What is the appropriate test for significance when comparing a subset of the population to the whole? I am looking for discrepancies in performance evaluation data. For example, let's say I have a population that has 240,000 men and 86,000 women. So 326,000 individuals and 73.6% male/26.4% female. Then I look at the portion that received the highest rating on the evaluation and find that there are 197,395 individuals and they break down 74.4% male and 25.6% female. So it's been put to me that men are over-represented and women under-represented in this evaluation category.
What is the appropriate test to say whether those differences in percentages are significant? If I were to perform a two proportion Z-test would I not need to back the top evaluation category individuals out of the rest of the population?
 A: Sampling without replacement can be modelled using the hypergeometric distribution. Using the notation from the Wikipedia article, denote $N$ the population size (326,000), $K$ the number of males in the population (240,000), $n$ the number of draws (197,395) and $k$ the number of observed males in the sample (146,862).
Using the upper tail of the CDF of the hypergeometric distribution we can directly answer the question

Assuming random sampling without replacement from the population
containing 240,000 men, what's the probability that we end up with
146,862 or more men in a sample of size 197,395?

The answer is found using software. Here, I'm using R (the parametrization is a bit different compared to the Wikipedia article):
m <- 240000 # Number of men in the population
n <- 86000  # Number of women in the population
k <- 197395 # Sample size
x <- 146862 # Number of men in the sample

# Calculate the upper tail probabiltiy of the hypergeometric distribution
phyper(x - 1, m, n, k, lower.tail = FALSE)
[1] 3.592856e-36

The probability is essentially zero.
A normal approximation also works well here and simplifies the calculations considerably. The mean of the hypergeometric distribution is $n\frac{K}{N}=145321.5$ and the variance is $n{K\over N}{(N-K)\over N}{N-n\over N-1} = 15123.5$. The corresponding probability based on the normal approximation is (again in R):
p <- m/(m + n)
mu_approx <- k*p
sd_approx <- sqrt(k*p*(1 - p)*(m + n - k)/(m + n - 1))

# Calculate the upper tail probabiltiy of the hypergeometric distribution
pnorm(trueval - 1, mean = mu_approx, sd = sd_approx, lower.tail = FALSE)
2.947386e-36

A simple simulation confirms the above considerations:
# The population
pop <- rep(0:1, times = c(86000, 240000))
# Reproducibility
set.seed(142857)
# Simulation
res <- replicate(1e4, {
  sum(sample(pop, size = 197395, replace = FALSE))
})
# Histogram of the number of males (not shown)
hist(res, breaks = 100)
# Mean and standard deviation
mean(res)
[1] 145322.1
sd(res)
[1] 123.8935

