R软件计算题--统计学专业.ppt

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1、例4.15P179（一个正态总体的区间估计）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:truemeanisnoteq

2、ualto095percentconfidenceinterval:9.87722510.222775sampleestimates:meanofx10.05得到的区间估计为：9.88,10.22例4.18P185(均值均值差的区间估计）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，标准差分别为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)两样本方差不同两样本方差不同t.test(x,y)程序结果：

3、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方差相同方差相同t.test(x,y,var.equal=TRUE)例4.19P186(配对数据情形下均值差的区间估

4、计）配对数据情形下均值差的区间估计）抽查患者10名。记录下治疗前后血红蛋白的含量数据。试求求治疗前后变化的区间估计。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:truemeanisnotequalto095

5、percentconfidenceinterval:-1.85728810.4972881sampleestimates:meanofx-0.68治疗前后变化的区间估计为-1.86,0.497例4.22 P193（一个总体求均值的单侧置信区间估计）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.求灯求灯泡寿命平均值的置信度为泡寿命平均值的置信度为0.95的单侧置信下限的单侧置信下限x-c(1050,1100,1120,1250,1280)t.test(x,alternative=great

6、er)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两稻种产量的期望差差u

7、1-u2的的置信区间置信区间（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:7.53626120.063739samp

8、leestimates:meanofxmeanofy140.6126.8置信区间为7.536261,20.063739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，试求求u1-u2的的置信系数为置信系数为0.95的区间估计。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

9、-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：0.00204,0.0020

10、5例5.2P209（单个正态总体均值的假设检验）（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问是否有理由认为元件的平均寿命大于255小时。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=0.257alternativehypot

11、hesis: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小时例5.6P221（二项分布总体的假设检验（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:445and500n

12、umberofsuccesses=445,numberoftrials=500,p-value=0.01207alternativehypothesis:trueprobabilityofsuccessisnotequalto0.8595percentconfidenceinterval:0.85923420.9160509sampleestimates:probabilityofsuccess0.89P值=0.012070.05，拒绝原假设，认为种衣剂对种子发芽率有显著效果。习题5.1P249正常男子血小板计数均值为225*109/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

13、人的血小板计数与正常成年男子有无差异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:172.274

14、3211.3257sampleestimates:meanofx191.8P值=0.0020960.05,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。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表5.1所示，问两种方法治疗后的患者血红蛋问两种方法治疗后的患者血红蛋白有无差异白有无差异.x-c(113,120,138,120,100,118,138,123)y0.05,接受原假设，两种方法治疗后的患者血红蛋白无差异例6.2P257（回归方程的显

15、著性检验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|t|)(Intercept)28.4931.58018.045.88e-09*x130.8359.68313.519.50e-08*-Signif.codes:0*0.001*0.01*0.05.0.11Residualstandarderror:1.319

16、on10degreesoffreedomMultipleR-squared:0.9481,AdjustedR-squared:0.9429F-statistic:182.6on1and10DF,p-value:9.505e-08例6.4P260（预测预测）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

17、46.37,52.49例6.5P261（全面展示一元回归模型的计算过程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

18、.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)forbes-data.frame(X)plot(forbes$F,fo

19、rbes$log100)程序结果是出现散点图散点图lm.sol|t|)(Intercept)-42.130873.33895-12.622.17e-09*F0.895460.0164554.452e-16*-Signif.codes:0*0.001*0.01*0.05.0.11Residualstandarderror:0.3789on15degreesoffreedomMultipleR-squared:0.995,AdjustedR-squared:0.9946F-statistic:2965on1and15DF,p-value:2.2e-16abline(lm.sol)程序结果：得到散点

20、图和相应的回归直线散点图和相应的回归直线y.res-residuals(lm.sol);plot(y.res)text(12,y.res12,labels=12,adj=1.2)程序结果：将第第12号残差点标出号残差点标出lm12|t|)(Intercept)-41.301801.00038-41.295.01e-16*F0.890960.00493180.732e-16*-Signif.codes:0*0.001*0.01*0.05.0.11Residualstandarderror:0.1133on14degreesoffreedomMultipleR-squared:0.9996,Adj

21、ustedR-squared:0.9995F-statistic:3.266e+04on1and14DF,p-value:2.2e-16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通过这些数据建立Y与与X的关系。的关系。X-scan()67929210124935821156997218910972078181817007472030164341435412767454355408741543102971014348371748138114

22、28125517773702316113046377072480879078340612426581746468111441317873560149522211526Y-scan()0.790.440.560.792.703.644.739.505.346.855.845.213.254.433.160.500.171.880.771.390.561.565.280.644.000.314.204.883.487.582.634.990.598.194.790.511.744.103.940.963.290.443.242.145.710.641.900.518.3314.945.113.85

23、3.93lm.sol|t|)(Intercept)-0.83130370.4416121-1.8820.0655.X0.00368280.000333911.0304.11e-15*-Signif.codes:0*0.001*0.01*0.05.0.11Residualstandarderror:1.577on51degreesoffreedomMultipleR-squared:0.7046,AdjustedR-squared:0.6988F-statistic:121.7on1and51DF,p-value:4.106e-15y.rst-rstandard(lm.sol);y.fit-pr

24、edict(lm.sol)plot(y.rsty.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标准化后的残差图lm.new|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*0.05.0.11Residualstandarderror:0.464on51degreesoffreedomMultipleR-squared:0.6485,AdjustedR-squared:0.

25、6416F-statistic:94.08on1and51DF,p-value:3.614e-13yn.rst-rstandard(lm.new);yn.fit-predict(lm.new)plot(yn.rstyn.fit)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

26、.6,8.0,6.4)1）plot(yx,data=snow)#生成散点图，据此得出变量之间的关系生成散点图，据此得出变量之间的关系lm.sol|t|)x385.515.0975.746.17e-14*-Signif.codes:0*0.001*0.01*0.05.0.1 1Residualstandarderror:104.6on9degreesoffreedomMultipleR-squared:0.9984,AdjustedR-squared:0.9983F-statistic:5736on1and9DF,p-value:6.169e-14现测得今年的数据是X=7m，给出今年灌溉面积的预

27、测值与相预测值与相应的区间估计应的区间估计。（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：fitlwrupr12698.6032448.7182948.488例7.2P339(方差分析表方差分析表的计算）（续例7.1)lamp-data.frame(X=c(1600,1610,1650,1680,1700,1700,1780,1500,640,1400,1700,1750,1640,1550,1600,1620,1640,1

28、600,1740,1800,1510,1520,1530,1570,1640,1600),A=factor(c(rep(1,7),rep(2,5),rep(3,8),rep(4,6)lamp.aovF)A349212164042.1660.121Residuals221666227574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

29、),A=factor(c(rep(1,11),rep(2,10),rep(3,12)mouse.aovF)A294.2647.138.4840.0012*Residuals30166.655.56-Signif.codes:0*0.001*0.01*0.05.0.11P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(

30、1:3,c(4,4,4)1）对数据作方差分析，判断对数据作方差分析，判断3个厂生产的产品的零件强个厂生产的产品的零件强度是否有显著差异度是否有显著差异lamp.aovF)A21304652.04.9230.0359*Residuals91192132.4-Signif.codes:0*0.001*0.01*0.05.0.11P值大于0.01，接受原假设。没有显著差异（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-testdata: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)