Practice a College Board-style free response question on a one-sample confidence interval for a proportion. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 3 · Confidence Interval for a Proportion
A polling organization takes a random sample of 200 adults from a large city and finds that 118 of them support a proposed policy. The organization wants to estimate p, the proportion of all adults in the city who support the policy.
A
Identify the appropriate inference procedure and verify that its conditions are met.
✓ Model answer
The appropriate procedure is a one-sample z-interval for a population proportion. Random: the 200 adults are a random sample. 10%: 200 is less than 10% of all adults in a large city, so observations are approximately independent. Large Counts: np̂ = 118 ≥ 10 and n(1 − p̂) = 82 ≥ 10. All conditions are met.
Why it scores: Names the correct procedure AND checks all three conditions (Random, 10%, Large Counts) with the actual numbers 118 and 82. Skipping a condition or checking with p₀ instead of p̂ would lose credit.
B
Construct a 95% confidence interval for p. Show the formula and the values you use, then interpret the interval in context.
✓ Model answer
p̂ = 118/200 = 0.59. The interval is p̂ ± z*·√(p̂(1 − p̂)/n) = 0.59 ± 1.96·√(0.59·0.41/200) = 0.59 ± 1.96(0.0348) = 0.59 ± 0.068, giving (0.522, 0.658). Interpretation: we are 95% confident that the interval from 0.522 to 0.658 captures the true proportion of all adults in the city who support the policy.
Why it scores: Shows the formula with z* = 1.96, computes the standard error and margin of error, gives the interval, AND interprets it as capturing the true population proportion in context. A bare interval with no interpretation would lose the final point.
C
Based on your interval, is there convincing evidence that more than half of all adults in the city support the policy? Justify your answer using the interval.
✓ Model answer
Yes. The entire 95% confidence interval (0.522 to 0.658) lies above 0.50. Since all plausible values of the true proportion are greater than one-half, there is convincing evidence that more than half of all adults in the city support the policy.
Why it scores: Makes a decision (yes) and justifies it specifically — the whole interval is above 0.50 — rather than just restating the interval. Saying 'yes because 0.59 > 0.50' without referencing the interval would earn less.
How to score points on AP Statistics FRQs
Name the procedure. Say "one-sample z-interval for a proportion" — the correct name signals you know which conditions and formula apply.
Check every condition with numbers. Random, 10%, and Large Counts (np̂ and n(1 − p̂)), using the actual counts.
Show the formula and substitution. A bare answer from a calculator can lose method points.
Interpret intervals about the parameter, in context. Refer to all adults in the city and the specific policy.
Justify decisions from the interval. Compare the whole interval to the value in question (here, 0.50).