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问题描述

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 i.e. 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?

标签: rfunctiondifferencesample-size

解决方案


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.


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