How do you conclude a chi-square investigation?
You will be able to: Compute the right-tail probability and make a qualified contextual claim.
How do you conclude a chi-square investigation?
A random sample of 60 from a large school is classified by session and transport mode. Rows (15,10,5) and (5,10,15) give χ²=10 with df=2.
A useful starting point: What counts would the null model predict? →
Words and symbols before equations
- Reference distribution
- The approximate chi-square model under H₀ with the correct df.
- Significance decision
- Comparison of right-tail p-value with a prechosen α.
- Contextual conclusion
- A statement about the categorical variables and population, limited by the design.
What this picture assumes
Two rows A/B and three travel categories: Walk, Bus, Car. Fixed equal row and column margins; all expected cells equal the chosen count. Independent random sampling and 10% conditions from large populations are assumed. This Fall 2026 CED requires every expected count >5. Each 3D block is one synthetic person.
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.
- χ²=10; df=2. Right-tail p=0.006738. At α=0.05, reject independence. The design determines the population interpretation.
- 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 H₀: session and transport are independent in the school population, versus association. Verify the random sample, n≤10% of the population, independent units and expected counts 10 in every cell.
The statistic is 10 with df=2. Technology gives P(χ²₂≥10)≈.00674. At prechosen α=.05, reject independence.
Assuming independence and the design model, a statistic at least 10 would occur about .674% of the time. There is convincing evidence of an association in the population. The test does not establish causation, identify every differing category or measure association strength by the p-value alone.
A worked example, step by step
A valid homogeneity test comparing three population response distributions has p=.12 and α=.05. Conclude.
- The null states the response distributions are the same across the three populations.
- The alternative says at least one differs.
- Because .12>.05, fail to reject H₀.
- There is insufficient evidence that the population distributions differ; equality has not been proved.
A significant omnibus test does not show that every group differs from every other group.
Does a small p-value prove a cause?
Compare with an explanation
No. The collection and assignment design governs causal interpretation.
Predict. Change one thing. Explain.
Move the observed table toward equal distributions while keeping margins fixed. Explain how χ² and the right-tail p-value change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
χ²=10; df=2. Right-tail p=0.006738. At α=0.05, reject independence. The design determines the population interpretation.
| Cell | Observed O | Expected E | (O−E)²/E |
|---|---|---|---|
| A / Walk | 15 | 10 | 2.5 |
| A / Bus | 10 | 10 | 0 |
| A / Car | 5 | 10 | 2.5 |
| B / Walk | 5 | 10 | 2.5 |
| B / Bus | 10 | 10 | 0 |
| B / Car | 15 | 10 | 2.5 |
Two rows A/B and three travel categories: Walk, Bus, Car. Fixed equal row and column margins; all expected cells equal the chosen count. Independent random sampling and 10% conditions from large populations are assumed. This Fall 2026 CED requires every expected count >5. Each 3D block is one synthetic person.
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 independence test gives χ²=10,df=2,p=.00674 at α=.01. Interpret and decide.
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Compare with the answer and four-point rubric
- 1 point: Under population independence, the chance of χ² at least 10 is approximately .00674.
- 1 point: Since .00674<.01, reject independence.
- 1 point: There is convincing evidence of a population association between the named variables.
- 1 point: This does not establish a cause or the probability that H₀ 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 belongs in a chi-square p-value interpretation?
The null association/distribution claim, at-least-as-large statistic and population context.
RECALL 2What does failing to reject establish?
Insufficient evidence against the null, not proof of sameness.
RECALL 3Why report design limitations?
They determine what population and causal claims the results support.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you conclude a chi-square investigation?
- Use the right-tail probability with df=(r−1)(c−1).
- Compare p-value to prechosen α.
- Name the population(s), variables and alternative in the conclusion.
Remember: A significant omnibus test does not show that every group differs from every other group.
Conditions: Two rows A/B and three travel categories: Walk, Bus, Car. Fixed equal row and column margins; all expected cells equal the chosen count. Independent random sampling and 10% conditions from large populations are assumed. This Fall 2026 CED requires every expected count >5. Each 3D block is one synthetic person.
Refresh Kid · AP Statistics Unit 3 · Objectives 3.15.B, 3.15.C, 3.15.D · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.15, objectives 3.15.B, 3.15.C, 3.15.D. 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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