lect19 - IE 410 Lecture 19: Chapter 7 Confounding in 2...

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IE 410 Lecture 19: Chapter 7 Confounding in 2 k Designs Or how to run 2 k designs in blocks
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MOTIVATION I have referred previously to a process of experimentation for factor screening. Running 2 k designs in blocks is an important part of this process. In the process we may collect additional data after the first set of experiments provides initial results.
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Thus we will have two (or more) sets of data collected at different points in time. Clearly these datasets cannot simply be combined and analyzed as if it were one big dataset. WHY?
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Because it is not properly randomized. The two datasets can be thought of as two blocks, and can then be properly analyzed.
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Suppose we have a 2 2 design with n=1 replicate. The expt involves chemicals that come in batches, and a batch contains only enough raw material for 2 runs. Thus our 2 2 must be run in two blocks.
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Suppose we did the following: A 0 1 ----------------- 0 | 5 6 | Block 1 ----------------- B ----------------- 1 | 21 20 | Block 2 -----------------
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Is there a B main effect? B = (21+20-5-6)/2 = 15 It looks like B is important. Or is the effect due to raw material batches? Clearly the B main effect and blocks are completely confounded.
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We could do this instead: B 0 1 ----------------- 0 | 5 21 | Block 1 ----------------- A ----------------- 1 | 6 20 | Block 2 -----------------
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However, now the A main effect is completely confounded. We have 2 blocks that have an associated 1 df. Blocks will always be confounded with one of the effects.
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Given that this is true, we should confound with a high order interaction that we care little or nothing about! In a 2 2 , it seems risky. Even so, we can still estimate the main effects.
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In a 2 5 , confounding blocks with the ABCDE interaction seems like a good idea. How do we determine the appropriate design? I think the easiest method is to use the contrast coefficient table, i.e. the so called "table of +'s and -'s"
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Example: 2 2 cell A B AB Block ------------------- ----- (1) - - + 1 a + - - 2 b - + - 2 ab + + + 1
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with the AB interaction. We can tell because they are identical
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This note was uploaded on 08/06/2008 for the course IE 410 taught by Professor Storer during the Fall '04 term at Lehigh University .

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lect19 - IE 410 Lecture 19: Chapter 7 Confounding in 2...

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