Exam 1

  1. Question

    The daily expenses of summer tourists in Vienna are analyzed. A survey with 7171 tourists is conducted. This shows that the tourists spend on average 130130 EUR. The sample variance sn12s^2_{n-1} is equal to 83.283.2.

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


    1. What is the lower confidence bound?
    2. What is the upper confidence bound?

    Solution

    The 95%95\% confidence interval for the average expenses μ\mu is given by: [y1.96sn12n,y+1.96sn12n]=[1301.9683.271,130+1.9683.271]=[127.878,132.122]. \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}


    1. The lower confidence bound is 127.878127.878.
    2. The upper confidence bound is 132.122132.122.