Question
========

The daily expenses of summer tourists in Vienna are analyzed. A
survey with $71$ tourists is conducted. This shows that the
tourists spend on average $130$ EUR. The sample variance
$s^2_{n-1}$ is equal to $83.2$.

Determine a $95\%$ confidence interval for the average daily
expenses (in EUR) of a tourist.

Answerlist
----------
* What is the lower confidence bound?
* What is the upper confidence bound?

Solution
========

The $95\%$ confidence interval for the average expenses $\mu$ is
given by:
$$
\begin{aligned}
&   & \left[\bar{y} \, - \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}, \; 
  \bar{y} \, + \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}\right] \\
& = & \left[ 130 \, - \, 1.96\sqrt{\frac{83.2}{71}}, \;
             130 \, + \, 1.96\sqrt{\frac{83.2}{71}}\right] \\
& = & \left[127.878, \, 132.122\right].
\end{aligned}
$$

Answerlist
----------
* The lower confidence bound is $127.878$.
* The upper confidence bound is $132.122$.

Meta-information
============
extype: cloze
exclozetype: num|num
exsolution: 127.878|132.122
exname: Confidence interval
extol: 0.01