Chi-Square Test of Independence — FRQ Walkthrough
A free-response question in exam style, fully worked — then the scoring logic behind every point.
The question · original, exam-style
A study café surveys a random sample of 200 customers, recording each customer's preferred drink (coffee, tea, or neither) and whether they stayed longer than one hour. The observed counts are — Coffee: stayed 54, did not 46; Tea: stayed 24, did not 36; Neither: stayed 12, did not 28. Is there convincing evidence, at the α = 0.05 level, that drink preference and staying longer than one hour are associated for this café's customers? Carry out the appropriate test.
Written by Marlio as an analog of the exam format. Not a College Board question.
Model response
Hypotheses: H₀: drink preference and staying longer than one hour are independent for this café's customers. Hₐ: drink preference and staying longer than one hour are associated.
Plan: Chi-square test of independence (one random sample, two categorical variables). Expected counts, using (row total × column total) / 200, are Coffee 45 and 55, Tea 27 and 33, Neither 18 and 22. Conditions: random sample (stated); 10% (200 < 10% of all customers); every expected count is at least 5 (the smallest is 18). Conditions met.
Do: χ² = (54−45)²/45 + (46−55)²/55 + (24−27)²/27 + (36−33)²/33 + (12−18)²/18 + (28−22)²/22 = 1.80 + 1.47 + 0.33 + 0.27 + 2.00 + 1.64 ≈ 7.52, with df = (3 − 1)(2 − 1) = 2. The p-value is P(χ² > 7.52) ≈ 0.023.
Conclude: Because 0.023 < α = 0.05, we reject H₀. There is convincing evidence that drink preference and staying longer than one hour are associated among this café's customers.
How it’s scored — point by point
- POINT 1
States the hypotheses in context.
H₀ and Hₐ must be about association versus independence between the two named variables for this population. Writing hypotheses about a proportion, or phrasing them as cause and effect, does not earn the point.
- POINT 2
Names the correct procedure and verifies conditions with expected counts.
Name the chi-square test of independence and show the expected counts, confirming each is at least 5 — computing the expected counts is itself part of the credit, not just naming the condition.
- POINT 3
Correct test statistic and degrees of freedom.
χ² ≈ 7.52 on df = 2, leading to a p-value near 0.023. A wrong df — a frequent error is using the number of cells — changes the p-value and loses this point.
- POINT 4
Conclusion in context, tied to α.
State the decision (reject H₀), link it to the p-value versus α, and answer in terms of drink preference and time stayed. A conclusion claiming one variable causes the other is wrong: the test shows association only.
Common point-losers
- Using degrees of freedom equal to the number of cells (6) instead of (rows − 1)(columns − 1) = 2.
- Concluding that drink choice causes people to stay longer — chi-square shows association, not causation.
- Running the test on row or column percentages instead of the raw observed counts.
- Never computing the expected counts before trusting the test.
- A conclusion with no context, e.g. "reject H₀," with no mention of the two variables.
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