1. **State the problem:** We are asked to test whether the two teachers, Ms. Faith and Mr. Omar, are equally effective based on their students' test scores. We use a 0.05 significance level for the hypothesis test. 2. **Set up hypotheses:** - Null hypothesis $H_0$: The two teachers have the same mean student scores, so $\mu_1 = \mu_2$. - Alternative hypothesis $H_a$: The two teachers have different mean student scores, so $\mu_1 \neq \mu_2$. 3. **Identify sample statistics:** - Ms. Faith: $n_1 = 12$, mean $\bar{x}_1 = 72.4$, standard deviation $s_1 = 9.8$. - Mr. Omar: $n_2 = 15$, mean $\bar{x}_2 = 68.5$, standard deviation $s_2 = 10.7$. 4. **Determine the test to use:** Since samples are independent and population standard deviations unknown, use the two-sample t-test for difference in means assuming unequal variances. 5. **Calculate the test statistic:** The test statistic is $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$ Compute numerator: $$72.4 - 68.5 = 3.9$$ Compute denominator: $$\sqrt{\frac{9.8^2}{12} + \frac{10.7^2}{15}} = \sqrt{\frac{96.04}{12} + \frac{114.49}{15}} = \sqrt{8.003 + 7.633} = \sqrt{15.636} \approx 3.954$$ Calculate $t$: $$t = \frac{3.9}{3.954} \approx 0.987$$ 6. **Degrees of freedom approximation:** Using Welch-Satterthwaite equation: $$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} = \frac{15.636^2}{\frac{(8.003)^2}{11} + \frac{(7.633)^2}{14}} = \frac{244.44}{\frac{64.05}{11}+\frac{58.24}{14}} = \frac{244.44}{5.82 +4.16} = \frac{244.44}{9.98} \approx 24.49$$ Use $df \approx 24$. 7. **Critical value and decision:** At $\alpha = 0.05$ and two-tailed test with $df=24$, critical t-value $t_{crit} \approx \pm 2.064$. Since our calculated $t=0.987$ lies between $-2.064$ and $2.064$, we **fail to reject** the null hypothesis. 8. **Conclusion:** There is insufficient evidence at the 0.05 significance level to conclude a difference in the effectiveness of Ms. Faith and Mr. Omar based on their students' test scores.