The daily expenses of summer tourists in Vienna are analyzed. A survey with $132$ tourists is conducted. This shows that the tourists spend on average $168.8$ EUR. The sample variance $s^2_{n-1}$ is equal to $230.6$.
Determine a $95\%$ confidence interval for the average daily expenses (in EUR) of a tourist.
The $95\%$ confidence interval for the average expenses $\mu$ is given by: $\begin{aligned} & & \left[\bar{y} \, - \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ & = & \left[ 168.8 \, - \, 1.96\sqrt{\frac{230.6}{132}}, \; 168.8 \, + \, 1.96\sqrt{\frac{230.6}{132}}\right] \\ & = & \left[166.209, \, 171.391\right]. \end{aligned}$