区间估计）现从生产线上随机抽取样本x1,x2，···，x12和y1,y2，···，y17，都服从正态分布，其均值分别为u1=201.1,u2=499.7，差分别为2.4,4.7。给定置信系数0.95，试求u1-u2的区间估计。x<-rnorm(12,501.1,2.4)y<-rnorm(17,499.7,4.7)①两样本方差不同t.test(x,y)程序结果：WelchTwoSamplet-testdata:xandyt=-0.6471,df=25.304,p-value=0.5234alternativehypothesis:truedifferenceinmeansisnotequalto095percentconfidenceinterval:-3.6571211.907620sampleestimates:meanofxmeanofy500.7888501.6635u1-u2的置信系数为0.95的区间估计为[-3.66,1.91]②方差相同t.test(x,y,var.equal=TRUE)22021/3/11例4.19P186(配对数据情形下均值差的区间估计）抽查患者10名。下治疗前后血红蛋白的含量数据。试求治疗前后变化的区间估计。（a=0.05）。x<-c(11.3,15.0,15.0,13.5,12.8,10.0,11.0,12.0,13.0,12.3)y<-c(14.0,13.8,14.0,13.5,13.5,12.0,14.7,11.4,13.8,12.0)t.test(x-y)程序结果：OneSamplet-testdata:x-yt=-1.3066,df=9,p-value=0.2237alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:-1.85728810.4972881sampleestimates:meanofx-0.68治疗前后变化的区间估计为[-1.86,0.497]32021/3/11例4.22P193（一个总体求均值的单侧置信区间估计）从一批灯泡中随机地取5只作寿命试验测得寿命以小时计为10501100112012501280设灯泡的寿命服从正态分布.求灯泡寿命平均值的置信度为0.95的单侧置信下限x<-c(1050,1100,1120,1250,1280)t.test(x,alternative="greater")程序结果：OneSamplet-testdata:xt=26.003,df=4,p-value=6.497e-06alternativehypothesis:truemeanisgreaterthan095percentconfidenceinterval:1064.9Infsampleestimates:meanofx116095%的灯泡寿命在1064.9小时以上42021/3/11习题4.6P201甲、乙两种稻种分布播种在10块试验田中，每块试验田甲、乙稻种各种一半，假设两稻种产量 X,Y均服从正态分布，且方差相等，收获后10块试验田的产量如下所示（单位：千克）。求出两稻种产量的期望差u1-u2的置信区间（a=0.05).x<-c(140,137,136,140,145,148,140,135,144,141)y<-c(135,118,115,140,128,131,130,115,131,125)t.test(x,y,var.equal=T)程序结果TwoSamplet-testdata:xandyt=4.6287,df=18,p-value=0.0002087alternativehypothesis:truedifferenceinmeansisnotequalto095percentconfidenceinterval:7.53626120.063739sampleestimates:meanofxmeanofy140.6126.8置信区间为[7.536261,20.063739]52021/3/11习题4.7甲、乙两组生产同种导线，现从甲组生产的导线中随机抽取4根，从乙组生产的导线中随机抽取5根，它们的电阻值分别为：甲：0.143,0.142,0.143,0.137；乙：0.140,0.142,0.136,0.138,0.140；假设两组电阻值分别服从正态分布，方差相同但未知，试求 u1-u2的置信系数为0.95的区间估计。x<-c(0.143,0.142,0.143,0.137)y<-c(0.140,0.142,0.136,0.138,0.140)a<-rnorm(4,mean(x),var(x))b<-rnorm(5,mean(y),var(y))t.test(a,b)程序结果：WelchTwoSamplet-testdata:aandbt=636.28,df=5.788,p-value=3.028e-15alternativehypothesis:truedifferenceinmeansisnotequalto095percentconfidenceinterval:0.0020414400.002057343sampleestimates:meanofxmeanofy0.14124940.1392000区间为：[0.00204,0.00205]62021/3/11例5.2P209（单个正态总体均值的假设检验）某种元件的寿命X（小时），服从正态分布,其中f方差和均值均未知，16只元件的寿命如下：问是否有理由认为元件的平均寿命大于255小时。x<-c(159,280,101,212,224,379,179,264,222,362,168,250,149,260,485,170)t.test(x,alternative="greater",mu=225)程序结果：OneSamplet-testdata:xt=0.66852,df=15,p-value=0.257alternativehypothesis:truemeanisgreaterthan22595percentconfidenceinterval:198.2321Infsampleestimates:meanofx241.5计算出P值为0.257大于0.05，所以，接受原假设，即认为元件的平均寿命不大于255小时72021/3/11例5.6P221（二项分布总体的假设检验）有一批蔬菜种子的平均发芽率为P=0.85,现在随机抽取500粒，用种衣剂进行浸种处理，结果有445粒发芽，问种衣剂有无效果。binom.test(445,500,p=0.85)程序结果：Exactbinomialtestdata:445and500numberofsuccesses=445,numberoftrials=500,p-value=0.01207alternativehypothesis:trueprobabilityofsuccessisnotequalto0.8595percentconfidenceinterval:0.85923420.9160509sampleestimates:probabilityofsuccess0.89P值=0.01207<0.05，拒绝原假设，认为种衣剂对种子发芽率有显著效果。82021/3/11习题5.1P249正常男子血小板计数均值为225*10^9/L,今测得20名男性油漆作业工人的血小板计数值如下。问油漆工人的血小板计数与正常成年男子有无差异？x<-c(220,188,162,230,145,160,237,188,247,113,126,245,164,231,250,183,190,158,224,175)t.test(x,alternative="two.side",mu=225)程序结果：OneSamplet-testdata:xt=-3.5588,df=19,p-value=0.002096alternativehypothesis:truemeanisnotequalto22595percentconfidenceinterval:172.2743211.3257sampleestimates:meanofx191.8P值=0.002096<0.05,拒绝原假设，认为油漆工人的血小板计数与正常成年男子有差异。92021/3/11习题5.3为研究某铁剂治疗和饮食治疗营养性缺铁性贫血的效果，将16名患者按年龄、体重、病程和病情相近的原则配成8对，分别使用饮食疗法和补充铁剂治疗的方法，三个月后测得两种患者血红蛋白如5.1所示，问两种方法治疗后的患者血红蛋白有无差异.x<-c(113,120,138,120,100,118,138,123)y<-c(138,116,125,136,110,132,130,110)t.test(x-y)程序结果：OneSamplet-testdata:x-yt=-0.65127,df=7,p-value=0.5357alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:-15.6288918.878891sampleestimates:meanofx-3.375P=0.537>0.05,接受原假设，两种方法治疗后的患者血红蛋白无差异102021/3/11例6.2P257（回归方程的显著性检验）求例6.1的回归方程，并对相应的方程做检验。x<-c(0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.20,0.21,0.23)y<-c(42.0,43.5,45.0,45.5,45.0,47.5,49.0,53.0,50.0,55.0,55.0,60.0)lm.sol<-lm(y~1+x)summary(lm.sol)程序结果见下一张PPT回归方程为：从回归结果可以看出，回归方程通过了回归参数的检验与回归方程的检验。112021/3/11例6.2的程序结果程序结果：Call:lm(formula=y~1+x) Residuals:Min1QMedian3QMax-2.0431-0.70560.16940.66332.2653 Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)28.4931.58018.045.88e-09***x130.8359.68313.519.50e-08***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:1.319on10degreesoffreedomMultipleR-squared:0.9481,AdjustedR-squared:0.9429F-statistic:182.6on1and10DF,p-value:9.505e-08122021/3/11例6.4P260（预测）求例6.1中X=x0=0.16时相应的Y的概率为0.95的预测区间new<-data.frame(x=0.16)lm.pred<-predict(lm.sol,new,interval="prediction",level=0.95)lm.pred程序结果：fitlwrupr49.4263946.3662152.48657预测值为49.43，预测区间[46.37,52.49]132021/3/11例6.5P261（全面展示一元回归模型的计算过程）Forbes数据X<-matrix(c(194.5,20.79,1.3179,131.79,194.3,20.79,1.3179,131.79,197.9,22.40,1.3502,135.02,198.4,22.67,1.3555,135.55,199.4,23.15,1.3646,136.46,199.9,23.35,1.3683,136.83,200.9,23.89,1.3782,137.82,201.1,23.99,1.3800,138.00,201.4,24.02,1.3806,138.06,201.3,24.01,1.3805,138.05,203.6,25.14,1.4004,140.04,204.6,26.57,1.4244,142.44,209.5,28.49,1.4547,145.47,208.6,27.76,1.4434,144.34,210.7,29.04,1.4630,146.30,211.9,29.88,1.4754,147.54,212.2,30.06,1.4780,147.80),ncol=4,byrow=T,dimnames=list(1:17,c("F","h","log","log100")))①forbes<-data.frame(X)plot(forbes$F,forbes$log100)程序结果是出现散点图142021/3/11②lm.sol<-lm(log100~F,data=forbes)summary(lm.sol)程序结果：Call:lm(formula=log100~F,data=forbes) Residuals:Min1QMedian3QMax-0.32261-0.14530-0.067500.021111.35924 Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)-42.130873.33895-12.622.17e-09***F0.895460.0164554.45<2e-16***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:0.3789on15degreesoffreedomMultipleR-squared:0.995,AdjustedR-squared:0.9946F-statistic:2965on1and15DF,p-value:<2.2e-16152021/3/11③abline(lm.sol)程序结果：得到散点图和相应的回归直线④y.res<-residuals(lm.sol);plot(y.res)text(12,y.res[12],labels=12,adj=1.2)程序结果：将第12号残差点标出⑤lm12<-lm(log100~F,data=forbes,subset=-12)summary(lm12)程序结果：Call:lm(formula=log100~F,data=forbes,subset=-12) Residuals:Min1QMedian3QMax-0.21175-0.061940.015900.090770.13042 Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)-41.301801.00038-41.295.01e-16***F0.890960.00493180.73<2e-16***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:0.1133on14degreesoffreedomMultipleR-squared:0.9996,AdjustedR-squared:0.9995F-statistic:3.266e+04on1and14DF,p-value:<2.2e-16162021/3/11例6.14P292某公司为了研究产品的营销策略，对产品的销售情况进行了调查，设Y为某地区该产品的家庭人均购买量（单位：元），X为家庭收入（单位：元），表6.8给出了53个家庭的数据。试通过这些数据建立Y与X的关系。X<-scan()6792921012493582115699721891097207818181700747203016434143541276745435540874154310297101434837174813811428125517773702316113046377072480879078340612426581746468111441317873560149522211526Y<-scan()0.790.440.560.792.703.644.739.505.346.855.845.213.254.433.160.500.171.880.771.390.561.565.280.644.000.314.204.883.487.582.634.990.598.194.790.511.744.103.940.963.290.443.242.145.710.641.900.518.3314.945.113.853.93172021/3/11①lm.sol<-lm(Y~X);summary(lm.sol)程序结果：Call:lm(formula=Y~X) Residuals:Min1QMedian3QMax-4.1399-0.8275-0.19341.23763.1522 Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)-0.83130370.4416121-1.8820.0655.X0.00368280.000333911.0304.11e-15***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:1.577on51degreesoffreedomMultipleR-squared:0.7046,AdjustedR-squared:0.6988F-statistic:121.7on1and51DF,p-value:4.106e-15②y.rst<-rstandard(lm.sol);y.fit<-predict(lm.sol)plot(y.rst~y.fit)abline(0.1,0.5);abline(-0.1,-0.5)程序结果：画出了标准化后的残差图182021/3/11③lm.new<-update(lm.sol,sqrt(.)~.);summary(lm.new)程序结果：Call:lm(formula=sqrt(Y)~X) Residuals:Min1QMedian3QMax-1.39185-0.30576-0.038750.253780.81027 Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)5.822e-011.299e-014.4814.22e-05***X9.529e-049.824e-059.6993.61e-13***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:0.464on51degreesoffreedomMultipleR-squared:0.6485,AdjustedR-squared:0.6416F-statistic:94.08on1and51DF,p-value:3.614e-13④yn.rst<-rstandard(lm.new);yn.fit<-predict(lm.new)plot(yn.rst~yn.fit)192021/3/11习题6.1P331为估计山上积雪融化后对下游灌溉的影响，在山上建立一个观测站，测量最大积雪深度X（米）与当年灌溉面积Y（公颂），测得连续10年的数据如表6.1所示。snow<-data.frame(y=c(1907,1287,2700,2373,3260,3000,1974,2273,3113,2493),x=c(5.1,3.5,7.1,6.2,8.8,7.8,4.5,5.6,8.0,6.4))1）plot(y~x,data=snow)##生成散点图，据此得出变量之间的关系lm.sol<-lm(y~x-1,data=snow)##生成过原点的直线summary(lm.sol)##提取模型中包含的信息abline(lm.sol)##生成拟合曲线程序结果：Call:lm(formula=y~x-1,data=snow) Residuals:Min1QMedian3QMax-132.53-53.63-12.1028.09239.18 Coefficients:EstimateStd.ErrortvaluePr(>|t|)x385.515.0975.746.17e-14***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1 Residualstandarderror:104.6on9degreesoffreedomMultipleR-squared:0.9984,AdjustedR-squared:0.9983F-statistic:5736on1and9DF,p-value:6.169e-14202021/3/11现测得今年的数据是X=7m，给出今年灌溉面积的预测值与相应的区间估计。（a=0.05）4）X<-data.frame(x=7)lm.pred<-predict(lm.sol,X,interval="prediction",level=0.95)lm.pred##拟合新数据，并生成置信区间程序结果：fitlwrupr12698.6032448.7182948.488212021/3/11例7.2P339(方差分析表的计算）（续例7.1)lamp<-data.frame(X=c(1600,1610,1650,1680,1700,1700,1780,1500,640,1400,1700,1750,1640,1550,1600,1620,1640,1600,1740,1800,1510,1520,1530,1570,1640,1600),A=factor(c(rep(1,7),rep(2,5),rep(3,8),rep(4,6))))lamp.aov<-aov(X~A,data=lamp)summary(lamp.aov)程序结果：DfSumSqMeanSqFvaluePr(>F)A349212164042.1660.121Residuals221666227574222021/3/11例7.3P341小白鼠在接种了3种不同菌型的伤寒杆菌后的存活天数如表所示，判断小白鼠被注射3种菌型后的平均存活天数有无显著差异？mouse<-data.frame(X=c(2,4,3,2,4,7,7,2,2,5,4,5,6,8,5,10,7,12,12,6,6,7,11,6,6,7,9,5,5,10,6,3,10),A=factor(c(rep(1,11),rep(2,10),rep(3,12))))mouse.aov<-aov(X~A,data=mouse)summary(mouse.aov)程序结果：DfSumSqMeanSqFvaluePr(>F)A294.2647.138.4840.0012**Residuals30166.655.56---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1P值远小于0.01应拒绝原假设，即认为小白鼠被注射3种菌型后的平均存活天数有显著差异232021/3/11习题7.1P3713个工厂生产同一种零件。现从各厂产品中选取4件产品做检测，其检测程度如表所示：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))))1）对数据作方差分析，判断3个厂生产的产品的零件强度是否有显著差异lamp.aov<-aov(X~A,data=lamp);summary(lamp.aov)程序结果：DfSumSqMeanSqFvaluePr(>F)A21304652.04.9230.0359*Residuals91192132.4---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1P值大于0.01，接受原假设。没有显著差异242021/3/11（2）对每个工厂生产的产品零件强度的均值，作出相应的区间估计甲的区间估计a<-c(115,116,98,83)t.test(a)程序结果：OneSamplet-test data:at=13.134,df=3,p-value=0.0009534alternativehypothesis:truemeanisnotequalto095percentconfidenceinterval:78.04264127.95736sampleestimates:meanofx103均值为103，区间估计为（78.04264，127.9574） 同理，乙的均值为111，区间估计为（99.59932，122.4007）3）多重t检验attach(lamp)pairwise.t.test(X,A,p.adjust.method="none")252021/3/11个人观点供参考，欢迎讨论