具有置信区间的二项式概率质量函数

Question

The following question we need to solve.

Consider the following binomial probability mass function (pmf) :

f(x;m,p) = (m¦x) p^x * (1-p)^(mx) , for x = 0, 1, 2,.....,m, and otherwise equal to 0 . Let X_1, X_2,....,Xn be independent and identically distributed random samples from f(x;m = 20; p = 0:45) .

1) Assume n = 15 and calculate the 95% confidence interval on p using the p-hat = Σ_(i=1)^n X_i/mn (an estimator of p). Simulate these confidence intervals 10000 times and count how often the parameter value p lies within these 10000 confidence intervals.

m <- 20
p <- 0.45
n <- 15
x <- m
nsim <- 10000
counter <- 0

for (i in 1:nsim) {
  bpmf <- rbinom(x,m,p)
  esti_p <- bpmf/(m*n)
  var_bpmf <- var(bpmf) 
  CI_lower <- esti_p - qnorm(0.975)*sqrt(var_bpmf/n) 
  CI_upper <- esti_p + qnorm(0.975)*sqrt(var_bpmf/n) 
  if ((CI_lower<p) & (CI_upper>p)) counter <- counter + 1
}    

It doesn't work properly and I don't see what I'm doing wrong. Is there anyone who can help me with this?

When I run my code, I believe the answer now is right, but it gives the following sentence: "There were 50 or more warnings (use warnings() to see the first 50)" When I run this it will give:

"1: In if ((CI_lower < p) & (CI_upper > p)) counter <- counter +  ... :
the condition has length > 1 and only the first element will be used".

Also I don't know for sure if;

 CI_lower <- esti_p - qnorm(0.975)*sqrt(var_bpmf/n) 
 CI_upper <- esti_p + qnorm(0.975)*sqrt(var_bpmf/n) 

is the right formula to calculate the confidence interval.

Answer 1

m <- 20
p <- 0.45
nsim <- 10000

  bpmf <- rbinom(size=m,prob=p,n=nsim)
  esti_p <- bpmf/m
  var_bpmf <- esti_p*(1-esti_p)/m 
  CI_lower <- esti_p - qnorm(0.975)*sqrt(var_bpmf) 
  CI_upper <- esti_p + qnorm(0.975)*sqrt(var_bpmf) 
  counter <-((CI_lower<p) & (CI_upper>p))
table(counter)