Choose a One-Proportion Procedure

Start With the Outcome and the Question

A one-proportion procedure is appropriate when the data come from one population and each observational unit has a categorical outcome that can be classified in one of two ways: success or failure. “Success” is a label for the outcome being counted; it does not mean that the outcome is desirable. For example, a study might define success as a patient experiencing a side effect, a device passing inspection, or a voter supporting a proposal.

The question then determines which one-proportion procedure is relevant. If the goal is to estimate the population’s success proportion, use a confidence interval. If the goal is to assess a claim that the population proportion equals a stated benchmark, use a hypothesis test. When the normal approximation needed for a z procedure is not credible, an exact binomial test may be appropriate for testing a benchmark.

Definition: Let \(p\) denote the proportion of all units in one population that have a specified binary outcome, designated “success.” In a sample of \(n\) units, let \(x\) be the number of successes. The sample proportion is \(\hat{p}=x/n\), an estimate of \(p\).

The population and the success category must be stated clearly. “What fraction of surveyed patients reported nausea?” is a one-proportion question if each patient is classified as reporting nausea or not. “Does nausea occur more often under treatment A than treatment B?” compares two groups and is not a one-proportion question. Likewise, a variable with several categories is not automatically a binary outcome: a category must be selected as the success of interest, and the remaining outcomes grouped as failure, if that definition suits the research question.

A Practical Selection Sequence

Identify the observational units and outcome.
There should be one group of units, each classified as success or failure for the outcome of interest.

Name the parameter.
The target is the population proportion \(p\), not a mean or a difference between proportions.

Match the question to the goal.
Use an interval to estimate \(p\). Use a test to assess a specified benchmark \(p_0\).

Check whether the chosen method is credible.
Consider how the sample was obtained, whether observations can reasonably be treated as independent, and whether the expected success and failure counts support a normal approximation. If expected counts are small, consider an exact binomial test for a benchmark.

This sequence is a selection aid, not a substitute for checking the study design. A large sample cannot repair serious selection bias, and a convenient sample does not become representative merely because it contains many observations. If the same person contributes multiple measurements, the observations may not be independent; a basic one-proportion procedure may then be inappropriate without accounting for that dependence.

One-Proportion z Procedures

For a test, the usual one-proportion z procedure compares the observed sample proportion with a benchmark \(p_0\). Under the null hypothesis, the number of successes \(X\) is modeled as binomial with sample size \(n\) and success probability \(p_0\). When the expected numbers of successes and failures are sufficiently large, the sampling distribution of \(\hat{p}\) under \(H_0\) is approximated by a normal distribution.

One-proportion z test: For a null hypothesis \(H_0:p=p_0\), the test statistic is

z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}.

The denominator is the standard error calculated under the null hypothesis. The p-value is the probability, assuming \(H_0\), of a test statistic at least as extreme in the direction or directions specified by \(H_a\).

For an interval estimate, a common introductory method is the one-proportion z confidence interval:

One-proportion z confidence interval: When the normal approximation is suitable, a confidence interval for \(p\) is

\hat{p}\ \pm\ z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},

where \(z^*\) is the standard-normal critical value for the chosen confidence level. Here the standard error uses \(\hat{p}\), because the aim is to estimate the unknown population proportion rather than calculate variability under a specified null value.

The test and interval formulas look similar, but they answer different questions and use different standard errors. In the test, variability is calculated using the benchmark \(p_0\), as required by the null model. In this interval, variability is estimated from \(\hat{p}\). A p-value is not the probability that the null hypothesis is true, and a 95% confidence interval does not mean there is a 95% probability that this particular interval contains \(p\). Its 95% confidence level describes the long-run coverage of the method under its assumptions.

Why the Test Standard Error Uses the Benchmark

The distinction in standard errors follows from the binomial model. Suppose \(X\) counts successes among \(n\) independent observations, each with success probability \(p\). Then \(X\) is the sum of \(n\) binary indicators. Each indicator has mean \(p\) and variance \(p(1-p)\); independence makes the variances add. Dividing the count by \(n\) gives the sample proportion.

E(\hat{p})=\frac{E(X)}{n}=\frac{np}{n}=p,\qquad \operatorname{Var}(\hat{p})=\frac{\operatorname{Var}(X)}{n^2} =\frac{np(1-p)}{n^2} =\frac{p(1-p)}{n}.

Under \(H_0:p=p_0\), this variance becomes \(p_0(1-p_0)/n\). For sufficiently large expected counts, the binomial distribution is approximated by a normal distribution, so standardizing the observed \(\hat{p}\) by its null standard error yields the z statistic. For an interval, \(p\) is unknown, so the z interval estimates its standard error by substituting \(\hat{p}\) for \(p\).

Worked Examples

Worked Example: Estimating a Population Proportion

Suppose a hypothetical random sample of 120 patients from a large clinic population is asked whether they experienced a specified symptom during the previous week. Each patient is recorded as Yes or No, and 78 answer Yes. The question is to estimate the population proportion who would answer Yes.

Choose the procedure. There is one population, one binary outcome, and the goal is estimation of its success proportion \(p\). A one-proportion confidence interval is the relevant procedure. The sample is described as random, and assume patients’ responses are independent; if sampling without replacement, the population is large enough to make that dependence negligible.

The observed proportion is \(\hat{p}=78/120=0.650\). The observed success and failure counts are 78 and \(120-78=42\), both comfortably above 10, so the normal approximation used by the introductory z interval is reasonable here.

For a 95% interval, use \(z^*=1.96\):

0.650\ \pm\ 1.96\sqrt{\frac{0.650(1-0.650)}{120}} =0.650\ \pm\ 1.96\sqrt{0.0018958} =0.650\ \pm\ 0.0854.

The interval is approximately \((0.565,\ 0.735)\). In context, the data give a 95% confidence interval of about 56.5% to 73.5% for the proportion of patients in this clinic population who would report the symptom. The interval describes uncertainty in the estimate; it does not establish that the symptom was caused by any particular factor.

Worked Example: Testing a Stated Benchmark

In a hypothetical quality check, 118 of 200 randomly selected units pass a specified inspection. The manufacturer’s benchmark is that 50% of units pass. The question is whether the population pass proportion differs from 50%.

Choose the procedure. This is one population, one binary outcome, and a question about a stated benchmark. Use a one-proportion z test. The sample is assumed random and its observations independent. Under the null, the expected success and failure counts are \(200(0.50)=100\) each, so the normal approximation is suitable.

The hypotheses are \(H_0:p=0.50\) and \(H_a:p\ne0.50\). The sample proportion is \(\hat{p}=118/200=0.590\). The null standard error and test statistic are:

SE_0=\sqrt{\frac{0.50(1-0.50)}{200}} =\sqrt{0.00125}=0.03536, \qquad z=\frac{0.590-0.500}{0.03536}=2.545.

For a two-sided test, the p-value is \(2P(Z\ge 2.545)\), approximately \(0.0109\), where \(Z\) has a standard normal distribution. At significance level \(\alpha=0.05\), this p-value is smaller than 0.05. The sample therefore provides evidence that the population pass proportion differs from 50%. It does not show that the benchmark is certainly false, nor does it quantify the practical importance of the difference.

Worked Example: Choosing an Exact Binomial Test

A hypothetical pilot sample contains 12 independently selected units. Four have a specified rare outcome. The research question is whether the population proportion with that outcome is greater than 10%.

Choose the procedure. There is one binary outcome and one population benchmark, so the question calls for a one-proportion test. However, under \(H_0:p=0.10\), the expected number of successes is \(12(0.10)=1.2\), which is too small for a normal approximation to be dependable. Use an exact binomial test instead. Assume the sample is random and the observations are independent.

Under the null, \(X\), the number of outcomes among the 12 units, has a binomial distribution with \(n=12\) and \(p_0=0.10\). For the upper-tailed alternative, the exact p-value is the probability of observing at least four outcomes under this null model:

P(X\ge4\mid p=0.10) =1-P(X\le3\mid p=0.10) =1-\sum_{k=0}^{3}{12\choose k}(0.10)^k(0.90)^{12-k} \approx 0.0256.

At \(\alpha=0.05\), the exact p-value is below the significance level, so these data provide evidence that the population proportion exceeds 10%. The exact calculation uses the binomial distribution directly rather than relying on a normal approximation. It still depends on the sampling and independence assumptions; “exact” does not mean immune to poor study design.

Common Mistakes and Why the Choice Matters

Choosing a procedure from the word “proportion” alone. First identify whether there is one population or a comparison between populations. A single population proportion is not the parameter for comparing two groups.
Using the interval standard error in a test. Under \(H_0:p=p_0\), the test standard error is \(\sqrt{p_0(1-p_0)/n}\), not \(\sqrt{\hat{p}(1-\hat{p})/n}\). The null hypothesis supplies the probability used to model the test statistic.
Treating every categorical outcome as binary. A procedure for one proportion requires a defined success-versus-failure outcome. If several categories are important separately, collapsing them may not answer the research question.
Applying a z procedure just because a sample proportion can be calculated. The normal approximation depends on the design and expected counts, not merely on whether \(\hat{p}\) exists. With small expected counts, consider an exact binomial test for a benchmark.
Interpreting a small p-value as the probability that \(H_0\) is true. The p-value is calculated assuming \(H_0\) and describes how unusual the observed result, or a more extreme one, would be under that assumption.

Key takeaway: Choose a one-proportion method when one population’s binary outcome is summarized by \(x\) successes among \(n\) observations. Estimate \(p\) with a suitable confidence interval, test a stated benchmark with a one-proportion test, and avoid a normal approximation when its conditions are not credible.

Check Your Understanding

For each situation, identify whether a one-proportion procedure fits and, if it does, whether the question calls for an interval or a test.

A random sample of households is asked whether anyone in the household owns an electric vehicle. The goal is to estimate the population percentage answering Yes. What parameter and procedure fit?
A clinic compares the percentage of patients reporting a side effect under two different treatments. Is a one-proportion procedure appropriate? Explain.
A factory tests whether its pass rate differs from a stated 92% benchmark using a large random sample. What procedure and null parameter would be relevant?
In a sample of 10 units, only one has a rare outcome. Why might a one-proportion z test be a poor choice for testing a small benchmark?
In a one-proportion z test, why is the standard error calculated using \(p_0\) rather than \(\hat{p}\)?