## 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



flaviobarros/exams documentation built on May 28, 2019, 8:39 p.m.