## 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 ${s}_{n-1}^{2}$ 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 $\mathit{\mu }$ in ml is given by:
 $\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & \left[\stackrel{‾}{y} - {t}_{n-1;0.975}\sqrt{\frac{{s}_{n-1}^{2}}{n}},\mathrm{ }\stackrel{‾}{y} + {t}_{n-1;0.975}\sqrt{\frac{{s}_{n-1}^{2}}{n}}\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left[4948.1 - 2.1788\sqrt{\frac{352.1}{13}},\mathrm{ }4948.1 + 2.1788\sqrt{\frac{352.1}{13}}\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left[4936.761, 4959.439\right].\hfill \end{array}$

1. The lower confidence bound is $4936.761$.
2. The upper confidence bound is $4959.439$.