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Stat 470-14 Today: Finish Chapter 3; Start Chapter 4 Assignment 3: 3.14 a, b (do normal qq-plots only),c, 3.16, 3.17 Additional questions: 3.14 b (also.

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Presentation on theme: "Stat 470-14 Today: Finish Chapter 3; Start Chapter 4 Assignment 3: 3.14 a, b (do normal qq-plots only),c, 3.16, 3.17 Additional questions: 3.14 b (also."— Presentation transcript:

1 Stat 470-14 Today: Finish Chapter 3; Start Chapter 4 Assignment 3: 3.14 a, b (do normal qq-plots only),c, 3.16, 3.17 Additional questions: 3.14 b (also use the IER version of Lenth’s method and compare to the qq-plot conclusions), 3.19 Important Sections in Chapter 3: –3.1-3.7 (Read these) –3.8 (don’t bother reading), –3.11-3.13 (Read these)

2 Blocking in 2 k Experiments The factorial experiment is an example of a completely randomized design Often wish to block such experiments As an example, you may wish to use the same paper helicopter for more than one trial But which treatments should appear together in a block?

3 Blocking in 2 k Experiments Consider a 2 3 factorial experiment in 2 blocks

4 Blocking in 2 k Experiments Presumably, blocks are important. Effect hierarchy suggest we sacrifice higher order interactions Which is better: –b=ABC –b=AB Can write as:

5 Blocking in 2 k Experiments Suppose wish to run the experiment in 4 blocks b 1 =AB and b 2 =BC These imply a third relation Group is called the defining contrast sub-group

6 Blocking in 2 k Experiments Identifies which effects are confounded with blocks Cannot tell difference between these effects and the blocking effects

7 Which design is better? Suppose wish to run the 2 3 experiment in 4 blocks b 1 =AB and b 2 =BC I=ABb 1 =BCb 2 =ACb 1 b 2 Suppose wish to run the 2 3 experiment in 4 blocks b 1 =ABC and b 2 =BC I=ABCb 1 =BCb 2 =Ab 1 b 2

8 Ranking the Designs Let D denote a blocking design g i (D) is the number of i-factor interactions confounded in blocks (i=1,2,…k) For any 2 blocking schemes (D 1 and D 2 ), let r be the smallest i such that

9 Ranking the Designs Effect hierarchy suggests that the design that confounds the fewest lower order terms is best So, if then D 1 has less aberration A design has minimum aberration (MA) if no design has less aberration

10 Fractional Factorial Designs at 2-Levels 2 k factorial experiments can be very useful in exploring a relatively large number of factors in relatively few trials When k is large, the number of trials is large Suppose have enough resources to run only a fraction of the 2 k unique treatments Which sub-set of the 2 k treatments should one choose?

11 Example Suppose have 5 factors, each at 2-levels, but only enough resources to run 16 trials Can use a 16-run full factorial to design the experiment Use the 16 unique treatments for 4 factors to set the levels of the first 4 factors (A-D) Use an interaction column from the first 4 factors to set the levels of the 5 th factor

12 Example

13 Fractional Factorial Designs at 2-Levels Use a 2 k-p fractional factorial design to explore k factors in 2 k-p trials In general, can construct a 2 k-p fractional factorial design from the full factorial design with 2 k-p trials Set the levels of the first (k-p) factors similar to the full factorial design with 2 k-p trials Next, use the interaction columns between the first (k-p) factors to set levels of the remaining factors

14 Example Suppose have 7 factors, each at 2-levels, but only enough resources to run 16 trials Can use a 16-run full factorial to design the experiment Use the 16 unique treatments for 4 factors to set the levels of the first 4 factors (A-D) Use interaction columns from the first 4 factors to set the levels of the remaining 3 factors

15 Example The 3 relations imply other relations Words Defining contrast sub-group Word-length pattern

16 Example Would like to have as few short words as possible Why?

17 How can we compare designs? Resolution

18 Minimum aberration:

19 Example Suppose have 7 factors, each at 2-levels, but only enough resources to run 32 trials Can use a 2 7-2 fractional factorial design Which one is better? –D 1 : I=ABCDF=ABCEG=DEFG –D 2 : I=ABCF=ADEG=BCDEG


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