Power of two-sample test of binomial proportions

Suppose I take Success to mean Female. Then the number of Females in a random sample from University A is $X sim mathsf{Binom}(n=100,p=0.7)$ and the number of Females in a random sample from University B is $Y sim mathsf{Binom}(n=100,p=0.3).$

Try one test. Let's try using prop.test in R to analyze one such experiment with 200 students:

set.seed(2020)
x = rbinom(1, 100, .7);  y = rbinom(1, 100, .3)
 x; y
[1] 68
[1] 32
prop.test(c(x,y),c(100,100), cor=F)

        2-sample test for equality of proportions 
              without continuity correction

data:  c(x, y) out of c(100, 100)
X-squared = 25.92, df = 1, p-value = 3.559e-07
alternative hypothesis: two.sided
95 percent confidence interval:
 0.2307018 0.4892982
sample estimates:
prop 1 prop 2 
  0.68   0.32 

So in this particular experiment, the test finds very strong evidence to reject $H_0: p_a = p_b$ with P-value very near $0.$ [Use of a continuity correction is not useful for samples of size 100, so I used parameter cor=F in prop.test to disallow continuity correction.]

Then the question is whether I somehow got an outrageously atypical pair of samples in the example above, or whether prop.test really does have good power to detect the large difference in the proportions of Female students at the two universities, based on samples of $n_a = n_b = 100$ from each university.

Simulate 100,000 tests to estimate power. By doing the experiment 100,000 times, I can closely estimate the power of this test. [Computations in R.]

set.seed(722)
pv = replicate(10^5, prop.test(c(rbinom(1,100,.3),
                                 rbinom(1,100,.7)), c(100,100),cor=F)$p.val)
mean(pv <= .05)
[1] 0.99996

The answer is that the power of the test to detect the difference in proportions (at the 5% level) is above 99%. So it would be extremely rare for such an experiment not to show a difference in proportions. Specifically, the answer is 'a probability of almost 1'.

There are several versions of this test (depending on whether a normal approximation is involved, whether a continuity correction is used, and whether the test uses a 'pooled' standard error (under the null hypothesis that proportions are equal). Not knowing the version of the test you will use, I can't give an algebraic solution. (Also, this is a 'self-study' problem and you have not shown what you have tried, so I have no way to guess what approach you might be planning/expected to use.)

Lower bound on power. Here is one possible approach that does not use simulation: If we have $X=60, Y=40,$ then prop.test rejects, so it will also reject for more extreme differences such as $X=61, Y=39,$ and so on. [You might use your favorite test here instead of R's implementation of prop.test.]

prop.test(c(40,60), c(100,100), cor=F)$p.val
[1] 0.004677735

However the exact binomial probability of $P(X ge 60, Y le 40) = P(X ge 50)P(Y le 40) = 0.9875.$ So that gives a pretty good idea that rejection is nearly certain.

pbinom(40, 100, .3)*(1-pbinom(40, 100, .7))
[1] 0.9875016

The plot below shows that PDFs of $mathsf{Binom}(100, 0.3)$ and $mathsf{Binom}(100, 0.7)$ hardly overlap.

x = 0:100;  pdf.x = dbinom(x, 100, .7)
y = 0:100;  pdf.y = dbinom(y, 100, .3)
hdr="PDFs of BINOM(100,.3) [left] and BINOM(100,.7)"
plot(x-.1, pdf.x, type="h", col="blue", lwd=2, 
     ylab="PDF", xlab="Nr of Females", main=hdr)
  points(y+.1, pdf.y, type="h", col="brown", lwd=2)

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Addendum, per Comment: The answer is about 12% power.

set.seed(723)
pv = replicate(10^5, prop.test(c(rbinom(1,100,.48),
                                 rbinom(1,100,.53)), c(100,100),cor=F)$p.val)
mean(pv <= .05)
[1] 0.11845