Why does a test use expected counts under the null?
You will be able to: Check the conditions for a one-sample z-test using p₀.
Why does a test use expected counts under the null?
A random sample contains only 8 yes responses among 100. If the test asks about p₀=.20, the normal-count check uses 20 expected yes responses, not the observed 8.
A useful starting point: How do you turn a question into hypotheses? →
Words and symbols before equations
- Null distribution
- Sampling distribution assuming H₀ and the design model are true.
- Null standard error
- √[p₀(1−p₀)/n], the reference spread for a one-proportion z-test.
- Expected count under H₀
- np₀ or n(1−p₀).
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
The test asks whether data look unusual if p=p₀. Therefore the count check and reference SD use p₀; a confidence interval estimates spread with p̂.
Check random sampling and, without replacement, n≤.10N. Then verify np₀≥10 and n(1−p₀)≥10.
Passing numerical checks cannot establish random selection or remove nonresponse bias. If conditions fail, do not report the ordinary z-test as justified; an exact or simulation-based method may be appropriate after checking the design.
| Feature | Confidence interval | Hypothesis test |
|---|---|---|
| Purpose | Estimate unknown p | Assess a stated p₀ |
| SE uses | Observed p̂ | Null p₀ |
| Normal count check | Observed x and n−x | Expected np₀ and n(1−p₀) |
A worked example, step by step
An SRS of n=100 from N=5000 tests H₀:p=.20; x=8. Check conditions.
- The stated SRS supports randomization.
- 100≤500 supports approximate independence without replacement.
- Under H₀, expected counts are 20 and 80, both at least 10.
- The usual z-test count check passes even though the observed success count is 8; an ordinary Wald interval would use a different count check.
Observed counts for an interval and null expected counts for a test are different checks.
Can a huge convenience sample justify population inference automatically?
Compare with an explanation
No. The data collection process remains a separate condition.
Predict. Change one thing. Explain.
Set p₀=.05 and n=100, then compare the warning with p₀=.50. Explain why changing the null affects the reference model.
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 questionAn SRS of 200 from 10000 tests p₀=.02. Explain why a normal test is not supported by the usual count condition.
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Compare with the answer and four-point rubric
- 1 point: Random sampling is stated.
- 1 point: 200≤1000 passes the 10% check.
- 1 point: Expected successes are 200×.02=4, below 10; expected failures are 196.
- 1 point: Do not claim the normal z approximation is justified; consider a suitable exact or simulation method with the design checked.
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 1Why use p₀ for a test count check?
The reference model assumes the null.
RECALL 2Does a count check verify a sampling method?
No.
RECALL 3What should happen when a condition fails?
State the limitation and use an appropriate alternative method or collect suitable data.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does a test use expected counts under the null?
- Test: np₀≥10 and n(1−p₀)≥10.
- Interval: x≥10 and n−x≥10.
- Design conditions must be checked separately.
Remember: Observed counts for an interval and null expected counts for a test are different checks.
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 3.5.C · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.5, objectives 3.5.C. 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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