1. **State the problem:** We have a sample of 30 emergency room patients with an average waiting time of $174.3$ minutes and a population standard deviation of $46.5$ minutes. We want to find: - The best point estimate of the population mean - The 99% confidence interval for the population mean. 2. **Best point estimate:** The best estimate of the population mean \(\mu\) is the sample mean \(\bar{x}\), which is given as \[\bar{x} = 174.3\]\ 3. **Confidence interval formula:** Since the population standard deviation \(\sigma\) is known and the sample size is \(n=30\), the 99% confidence interval for the mean is calculated using the z-distribution: $$\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ where \(z_{\alpha/2}\) is the z-score corresponding to a 99% confidence level, \(\sigma = 46.5\), and \(n=30\). 4. **Find the z-score:** The critical z-value for a 99% confidence level (two-tailed) is approximately \(z_{0.005} = 2.576\). 5. **Calculate the margin of error:** $$ E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 2.576 \times \frac{46.5}{\sqrt{30}} $$ Calculate the denominator: \(\sqrt{30} \approx 5.477\), thus $$ E = 2.576 \times \frac{46.5}{5.477} = 2.576 \times 8.49 \approx 21.87 $$ 6. **Construct the confidence interval:** $$ 174.3 \pm 21.87 $$ Lower limit: \(174.3 - 21.87 = 152.43\) Upper limit: \(174.3 + 21.87 = 196.17\) 7. **Final answer:** The best point estimate of the population mean waiting time is \(174.3\) minutes. The 99% confidence interval for the population mean is $$ (152.43, 196.17) \text{ minutes} $$