Sampling Distribution of a Sample Proportion

From One Sample to a Sampling Distribution

A sample proportion describes the fraction of observations in a sample that meet a specified condition. Its value depends on which observations happen to be selected. If we repeatedly take samples of the same size under the same sampling model, we can get different proportions. The probabilities of those possible results make up the sampling distribution of the sample proportion.

For example, suppose a binary outcome is whether a randomly selected item passes inspection. In one sample, every item might pass; in another, some might fail. The sample proportion is therefore not just a fixed summary of one collected sample. Before the sample is drawn, it is a random quantity whose possible values and probabilities can be described.

Definition: For a sample of size \(n\), let \(X\) be the number of observations with a specified binary outcome, called a success. The sample proportion is \(\hat{p}=X/n\). Its sampling distribution is the probability distribution of \(\hat{p}\) over all possible samples generated by the specified sampling model.

A binary outcome has two categories: success and failure. The names are labels for the outcome of interest; “success” need not mean that the outcome is desirable. The population proportion \(p\) is the probability of success on any one draw in the model considered here.

The Sampling Model and Its Possible Values

The exact distribution in this tutorial uses a model in which a sample consists of \(n\) independent draws, each with the same probability \(p\) of success. Equivalently, each observation is a Bernoulli trial with success probability \(p\). The sample size \(n\) is fixed before the draws are made.

Under this model, the number of successes \(X\) can be any whole number from 0 through \(n\). Since \(\hat{p}=X/n\), the possible sample proportions are

0,\ \frac{1}{n},\ \frac{2}{n},\ \ldots,\ \frac{n-1}{n},\ 1.

These are the possible values, or support, of \(\hat{p}\). Each increase of one success changes the sample proportion by \(1/n\). The values are not generally every number between 0 and 1: for example, a sample of size 4 can produce proportions 0, 0.25, 0.50, 0.75, or 1, but not 0.60.

The probabilities of these values depend on both \(n\) and \(p\). A small sample can be described by listing each possible sequence of outcomes. For larger samples, it is more efficient to count how many sequences produce each possible number of successes.

Worked Example: Listing Outcomes for a Sample of Two

Suppose each independent draw has probability \(p=0.5\) of success, and the sample size is \(n=2\). List the possible outcomes and find the sampling distribution of \(\hat{p}\).

There are four equally likely ordered sequences: failure-failure, failure-success, success-failure, and success-success. Each sequence has probability \(0.5 \times 0.5=0.25\), by independence.

The first sequence has \(X=0\) and \(\hat{p}=0/2=0\). The next two have \(X=1\) and \(\hat{p}=1/2=0.5\). The final sequence has \(X=2\) and \(\hat{p}=2/2=1\). Thus the probabilities of the three distinct sample proportions are

P(\hat{p}=0)=0.25,\qquad P(\hat{p}=0.5)=0.25+0.25=0.50,\qquad P(\hat{p}=1)=0.25.

The proportion 0.5 occurs in two different sequences, so its probability is the sum of their probabilities. The probabilities add to \(0.25+0.50+0.25=1\), as they must for a complete distribution.

Calculating the Exact Probabilities

Let \(X\) be the number of successes in \(n\) independent draws with common success probability \(p\). To have exactly \(k\) successes, first choose which \(k\) of the \(n\) positions are successes. There are \(\binom{n}{k}\) such choices. Each resulting sequence has \(k\) successes, with combined probability \(p^k\), and \(n-k\) failures, with combined probability \((1-p)^{n-k}\). Independence means the probability of that sequence is the product \(p^k(1-p)^{n-k}\). Summing over all sequences with \(k\) successes gives the probability for that count.

Result: Under independent binary sampling with fixed sample size \(n\) and success probability \(p\), the success count has a binomial distribution, \(X\sim\operatorname{Binomial}(n,p)\). Since \(\hat{p}=X/n\), its sampling distribution is

P\left(\hat{p}=\frac{k}{n}\right) = P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n.

Here \(\binom{n}{k}=n!/[k!(n-k)!]\) counts the different positions the \(k\) successes could occupy. The formula gives a probability for each possible sample proportion \(k/n\). It relies on a fixed sample size and independent draws with the same success probability; changing the sampling model can change the resulting distribution.

Worked Example: A Sample of Three With Success Probability 0.4

Suppose a hypothetical process produces independent binary outcomes, with probability \(p=0.4\) of success on each draw. For a sample of \(n=3\), find the probability of every possible sample proportion, and then find the probability that \(\hat{p}\) is at least \(2/3\).

The possible values are \(0,1/3,2/3,\) and \(1\). Applying the binomial formula for each number \(k\) of successes:

For \(k=0\), \(\hat{p}=0\), and \(P(\hat{p}=0)=\binom{3}{0}(0.4)^0(0.6)^3=1(1)(0.216)=0.216\).

For \(k=1\), \(\hat{p}=1/3\), and \(P(\hat{p}=1/3)=\binom{3}{1}(0.4)^1(0.6)^2=3(0.4)(0.36)=0.432\).

For \(k=2\), \(\hat{p}=2/3\), and \(P(\hat{p}=2/3)=\binom{3}{2}(0.4)^2(0.6)^1=3(0.16)(0.6)=0.288\).

For \(k=3\), \(\hat{p}=1\), and \(P(\hat{p}=1)=\binom{3}{3}(0.4)^3(0.6)^0=1(0.064)(1)=0.064\).

The probabilities total \(0.216+0.432+0.288+0.064=1.000\). The event \(\hat{p}\geq 2/3\) includes the outcomes with two or three successes, so

P(\hat{p}\geq 2/3) = P(\hat{p}=2/3)+P(\hat{p}=1) = 0.288+0.064 = 0.352.

Thus, under this model, 35.2% of samples of three would have a sample proportion of at least \(2/3\). This is a probability about the result of repeated samples, not a claim that any one observed sample is partly successful.

How Repeated Samples Produce the Distribution

A sampling distribution describes a process of repetition. Imagine repeatedly drawing a fresh sample of size \(n\) according to the same model, calculating \(\hat{p}\) each time, and recording the result. The observed proportions would vary from sample to sample, but only among the permitted values \(0,1/n,\ldots,1\). Over many repetitions, the proportion of repetitions with a given result would tend to be close to that result’s probability under the model.

This distinction is important: the sampling distribution is not the distribution of individual binary observations, and it is not a list of the values in one particular sample. It is a probability distribution for a statistic calculated from repeated samples. The model specifies which samples are possible and how likely they are; the statistic assigns a sample proportion to each one.

Worked Example: A Sample of Four With Success Probability 0.2

Consider a hypothetical sequence of independent binary draws with success probability \(p=0.2\). For samples of size \(n=4\), find the sampling distribution and the probability that exactly half the observations are successes.

The possible proportions are \(0,0.25,0.50,0.75,\) and \(1\). For \(k=0\), the probability is \(\binom{4}{0}(0.2)^0(0.8)^4=1(1)(0.4096)=0.4096\).

For \(k=1\), it is \(\binom{4}{1}(0.2)^1(0.8)^3=4(0.2)(0.512)=0.4096\). For \(k=2\), it is \(\binom{4}{2}(0.2)^2(0.8)^2=6(0.04)(0.64)=0.1536\).

For \(k=3\), it is \(\binom{4}{3}(0.2)^3(0.8)^1=4(0.008)(0.8)=0.0256\). For \(k=4\), it is \(\binom{4}{4}(0.2)^4(0.8)^0=1(0.0016)(1)=0.0016\).

These probabilities add to \(0.4096+0.4096+0.1536+0.0256+0.0016=1.0000\). Exactly half the observations are successes when \(k=2\), which means \(\hat{p}=2/4=0.5\). Therefore, \(P(\hat{p}=0.5)=0.1536\). Across many repetitions, about 15.36% of samples would have exactly two successes under the stated model.

Why the Exact Distribution Matters

The sampling distribution makes sample-to-sample variation explicit. It lets us answer questions such as “What values could the sample proportion take?” and “How likely is a sample proportion in a specified set?” These answers depend on the sample size and the population success probability, as well as on the sampling assumptions. A computed probability is therefore meaningful only when the model is appropriate.

The binomial formula also provides a precise starting point for later study of the center and variability of sample proportions and for approximations to their distributions. Here, the key idea is the exact distribution: each allowed proportion corresponds to a success count, and the binomial probability gives the chance of that count under independent binary sampling.

Common Mistake

A common error is to assign probability \(p\) directly to a sample proportion. For example, if \(p=0.4\), it is incorrect to say that \(P(\hat{p}=0.4)=0.4\) without checking the sample size and the possible values. With \(n=3\), the sample proportion can only be \(0,1/3,2/3,\) or \(1\); 0.4 is not possible at all. The number \(p\) is the success probability for one draw, whereas \(\hat{p}\) is a statistic calculated from an entire sample. Find its possible values from \(k/n\), then use the appropriate probability for each value.

Key takeaway: In a fixed-size sample of independent binary outcomes with success probability \(p\), the sample proportion is the success count divided by \(n\). Its possible values are \(k/n\), and their probabilities are \(\binom{n}{k}p^k(1-p)^{n-k}\).

Check Your Understanding

Use the independent binary sampling model where specified.

For a sample size of 5, list all possible values of the sample proportion.
If \(n=2\) and \(p=0.3\), calculate the probability that the sample proportion is 0.5.
For \(n=4\) and \(p=0.5\), what number of successes corresponds to \(\hat{p}=0.75\)?
Explain why a sampling distribution is not the same thing as the distribution of individual outcomes in one sample.
State the assumptions behind the binomial formula for the sampling distribution of a sample proportion.