卡方拟合优度检验是一种非参数检验，用于检验落入两类或更多类的观察值是否遵循特定的非分布。可以说，它会将观察到的比例与预期机会进行比较。在R中，我们可以使用chisq.test函数执行此测试。查看以下示例以了解其操作方法。
> x1<-sample(0:9,200,replace=TRUE) > x1
输出结果
[1] 9 4 1 9 6 6 1 6 0 0 5 8 8 3 7 8 0 3 3 9 6 0 3 8 2 0 8 5 9 1 3 4 6 7 0 1 4 [38] 5 4 8 1 7 2 1 1 3 4 2 5 6 3 4 4 5 6 8 6 4 6 2 0 0 5 2 0 1 6 9 3 0 5 1 3 9 [75] 8 0 9 5 9 4 2 5 9 2 2 0 6 9 1 8 0 1 7 8 4 0 0 2 5 7 1 0 6 7 0 8 8 5 4 3 4 [112] 6 7 4 7 2 1 4 4 4 2 8 4 4 5 6 5 0 5 7 1 5 7 3 4 1 7 9 1 3 9 0 7 1 5 7 7 5 [149] 6 3 4 8 1 8 2 6 8 8 8 8 1 0 9 3 1 6 9 1 5 4 9 3 4 2 6 8 1 6 5 1 0 8 5 0 7 [186] 2 5 8 0 3 6 3 6 7 7 8 4 0 8 3
> x1_table<-table(x1) > x1_table
输出结果
x1 0 1 2 3 4 5 6 7 8 9 24 23 14 18 23 21 21 17 24 15 > chisq.test(x1_table,p=rep(0.1,10)) Chi-squared test for given probabilities data: x1_table X-squared = 6.3, df = 9, p-value = 0.7096
> x2<-c(14,25,17,14) > p<-c(0.25,0.25,0.25,0.25) > chisq.test(x2,p=p)
输出结果
Chi-squared test for given probabilities data: x2 X-squared = 4.6286, df = 3, p-value = 0.2011
> x3<-rpois(200,5) > x3
输出结果
[1] 3 2 4 4 5 4 9 5 8 8 2 9 5 0 7 3 3 4 8 4 3 7 3 7 3 [26] 2 4 2 7 5 7 5 2 5 3 6 4 6 4 5 7 7 6 7 5 9 6 6 4 1 [51] 6 4 5 7 8 7 3 3 2 7 3 6 7 7 1 2 1 3 7 6 5 5 3 4 5 [76] 2 5 5 3 5 5 7 5 3 10 8 6 3 6 10 6 3 2 3 3 7 4 6 2 5 [101] 3 5 3 2 4 4 3 4 7 5 6 7 9 4 4 6 4 10 4 2 4 0 4 3 6 [126] 5 5 1 4 5 5 6 6 5 1 7 2 4 6 6 5 2 2 5 7 2 6 5 3 8 [151] 2 5 4 4 4 3 4 4 9 4 7 2 6 2 3 5 5 3 8 5 5 9 4 4 7 [176] 5 6 6 5 6 3 3 8 5 5 9 6 9 8 4 8 3 2 6 6 4 6 6 7 6
> x3_table<-table(x3) > x3_table
输出结果
x3 0 1 2 3 4 5 6 7 8 9 10 2 5 20 29 33 37 30 23 10 8 3 > chisq.test(x3_table,p=rep(1/11,11)) Chi-squared test for given probabilities data: x3_table X-squared = 93.15, df = 10, p-value = 1.268e-15
> x4<-c(24,98,30,35,27,28) > chisq.test(x4)
输出结果
Chi-squared test for given probabilities data: x4 X-squared = 100.6, df = 5, p-value < 2.2e-16
> x5<-c(12,15,17,15,9,14) > p<-c(0.1,0.1,0.2,0.2,0.1,0.1) > chisq.test(x5,p)
输出结果
Pearson's Chi-squared test data: x5 and p X-squared = 3.75, df = 4, p-value = 0.4409 Warning message: In chisq.test(x5, p) : Chi-squared approximation may be incorrect
> x6<-c(36,27,25,84,14,25,36,27,29) > chisq.test(x6,p=rep(1/9,9))
输出结果
Chi-squared test for given probabilities data: x6 X-squared = 94.812, df = 8, p-value < 2.2e-16