来自 fisher.test() 的 p 值与 phyper() 不匹配 [英] p-value from fisher.test() does not match phyper()

问题描述

Fisher's Exact Test 与超几何分布有关，我希望这两个命令会返回相同的 pvalues.谁能解释我做错了什么，他们不匹配?

The Fisher's Exact Test is related to the hypergeometric distribution, and I would expect that these two commands would return identical pvalues. Can anyone explain what I'm doing wrong that they do not match?

#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

推荐答案

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

In this case, the actual call to phyper that is relevant is phyper(x - 1, m, n, k, lower.tail = FALSE). Look at the source code for fisher.test relevant to your call of fisher.test(matrix(c(x, m-x, k-x, n-(k-x)),2,2), alternative='greater'). At line 138, PVAL is set to:

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 被设置为 pnhyper(x, or, upper.tail = TRUE).可以看到第122行定义了pnhyper.这里，or = 1，传递给ncp，所以调用的是phyper(x - 1, m, n, k,lower.tail = FALSE)

Since alternative = 'greater', PVAL is set to pnhyper(x, or, upper.tail = TRUE). You can see pnhyper defined on line 122. Here, or = 1, which is passed to ncp, so the call is phyper(x - 1, m, n, k, lower.tail = FALSE)

带着你的价值观:

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

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