Solution to Exercise 10.16.

  (a) The P-value from the ANOVA is 0.000. If the experiment was designed correctly (i.e., so the independent random samples assumption, the major assumption of ANOVA), then it is absolutely not plausible that the five axial stiffness index distributions are identical.

(b) The P-value of 0.000 quoted above is less than anyone's reasonable level of significance (0.05, 0.01, 0.005, .0.001). We reject the null hypothesis of equal mean axial stiffness index.

(c) The factor being tested is plate length, with levels 4, 6, 8, 10, 12. The observed means (333.21, 368.06, 375.13, 407.36, and 437.17, respectively) are increasing, so we just list the factor levels in the same order:

  4   6   8  10  12
 -------
 ----------
    -------
    -----------
        -------
            -------

Tukey's intervals tell us that 4 is less than 10 and 12, 6 is less than 12, 8 is less than 12, and that's all.

Note: We can say one level of the factor A is less than another factor B (meaning $\mu_A < \mu_B$) if all of the confidence interval values (meaning, both endpoints) are less than . The ``Family error rate = 0.0500'' means that the chance of rejecting any of the 10 possible null hypotheses $H_{0}: \mu_i = \mu_j$ is controlled at the 0.05 level of significance.