One Factor Comparison Design: Difference between revisions

From ReliaWiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(17 intermediate revisions by one other user not shown)
Line 1: Line 1:
{{Reference Example|{{Banner DOE Reference Examples}}}}
{{Reference Example|{{Banner DOE Reference Examples}}}}
Validate the calculation of one factor comparison design.
This example validates the calculation of the one factor comparison design in Weibull++.


{{Reference_Example_Heading1}}
{{Reference_Example_Heading1}}
Data is from Example 11-1 on page 71 in 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.


{{Reference_Example_Heading2}}
{{Reference_Example_Heading2}}
To test hypothesis:
 
To test the hypothesis:


::<math>\begin{align}
::<math>\begin{align}
H_0:\mu_1=\mu_2=...=\mu_5 \\
H_0:\mu_1=\mu_2=...=\mu_5 \\


H_1:\mu_1=\mu_2=...=\mu_5
H_1:\mu_1\ne\mu_2\ne...\ne\mu_5
\end{align}</math>
\end{align}</math>


{| {{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>)
|-
|-
!style="background:#f0f0f0;"|1
!style="background:#f0f0f0;"|1
Line 38: Line 39:


{{Reference_Example_Heading3}}
{{Reference_Example_Heading3}}
From the book the ANOVA table is:
From the book the ANOVA table is:


Line 59: Line 61:
|}
|}


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.  
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.  




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


Since 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>.

Latest revision as of 21:09, 18 September 2023

DOE Reference Examples Banner.png


New format available! This reference is now available in a new format that offers faster page load, improved display for calculations and images and more targeted search.

As of January 2024, this Reliawiki page will not continue to be updated. Please update all links and bookmarks to the latest references at DOE examples and DOE reference examples.




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].