Finding the finite correction factor for the margin of error I have the following question in exercise 5.3.21 in Mathematical Statistics and Application by Hogg, that asks to redefine the Margin of error equation to cover the case where a finite correction factor needs to be included.
I was confused by the initial wording of this equation, and after having a look at the solution, it starts by defining the variance of $\frac{k}{n}$, which is given as:
$$\frac{p(1-p)}{n}\frac{N-n}{N-1}$$
To give at the 95% confidence interval:
$$\frac{1.96}{2\sqrt{n}}\sqrt{\frac{N-n}{N-1}}$$
Where $\frac{N-n}{N-1}$ is taken from a hypergeometric distribution I beleive, although I am unsure. What is a proof behind the derivation for this?
 A: People often say that the sample size must be less than 10% (or 20%) for the finite population correction factor $\frac{N-n}{N-1}$ to be necessary. This is more a rough rule of thumb than a theorem.
The idea is that the depletion of the population under sampling without replacement is not very noticeable unless more than about 10% of the population is missing. Most simply, note that
$\frac{N-n}{N-1} = \frac{100 - 10}{100 - 1} \approx 0.9;$ $\frac{250 - 25}{249}\approx 0.9,$ and so on.
Sampling with replacement (or sampling from an infinite population) is often modeled by the binomial distribution and sampling
without replacement is often modeled by the hypergeometric
distribution. However, as a practical matter, there is not
much difference between $\mathsf{Binom}(n = 10, p = .2)$ and the
corresponding hypergeometric distribution with $N = 100.$
    x = 0:10; PDF = dbinom(x, 10, .2)
    pdf.h = dhyper(x, 20, 80, 10)
    round(cbind(x,PDF,pdf.h),3)
           x   PDF pdf.h
     [1,]  0 0.107 0.095
     [2,]  1 0.268 0.268
     [3,]  2 0.302 0.318
     [4,]  3 0.201 0.209
     [5,]  4 0.088 0.084
     [6,]  5 0.026 0.022
     [7,]  6 0.006 0.004
     [8,]  7 0.001 0.000
     [9,]  8 0.000 0.000
    [10,]  9 0.000 0.000
    [11,] 10 0.000 0.000

In the figure below binomial probabilities are represented by blue bars and the corresponding hypergeometric probabilities by brown bars. [R code follows figure.]

    plot(x+.05, PDF, type="h", lwd=2, col="blue")
     lines(x-.05, pdf.h, type="h", lwd=2, col="brown")
      abline(h = 0, col="green2")
      abline(v = 0, col="green2")

Note: The formal derivation of the variance is shown here and on stackexchange here. Google for yet other proofs. (A problem in Wackerly, Mendenhall, Scheaffer outlines the derivation step-by-step, using properties of binomial coefficients).
