Exact Binomial Test
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
Binomial Model
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
Examples
Example 1: Medical Consultant (One-Sample)
- 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 |