统计建模与R软件第五章习题…
发布日期:2021-10-16 07:12:21 浏览次数:0 分类:技术文章
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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。


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