R软件课后习题解第四章

1、习题4.1计算代码如下（极大似然估计）：x-c(0.1,0.2,0.9,0.8,0.7,0.7)n- length(x)f-function(a)n/(a+1)+sum(log(x)uniroot(f,c(0,1)$root1 0.$f.root1 -3.e-05$iter1 5$estim.prec1 6.e-05习题4.2（指数分布）x-c(5,15,25,35,45,55,65)v-c(365,245,150,100,70,45,25)y-x*vf-function(k)1000/k-sum(y)uniroot(f,c(0,100)$root1 0.$f.root1 -10.46586$

2、iter1 14$estim.prec1 6.e-05估计值为0.习题4.3（极大似然估计）(作业)解：设为字样的一组观测值，所以似然函数为取对数，所以 求偏导，并令其等于0，解得，所以的极大似然估计量为.x-rep(0:6,c(17,20,10,2,1,0,0)mean(x)1 1习题4.4y-function(x)f-c(-13+x1+(5-x2)*x2-2)*x2,-29+x1+(x2+1)*x2-14)*x2);sum(f2)x0-c(0.5,-2)nlm(y,x0)$minimum1 48.98425$estimate1 11. -0.$gradient1 1.e-08 -1.e-0

3、7$code1 1$iterations1 16习题4.5X- c(54,67,68,78,70,66,67,70,65,69)t.test(X) One Sample t-testdata: X t = 35.947, df = 9, p-value = 4.938e-11alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: 63.1585 71.6415 sample estimates:mean of x 67.4因此,10名患者平均脉搏在95%的置信区间为63.16,71.

4、64 10个人的平均脉搏为67.4，所以这10名患者的平均脉搏属不低于正常人的平均脉搏习题4.6(作业)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) Welch Two Sample t-testdata: x and y t = 4.6287, df = 13.014, p-value = 0.alternative hypothesis: true difference in means is not equal to 0 95 perc

5、ent confidence interval: 7. 20. sample estimates:mean of x mean of y 140.6 126.8所以u1-u2的置信区间为7.53626 ,20.06374习题4.7x-c(0.143,0.142,0.143,0.137)y-c(0.140,0.142,0.136,0.138,0.140)t.test(x,y,var.equal=TRUE) # 注意：如果方差相同，需要声明var.equal=TRUE Two Sample t-testdata: x and y t = 1.198, df = 7, p-value = 0.269

6、9alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0. 0. sample estimates:mean of x mean of y 0.14125 0.13920习题4.8(作业)x-c(140,137,136,140,145,148,140,135,144,141)y-c(135,118,115,140,128,131,130,115,131,125)var.test(x,y) F test to compare two variance

7、sdata: x and y F = 0.2353, num df = 9, denom df = 9, p-value = 0.04229alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0. 0. sample estimates:ratio of variances 0. t.test(x,y) Welch Two Sample t-testdata: x and y t = 4.6287, df = 13.014, p-value = 0.a

8、lternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 7. 20. sample estimates:mean of x mean of y 140.6 126.8所以所求置信区间为7.,20.习题4.9X-rep(0:6,c(7,10,12,8,3,2,0)t.test(X) One Sample t-testdata: X t = 9.0895, df = 41, p-value = 2.238e-11alternative hypothesis:

9、true mean is not equal to 0 95 percent confidence interval: 1. 2. sample estimates:mean of x 1. 所以估计值为1.，置信区间为1.,2.习题4.10(作业)X-c(1067,919,1196,785,1126,936,918,1156,920,948)t.test(X,alternative=greater) One Sample t-testdata: X t = 23.9693, df = 9, p-value = 9.148e-10alternative hypothesis: true mea

10、n is greater than 0 95 percent confidence interval: 920.8443 Inf sample estimates:mean of x 997.1所以有95%的灯泡寿命在920.8443h以上.习题5.1 x-c(220, 188, 162, 230, 145, 160, 238, 188, 247, 113, 126, 245, 164, 231, 256, 183, 190, 158, 224, 175) t.test(x,mu=225) One Sample t-testdata: x t = -3.4783, df = 19, p-val

11、ue = 0.alternative hypothesis: true mean is not equal to 225 95 percent confidence interval: 172.3827 211.9173 sample estimates:mean of x 192.15原假设：油漆工人的血小板计数与正常成年男子无差异。备择假设：油漆工人的血小板计数与正常成年男子有差异。p值小于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。上述检验是双边检验。也可采用单边检验。备择假设：油漆工人的血小板计数小于正常成年男子。t.test(x,mu=225,alternat

12、ive=less) One Sample t-testdata: x t = -3.4783, df = 19, p-value = 0.alternative hypothesis: true mean is less than 225 95 percent confidence interval: -Inf 208.4806 sample estimates:mean of x 192.15同样可得出油漆工人的血小板计数小于正常成年男子的结论。习题5.2x-c(1067,919,1196,785,1126,936,918,1156,920,948)pnorm(1000,mean(x),sd

13、(x)1 0.x=1000的概率为0.509,故x大于1000的概率为0.491.要点：pnorm计算正态分布的分布函数。在R软件中，计算值均为下分位点。习题5.3A-c(113,120,138,120,100,118,138,123)B-c(138,116,125,136,110,132,130,110)t.test(A,B,paired=TRUE) Paired t-testdata: A and B t = -0.6513, df = 7, p-value = 0.5357alternative hypothesis: true difference in means is not eq

14、ual to 0 95 percent confidence interval: -15. 8. sample estimates:mean of the differences -3.375习题5.4(作业)（1）正态性W检验：x-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40)a-shapiro.test(x);a Shapiro-Wilk normality testdata: x W = 0.9699, p-value =

15、 0.7527y-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00)b-shapiro.test(y);b Shapiro-Wilk normality testdata: y W = 0.971, p-value = 0.7754Kolmogorov-Smirnor检验：（K-S检验）ks.test(x,pnorm,mean(x),sd(x) One-sample Kolmogorov-Smirnov testdata: x D = 0

16、.1065, p-value = 0.9771alternative hypothesis: two-sided 警告信息：In ks.test(x, pnorm, mean(x), sd(x) : Kolmogorov - Smirnov检验里不应该有连结ks.test(y,pnorm,mean(y),sd(y) One-sample Kolmogorov-Smirnov testdata: y D = 0.1197, p-value = 0.9368alternative hypothesis: two-sided 警告信息：In ks.test(y, pnorm, mean(y), sd

17、(y) : Kolmogorov - Smirnov检验里不应该有连结Pearson拟合优度卡方检验：对x有：sort(x) 1 -5.6 -1.6 -1.4 -0.7 -0.5 0.4 0.7 1.7 2.0 2.5 2.5 2.8 3.0 3.5 4.0 4.5 4.6 5.8 6.0 7.1x1-table(cut(x,br=c(-6,-3,0,3,6,9)p-pnorm(c(-3,0,3,6,9),mean(x),sd(x);p1 0. 0. 0. 0. 0.p-c(p1,p2-p1,p3-p2,p4-p3,1-p4);p1 0. 0. 0. 0. 0.chisq.test(x1,p=

18、p) Chi-squared test for given probabilitiesdata: x1 X-squared = 0.5639, df = 4, p-value = 0.967警告信息：In chisq.test(x1, p = p) : Chi-squared近似算法有可能不准对y有：对x有,p值为0.967，接受原假设，x符合正态分布。对y同理，所以，两组数据均为正态分布。（2）方差相同模型t检验：x-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.

19、60,2.50,6.00,-1.40)y-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00)t.test(x,y,var.equal=TRUE) Two Sample t-testdata: x and y t = -0.6419, df = 38, p-value = 0.5248alternative hypothesis: true difference in means is not equal to 0 95 percent c

20、onfidence interval: -2. 1. sample estimates:mean of x mean of y 2.065 2.625 方差不同模型t检验：x-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40)y-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00

21、)t.test(x,y) Welch Two Sample t-testdata: x and y t = -0.6419, df = 36.086, p-value = 0.525alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -2.32926 1.20926 sample estimates:mean of x mean of y 2.065 2.625成对t检验：x-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.0

22、0,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40)y-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00)t.test(x,y,paired=TRUE) Paired t-testdata: x and y t = -0.6464, df = 19, p-value = 0.5257alternative hypothesis: true differ

23、ence in means is not equal to 0 95 percent confidence interval: -2. 1. sample estimates:mean of the differences -0.56以上P值均大于0.05，故均值无差异。（三种检验的结果都显示两组数据均值无差异。）(3)方差检验：x-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40)y-c(3.70,6.50,5.00,5.20,0

24、.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00)var.test(x,y) F test to compare two variancesdata: x and y F = 1.5984, num df = 19, denom df = 19, p-value = 0.3153alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0. 4.

25、sample estimates:ratio of variances 1.P值大于0.05，接受原假设，所以方差无差异。习题5.5（1）正态性检验，采用ks检验：a - c(126,125,136,128,123,138,142,116,110,108,115,140)b - c(162,172,177,170,175,152,157,159,160,162)ks.test(a,pnorm,mean(a),sd(a) One-sample Kolmogorov-Smirnov testdata: a D = 0.1464, p-value = 0.9266alternative hypoth

26、esis: two-sided ks.test(b,pnorm,mean(b),sd(b) One-sample Kolmogorov-Smirnov testdata: b D = 0.2222, p-value = 0.707alternative hypothesis: two-sided 警告信息：In ks.test(b, pnorm, mean(b), sd(b) : Kolmogorov - Smirnov检验里不应该有连结a和b都服从正态分布。（2）方差齐性检验：var.test(a,b) F test to compare two variancesdata: a and b

27、 F = 1.9646, num df = 11, denom df = 9, p-value = 0.32alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0. 7. sample estimates:ratio of variances 1.可认为a和b的方差相同。（3）选用方差相同模型t检验：t.test(a,b,var.equal=TRUE) Two Sample t-testdata: a and b t = -8.8148, df = 2

28、0, p-value = 2.524e-08alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -48.24975 -29.78358 sample estimates:mean of x mean of y 125.5833 164.6000 可认为两者有差别。习题5.6（作业）直接调用binom.test()函数可得：binom.test(57, 400, p = 0.147) Exact binomial testdata: 57 and 40

29、0 number of successes = 57, number of trials = 400, p-value = 0.8876alternative hypothesis: true probability of success is not equal to 0.147 95 percent confidence interval: 0. 0. sample estimates:probability of success 0.1425 P值大于0.05,所以支持结果。习题5.7二项分布总体的假设检验：binom.test(178,328,p=0.5,alternative=gre

30、ater) Exact binomial testdata: 178 and 328 number of successes = 178, number of trials = 328, p-value = 0.06794alternative hypothesis: true probability of success is greater than 0.5 95 percent confidence interval: 0. 1. sample estimates:probability of success 0.不能认为这种处理能增加母鸡的比例。习题5.8利用pearson卡方检验是否

31、符合特定分布：chisq.test(c(315,101,108,32),p=c(9,3,3,1)/16) Chi-squared test for given probabilitiesdata: c(315, 101, 108, 32) X-squared = 0.47, df = 3, p-value = 0.9254接受原假设，符合自由组合定律。习题5.9（作业）X-0:5;Y-c(92,68,28,11,1,0)q-ppois(X,mean(rep(X,Y);n-length(Y);p-numeric(n)p1-q1;pn-1-qn-1for(i in 2:(n-1)pi-qi-qi-

32、1chisq.test(Y,p=p) Chi-squared test for given probabilitiesdata: Y X-squared = 2.1596, df = 5, p-value = 0.8267警告信息：In chisq.test(Y, p = p) : Chi-squared近似算法有可能不准以下是矫正后的结果:X-0:3;Z-c(92,68,28,12)q-ppois(X,mean(rep(X,Z);n-length(Z);p-numeric(n)p1-q1;pn-1-qn-1for(i in 2:(n-1)pi0.1,所以,可以认为每分钟顾客数X服从Poiss

33、on分布。习题5.10ks检验 两个分布是否相同：x-c(2.36,3.14,752,3.48,2.76,5.43,6.54,7.41)y-c(4.38,4.25,6.53,3.28,7.21,6.55)ks.test(x,y) Two-sample Kolmogorov-Smirnov testdata: x and y D = 0.375, p-value = 0.6374alternative hypothesis: two-sided 习题5.11列联数据的独立性检验：x- c(358,2492,229,2745)dim(x)-c(2,2)chisq.test(x) Pearsons

34、Chi-squared test with Yates continuity correctiondata: x X-squared = 37.4143, df = 1, p-value = 9.552e-10P 值0.05 ，拒绝原假设，有影响。习题5.12（作业）列联数据的独立性检验：x-c(45,46,28,11,12,20,23,12,10,28,30,35)dim(x)-c(4,3)chisq.test(x,correct = FALSE) Pearsons Chi-squared testdata: x X-squared = 40.401, df = 6, p-value = 3

35、.799e-07P值小于0.05,所以B与C不独立。习题5.13因有的格子的频数小于5，故采用fiser确切概率法检验独立性。x-c(3,6,4,4);dim(x)-c(2,2)fisher.test(x) Fishers Exact Test for Count Datadata: x p-value = 0.6372alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0. 5. sample estimates:odds ratio 0.p值大于0.05，两变量独

36、立，两种工艺对产品的质量没有影响。习题5.14由于是在相同个体上的两次试验，故采用McNemar检验。x-c(58,1,8,2,42,9,3,7,17);dim(x)=14.6;H1: 中位数14.6x14.6,length(x),al=l) Exact binomial testdata: sum(x) 14.6 and length(x) number of successes = 1, number of trials = 10, p-value = 0.01074alternative hypothesis: true probability of success is less th

37、an 0.5 95 percent confidence interval: 0. 0. sample estimates:probability of success 0.1拒绝原假设，中位数小于14.6（2）Wilcoxon符号秩检验：wilcox.test(x,mu=14.6,al=l,exact=F) Wilcoxon signed rank test with continuity correctiondata: x V = 4.5, p-value = 0.01087alternative hypothesis: true location is less than 14.6 拒绝

38、原假设，中位数小于14.6习题5.16（作业）（1）采用成对符号检验，输入数据，调用binom.test()作检验:x-c(48.0,33.0,37.5,48.0,42.5,40.0,42.0,36.0,11.3,22.0,36.0,27.3,14.2,32.1,52.0,38.0,17.3,20.0,21.0,46.1)y-c(37.0,41.0,23.4,17.0,31.5,40.0,31.0,36.0,5.7,11.5,21.0,6.1,26.5,21.3,44.5,28.0,22.6,20.0,11.0,22.3)binom.test(sum(xy),length(x) Exact b

39、inomial testdata: sum(x y) and length(x) number of successes = 3, number of trials = 20, p-value = 0.alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval: 0. 0. sample estimates:probability of success 0.15P值0.0.05，所以，2种方法有显著差异。（2）Wilcoxon符号秩检验：输入数据，调

40、用wilcox.test()函数:x-c(48.0,33.0,37.5,48.0,42.5,40.0,42.0,36.0,11.3,22.0,36.0,27.3,14.2,32.1,52.0,38.0,17.3,20.0,21.0,46.1)y-c(37.0,41.0,23.4,17.0,31.5,40.0,31.0,36.0,5.7,11.5,21.0,6.1,26.5,21.3,44.5,28.0,22.6,20.0,11.0,22.3)wilcox.test(x,y,alternative=greater,paired=TRUE) Wilcoxon signed rank test wi

41、th continuity correctiondata: x and y V = 136, p-value = 0.alternative hypothesis: true location shift is greater than 0 警告信息：1: In wilcox.test.default(x, y, alternative = greater, paired = TRUE) : 无法精確計算带连结的p值2: In wilcox.test.default(x, y, alternative = greater, paired = TRUE) : 有0时无法計算精確的p值P值0.0.

42、05，所以，2种方法有显著差异。（3）Wilcoxon秩和检验：输入数据，调用wilcox.test()函数:x-c(48.0,33.0,37.5,48.0,42.5,40.0,42.0,36.0,11.3,22.0,36.0,27.3,14.2,32.1,52.0,38.0,17.3,20.0,21.0,46.1)y0.05，所以，2种方法有显著差异。（4）正态性检验：ks.test(x,pnorm,mean(x),sd(x) One-sample Kolmogorov-Smirnov testdata: x D = 0.1407, p-value = 0.8235alternative hypothesis: two-sided 警告信息：In ks.test(x, pnorm, mean(x), sd(x) : Kolmogor