我想进行以下配对t检验:
str1<-' ENSEMBLE 0.934 0.934 0.934 0.934 '
str2<-' J48 0.934 0.934 0.934 0.934 '
df1 <- read.table(text=scan(text=str1, what='', quiet=TRUE), header=TRUE)
df2 <- read.table(text=scan(text=str2, what='', quiet=TRUE), header=TRUE)
t.test ( df1$ENSEMBLE, df2$J48, mu=0 , alt="two.sided", paired = T, conf.level = 0.95)
我得到以下结果:
Paired t-test
data: df1$ENSEMBLE and df2$J48
t = NaN, df = 3, p-value = NA
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
NaN NaN
sample estimates:
mean of the differences
0
为什么我会得到它?
答案 0 :(得分:5)
这是因为数据集完全相同。
df2[1,1] <- .935
t.test ( df1$ENSEMBLE, df2$J48, mu=0 , alt="two.sided", paired = T, conf.level = 0.95)
Paired t-test
data: df1$ENSEMBLE and df2$J48
t = -1, df = 3, p-value = 0.391
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.0010456116 0.0005456116
sample estimates:
mean of the differences
-0.00025
答案 1 :(得分:2)
你的两个向量完全一样。两组都没有差异,因此没有标准错误。所以你的答案是未定义的