1. **Problem statement:** We want to test if the mean salary of students at a certain college is lower than the population mean salary of 8600 NIS, given a sample of 38 students with an average salary of 7300 NIS and population standard deviation 2500 NIS. 2. **Hypotheses:** - Null hypothesis $H_0: \mu = 8600$ - Alternative hypothesis $H_1: \mu < 8600$ 3. **Significance level:** $\alpha = 0.01$ 4. **Test statistic and critical value:** Since population standard deviation is known, use the z-test. Calculate the critical value $x_c$ using the formula: $$x_c = \mu_0 - z_{1-\alpha} \cdot \frac{\sigma}{\sqrt{n}}$$ where $z_{1-\alpha}$ is the z-score for confidence level $1-\alpha$. 5. **Calculate $z_{1-\alpha}$:** For $\alpha=0.01$, $z_{0.99} = 2.33$ (from z-tables). 6. **Calculate $x_c$:** $$x_c = 8600 - 2.33 \times \frac{2500}{\sqrt{38}} = 8600 - 2.33 \times 405.54 = 8600 - 945.91 = 7654.09$$ 7. **Decision rule:** Reject $H_0$ if sample mean $\bar{x} < x_c$. Given $\bar{x} = 7300 < 7654.09$, we reject $H_0$ at $\alpha=0.01$. 8. **Answer to part b:** - For significance levels higher than 0.01 (e.g., 0.05), the critical value $x_c$ will be higher (less strict), so since $7300 < x_c$ still holds, we can reject $H_0$ without further calculations. - For significance levels lower than 0.01 (e.g., 0.001), $x_c$ will be lower (more strict), so we cannot conclude without calculation. 9. **Confidence interval (CI) at 95%:** Use formula: $$CI = \bar{x} \pm z_{1-\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}}$$ For 95%, $z_{0.975} = 1.96$. Calculate margin of error: $$ME = 1.96 \times \frac{2500}{\sqrt{38}} = 1.96 \times 405.54 = 795$$ CI: $$7300 \pm 795 = (6505, 8095)$$ 10. **Minimum sample size to reduce CI width by 15%:** Current CI width: $$W = 2 \times 795 = 1590$$ New width: $$W_{new} = 0.85 \times 1590 = 1351.5$$ Since width $W = 2 z \frac{\sigma}{\sqrt{n}}$, solve for $n$: $$n = \left( \frac{2 z \sigma}{W} \right)^2$$ Calculate new $n$: $$n = \left( \frac{2 \times 1.96 \times 2500}{1351.5} \right)^2 = \left( \frac{9800}{1351.5} \right)^2 = (7.25)^2 = 52.56$$ Minimum sample size is $53$ (round up). **Final answers:** - a) Reject $H_0$ at $\alpha=0.01$. - b) Yes, reject $H_0$ for $\alpha > 0.01$ without further calculations; no conclusion without calculation for $\alpha < 0.01$. - c) 95% CI is $(6505, 8095)$. - d) Minimum sample size to reduce CI width by 15% is 53.