非独立样本的t检验假定组间的差异呈正态分布。对于本例,检验的调用格式为:
t.test(y1, y2, paired=TRUE)
其中的y1和y2为两个非独立组的数值向量。结果如下:
> library(MASS)
> sapply(UScrime[c("U1","U2")], function(x)(c(mean=mean(x),sd=sd(x))))
U1 U2
mean 95.5 33.98
sd 18.0 8.45
> with(UScrime, t.test(U1, U2, paired=TRUE))
Paired t-test
data: U1 and U2
t = 32.4066, df = 46, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
57.67003 65.30870
sample estimates:
mean of the differences
61.48936
差异的均值(61.5)足够大,可以保证拒绝年长和年轻男性的平均失业率相同的假设。年轻
男性的失业率更高。事实上,若总体均值相等,获取一个差异如此大的样本的概率小于0.000 000
000 000 000 22(即2.2e–16)