One Factor Comparison Design: Difference between revisions

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!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<math>^2</math>)
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!Degrees of Freedom
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!Mean Square
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!F_0
!<math>F_0\,\!</math>
!P value
!P value
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Revision as of 21:45, 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[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.


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