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 14 canisters is measured exactly. This shows that the canisters contain on average 4957.4 ml. The sample variance sn-1 2 is equal to 1176.3.
    Determine a 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% confidence interval for the average content μ in ml is given by:
    [ y - tn-1;0.975 sn-1 2 n ,   y + tn-1;0.975 sn-1 2 n ] = [4957.4-2.1604 1176.3 14 ,  4957.4+2.1604 1176.3 14 ] = [4937.597,4977.203].


    1. The lower confidence bound is 4937.597.
    2. The upper confidence bound is 4977.203.