Exam-Style Free Response on a One-Proportion Interval

Turn the Four Steps into a Timed Response

In Complete Four-Step Confidence Interval Procedure for a Proportion, you practiced connecting the population parameter, conditions, calculation, and conclusion. A timed free-response question asks you to do the same work while responding to several parts, which may ask for the interval, its interpretation, and what it suggests about a benchmark.

The key is to organize the reasoning rather than rush straight to the calculator. Read every part first, identify what the interval is meant to estimate, and keep the response tied to the stated population and characteristic. As in earlier tutorials, the one-proportion \(z\)-interval is centered at \(\hat p=x/n\), and its estimated standard error uses \(\hat p\), not the unknown population proportion \(p\).

Key strategy: Treat each part as a link in one chain: identify \(p\), justify the one-proportion \(z\)-interval, calculate it, interpret it, and then use the interval to answer any follow-up question. Do not repeat the entire calculation when a later part only asks what the endpoints imply.

A Quick Plan for Exam Timing

For a question with several parts, spend a brief moment mapping the task before writing. A useful approach for a 10- to 12-minute item is to use about one minute to read and mark the requests, a few minutes for the parameter and conditions, several minutes for the calculation and interpretation, and the remaining time to answer follow-up parts and check that each conclusion is in context. Adjust the pace to the question; the goal is not to follow a rigid schedule but to avoid leaving a requested part unanswered.

Circle the target. Identify the population and the yes/no characteristic. This determines what \(p\) means.
Mark the requested confidence level. The critical value depends on it. Do not default to 95% if the question specifies another level.
Check conditions before calculating. Use the sampling description, population size if given, and observed success and failure counts.
Keep unrounded values during the calculation. Round the final endpoints consistently, usually to three decimal places unless the question specifies otherwise.
Read all follow-up parts. A question may ask whether a target value is plausible or what the interval implies for a population count.

A confidence interval does not prove a claim merely because the claim falls inside it. As covered in Using an Interval to Evaluate a Claimed Proportion, an included value is plausible according to the interval; an excluded value is evidence against it at the corresponding confidence level. State the strength and limits of that comparison accurately.

Worked Examples

Worked Example: A Timed Survey About Bike-Lane Support

A fictional city has 3,200 adult residents. A simple random sample of 200 adults is asked whether they support adding a protected bike lane on a major street. Of those sampled, 120 say yes. A free-response question asks you to: (a) define the parameter and construct a 95% confidence interval, (b) interpret the interval, and (c) assess whether the results provide evidence that more than half of the city’s adults support the proposal.

State. Let \(p\) be the true proportion of all adult residents of this city who support adding the protected bike lane. We want a 95% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The sample is a simple random sample, so the random condition is met. The sample was taken without replacement from a finite population. For the 10% condition, \(0.10(3{,}200)=320\), and \(200\leq320\), so the sample is no more than 10% of the population. There are \(120\) observed successes and \(200-120=80\) observed failures; both counts are at least 10, so the Large Counts condition is met. These checks support using the interval.

Do. The sample proportion is \(\hat p=120/200=0.60\). For 95% confidence, use \(z^*=1.96\). The estimated standard error and margin of error are \(\sqrt{0.60(0.40)/200}=\sqrt{0.0012}\approx0.03464\) and \(1.96\sqrt{0.60(0.40)/200}\approx0.06790\), respectively. Keeping the unrounded standard error in the margin calculation gives the displayed result. Thus,

\hat p\pm z^*\sqrt{\frac{\hat p(1-\hat p)}{n}} =0.60\pm1.96\sqrt{\frac{0.60(0.40)}{200}} =0.60\pm0.06790 \approx(0.5321,\ 0.6679)

Conclude. We are 95% confident that between about 53.21% and 66.79% of all adult residents of this city support adding the protected bike lane.

Use the interval. The value \(0.50\) is below the interval’s lower endpoint, so it is not a plausible value for \(p\) according to this 95% interval. The interval provides evidence that more than half of the city’s adults support the proposal. It does not establish that every adult supports it or explain why residents hold their views.

This example shows why a free-response answer needs both the calculation and the requested follow-up. The interval itself does not say “more than half” until you compare its endpoints with \(0.50\). Because the whole interval is above \(0.50\), the comparison supports that conclusion. The 95% confidence level describes the long-run capture rate of the method, not a probability that the fixed \(p\) is in this particular interval.

Worked Example: Applying a 90% Interval to a Household Estimate

A fictional town has 2,000 households. A random sample of 140 households is asked whether anyone in the household used a public library during the past month. Eighty-four households answer yes. Construct and interpret a 90% confidence interval for the proportion of town households with library use. Then use the interval to estimate a range for the number of households with library use and assess whether a town planning benchmark of 55% is plausible.

State. Let \(p\) be the true proportion of the town’s 2,000 households in which someone used a public library during the past month. We want a 90% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The sample is random. Since households were sampled without replacement, check the 10% condition: \(0.10(2{,}000)=200\), and \(140\leq200\). The observed success count is \(84\), and the failure count is \(140-84=56\); both are at least 10. The random, 10%, and Large Counts conditions support the interval.

Do. Here, \(\hat p=84/140=0.60\). For a 90% confidence level, \(z^*\approx1.645\). The estimated standard error is \(\sqrt{0.60(0.40)/140}=\sqrt{0.001714286}\approx0.04140\). Using the unrounded standard error, the margin of error is \(1.645\sqrt{0.60(0.40)/140}\approx0.06811\).

0.60\pm1.645\sqrt{\frac{0.60(0.40)}{140}} =0.60\pm0.06811 \approx(0.5319,\ 0.6681)

Conclude and apply. We are 90% confident that between about 53.19% and 66.81% of the town’s households had someone use a public library during the past month. Multiplying the interval endpoints by 2,000 gives an estimated range of \(0.5319(2{,}000)=1{,}063.8\) to \(0.6681(2{,}000)=1{,}336.2\), or approximately 1,064 to 1,336 households. The 55% benchmark is inside the interval, so it is plausible according to these data; the interval does not show that the true proportion equals exactly 55%.

When converting an interval for a proportion into a range of population counts, keep the target population and unit clear. Here the unit is a household, not an individual library user. The count range is an approximate way to express the interval in the town’s household units; it is not a separate interval procedure.

Worked Example: A Wide 99% Interval and a Close Benchmark

A fictional regional transit office takes a random sample of 250 riders and asks whether they used a mobile fare pass during the previous week. Of the sampled riders, 145 say yes. Construct and interpret a 99% confidence interval for the proportion of all riders in the region who used the pass. A report also claims that 50% of regional riders used it. Does this interval provide convincing evidence against that claim?

State. Let \(p\) be the true proportion of all riders in the region who used a mobile fare pass during the previous week. We want a 99% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The riders were randomly sampled. The regional population has 5,000 riders, so the 10% check is \(0.10(5{,}000)=500\), and \(250\leq500\). There are \(145\) observed successes and \(250-145=105\) observed failures; both counts are at least 10. The conditions support using the interval.

Do. The sample proportion is \(\hat p=145/250=0.58\). For 99% confidence, \(z^*\approx2.576\). The estimated standard error is \(\sqrt{0.58(0.42)/250}=\sqrt{0.0009744}\approx0.03122\). Using the unrounded standard error, the margin of error is \(2.576\sqrt{0.58(0.42)/250}\approx0.08041\). Therefore,

0.58\pm2.576\sqrt{\frac{0.58(0.42)}{250}} =0.58\pm0.08041 \approx(0.4996,\ 0.6604)

Conclude and assess the claim. We are 99% confident that between about 49.96% and 66.04% of all riders in the region used a mobile fare pass during the previous week. The claimed value \(0.50\) is inside the interval, just above its lower endpoint. Therefore, this interval does not provide convincing evidence against a 50% population proportion. Being inside the interval does not prove the report’s claim is exactly correct.

How to Make the Answer Easy to Score

A strong free-response answer makes each part visible. Use the question’s labels when possible, and give enough detail to show why the procedure applies. The condition checks are not decoration: the random sample supports generalizing to the stated population, the 10% condition supports treating observations as independent when sampling without replacement, and the observed counts support the Normal-based interval.

The calculation should also be auditable. Show \(\hat p=x/n\), name the critical value, display the interval formula, and report endpoints with a consistent level of precision. If you show a rounded standard error and multiply it by \(z^*\), the resulting margin should agree with that displayed multiplication. To avoid rounding mismatches, show the margin as \(z^*\sqrt{\hat p(1-\hat p)/n}\) and report its rounded result.

Do not omit the failure count. For Large Counts, check \(x\geq10\) and \(n-x\geq10\), not just the number of successes.
Do not use the wrong standard error. A one-proportion interval estimates its standard error with \(\hat p\), so use \(\sqrt{\hat p(1-\hat p)/n}\).
Do not interpret the interval as describing individuals. Its endpoints estimate a population proportion, not the range of responses among residents, households, or riders.
Do not call an included benchmark proven. Say that it is plausible according to the interval. If it is outside the interval, say the interval provides evidence against it, rather than saying it is impossible.
Do not skip a requested part. A correct interval alone does not answer a question that also asks for an interpretation or a benchmark comparison.
Do not overstate the conclusion. A confidence interval measures sampling uncertainty under its conditions; it does not account for every possible source of bias, as explained in What Margin of Error Does Not Account For.

AP Exam Tip: A concise full-credit response can still be complete: state \(p\), name the one-proportion \(z\)-interval, support the conditions with evidence, show the calculation, interpret the endpoints in context, and answer each follow-up by comparing the relevant value with the interval.

Key Takeaway

A timed interval question is a sequence of connected decisions, not just a calculator task. Identify what population proportion is being estimated, justify the method, calculate and interpret the interval, and then use its endpoints to address any benchmark or population-count question. Keep the conclusion proportional to what the interval supports.

Key takeaway: In a multi-part response, make the parameter, conditions, interval, contextual interpretation, and follow-up conclusion easy to locate. A benchmark inside the interval is plausible, not proven; a benchmark outside it is evidence against that value, not proof of impossibility.

Check Your Understanding

Use the four-step structure and answer each follow-up in context.

A random sample of 160 residents from a town of 2,400 includes 104 who support a new recreation center. Identify \(p\), check the 10% condition, and calculate the observed success and failure counts.
For the sample in Question 1, state whether the Large Counts condition is met. Give the two counts that justify your answer.
A 95% confidence interval for the proportion of residents who support the center is \((0.58,0.72)\). What does it imply about a benchmark of 60%? Explain without claiming the benchmark is proven.
A student reports a 90% confidence interval as \(0.60\pm1.645(0.04)\), then gives a margin of error of 0.065. What arithmetic inconsistency should the student correct?
Explain why a range of plausible household counts obtained by multiplying interval endpoints by the number of households is still about households, not individual residents.