1. **State the problem:** A sociologist wants to test if the distribution of preferred study methods among college students is equally likely using a Chi-square Goodness of Fit test with significance level $\alpha = 0.05$. 2. **Hypotheses:** - Null hypothesis $H_0$: The distribution of preferred study methods is uniform (all methods equally likely). - Alternative hypothesis $H_a$: The distribution of preferred study methods is not uniform. 3. **Expected probabilities:** Since there are 6 study methods, each is expected to have probability $\frac{1}{6}$. 4. **Observed values:** Roll a six-sided die 30 times to simulate observed counts for each method. Assume the following observed counts (example): - Group Study (1): $O_1 = 5$ - Solo Study (2): $O_2 = 4$ - Online Tutoring (3): $O_3 = 6$ - In-Person Tutoring (4): $O_4 = 7$ - Library Study (5): $O_5 = 3$ - Peer Teaching (6): $O_6 = 5$ 5. **Expected counts:** $$E_i = n \times p_i = 30 \times \frac{1}{6} = 5$$ for each category $i$. 6. **Create observed vs expected table:** | Method | Observed ($O_i$) | Expected ($E_i$) | |--------|------------------|------------------| | 1 | 5 | 5 | | 2 | 4 | 5 | | 3 | 6 | 5 | | 4 | 7 | 5 | | 5 | 3 | 5 | | 6 | 5 | 5 | 7. **Calculate Chi-square test statistic:** $$\chi^2 = \sum_{i=1}^6 \frac{(O_i - E_i)^2}{E_i} = \frac{(5-5)^2}{5} + \frac{(4-5)^2}{5} + \frac{(6-5)^2}{5} + \frac{(7-5)^2}{5} + \frac{(3-5)^2}{5} + \frac{(5-5)^2}{5}$$ $$= 0 + \frac{1}{5} + \frac{1}{5} + \frac{4}{5} + \frac{4}{5} + 0 = \frac{10}{5} = 2$$ 8. **Degrees of freedom:** $$df = k - 1 = 6 - 1 = 5$$ 9. **Find p-value:** Using Chi-square distribution with $df=5$, $\chi^2=2$ gives a p-value approximately $0.85$ (high p-value). 10. **Conclusion:** Since $p = 0.85 > \alpha = 0.05$, we fail to reject the null hypothesis. **Interpretation:** There is not enough evidence to conclude that the preferred study methods are distributed unequally. The sociologist can say students use all study methods equally based on this test.