Solved – power analysis for binomial test

In order to find 'power', you need to have a specific alternative in mind. Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where $p = P(mathrm{Female}).$ Also suppose you have $n = 64$ and you want the power of a test at level $alpha = 0.05$ against the specific alternative $p = 0.6.$

For an exact binomial test, you need to find the critical value $c$ such that $P(X ge c,|,n=64, p=.5)$ is maximized, but still below $0.05.$ In R, where dbinom, pbinom, and qbinom denote binomial PDF, CDF, and quantile function (inverse CDF), respectively, we see that the critical value is $c = 40.$ Notice that, because of the discreteness of binomial distributions, the so-called `5% level' actually rejects with probability $P(mathrm{Rej}, H_0 | H_0, mathrm{True}) approx 3%.$

qbinom(.95, 64, .5) [1] 39 sum(dbinom(39:64, 64, .5)) [1] 0.05171094 sum(dbinom(40:64, 64, .5)) [1] 0.02997059 1 - pbinom(39, 64, .5) [1] 0.02997059 

Then the power of this test against alternative value $p = 0.6$ is given by $P(X ge 40,|,n=64, p=0.6) = 0.3927.$

1 - pbinom(39, 64, .6) [1] 0.392654 

We can make a 'power curve' for this test by looking at a sequence of alternative values p.a between $0.5$ and $.75.$ The first block of R code below makes the solid black line in the plot below.

p.a = seq(.50, .75, by=.01) p.rej = 1 - pbinom(39, 64, p.a) plot(p.a, p.rej, type="l", main="Power Curve")  abline(h=c(.03,1), col="green2") 

If we look at a level $alpha = 0.05$ test of $H_0: p = 0.5$ vs $H_a: p > 0.5$ with $n = 256$ subjects, then the critical value is $c = 141,$ the rejection probability when $H_0$ is true is $0.046,$ and the power against various alternative values of $p$ is greater, as shown by the dotted blue line in the figure.

c.256 = qbinom(.95, 256, .5); c.256 [1] 141 1 - pbinom(c.256, 256, .5) [1] 0.04565604 p.rej.256 = 1 - pbinom(c.256, 256, p.a) lines(p.a, p.rej.256, col="blue", lty="dotted") 

Notes: Because $n = 64$ is sufficiently large to use normal approximations, you might want to try using normal approximations. A disadvantage is that this ignores the issue of discreteness, so it may appear that your test rejects exactly 5% of the time when $H_0$ is true. Also, you'd need to use a continuity correction for best results.

One relevant computation for the significance level in R is:

1 - pnorm(39.5, 32, 4) [1] 0.03039636 

(Approximate) power is $0.3895:$

mu.a = 64*.6;  sg.a = sqrt(64*.6*.4) mu.a; sg.a [1] 38.4 [1] 3.919184  1 - pnorm(39.5, mu.a, sg.a)     # Using NORM(mu.a, sg.a) [1] 0.3894815 1 - pnorm((39.5 - mu.a)/sg.a)   # Standardizing and using NORM(0,1). [1] 0.3894815