Setting Up a Test for the Difference Between Two Population Proportions: Study Guide (Statistics 3.12)

When the question compares two proportions

In Topic 3.11 you used a confidence interval to judge whether two population proportions differ. A significance test answers the same kind of question in a different way. It starts by assuming there is no difference, then asks how surprising the observed difference in sample proportions would be if that were true.

Use a two-sample z-test for the difference between two population proportions when all of these hold:

• There are two groups: two populations, each with its own random sample, or two treatments in a randomized experiment.
• The response is categorical with two outcomes, so each group gives a proportion of “successes”.
• The question asks whether there is convincing evidence of a difference (or of a difference in a stated direction), not for an estimate of its size.

Compare the procedures you know so far:

The question…	Procedure
estimates one proportion	one-sample z-interval for p
tests a claimed value of one proportion	one-sample z-test for p
estimates a difference between two proportions	two-sample z-interval for p₁ − p₂
asks for evidence of a difference between two proportions	two-sample z-test for p₁ − p₂

Write the name in full. “Two-sample z-test for the difference between two population proportions” is clear; “z-test” alone is not.

As in Topic 3.5, a full test has four parts: hypotheses, method and conditions, calculations, conclusion. This topic covers the first two. Topic 3.13 does the calculations and the conclusion.

Defining the parameters

The hypotheses are about two parameters, so you define both. Each definition names:

• that it is a true (population) proportion;
• the response variable (what counts as a success);
• the population (or, in an experiment, the treatment) it belongs to.

Weak: “p₁ = proportion of Seabrook people, p₂ = proportion of Thornby people.” Strong: “p₁ = the true proportion of all adults in Seabrook who can swim 50 metres; p₂ = the true proportion of all adults in Thornby who can.”

For an experiment, describe the treatment: “p₁ = the true proportion of learners like these who would pass first time after the simulator course.”

Subscripts that match the groups (p_S and p_T) are often clearer than 1 and 2. Whatever you choose, keep it for the whole answer.

Writing the hypotheses

The null hypothesis says there is no difference:

H₀: p₁ = p₂, or equivalently H₀: p₁ − p₂ = 0.

The alternative hypothesis says what you are looking for evidence of. Read it from the question’s wording, before you look at the data.

Question wording	Alternative (two equivalent forms)
“Is p₁ greater, higher, more likely…?”	Hₐ: p₁ > p₂, or Hₐ: p₁ − p₂ > 0
“Is p₁ less, lower, reduced…?”	Hₐ: p₁ < p₂, or Hₐ: p₁ − p₂ < 0
“Is there a difference? Does the treatment affect…?”	Hₐ: p₁ ≠ p₂, or Hₐ: p₁ − p₂ ≠ 0

Hypotheses are always about the parameters, p₁ and p₂, never the statistics p̂₁ and p̂₂. You already know the sample proportions; the test asks about the populations.

The pooled proportion

The test is calculated assuming H₀ is true. If p₁ = p₂, both samples are estimating the same proportion, so the best estimate of it combines them. This is the combined (pooled) proportion:

p̂c = (x₁ + x₂) / (n₁ + n₂) = (n₁p̂₁ + n₂p̂₂) / (n₁ + n₂)

Here x₁ and x₂ are the numbers of successes. In words: total successes divided by total individuals. Do not simply average p̂₁ and p̂₂ unless n₁ = n₂.

The three conditions

Condition	What to check	Notes
Randomization	Two independent random samples, one from each population, or a randomized experiment.	Independent means the two samples do not affect each other: no person in both, no pairing.
10%	When sampling without replacement: n₁ ≤ 10% of N₁ and n₂ ≤ 10% of N₂.	Not needed for a randomized experiment.
Normality	n₁p̂c, n₁(1 − p̂c), n₂p̂c and n₂(1 − p̂c) are all at least 10.	Uses the pooled proportion, because the test assumes H₀ is true.

The normality check is the one that differs from Topic 3.10. For an interval you checked the observed successes and failures in each group. For a test you check expected counts based on p̂c. The values need not be whole numbers; 44.4 is fine.

Show the numbers for every condition. “Conditions are met” with no evidence earns nothing.

Worked example 1: independent random samples, one-sided

Context (fictional). A swimming charity compares two coastal towns, Seabrook (about 8,000 adults) and Thornby (about 11,000 adults). Its question: Is the proportion of adults who can swim 50 metres lower in Thornby than in Seabrook? It takes independent random samples: 124 of 160 Seabrook adults and 136 of 200 Thornby adults can swim 50 metres. Set up the test at α = 0.05.

1. Hypotheses. Let p_S = the true proportion of all Seabrook adults who can swim 50 metres, and p_T = the true proportion of all Thornby adults who can.

• H₀: p_S = p_T
• Hₐ: p_S > p_T (the question asks whether Thornby’s proportion is lower)
• Equivalent form: H₀: p_S − p_T = 0 and Hₐ: p_S − p_T > 0.
• α = 0.05

2. Method. Two-sample z-test for the difference between two population proportions.

3. Conditions.

1. Randomization: independent random samples of adults from each town.
2. 10%: 160 ≤ 10% of 8,000 = 800, and 200 ≤ 10% of 11,000 = 1,100.
3. Normality: p̂c = (124 + 136) ÷ (160 + 200) = 260 ÷ 360 ≈ 0.7222. Then n_Sp̂c ≈ 115.6, n_S(1 − p̂c) ≈ 44.4, n_Tp̂c ≈ 144.4 and n_T(1 − p̂c) ≈ 55.6. All four are at least 10.

All three conditions are met, so the test can go ahead (Topic 3.13).

Check. The sample proportions are 0.775 and 0.68, but they do not appear in the hypotheses. The direction came from the word “lower” in the question. If you prefer p_T first, write Hₐ: p_T < p_S; it is the same hypothesis.

Worked example 2: a randomized experiment, two-sided

Context (fictional). A driving school has 180 learner drivers who agree to take part in a study. It randomly assigns 90 to a course that includes simulator lessons and 90 to its standard course. Of the simulator group, 58 pass their test at the first attempt; of the standard group, 49 do. The question: Is there a difference in first-attempt pass rates between the two courses? Set up the test at α = 0.05.

1. Hypotheses. Let p_Sim = the true proportion of learners like these who would pass at the first attempt after the simulator course, and p_Std = the true proportion who would pass after the standard course.

• H₀: p_Sim − p_Std = 0
• Hₐ: p_Sim − p_Std ≠ 0 (the question asks about a difference, with no direction)
• α = 0.05

2. Method. Two-sample z-test for the difference between two population proportions.

3. Conditions.

1. Randomization: the two courses were randomly assigned to the learners.
2. 10%: not needed. The learners are volunteers in an experiment, not a random sample from a population.
3. Normality: p̂c = (58 + 49) ÷ 180 = 107 ÷ 180 ≈ 0.5944. Both groups have n = 90, so the counts are 90 × 0.5944 ≈ 53.5 successes and 90 × 0.4056 ≈ 36.5 failures in each group. All at least 10.

Check. Because the learners volunteered, any conclusion applies to learners like these, not to all learners. Because the courses were randomly assigned, a significant result could be read as cause and effect.

Worked example 3: when the two-sample z-test is not appropriate

(a) Expected counts too small (fictional). A consumer group selects random samples of 120 bulbs from each of two brands’ large production runs. Within a year, 4 Brand A bulbs and 9 Brand B bulbs fail. Is the normality condition met for a test of H₀: p_A = p_B?

p̂c = (4 + 9) ÷ 240 = 13 ÷ 240 ≈ 0.0542. Then n_Ap̂c = n_Bp̂c = 120 × 13/240 = 6.5, which is less than 10. The normality condition fails. The sampling distribution of p̂_A − p̂_B may not be approximately normal, so a z-test is not appropriate. Say so, and do not carry on with the calculation. Larger samples would be needed.

(b) Samples not independent (fictional). A college asks the same 150 students, before and after a careers talk, whether they plan to apply for an apprenticeship. It wants to compare the “before” and “after” proportions. The two sets of answers come from the same students, so they are not independent samples: a student’s “after” answer is linked to their “before” answer. The randomization condition for a two-sample z-test is not met. A method for paired data would be needed, and that is beyond this topic.

Common misconceptions

• Writing hypotheses with p̂₁ and p̂₂. Hypotheses are about the population proportions.
• Choosing Hₐ from the data. “p̂₁ is bigger, so Hₐ: p₁ > p₂” is wrong. The direction comes from the question, decided in advance.
• Putting “≠” or “<” in H₀. The null hypothesis is always no difference.
• Defining the parameters with the samples. “The proportion of the 160 adults sampled” is a statistic.
• Checking normality with observed counts. For the test, use p̂c: n₁p̂c, n₁(1 − p̂c), n₂p̂c and n₂(1 − p̂c).
• Averaging p̂₁ and p̂₂ to get p̂c. Add the successes and add the sample sizes.
• Checking only two counts, or using n ≥ 30. Check all four counts against 10.
• Applying the 10% condition to an experiment. It is only for sampling without replacement.
• Treating “before and after” data on the same people as two independent samples.

Where this leads

Next, Topic 3.13 uses the pooled proportion to calculate the test statistic and p-value and to write the conclusion: see Carrying Out a Test for the Difference Between Two Population Proportions. Try the practice questions now, then use the revision notes and the checklist. To compare this with the confidence-interval approach, revisit Topic 3.11; for the one-proportion set-up, see Topic 3.5.