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One Factor Comparison Design
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This example validates the calculation of the one factor comparison design in Weibull++.
Reference Case
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.
Data
To test the hypothesis:
- [math]\displaystyle{ \begin{align}
H_0:\mu_1=\mu_2=...=\mu_5 \\
H_1:\mu_1\ne\mu_2\ne...\ne\mu_5
\end{align} }[/math]
| Weight percentage of cotton
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Observed tensile strength (lb/in[math]\displaystyle{ ^2 }[/math])
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| 1
|
2
|
3
|
4
|
5
|
| 15 |
7 |
7 |
15 |
11 |
9
|
| 20 |
12 |
17 |
12 |
18 |
18
|
| 25 |
14 |
18 |
18 |
19 |
19
|
| 30 |
19 |
25 |
22 |
19 |
23
|
| 35 |
7 |
10 |
11 |
15 |
11
|
Result
From the book the ANOVA table is:
| Source of Variation
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Sum of Squares
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Degrees of Freedom
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Mean Square
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[math]\displaystyle{ F_0\,\! }[/math]
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P value
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| Cotton weight percentage |
475.76 |
4 |
118.94 |
14.76 |
<0.01
|
| Error |
161.20 |
20 |
8.06 |
|
|
| Total |
636.96 |
24 |
|
|
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Suppose that the experimenter selects [math]\displaystyle{ \alpha=0.05\,\! }[/math]. Since [math]\displaystyle{ F_{0.05,4,20}=2.87\,\! }[/math], and [math]\displaystyle{ 14.76 \gt 2.87 \,\! }[/math], the engineer rejects [math]\displaystyle{ H_0\,\! }[/math] and concludes that treatment means differ.
Results in DOE++
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 [math]\displaystyle{ H_0\,\! }[/math].
