Fractional Factorials

```2k  p _ Designs
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2k-p Designs
 k factors
 two levels for each factor
 will only run 2-p of the possible factor combinations
 will only run 2k-p observations total
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 will only have 2k-p-1 degrees of freedom total
 can only estimate that many terms in the model
(actually less if we are to have degrees of freedom for
error)
 many terms are confounded with each other
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What do we need to do?
 determine which terms we are most interested in (Main
Effects, two-way Interactions, etc.)
 figure out which treatment combinations to run
 then figure out which terms in the Model are
confounded with each other
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Let’s start
 The minimum we want out of any fractional factorial is
to retrieve Main Effects
 Pick an Identity to generate incomplete Blocks for the
fractional factorial
 Use Orthogonal Contrasts to get the Identity!!!
 (knew Dr. Tom was going over those for some reason
:p)
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Easiest case, 2k-1, a half fraction
 Recall the three factor design with A, B and C
 There were eight treatment combinations
 Lets look at the contrast for ABC
ABC
(1) a b ab c ac bc abc
- + + - + +
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This Identity gives us two (incomplete) Blocks
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How do we know which treatment combinations go in
the minus (-) block and which in the plus (+) block?
 See what factors are in the experiment
 For the 23-1 it is Factors A, B and C
 List the treatment combinations, i.e. (1), a, b, ab, c, ac,
bc, abc
 If I block on ABC, then count the number of letters in a
treatment combination whose upper cases are in the
blocking term, here it is ABC
 If that number is odd, it goes into the one block,
otherwise it goes into the other block (useful for the
next HW problem)
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The Identity
 I=ABC is the “Identity”
 The Identity gives us the incomplete blocks
 The Identity tells us the pattern of confounding
 A=A2BC=BC, for example
 and how do we know that, pray tell?
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It’s all Group theory, of course! A semigroup has
an operator * such that:
*: G  G  G
(a * b) * c  a *(b * c)
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For fractional factorials, the Contrasts for the
Effects (plus a vector of 1’s) form a semigroup
If G is our semigroup, then the members of G include
the contrasts:
(-1,+1,-1,+1,-1,+1,-1,+1) for A
(-1,-1,+1,+1,-1,-1,+1,+1) for B
(-1,-1,-1,-1,+1,+1,+1,+1) for C
and A*B (which we will denote as AB from now on) gives
(+1,-1,-1,+1,+1,-1,-1,+1)
as long as * means the pair-wise product of elements for
the vectors. We could have done this in terms of
Matrices but we’ll stick to the shorthand notation.
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Is our semigroup, G a Group?
Yes, as long as we include:
AA=A2=I and
I=(1,1,1,1,1,1,1,1).
In Group theory, I is referred to as the identity,
although for Fractional Factorials this term has a
double meaning for determining “aliases”. Aliases tell us
which terms are confounded with which other terms.
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Back to the 23-1
 For the 23-1 design, we run one incomplete block
 The Identity, I=ABC, gave us our incomplete blocks, and we
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run one of them (doesn’t matter which for the analysis)
The Identity also tells us which terms are confounded with
each other
So A=BC, B=AC and C=BC
So at least the Main Effects are not confounded with each
other
This is the minimum we want out of any fractional factorial
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 A 23-1 design is not too useful since we have no degrees
of freedom for error (though you can at least estimate
the Main Effects)
 For more factors you just need the Identity to specify
the design (很容易！！)
 For half fractions we always use the highest order
interaction
 So for a 25-1 design we use I=ABCDE for the Identity to
generate the design
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What if….
 My client comes in with six factors
 I recommend a 26-1 fractional factorial
 My client asks how many observations
 You say 32
 Client says that will use up the rest of the research
money
 So how about a 26-2 fractional factorial?
 Your client can afford 16 observations and have some
money left over for further experiments!
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How to set up a 26-2 fractional factorial?
 We need to divide up our treatment combinations in
half, then divide those in half again
 This gives us a &frac14; fraction
 We need to choose two terms in the Model to set up two
incomplete blocks
 How about ABCD and CDEF?
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What are the incomplete blocks that my
treatment combinations go to?
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What is the pattern of confounding?
 Well, if we ran only &frac14; of the possible combinations, you can
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show that terms must be confounded in groups of four
I=ABCD=CDEF doesn’t give all of the confounded terms, but
if we include one more term…
I=ABCD=CDEF=ABEF
Where did ABEF come from?
ABCDCDEF=ABC2D2EF=ABEF
ABEF is called the Generalized Interaction of ABCD and
CDEF
That part was not intuitive (unless you are a Math head),
but everything is accounted for now
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Review of the basics:
 We generate the design by choosing one or more terms
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in the model to set up our incomplete blocks
The number of terms we chose determines the degree of
the fractional factorial
One term gives a 2k-1, two terms gives a 2k-2, three terms
gives a 2k-3 and so forth
The terms, along with the generalized interactions, give
the Identity
The Identity tells us what is confounded with what
Whew!!
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Back to the 2k-2
The Identity was I=ABCD=CDEF=ABEF
This tells us that:
 A=BCD=ACDEF=BEF
 B=ACD=BCDEF=AEF
 AB=CD=ABCDEF=EF
 etc.
 So main effects are not confounded with each other, but
two ways are confounded with each other
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How many fractional factorials are there and how
do I choose which one to use?
 Lots of possibilities exist
 You decide based upon which terms you are interested
in estimating/testing
 Generally, this means you want Main effects at the
minimum
 You don’t want Main effects confounded with things
that are likely to be significant
 Low order terms are likely to be significant, high order
terms are less likely to be insignificant
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Is there an easy way to tell?
 For once, the answer is yes (WooHoo!, 太好了！, Super!)
 Look at the Identity (example: I=ABCD=CDEF=ABEF)
 Count the number of letters in the shortest string of
letters (other than I), that is the Resolution of the
design (for the example above it is Resolution IV)
 The higher the Resolution the better
 Most Statistical Software packages will actually
generate the designs for you if you ask for designs of a
certain Resolution for a given number of factors (they
give you the treatment combinations to run also)
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For example:
 For a 23-1 with I=ABC, the Resolution is III
 For a 24-1 with I=ABCD, the Resolution is IV
 For a 26-2 with I=ABCD=CDEF=ABEF, the Resolution is
IV
 For a 26-2 with I=ABC=DEF=ABCDEF, the Resolution is
III
 Resolution IV is better then Resolution III since Main
effects are confounded with three way and higher
terms, while the Resolution III design confounds Main
effects with two-way interactions which are more likely
to be significant than higher order terms
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Does this get more complicated?
 Of course
 You may need to run the experiment in Blocks
 You may have more than two levels in some of your factors
 You may want to fit a curvilinear response if most variables
are quantitative (Chemical Engineers do this all the time,
it’s called Response Surface Methodology)
 Go see a Statistical consultant who understands this stuff,