## 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 $13$ canisters is measured exactly. This shows that the canisters contain on average $4948.1$ ml. The sample variance $s^2_{n-1}$ is equal to $352.1$.

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 $\mu$ in ml is given by: \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 $4936.761$.
2. The upper confidence bound is $4959.439$.