两个配对的t检验p值= NA和t = NaN

时间:2015-12-22 19:39:34

标签: r p-value

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?

2 个答案:

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