An airline transports over 10,000 passengers daily, and the airline is curious what proportion of their passengers use mobile boarding passes instead of paper passes. They take an SRS of 80 passengers and find that 60 of them use mobile boarding passes. Based on this sample, which of the following is a \( 95 \% \) confidence interval for the proportion of passengers who use mobile boarding passes? Choose 1 answer: (A) \( (0.625,0.875) \) (B) \( (0.637,0.863) \) (C) \( (0.655,0.845) \) (D) \( (0.670,0.830) \)

To calculate the \( 95\% \) confidence interval for the proportion of passengers using mobile boarding passes, we use the formula for the confidence interval for a proportion: \[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Where: - \(\hat{p}\) is the sample proportion (60 out of 80), - \(n\) is the sample size (80), - \(z^*\) is the critical value for \(95\%\) confidence (approximately \(1.96\)). First, calculate \(\hat{p} = \frac{60}{80} = 0.75\). Next, calculate the standard error: \[ SE = \sqrt{\frac{0.75(1 - 0.75)}{80}} = \sqrt{\frac{0.75 \times 0.25}{80}} = \sqrt{\frac{0.1875}{80}} \approx 0.0486 \] Now, calculate the margin of error: \[ ME = z^* \cdot SE = 1.96 \cdot 0.0486 \approx 0.095 \] Finally, construct the confidence interval: \[ 0.75 \pm 0.095 \Rightarrow (0.655, 0.845) \] Thus, the correct answer is: (C) \( (0.655,0.845) \)