r - 为什么 chisq.test 的 2 个输出在 R 中不同

Question

以下，当数据实际上相同时，为什么 2 chisq.test 的输出不同:

> df1
  count position
1     1       11
2     6       12
3    12       13
4    23       14
5    27       15
> df2
  count position
1     1       11
2     4       12
3     9       13
4    24       14
5    24       15
> mm = merge(df1, df2,  by='position')
> mm
  position count.x count.y
1       11       1       1
2       12       6       4
3       13      12       9
4       14      23      24
5       15      27      24

第一种方法:

> chisq.test(mm[2:3])

        Pearson's Chi-squared test

data:  mm[2:3]
X-squared = 0.6541, df = 4, p-value = 0.9569

Warning message:
In chisq.test(mm[2:3]) : Chi-squared approximation may be incorrect

第二种方法:

> chisq.test(df1$count, df2$count)

        Pearson's Chi-squared test

data:  df1$count and df2$count
X-squared = 15, df = 12, p-value = 0.2414

Warning message:
In chisq.test(df1$count, df2$count) :
  Chi-squared approximation may be incorrect
> 

编辑:回复评论:以下看起来相同:

> mm[2:3]
  count.x count.y
1       1       1
2       6       4
3      12       9
4      23      24
5      27      24
> 

> mm[,2:3]
  count.x count.y
1       1       1
2       6       4
3      12       9
4      23      24
5      27      24

数据:

> dput(df1)
structure(list(count = c(1L, 6L, 12L, 23L, 27L), position = 11:15), .Names = c("count", 
"position"), class = "data.frame", row.names = c(NA, -5L))
> dput(df2)
structure(list(count = c(1L, 4L, 9L, 24L, 24L), position = 11:15), .Names = c("count", 
"position"), class = "data.frame", row.names = c(NA, -5L))

Answer 1

参见 ?chisq :在第一种情况下，将 mm[2:3] 作为列联表，在第二种情况下，计算列联表。

chisq.test(table(df1$count, df2$count))

        Pearson's Chi-squared test

data:  table(df1$count, df2$count)
X-squared = 15, df = 12, p-value = 0.2414

Warning message:
In chisq.test(table(df1$count, df2$count)) :
  Chi-squared approximation may be incorrect

所以，真的，你是这个表的计算 chisq :

     1 4 9 24
  1  1 0 0  0
  6  0 1 0  0
  12 0 0 1  0
  23 0 0 0  1
  27 0 0 0  1