This is a Statistical Test for proportions that uses the Binomial Distribution as the null (sampling) distribution.

It doesn't use the Normal Approximation

- because sometimes it's possible to use the Binomial model directly
- or because it's not possible to use the Normal Model: some conditions are not met

Recall the formula:

- $P(\text{success}) = { n \choose k } p^k (1 - p)^{n - k}$
- this is the null distribution of our test

Test

- the tail area of the null distribution:
- add up the probabilities (using the formula) for all $k$ that support the alternative hypothesis $H_A$

- one-sided test - use single tail area
- two-sided - compute single tail and double it

- medical consultant helps patients
- he claims that with his help the ratio of complications is lower than usually
- (i.e. lower than 0.10)

- is it true?

We want to test a hypothesis:

- $H_0: p_A = 0.10$ - ratio of complications without a specialist
- $H_A: p_A < 0.10$ - specialist helps, the complications ratio is lower than usual

Observed data:

- 3 complications in 62 cases
- $\hat{p} = 0.048$
- is it only due to chance?

Normal Model

- the Success-Failure condition is not met: $p_A \cdot 62 = 0.10 \approx 6.2 < 10$
- under $H_0$ we'd expect to see only 6.2 complications

- thus cannot use Normal Approximation and perform a Binomial Proportion Test

Apply the Binomial Model:

- $p\text{-val} = \sum_{j = 0}^3 { n \choose j } p^j (1 - p)^{n - j} = 0.0015 + 0.01 + 0.034 + 0.0355 = 0.121$
- we don't reject the $H_0$ at $\alpha = 0.05$

check! sim got 0.04