17. Many parents allow their underage children to drinkalcohol in limited situations when an adult is presentto supervise. The idea is that teens will learn responsi-ble drinking habits if they first experience alcohol in acontrolled environment. Other parents take a strict no-drinking approach with the idea that they are sendinga clear message about what is right and what is wrong.Recent research, however, suggests that the more per-missive approach may actually result in more negativeconsequences (McMorris et al., 2011). The research-ers surveyed a sample of 150 students each year fromages 14 to 17. The students were asked about theiralcohol use and about alcohol-related problems suchas binge drinking, fights, and blackouts. The followingtable shows data similar to the results from the study.Not Allowed to DrinkAllowed to DrinkExperience with Alcohol-Related ProblemsNo7189160Yes9314080120n = 200 17. Many parents allow their underage children to drinkalcohol in limited situations when an adult is presentto supervise. The idea is that teens will learn responsi-ble drinking habits if they first experience alcohol in acontrolled environment. Other parents take a strict no-drinking approach with the idea that they are sendinga clear message about what is right and what is wrong.Recent research, however, suggests that the more per-missive approach may actually result in more negativeconsequences (McMorris et al., 2011). The research-ers surveyed a sample of 150 students each year fromages 14 to 17. The students were asked about theiralcohol use and about alcohol-related problems suchas binge drinking, fights, and blackouts. The followingtable shows data similar to the results from the study.Not Allowed to DrinkAllowed to DrinkExperience with Alcohol-Related ProblemsNo7189160Yes9314080120n = 200

Part (a) : To test the relationship we will use the chi-square test of independence. The null and alternative hypotheses are given by — Null hypothesis : H0 : There is no relationship between parent's rule about alcohol and alcohol related problems. Alternative hypothesis : Ha : There exists a relationship between parent's rule about alcohol and alcohol related problems. The chi-square test statistic is calculated using the formula : χ2 = Σ((Oi–Ei)2/Ei ) Where, Oi = Observed frequency, and Ei = Expected frequency. The expected frequency for each cell is calculated as follows : Eij = (Ri × Cj)/N Where, Eij = Expected frequency of cell (i, j), Ri = ith row total, Cj = jth column total. Cell (1,1) E11 = (R1 × C1)/N E11 = (80 × 160)/200 = 64 Cell (1,2) E12 = (R1 × C1)/N E12 = (80 × 40)/200 = 16 Cell (2,1) E21 = (R2 × C1)/N E21 = (120 × 160)/200 = 96 Cell (2,2) E22 = (R2 × C1)/N E22 = (120 × 40)/200 = 24Writing the data in a tabular form we have, Cell (i, j) observed frequency (Oi) Expected frequency (Ei ) (Oi–Ei) (Oi–Ei)2 (Oi–Ei)2/Ei Cell (1,1) 71 64 7 49 0.765625 Cell (1,2) 9 16 –7 49 3.062500 Cell (2,1) 89 96 –7 49 0.510417 Cell (2,2) 31 24 7 49 2.041667 Σ((Oi–Ei)2/Ei )= 6.380209 The chi-square test statistic is given by, χ2 = Σ((Oi–Ei)2/Ei ) Substituting the value from the table, χ2 = 6.380209 Conclusion : Critical value approach – If the test-statistic is less than the critical value we fail to reject the null hypothesis. If the test - statistic is greater than the critical value we reject the null hypothesis. The critical value of chi-square for k degree of freedom at given level of significance (α) : Degree of freedom (k) = (r–1)(c–1) Where, r = Number of rows, and c = Number of columns Degree of freedom (k) = (2–1)(2–1) = 1 Level of significance (α) = 0.05 Using the chi-square distribution table the critical value of chi-square for 1 degree of freedom at 0.05 level of significance is 3.84 Since, the value of chi-square test-statistic is greater than the critical value of chi-square i.e., 6.380209 > 3.84 . Hence, we reject the null hypothesis that there exist no relationship between the two variables. Part (b) : Cramer's V is calculated by using the formula : Cramer's V = √(χ2/Nm) Where, N = grand total m = min(r–1)(c–1). r = Number of rows and c = Number of columns m = min(2–1, 2–1) m = min(1, 1) Minimum of (1,1) is 1. So we have, m = 1 Substituting the values we get, Cramer's V = √(6.380209/(200×1)) Cramer's V = √(6.380209/200) Cramer's V = 0.1786 Interpretation : Degree of freedom Weak Moderate Strong 1 <0.3 < 0.5 0.5 < Since, the Cramer's V is 0.1786 hence, there exist a weak relationship between the two variables.