Question

我的数据和代码是这样的：

my_vector <- rnorm(150)
my_factor1 <- gl(3,50)
my_factor2 <- gl(2,75)

tapply(my_vector, my_factor1, function(x)
  t.test(my_vector~my_factor2, paired=T))

我想对每个级别的my_factor1进行单独的t检验，以测试my_vector的两个级别的my_vector。

但是，使用我的代码时，t-test不会分割my_factor1的级别，并且每个级别的结果都相同，因为my_vector完全包含在每个t.test中。

这是我的代码的输出：

$`1`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775 


$`2`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775 


$`3`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775

我错过了什么或做错了什么？

Answer 1

您的示例稍有问题，因为如果您设置：

df <- data.frame(my_vector = rnorm(150),
                 my_factor1 = gl(3,50),
                 my_factor2 = gl(2,75)
                )

my_factor2 = 1或3时，my_factor1只有一个唯一值，因为重复的重叠方式。见?gl。所以：

df <- data.frame(my_vector = rnorm(150),
                 my_factor1 = gl(3,1,150),
                 my_factor2 = gl(2,1,150)
                )
with(df,
       by(df, my_factor1,
          function(x) t.test(my_vector ~ my_factor2, data=x)
       )
     )

这似乎可以产生您想要的输出。

作为旁注 - 考虑更正多重比较：https://stats.stackexchange.com/questions/16779/when-is-multiple-comparison-correction-necessary

如何使用tapply对每个级别的因子执行t检验

1 个答案: