r - 寻找幂函数之间的最大差异

Question

如何估计幂函数之间的最大差异？ (CLT 检验功效和二项式检验功效，n=40)

power.z = function(mu,n) {
  power = 1 - pnorm( 1.645 - (mu - 75)/(2.5/sqrt(n)) , 0, 1)
  return(power)
}

power.bin = function(mu,n) {
  p = 1 - pnorm(75, mu, 2.5)
  power = 1 - pnorm( 1.645*sqrt(.25/(p*(1-p))) - (p-.5)/sqrt(p*(1-p)/n), 0, 1)
  return(power)
}

power.z(75.8, 40)   ## 0.6476032
power.bin(75.8, 40) ## 0.4763833

Answer 1

这是基于@Khashaa建议的解决方案:

power.z = function(mu,n) {
  power = 1 - pnorm( 1.645 - (mu - 75)/(2.5/sqrt(n)) , 0, 1)
  return(power)
}

power.bin = function(mu,n) {
  p = 1 - pnorm(75, mu, 2.5)
  power = 1 - pnorm( 1.645*sqrt(.25/(p*(1-p))) - (p-.5)/sqrt(p*(1-p)/n), 0, 1)
  return(power)
}

fnDiffZBin = function(mu, n) abs(power.z(mu, n) - power.bin(mu, n))
optimize(fnDiffZBin, interval = c(75, 80), n = 40, maximum = TRUE)

给出:

> optimize(fnDiffZBin, interval = c(75, 80), n = 40, maximum = TRUE)
$maximum
[1] 75.68248

$objective
[1] 0.1755956