1. **Problem statement:** We want to test if the mean salary of students at a certain college is lower than the population mean of 8600 with a sample mean of 7300, sample size 38, and population standard deviation 2500. 2. **Hypothesis test:** - Null hypothesis $H_0$: $\mu = 8600$ - Alternative hypothesis $H_a$: $\mu < 8600$ (claim that mean is lower) - Significance level $\alpha = 0.01$ 3. **Test statistic formula:** $$ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} $$ where $\bar{x} = 7300$, $\mu_0 = 8600$, $\sigma = 2500$, $n = 38$ 4. **Calculate test statistic:** $$ z = \frac{7300 - 8600}{2500 / \sqrt{38}} = \frac{-1300}{2500 / 6.1644} = \frac{-1300}{405.16} \approx -3.21 $$ 5. **Decision rule:** At $\alpha=0.01$, the critical z-value for a left-tailed test is approximately $-2.33$. Since $z = -3.21 < -2.33$, we reject $H_0$. 6. **Conclusion:** There is sufficient evidence at the 0.01 significance level to support the claim that the mean salary of students is lower than the population mean. 7. **Part b:** - For significance levels higher than 0.01 (e.g., 0.05), the critical value is less extreme (e.g., -1.645), so we would also reject $H_0$ without further calculations. - For significance levels lower than 0.01 (e.g., 0.001), the critical value is more extreme (e.g., -3.09), and since $z = -3.21 < -3.09$, we would still reject $H_0$. 8. **Part c: Confidence interval at 95%:** Formula for confidence interval when population standard deviation is known: $$ \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} $$ At 95%, $z_{0.025} = 1.96$ Calculate margin of error: $$ ME = 1.96 \times \frac{2500}{\sqrt{38}} = 1.96 \times 405.16 = 794.12 $$ Confidence interval: $$ 7300 \pm 794.12 = (6505.88, 8094.12) $$ 9. **Part d: Minimum sample size to reduce CI width by 15%:** Current margin of error $ME = 794.12$ New margin of error $ME_{new} = 0.85 \times 794.12 = 674.00$ Formula for margin of error: $$ ME = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} $$ Solve for $n$: $$ n = \left( \frac{z_{\alpha/2} \times \sigma}{ME} \right)^2 $$ Calculate new $n$: $$ n = \left( \frac{1.96 \times 2500}{674.00} \right)^2 = \left(7.27\right)^2 = 52.83 $$ Minimum sample size needed is $53$ (round up). **Final answers:** - a) Reject $H_0$ at 0.01 significance level. - b) Same conclusion for significance levels higher or lower than 0.01. - c) 95% confidence interval: $(6505.88, 8094.12)$. - d) Minimum sample size to reduce CI width by 15% is 53.