Exam 1

  1. Question

    It is suspected that a supplier systematically underfills 5 l canisters of detergent. The filled volumes are assumed to be normally distributed. A small sample of 1313 canisters is measured exactly. This shows that the canisters contain on average 4948.14948.1 ml. The sample variance sn−12s^2_{n-1} is equal to 352.1352.1.

    Determine a 95%95\% confidence interval for the average content of a canister (in ml).


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

    Solution

    The 95%95\% confidence interval for the average content μ\mu in ml is given by: [y‾−tn−1;0.975sn−12n,y‾+tn−1;0.975sn−12n]=[4948.1−2.1788352.113,4948.1+2.1788352.113]=[4936.761,4959.439]. \begin{aligned} & \left[\bar{y} \, - \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, t_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ &= \left[ 4948.1 \, - \, 2.1788\sqrt{\frac{352.1}{13}}, \; 4948.1 \, + \, 2.1788\sqrt{\frac{352.1}{13}}\right] \\ &= \left[4936.761, \, 4959.439\right]. \end{aligned}


    1. The lower confidence bound is 4936.7614936.761.
    2. The upper confidence bound is 4959.4394959.439.