AP Syllabus focus:
‘For combined samples, calculate the pooled proportion (pc) which assumes H0 is true (p1 = p2). - Check if the expected number of successes (npc) and failures [n(1-pc)] for each sample are both greater than or equal to a certain value, typically 5 or 10, to ensure that the sampling distribution of (p̂1 - p̂2) is approximately normal. This check validates the use of the z-test for the difference in proportions.’
Assessing whether the sampling distribution of p̂₁ − p̂₂ is approximately normal is essential before performing a two-sample z-test for population proportions, ensuring valid statistical inference.
Assessing Normality for Inference
Before carrying out a significance test for the difference between two population proportions, AP Statistics requires verifying that the sampling distribution of the statistic p̂₁ − p̂₂ behaves approximately normally. This condition ensures that the z-test is appropriate and that p-values derived from the standard normal distribution are meaningful. The normality check relies directly on the assumption that the null hypothesis (H₀: p₁ = p₂) is true, which allows the use of a pooled proportion, a central tool in evaluating expected counts. The goal is to confirm that each sample contains enough expected successes and failures for the normal approximation to hold.
Using the Pooled Proportion Under the Null Hypothesis
The sampling distribution of p̂₁ − p̂₂ depends on the true population proportions. Because the null hypothesis states p₁ = p₂, the inference procedure substitutes a combined or pooled estimate of the shared proportion. This pooled value forms the basis for calculating expected counts in each sample.
Pooled Proportion (): A combined estimate of the common population proportion under the assumption that , computed from total successes and total observations across both samples.
Once the pooled proportion is obtained, it becomes the reference point for evaluating whether each sample has sufficiently large expected successes and failures.
Expected Successes and Failures
To justify the normal approximation, both samples must meet minimum expected-frequency thresholds. These thresholds help ensure that the central limit effect adequately stabilizes the sampling distribution.
EQUATION
= sample size
= pooled proportion
= complement of the pooled proportion
These values must be checked separately for each sample so that all expected counts reflect the combined estimate of proportion implied by the null hypothesis.
Normality is considered sufficiently justified when each expected value meets commonly used cutoffs.
Minimum Cutoffs for Normality
The AP syllabus emphasizes that expected successes and failures should be at least 5 or 10. This range acknowledges variation in teacher choice and textbook conventions but rests on the same principle: the more expected observations contributing to each category, the more stable the sampling distribution of p̂₁ − p̂₂.
Key expectations for both samples:
Expected successes should be ≥ 5–10.
Expected failures should be ≥ 5–10.
All conditions must be evaluated using p₁ = p₂, which requires the pooled proportion.
Why Expected Counts Rely on the Pooled Proportion
The use of p_c is essential because the sampling distribution being evaluated presumes the null hypothesis is true. Unlike conditions used for one-sample tests or confidence intervals, which use the observed sample proportions, the two-proportion significance test must assess normality under the assumption of equal population proportions. This allows:
A unified estimate that reflects the null-hypothesized equality.
Expected counts that match the distribution used for computing the test statistic.
A consistent foundation for determining whether the standard normal model is suitable.
This alignment between assumptions and calculations ensures that the test statistic’s distribution corresponds to the model used in hypothesis testing.
Layered Structure of the Normality Check
To maintain clarity, the normality condition for two-sample proportion inference can be viewed as a multi-step logic chain grounded in the syllabus:
Assume H₀ is true (p₁ = p₂).
Compute the pooled proportion, incorporating total successes and total sample size.
Calculate expected counts for each sample:
Expected successes: n p_c
Expected failures: n(1 − p_c)
Evaluate minimum thresholds (≥ 5 or 10 for both categories).
Confirm approximate normality, allowing the z-test to proceed.
Importance for the Two-Sample z-Test
The sampling distribution of p̂₁ − p̂₂ approximates normality only when both samples contain enough expected observations in each category. If either sample falls below the minimum threshold for successes or failures, the resulting distribution may be skewed, violating the assumptions of the z-test. When conditions are met, the inference procedure can safely rely on the properties of the standard normal distribution to compute a valid test statistic and p-value.
Under these conditions, the sampling distribution of (p̂₁ − p̂₂) is approximately normal, centered at the true difference in population proportions and with spread given by its standard error.

Normal distribution model for the sampling distribution of p^1−p^2\hat p_1 - \hat p_2p^1−p^2 when the success–failure and independence conditions are satisfied. The curve is centered at zero under the null hypothesis, with shaded tails illustrating extremity relative to the test statistic. The shading represents p-value regions, which go slightly beyond this subsubtopic’s focus but clarify how the normal model functions in inference. Source.
Summary of Key Terminology
To reinforce the conceptual structure:
Pooled proportion: Combines successes across both samples assuming p₁ = p₂.
Expected successes/failures: Computed using sample sizes and the pooled proportion.
Normality condition: Requires sufficiently large expected counts to approximate a normal distribution.
This success–failure condition ensures that each underlying binomial distribution is close enough to symmetric and bell-shaped for the normal approximation to be reasonable.

Binomial sampling distributions becoming more symmetric and approximately normal as sample sizes increase. This illustrates why expected successes and failures must exceed minimum thresholds to justify a normal model. Although drawn for a single proportion rather than a difference of two proportions, it conveys the same success–failure logic supporting normality conditions. Source.
FAQ
When the pooled proportion is near 0 or 1, expected successes or failures may fall below the minimum threshold even with large samples.
In such cases, the normal approximation becomes unreliable because:
The binomial distribution becomes heavily skewed.
The standard error is underestimated or unstable.
Researchers may need to:
Increase the sample sizes.
Consider alternative, non-normal-based methods if available in more advanced settings.
For a two-proportion significance test, no. The expected counts must reflect the assumption that the two population proportions are equal.
Using observed proportions would:
Violate the logic of assessing normality under the null hypothesis.
Produce inconsistent expected counts for the test statistic and normal model.
Only the pooled proportion aligns all components of the test with the null hypothesis.
Normality must hold for each sample’s contribution to the statistic p̂1 − p̂2.
A jointly adequate total does not guarantee:
That one group does not have too few expected successes.
That the other does not have too few expected failures.
If either sample fails the threshold, the resulting sampling distribution may be skewed, making the z-test invalid.
The threshold of 5 is more permissive, while 10 is more conservative.
In practice:
Using 10 reduces the risk of the normal approximation breaking down, especially with moderate skew.
Using 5 may be acceptable with balanced samples and pooled proportions near 0.5.
Teachers often choose the stricter rule to avoid borderline cases where inference becomes unreliable.
Yes, unequal sample sizes can influence expected counts and therefore affect normality.
Situations where issues arise:
A small sample paired with a moderate pooled proportion may fail the success–failure condition even if the larger sample passes easily.
The validity of the z-test hinges on both samples meeting the threshold.
Thus, unequal sample sizes require especially careful checking of expected counts in each group.
Practice Questions
Question 1 (1–3 marks)
A researcher plans to carry out a two-sample z-test comparing two population proportions. The null hypothesis states that p1 = p2. The researcher calculates a pooled proportion and finds the expected number of successes and failures for each sample.
Explain why the pooled proportion must be used when checking the normality condition for this test.
Question 1
1 mark: States that the pooled proportion is required because the null hypothesis assumes the population proportions are equal.
1 mark: Explains that expected successes and failures must be calculated using this assumption.
1 mark: States that this ensures the sampling distribution used for the test matches the conditions under which the test statistic is derived.
Question 2 (4–6 marks)
A school investigates whether two year groups have different proportions of students who regularly complete their homework. A random sample of 80 students from Year 10 and 70 students from Year 11 is taken. Using the null hypothesis that the population proportions are equal, the pooled proportion is calculated as 0.55.
(a) Calculate the expected numbers of successes and failures for each year group under the null hypothesis.
(b) Assess whether the normality condition for the two-sample z-test is met.
(c) Explain why this condition must be satisfied before using the z-test.
Question 2
(a)
1 mark: Correct expected successes for Year 10: 80 × 0.55 = 44.
1 mark: Correct expected failures for Year 10: 36.
1 mark: Correct expected successes for Year 11: 70 × 0.55 = 38.5.
1 mark: Correct expected failures for Year 11: 31.5.
(b)
1 mark: States that all expected counts exceed the required threshold (typically 5 or 10).
1 mark: Correctly concludes that the normality condition is met.
(c)
1 mark: Explains that the condition ensures the sampling distribution of the difference in sample proportions is approximately normal.
1 mark: States that this justifies the use of the z-test, which relies on the normal model for valid inference.
