Find a \( 95 \% \) confidence interval for the proportion of graduating students who took Chemistry if \( 35 \% \) of a simple random sample of 870 graduates took Chemistry. \( (0.02927,0.05119) \) \( (0.02717,0.05329) \) \( (0.31887,0.38228) \) \( (0.32397,0.37718) \)

We are given that in a sample of \( n = 870 \) graduates, \( 35\% \) took Chemistry. Define the sample proportion as \[ \hat{p} = 0.35. \] The formula for a \( 95\% \) confidence interval for a proportion is \[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \] where \( z^* \) is the critical value for \( 95\% \) confidence, which is approximately \( 1.96 \). **Step 1. Compute the Standard Error (SE):** \[ \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.35 \times 0.65}{870}}. \] Calculate the product in the numerator: \[ 0.35 \times 0.65 = 0.2275. \] Then, \[ \text{SE} = \sqrt{\frac{0.2275}{870}} = \sqrt{0.0002615} \approx 0.01617. \] **Step 2. Compute the Margin of Error (ME):** \[ \text{ME} = z^* \times \text{SE} = 1.96 \times 0.01617 \approx 0.0317. \] **Step 3. Determine the Confidence Interval:** \[ \hat{p} \pm \text{ME} = 0.35 \pm 0.0317. \] Calculate the endpoints: \[ \text{Lower limit} = 0.35 - 0.0317 \approx 0.3183, \] \[ \text{Upper limit} = 0.35 + 0.0317 \approx 0.3817. \] Among the given choices, the interval closest to \((0.3183, 0.3817)\) is \[ (0.31887, 0.38228). \] Thus, the \( 95\% \) confidence interval for the proportion of graduating students who took Chemistry is \[ (0.31887,0.38228). \]