By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
A confidence interval for one proportion estimates the true proportion (p) of a population based on a sample proportion (p?). It answers questions like: "What percentage of voters support a policy?" or "What’s the defect rate in a factory?" The AP exam tests this concept in multiple-choice and FRQs, often requiring you to construct, interpret, and justify intervals. Example: A pollster surveys 500 voters and finds 60% support a candidate—what’s the plausible range for the true support level?
invNorm(area to left)
invNorm(0.975)
For a typical FRQ (e.g., "Construct and interpret a 95% CI for the proportion of students who prefer online learning"):
Calculate p? = (number of successes) / n.
Check conditions (BINS):
SRS: Assume random sampling unless stated otherwise.
Calculate the interval:
CI = p? ± z* × SE.
Interpret in context:
"We are 95% confident that the true proportion of students who prefer online learning is between [lower bound] and [upper bound]."
Address follow-ups (if asked):
Correction: Always verify n-10% of the population when sampling without replacement. The AP exam loves to test this!
Mistake: Using p instead of p? in the SE formula.
Correction: SE = ?(p?(1 – p?) / n). p is unknown (that’s why we’re estimating it!).
Mistake: Misinterpreting the confidence level (e.g., "There’s a 95% chance p is in the interval").
Correction: The correct interpretation is about the method, not the interval: "We are 95% confident that the interval captures p."
Mistake: Rounding p? too early (e.g., using p? = 0.6 for n = 50, x = 30).
Correction: Keep p? as 30/50 = 0.6 exactly until the final step to avoid rounding errors.
Mistake: Using t instead of z for proportions.
1-PropZInt
invNorm
(D) SRS Answer: (B) The 10% condition may not be met if the population of teens is small (e.g., < 4,000).
FRQ Part: A poll of 1,200 voters finds 52% support Candidate A. Construct a 99% CI for the true proportion of supporters.
invNorm(0.995)
Interpretation: We are 99% confident the true proportion of voters supporting Candidate A is between 48.3% and 55.7%.
Multiple Choice: Which change would decrease the margin of error for a 95% CI for p?
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