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One Factor Comparison Design |
Validate the calculation of one factor comparison design.
Reference Case
Data is from Example 11-1 on page 71 in book “Design and Analysis of Experiments” by Douglas C. Montgomery, John Wiley & Sons, 2001.
Data To test hypothesis:
| weight percentage of cotton | Observed tensile strength ( lb/in[math]\displaystyle{ ^2 }[/math]) | ||||
|---|---|---|---|---|---|
| 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 | Sum of Squares | Degrees of Freedom | Mean Square | [math]\displaystyle{ F_0\,\! }[/math] | P value |
|---|---|---|---|---|---|
| Cotton weight percentage | 475.76 | 4 | 118.94 | 14.76 | <0.01 |
| Error | 161.20 | 20 | 8.06 | ||
| Total | 636.96 | 24 |
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. The ANOVA table is:
Since P Value =9.13E-6 < 0.01, we reject [math]\displaystyle{ H_0\,\! }[/math].