1. 1 Chapter 6 The 2 k Factorial Design

2. 2 6.1 Introduction The special cases of the general factorial design (Chapter 5) k factors and each factor has only two levels Levels: –quantitative (temperature, pressure,…), or qualitative (machine, operator,…) –High and low –Each replicate has 2    2 = 2 k observations

3. 3 Assumptions: (1) the factor is fixed, (2) the design is completely randomized and (3) the usual normality assumptions are satisfied Wildly used in factor screening experiments

4. 4 6.2 The 2 2 Factorial Design Two factors, A and B, and each factor has two levels, low and high. Example: the concentration of reactant v.s. the amount of the catalyst (Page 208)

5. 5 “-” And “+” denote the low and high levels of a factor, respectively Low and high are arbitrary terms Geometrically, the four runs form the corners of a square Factors can be quantitative or qualitative, although their treatment in the final model will be different

6. 6 Average effect of a factor = the change in response produced by a change in the level of that factor averaged over the levels if the other factors. (1), a, b and ab: the total of n replicates taken at the treatment combination. The main effects:

7. 7 The interaction effect: In that example, A = 8.33, B = -5.00 and AB = 1.67 Analysis of Variance The total effects:

8. 8 Sum of squares:

9. 9 Response:Conversion ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum ofMeanF SourceSquaresDFSquareValueProb > F Model291.67397.2224.820.0002 A208.331208.3353.19< 0.0001 B75.00175.0019.150.0024 AB8.3318.332.130.1828 Pure Error31.3383.92 Cor Total323.0011 Std. Dev.1.98R-Squared 0.9030 Mean27.50Adj R-Squared0.8666 C.V.7.20Pred R-Squared0.7817 PRESS70.50Adeq Precision11.669 The F-test for the “model” source is testing the significance of the overall model; that is, is either A, B, or AB or some combination of these effects important?

10. 10 Table of plus and minus signs: IABAB (1)+––+ a++–– b+–+– ab++++

11. 11 The regression model: –x 1 and x 2 are coded variables that represent the two factors, i.e. x 1 (or x 2 ) only take values on – 1 and 1.

12. –Use least square method to get the estimations of the coefficients –For that example, –Model adequacy: residuals (Pages 213~214) 12

13. 13 Response surface plot: –Figure 6.3

14. 14 6.3 The 2 3 Design Three factors, A, B and C, and each factor has two levels. (Figure 6.4 (a)) Design matrix (Figure 6.4 (b)) (1), a, b, ab, c, ac, bc, abc 7 degree of freedom: main effect = 1, and interaction = 1

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16. 16 Estimate main effect: Estimate two-factor interaction: the difference between the average A effects at the two levels of B

17. 17 Three-factor interaction: Contrast: Table 6.3 –Equal number of plus and minus –The inner product of any two columns = 0 –I is an identity element –The product of any two columns yields another column –Orthogonal design Sum of squares: SS = (Contrast) 2 /8n

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19. 19 Factorial Effect Treatment Combination I ABABCACBC ABC (1)+ –– + – ++ – a ++ –––– ++ b + – + –– + – + ab++++ –––– c+ –– ++ –– + ac++ –– ++ –– bc+ – + – + – + – abc++++++++ Contrast 2418614244 Effect 3.002.250.751.750.250.50 Table of – and + Signs for the 2 3 Factorial Design (pg. 218)

20. 20 Example 6.1 A = gap, B = Flow, C = Power, y = Etch Rate

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22. 22 The regression model and response surface: –The regression model: –Response surface and contour plot (Figure 6.7)

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25. 25 6.4 The General 2 k Design k factors and each factor has two levels Interactions The standard order for a 2 4 design: (1), a, b, ab, c, ac, bc, abc, d, ad, bd, abd, cd, acd, bcd, abcd

26. 26 The general approach for the statistical analysis: –Estimate factor effects –Form initial model (full model) –Perform analysis of variance (Table 6.9) –Refine the model –Analyze residual –Interpret results

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28. 28 6.5 A Single Replicate of the 2 k Design These are 2 k factorial designs with one observation at each corner of the “cube” An unreplicated 2 k factorial design is also sometimes called a “single replicate” of the 2 k If the factors are spaced too closely, it increases the chances that the noise will overwhelm the signal in the data

29. 29 Lack of replication causes potential problems in statistical testing –Replication admits an estimate of “pure error” (a better phrase is an internal estimate of error) –With no replication, fitting the full model results in zero degrees of freedom for error Potential solutions to this problem –Pooling high-order interactions to estimate error (sparsity of effects principle) –Normal probability plotting of effects (Daniels, 1959)

30. 30 Example 6.2 (A single replicate of the 2 4 design) –A 2 4 factorial was used to investigate the effects of four factors on the filtration rate of a resin –The factors are A = temperature, B = pressure, C = concentration of formaldehyde, D= stirring rate

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32. 32 Estimates of the effects TermEffectSumSqr% Contribution Model Intercept Error A21.6251870.5632.6397 Error B3.12539.06250.681608 Error C9.875390.0626.80626 Error D14.625855.56314.9288 Error AB0.1250.06250.00109057 Error AC-18.1251314.0622.9293 Error AD16.6251105.5619.2911 Error BC2.37522.56250.393696 Error BD-0.3750.56250.00981515 Error CD-1.1255.06250.0883363 Error ABC1.87514.06250.245379 Error ABD4.12568.06251.18763 Error ACD-1.62510.56250.184307 Error BCD-2.62527.56250.480942 Error ABCD1.3757.56250.131959 Lenth's ME6.74778 Lenth's SME13.699

33. 33 The normal probability plot of the effects

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36. 36 B is not significant and all interactions involving B are negligible Design projection: 2 4 design => 2 3 design in A,C and D ANOVA table (Table 6.13)

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38. 38 Response:Filtration Rate ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum ofMeanF SourceSquaresDFSquareValueProb >F Model5535.8151107.1656.74< 0.0001 A1870.5611870.5695.86< 0.0001 C390.061390.0619.990.0012 D855.561855.5643.85< 0.0001 AC1314.0611314.0667.34< 0.0001 AD1105.5611105.5656.66< 0.0001 Residual195.121019.51 Cor Total5730.9415 Std. Dev.4.42R-Squared0.9660 Mean70.06Adj R-Squared0.9489 C.V.6.30Pred R-Squared0.9128 PRESS499.52Adeq Precision20.841

39. 39 The regression model: Residual Analysis (P. 235) Response surface (P. 236) Final Equation in Terms of Coded Factors: Filtration Rate = +70.06250 +10.81250 * Temperature +4.93750 * Concentration +7.31250 * Stirring Rate -9.06250 * Temperature * Concentration +8.31250 * Temperature * Stirring Rate

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42. 42 Half-normal plot: the absolute value of the effect estimates against the cumulative normal probabilities.

43. 43 Example 6.3 (Data transformation in a Factorial Design) A = drill load, B = flow, C = speed, D = type of mud, y = advance rate of the drill

44. 44 The normal probability plot of the effect estimates

45. 45 Residual analysis

46. 46 The residual plots indicate that there are problems with the equality of variance assumption The usual approach to this problem is to employ a transformation on the response In this example,

47. 47 Three main effects are large No indication of large interaction effects What happened to the interactions?

48. 48 Response:adv._rate Transform: Natural log Constant: 0.000 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum ofMeanF SourceSquaresDFSquareValueProb > F Model7.1132.37164.82< 0.0001 B5.3515.35371.49< 0.0001 C1.3411.3493.05< 0.0001 D0.4310.4329.920.0001 Residual0.17120.014 Cor Total7.2915 Std. Dev.0.12R-Squared 0.9763 Mean1.60Adj R-Squared0.9704 C.V.7.51Pred R-Squared0.9579 PRESS0.31Adeq Precision34.391

49. 49 Following Log transformation Final Equation in Terms of Coded Factors: Ln(adv._rate) = +1.60 +0.58 * B +0.29 * C +0.16 * D

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51. 51 Example 6.4: –Two factors (A and C) affect the mean number of defects –A third factor (B) affects variability –Residual plots were useful in identifying the dispersion effect –The magnitude of the dispersion effects: –When variance of positive and negative are equal, this statistic has an approximate normal distribution

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56. 6.7 2 k Designs are Optimal Designs Consider 2 2 design with one replication. Fit the following model: Matrix form: 56

57. The LS estimation: D-optimal criterion, |X’X|: the volumn of the joint confidence region that contains all coefficients is inversely proportional to the square root of |X’X|. G-optimal design: 57

58. 58 6.8 The Addition of Center Points to the 2 k Design Based on the idea of replicating some of the runs in a factorial design Runs at the center provide an estimate of error and allow the experimenter to distinguish between two possible models:

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60. 60 The hypotheses are: This sum of squares has a single degree of freedom To detect the possibility of the quadratic effects: add center points

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62. 62 Example 6.6 Refer to the original experiment shown in Table 6.10. Suppose that four center points are added to this experiment, and at the points x1=x2 =x3=x4=0 the four observed filtration rates were 73, 75, 66, and 69. The average of these four center points is 70.75, and the average of the 16 factorial runs is 70.06. Since are very similar, we suspect that there is no strong curvature present. Usually between 3 and 6 center points will work well Design-Expert provides the analysis, including the F-test for pure quadratic curvature

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65. 65 If curvature is significant, augment the design with axial runs to create a central composite design. The CCD is a very effective design for fitting a second-order response surface model