By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
A hypothesis test for one proportion determines whether a sample proportion (p?) provides convincing evidence that a population proportion (p) differs from a claimed value (p?). This is essential on the AP exam because it’s a foundational inference procedure, often tested in Free-Response Questions (FRQs) and multiple-choice. Real-world applications include: - Testing if a new drug’s success rate exceeds the current standard (e.g., does a vaccine work better than 70% efficacy?). - Determining if a factory’s defect rate is higher than advertised (e.g., is the proportion of faulty lightbulbs > 5%?). - Evaluating if a political candidate’s support has changed since the last poll.
normalcdf(lower, upper, 0, 1)
normalcdf(z, 1E99, 0, 1)
normalcdf(-1E99, z, 0, 1)
2 * normalcdf(|z|, 1E99, 0, 1)
1-PropZTest(p?, x, n, H?)
?p?
<p?
>p?
Follow these steps for every one-proportion z-test on the AP exam:
Write H? and H? in context (define p clearly).
Check Conditions (BINS)
SRS? Assume random sampling unless stated otherwise.
Calculate the Test Statistic
OR use 1-PropZTest on TI-84 (faster and less error-prone).
1-PropZTest
Find the P-value
normalcdf
For two-sided tests, double the tail area.
Make a Conclusion in Context
Never say "accept H?"—only "fail to reject."
Interpret the P-value (if asked)
Correction: Always verify n-0.10N (even if the problem doesn’t mention it). The 10% condition ensures independence.
Mistake: Using the sample proportion (p?) instead of p? in the standard error formula.
Correction: The standard error is ?(p?(1–p?)/n), not ?(p?(1–p?)/n). The null hypothesis assumes p = p?.
Mistake: Misinterpreting the p-value as the probability that H? is true.
Correction: The p-value is the probability of the observed data (or more extreme), assuming H? is true. It does not measure the probability of H? itself.
Mistake: Skipping the Normal condition check for small samples.
Correction: If np? < 10 or n(1–p?) < 10, the sampling distribution of p? is not normal, and the z-test is invalid.
Mistake: Writing conclusions without context.
Partial credit is given for correct hypotheses and conditions, even if the test statistic is wrong.
Tricky Distinctions:
Confidence intervals vs. hypothesis tests: A CI estimates p, while a hypothesis test tests a claim about p. They’re related but not the same.
Calculator Pitfalls:
1-PropZInt
Two-sided p-values: The calculator gives the one-tailed p-value for ?p?. You must double it for the correct two-sided p-value.
Common FRQ Setups:
Answer: (A) Explanation: The standard error uses p? (0.05), not p? (0.08), and this is a z-test for proportions (not a t-test).*
Answer: - a. H?: p = 0.70 (true pass rate with new method = historical rate) H?: p > 0.70 (true pass rate with new method > historical rate) - b. Conditions: - Binary: Pass/fail data (yes). - Independent: 50-0.10N (assume N-500, so yes). - Normal: np? = 50(0.70) = 35-10; n(1–p?) = 50(0.30) = 15-10 (yes). - SRS: Random sample (assumed). - c. p? = 40/50 = 0.80 z = (0.80 – 0.70) / ?(0.70(0.30)/50)-1.54 p-value = normalcdf(1.54, 1E99, 0, 1)-0.0618
normalcdf(1.54, 1E99, 0, 1)
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