Why combine the shares under an equality null?
You will be able to: Calculate a pooled estimate and distinguish test and interval standard errors.
Why combine the shares under an equality null?
Group 1 has 30 successes in 50; group 2 has 40 in 100. Under equal population shares, both samples estimate one common probability.
A useful starting point: Are two groups really independent comparisons? →
Words and symbols before equations
- Pooled estimate
- p̂c=(x₁+x₂)/(n₁+n₂).
- Weighted average
- Each group’s sample proportion receives weight proportional to its sample size.
- Null SE
- √[p̂c(1−p̂c)(1/n₁+1/n₂)] for an equality test.
What this picture assumes
Synthetic study: independent SRSs from two separate populations, each N=100000. The chosen sizes satisfy both 10% conditions. Paired responses require a different method. Equality null p₁=p₂. The normal test uses the pooled share and four pooled expected counts of at least 10. Zero estimated SE makes the z-test unavailable.
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.
- D=0.2; z=2.8284; p-value=0.002339. At α=0.05, reject equal population shares. Alternative p₁ > p₂.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Combine counts, not unweighted percentages. Here p̂c=70/150≈.4667. Averaging .60 and .40 would give .50, wrongly giving a 50-person sample as much weight as a 100-person sample.
The equality-null reference spread uses the common pooled estimate. By contrast, the ordinary confidence interval uses each observed share separately.
Check the four expected counts under that common probability. If the pooled estimate is 0 or 1, its standard error is zero and the usual z-test is not usable; do not divide by zero.
| Feature | Difference interval | Equality-null test |
|---|---|---|
| Assumption | p₁ and p₂ need not be equal | p₁=p₂ |
| SE estimates | Separate observed shares | One pooled share |
| Count check | Four observed counts | Four pooled expected counts |
A worked example, step by step
For x₁=30,n₁=50,x₂=40,n₂=100, find pooled counts and null SE.
- p̂c=70/150=.46667.
- Expected counts are 23.333,26.667 in group 1 and 46.667,53.333 in group 2.
- SE₀=√[(.46667)(.53333)(1/50+1/100)].
- SE₀≈.08641; all four expected counts exceed 10.
Pooling is not averaging two percentages unless their denominators are equal.
Why is pooling defensible in the test?
Compare with an explanation
The null assumes both populations share a common proportion, estimated using both samples.
Predict. Change one thing. Explain.
Give the groups unequal sizes and compare their shares with the pooled estimate. Explain why the larger group has greater weight.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
D=0.2; z=2.8284; p-value=0.002339. At α=0.05, reject equal population shares. Alternative p₁ > p₂.
| Group | Successes | Failures | Actual share |
|---|---|---|---|
| 1 | 60 | 40 | 0.6 |
| 2 | 40 | 60 | 0.4 |
| Null quantity | Value |
|---|---|
| Pooled share | 0.5 |
| Four expected counts | 50, 50, 50, 50 |
| Pooled SE | 0.07071 |
Synthetic study: independent SRSs from two separate populations, each N=100000. The chosen sizes satisfy both 10% conditions. Paired responses require a different method. Equality null p₁=p₂. The normal test uses the pooled share and four pooled expected counts of at least 10. Zero estimated SE makes the z-test unavailable.
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 questionGroups have 20/40 and 60/120 successes. Calculate the pooled estimate and all four expected counts.
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Compare with the answer and four-point rubric
- 1 point: p̂c=80/160=.50.
- 1 point: Group 1 expected counts are 20 and 20.
- 1 point: Group 2 expected counts are 60 and 60.
- 1 point: All exceed 10; randomization, independence and any 10% conditions still require separate verification.
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 1Which data enter pooling?
Success counts and sample sizes from both groups.
RECALL 2Why not pool for the ordinary interval?
Equality is not assumed when estimating the difference.
RECALL 3What if all observations are successes?
The usual normal equality test has zero estimated spread and fails its count conditions.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why combine the shares under an equality null?
- p̂c=(x₁+x₂)/(n₁+n₂).
- SE₀=√[p̂c(1−p̂c)(1/n₁+1/n₂)].
- Use separate observed shares for a difference interval.
Remember: Pooling is not averaging two percentages unless their denominators are equal.
Conditions: Synthetic study: independent SRSs from two separate populations, each N=100000. The chosen sizes satisfy both 10% conditions. Paired responses require a different method. Equality null p₁=p₂. The normal test uses the pooled share and four pooled expected counts of at least 10. Zero estimated SE makes the z-test unavailable.
Refresh Kid · AP Statistics Unit 3 · Objectives 3.12.C · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.12, objectives 3.12.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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