统计建模与R软件第五章习题…

原文地址：统计建模与R软件第五章习题答案(假设检验) 作者：蘓木柒 Ex5.1 > 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-test data:  x t = -3.4783, df = 19, p-value = 0.002516 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,alternative="less")         One Sample t-test data:  x t = -3.4783, df = 19, p-value = 0.001258 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。 Ex5.2 > pnorm(1000,mean(x),sd(x)) [1] 0.5087941 > x  [1] 1067  919 1196  785 1126  936  918 1156  920  948 > pnorm(1000,mean(x),sd(x)) [1] 0.5087941 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。 Ex5.3 > A<-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-test data:  A and B t = -0.6513, df = 7, p-value = 0.5357 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:  -15.62889   8.87889 sample estimates: mean of the differences                  -3.375 p值大于0.05，接受原假设，两种方法治疗无差异。 Ex5.4 （1） 正态性W检验： >x<-c(-0.7,-5.6,2,2.8,0.7,3.5,4,5.8,7.1,-0.5,2.5,-1.6,1.7,3,0.4,4.5,4.6,2.5,6,-1.4) >y<-c(3.7,6.5,5,5.2,0.8,0.2,0.6,3.4,6.6,-1.1,6,3.8,2,1.6,2,2.2,1.2,3.1,1.7,-2)                     > shapiro.test(x)         Shapiro-Wilk normality test data:  x W = 0.9699, p-value = 0.7527 > shapiro.test(y)         Shapiro-Wilk normality test data:  y W = 0.971, p-value = 0.7754 ks检验： > ks.test(x,"pnorm",mean(x),sd(x))         One-sample Kolmogorov-Smirnov test data:  x D = 0.1065, p-value = 0.977 alternative hypothesis: two-sided Warning message: In ks.test(x, "pnorm", mean(x), sd(x)) :   cannot compute correct p-values with ties > ks.test(y,"pnorm",mean(y),sd(y))         One-sample Kolmogorov-Smirnov test data:  y D = 0.1197, p-value = 0.9368 alternative hypothesis: two-sided Warning message: In ks.test(y, "pnorm", mean(y), sd(y)) :   cannot compute correct p-values with ties pearson拟合优度检验 ，以x为例。 > 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 [16]  4.5  4.6  5.8  6.0  7.1 > x1<-table(cut(x,br=c(-6,-3,0,3,6,9))) > p<-pnorm(c(-3,0,3,6,9),mean(x),sd(x)) > p [1] 0.04894712 0.24990009 0.62002288 0.90075856 0.98828138 > p<-c(p[1],p[2]-p[1],p[3]-p[2],p[4]-p[3],1-p[4]);p [1] 0.04894712 0.20095298 0.37012278 0.28073568 0.09924144 > chisq.test(x1,p=p)         Chi-squared test for given probabilities data:  x1 X-squared = 0.5639, df = 4, p-value = 0.967 Warning message: In chisq.test(x1, p = p) : Chi-squared approximation may be incorrect p值为0.967，接受原假设，x符合正态分布。 （2） 方差相同模型t检验： > t.test(x,y,var.equal=TRUE)         Two Sample t-test data:  x and y t = -0.6419, df = 38, p-value = 0.5248 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:  -2.326179  1.206179 sample estimates: mean of x mean of y     2.065     2.625 方差不同模型t检验： > t.test(x,y)         Welch Two Sample t-test data:  x and y t = -0.6419, df = 36.086, p-value = 0.525 alternative 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 配对t检验： > t.test(x,y,paired=TRUE)         Paired t-test data:  x and y t = -0.6464, df = 19, p-value = 0.5257 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:  -2.373146  1.253146 sample estimates: mean of the differences                   -0.56 三种检验的结果都显示两组数据均值无差异。 （3） 方差检验： > var.test(x,y)         F test to compare two variances data:  x and y F = 1.5984, num df = 19, denom df = 19, p-value = 0.3153 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval:  0.6326505 4.0381795 sample estimates: ratio of variances           1.598361 接受原假设，两组数据方差相同。 Ex5.5 > 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) 正态性检验，采用ks检验： > ks.test(a,"pnorm",mean(a),sd(a))         One-sample Kolmogorov-Smirnov test data:  a D = 0.1464, p-value = 0.9266 alternative hypothesis: two-sided > ks.test(b,"pnorm",mean(b),sd(b))         One-sample Kolmogorov-Smirnov test data:  b D = 0.2222, p-value = 0.707 alternative hypothesis: two-sided Warning message: In ks.test(b, "pnorm", mean(b), sd(b)) :   cannot compute correct p-values with ties a和b都服从正态分布。 方差齐性检验： > var.test(a,b)         F test to compare two variances data:  a and b F = 1.9646, num df = 11, denom df = 9, p-value = 0.3200 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval:  0.5021943 7.0488630 sample estimates: ratio of variances           1.964622 可认为a和b的方差相同。 选用方差相同模型t检验： > t.test(a,b,var.equal=TRUE)         Two Sample t-test data:  a and b t = -8.8148, df = 20, p-value = 2.524e-08 alternative 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。 Ex5.6 二项分布总体的假设检验： > binom.test(57,400,p=0.147)         Exact binomial test data:  57 and 400 number of successes = 57, number of trials = 400, p-value = 0.8876 alternative hypothesis: true probability of success is not equal to 0.147 95 percent confidence interval:  0.1097477 0.1806511 sample estimates: probability of success                 0.1425 P 值>0.05，故接受原假设，表示调查结果支持该市老年人口的看法 Ex5.7 二项分布总体的假设检验： > binom.test(178,328,p=0.5,alternative="greater")         Exact binomial test data:  178 and 328 number of successes = 178, number of trials = 328, p-value = 0.06794 alternative hypothesis: true probability of success is greater than 0.5 95 percent confidence interval:  0.4957616 1.0000000 sample estimates: probability of success              0.5426829 不能认为这种处理能增加母鸡的比例。 Ex5.8 利用pearson卡方检验是否符合特定分布： > chisq.test(c(315,101,108,32),p=c(9,3,3,1)/16)         Chi-squared test for given probabilities data:  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。 Ex5.9 利用pearson卡方检验是否符合泊松分布： > n<-length(z) > y<-c(92,68,28,11,1,0) > x<-0:5 > q<-ppois(x,mean(rep(x,y)));n<-length(y) > p[1]<-q[1];p[n]=1-q[n-1] > chisq.test(y,p=p)         Chi-squared test for given probabilities data:  y X-squared = 2.1596, df = 5, p-value = 0.8267 Warning message: In chisq.test(y, p = p) : Chi-squared approximation may be incorrect 重新分组，合并频数小于5的组： > z<-c(92,68,28,12) > n<-length(z);p<-p[1:n-1];p[n]<-1-q[n-1] > chisq.test(z,p=p)         Chi-squared test for given probabilities data:  z X-squared = 0.9113, df = 3, p-value = 0.8227 可认为数据服从泊松分布。 Ex5.10 ks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 test data:  x and y D = 0.375, p-value = 0.6374 alternative hypothesis: two-sided Ex5.11 列联数据的独立性检验： > x <- c(358,2492,229,2745) > dim(x)<-c(2,2) > chisq.test(x)         Pearson's Chi-squared test with Yates' continuity correction data:  x X-squared = 37.4143, df = 1, p-value = 9.552e-10 P 值<0.05 ，拒绝原假设，有影响。 Ex5.12 列联数据的独立性检验： > y      [,1] [,2] [,3] [1,]   45   12   10 [2,]   46   20   28 [3,]   28   23   30 [4,]   11   12   35 > chisq.test(y)         Pearson's Chi-squared test data:  y X-squared = 40.401, df = 6, p-value = 3.799e-07 P 值<0.05 ，拒绝原假设，不独立，有关系。 Ex5.13 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。 > fisher.test(x)         Fisher's Exact Test for Count Data data:  x p-value = 0.6372 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval:  0.04624382 5.13272210 sample estimates: odds ratio   0.521271 p值大于0.05，两变量独立，两种工艺对产品的质量没有影响。 Ex5.14 由于是在相同个体上的两次试验，故采用McNemar检验。 > mcnemar.test(x)         McNemar's Chi-squared test data:  x McNemar's chi-squared = 2.8561, df = 3, p-value = 0.4144 p值大于0.05，不能认定两种方法测定结果不同。 Ex5.15 符号检验 ： H0：中位数>=14.6; H1: 中位数<14.6 > x<-c(13.32,13.06,14.02,11.86,13.58,13.77,13.51,14.42,14.44,15.43) > binom.test(sum(x)>14.6,length(x),al="l")         Exact binomial test data:  sum(x) > 14.6 and length(x) number of successes = 1, number of trials = 10, p-value = 0.01074 alternative hypothesis: true probability of success is less than 0.5 95 percent confidence interval:  0.0000000 0.3941633 sample estimates: probability of success                    0.1 拒绝原假设，中位数小于14.6 Wilcoxon符号秩检验： > wilcox.test(x,mu=14.6,al="l",exact=F)         Wilcoxon signed rank test with continuity correction data:  x V = 4.5, p-value = 0.01087 alternative hypothesis: true location is less than 14.6 拒绝原假设，中位数小于14.6 Ex5.16 符号检验法： > x<-c(48,33,37.5,48,42.5,40,42,36,11.3,22,36,27.3,14.2,32.1,52,38,17.3,20,21,46.1) > y<-c(37,41,23.4,17,31.5,40,31,36,5.7,11.5,21,6.1,26.5,21.3,44.5,28,22.6,20,11,22.3) > binom.test(sum(x>y),length(x))         Exact binomial test data:  sum(x > y) and length(x) number of successes = 14, number of trials = 20, p-value = 0.1153 alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval:  0.4572108 0.8810684 sample estimates: probability of success                    0.7 接受原假设，无差别。 Wilcoxon符号秩检验： > wilcox.test(x,y,paired=TRUE,exact=FALSE)         Wilcoxon signed rank test with continuity correction data:  x and y V = 136, p-value = 0.005191 alternative hypothesis: true location shift is not equal to 0 拒绝原假设，有差别。 Wilcoxon秩和检验： > wilcox.test(x,y,exact=FALSE)         Wilcoxon rank sum test with continuity correction data:  x and y W = 274.5, p-value = 0.04524 alternative hypothesis: true location shift is not equal to 0 拒绝原假设，有差别。 正态性检验： > ks.test(x,"pnorm",mean(x),sd(x))         One-sample Kolmogorov-Smirnov test data:  x D = 0.1407, p-value = 0.8235 alternative hypothesis: two-sided Warning message: In ks.test(x, "pnorm", mean(x), sd(x)) :   cannot compute correct p-values with ties > ks.test(y,"pnorm",mean(y),sd(y))         One-sample Kolmogorov-Smirnov test data:  y D = 0.1014, p-value = 0.973 alternative hypothesis: two-sided 两组数据均服从正态分布。 方差齐性检验： > var.test(x,y)         F test to compare two variances data:  x and y F = 1.1406, num df = 19, denom df = 19, p-value = 0.7772 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval:  0.4514788 2.8817689 sample estimates: ratio of variances           1.140639 可认为两组数据方差相同。 综上，该数据可做t检验。 t检验： > t.test(x,y,var.equal=TRUE)         Two Sample t-test data:  x and y t = 2.2428, df = 38, p-value = 0.03082 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:   0.812553 15.877447 sample estimates: mean of x mean of y    33.215    24.870 拒绝原假设，有差别。 综上所述，Wilcoxon符号秩检验的差异检出能力最强，符号检验的差异检出最弱。 Ex5.17 spearman秩相关检验： > x<-c(24,17,20,41,52,23,46,18,15,20) > y<-c(8,1,4,7,9,5,10,3,2,6) > cor.test(x,y,method="spearman",exact=F)         Spearman's rank correlation rho data:  x and y S = 9.5282, p-value = 4.536e-05 alternative hypothesis: true rho is not equal to 0 sample estimates:       rho 0.9422536 kendall秩相关检验： > cor.test(x,y,method="kendall",exact=F)         Kendall's rank correlation tau data:  x and y z = 3.2329, p-value = 0.001225 alternative hypothesis: true tau is not equal to 0 sample estimates:       tau 0.8090398 二者有关系，呈正相关。 Ex5.18 > x<-rep(1:5,c(0,1,9,7,3));y<-rep(1:5,c(2,2,11,4,1)) > wilcox.test(x,y,exact=F)         Wilcoxon rank sum test with continuity correction data:  x and y W = 266, p-value = 0.05509 alternative hypothesis: true location shift is not equal to 0 p值大于0.05，不能拒绝原假设，尚不能认为新方法的疗效显著优于原疗法。