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:
   

   ${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & [\bar{y}\hspace{0.5em}-\hspace{0.5em}{t}_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}},\mathrm{\hspace{0.5em}\hspace{0.5em}}\bar{y}\hspace{0.5em}+\hspace{0.5em}{t}_{n-1;0.975}\sqrt{\frac{s_{n-1}^2}{n}}]\hfill \\ \multicolumn{1}{c}{}& =\hfill & [4948.1\hspace{0.5em}-\hspace{0.5em}2.1788\sqrt{\frac{352.1}{13}},\mathrm{\hspace{0.5em}\hspace{0.5em}}4948.1\hspace{0.5em}+\hspace{0.5em}2.1788\sqrt{\frac{352.1}{13}}]\hfill \\ \multicolumn{1}{c}{}& =\hfill & [4936.761,\hspace{0.5em}4959.439].\hfill \end{array}}$

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