Find a \( 90 \% \) confidence interval for the proportion of graduating students who took Statistics if 218 out of a simple, random sample of 260 graduates took Statistics. \( (0.79373,0.8832) \) \( (0.78537,0.89155) \) \( (0.80092,0.876) \) \( (0.7752,0.90173) \)

Given: - Sample size: \( n = 260 \) - Number of successes: \( X = 218 \) - Confidence level: \( 90\% \) (which implies \(\alpha = 0.10\) and \(\alpha/2 = 0.05\)) **Step 1. Calculate the sample proportion \(\hat{p}\):** \[ \hat{p} = \frac{X}{n} = \frac{218}{260} \approx 0.83846 \] **Step 2. Determine the critical value \(z_{\alpha/2}\):** For a \(90\%\) confidence level, \(z_{0.05} \approx 1.645\). **Step 3. Compute the standard error (SE) of \(\hat{p}\):** \[ \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.83846 \times (1-0.83846)}{260}} \] Calculate the product in the numerator: \[ 0.83846 \times 0.16154 \approx 0.13538 \] Then divide by \( n \): \[ \frac{0.13538}{260} \approx 0.0005207 \] Taking the square root: \[ \text{SE} \approx \sqrt{0.0005207} \approx 0.02283 \] **Step 4. Compute the margin of error (ME):** \[ \text{ME} = z_{\alpha/2} \times \text{SE} \approx 1.645 \times 0.02283 \approx 0.03753 \] **Step 5. Construct the \(90\%\) confidence interval:** \[ \hat{p} \pm \text{ME} = 0.83846 \pm 0.03753 \] This gives us: \[ (0.83846 - 0.03753, \; 0.83846 + 0.03753) \approx (0.80093, \; 0.87599) \] Rounded appropriately, we have: \[ (0.80092, 0.876) \] Thus, the correct \(90\%\) confidence interval is: \[ \boxed{(0.80092, \; 0.876)} \]