Exam-Style Free Response on a One-Proportion Test

Turning a Multi-Part Prompt into One Connected Response

A free-response question may ask for hypotheses, conditions, a calculation, and a conclusion in separate lettered parts. Those parts are not separate statistical tasks: each one sets up the next. A clear response follows the State, Plan, Do, Conclude structure from Four-Step Process for a One-Proportion z-Test, while making sure every requested part is easy to find.

The new skill here is managing the prompt as you work. Before calculating, identify the population and characteristic, the benchmark, the direction of the research question, and the significance level. Then use those decisions consistently. In particular, choose the alternative from the wording of the question—not from whether the sample proportion happens to be high or low.

Key idea: Treat the lettered parts as a checklist, but write them as a connected argument. The hypotheses identify the question, the conditions justify the procedure, the calculation measures the evidence, and the conclusion answers the question.

A Timed Response Plan

Under time pressure, a brief outline can prevent a missing condition or an unsupported conclusion. A practical approach is to spend a short moment marking the prompt before writing: underline the population, circle the benchmark, mark words such as “higher,” “lower,” or “different,” and locate \(\alpha\). Then work through the steps below. Adjust the time to the length and point value of the question; do not sacrifice the condition checks simply to reach the calculator faster.

State the target.
Define \(p\) in context, write \(H_0:p=p_0\), choose \(H_a\) to match the question, and record \(\alpha\).

Plan the test.
Name a one-proportion \(z\)-test. Justify the Random condition, the 10% condition when sampling without replacement from a finite population, and the test’s Large Counts condition.

Do the calculation.
Find \(\hat{p}\), calculate the null standard error and \(z\), and use the tail or tails specified by \(H_a\) to obtain the p-value.

Conclude in context.
Compare the p-value with \(\alpha\), state whether you reject or fail to reject \(H_0\), and answer the research question about the population proportion.

A complete condition check gives evidence, not just a label. For a one-proportion \(z\)-test, use \(p_0\) for the Large Counts condition: \(np_0\geq10\) and \(n(1-p_0)\geq10\). If observations are sampled without replacement from a finite population of size \(N\), check \(n\leq0.10N\). The Random condition concerns how the data were obtained and the population the sampling process represents. These distinctions were developed in earlier tutorials, including Checking the Success-Failure Condition for Tests and Independence and the 10 Percent Condition.

Formula: For a one-proportion \(z\)-test of \(H_0:p=p_0\), calculate:

\hat{p}=\frac{x}{n} \qquad SE_0=\sqrt{\frac{p_0(1-p_0)}{n}} \qquad z=\frac{\hat{p}-p_0}{SE_0}

The alternative determines the p-value: use the lower tail for \(H_a:p<p_0\), the upper tail for \(H_a:p>p_0\), and both tails for \(H_a:p\ne p_0\).

Worked Examples

Worked Example: A Multi-Part Question About Reusable Bottles

A fictional environmental group asks whether more than 60% of visitors to a nature center bring a reusable water bottle. A random sample of 200 visitors from a population of 6,000 includes 132 who brought one. Test the group’s question at \(\alpha=0.05\). In a timed response, the parts ask you to state hypotheses, check conditions, calculate the test statistic and p-value, and give a conclusion.

State: Let \(p\) be the true proportion of visitors to this nature center who bring a reusable water bottle. The question asks whether the proportion is greater than 0.60, so \(H_0:p=0.60\) and \(H_a:p>0.60\). The significance level is \(\alpha=0.05\).

Plan: Use a one-proportion \(z\)-test. The visitors were randomly sampled, meeting the Random condition for the population represented by the visitor list. The sample was selected without replacement from 6,000 visitors, and \(200\leq0.10(6{,}000)=600\), so the 10% condition is met. Under \(H_0\), the expected success count is \(np_0=200(0.60)=120\), and the expected failure count is \(n(1-p_0)=200(0.40)=80\). Both are at least 10, so the Large Counts condition is met.

Do: Here \(x=132\) and \(n=200\). The sample proportion and null standard error are:

\hat{p}=\frac{132}{200}=0.66 \qquad SE_0=\sqrt{\frac{0.60(0.40)}{200}} =\sqrt{0.0012} \approx0.03464

Using the unrounded standard error:

z=\frac{0.66-0.60}{\sqrt{0.60(0.40)/200}} \approx1.7321

Because \(H_a\) is right-tailed, the p-value is \(\text{normalcdf}(1.7321,1\text{E}99,0,1)\approx0.0416\), rounded. Since \(0.0416<0.05\), reject \(H_0\).

Conclude: The sample provides convincing evidence that more than 60% of visitors to this nature center bring a reusable water bottle. The conclusion concerns the true proportion of visitors, not just the 66% observed in this sample.

Worked Example: A Lower Proportion of Late Bus Arrivals

A fictional school district is evaluating a revised bus schedule. Staff want to know whether fewer than 25% of students on a particular route arrive late. A random sample of 240 students from a population of 4,000 on that route includes 48 who arrived late. Test the question at \(\alpha=0.05\).

State: Let \(p\) be the true proportion of students on this route who arrive late under the revised schedule. The question asks whether this proportion is below 0.25, so \(H_0:p=0.25\), \(H_a:p<0.25\), and \(\alpha=0.05\).

Plan: Use a one-proportion \(z\)-test. The students were randomly sampled, meeting the Random condition for students on this route. The sample was selected without replacement from 4,000 students, and \(240\leq0.10(4{,}000)=400\), so the 10% condition is met. Under \(H_0\), the expected late count is \(np_0=240(0.25)=60\), and the expected not-late count is \(n(1-p_0)=240(0.75)=180\). Both counts are at least 10, so the Large Counts condition is met.

Do: The sample proportion and null standard error are:

\hat{p}=\frac{48}{240}=0.20 \qquad SE_0=\sqrt{\frac{0.25(0.75)}{240}} =\sqrt{0.00078125} \approx0.02795

Using the unrounded standard error:

z=\frac{0.20-0.25}{\sqrt{0.25(0.75)/240}} \approx-1.7889

The alternative is left-tailed, so the p-value is \(\text{normalcdf}(-1\text{E}99,-1.7889,0,1)\approx0.0368\), rounded. Since \(0.0368<0.05\), reject \(H_0\).

Conclude: The sample provides convincing evidence that fewer than 25% of students on this route arrive late under the revised schedule. This test supports a conclusion about the route’s population proportion, but it does not by itself establish that the revised schedule caused a change.

Worked Example: Does a Packaging Rate Differ From Its Target?

A fictional snack company wants to know whether the proportion of packages containing at least the labeled amount differs from 50%. A random sample of 225 packages from a production period includes 126 that meet or exceed the labeled amount. Test the question at \(\alpha=0.05\).

State: Let \(p\) be the true proportion of packages from this production period that contain at least the labeled amount. Because the question asks whether the proportion differs from 0.50 in either direction, \(H_0:p=0.50\), \(H_a:p\ne0.50\), and \(\alpha=0.05\).

Plan: Use a one-proportion \(z\)-test. The packages were randomly sampled from the production period, meeting the Random condition for that period. The sample was selected without replacement from a population of 5,000 packages, and \(225\leq0.10(5{,}000)=500\), so the 10% condition is met. Under \(H_0\), the expected success count is \(np_0=225(0.50)=112.5\), and the expected failure count is \(n(1-p_0)=225(0.50)=112.5\). Both are at least 10, so the Large Counts condition is met.

Do: Calculate the sample proportion and null standard error:

\hat{p}=\frac{126}{225}=0.56 \qquad SE_0=\sqrt{\frac{0.50(0.50)}{225}} =\sqrt{\frac{0.25}{225}} =\frac{1}{30} \approx0.03333

Using the unrounded standard error, the test statistic is:

z=\frac{0.56-0.50}{\sqrt{0.50(0.50)/225}} =1.8

For a two-sided alternative, results at least as far from zero as \(1.8\) count in either direction. The p-value is \(2\text{normalcdf}(1.8,1\text{E}99,0,1)\approx0.0719\), rounded. Since \(0.0719>0.05\), fail to reject \(H_0\).

Conclude: The sample does not provide convincing evidence that the proportion of packages from this production period containing at least the labeled amount differs from 50%. Failing to reject does not establish that the true proportion is exactly 50%.

Common Mistakes and AP Exam Tips

On a multi-part item, a correct calculator result cannot repair a mismatched hypothesis or an unsupported condition check. Before moving to the next part, check that each answer uses the same population, characteristic, benchmark, and alternative.

Writing hypotheses about the sample. Hypotheses describe the population proportion \(p\), not the sample proportion \(\hat{p}\). Define \(p\) in context before writing \(H_0\) and \(H_a\).
Choosing a direction from the sample result. The research question determines the alternative. “More than” means \(p>p_0\), “fewer than” means \(p<p_0\), and “differs” means \(p\ne p_0\).
Writing “conditions are met” without evidence. Name the random process, show the 10% comparison when relevant, and calculate both expected counts using \(p_0\). The test’s Large Counts check is not based on \(x\) and \(n-x\).
Using a tail that does not match \(H_a\). The alternative—not the sign of \(z\)—determines which area counts toward the p-value. A two-sided test includes both tails.
Stopping after the p-value. State the decision at the given \(\alpha\), then answer the question in context. Use “reject \(H_0\)” or “fail to reject \(H_0\),” not “accept \(H_0\).”
Overstating the conclusion. A significant test gives convincing evidence for the alternative; it does not prove the alternative. A nonsignificant result does not prove the null.

AP Exam Tip: Make the response scannable. Use labels such as State, Plan, Do, and Conclude, or answer each lettered part directly. A strong response defines the parameter, states both hypotheses, justifies every relevant condition, shows the statistic and p-value, and ends with a decision and a contextual conclusion.

Key Takeaway

A timed multi-part test question is best handled as one connected argument. Set up the population question first, justify the test before calculating, use the alternative to select the p-value tail, and finish by interpreting the decision in context.

Key takeaway: Answer every requested part, but keep the logic linked: hypotheses identify the claim, conditions support the procedure, the p-value measures evidence under the null, and the conclusion addresses the population question.

Check Your Understanding

Use the response structure from this tutorial to answer each question.

A prompt asks whether fewer than 18% of residents use a community composting site. What wording determines the direction of \(H_a\), and what should \(p\) describe?
A random sample of 180 items is selected without replacement from a population of 2,000. Check the 10% condition by showing the comparison.
For a test of \(H_0:p=0.40\) with \(n=150\), calculate the two expected counts for the Large Counts condition and state whether it is met.
A test has \(H_a:p\ne p_0\) and \(z=-2.05\). Which tail or tails are used to find its p-value?
A test at \(\alpha=0.05\) produces a p-value of 0.12. State the decision and describe what a careful conclusion should—and should not—claim.