REG-4 Solutions: Chi-Square Test of Independence

Problem 1 — Expected Frequencies

Variant 0 (Smoking/Exercise):

(a) \(E = 50 \times 40 / 100 = 20.0\)   (b) \(E = 50 \times 60 / 100 = 30.0\)

Variant 1 (Age/Vaccination):

(a) \(E = 100 \times 110 / 200 = 55.0\)   (b) \(E = 100 \times 90 / 200 = 45.0\)

Variant 2 (Education/Newspaper):

(a) \(E = 100 \times 70 / 200 = 35.0\)   (b) \(E = 100 \times 70 / 200 = 35.0\)

Variant 3 (Stress/Sleep):

(a) \(E = 80 \times 70 / 140 = 40.0\)   (b) \(E = 60 \times 70 / 140 = 30.0\)

Variant 4 (Diet/BMI):

(a) \(E = 50 \times 60 / 100 = 30.0\)   (b) \(E = 50 \times 40 / 100 = 20.0\)

---

Problem 2 — Conditions and df

Variant	Min E	Conditions	df
0 (Smoking/Exercise, 2×2)	20.0	Yes — all E ≥ 5	1
1 (Conditions violated, 2×2)	2.0	No — E₁₂ = 2.0 < 5	1
2 (Gender/Transport, 2×3)	12.5	Yes	2
3 (Conditions violated, 2×3)	4.2	No — E₁₁ = E₁₃ = 4.2 < 5	2
4 (3×2 table)	15.0	Yes	2

---

Problem 3 — Reading the Chi-Square Table

(a) \(\chi^2_0.05(1) = 3.841\)

(b) \(\chi^2_0.05(2) = 5.991\)

(c) \(\chi^2 = 6.8\) > 5.991 = \(\chi^2_0.05(2)\) → reject H₀.

---

Problem 4 — Full Five-Step Test (Generator)

Solutions vary by dataset selected. The generator's "Show Solution" button displays the complete step-by-step answer including all E values, χ² computation, and decision.

Problem 1 — Full Analysis Chain

Variant 0 (Smoking/Exercise, reject):

\(\chi^2 = 4.167\), df = 1, reject H₀ (\(4.167 > 3.841\)). There is sufficient evidence that smoking status and exercise frequency are not independent. \(V = 0.204\) (small).

Variant 1 (Age/Vaccination, reject):

\(\chi^2 = 8.081\), df = 1, reject H₀. There is sufficient evidence that age group and vaccine uptake are not independent. \(V = 0.201\) (small).

Variant 2 (Education/Newspaper, fail to reject):

\(\chi^2 = 2.197\), df = 1, fail to reject H₀ (2.197 < 3.841). There is insufficient evidence to conclude that education level and daily newspaper reading are not independent.

Variant 3 (Stress/Sleep, reject):

\(\chi^2 = 11.667\), df = 1, reject H₀. There is sufficient evidence that stress level and sleep quality are not independent. \(V = 0.289\) (small–medium).

Variant 4 (Commute/Satisfaction, fail to reject):

\(\chi^2 = 2.420\), df = 1, fail to reject H₀. There is insufficient evidence to conclude that commute length and job satisfaction are not independent.

---

Problem 2 — Full 2×3 Test with Cramér's V (Generator)

Solutions vary by dataset. Generator solutions cover V6–V9 (chi² = 3.556, 16.333, 11.429, 2.052 respectively).

---

Problem 3 — Find the Error

Variant	Error	Correct approach
0 (Wrong df)	Used df = n−1 = 89 instead of (r−1)(c−1) = 2	df = (2−1)(3−1) = 2 for a 2×3 table
1 (O vs. E)	Divided by O instead of E in the χ² formula	Denominator must be E — the reference under H₀
2 (Fail to reject = independent)	"Variables are independent" after failing to reject H₀	"Insufficient evidence to conclude non-independence"
3 (Causation)	Concluded ice cream causes drowning from a significant χ²	Lurking variable (summer heat) drives both; chi-square shows association only
4 (Conditions violated)	Ran test with E = 2.3 < 5; conditions violated	Combine categories or use exact test; p-value is unreliable

---

Problem 4 — Cramér's V (Generator)

Solutions vary by pair selected. Key formula: \(V = \sqrt{\chi^2 / (n \cdot k)}\), where \(k = \min(r-1, c-1)\). For all W0–W6 pairs, k = 1.

---

Problem 5 — Physical Activity vs. Stress (Synthesis)

(a) H₀: Physical activity and stress level are independent in the worker population. Hₐ: Not independent.

(b) Expected frequencies (all ≥ 5 ✓):

	None	Moderate	Vigorous
Low stress	21.538	26.923	21.538
High stress	18.462	23.077	18.462

(c) \(\chi^2 = 1.985 + 0.352 + 0.556 + 2.316 + 0.410 + 0.649 = 6.268\)

(d) df = 2. \(\chi^2_0.05(2) = 5.991\). Since 6.268 > 5.991 → reject H₀ (\(p \approx 0.043\)).

(e) \(V = \sqrt{6.268/130} = 0.220\) — small effect. Statistically real but weak.

(f) Error 1: Chi-square shows association, not causation — mandating exercise may not reduce stress. Error 2: \(V = 0.22\) is small; even if causal, the effect is modest and unlikely to justify a blanket policy.

Feynman Prompt

A significant p-value means the association is unlikely to be due to chance in the sample — not that it is strong. With large n, even trivially weak associations become statistically significant. The colleague should compute Cramér's V: if V < 0.1, the association is negligible despite the significance; if V ≥ 0.3, it is at least moderate.

Apply (Study location / Year of study)

(a) \(\chi^2 = 9.2\) > 5.991 = \(\chi^2_0.05(2)\) → reject H₀.

(b) There is sufficient evidence at the 0.05 level that preferred study location and year of study are not independent in the population.

Analyze the Error

The researcher wrote "pet owners are just as happy as non-pet-owners" — this confuses failing to reject H₀ with proving independence. Correct: "There is insufficient evidence to conclude that pet ownership and happiness are not independent in the population."

Path A — The Analyst (Caffeine/Sleep)

Expected frequencies: All cells E = 22.5, 20.0, or 17.5 — all ≥ 5. df = 2.

\[\chi^2 = 2.500 + 0 + 3.214 + 2.500 + 0 + 3.214 = 11.429\]

\(\chi^2_0.05(2) = 5.991\). Since 11.429 > 5.991 → reject H₀ (\(p \approx 0.003\)).

\(V = \sqrt{11.429/120} = 0.309\) — medium effect.

Research summary: See lesson — one-paragraph summary using correct language (no causation, includes V, states decision).

Path B — The Communicator

• Report 1: Correct. Proper language, V reported, no causation claim.
• Report 2: Error — "confirms independence." Correct: "insufficient evidence to conclude non-independence."
• Report 3: Error — causation from association. Lurking variable (heat) drives both. Correct: state association with appropriate hedging about confounders.

Challenge 1 — Sample Size Effect

n	χ²	Decision	V
100	4.167	Reject H₀	0.204
200	8.333	Reject H₀	0.204
400	16.667	Reject H₀	0.204
800	33.333	Reject H₀	0.204

Lesson: χ² scales with n. V stays constant. With large n, even trivial associations become highly significant. Always report V alongside χ².

---

Challenge 2 — 3×3 Table

All E = 33.333 ≥ 5. df = (3−1)(3−1) = 4.

\(\chi^2 = 30.000\). \(\chi^2_0.05(4) = 9.488\). Reject H₀ strongly.

\(k = 2\), \(V = \sqrt{30/(300 \times 2)} = \sqrt{0.050} = 0.224\) — small to medium effect despite very significant p-value.

---

Challenge 3 — Simpson's Paradox

(a) Treatment A: 78% vs. B: 73% — A appears better overall.

(b) Mild: both 90%. Tied.

(c) Severe: A = 30%, B = 56%. Treatment B is better.

(d) Treatment A was assigned to mostly mild cases (80/100), which have inherently higher recovery. Disease severity is the lurking variable (confounder).

(e) Recommending A based on combined data ignores the confounding by severity — Treatment B is at least as good for mild cases and clearly better for severe cases.

(f) Simpson's Paradox: a combined-table association can mask or reverse stratum-specific patterns. Always stratify by potential confounders before drawing conclusions.