Proportions and the binomial distribution

A proportion is a count divided by a total. Out of 1,000 vaccinated infants, 23 ran a fever afterwards: the proportion is 0.023. That single number carries sampling error, the same way a sample mean does. Run the survey again on a fresh 1,000 infants and you would not see exactly 23. This lesson works out how much a proportion bounces from sample to sample, and how to turn that spread into a confidence interval you can report.

The chance model behind a proportion is the binomial distribution. Once you have it, the standard error, the confidence interval, and the z-test all follow from it. We write (the Greek letter pi) for the true population proportion, and for the proportion you measured in your sample. Here has nothing to do with 3.14159; it is just the symbol for the underlying risk.

The sampling distribution of a proportion

Suppose the true risk of a fever after a vaccine is . You vaccinate 1,000 infants and count the fevers. You will not get exactly 20 every time. One sample gives 17, the next gives 26. The full list of counts you could observe, each tagged with its probability, is the sampling distribution of the count. Divide every count by and you have the sampling distribution of the proportion. They are one distribution drawn on two scales.

That distribution is discrete. For a sample of 1,000 the proportion can be 17/1000 or 18/1000 but nothing in between, so it steps rather than flows. Its shape has a name and a formula. It is the binomial distribution, fixed by two numbers: the sample size and the true proportion .

The binomial distribution and its assumptions

The binomial distribution gives the probability of getting exactly events out of people. It holds when four things are true.

• Fixed n. You decide the number of people before you look at any outcome.
• Two outcomes. Each person is an event or not: fever or no fever, dead or alive, positive or negative.
• Constant probability. Every person shares the same probability of being an event.
• Independence. One person's outcome tells you nothing about the next person's.

The probability of seeing exactly events is:

The first part counts the number of different orders in which events could fall among people. The exclamation mark is the factorial: , and is defined as 1. The second part is the probability of any one such order: events each with probability , and non-events each with probability .

Worked example 1: a four-child family

Beta-thalassaemia carrier status is common in parts of Malaysia and the wider region. Suppose both parents carry the trait. By Mendel's rule, each child has probability of inheriting thalassaemia major, independently of the other children. A couple has four children. What is the chance that 0, 1, 2, 3, or 4 of them have the major form?

Take . Plug , into the formula:

Repeat for every value of and you get the whole distribution. The number of orders is the binomial coefficient : 1, 4, 6, 4, 1.

The probabilities add to 1, as they must, because one of the five outcomes has to happen. The most likely single result is one affected child, not zero, even though no child is the commonest individual genotype.

n  <- 4
pi <- 0.25
d  <- 0:4

prob <- # fill in with dbinom()

prob <- dbinom(d, n, pi)
barplot(prob, names.arg = d,
        xlab = "No. of children with thalassaemia major",
        ylab = "Probability")
round(prob, 4)
# 0.3164 0.4219 0.2109 0.0469 0.0039

import numpy as np
from scipy.stats import binom
import matplotlib.pyplot as plt
n, pi = 4, 0.25
d = np.arange(0, 5)

prob = # fill in with binom.pmf()

prob = binom.pmf(d, n, pi)
plt.bar(d, prob)
plt.xlabel("No. of children with thalassaemia major")
plt.ylabel("Probability")
plt.show()
print(np.round(prob, 4))
# [0.3164 0.4219 0.2109 0.0469 0.0039]

Standard error of a proportion

Two facts about the binomial drive everything that follows. Its mean is the count you expect, . Its standard deviation, read as a sampling distribution, is the standard error of the count.

Divide by to move from the count to the proportion. The mean of is , and its standard error, the spread of the proportion across repeated samples, is:

You almost never know . For a confidence interval you plug in your observed , giving the estimated standard error . Two things to notice. The spread is widest when and shrinks toward the ends. And it falls with , so to halve the standard error you have to quadruple the sample.

Normal approximation to the binomial

Computing exact binomial tail probabilities by hand is slow once is large. The shortcut is that as grows, the binomial gets close to a normal distribution with the same mean and standard error. So you can borrow the z-based machinery you already use for means.

The approximation is good enough when both tails carry enough probability. The working rule is that and are both 10 or more. When you are estimating from data, check the same condition with : both and at least 10.

Confidence interval using the normal approximation

When the conditions hold, the confidence interval for a proportion takes the same form as the one for a mean: the estimate plus or minus a multiple of its standard error.

Here is the percentage point of the standard normal distribution. For a 95% interval, . This is the Wald interval, the one most software reports by default.

Worked example 2: a vaccine adverse-reaction rate

In a district immunisation programme, 23 of 1,000 infants showed an adverse reaction within 24 hours of a new vaccine. The observed proportion is , or 2.3%.

1. Check the conditions: and , both well above 10, so the normal approximation is fine.
2. Standard error: .
3. Margin: .
4. Interval: .

With 95% confidence, the true adverse-reaction rate is between 1.37% and 3.23%. You would tell parents the rate is about 2.3%, with that range as the honest uncertainty around it.

x <- 23
n <- 1000

p  <- # fill in
se <- # fill in

p  <- x / n
se <- sqrt(p * (1 - p) / n)
ci <- p + c(-1, 1) * 1.96 * se
round(c(p = p, se = se, lower = ci[1], upper = ci[2]), 5)
#       p      se   lower   upper
# 0.02300 0.00474 0.01371 0.03229

import numpy as np
x, n = 23, 1000

p  = # fill in
se = # fill in

p  = x / n
se = np.sqrt(p * (1 - p) / n)
ci = p + np.array([-1, 1]) * 1.96 * se
print(round(p, 5), round(se, 5), np.round(ci, 5))
# 0.023 0.00474 [0.01371 0.03229]

Exact binomial confidence interval

When or drops below 10, the normal approximation is unreliable and the Wald interval can spill below 0 or above 1, which is nonsense for a proportion. The fix is an exact binomial confidence interval, also called the Clopper-Pearson interval. It is built directly from the binomial probabilities, so it is valid for any sample size. The price is that it is conservative: its real coverage is at least 95%, often a little more, so the interval runs slightly wide.

Take a small phase-2 trial where 9 of 12 patients responded to a drug, so . Here , below 10, so the normal approximation is out. The exact 95% interval runs from about 0.428 to 0.945. That is wide, and honestly so: twelve patients cannot pin a response rate down tightly.

x <- 9
n <- 12

bt <- # call binom.test()

bt <- binom.test(x, n)
round(bt$conf.int, 4)
# 0.4281 0.9450
# attr 95 percent confidence interval

from scipy.stats import beta
x, n = 9, 12

lower = # fill in
upper = # fill in

lower = beta.ppf(0.025, x, n - x + 1)
upper = beta.ppf(0.975, x + 1, n - x)
print(round(lower, 4), round(upper, 4))
# 0.4281 0.9451

Continuity correction

One more wrinkle joins the discrete binomial to the continuous normal. The binomial count can be 9 or 10 but nothing between, while the normal curve flows smoothly. To approximate a discrete probability like with the normal, you treat each whole number as a bar one unit wide, running from to . So becomes the normal area above 8.5, not above 9. That half-unit shift is the continuity correction.

See it in a case where the normal approximation is stretched: , , so the mean is 6 and the standard error is . The exact binomial probability of 9 or more events is 0.0730. Compare the two normal approximations.

Without the correction the normal undershoots badly, 0.0416 against the true 0.0730. With it, the approximation lands close at 0.0745. The continuity correction matters most when is small and the normal fit is rough.

n  <- 12
pi <- 0.5

exact <- # P(X >= 9) from pbinom

exact   <- pbinom(8, n, pi, lower.tail = FALSE)   # P(X >= 9)
mu      <- n * pi
sigma   <- sqrt(n * pi * (1 - pi))
no_cc   <- pnorm((9   - mu) / sigma, lower.tail = FALSE)
with_cc <- pnorm((8.5 - mu) / sigma, lower.tail = FALSE)
round(c(exact = exact, no_cc = no_cc, with_cc = with_cc), 4)
#   exact   no_cc with_cc
#  0.0730  0.0416  0.0745

import numpy as np
from scipy.stats import binom, norm
n, pi = 12, 0.5

exact = # P(X >= 9)

exact   = binom.sf(8, n, pi)            # P(X >= 9)
mu      = n * pi
sigma   = np.sqrt(n * pi * (1 - pi))
no_cc   = norm.sf((9   - mu) / sigma)
with_cc = norm.sf((8.5 - mu) / sigma)
print(round(exact, 4), round(no_cc, 4), round(with_cc, 4))
# 0.073 0.0416 0.0745

Common mistakes

• Using the Wald interval when np is small. A rare event with a modest sample fails the np ≥ 10 check. The normal interval then runs too narrow and can dip below 0. Switch to the exact binomial interval.
• Checking only one of the two conditions. Both and must reach 10. A common proportion can pass on one side and fail on the other when the event is rare or near-universal.
• Confusing the count's standard error with the proportion's. The count uses ; the proportion divides inside the root, . Reporting one where you needed the other is off by a factor of .
• Quoting a confidence interval outside 0 to 1. If the Wald formula returns a bound like −0.01 or 1.04, that is the approximation breaking, not a real result. It signals you should have used an exact interval.
• Forgetting the binomial assumptions. Clustered or repeated observations on the same people break independence, so the plain binomial standard error understates the true spread. Outbreak data and repeated measures need a different model.

Tips

• Before any normal-based interval, compute and . If either is below 10, reach for binom.test or the Clopper-Pearson interval without hesitation.
• For sample-size planning, the standard error is largest at . Use 0.5 as a worst-case guess and you will never under-power.
• Keep for the truth and for your data. Use the observed in the standard error for a confidence interval, but the hypothesised in the standard error for a z-test.
• The width of a proportion's interval falls with . To halve it, plan for four times the sample, not twice.
• Report exact intervals for small studies and rare outcomes. They stay inside 0 to 1 and they do not over-promise precision you do not have.