AP Syllabus focus:
‘Normality is checked by verifying that the conditions for the expected number of successes (np0) and failures [n(1 - p0)], under the assumption that the null hypothesis is correct, are met. This step is crucial to justify the use of the z-test for a population proportion.’
Students must verify that the sampling distribution of the sample proportion is approximately normal when performing a one-sample z-test for a population proportion.
Understanding the Purpose of Normality Checks
Checking normality for inference ensures that the test statistic used in a one-sample z-test behaves according to the standard normal distribution. This requirement matters because the z-test relies on approximating the distribution of the sample proportion with a normal model. When this approximation is justified, probability calculations and p-values accurately reflect the behavior of repeated samples under the null hypothesis.
The Role of the Null Hypothesis in Normality Assessment
Normality is assessed under the assumption that the null hypothesis is true, meaning the population proportion equals a specified value, p₀. Because inference is anchored to this assumption, expected counts of successes and failures must be evaluated using p₀, not the sample proportion. This aligns the model with the hypothesized distribution forming the basis of the test.
Conditions for Approximate Normality
The sampling distribution of the sample proportion is considered approximately normal when two fundamental criteria are met:
Minimum Successes and Failures
To justify normality, both expected successes and expected failures must reach a sufficient minimum. This ensures that the binomial distribution underlying the sample proportion is well-approximated by a normal model.
Expected successes: np₀ ≥ 10
Expected failures: n(1 – p₀) ≥ 10
These thresholds guarantee that the distribution is not excessively skewed and that the z-test yields reliable probability estimates.

This figure compares a binomial distribution to its normal approximation. The bars represent binomial probabilities, while the curve shows the normal density used as an approximation. It includes minor extra detail by displaying counts of successes rather than proportions, though the concept underlying approximate normality is the same. Source.
EQUATION
= sample size
= hypothesized population proportion
= sample size
= hypothesized failure probability
Because these values depend entirely on the null hypothesis, they remain fixed for any given test design. This provides a consistent framework for assessing whether the z-test can be legitimately applied.
Why Expected Counts Matter
Expected successes and failures reflect how often each outcome is anticipated if the null hypothesis were true. When these expected counts are too small, the distribution of the sample proportion becomes noticeably non-normal. This undermines the assumptions behind both the z-statistic and its corresponding p-value. When counts meet the required minimums, however, sampling variability behaves predictably, allowing the normal model to serve as a reasonable approximation.
Evaluating Whether the Conditions Are Met
Students should systematically evaluate the normality condition using a structured approach:
Step-by-Step Normality Check
Identify the sample size, n.
Identify the hypothesized population proportion, p₀, from the null hypothesis.
Compute the expected number of successes using np₀.
Compute the expected number of failures using n(1 − p₀).
Compare the results to the required minimum thresholds of 10 for each category.
If either expected count falls below 10, the normal approximation is not reliable, and the z-test should not be used.
Key Terminology Supporting Normality Assessment
When introducing the z-test, the term sampling distribution arises frequently.
Sampling Distribution: The distribution of a statistic, such as a sample proportion, over all possible samples of the same size drawn from the same population.
Understanding this definition clarifies why the normal model is needed: it predicts the behavior of the statistic across repeated samples.

This figure illustrates simulated sampling distributions of the sample proportion for increasing sample sizes, each overlaid with a normal curve. It demonstrates how larger samples produce distributions that are more tightly clustered and more closely approximate normality. The image adds minor additional detail by also showing the reduction in spread as n increases. Source.
Between expected successes, expected failures, and the sampling distribution framework, students develop a coherent understanding of why normality checks are essential for valid inference. Ensuring compliance with these criteria gives meaning and reliability to the test statistic and the resulting p-value that underpins statistical decision-making.
Interpreting the Outcome of the Normality Check
Meeting the normality conditions signals that the z-test produces trustworthy results for the given context. Failing the conditions, on the other hand, suggests that the binomial distribution cannot be safely approximated by a normal curve. In such cases, alternative methods should be considered, though these fall outside the scope of this subsubtopic. The primary objective is recognizing that normality verification is not optional but a foundational step in performing valid inference for a population proportion.
FAQ
The purpose of the check is to determine whether the sampling distribution of the test statistic will follow the theoretical model assumed under the null hypothesis. Because the z-test is built on the assumption that the null hypothesis is true, all expected counts must be computed using that hypothesised value.
Using the sample proportion instead would incorrectly assess normality based on observed data rather than the model being tested.
A slight miss does not automatically invalidate the test, but it weakens the reliability of the normal approximation.
In such borderline cases, statisticians may:
• Acknowledge the limitation but proceed cautiously.
• Consider alternative inference methods less sensitive to small sample sizes.
• Evaluate whether the sample size or hypothesised proportion could reasonably be adjusted in study design.
A large sample size greatly increases the likelihood of normality, but it does not guarantee it.
If the hypothesised proportion is extremely close to 0 or 1, even a large sample may not produce sufficiently large expected successes and failures. In such cases, the distribution may remain skewed despite the high sample size.
For proportions, the underlying population distribution is always categorical, so its shape does not influence the normality of the sampling distribution.
Normality depends instead on the binomial behaviour of the count of successes. When expected successes and failures are sufficiently large, the binomial distribution becomes approximately symmetric, allowing normal modelling regardless of the original population.
When expected successes and failures are both large, the binomial distribution becomes less skewed.
• If expected successes are small, the distribution is skewed right.
• If expected failures are small, the distribution is skewed left.
• When both are large, the distribution becomes roughly symmetric.
This symmetry is crucial because the z-test assumes a bell-shaped sampling distribution centred at the hypothesised proportion.
Practice Questions
Question 1 (1–3 marks)
A researcher tests whether the proportion of customers who prefer a new product differs from 0.40. Before conducting a one-sample z-test for a population proportion, the researcher checks the normality condition. The sample size is 80.
a) State the values that must be calculated to check the normality condition.
b) Determine whether the normality condition is met when the null hypothesis proportion p0 = 0.40.
Question 1
a) (1 mark)
• Identifies the two expected counts required: expected successes np0 and expected failures n(1 - p0).
b) (2 marks)
• Correctly calculates expected successes: 80 × 0.40 = 32 (1 mark).
• Correctly calculates expected failures: 80 × 0.60 = 48 (1 mark).
Award 1 mark for stating that both are at least 10 and therefore the normality condition is satisfied.
Question 2 (4–6 marks)
A school administrator claims that 70% of students regularly complete their homework. A random sample of 120 students is selected to test this claim using a one-sample z-test for a population proportion.
a) Explain why the normality condition must be checked before carrying out the test.
b) Using the null hypothesis proportion p0 = 0.70, show whether the expected numbers of successes and failures satisfy the normality requirement.
c) Comment on whether a one-sample z-test is appropriate based on your results.
Question 2
a) (1 mark)
• States that the sampling distribution of the sample proportion must be approximately normal for the z-test to be valid.
b) (2–3 marks)
• Correctly calculates expected successes: 120 × 0.70 = 84 (1 mark).
• Correctly calculates expected failures: 120 × 0.30 = 36 (1 mark).
• States that both expected counts exceed 10 and thus meet the normality requirement (1 mark).
c) (1–2 marks)
• States that the one-sample z-test is appropriate because the normality condition is met (1 mark).
• Provides a clear justification referencing the adequacy of expected successes and failures (1 mark).
