The binomial test

Linguistic Data: Quantitative Analysis and Visualisation: linguistic theory

Binomial test

dbinom(0, 5, 0.5)

[1] 0.03125

1/(2**5)

[1] 0.03125

Probability to get 0 heads in 5 tosses, fair coin (prob. of head is 0.5)

dbinom(1, 5, 0.5)

[1] 0.15625

Probability to get 1 head in 5 tosses, fair coin.

5 / 2 ** 5

[1] 0.15625

pbinom(1, 5, 0.5)

[1] 0.1875

Probability to get 1 or less heads in 5 tosses, fair coin.

1 / 2 ** 5 + 5 / 2 ** 5

[1] 0.1875

plot(dbinom(0:5, 5, 0.5))

plot(dbinom(0:20, 20, 0.5))

Assume we have a fair coin. Magician says he will make this coin flip only with heads. We flipped coin 12 times and obtained 10 heads. The magician says this two times where coin flipped tail was a mistake, and the rest of results confirm that he has magical abilities. Let’s test it.

Let \(p\) be the probability to get head after magician’s intervention.

\[H_0\colon p=1/2\] Under null hypethis, the magician doesn’t have any abilities, and we have just a fair coin.

\[H_1\colon p > 1/2\] Under alternative hypothesis, magician’s intervention increases probability to flip coin at least slighly.

p-value is probability to get the result that we actually obtained (e.g. 10 heads) or more convincing (e.g. 11 heads or 12 heads) under condition that null hypothesis holds (i.e. no magical abilities). Small p-values are argument in favor of alternative.

Let’s find p-value:

dbinom(10, 12, 0.5) + dbinom(11, 12, 0.5) + dbinom(12, 12, 0.5)

[1] 0.01928711

Shorter version:

dbinom(10:12, 12, 0.5)
sum(dbinom(10:12, 12, 0.5))

We can also use pbinom

1 - pbinom(9, 12, 0.5)

[1] 0.01928711

And finally we can use the specialised function binom.test:

binom.test(10, 12, 0.5, alternative="greater")

    Exact binomial test

data:  10 and 12
number of successes = 10, number of trials = 12,
p-value = 0.01929
alternative hypothesis: true probability of success is greater than 0.5
95 percent confidence interval:
 0.5618946 1.0000000
sample estimates:
probability of success 
             0.8333333