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Categorical

Chi-Square Test of Independence — Smoking vs. Cancer

Test association between smoking and cancer in a 2×2 contingency table. Expected frequencies, χ², Cramér's V=.44.

📊 χ²(1) = 19.23, p < .001, V = .44
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🗂️ 7 · Chi-Square Test of Independence

Smoking Status vs Cancer Diagnosis
Categorical
Research Question: Is there an association between smoking status and cancer diagnosis in a sample of 200 patients?
Cancer: YesCancer: NoRow Total
Smoker50 (E=32)30 (E=48)80
Non-Smoker30 (E=48)90 (E=72)120
Column Total80120200
Expected frequencies: E = (Row Total × Col Total) / N
$$E_{11} = \frac{80 \times 80}{200} = 32, \quad E_{12} = \frac{80 \times 120}{200} = 48$$ $$E_{21} = \frac{120 \times 80}{200} = 48, \quad E_{22} = \frac{120 \times 120}{200} = 72$$
χ²
$$\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(50-32)^2}{32} + \frac{(30-48)^2}{48} + \frac{(30-48)^2}{48} + \frac{(90-72)^2}{72}$$ $$= \frac{324}{32} + \frac{324}{48} + \frac{324}{48} + \frac{324}{72} = 10.125 + 6.75 + 6.75 + 4.5 = 28.125$$
Cramér's V
$$V = \sqrt{\frac{\chi^2}{N \cdot \min(r-1,c-1)}} = \sqrt{\frac{28.125}{200 \times 1}} = \sqrt{0.1406} = 0.375 \quad (\text{medium-large})$$
28.125
χ²
1
df
<.001
p
0.375
Cramér's V
🔴 Reject H₀ — Smoking and cancer are not independent, χ²(1)=28.13, p<.001, V=.38.
APA-7
A chi-square test of independence indicated a significant association between smoking status and cancer diagnosis, χ²(1, N = 200) = 28.13, p < .001, V = .375. Smokers had a higher cancer rate (62.5%) compared to non-smokers (25.0%).

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