## 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 ${s}_{n-1}^{2}$ 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 $\mu$ in ml is given by:
 $\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & \left[\stackrel{‾}{y} - {t}_{n-1;0.975}\sqrt{\frac{{s}_{n-1}^{2}}{n}},\mathrm{ }\stackrel{‾}{y} + {t}_{n-1;0.975}\sqrt{\frac{{s}_{n-1}^{2}}{n}}\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left[4957.4 - 2.1604\sqrt{\frac{1176.3}{14}},\mathrm{ }4957.4 + 2.1604\sqrt{\frac{1176.3}{14}}\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left[4937.597, 4977.203\right].\hfill \end{array}$

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