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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