91Ó°ÊÓ

In Example 8.2.3, suppose that we will observe n = 20 cheese chunks with lactic acid concentrations\({X_1},...,{X_{20}}\). We need to find a number c so that \(P\left( {{{\bar X}_{20}} \le \mu + c\sigma '} \right) = 0.95\)

\(\begin{align}P\left( {{{\bar X}_{20}} \le \mu + c\sigma '} \right)\\ = P\left( {\frac{{\sqrt {20} \left( {{{\bar X}_{20}} - \mu } \right)}}{{\sigma '}} \le c\sqrt {20} } \right)\\\end{align}\)

Let the distribution of \(U = \frac{{{{\left( {20} \right)}^{\frac{1}{2}}}\left( {{{\bar X}_{20}} - \mu } \right)}}{{\sigma '}}\) is a t-distribution with 19 degrees of freedom. From the table of t-distribution with 19 degrees of freedom the value of 0.95 quantile is 1.729. Then,

\(\begin{align}P\left( {X \le 1.729} \right) &= 0.95\\ \Rightarrow c\sqrt {20} &= 1.729\\ \Rightarrow c &= 0.387\end{align}\)