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 1414 canisters is measured exactly. This shows that the canisters contain on average 4957.44957.4 ml. The sample variance sn12s^2_{n-1} is equal to 1176.31176.3.

    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: [ytn1;0.975sn12n,y+tn1;0.975sn12n]=[4957.42.16041176.314,4957.4+2.16041176.314]=[4937.597,4977.203]. \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[ 4957.4 \, - \, 2.1604\sqrt{\frac{1176.3}{14}}, \; 4957.4 \, + \, 2.1604\sqrt{\frac{1176.3}{14}}\right] \\ &= \left[4937.597, \, 4977.203\right]. \end{aligned}


    1. The lower confidence bound is 4937.5974937.597.
    2. The upper confidence bound is 4977.2034977.203.