如何在 R 中执行 Fisher 检验?


Fisher 检验帮助我们了解分类变量之间是否存在显著的非随机关系。它应用于列联表,因为这些表用于表示分类变量的频率,并且我们既可以将其应用于矩阵,也可以将其应用于行矩阵,因为它们具有类似的形式。在 R 中,我们可以使用 fisher.test 函数来执行 Fisher 检验。

范例

 现场演示

M1<-matrix(1:9,ncol=3)
M1

输出

   [,1] [,2] [,3]
[1,] 1    4    7 
[2,] 2    5    8
[3,] 3    6    9

fisher.test(M1)

Fisher's Exact Test for Count
Data
data: M1
p-value = 0.9888
alternative hypothesis: two.sided

范例

 现场演示

M2<-matrix(1:16,ncol=4)
M2

输出

     [,1] [,2] [,3] [,4]
[1,]   1    5    9    13
[2,]   2    6    10   14
[3,]   3    7    11   15
[4,]   4    8    12   16

fisher.test(M2)

Fisher's Exact Test for Count Data
data: M2
p-value = 0.9993
alternative hypothesis: two.sided

范例

 现场演示

M3<-matrix(sample(0:4,9,replace=TRUE),nrow=3)
M3

输出

   [,1] [,2] [,3]
[1,] 0    0    4
[2,] 4    0    4
[3,] 1    2    3

fisher.test(M3)

Fisher's Exact Test for Count
Data
data: M3
p-value = 0.5567
alternative hypothesis: two.sided

范例

 现场演示

M4<-matrix(c(14,27,15,24,27,17,39,19,24),nrow=3)
M4

输出

    [,1] [,2] [,3]
[1,] 14   24   39
[2,] 27   27   19
[3,] 15   17   24

fisher.test(M4)

Fisher's Exact Test for Count Data
data: M4
p-value = 0.02126
alternative hypothesis: two.sided

fisher.test(M4,alternative="greater")

Fisher's Exact Test for Count Data
data: M4
p-value = 0.02126
alternative hypothesis: greater

fisher.test(M4,alternative="less")

Fisher's Exact Test for Count Data
data: M4
p-value = 0.02126
alternative hypothesis: less

范例

 现场演示

M5<-matrix(sample(c(545,501,576),4,replace=TRUE),nrow=2)
M5

输出

    [,1] [,2]
[1,] 545 545
[2,] 545 545

fisher.test(M5)

Fisher's Exact Test for Count Data
data: M5
p-value = 1
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval: 0.8391933 1.1916205
sample estimates:
odds ratio
1

fisher.test(M5,alternative="greater")

Fisher's Exact Test for Count Data
data: M5
p-value = 0.5175
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval: 0.8626582 Inf
sample estimates:
odds ratio
1

fisher.test(M5,alternative="less")

Fisher's Exact Test for Count Data
data: M5 p-value = 0.5175
alternative hypothesis: true
odds ratio is less than 1
95 percent confidence interval: 0.000000 1.159208
sample estimates:
odds ratio
1

更新于:19-Oct-2020

884 次浏览

开启你的 职业

通过完成课程获得认证

立即开始
广告