1. **Problem statement:** We are given a sample of 50 patients with a mean blood sugar level of 130 mg/dl and a standard deviation of 4.6 mg/dl. We want to find a 90% confidence interval for the mean blood sugar level of all patients. 2. **Formula for confidence interval:** For a large sample size ($n \geq 30$), the confidence interval for the population mean $\mu$ is given by: $$\bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean, - $s$ is the sample standard deviation, - $n$ is the sample size, - $z_{\alpha/2}$ is the critical z-value for the desired confidence level. 3. **Find the critical z-value:** For a 90% confidence interval, $\alpha = 1 - 0.90 = 0.10$, so $\alpha/2 = 0.05$. From the standard normal distribution table, the z-value corresponding to $P(Z \leq z) = 0.95$ is approximately $z_{0.05} = 1.645$. 4. **Calculate the margin of error:** $$\text{Margin of error} = z_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.645 \times \frac{4.6}{\sqrt{50}}$$ Calculate $\sqrt{50}$: $$\sqrt{50} = 7.071$$ So: $$1.645 \times \frac{4.6}{7.071} = 1.645 \times 0.650 = 1.069$$ 5. **Construct the confidence interval:** $$130 \pm 1.069$$ This gives: - Lower limit: $130 - 1.069 = 128.931$ - Upper limit: $130 + 1.069 = 131.069$ 6. **Interpretation:** We are 90% confident that the true mean blood sugar level of all patients in the hospital lies between 128.931 mg/dl and 131.069 mg/dl.