How do you test a difference between two shares?
You will be able to: Calculate a pooled z-test, interpret its p-value and conclude in context.
How do you test a difference between two shares?
Independent random samples from two large schools give 60/100 and 40/100 supporters. A pre-specified two-sided test asks whether the population shares differ.
A useful starting point: Why combine the shares under an equality null? →
Words and symbols before equations
- Observed gap D
- p̂₁−p̂₂.
- Pooled z statistic
- D divided by the equality-null SE.
- Two-sided evidence
- Differences at least as far from zero in either direction.
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
After verifying design and count conditions, pool to p̂c=100/200=.50. Null SE=√[.25(.01+.01)]≈.07071.
The observed gap is .20, giving z=.20/.07071≈2.828. Two-sided p≈.00468. At prechosen α=.05, reject equal population shares.
Assuming equal shares and the sampling model, the approximate probability of a standardized difference at least this extreme is .00468. There is convincing evidence of a population difference; the sample gap is 20 percentage points. Observational samples alone do not identify its cause.
A worked example, step by step
Valid independent samples give 55/100 and 45/100. Test equality against a two-sided alternative at α=.05.
- p̂c=.50 and D=.10.
- Null SE=√.005≈.07071.
- z≈1.414 and two-sided p≈.1573.
- Fail to reject equality: evidence of a population difference is insufficient at .05, not proof the shares are equal.
Do not use a one-sided p-value for a two-sided research question.
What is assumed when interpreting this p-value?
Compare with an explanation
Equal population shares and the specified design/model assumptions.
Predict. Change one thing. Explain.
Hold α fixed and move the group counts closer together. Explain the changes in z, p-value and the strength of evidence.
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 questionA valid equality test gives z=2.0 for Hₐ:p₁>p₂ and α=.05. Interpret p≈.0228 and decide.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Under equal population shares, the upper-tail probability for a statistic at least 2 is about .0228.
- 1 point: Since .0228<.05, reject H₀.
- 1 point: There is convincing evidence that population 1’s share is higher, in the defined context.
- 1 point: The conclusion remains conditional on the design and model; p is not the chance that equality is true.
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 centers the equality-null distribution?
Zero difference.
RECALL 2Why is the denominator pooled?
H₀ assumes a common population proportion.
RECALL 3What should accompany statistical significance?
Effect magnitude, context and study limitations.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you test a difference between two shares?
- z=[(p̂₁−p̂₂)−0]/SE₀.
- Two-sided p=2[1−Φ(
- z
- )].
- Conclude about the population proportions and match the design.
Remember: Do not use a one-sided p-value for a two-sided research question.
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 3.13.A, 3.13.B, 3.13.C · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.13, objectives 3.13.A, 3.13.B, 3.13.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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