Lec 5: Factorial Experiment

Transcription

1 November 21, 2011

2 Example Study of the battery life vs the factors temperatures and types of material. A: Types of material, 3 levels. B: Temperatures, 3 levels.

3 Example Study of the battery life vs the factors temperatures and types of material. A: Types of material, 3 levels. B: Temperatures, 3 levels. In general, factorial designs are most efficient for the study of the effects of two or more factors.

4 Factorial Design In each complete trial or replication of the experiment all possible combinations of the levels of the factors are investigated.

5 Factorial Design In each complete trial or replication of the experiment all possible combinations of the levels of the factors are investigated. Question What s the advantage of the factorial design comparing with two-single factor experient?

6 Advantages of factorial design The two-factor factorial gives estimates with higher precision (given the same number of experimental units).

7 Advantages of factorial design The two-factor factorial gives estimates with higher precision (given the same number of experimental units). The two-factor factorial experiment makes it possible to detect interactions.

8 Example

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11 Manual Calculation

12 Results for the example

13 15 o F 70 o F 125 o F Type Type Type average life average life Temperature type 1 type 2 type 3 Material type

14 15 o F 70 o F 125 o F Type Type Type

15 Multiple Comparison When interaction is significant, comparisons between the means of one factor (A) may be obscured by the interaction (AB). One approach to this situation is to fix factor B at a specific level and apply Tukey s test to the means of factor A at that level.

16 Multiple Comparison When interaction is significant, comparisons between the means of one factor (A) may be obscured by the interaction (AB). One approach to this situation is to fix factor B at a specific level and apply Tukey s test to the means of factor A at that level. Example MSE MSE T 0.05 = q 0.05 (a, f ) = q 0.05 (3, 27) = n 4 ȳ 13. ȳ 23. = 8 ȳ 33. ȳ 23. = 36 ȳ 33. ȳ 13. = 28

17 No interaction in a two-factor model

18 No interaction in a two-factor model

19 Model adequacy checking

20 One observation per cell case Model 1 y ij = µ + τ i + β j + (τβ) ij + ɛ ij i = 1, 2,..., a, j = 1, 2,..., b

21 One observation per cell case Model 1 y ij = µ + τ i + β j + (τβ) ij + ɛ ij i = 1, 2,..., a, j = 1, 2,..., b

22 One observation per cell case Model 1 y ij = µ + τ i + β j + (τβ) ij + ɛ ij i = 1, 2,..., a, j = 1, 2,..., b

23 One observation per cell case Model 1 y ij = µ + τ i + β j + (τβ) ij + ɛ ij i = 1, 2,..., a, j = 1, 2,..., b Model 2 y ij = µ + τ i + β j + ɛ ij i = 1, 2,..., a, j = 1, 2,..., b

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25 Tukey single-degree-of-freedom test It is helpful in determining whether interaction is present or not. The procedure assumes (τβ) ij = γτ i β j. Then the test partitions the residual sum of squares into two part. One is SS N = [ a i=1 with 1 degree of freedom, and b j=1 y ijy i. y.j y.. (SS A + SS B + y 2.. ab )]2 abss A SS B, SS Error = SS residual SS N, with (a 1)(b 1) 1 degrees of freedom. To test the present of interaction, we compute F 0 = SS N SS Error /[(a 1)(b 1) 1] If F 0 > F α,1,(a 1)(b 1) 1, the hypothesis of no interaction must be rejected.

26 Question Is the two-factor factorial model with one observation per cell identical to the randomized complete block model?

27 Blocking in a Factorial Design Example Study of the battery life vs the factors temperatures and types of material. A: Types of material, 3 levels. B: Temperatures, 3 levels. 9 combinations of A and B. 4 observations per combination. In total 36 observations.

28 Example Study of the battery life vs the factors temperatures and types of material. A: Types of material, 3 levels. B: Temperatures, 3 levels. 9 combinations of A and B. 4 observations per combination. In total 36 observations. 4 blocks.

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30 Source DF Anova SS MeanSquare F Value Pr > F temperature <.0001 material Interaction block Error Total

31 Three factorial experiment y ijkl = µ + τ i + β j + γ r + (τβ) ij + (τγ) ik + (βγ) jk + (τβγ) ijk + ɛ ijkl i = 1, 2,, a j = 1, 2,, b k = 1, 2,, c l = 1, 2,, n

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33 Example Soft drink bottling Fill correct volume Response variable: difference from correct volume Carbonation (10%, 12%, 14% ) Pressure (25psi, 30psi) Line speed (200bpm, 250bpm)

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