
\begin{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.

\begin{answerlist}
\item What is the lower confidence bound?
\item What is the upper confidence bound?
\end{answerlist}
\end{question}

%% SOLUTION
\begin{solution}
The $95\%$ confidence interval for the average expenses $\mu$ is
given by:
\begin{eqnarray*}
&   & \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{eqnarray*}

\begin{answerlist}
\item The lower confidence bound is $127.878$.
\item The upper confidence bound is $132.122$.
\end{answerlist}
\end{solution}

%% META-INFORMATION
%% \extype{cloze}
%% \exclozetype{num|num}
%% \exsolution{127.878|132.122}
%% \exname{Confidence interval}
%% \extol{0.01}
