来自 fisher.test() 的 p 值与 phyper() 不匹配

Question

Fisher 精确检验与超几何分布有关，我希望这两个命令会 return 相同的 p 值。谁能解释我做错了什么，他们不匹配？

#data (variable names chosen to match dhyper() argument names)
x = 14
m = 20
n = 41047
k = 40

#Fisher test, alternative = 'greater'
(fisher.test(matrix(c(x, m-x, k-x, n-(k-x)),2,2), alternative='greater'))$p.value 
#returns 2.01804e-39

#geometric distribution, lower.tail = F, i.e. P[X > x]
phyper(x, m, n, k, lower.tail = F, log.p = F)
#returns 5.115862e-43

Answer 1

在这种情况下，对 phyper 的实际调用是 phyper(x - 1, m, n, k, lower.tail = FALSE)。查看与您调用 fisher.test(matrix(c(x, m-x, k-x, n-(k-x)),2,2), alternative='greater') 相关的 fisher.test 的源代码。在第 138 行，PVAL 设置为：

switch(alternative, less = pnhyper(x, or), 
    greater = pnhyper(x, or, upper.tail = TRUE), 
    two.sided = {
      if (or == 0) as.numeric(x == lo) else if (or == 
        Inf) as.numeric(x == hi) else {
        relErr <- 1 + 10^(-7)
        d <- dnhyper(or)
        sum(d[d <= d[x - lo + 1] * relErr])
      }
    })

由于 alternative = 'greater'，PVAL 设置为 pnhyper(x, or, upper.tail = TRUE)。可以看到在122行定义了pnhyper，这里，or = 1传递给了ncp，所以调用的是phyper(x - 1, m, n, k, lower.tail = FALSE)

你的价值观：

x = 14
m = 20
n = 41047
k = 40
phyper(x - 1, m, n, k, lower.tail = FALSE)
# [1] 2.01804e-39

来自 fisher.test() 的 p 值与 phyper() 不匹配

p-value from fisher.test() does not match phyper()

statistics

r

contingency

p-value

hypothesis-test