Build the method
For a one-proportion confidence interval, the usual success/failure check uses observed counts. Five successes and eighty-five failures do not provide ten successes, even though the total sample is ninety. A rare outcome can require a much larger sample than a balanced outcome.
Worked example
For a test with H0:p=0.05 and n=120, use null expected successes 120(0.05)=6 and expected failures 114. Twenty observed successes would not replace the null count six in this check. The test evaluates the sampling distribution specified by the null hypothesis.
Compare the alternatives
For a random sample of forty without replacement from one thousand, the sampling fraction is 40/1000=0.04. This meets the conventional ten-percent check for a negligible dependence approximation. It does not prove that the sampling frame was complete or that responses were measured without error.
Avoid a specific trap
Hypotheses concern population parameters. H0:p=0.30 describes a claim about the target population proportion, whereas an equation about p-hat describes the observed statistic. Define p in context so the equation cannot be mistaken for an abstract number without a population.
Transfer check with explanation
Transfer: a school surveys everyone in one convenient classroom. Even if success and failure counts exceed ten and the class is below ten percent of school enrollment, a random sample of the school was not taken. Numerical conditions cannot repair the convenience selection.
A random sample has 5 successes among 90 trials. For the ordinary one-proportion normal confidence interval, do the observed success/failure counts meet the at-least-ten guideline?
- For an interval, count observed successes and failures separately.
- Successes=5<10; failures=85>10. The normal interval condition fails on successes.
- Why an alternative fails: A large total does not ensure enough successes for this normal interval.
Compare the other answer choices
A large total does not ensure enough successes for this normal interval.
A confidence-interval large-count check uses the observed estimate, not an invented null hypothesis.
The counts need to be sufficiently large, not equal.
Explain your thinking.
Optional written extension. Self-assessed only; it does not change your automatic practice result. These are teaching criteria, not official awarded marks.
A random sample of 90 from 2,000 has 5 successes. (a) check the sampling fraction; (b) check both observed counts; (c) state whether the ordinary normal proportion interval is justified by these checks.
This draft stays on this page only. Use paper if you want to keep your working.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Statistics Course and Exam Description, effective Fall 2026