如何在 R 语言中执行双向方差分析的方差齐性检验?
r programmingserver side programmingprogramming更新于 2025/4/11 23:07:17
一般来说，方差齐性检验是比较两个或多个变量的方差，并找出它们之间是否存在显著差异的检验类型。对于双向方差分析，最常用的方差齐性检验之一是 Levene 检验，它可以借助 R 基础库中 car 包的 leveneTest 函数轻松完成。
考虑以下数据框 −
示例
set.seed(151)
x1<-sample(c("C1","C2","C3"),20,replace=TRUE)
x2<-sample(c("S1","S2","S3","S4","S5"),20,replace=TRUE)
y<-rnorm(20,5,2)
df1<-data.frame(x1,x2,y)
df1
输出
x1 x2 y 1 C2 S2 2.255857 2 C3 S5 1.726474 3 C3 S4 4.280697 4 C2 S3 7.402230 5 C2 S3 3.708252 6 C2 S4 3.978782 7 C2 S1 3.801754 8 C3 S3 6.091206 9 C2 S3 4.017412 10 C3 S3 5.383071 11 C3 S1 3.882945 12 C1 S5 6.845399 13 C1 S1 7.307996 14 C3 S4 2.255179 15 C1 S5 7.580363 16 C2 S5 7.309804 17 C2 S4 7.891359 18 C2 S3 5.522026 19 C3 S4 8.858292 20 C1 S1 3.800228
加载 car 包并对 df1 执行 Levene 测试 −
library(car) leveneTest(y~x1*x2,data=df1) Levene's Test for Homogeneity of Variance (center = median) Df F value Pr(>F) group 9 1.5987 0.2374 10
我们来看另一个例子 −
示例
Age_group<-sample(c("First","Second"),20,replace=TRUE)
Ethnicity<-sample(c("Asian","NorthAmerican","Chinese","Japanese"),20,replace=TRUE)
Salary<-sample(20000:50000,20)
df2<-data.frame(Age_group,Ethnicity,Salary)
df2
输出
Age_group Ethnicity Salary 1 Second NorthAmerican 25678 2 Second Asian 34597 3 Second Chinese 49861 4 Second Chinese 37386 5 First Japanese 38426 6 Second NorthAmerican 45889 7 Second Asian 35033 8 Second NorthAmerican 46098 9 First Japanese 34070 10 Second Japanese 33618 11 First Japanese 35760 12 Second Chinese 33376 13 Second NorthAmerican 30630 14 First Asian 23820 15 Second Asian 40899 16 First Asian 35095 17 Second Chinese 43439 18 First Japanese 35641 19 Second Asian 41754 20 Second NorthAmerican 35337
在 df2 上执行 Levene 检验 −
leveneTest(Salary~Age_group*Ethnicity,data=df2) Levene's Test for Homogeneity of Variance (center = median) Df F value Pr(>F) group 6 0.6593 0.6835 13
