## 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)
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.
The $95\%$ confidence interval for the average expenses $\mu$ is
given by:
$$
\begin{aligned}
& & \left[\bar{y} \, - \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}, \;
\bar{y} \, + \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}\right] \
& = & \left[ r Mean
\, - \, 1.96\sqrt{\frac{r Var
}{r n
}}, \;
r Mean
\, + \, 1.96\sqrt{\frac{r Var
}{r n
}}\right] \
& = & \left[r LB
, \, r UB
\right].
\end{aligned}
$$
r LB
$.r UB
$.extype: cloze
exclozetype: num|num
exsolution: r LB
|r UB
exname: Confidence interval
extol: 0.01
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