Exam-Style Free Response on Two-Proportion Tests

Turn a Long Prompt into a Clear Response

A multi-part free-response question may ask for hypotheses, conditions, calculations, and a conclusion in separate lettered parts. The parts are connected: your parameter definitions determine the hypotheses, the alternative determines the p-value tail, and the study design determines what your conclusion can say. A useful exam skill is to map the prompt to those tasks before calculating.

In “Four-Step Two-Proportion \(z\)-Test Walkthrough,” you learned to organize a test as State, Plan, Do, Conclude. Here, the emphasis is on answering a timed, multi-part item efficiently: identify what each part requests, make each response explicit, and avoid repeating the same explanation in every part. The earlier tutorials “Hypotheses for Comparing Two Population Proportions,” “Conditions for a Two-Proportion \(z\)-Test,” and “Writing a Conclusion for a Two-Proportion Test” provide the details this practice uses.

Key idea: Under time pressure, make a brief response map first: parameters and hypotheses; design and conditions; pooled calculation and p-value; decision and contextual conclusion. Then answer each requested part directly, keeping the group order consistent.

A Timed Response Map

Suppose an item has four parts. Mark the task each part requires before starting: define and test parameters, justify the procedure, calculate the test results, and interpret the evidence. Do not spend time writing a long introduction. A direct, well-supported sentence for each condition is more useful than a broad claim that “the conditions are met.”

State the comparison.
Define each population proportion for the same outcome and write \(H_0:p_1=p_2\). Match the alternative to the question, not to the direction of the observed sample difference.

Justify the test.
Name the two-proportion \(z\)-test. Cite the sampling or assignment method, establish that the groups are independent, check the 10% condition when sampling without replacement, and report the four pooled expected counts for Large Counts.

Show the evidence calculation.
Report the sample proportions and their difference, calculate the pooled proportion and pooled standard error, then give \(z\) and the p-value for the stated alternative.

Answer the research question.
Compare the p-value with \(\alpha\), state whether you reject or fail to reject \(H_0\), and describe the evidence about the population proportions in context.

A compact condition check should include facts, not just condition names. For example, “The two groups are independent random samples, and each sample is no more than 10% of its population” supports the design conditions if the problem gives evidence for those statements. For Large Counts, show the pooled estimate and expected successes and failures in both groups. As explained in “Large Counts Using Expected Successes and Failures in Each Group,” those counts come from the null model of equal proportions.

Formula: For a test of \(H_0:p_1=p_2\), calculate \(\hat{p}_c=(x_1+x_2)/(n_1+n_2)\), then \(SE_{\text{pooled}}=\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1+1/n_2)}\), and \(z=(\hat{p}_1-\hat{p}_2)/SE_{\text{pooled}}\). Use the alternative hypothesis to choose the p-value tail.

In a timed answer, arithmetic should be visible enough to audit. You do not need to write every calculator keystroke, but include the pooled estimate, the standard error substitution, the \(z\)-statistic, and the p-value. A calculator result alone does not explain why that result answers the question.

Worked Multi-Part Responses

Worked Example: Comparing Bike-Helmet Use

Question: A fictional transportation team takes independent random samples of 140 cyclists from each of two large towns. In Town 1, 91 sampled cyclists report wearing a helmet on their most recent ride; in Town 2, 77 do. The team wants to know whether helmet use is more common in Town 1. Each town has at least 1,400 cyclists. At \(\alpha=0.05\), answer: (a) define parameters and state hypotheses; (b) check conditions; (c) calculate the test statistic and p-value; (d) conclude.

(a) State: Let \(p_1\) be the true proportion of cyclists in Town 1 who wore a helmet on their most recent ride, and let \(p_2\) be the corresponding true proportion in Town 2. Test \(H_0:p_1=p_2\) against \(H_a:p_1>p_2\). The question asks whether Town 1’s population proportion is higher.

(b) Plan and conditions: Use a two-proportion \(z\)-test. The problem says each group is an independent random sample, and cyclists are counted in only one town. Each sample is at most 10% of its population because \(140\le0.10(1{,}400)=140\). Pool the successes to check the null-model counts:

\hat{p}_c=\frac{91+77}{140+140} =\frac{168}{280} =0.60

For either group, the expected number of successes is \(140(0.60)=84\), and the expected number of failures is \(140(0.40)=56\). The four expected counts are 84, 56, 84, and 56; each is at least 10. The conditions support the test.

(c) Do: The sample proportions are \(\hat{p}_1=91/140=0.65\) and \(\hat{p}_2=77/140=0.55\), so their difference is \(0.10\). The pooled standard error is:

SE_{\text{pooled}} =\sqrt{0.60(0.40)\left(\frac{1}{140}+\frac{1}{140}\right)} =\sqrt{0.24\left(\frac{2}{140}\right)} =\sqrt{0.00342857} \approx0.05855

Thus:

z=\frac{0.65-0.55}{0.05855} \approx1.708

Because \(H_a:p_1>p_2\), use the upper-tail probability: \(P(Z\ge1.708)\approx0.0438\), rounded to four decimal places.

(d) Conclude: Since \(0.0438<0.05\), reject \(H_0\). The data provide convincing evidence that the true proportion of cyclists who wore a helmet on their most recent ride is higher in Town 1 than in Town 2.

This response makes each part easy to locate. It also keeps the parameter order, observed difference, alternative, and conclusion aligned from beginning to end.

Worked Example: Testing for a Difference in Composting

Question: A fictional county takes independent random samples of 100 households from one large neighborhood and 120 from another. In the first sample, 42 households report composting food scraps; in the second, 60 do. Each neighborhood has at least ten times its sample size. At \(\alpha=0.05\), test whether the true composting proportions differ.

State: Let \(p_1\) be the true proportion of households in the first neighborhood that compost food scraps, and \(p_2\) the true proportion in the second neighborhood. Test \(H_0:p_1=p_2\) against \(H_a:p_1\ne p_2\), since the question asks whether the proportions differ in either direction.

Plan: The two groups are independent random samples, and a household appears in only one group. Each sample is at most 10% of its neighborhood because of the stated population sizes. The pooled proportion is:

\hat{p}_c=\frac{42+60}{100+120} =\frac{102}{220} \approx0.4636

Under the null model, the expected successes are \(100(0.4636)\approx46.36\) and \(120(0.4636)\approx55.64\). The expected failures are \(100(1-0.4636)\approx53.64\) and \(120(1-0.4636)\approx64.36\). All four expected counts exceed 10, so the Large Counts condition is met.

Do: The sample proportions are \(\hat{p}_1=42/100=0.42\) and \(\hat{p}_2=60/120=0.50\); the observed difference is \(-0.08\). Use the pooled standard error:

SE_{\text{pooled}} =\sqrt{0.4636(1-0.4636)\left(\frac{1}{100}+\frac{1}{120}\right)} \approx\sqrt{0.0045595} \approx0.06752

Then \(z=(-0.08)/0.06752\approx-1.185\). For a two-sided alternative, outcomes at least as far from zero as this statistic count in both tails. Therefore, \(p\text{-value}=2P(Z\le-1.185)\approx0.2361\), rounded to four decimal places.

Conclude: Since \(0.2361>0.05\), fail to reject \(H_0\). The samples do not provide convincing evidence that the true proportions of households composting food scraps differ between the two neighborhoods. This does not prove the proportions are equal.

Worked Example: A Randomized Trial of a Reminder

Question: A fictional clinic randomly assigns 200 patients to receive a text reminder before an appointment and 180 patients to receive the usual notification. Of the reminder group, 85 attend; of the usual-notification group, 58 attend. Does the reminder increase the attendance proportion? Test at \(\alpha=0.05\). Each group contains different patients.

State: Let \(p_T\) be the true proportion of patients like those in this experiment who would attend under the text-reminder condition, and let \(p_C\) be the corresponding proportion under the usual-notification condition. Test \(H_0:p_T=p_C\) against \(H_a:p_T>p_C\). The claim is that the reminder increases attendance.

Plan: Use a two-proportion \(z\)-test. The patients were randomly assigned to the two conditions, so the assignment is a chance-based process; the groups contain different patients and are independent. Since the study uses random assignment rather than sampling without replacement from a population, the 10% sampling condition is not the relevant justification here. Pool the successes:

\hat{p}_c=\frac{85+58}{200+180} =\frac{143}{380} \approx0.3763

Under \(H_0\), the reminder group has \(200(0.3763)\approx75.26\) expected successes and \(200(0.6237)\approx124.74\) expected failures. The usual-notification group has \(180(0.3763)\approx67.74\) expected successes and \(180(0.6237)\approx112.26\) expected failures. All four are at least 10.

Do: The sample proportions are \(\hat{p}_T=85/200=0.425\) and \(\hat{p}_C=58/180\approx0.3222\), giving a difference of approximately \(0.1028\). The pooled standard error and statistic are:

SE_{\text{pooled}} =\sqrt{0.3763(1-0.3763)\left(\frac{1}{200}+\frac{1}{180}\right)} \approx\sqrt{0.0024774} \approx0.04977

z=\frac{0.425-0.3222}{0.04977} \approx2.065

The alternative is upper-tailed, so \(p\text{-value}=P(Z\ge2.065)\approx0.0195\), rounded to four decimal places.

Conclude: Since \(0.0195<0.05\), reject \(H_0\). The experiment provides convincing evidence that the text-reminder condition leads to a higher attendance proportion than the usual-notification condition for patients like those in the experiment. Random assignment supports a cause-and-effect conclusion for the experimental units; without random sampling, do not claim the result necessarily generalizes to all patients.

Common Mistakes and Full-Credit Communication

A multi-part prompt can make students rush from one calculation to the next. Keep an eye out for these specific errors:

Using sample proportions in the hypotheses: Hypotheses describe population parameters. Write \(H_0:p_1=p_2\), not a hypothesis about \(\hat{p}_1\) and \(\hat{p}_2\).
Choosing the alternative after seeing the data: The research question sets the direction. A sample difference that points the other way does not change a one-sided alternative.
Writing “conditions are met” without evidence: Identify the sampling or assignment method, explain group independence, address the 10% condition when it applies, and show all four pooled expected counts.
Using the interval standard error: A test of equal proportions uses the pooled estimate in its standard error. The unpooled standard error is used for a two-proportion interval.
Doubling the wrong probability: For a two-sided test, double one tail area beyond the absolute value of the observed \(z\). Do not report the single-tail area as the two-sided p-value.
Ending with a decision only: “Reject \(H_0\)” is incomplete by itself. State what the evidence indicates about the population proportions, in context. If you fail to reject, do not claim that the proportions are equal.
Overstating an experiment’s reach: Random assignment can support a causal conclusion about the conditions, but it does not by itself make the experimental units representative of a wider population.

AP Exam Tip: Use a final consistency check: Are the parameter definitions in the same order as the subtraction? Does the alternative match the question? Do the conditions refer to the actual design? Does the p-value use the correct tail? Does the conclusion name the outcome, groups, decision, and evidence?

Key Takeaway

A timed free response is strongest when every requested part is answered explicitly and the parts work together. Spend effort on the details that communicate statistical reasoning: defined parameters, specific condition checks, a traceable pooled calculation, the correct p-value, and a contextual conclusion.

Key takeaway: Map the prompt before calculating, keep the group order fixed, and let the research question determine the alternative. A complete answer connects the study design and null model to the test, then connects the p-value decision to a careful conclusion in context.

Check Your Understanding

For each prompt, focus on what a complete timed response should include.

A question asks whether Group 1 has a lower proportion than Group 2. Write the alternative using \(p_1\) and \(p_2\), and identify the p-value tail.
For \(x_1=36\), \(n_1=90\), \(x_2=48\), and \(n_2=120\), calculate the pooled proportion and the four expected success and failure counts. Does the Large Counts condition hold?
What specific information should a response give when it checks the random and independence conditions?
A two-sided test has \(z=-1.50\). Explain which standard normal areas belong in its p-value.
A test gives \(p\text{-value}=0.03\) at \(\alpha=0.05\). State the decision and describe the evidence a conclusion should report.