Your solution is **correct** and well-organized! Here’s a recap and slight clarification for each step, for clarity and completeness: --- ### **Step 1: Standard Error (SE)** The **standard error** of the sample proportion is calculated as: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.0244}{\sqrt{10000}} = \frac{0.0244}{100} = 0.000244 \] --- ### **Step 2: Margin of Error (ME)** The **margin of error** at 95% confidence (using \( z = 1.96 \)) is: \[ ME = z \cdot SE = 1.96 \times 0.000244 = 0.00047824 \] --- ### **Step 3: Confidence Interval (CI)** The **confidence interval** for the proportion is: \[ \text{CI} = \bar{p} \pm ME \] - **Lower bound:** \( 0.1518 - 0.00047824 = 0.15132176 \approx 0.1513 \) - **Upper bound:** \( 0.1518 + 0.00047824 = 0.15227824 \approx 0.1523 \) --- ## **✅ Final 95% Confidence Interval:** \[ \boxed{(0.1513,\ 0.1523)} \] --- **Interpretation:** We are 95% confident that the true proportion of OSU faculty, staff, and students wearing orange on Fridays during football season is between **15.13% and 15.23%**. Let me know if you’d like the calculation visualized with a diagram or further explanation of any step!