Interpreting the Confidence Interval for a Proportion

Turn the Interval into a Sentence

In Using 1-PropZInt on the Calculator, you used the sample information to find a confidence interval for a population proportion. The calculator provides two endpoints, but those numbers do not explain themselves. A complete interpretation identifies the population, names the characteristic being estimated, and states the confidence level.

The interval estimates the fixed but usually unknown population proportion \(p\). Its endpoints are proportions, not counts of people, and they do not describe only the sample. The sample proportion \(\hat{p}\) is the estimate calculated from the observed sample; the interval gives a range of plausible values for the population proportion \(p\), using the method and confidence level specified.

Interpretation template: We are \(C\)% confident that the true proportion of [specified population] who [have the characteristic of interest] is between [lower endpoint] and [upper endpoint].

Fill in every part of the sentence. “We are 95% confident that the true proportion is between 0.32 and 0.47” is incomplete if the reader cannot tell which population or characteristic the proportion refers to. A strong answer supplies both. If the endpoints are decimals, you may state them as percentages instead, as long as you convert both correctly.

The word true refers to the population proportion \(p\), not the sample proportion \(\hat{p}\). The interval is based on a sample, but its purpose is to estimate a proportion in a larger population. In the next tutorial, we will examine more closely what the stated confidence level means. For now, use the confidence level given for the interval and avoid describing it as a probability that the fixed population proportion changes or has a particular value.

A Quick Translation Check

Before writing an interpretation, check the context and the two endpoints. These questions help keep the sentence precise:

Who is the population? Name the people or other individuals the study is intended to represent.
What counts as the characteristic? State what “success” means in ordinary language, not just as a label.
What are the units? The endpoints are proportions. Use decimals or convert them consistently to percentages.
What does the interval estimate? Name the true population proportion, rather than the number or percentage observed in the sample.

As in Constructing a One-Proportion z-Interval by Hand and Using 1-PropZInt on the Calculator, the interval method also requires appropriate conditions. In each example, we will identify the population proportion, check the random, 10%, and Large Counts conditions, and then interpret the interval. The calculator’s ability to produce endpoints does not replace those checks.

Conditions to recall:

Random: The data come from a random sample or a suitable random process.
10% condition: If sampling without replacement from a finite population of size \(N\), verify \(n\leq0.10N\).
Large Counts: The observed numbers of successes and failures are each at least 10: \(x\geq10\) and \(n-x\geq10\).

Worked Example: A Community Health Survey

Consider a hypothetical survey intended to estimate the proportion of adults in a town who get a seasonal flu vaccine. A random sample of 180 adults is selected from 2,400 adults; 72 say they got the vaccine. Find and interpret a 95% one-proportion \(z\)-interval.

State.
Let \(p\) be the true proportion of all adults in this town who got a seasonal flu vaccine. We will estimate \(p\) with a 95% confidence interval.

Plan and check conditions.
The sample is stated to be random. Since sampling is without replacement, the 10% limit is \(0.10(2400)=240\), and \(180\leq240\). There are 72 successes and \(180-72=108\) failures, both at least 10. The conditions support using a one-proportion \(z\)-interval.

Do.
Calculate the interval from \(x=72\), \(n=180\), and \(z^*=1.96\). The sample proportion is \(72/180=0.40\). The endpoints are approximately \(0.3284\) and \(0.4716\).

Conclude.
We are 95% confident that the true proportion of all adults in this town who got a seasonal flu vaccine is between \(0.3284\) and \(0.4716\), or about 32.8% and 47.2%.

To verify the endpoints, use the one-proportion interval formula. The estimated standard error is:

SE_{\hat{p}} =\sqrt{\frac{(0.40)(1-0.40)}{180}} =\sqrt{0.001333333\ldots} \approx0.0365148

Using the unrounded standard error, the margin of error is approximately \(1.96(0.0365148\ldots)=0.0715691\). Thus the endpoints are \(0.40-0.0715691\ldots\approx0.3284\) and \(0.40+0.0715691\ldots\approx0.4716\). The interpretation refers to all adults in the town, not only to the 180 adults surveyed.

Keep the Context Attached to the Numbers

A useful habit is to say what the interval endpoints mean before inserting them into the interpretation sentence. For example, if the interval is \((0.30,0.42)\), the lower endpoint does not mean that 30 individuals in the sample had the characteristic. It means that 0.30 is the lower endpoint of the estimated range for the population proportion. As a percentage, that endpoint is 30%.

Also watch the boundary of the population. A sample from one school can support an estimate about that school’s students if the sampling method represents them; it does not automatically estimate the proportion among all students in a city or country. The interpretation should not claim more than the design justifies.

Worked Example: Households with a Home Compost Bin

In a hypothetical city survey, a random sample of 150 households is selected from 2,200 households. Of those sampled, 54 report having a home compost bin. Interpret the 90% one-proportion \(z\)-interval.

Let \(p\) be the true proportion of all households in this city that have a home compost bin. The sample is random. For sampling without replacement, \(0.10(2200)=220\), and \(150\leq220\), so the 10% condition is met. There are \(54\) successes and \(150-54=96\) failures, both at least 10. The conditions support the interval method.

The sample proportion is \(\hat{p}=54/150=0.36\). For a 90% confidence interval, \(z^*\approx1.6449\). The estimated standard error is:

SE_{\hat{p}} =\sqrt{\frac{(0.36)(1-0.36)}{150}} =\sqrt{0.001536} \approx0.0391918

Using the unrounded standard error, the margin of error is approximately \(1.6449(0.0391918\ldots)=0.064465\). The endpoints are:

0.36-0.0644648\ldots\approx0.2955, \qquad 0.36+0.0644648\ldots\approx0.4245

The interpretation is: We are 90% confident that the true proportion of all households in this city that have a home compost bin is between 0.2955 and 0.4245. In percentage terms, we are 90% confident that between about 29.6% and 42.4% of the city’s households have a home compost bin. Both forms describe the same interval; the percentages are the decimal endpoints multiplied by 100.

When the Population or Characteristic Changes

The same interpretation structure works across settings, but the context in the sentence must change with the study. Do not reuse a population name from a previous problem or assume “success” has its ordinary everyday meaning. In a survey, success is simply the category being counted—for example, students who report using a particular transit option.

A helpful final check is to compare the sentence with the original question. If the question asks about the proportion of students who use a campus shuttle at least once a week, the interpretation should say exactly that. “Students who use transportation” is broader, while “survey respondents who said yes” describes the sample rather than the target population.

Worked Example: Weekly Use of a Campus App

Suppose a campus technology team wants to estimate the proportion of students who use a campus navigation app at least once a week. In a hypothetical random sample of 240 students from a campus population of 10,000, 126 report weekly use. Interpret a 99% one-proportion \(z\)-interval.

Let \(p\) be the true proportion of all students on this campus who use the navigation app at least once a week. The sample is random. The 10% limit is \(0.10(10{,}000)=1{,}000\), and \(240\leq1{,}000\). There are 126 successes and \(240-126=114\) failures, both at least 10. The conditions support using a one-proportion \(z\)-interval.

The sample proportion is \(126/240=0.525\). For a 99% confidence interval, \(z^*\approx2.5758\). Calculate the estimated standard error:

SE_{\hat{p}} =\sqrt{\frac{(0.525)(1-0.525)}{240}} =\sqrt{0.0010390625} \approx0.0322345

Using the unrounded standard error, the margin of error is approximately \(2.5758(0.0322345\ldots)=0.0830296\). The endpoints are:

0.525-0.0830296\ldots\approx0.4420, \qquad 0.525+0.0830296\ldots\approx0.6080

We are 99% confident that the true proportion of all students on this campus who use the navigation app at least once a week is between \(0.4420\) and \(0.6080\). Equivalently, we are 99% confident that about 44.2% to 60.8% of all students on this campus use the app at least once a week. The upper endpoint being above 0.50 does not mean that the sample contained more than 100% of students; it is a proportion estimate expressed on a scale from 0 to 1.

Common Mistakes and What a Full-Credit Answer Says

Describing only the sample. “Between 32.8% and 47.2% of the surveyed adults got vaccinated” does not state what the interval estimates. Name the target population and say “the true proportion of all adults in the town.”
Leaving out the characteristic. “We are 95% confident the proportion is between 0.3284 and 0.4716” may not be meaningful without context. Say what the adults did or had.
Calling an endpoint a count. An endpoint such as \(0.40\) is a proportion, or 40%, not 40 people. Attach it to the population proportion and the characteristic.
Mixing decimals and percentages. If \(0.3284\) is converted to 32.84%, convert the other endpoint too. Do not write “between 0.3284 and 47.16%.”
Changing the population. A sample from one campus does not, by itself, represent all college students. Keep the conclusion within the population described by the sampling plan.
Claiming the sample proportion is the true proportion. The sample proportion is the observed estimate. The interval reflects uncertainty about the population proportion.
Writing that a percentage of individuals are in the interval. The endpoints bound an estimate of a proportion; they do not identify which individuals are included.
Skipping the conditions because a calculator returned endpoints. A complete procedure checks random sampling, the 10% condition when appropriate, and the observed success and failure counts.

AP Exam Tip: A full-credit interpretation identifies the confidence level, the true proportion, the population, the characteristic, and both endpoints. Make each endpoint a proportion or percentage, and keep the population and characteristic consistent with the question.

Key Takeaway

A one-proportion confidence interval is a range of plausible values for a population proportion. Interpret it with a complete sentence that names the confidence level, the population, and the characteristic being estimated. Report both endpoints in matching units and avoid substituting the sample for the population.

Key takeaway: Say, “We are \(C\)% confident that the true proportion of [population] who [characteristic] is between [lower endpoint] and [upper endpoint].” The interval estimates the population proportion \(p\), not the sample proportion \(\hat{p}\).

Check Your Understanding

For each question, focus on what the interval estimates and how to express its endpoints in context.

A random sample of 100 residents is used to estimate the proportion of all residents in a town who own an electric bicycle. Write an interpretation template that leaves blanks for the confidence level and endpoints.
An interval for the proportion of students at a school who participate in a music group is \((0.18,0.31)\). Express the endpoints as percentages and identify the population proportion being estimated.
A survey sampled customers from one grocery store. Why would an interpretation about all shoppers in the region require evidence that the sample represents that larger population?
A student writes, “We are 95% confident that 40% of the 150 people surveyed have the characteristic.” What is wrong with this sentence if the interval’s endpoints are 0.32 and 0.48?
A 90% interval for the proportion of households in a neighborhood that use a rain barrel is \((0.27,0.39)\). Write a complete interpretation in context using percentages.