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Multi-Factor Studies

Multi-Factor Studies. Stat 701 Lecture E. Pena. An Example with Two Factors. To study effects of carbon content and tempering temperature on the strength of steel. Factor A: Carbon Content Factor A Levels: a 1 = Low Carbon level; a 2 = High Carbon level Factor B: Tempering Temperature

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Multi-Factor Studies

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  1. Multi-Factor Studies Stat 701 Lecture E. Pena

  2. An Example with Two Factors • To study effects of carbon content and tempering temperature on the strength of steel. • Factor A: Carbon Content • Factor A Levels: • a1 = Low Carbon level; a2 = High Carbon level • Factor B: Tempering Temperature • Factor B Levels: • b1 = Low temperature; b2 = High temperature • This will be an example of a 22-factorial design

  3. Factor Level Combinations Possible Treatment Combinations: Treatment 1: a1b1 = Low Carbon and Low Temperature Treatment 2: a1b2 = Low Carbon and High Temperature Treatment 3: a2b1 = High Carbon and Low Temperature Treatment 4: a2b2 = High Carbon and High Temperature Possible Experimental Approaches: • Just use a completely randomized design with the four • treatments above. What will be the disadvantages? • Employ a factorial experiment where the two factors are delineated so their main effects and interaction effects could be ascertained.

  4. Why Multi-Factor Studies? Advantages: • Efficiency and economy • Informativeness • Validity of findings Disadvantages: • Could be costly if not properly designed. • If interaction effects are strong, will be hard to determine or interpret main effects of factors. • May need to employ fractional designs where only a subset of all possible treatments are included.

  5. Two-Factor Factorial Model(Balanced Design) Let there be two factors: A and B Factor A has levels: a1, a2, …, aA Factor B has levels: b1, b2, …, bB Number of treatment combinations: AB In a balanced design, the same number ofexperimental units per treatment combination is allocated. Let k denote the number of eu’s allocated to each treatment combination. Then n = ABk is the total number of eu’s in the study. Note: In allocating eu’s, the principle of randomization should be observed.

  6. Model for Two-Factor Factorial Analysis Yijl = lth observation in treatment (ai,bj) Yijl = mij + eijl mij is the mean response of eu’s assigned (aibj) eijl is the error component for the lth unit in (aibj) • Assumptions (fixed effects model): • mij are fixed (but unknown) quantities • Errors each have mean zero • Errors are uncorrelated (independent) from each other • Errors have equal variances • Errors have normal distributions

  7. Interpretation of Parameters Consider a two-factor study with Factor A having 2 levels and Factor B having 3 levels. For each treatment combination, we have the (population) mean response mij. We may summarize this in a table of form: Factor A Main Effects: ai = mi. - m.. for i=1,2,…,A Factor B Main Effects: bj = m.j - m.. for j=1,2,…,B Interaction Effects: (ab)ij = mij - mi. - m.j + m.. for i=1,…,A; j=1,..,B Model: mij = m.. + ai + bj + (ab)ij for i=1,…,A; j=1,…,B.

  8. Interpretation … continued Identities Case 1: No interaction effects is when (ab)ij = 0 for all i,j. What happens in this case? Pictorial representation. Case 2: Interaction effects are present. Pictorial representation. Are main effects meaningful? Strong and weak interactions.

  9. Two Examples Example 1: A model without interaction effects. Factor A Main Effects: a1 = 4 - 5 = -1; a2 = 6 - 5 = +1 Factor B Main Effects: b1 = 5 - 5 = 0; b2 = 8 - 5 = +3; b3 = 2 - 5 = -3 Interaction Effects: (ab)11 = 4 - 4 - 5 + 5 = 0; all of them are zeros. A Pictorial Representation of these Treatment Means?

  10. Examples … continued Example 2: A model with interaction effects. Factor A Main Effects: a1 = 5 - 5 = 0; a2 = 5 - 5 = 0 Factor B Main Effects: b1 = 5 - 5 = 0; b2 = 8 - 5 = +3; b3 = 2 - 5 = -3 Interaction Effects: (ab)11 = 4 - 5 - 5 + 5 = -1; (ab)12 = 9 - 5 - 8 + 5 = +1; (ab)13 = 2 - 5 - 2 + 5 = 0; (ab)21 = 6 - 5 - 5 + 5 = +1; (ab)22 = 7 - 5 - 8 + 5 = -1; (ab)23 = 2 - 5 - 2 + 5 = 0. Note that the sum of these effects is 0. Would appear that there are no differences in Factor A levels! Pictorial Representation?

  11. Example: Drug Development for Hay Fever

  12. Tabular Presentation of Factorial Data with Totals and Means

  13. Estimates of Main and Interaction Effects • Factor A Main Effects • A1: 3.88 - 7.18 = -3.3 • A2: 7.83 - 7.18 = 0.65 • A3: 9.83 - 7.18 = 2.65 • Factor B Main Effects • B1: 4.63 - 7.18 = -2.55 • B2: 7.93 - 7.18 = 0.75 • B3: 8.98 - 7.18 = 1.80 Interaction Effects (A Level, B Level) (A1, B1): 2.475 - 3.88 - 4.63 + 7.18 = 1.145 (A1, B2): 4.6 - 3.88 - 7.93 + 7.18 = -0.03 (A1, B3): 4.575 - 3.88 - 8.98 + 7.18 = -1.105 (A2, B1): 5.45 - 7.83 - 4.63 + 7.18 = 0.17 (A2, B2): 8.925 - 7.83 - 7.93 + 7.18 = 0.345 (A2, B3): 9.125 - 7.83 - 8.98 + 7.18 = -.505 (A3, B1): 5.975 - 9.83 - 4.63 + 7.18 = -1.305 (A3, B2): 10.275 - 9.83 - 7.93 + 7.18 = -.305 (A3, B3): 13.25 - 9.83 - 8.98 + 7.18 = 1.62 Estimates of Interaction effects far from zeros, so indicative of interactions between A, B

  14. Three Dimensional Plot of the Estimated Cell Means

  15. Plot of the Estimated Cell Means in Two-Dimensions

  16. Treatment Means and Factor Level Means

  17. Results of Two-Factor Analysis using Minitab Conclusions?

  18. Histogram of Residuals

  19. Normal Probability Plot

  20. Plot of Residuals vs Order

  21. Residuals vs Predicted Values

  22. SAS Program to Perform Two-Way Analysis with Interaction Together with Analysis of Means /* Hay Fever Drug Development */ data hay; input relief FactorA $ FactorB $ RepNum @@; cards; (Insert the data set here) ; procprint; procanova; class FactorA FactorB; model relief = FactorA FactorB FactorA*FactorB; means FactorA FactorB FactorA*FactorB / tukeybon; run; The “tukey” and “bon” keywords are for performing theTukey multiple comparisons procedure, while “bon” is for the Bonferroni procedure. One may also use the PROC GLM command above instead of the PROC ANOVA command. The former command is recommended for unbalanced designs.

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