Determining if awards are fair

Question

Imagine a population of people where $x\%$ have blue hats and $(100-x)\%$ have green hats where every year some people are awarded prizes. The rule is that everyone should have an equal chance of being awarded a prize every year. The village bosses would like to decide if this rule is being followed as they suspect there may be some prejudice based on hat color. They count the number of people with blue hats who get a prize and call it $A$ and also the number of people with green hats who get a prize and call that $B$. The total population size of green and blue hats is $P$.

What is a good method to determine if the prize awarding has indeed been fair?

Answer 1

You will have a two-way contingency table and the classic chi-square test of independence will provide a good answer.

Answer 2

You could consider using a hypergeometric distribution.

Let $Y$ denote the number of people with blue hats that receive prizes. For each year, let $a$, $b$, and $p$ be the observed values of $A$, $B$, and $P$. Based upon your description, it is assumed that $x$ is constant across all years. Then: $$Y \sim Hypergeometric(N=p, K=x*p, n=a+b) $$ $$P(Y=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$ You would then calculate the probability that $Y = a$ for that year: $$P(Y=a) = \frac{\binom{x*p}{a}\binom{p-x*p}{a+b-a}}{\binom{p}{a+b}}\\ = \frac{\binom{x*p}{a}\binom{(1-x)*p}{b}}{\binom{p}{a+b}}$$ If that probability is too low (based on some threshold), you would know it was rigged.

Answer 3

$x/100 = A/(A+B)$?

Within limits of random error, of course. How confident does the boss want to be that that a prejudice is indeed a prejudice?

There's probably a good theoretical way of getting the confidence (p-values?) but personally lazy me would just simulate many populations of size $P$, then draw $(A+B)$ number of prize-winners randomly (without replacement) and finally tally the $A$ and $B$ values observed to get my confidence limits.