Chapter 7 类别变量分析 / 拟合优度检验

0.1 一个类别变量的拟合优度检验

0.1.1 期望频数相等情况

\[H_0:观察频数与期望频数无显著差异 \quad v.s. \quad H_1:观察频数与期望频数有显著差异\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch7\\example7_1.RData")
chisq.test(example7_1$人数)

## 
##  Chi-squared test for given probabilities
## 
## data:  example7_1$人数
## X-squared = 12.1, df = 3, p-value = 0.007048

0.1.2 期望频数不相等情况（最经典例子：孟德尔杂交实验）

\[H_0:不同受教育程度的离婚家庭与期望频数无显著差异 \quad v.s. \quad H_1:不同受教育程度的离婚家庭与期望频数有显著差异\]

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch7\\example7_2.RData")
chisq.test(example7_2$离婚家庭数,p=example7_2$期望比例)

## 
##  Chi-squared test for given probabilities
## 
## data:  example7_2$离婚家庭数
## X-squared = 19.586, df = 4, p-value = 0.0006028

0.2 两个类别变量的\(\chi^2\)独立性检验

\[H_0:满意度与地区独立 \quad v.s. \quad H_1:满意度与地区不独立\]

    #生成列联表下检验
x=c(126,158,35,34,82,65)
M=matrix(x,nr=2,nc=3,byrow=TRUE,dimnames=list(c('满意','不满意'),c('东部','中部','西部')))
chisq.test(M)

## 
##  Pearson's Chi-squared test
## 
## data:  M
## X-squared = 51.827, df = 2, p-value = 5.572e-12

    #原始数据检验
load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch7\\example7_3.RData")
count=table(example7_3)
chisq.test(count)

## 
##  Pearson's Chi-squared test
## 
## data:  count
## X-squared = 51.827, df = 2, p-value = 5.572e-12

0.3 两个类别变量的相关性度量：\(\phi\)系数、Cramer’s V系数、列联系数

load("D:\\New_Folder\\Study_Programming\\R_Programme\\Applied Statistics\\datas - Copy\\example\\ch7\\example7_3.RData")
count=table(example7_3)
library(vcd)

## Loading required package: grid

assocstats(count)

##                     X^2 df   P(> X^2)
## Likelihood Ratio 51.326  2 7.1559e-12
## Pearson          51.827  2 5.5718e-12
## 
## Phi-Coefficient   : NA 
## Contingency Coeff.: 0.306 
## Cramer's V        : 0.322