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Unformatted text preview: What If There Are More Than Two Factor Levels? • There are lots of practical situations where there are either more than two levels of interest, or there are several factors of simultaneous interest • The ttest does not directly apply • The analysis of variance (ANOVA) is the appropriate analysis “engine” for these types of experiments – Chapter 3, textbook • Used extensively today for industrial experiments Chapter 3 Design & Analysis of Experiments 7E 2009 Montgomery 2 An Example (See pg. 61) • An engineer is interested in investigating the relationship between the RF power setting and the etch rate for this tool. The objective of an experiment like this is to model the relationship between etch rate and RF power, and to specify the power setting that will give a desired target etch rate. • The response variable is etch rate. • She is interested in a particular gas (C2F6) and gap (0.80 cm), and wants to test four levels of RF power: 160W, 180W, 200W, and 220W. She decided to test five wafers at each level of RF power. • The experimenter chooses 4 levels of RF power 160W, 180W, 200W, and 220W • The experiment is replicated 5 times – runs made in random order Chapter 3 Design & Analysis of Experiments 7E 2009 Montgomery 3 Chapter 3 Design & Analysis of Experiments 7E 2009 Montgomery 4 An Example (See pg. 62) Chapter 3 Design & Analysis of Experiments 7E 2009 Montgomery 5 • Does changing the power change the mean etch rate? • Is there an optimum level for power? • We would like to have an objective way to answer these questions • The ttest really doesn’t apply here – more than two factor levels Analysis of Variance (ANOVA) s 2 = 1 Σ (y i y gav ) 2 n  1 • A convenient and rapid calculation method for separating the variability possessed by a collection of observations into components or sources. • It is always connected with a mathematical model and with the type of experimental design employed. Example: Two different additives to a ceramic material are compared with respect to a measured response of the ceramic product and the following data are obtained: Treatment # 1 (9,7,13,7) Total= 36 ; n 1 = 4 Treatment # 2 (4,5,6,2,5,2) Total= 24 ; n 2 = 6 Construct the analysis of variance table for the above model and data and test the hypothesis that μ 1 μ 2 = 0 Treatment # 1 (9,7,13,7) Total= 36 ; n 1 = 4 Average: y av1 = 9 Treatment # 2 (4,5,6,2,5,2) Total= 24 ; n 2 = 6 Averages: y av2 = 4 y gav. = (Grand Average)= (36+24)/10 = 6 y gav. y av1 = 6 – 9 =  3 = Effect of Treatment 1 = τ 1 y gav. – y av2 = 6 – 4 = 2 = Effect of Treatment 2 = τ 2 Simple Calculations Sum of Degrees of Mean Square Squares Freedom Crude Sum of Squares: Σ y 2 ij 458 10 Correction Factor: n (y gav ) 2 = ( Σ y) 2 /n 360 1 Treatment Sum of Squares: Σ n j τ j 2 60 1 60.0 = s 2 σ 2 if τ j = 0 Residual Sum of Squares 38 8 S 2 = 38/8 = 4.75 y ĳ = μ + τ j + ε ĳ ; ε ĳ NID(0,σ 2 ) The ANOVA table can be written as follows: Treatment SSq...
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 Spring '10
 ElShall
 Normal Distribution, Regression Analysis, Variance, Analysis of Experiments

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