Let's solve the problem step by step: ### **Given Data** - Sample mean (\(\bar{x}\)): **3.1 days** - Sample standard deviation (s): **1.9 days** - Sample size (n): **22 hospitals** - Confidence level: **90%** ### **Step 1: Find the t-value for 90% confidence, df = n-1** Degrees of freedom (\(df\)) = \(n - 1 = 22 - 1 = 21\). For a 90% confidence interval: - Area in each tail = (1 - 0.90)/2 = 0.05. - We need the t-value for \(df = 21\) and area in right tail = 0.05. Using a t-table or calculator: - \(t_{0.05, 21} \approx 1.721\) ### **Step 2: Calculate Standard Error (SE)** \[ SE = \frac{s}{\sqrt{n}} = \frac{1.9}{\sqrt{22}} \] \[ \sqrt{22} \approx 4.690 \] \[ SE = \frac{1.9}{4.690} \approx 0.405 \] ### **Step 3: Calculate Margin of Error (E)** \[ E = t \times SE = 1.721 \times 0.405 \approx 0.697 \] ### **Step 4: Construct the Confidence Interval** \[ \text{Lower Limit} = \bar{x} - E = 3.1 - 0.697 = 2.403 \] \[ \text{Upper Limit} = \bar{x} + E = 3.1 + 0.697 = 3.797 \] ### **Final Answer** The 90% confidence interval for the population mean LOS for the state's hospitals is: \[ \boxed{(2.403, 3.797)} \] **(Rounded to three decimal places as needed.)**