How to calculate the power of a test that compares two proportions

Question

I'm trying to compute the power of a proportion test by hand. The null and alternative hypotheses are below and I'm using a significance level, $\alpha$ = 0.05

$H_0: p_1 = p_2$

$H_a: p_1 \neq p_2$

Let's assume the true value of $p_1$ = 0.69 and that the true value of $p_2$ = 0.68. Based on this, we can calculate the estimator of $p_i$ as:

$\hat{P}_i$ = $X_i/n$ where $X_i \sim~ B(n, p_i)$. Here $B$ is the Binomial distribution parameterized by the sample size, $n$ and the probability, $p_i$

I went ahead and simulated both distributions as follows:

p_1 <- 0.69
p_2 <- 0.68
num_samples <- 10000
sample_size <- 10000
set.seed(1231421)

p_1_hat <- rbinom(num_samples, size = sample_size, prob = p_1)/sample_size
p_2_hat <- rbinom(num_samples, size = sample_size, prob = p_2)/sample_size

Now to calculate power, I first find out the 0.025 and 0.975 quantile of $\hat{P}_1$ as follows:

> quantile(p_1_hat, c(0.025, 0.975))
  2.5%  97.5% 
0.6809 0.6990 

At this point, the power calculation should be straightforward:

> pnorm(quantile(p_1_hat, 0.025), p_2, sd(p_2_hat)) +
+ pnorm(quantile(p_1_hat, 0.975), p_2, sd(p_2_hat), lower.tail = FALSE) 
0.576146 

Unfortunately, this does not match with the output from the pwr library.

> library(pwr)
> pwr.2p.test(ES.h(p_1, p_2), n = sample_size)

     Difference of proportion power calculation for binomial distribution (arcsine transformation) 

              h = 0.0215284
              n = 10000
      sig.level = 0.05
          power = 0.3310592
    alternative = two.sided

NOTE: same sample sizes

Where am I going wrong? A plot of the way I'm trying to calculate power is shown below. I'm trying to find the area to the left of the vertical line near 0.68 and to the right of the vertical line near 0.70 (practically zero). Isn't that what power is?

Edit: Calculations using the arcsine transform based on gammer's comments.

p_1_hat_asin <- asin(sqrt(p_1_hat))
p_2_hat_asin <- asin(sqrt(p_2_hat))
> pnorm(quantile(p_1_hat_asin, 0.025), mean(p_2_hat_asin), sd(p_2_hat_asin))
0.5752689 

Answer 1

The power calculator you're using operates on the arcsin transformed scale (which is what the cohen's H is based on). So, you'd need to use the approximate normality of the arcsine transformed proportions and do the analytic calculation on that scale to recover the result of pwr.2p.test. I'll leave that analysis to you. It shouldn't be too complicated. Just use the fact that difference between two independent approximately normal variables is approximately normal and calculate areas under that distribution. Give it a shot and post the answer if you figure it out. Good luck.

As a sidebar, below is some code to estimate the power of the $\chi^2$ test of equality of proportions by simulation by calculating the proportion of times (out of a large number, 1000 in this case) the null hypothesis is rejected.

set.seed(1122)
p1 = 0.69 
p2 = 0.68 
n = 10000
n.reject = 0
n.rep = 1000
for(k in 1:n.rep)
{ 
  x1 = sum(runif(n)<p1)
  x2 = sum(runif(n)<p2)
  ptest = prop.test( c(x1,x2), c(n,n), correct=FALSE)
  if(  ptest$p.value < 0.05 ) n.reject = n.reject + 1
}
n.reject/n.rep
[1] 0.322

Answer 2

Addressing this part of the question first:

Where am I going wrong? A plot of the way I'm trying to calculate power is shown below. I'm trying to find the area to the left of the vertical line near 0.68 and to the right of the vertical line near 0.70 (practically zero). Isn't that what power is?

Power should be calculated relative to the alternate hypothesis, rather than the null hypothesis, as we're calculating the probability of correctly rejecting the null hypothesis assuming that the alternate hypothesis is true. So the plot should look more like this:

And we're looking to calculate the area shaded in Pink (to the left of the black dashed line that is $-Z_{\alpha/2}$, and to the right of the other black dashed line that is $Z_{\alpha/2}$, that as you've mentioned, is close to 0)

As for the calculations, there are 2 ways to calculate it from the graph:

1. We can find the equivalent Z values on the Null hypothesis curve: $Z = \frac{p_2 - p_1}{\sqrt{p_{pooled} * (1 - p_{pooled}) * (\frac{1}{n} + \frac{1}{n})}}$ = $\frac{0.68 - 0.69}{\sqrt{0.685 * (1 - 0.685) * (\frac{1}{10000} + \frac{1}{10000})}}$ = $-1.52224$
   
   $Z_{\beta L} = -Z_{\alpha/2} + Z = -1.95996 + 1.52224 = -0.43772$
   
   $Z_{\beta R} = Z_{\alpha/2} + Z = 1.95996 + 1.52224 = 3.4822$
   
   Power = Pink Area to the left + Pink Area to the right = scipy.stats.norm(0, 1).cdf(-0.43772) + (1 - scipy.stats.norm(0, 1).cdf(3.4822)) = 0.33104329

2. Or directly find the area under the Alternate hypothesis curve: scipy.stats.norm(-1.52224, 1).cdf(-1.95996) + (1 - scipy.stats.norm(-1.52224, 1).cdf(1.95996)) = 0.33104329

This is really close to R's pwr.2p.test answer of 0.3310592, and the reason for the difference is because of the methods used to calculate Power. As mentioned by @gammer, pwr uses the arcsin transformation, and calculates $Z_\beta$ with the following formula:

$Z_\beta = Z_\alpha - \frac{h\sqrt{n}}{\sigma}$,
where $h = 2arcsin(\sqrt{p_1} - 2arcsin(\sqrt{p_2})$,
and $n = n/2$ for equal $n$'s,
and $n = \frac{1}{\frac{1}{n_1}+\frac{1}{n_2}}$ for unequal $n$'s

Using this method:

$h = 0.0215$

$Z_{\beta L} = 1.96 - \frac{0.0215 * \sqrt{5000}}{1} = -0.437$

$Z_{\beta R} = -1.96 - \frac{0.0215 * \sqrt{5000}}{1} = 3.482$

Power = 1 - scs.norm(0, 1).cdf(3.482) + (scs.norm(0, 1).cdf(-0.437)) = 0.331059

Note: numbers have been truncated for brevity. But if you use the full number string, you would get the same numbers as pwr:

• h = 0.02152840303762038
• Z_alpha = ± 1.9599639845400545
• $Z_{\beta L}$ = 3.4822519621418984
• $Z_{\beta R}$ = -0.43767600693820996

Resources:
1. Documentation on pwr using the arcsin transformation
2. This other SE question outlining the 2 methods of calculating power
3. pwr source code: getAnywhere(pwr.2p.test)

Answer 3

I think that the n in the pwr.2p.test refers to the size of one of the two samples instead of the total size of the two samples.

If you double the sample size (i.e if you write: pwr.2p.test(ES.h(p_1, p_2), n = 2*sample_size) then you get a power of 0.5764 which is reasonably close to your other estimate.