1. **Stating the problem:** We are given three groups with sample sizes $n_1=4$, $n_2=5$, $n_3=4$ and variances $s_1^2=6.67$, $s_2^2=2.50$, $s_3^2=6.67$. We want to test the homogeneity of variances using Bartlett's test. 2. **Formula for pooled variance $S_p^2$:** $$S_p^2 = \frac{\sum_{i=1}^k (n_i - 1)s_i^2}{N - k}$$ where $N = \sum n_i$ and $k$ is the number of groups. 3. **Calculate pooled variance:** $$N = 4 + 5 + 4 = 13, \quad k = 3$$ $$S_p^2 = \frac{(4-1)6.67 + (5-1)2.50 + (4-1)6.67}{13 - 3} = \frac{20.01 + 10.00 + 20.01}{10} = \frac{50.02}{10} = 5.002$$ 4. **Formula for Bartlett's test statistic $\chi^2$:** $$\chi^2 = (N-k) \ln(S_p^2) - \sum_{i=1}^k (n_i - 1) \ln(s_i^2)$$ 5. **Calculate $\chi^2$:** $$\chi^2 = 10 \times \ln(5.002) - [3 \times \ln(6.67) + 4 \times \ln(2.50) + 3 \times \ln(6.67)]$$ Calculate each term: $$10 \times 1.609 = 16.09$$ $$3 \times 1.898 = 5.694$$ $$4 \times 0.916 = 3.664$$ $$3 \times 1.898 = 5.694$$ Sum inside bracket: $$5.694 + 3.664 + 5.694 = 15.052$$ So, $$\chi^2 = 16.09 - 15.052 = 1.038$$ 6. **Decision rule:** Compare $\chi^2$ calculated with $\chi^2$ table value at $\alpha=0.05$ and degrees of freedom $df = k-1 = 2$: $$\chi^2_{table} = 5.991$$ 7. **Conclusion:** Since $\chi^2_{calculated} = 1.038 < 5.991 = \chi^2_{table}$, we fail to reject $H_0$. **Interpretation:** The variances are homogeneous. **Final answer:** **B. Gagal tolak H₀, varians homogen karena χ² hitung < χ² tabel**