One Factor Comparison Design: Difference between revisions

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{{Reference Example|{{Banner DOE Reference Examples}}}}
{{Reference Example|{{Banner DOE Reference Examples}}}}
This example validates the calculation of one factor comparison design in DOE++.
This example validates the calculation of the one factor comparison design in Weibull++.


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{{Reference_Example_Heading1}}
The data is from Example 11-1 on page 71 in the book ''Design and Analysis of Experiments''  by Douglas C. Montgomery, John Wiley & Sons, 2001.
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.


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{{Reference_Example_Heading2}}
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{| {{Table}}
{| {{Table}}
!rowspan="2" | weight percentage of cotton
!rowspan="2" | Weight percentage of cotton
!colspan="5" | Observed tensile strength  (lb/in<math>^2</math>)
!colspan="5" | Observed tensile strength  (lb/in<math>^2</math>)
|-
|-
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{{Reference_Example_Heading4|DOE++}}
{{Reference_Example_Heading4|DOE++}}
The software results match the book results. The ANOVA table is:
The software results match the book results and the conclusions are matched. The ANOVA table is:


[[Image:one_factor_anova.png|center]]
[[Image:one_factor_anova.png|center]]


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

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One Factor Comparison Design

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 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 and the conclusions are matched. The ANOVA table is:

One factor anova.png

Since the P Value = 9.13E-6 < 0.01, we reject [math]\displaystyle{ H_0\,\! }[/math].

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