Binomial hypothesis test for $p_1 = p_2 = P$ without normal approximation

Question

I was testing whether the proportion of successes from two populations is the same. It made me realize I only know how to do this with a normal approximation.

In particular, I would like to solve Exercise 9.3 from this set of exercises using a binomial distribution.

I understand that you may not feel like checking out my GitHub, so here's my problem in more general terms: suppose I flip a coin 100 times to test the hypothesis that the coin is unbiased. Instead of comparing the observed number of successes against the expected number of successes under $H_0$, I want to compare the observed proportion of successes against the expected proportion of successes. How can I test this without a normal approximation?

I found a post that suggests using Fisher's exact test. This is helpful in that it skips the normal approximation, but can this be done using a binomial test?

Answer 1

You have $x = 705$ successes in $n = 929$ failures, and you want to test $H_0: p = 3/4$ against $H_a: p \ne 3/4,$ at the 5% level. If $H_0$ is true, you expect about $627$ successes in $929$ trials. You will reject $H_0$ if the number $x$ of observed successes is far from $627$ in either direction.

Under $H_0,$ you have $X \sim \mathsf{Binom}(n =929, p=3/4).$ For a test at about the 5% level, you seek critical values $c_1$ and $c_2$ such that $P(X \le c_1\,|\, H_0) + P(X \ge c_2 \,|\, H_0) \approx 0.05.$

From R, where pbinom is a binomial CDF and qbinom is a binomial quantile function (inverse CDF), we have $P(X \le 670\, |\,H_0) = 0.0243$ and $P(X \ge 723\, |\,H_0) = 0.0244,$ so $c_1 = 670, c_2 = 723$ are suitable critical values.

qbinom(c(.025,.975), 929, 3/4)
[1] 671 722
pbinom(670, 929, 3/4)
[1] 0.02433703
1 - pbinom(722, 929, 3/4)
[1] 0.02441926

Therefore, if we observe $x = 705$ successes, with $c_1 = 670 < 705 < 723 = c_2,$ we do not reject $H_0.$

x = 625:775;  PDF = dbinom(x, 929, 3/4)
hdr = "PDF of BINOM(929, 3/4) with Normal Approx."
plot(x, PDF, type = "h", col="blue", main=hdr)
 abline(h=0, col="green2")
 abline(v=c(670.5, 721.5), col="red")
  mu = 929*.75;  sg = sqrt(929*.75*.25)
  curve(dnorm(x, mu, sg), add=T, lwd=2)

The approximating normal distribution to $\mathsf{Binom}(n=929,p=3/4)$ is $\mathsf{Norm}(\mu=696.75, \sigma=13.20);$ it provides a very close fit. With an approximate normal test, we would have used $c_1=670,c_2=723.$ which are not substantially different from the critical values of the exact normal test.

qnorm(c(.025,.975), mu, sg)
[1] 670.8824 722.6176