One Factor Comparison Design: Difference between revisions

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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^{2})
!colspan="5" | Observed tensile strength  ( lb/in^2)
|-
|-
!style="background:#f0f0f0;"|1
!style="background:#f0f0f0;"|1
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{{Reference_Example_Heading3}}
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From the book the ANOVA table is:
From the book the ANOVA table is:
{| {{Table}}
!Source of Variation
!Sum of Squares
!Degrees of Freedom
!Mean Square
!F_0
!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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Revision as of 21:40, 8 July 2015

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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:

[math]\displaystyle{ \begin{align} H_0:\mu_1=\mu_2=...=\mu_5 \\ H_1:\mu_1=\mu_2=...=\mu_5 \end{align} }[/math]
weight percentage of cotton Observed tensile strength ( lb/in^2)
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 F_0 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.


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