Solution HW6

# Solution HW6 - Designed Experimentation Solutions for HW6...

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Designed Experimentation Solutions for HW6 Analysis of Variance for Result Source DF SS MS F P Process 2 640.50 320.25 1.41 0.293 Batch(Process) 9 2040.92 226.77 11.17 0.000 Error 24 487.33 20.31 Total 35 3168.75 S = 4.50617 R-Sq = 84.62% R-Sq(adj) = 77.57% First, because batch is a random variable, we must calculate the difference in variances and check the significance manually using an F table. (Note: since process is a fixed variable, we are evaluating the differences in mean response, so we can just directly evaluate this from its P value as we have done throughout the course. If this is confusing, I suggest that you check back and review Chapter 13 which explains the differences in evaluating random and fixed variables.) σ 2 (for batch) = [ MS(batch) – MS(error) ] / n = [226.77 – 20.31] / 3 = 68.82 σ 2 (for error) = MS(error) F 0 = σ 2 (for batch) / σ 2 (for error) = 68.82 / 20.31 = 3.39. Now, comparing that to F

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## This note was uploaded on 08/27/2011 for the course EIN 4905 taught by Professor Staff during the Spring '08 term at University of Florida.

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Solution HW6 - Designed Experimentation Solutions for HW6...

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