How do you finish a one-proportion test in context?
You will be able to: Calculate z, compare p-value with α and state a qualified conclusion.
How do you finish a one-proportion test in context?
An SRS of 100 from 5000 members finds 60 supporters. Before polling, organizers set α=.05 and ask whether the population support share exceeds .50.
A useful starting point: How can repeated chance trials estimate a p-value? →
Words and symbols before equations
- Test statistic z
- Observed-minus-null difference divided by its null standard error.
- Significance level α
- Prechosen Type I error threshold used for the decision.
- Reject H₀
- The data meet the chosen evidence threshold against H₀.
- Fail to reject
- Evidence is insufficient at that threshold, without proving equality.
What this picture assumes
Synthetic study: a simple random sample without replacement from N=100000 independent units. The chosen sizes satisfy the 10% condition. Real studies require their own design checks. Actual count x is rounded from the requested percentage. Normal z inference is withheld when either null expected count is below 10.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- z=2; p-value=0.02275. At α=0.05, reject H₀. Alternative: p > 0.5. This does not give P(H₀).
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Define p, write H₀:p=.50 and Hₐ:p>.50, and check random sampling, 10% and null counts 50/50. Then compute SE₀=√(.5×.5/100)=.05.
The standardized result is z=(.60−.50)/.05=2. Its approximate upper-tail p-value is .0228. Because .0228<.05, reject H₀.
Conclude there is convincing statistical evidence that more than half of the population of club members supports the proposal. The test does not establish causation or guarantee the exact effect size. Report the observed difference and study limitations as well.
A worked example, step by step
A valid two-sided test gives z=1.50 and α=.05. Complete the decision and interpretation.
- Two-sided p=2[1−Φ(1.50)]≈.1336.
- Compare .1336 with .05.
- Fail to reject H₀ because the p-value is larger.
- There is insufficient evidence of a population difference from the benchmark; do not conclude equality or “no effect.”
A significant result is not automatically a large or important difference. Statistical evidence and practical value require separate discussion.
If α changes after the calculation, does the p-value change?
Compare with an explanation
No. It depends on the observed statistic and null model, not the chosen threshold.
Predict. Change one thing. Explain.
Keep the same data and vary α. Explain why the p-value stays fixed even if the formal decision changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
z=2; p-value=0.02275. At α=0.05, reject H₀. Alternative: p > 0.5. This does not give P(H₀).
| Quantity | Value |
|---|---|
| Observed successes x | 60 |
| Observed failures n−x | 40 |
| Actual p̂=x/n | 0.6 |
| Null quantity | Value |
|---|---|
| Expected successes | 50 |
| Expected failures | 50 |
| SE under H₀ | 0.05 |
Synthetic study: a simple random sample without replacement from N=100000 independent units. The chosen sizes satisfy the 10% condition. Real studies require their own design checks. Actual count x is rounded from the requested percentage. Normal z inference is withheld when either null expected count is below 10.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the probability values, reference groups, graph scales or model assumptions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA valid one-sided test of p=.50 versus p>.50 has p̂=.60,n=100. Use α=.01.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Null SE=.05.
- 1 point: z=2 and upper-tail p≈.0228.
- 1 point: Since .0228>.01, fail to reject H₀.
- 1 point: At the 1% level there is insufficient evidence that the population share exceeds .50, despite the sample’s 60% share.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is in the denominator of z?
The null standard error.
RECALL 2Does failing to reject prove the null?
No.
RECALL 3Should α be chosen after seeing p?
No; set it in advance.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you finish a one-proportion test in context?
- z=(p̂−p₀)/√[p₀(1−p₀)/n].
- Reject when p-value≤α; otherwise fail to reject.
- Conclude in terms of Hₐ and the population.
Remember: A significant result is not automatically a large or important difference. Statistical evidence and practical value require separate discussion.
Conditions: Synthetic study: a simple random sample without replacement from N=100000 independent units. The chosen sizes satisfy the 10% condition. Real studies require their own design checks. Actual count x is rounded from the requested percentage. Normal z inference is withheld when either null expected count is below 10.
Refresh Kid · AP Statistics Unit 3 · Objectives 3.7.A, 3.7.B · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.7, objectives 3.7.A, 3.7.B. Framework effective Fall 2026, checked September 17, 2026. Unit 3 includes inference for one and two population proportions, errors and power, and chi-square homogeneity/independence tests; it is part of the revised five-unit course.
Examples and datasets are synthetic, independently authored teaching material. Inference requires a justified design and appropriate counts. Intervals use observed proportions; null tests use their reference-model proportions. The revised CED states chi-square expected counts should be greater than 5; this unit follows that wording even though some companion texts use at least 5. Normal and chi-square inference are approximate. The chi-square goodness-of-fit test is not included in this unit’s official scope.
The Organic Chemistry Tutor companion title and destination were located; the full video was not reviewed. Khan Academy’s destination was checked, but its lesson content was not fully readable by the research tool. OpenStax provides optional reference reading. No provider scripts, questions or graphics were copied. Refresh Kid is not affiliated with these providers.
GitHub’s 3D website collection informed optional spatial inspection. Our original categorical count grid uses self-hosted Three.js with its MIT license. Two rows and three columns organize synthetic people into categorical cells. Each stacked block represents one person. Camera rotation changes only the view; exact counts, expectations and contributions are always available in the 2D table. Inference curves and intervals remain 2D; use the exact table rather than apparent 3D size for comparisons. Complete labeled diagrams, count tables and explanations remain available without 3D.
Independent teacher review and observation of students remain pending. Technical checks do not certify statistical accuracy, accessibility or learning effectiveness. This is a review edition.
Released AP Statistics questions and scoring guides are optional. Older exams use the earlier framework, so check alignment before selecting parts. All practice on this page is original, not official AP material.
Learn → Explore → Practice → Review is informed by the IES learning guide. This implementation has not yet been evaluated with learners.
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