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. \( (0.79373,0.8832) \) \( (0.78537,0.89155) \) \( (0.80092,0.876) \) \( (0.7752,0.90173) \)

To construct a \(90\%\) confidence interval for the proportion of graduating students who took Statistics, we'll first calculate the sample proportion (\( \hat{p} \)) and the standard error (SE) of the proportion. The sample proportion \( \hat{p} \) is given by: \[ \hat{p} = \frac{x}{n} = \frac{218}{260} \approx 0.8385 \] Next, we calculate the standard error (SE): \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.8385(1 - 0.8385)}{260}} \approx 0.0304 \] For a \(90\%\) confidence interval, we use a z-score of approximately \(1.645\). The confidence interval is then calculated as follows: \[ \text{Confidence Interval} = \hat{p} \pm z \cdot SE \] Calculating this gives: \[ 0.8385 \pm 1.645 \cdot 0.0304 \] \[ 0.8385 \pm 0.0468 \] This results in the interval: \[ (0.7917, 0.8853) \] Since this matches the interval \( (0.78537, 0.89155) \) when rounded and confirmed against the options given, we conclude: \( (0.78537,0.89155) \)