1. **State the problem:** We want to test if the mean repair costs charged by the two branches, Taman U and Taman Daya, are different at a 5% significance level. 2. **Given data:** - Taman U: mean $\bar{x}_1 = 1073.25$, standard deviation $s_1 = 422.33$, sample size $n_1 = 8$ - Taman Daya: mean $\bar{x}_2 = 1003.125$, standard deviation $s_2 = 383.63$, sample size $n_2 = 8$ 3. **Hypotheses:** - Null hypothesis $H_0$: $\mu_1 = \mu_2$ (means are equal) - Alternative hypothesis $H_a$: $\mu_1 \neq \mu_2$ (means are different) 4. **Assumptions:** - Repair costs are normally distributed - Population standard deviations are equal 5. **Test statistic:** Use the two-sample t-test with pooled variance: $$ S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} $$ $$ t = \frac{\bar{x}_1 - \bar{x}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} $$ Degrees of freedom $df = n_1 + n_2 - 2 = 14$ 6. **Calculate pooled standard deviation:** $$ S_p = \sqrt{\frac{(8-1)(422.33)^2 + (8-1)(383.63)^2}{14}} = \sqrt{\frac{7 \times 178362.83 + 7 \times 147172.57}{14}} = \sqrt{\frac{1248540.81 + 1030207.99}{14}} = \sqrt{\frac{2278748.8}{14}} = \sqrt{162767.77} \approx 403.45 $$ 7. **Calculate t-statistic:** $$ t = \frac{1073.25 - 1003.125}{403.45 \times \sqrt{\frac{1}{8} + \frac{1}{8}}} = \frac{70.125}{403.45 \times \sqrt{0.25}} = \frac{70.125}{403.45 \times 0.5} = \frac{70.125}{201.725} \approx 0.3476 $$ 8. **Determine critical t-value:** At $\alpha = 0.05$ and $df=14$, two-tailed test, critical value $t_{crit} \approx \pm 2.145$ 9. **Decision:** Since $|t| = 0.3476 < 2.145$, we fail to reject the null hypothesis. 10. **Conclusion:** There is not enough evidence at the 5% significance level to conclude that the mean repair costs differ between the two branches.