```{r data generation, echo = FALSE, results = "hide"}
## DATA GENERATION
n <- sample(50:150, 1)
y <- rnorm(n, runif(1, 100, 200), runif(1, 10, 15))

## QUESTION/ANSWER GENERATION
Mean <- round(mean(y), digits = 1)
Var <- round(var(y), digits = 1)
sd <- sqrt(Var/n)
LB <- round(Mean - 1.96*sd, 3)
UB <- round(Mean + 1.96*sd, 3)
```

Question
========

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

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


Solution
========

The $95\%$ confidence interval for the average expenses $\mu$ is
given by:
$$
\begin{aligned}
&   & \left[\bar{y} \, - \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}, \; 
  \bar{y} \, + \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}\right] \\
& = & \left[ `r Mean` \, - \, 1.96\sqrt{\frac{`r Var`}{`r n`}}, \;
             `r Mean` \, + \, 1.96\sqrt{\frac{`r Var`}{`r n`}}\right] \\
& = & \left[`r LB`, \, `r UB`\right].
\end{aligned}
$$


Meta-information
============
extype: num
exsolution: `r LB`|`r UB`
exname: Confidence interval
extol: 0.01
