One Factor Comparison Design

This example validates the calculation of the one factor comparison design in Weibull++.

The data are from Example 11-1 on page 71 in the book Design and Analysis of Experiments by Douglas C. Montgomery, John Wiley & Sons, 2001.

To test the hypothesis:


 * $$\begin{align}

H_0:\mu_1=\mu_2=...=\mu_5 \\

H_1:\mu_1\ne\mu_2\ne...\ne\mu_5 \end{align}$$

From the book the ANOVA table is:

Suppose that the experimenter selects $$\alpha=0.05\,\!$$. Since $$F_{0.05,4,20}=2.87\,\!$$, and $$14.76 >2.87 \,\!$$, the engineer rejects $$H_0\,\!$$ and concludes that treatment means differ.

The software results match the book results and the conclusions are matched. The ANOVA table is:



Since the P Value = 9.13E-6 < 0.01, we reject $$H_0\,\!$$.