Women passing exams: Help with stats please

[tempguy]

Hello,

I am looking at the stats for a particular academic test:

- In 2007, 11 women out of 50 candidates took the test; there were only 3 women among the 16 successful candidates.
- In 2009, 12 women out of 52 candidates took the test; there was only 1 woman women among the 14 successful candidates.

The same results in percentages:

- Women accounted for 18.7% of successful candidates in 2007, which is more or less normal insofar as they represented 22% of the candidates taking the test that year (11/50).
- However, women accounted for 7% of successful candidates in 2009, whereas they constituted 23% of all candidates taking the test that year (12/52).

I would like to calculate the probability of this last result, assuming the following:

a) Women and men taking the exam are supposed to have equal chances of succeeding (also known as the anti-http://en.wikipedia.org/wiki/Lawrence_Summers#Differences_between_the_sexes"…).
b) The exams are supposed to be independent events (Bernoulli trials?).

How should I proceed please?

My educated guess is that binomial regression is the way to go but my stats skills are too low to actually perform the operation, or to analyse correctly its results.

If the probability of the 2009 results are very low, then one might reasonably assume that there is an anti-women bias at play (the test is basically a face-to-face interview).

Help much appreciated!

 

[Focus]

Well if you assume they have the same probability of passing (say p), then each person is a Bernoulli trial. So you have a binomial distribution (which is the sum of Bernoulli trials) for the amount of people that pass. You want \mathbb{P}(\text{someone passes the test, given they were a women})=\frac{\mathbb{P}(\text{they pass the the test and they are a woman})}{\mathbb{P}(\text{they are a woman})}.

If you want to check if they are biased then I suggest making a hypothesis test.