最小样本量n，以使差异不超过

Question

What is the minimum sample size n (or the length n = length(x) of the data vector x ) such that the difference D = 1 - statx4(x)/statx5(x) of the functions statx4 and statx5 is no more than 1/100 ie D ≤ 1/100 ?

And here are the functions:

statx4 <- function(x)  {
  numerator <- sum((x-mean(x))^2)
  denominator <- length(x)
  result <- numerator/denominator
  return(result)
}

statx5 <- function(x)  {
  numerator <- sum((x-mean(x))^2)
  denominator <- length(x)-1
  result <- numerator/denominator
  return(result)
}

I've been doing this exercise set for a while, but haven't managed to get anything valid on this question. Could you point me to right direction?

Answer 1

For the normal distribution, it is the following:

  statx4 <- function(x)  {
  numerator <- sum((x-mean(x))^2)
  denominator <- length(x)
  result <- numerator/denominator
  return(result)
}
statx5 <- function(x)  {
  numerator <- sum((x-mean(x))^2)
  denominator <- length(x)-1
  result <- numerator/denominator
  return(result)
}

D <- function(x){

  1-statx4(x)/statx5(x)
}


DD <- function(N=1111,seed =1){
  set.seed(seed)
  Logi <- vector()
  for (n in 1:N) {
    x<- rnorm(n)
    y <- D(x)
    Logi[n] <- (y  > 1/100) 
  }
  return(Logi)
}

 min  <- vector()
 for (seed in 1:100) {
   message(seed)
   DD(1000,seed)
   min[seed] <-  length(which(DD(1000) == TRUE))
 }

  Answer <- mean(min)+1
Answer 

Note that the function D evaluates the difference of the unbiased variance and the ordinal variance.

I think this problem should be more clear in mathematical sense.

Answer 2

I got solutions today and all you had to do was guess random values:

a <- rnorm(99); 1-statx4(a)/statx5(a)
a <- rnorm(100); 1-statx4(a)/statx5(a)
a <- rnorm(101); 1-statx4(a)/statx5(a)`

And correct answer is 100.

Thank you all for help.