91Ó°ÊÓ

Hypotheses are typically written in the form of if/then statements, such as if someone consumes a lot of sugar, they will develop cavities in their teeth.

The hypotheses of interest are

\({H_0}\,:\,{\mu _i} = {\mu _j},i \ne j\,\)

versus alternative hypothesis

\({H_0}\,:\,\) at least two of the are different.

F is ratio of the two mean squares

\(F = \frac{{MS{T_r}}}{{MSE}}.\)

The two values are given in the exercise

\(\begin{aligned}{l}MS{T_r} = 2673;\\MSE = 1094.2.\end{aligned}\)

Thus, the values of the statistic F is

\(\begin{aligned}{c}f = \frac{{MS{T_r}}}{{MSE}}\\ = \frac{{267.3}}{{1094.2}}\\ = 2.44.\end{aligned}\)

The degrees of freedom are \(l - 1 = 5 - 1 = 4\) and \(I\left( {J - 1} \right) = 5 \times 3 = 15.\)

Using degrees of freedom for \(\alpha = 0.05\) , the P value is

\(P = P\left( {F > 2.44} \right) = 0.09\)

This value was computed using software. You can estimate it using the table in the appendix of the book using software is recommended

Since \(P = 0.09 > 0.05,\) do not reject null hypothesis

At significance level \({\rm{0}}{\rm{.5}}{\rm{.}}\) From the given value it can be concluded that there is no statistical difference in the mean of tensile strengths of the given types.