--- title: "Chi-square: smokers, gender and starting age" subject: "Probability and Statistics" chapter: 7 key: "semester_5/probability_and_statistics:chi-square-smokers-gender-age" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-smokers-gender-age site: BCT Engineering --- # Chi-square: smokers, gender and starting age Probability and Statistics · Chapter 7: Inference concerning proportion - Asked in: 2074 Asoj, 5 marks - Video: [Chi Square Test Explained with Example | Calculate χ² & Test the Null Hypothesis](https://www.youtube.com/watch?v=f53nXHoMXx4) — Vectors Academy - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-smokers-gender-age - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7 ## Question A random sample of smokers was obtained and each individual was classified both with respect to gender and with respect to the age at which he/she first started smoking. | Age | Male | Female | |---|---|---| | <16 | 25 | 10 | | 16-17 | 24 | 32 | | 18-20 | 28 | 17 | | >20 | 19 | 34 | Carry out a test of hypotheses to decide whether there might be an association between gender and the age at which an individual first smokes? ## Short answer Chi-square test of independence gives χ² = 14.462 with 3 df. The critical value at 0.05 is 7.815. Since 14.462 > 7.815, reject H₀. There is significant association between gender and age at first smoking. ## Answer Chi-square test of independence between gender and the age at which smoking was started (4 × 2 table, N = 189). 1. H0: the two attributes are independent (no association). H1: the two attributes are associated (dependent). 2. Level of significance: α = 0.05 3. Test statistic: χ² = Σ (O − E)²/E, where E = (row total × column total)/grand total. Observed frequencies with totals: ```text Male Female Total <16 25 10 35 16-17 24 32 56 18-20 28 17 45 >20 19 34 53 Total 96 93 189 ``` Expected frequencies: ```text Male Female <16 17.78 17.22 16-17 28.44 27.56 18-20 22.86 22.14 >20 26.92 26.08 ``` Terms (O − E)²/E: ```text Male Female <16 2.934 3.029 16-17 0.694 0.717 18-20 1.157 1.194 >20 2.330 2.406 ``` χ² = Σ (O − E)²/E = **14.462** 4. Degrees of freedom = (r − 1)(c − 1) = (4 − 1)(2 − 1) = **3**. Critical value: χ²(0.05; 3) = **7.815**. Reject H0 if χ² > 7.815. 5. Decision: calculated χ² > tabulated χ² (14.462 > 7.815, p-value = 0.0023), so **reject H0**. **Conclusion:** there is a significant association between gender and the age at which a person first starts smoking. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the null hypothesis in the chi-square test for gender and age of starting smoking? A. Gender and age of starting smoking are associated B. Gender and age of starting smoking are independent C. The mean age is the same for males and females D. Males and females smoke equally often Answer: B. A test of independence always takes H0 as no association between the two attributes. (Key point: Null hypothesis) 2. How is the expected frequency of a cell calculated? A. (row total x column total) / grand total B. (row total + column total) / grand total C. Observed frequency / number of cells D. Grand total / number of cells Answer: A. E = (row total x column total)/grand total, which is the count expected if the attributes were independent. (Key point: Expected frequency) 3. For the 4 x 2 table, what are the degrees of freedom? A. 7 B. 4 C. 3 D. 1 Answer: C. d.f. = (r - 1)(c - 1) = (4 - 1)(2 - 1) = 3. (Key point: Degrees of freedom) 4. Calculated chi-square = 14.462 and the critical value at 5% with 3 d.f. is 7.815. What is the conclusion? A. Fail to reject H0; gender and starting age are independent B. Reject H0; no association exists C. Reject H0; gender and age of starting smoking are associated D. The test is inconclusive because the table is not 2 x 2 Answer: C. 14.462 > 7.815 puts the statistic in the rejection region, so H0 is rejected and an association is concluded. 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