Binomial → Normal → Wilson

Introduction

One of the questions that keeps coming up with students is the following.

What does the Wilson score interval represent, and how does it encapsulate the right way to calculate a confidence interval on an observed Binomial proportion?

In this blog post I will attempt to explain, in a series of hopefully simple steps, how we get from the Binomial distribution to the Wilson score interval. I have written about this in a more ‘academic’ style elsewhere, but I haven’t spelled it out in a blog post.

An unnatural probability?

Not everything that looks like a probability is one.

Just because a variable or function ranges from 0 to 1, it does not mean that it behaves like a unitary probability over that range.

Natural probabilities

What we might term a natural probability is a proper fraction of two frequencies, which we might write as p = f / n.

• Provided that f can be any value from 0 to n, p can range from 0 to 1.
• In this formula, f and n must also be natural frequencies, that is, n stands for the size of the set of all cases, and f the size of a true subset of these cases. The term ‘natural’ here refers to the mathematical sense of the set of positive integers.

Aside: In certain models, these frequencies could be obtained from the sum of a set of probability estimates, each representing the probability that the observation was genuinely independent from others in the sample. This might permit a ‘frequency’ to be observed that was not a natural number. But the principle is the same.

This natural probability is expected to be a Binomial variable, and the formulae for z tests, χ² tests, Wilson intervals, etc., as well as logistic regression and similar methods, may be legitimately applied to such variables. The Binomial distribution is the expected distribution of such a variable if each observation is drawn independently at random from the population (an assumption that is not strictly true with corpus data).

Another way of putting this is that a Binomial variable expresses the number of individual events of Type A in a situation where an outcome of either A or B are possible. If we observe, say that 8 out of 10 cases are of Type A, then we can say we have an observed probability of A being chosen, p(A | {A, B}), of 0.8. In this case, f is the frequency of A (8), and n the frequency of both A and B (10). See Wallis (2013a).

Comparing frequencies within a discrete distribution

Note:
This page explains how to compare observed frequencies f₁ and f₂ from the same distribution, F = {f₁, f₂,…}. To compare observed frequencies f₁ and f₂ from different distributions, i.e. where F₁ = {f₁,…} and F₂ = {f₂,…}, you need to use a chi-square or Newcombe-Wilson test.

Introduction

In a recent study, my colleague Jill Bowie obtained a discrete frequency distribution by manually classifying cases in a small sample drawn from a large corpus.

Jill converted this distribution into a row of probabilities and calculated Wilson score intervals on each observation, to express the uncertainty associated with a small sample. She had one question, however:

How do we know whether the proportion of one quantity is significantly greater than another?

We might use a Newcombe-Wilson test (see Wallis 2013a), but this test assumes that we want to compare samples from independent sources. Jill’s data are drawn from the same sample, and all probabilities must sum to 1. Instead, the optimum test is a dependent-sample test.

Example

A discrete distribution looks something like this: F = {108, 65, 6, 2}. This is the frequency data for the middle column (circled) in the following chart.

This may be converted into a probability distribution P, representing the proportion of examples in each category, by simply dividing by the total: P = {0.60, 0.36, 0.03, 0.01}, which sums to 1.

We can plot these probabilities, with Wilson score intervals, as shown below.

An example graph plot showing the changing proportions of meanings of the verb think over time in the US TIME Magazine Corpus, with Wilson score intervals, after Levin (2013). In this post we discuss the 1960s data (circled). The sum of each column probability is 1. Many thanks to Magnus for the data!

So how do we know if one proportion is significantly greater than another?

• When comparing values diachronically (horizontally), data is drawn from independent samples. We may use the Newcombe-Wilson test, and employ the handy visual rule that if intervals do not overlap they must be significantly different.
• However, probabilities drawn from the same sample (vertically) sum to 1 — which is not the case for independent samples! There are k−1 degrees of freedom, where k is the number of classes. It turns out that the relevant significance test we need to use is an extremely basic test, but it is rarely discussed in the literature.