0.1 一个总体均值的检验

0.1.1 大样本下

\[H_0:\mu \geq 81 \quad v.s. \quad H_1:\mu < 81\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_3.RData")
library(BSDA)
## Loading required package: lattice
## 
## Attaching package: 'BSDA'
## The following object is masked from 'package:datasets':
## 
##     Orange
z.test(example6_3$PM2.5值,mu=81,sigma.x=sd(example6_3$PM2.5值),alternative='less',conf.level=0.95)
## 
##  One-sample z-Test
## 
## data:  example6_3$PM2.5值
## z = -1.1856, p-value = 0.1179
## alternative hypothesis: true mean is less than 81
## 95 percent confidence interval:
##        NA 81.56174
## sample estimates:
## mean of x 
##     79.55
    #左侧检验只显示置信区间的上限
    #这里显然，总体方差未知

0.1.2 小样本下

\[H_0:\mu =5 \quad v.s. \quad H_1:\mu \neq 5\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_4.RData")
    #这里显然，总体方差未知
t.test(example6_4$厚度,mu=5)
## 
##  One Sample t-test
## 
## data:  example6_4$厚度
## t = -5.6273, df = 19, p-value = 1.998e-05
## alternative hypothesis: true mean is not equal to 5
## 95 percent confidence interval:
##  4.725612 4.874388
## sample estimates:
## mean of x 
##       4.8

0.1.2.1 计算单样本t检验的效应量:cohen的d统计量

\[d=\frac{|\overline{x}-\mu_0|}{s}\] cohen的d统计量表示样本均值与假设的总体均值的差异是多少个标准差。
当 \(d<0.20\) 时，效应量非常小，几乎为0；
当 \(0.20\leq d<0.50\) 时，效应量小；
当 \(0.50\leq d<0.80\) 时，效应量中；
当 \(d\geq 0.80\) 时，效应量大。

library(lsr)
cohensD(example6_4$厚度,mu=5)
## [1] 1.258306

0.2 两个总体均值之差的检验

0.2.1 独立大样本

\[H_0:\mu_1-\mu_2 =0 \quad v.s. \quad H_1:\mu_1-\mu_2 \neq 0\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_5.RData")
library(BSDA)
    #这里显然，总体方差未知
z.test(example6_5$男生上网时间,example6_5$女生上网时间,sigma.x = sd(example6_5$男生上网时间),sigma.y = sd(example6_5$女生上网时间),alternative = 'two.sided')
## 
##  Two-sample z-Test
## 
## data:  example6_5$男生上网时间 and example6_5$女生上网时间
## z = 1.1188, p-value = 0.2632
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.1712448  0.6268003
## sample estimates:
## mean of x mean of y 
##  3.058333  2.830556

0.2.2 独立小样本，假定均服从正态分布

\[H_0:\mu_1-\mu_2 =0 \quad v.s. \quad H_1:\mu_1-\mu_2 \neq 0\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_6.RData")
t.test(example6_6$甲企业,example6_6$乙企业,var.equal=TRUE)#若两个总体方差相等
## 
##  Two Sample t-test
## 
## data:  example6_6$甲企业 and example6_6$乙企业
## t = 3.4943, df = 38, p-value = 0.001225
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  135.24 507.76
## sample estimates:
## mean of x mean of y 
##    8487.5    8166.0
t.test(example6_6$甲企业,example6_6$乙企业,var.equal=FALSE)#若两个总体方差不相等
## 
##  Welch Two Sample t-test
## 
## data:  example6_6$甲企业 and example6_6$乙企业
## t = 3.4943, df = 33.683, p-value = 0.001353
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  134.4528 508.5472
## sample estimates:
## mean of x mean of y 
##    8487.5    8166.0

0.2.2.1 计算独立样本t检验的效应量

\[d=|t|\sqrt{\frac{n_1+n_2}{n_1n_2}}\]

library(lsr)
cohensD(example6_6$甲企业,example6_6$乙企业)
## [1] 1.104985

0.2.3 配对样本

\[H_0:\mu_1-\mu_2 =0 \quad v.s. \quad H_1:\mu_1-\mu_2 \neq 0\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_7.RData")
t.test(example6_7$旧款饮料,example6_7$新款饮料,paired=TRUE)
## 
##  Paired t-test
## 
## data:  example6_7$旧款饮料 and example6_7$新款饮料
## t = -2.7508, df = 9, p-value = 0.02245
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -2.3690538 -0.2309462
## sample estimates:
## mean of the differences 
##                    -1.3

0.2.3.1 计算配对样本t检验的效应量

\[d=\frac{\overline{d}}{s_d}=\frac{|t|}{\sqrt{n}}\]

library(lsr)
cohensD(example6_7$旧款饮料,example6_7$新款饮料,method='paired')
## [1] 0.8698945

0.3 总体比例的检验【本节并未在茆诗松课本上有体现，需注意】

0.3.1 一个总体比例的检验

\[H_0:\pi \leq 25\% \quad v.s. \quad H_1:\pi > 25\%\] 检验统计量： \[z=\frac{p-\pi_0}{\sqrt{\frac{\pi_0(1-\pi_0)}{n}}}\]

n=2000;p=450/2000;pi_0=0.25
z=(p-pi_0)/sqrt(pi_0*(1-pi_0)/n)
p_value=1-pnorm(z)
data.frame(z,p_value)
##           z   p_value
## 1 -2.581989 0.9950884

0.3.2 两个总体比例之差的检验

0.3.2.1 \(H_0:\pi_1-\pi_2 = 0\)

\[H_0:\pi_1-\pi_2 \geq 0 \quad v.s. \quad H_1:\pi_1-\pi_2 < 0\] 检验统计量： \[z=\frac{p_1-p_2}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}}\] 其中 \[p=\frac{p_1n_1+p_2n_2}{n_1+n_2}\]

n1=200;n2=200;p1=0.27;p2=0.35
p=(p1*n1+p2*n2)/(n1+n2)
z=(p1-p2)/sqrt(p*(1-p)*(1/n1+1/n2))
p_value=pnorm(z)
data.frame(z,p_value)
##           z    p_value
## 1 -1.729755 0.04183703

0.3.2.2 \(H_0:\pi_1-\pi_2 = d_0\)

\[H_0:\pi_1-\pi_2 \geq 8\% \quad v.s. \quad H_1:\pi_1-\pi_2 < 8\%\]

检验统计量： \[z=\frac{(p_1-p_2)-d_0}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\]

n1=300;n2=300;p1=33/300;p2=84/300;d_0=0.08
z=((p1-p2)-0.08)/sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)
p_value=pnorm(z)
data.frame(z,p_value)
##          z     p_value
## 1 -7.91229 1.26348e-15

0.4 总体方差的检验

0.4.1 一个总体方差

\[H_0:\sigma^2 \leq 16 \quad v.s. \quad H_1:\sigma^2 > 16\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_11.RData")
library(TeachingDemos)
## 
## Attaching package: 'TeachingDemos'
## The following object is masked from 'package:BSDA':
## 
##     z.test
sigma.test(example6_11$填装量,sigmasq=16,alternative='greater')
## 
##  One sample Chi-squared test for variance
## 
## data:  example6_11$填装量
## X-squared = 2.9741, df = 9, p-value = 0.9653
## alternative hypothesis: true variance is greater than 16
## 95 percent confidence interval:
##  2.812522      Inf
## sample estimates:
## var of example6_11$填装量 
##                  5.287222

0.4.2 两个总体方差比

\[H_0:\frac{\sigma_1^2}{\sigma_2^2} = 1 \quad v.s. \quad H_1:\frac{\sigma_1^2}{\sigma_2^2} \neq 1\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_6.RData")
var.test(example6_6[,1],example6_6[,2],alternative = 'two.sided')#和 $ 的选择是一样的
## 
##  F test to compare two variances
## 
## data:  example6_6[, 1] and example6_6[, 2]
## F = 0.47273, num df = 19, denom df = 19, p-value = 0.111
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.1871127 1.1943320
## sample estimates:
## ratio of variances 
##          0.4727311

0.5 正态性检验

0.5.1 图示法

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_3.RData")
    #绘制Q-Q图
par(mfrow=c(1,2),mai=c(0.7,0.7,0.2,0.1),cex=0.8)
qqnorm(example6_3$PM2.5值,xlab='期望正态值',ylab='观测值',datax=TRUE,main='正态Q-Q图')
qqline(example6_3$PM2.5值,datax=TRUE,col='red')
    #绘制P-P图
f=ecdf(example6_3$PM2.5值)
p1=f(example6_3$PM2.5值)
p2=pnorm(example6_3$PM2.5值,mean(example6_3$PM2.5值),sd(example6_3$PM2.5值))
plot(p1,p2,xlab='观测的累积概率',ylab='期望的累积概率',main='正态P-P图')
abline(a=0,b=1,col='red')

0.5.2 W检验 / Shapiro-Wilk正态性检验，\(8 \leq n < 50\)

\[H_0:厚度服从正态分布 \quad v.s. \quad H_1:厚度不服从正态分布\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_4.RData")
shapiro.test(example6_4$厚度)
## 
##  Shapiro-Wilk normality test
## 
## data:  example6_4$厚度
## W = 0.91377, p-value = 0.07522

0.5.3 K-S检验

\[H_0:F(x)=F_0(x)—总体分布与指定的理论分布无显著差异 \quad v.s. \\ H_1:F(x) \neq F_0(x)—总体分布与指定的理论分布有显著差异\] 检验统计量： \[D=\max \{|S(x_i)-F(x_i)|\} \quad \text{or} \quad D=\max \{(|S(x_i)-F(x_i)|),(|S(x_{i-1})-F(x_i)|)\}\]

小样本下，统计量D服从Kolmogorov分布；大样本下，用正态分布近似，此时统计量： \[z=\sqrt{n}D\] K-S检验要求正态总体已知，当参数未知时，用样本均值、样本方差代替。 \[H_0:厚度服从正态分布 \quad v.s. \quad H_1:厚度不服从正态分布\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_4.RData")
ks.test(example6_4$厚度,'pnorm',mean(example6_4$厚度),sd(example6_4$厚度))
## Warning in ks.test(example6_4$厚度, "pnorm", mean(example6_4$厚度),
## sd(example6_4$厚度)): ties should not be present for the Kolmogorov-Smirnov test
## 
##  One-sample Kolmogorov-Smirnov test
## 
## data:  example6_4$厚度
## D = 0.23538, p-value = 0.2178
## alternative hypothesis: two-sided

0.6 非参数检验

0.6.1 Wilcoxon符号秩检验

\[H_0:M =5 \quad v.s. \quad M \neq 5\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_4.RData")
wilcox.test(example6_4$厚度,m=5)
## Warning in wilcox.test.default(example6_4$厚度, m = 5): cannot compute exact p-
## value with ties
## Warning in wilcox.test.default(example6_4$厚度, m = 5): cannot compute exact p-
## value with zeroes
## 
##  Wilcoxon signed rank test with continuity correction
## 
## data:  example6_4$厚度
## V = 2, p-value = 0.0005812
## alternative hypothesis: true location is not equal to 5

0.6.2 Mann-Whitney检验 / Wilcoxon（符号）秩和检验

\[H_0:M_甲=M_乙 \quad v.s. H_1:M_甲 \neq M_乙\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_6.RData")
wilcox.test(example6_6$甲企业,example6_6$乙企业)
## 
##  Wilcoxon rank sum exact test
## 
## data:  example6_6$甲企业 and example6_6$乙企业
## W = 319, p-value = 0.0009334
## alternative hypothesis: true location shift is not equal to 0

0.6.3 配对样本的Wilcoxon符号秩检验

\[H_0:M_d=0 \quad (两个总体相同) \quad v.s. H_1:M_d \neq 0 \quad (两个总体不相同)\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch6\\example6_7.RData")
wilcox.test(example6_7$旧款饮料,example6_7$新款饮料,paired=TRUE)
## Warning in wilcox.test.default(example6_7$旧款饮料, example6_7$新款饮料, :
## cannot compute exact p-value with ties
## Warning in wilcox.test.default(example6_7$旧款饮料, example6_7$新款饮料, :
## cannot compute exact p-value with zeroes
## 
##  Wilcoxon signed rank test with continuity correction
## 
## data:  example6_7$旧款饮料 and example6_7$新款饮料
## V = 5.5, p-value = 0.04759
## alternative hypothesis: true location shift is not equal to 0