R - 为什么成对的 Fishers 测试对每个组合对 Fishers 产生不同的结果

Question

我正在尝试对 n x 2 数据框的组合进行 Fisher 精确检验，从我读到的内容来看，成对的渔民似乎是我想要使用的（参见 here）。但是，这样做会产生看起来不正确的 p 值结果，因此我决定手动检查组合并得到不同的结果。我已经包括了我希望是一个可重现的例子来突出我已经尝试过的东西。也许我在 R 代码上做错了什么，因为我仍然相对缺乏经验，或者我可能完全误解了成对测试的目的是计算什么——如果是这样，抱歉，如果这个问题不适合我，我可以删除它所以。

# Packages -----------------------------------------------------------

library("tidyverse")
library("janitor")
library("RVAideMemoire")
library("fmsb")

# Generate Data -----------------------------------------------------------

set.seed(1)
test <-
  tibble(
    "drug" = sample(
      c("Control", "Treatment1", "Treatment2"), 
      size = 300,
      prob = c(0.1, 0.4, 0.3),
      replace = TRUE),
    "country" = sample(
      c("Canada", "United States"),
      size = 300,
      prob = c(0.4, 0.6),
      replace = TRUE
    ),
    "selected" = sample(
      c(0, 1), 
      size = 300, 
      prob = c(0.1, 0.65), 
      replace = TRUE)
  )

test2 <- test %>%
  filter(selected == 1)

test2_tab <- test2 %>%
  tabyl(drug, country) %>%
  remove_rownames() %>%
  column_to_rownames(var = colnames(.[1])) %>%
  as.matrix()

当我运行以下成对测试时，我将其作为输出（我使用了 2 个包只是为了确保不是我只是错误地实施了一个包）。

# Pairwise ----------------------------------------------------------------

RVAideMemoire::fisher.multcomp(test2_tab, p.method = "bonferroni")
fmsb::pairwise.fisher.test(test2_tab, p.adjust.method = "bonferroni")

        Pairwise comparisons using Fisher's exact test for count data

data:  test2_tab

           Control Treatment1
Treatment1       1          -
Treatment2       1          1

P value adjustment method: bonferroni



    Pairwise comparisons using Pairwise comparison of proportions (Fisher) 

data:  test2_tab 

           Control Treatment1
Treatment1 1       -         
Treatment2 1       1         

P value adjustment method: bonferroni

但是，当我创建单独的表来执行单独的 Fisher 检验时，如下所示，我得到了不同的结果。

# Individual --------------------------------------------------------------

drug.groups2 <- unique(test2$drug)

# Just to check the correct 2x2 tables are produced
# combn(drug.groups2, 2, function(x) {
#   id <- test2$drug %in% x
#   cross_tabs <- table(test2$drug[id], test2$country[id])
# }, simplify = FALSE)


combn(drug.groups2, 2, function(x) {
  id <- test2$drug %in% x
  cross_tabs <- table(test2$drug[id], test2$country[id])
  fishers <- fisher.test(cross_tabs)
  fishers$data.name <-
    paste(
      unique(
        as.character(test2$drug[id])
      ),collapse="-")
  return(fishers)
}, simplify = FALSE)

[[1]]

    Fisher's Exact Test for Count Data

data:  Treatment1-Treatment2
p-value = 0.3357
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.7566901 2.4175206
sample estimates:
odds ratio 
  1.347105 


[[2]]

    Fisher's Exact Test for Count Data

data:  Treatment1-Control
p-value = 0.4109
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.2560196 1.6292583
sample estimates:
odds ratio 
 0.6637235 


[[3]]

    Fisher's Exact Test for Count Data

data:  Treatment2-Control
p-value = 1
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.3294278 2.3146386
sample estimates:
odds ratio 
 0.8940101

Answer 1

正如 Lukasz 和 StupidWolf 在评论中明确指出的那样，我忘记了我已经应用了 p.method = "bonferroni" 更正，结果与函数调用 p.method = "none" 相同......

Answer 2

这不是因为 Bonferroni 校正应用于成对比较而不应用于单个测试吗？

R - 为什么成对的 Fishers 测试对每个组合对 Fishers 产生不同的结果

R - Why pairwise Fishers test produces different results to Fishers on each combination

statistics

r

pairwise