One Factor Comparison Design: Difference between revisions

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


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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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To test the hypothesis:
 
::<math>\begin{align}
H_0:\mu_1=\mu_2=...=\mu_5 \\
 
H_1:\mu_1\ne\mu_2\ne...\ne\mu_5
\end{align}</math>
 
{| {{Table}}
!rowspan="2" | Weight percentage of cotton
!colspan="5" | Observed tensile strength  (lb/in<math>^2</math>)
|-
!style="background:#f0f0f0;"|1
!style="background:#f0f0f0;"|2
!style="background:#f0f0f0;"|3
!style="background:#f0f0f0;"|4
!style="background:#f0f0f0;"|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
 
|}
 


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From the book the ANOVA table is:
 
{| {{Table}}
 
!Source of Variation
!Sum of Squares
!Degrees of Freedom
!Mean Square
!<math>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>\alpha=0.05\,\!</math>. Since <math>F_{0.05,4,20}=2.87\,\!</math>,  and <math>14.76 >2.87 \,\!</math>, the engineer rejects <math>H_0\,\!</math> and concludes that treatment means differ.
 


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The software results match the book results and the conclusions are matched. The ANOVA table is:
 
[[Image:one_factor_anova.png|center]]
 
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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