1. **State the problem:** We want to test if there is an association between Age Group and Preferred Workout Type using a chi-square test of independence. 2. **State hypotheses:** - Null hypothesis $H_0$: Age Group and Preferred Workout Type are independent. - Alternative hypothesis $H_1$: Age Group and Preferred Workout Type are not independent. 3. **Choose significance level and calculate degrees of freedom:** - Given $\alpha = 0.05$. - Number of rows = 3 (Age Groups: 18–30, 31–50, 51+). - Number of columns = 3 (Workout Types: Gym, Home Workout, Outdoor). - Degrees of freedom $df = (3-1) \times (3-1) = 2 \times 2 = 4$. 4. **Find critical value:** - From chi-square distribution table for $df=4$ and $\alpha=0.05$, critical value $\chi^2_{crit} = 9.488$. 5. **Calculate expected frequencies $E$ for each cell:** Use formula $$E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}$$ | Age Group | Gym (E) | Home Workout (E) | Outdoor (E) | |---|---|---|---| | 18–30 | $\frac{100 \times 90}{270} = \frac{9000}{270} = 33.33$ | $\frac{100 \times 95}{270} = 35.19$ | $\frac{100 \times 85}{270} = 31.48$ | | 31–50 | $\frac{100 \times 90}{270} = 33.33$ | $35.19$ | $31.48$ | | 51+ | $\frac{70 \times 90}{270} = 23.33$ | $\frac{70 \times 95}{270} = 24.63$ | $\frac{70 \times 85}{270} = 22.04$ | 6. **Calculate chi-square statistic $\chi^2 = \sum \frac{(O - E)^2}{E}$:** Calculate for each cell: - For 18–30 Gym: $O=40$, $E=33.33$, $$\frac{(40 - 33.33)^2}{33.33} = \frac{(6.67)^2}{33.33} = \frac{44.49}{33.33} = 1.33$$ - 18–30 Home Workout: $O=25$, $E=35.19$, $$\frac{(25 - 35.19)^2}{35.19} = \frac{(-10.19)^2}{35.19} = \frac{103.83}{35.19} = 2.95$$ - 18–30 Outdoor: $O=35$, $E=31.48$, $$\frac{(35 - 31.48)^2}{31.48} = \frac{(3.52)^2}{31.48} = \frac{12.39}{31.48} = 0.39$$ - 31–50 Gym: $O=30$, $E=33.33$, $$\frac{(30 - 33.33)^2}{33.33} = \frac{(-3.33)^2}{33.33} = \frac{11.09}{33.33} = 0.33$$ - 31–50 Home Workout: $O=40$, $E=35.19$, $$\frac{(40 - 35.19)^2}{35.19} = \frac{(4.81)^2}{35.19} = \frac{23.14}{35.19} = 0.66$$ - 31–50 Outdoor: $O=30$, $E=31.48$, $$\frac{(30 - 31.48)^2}{31.48} = \frac{(-1.48)^2}{31.48} = \frac{2.19}{31.48} = 0.07$$ - 51+ Gym: $O=20$, $E=23.33$, $$\frac{(20 - 23.33)^2}{23.33} = \frac{(-3.33)^2}{23.33} = \frac{11.09}{23.33} = 0.48$$ - 51+ Home Workout: $O=30$, $E=24.63$, $$\frac{(30 - 24.63)^2}{24.63} = \frac{(5.37)^2}{24.63} = \frac{28.83}{24.63} = 1.17$$ - 51+ Outdoor: $O=20$, $E=22.04$, $$\frac{(20 - 22.04)^2}{22.04} = \frac{(-2.04)^2}{22.04} = \frac{4.16}{22.04} = 0.19$$ 7. **Sum all values:** $$\chi^2 = 1.33 + 2.95 + 0.39 + 0.33 + 0.66 + 0.07 + 0.48 + 1.17 + 0.19 = 7.57$$ 8. **Compare $\chi^2$ to critical value:** - $7.57 < 9.488$ so we **fail to reject** the null hypothesis. 9. **Interpretation:** There is not enough evidence at the 0.05 significance level to conclude that Age Group and Preferred Workout Type are dependent. They appear to be independent in this sample.