17. 2 factorials
Well start with a basic 2sign, where it is easy
to see what is going on. Also, these are very
widely used in industrial experiments.
Two factors (A and B), each at 2 levels - low
( ) and high (+). # of replicates =.
Example - investigate yield ( ) of a chemical process
when the concentration of a reactant (the primary
substance producing the yield) - factor A - and
amount of a catalyst (to speed up the reaction)
- factor B - are changed.E.g. nickel is used as
a catalyst, or a carrier of hydrogen in the hy-
drogenation of oils (the reactants) for use in the
manufacture of margarine.
Factor = 3 replicates
28 25 27 80 = (1)
+ 36 32 32 100 =
+ 18 19 23 60 =
+ + 31 30 29 90 = 126
(1) = sum of obsns at low levels of both factors,
= sum of obsns with A high and B low,
= sum of obsns with B high and A low,
= sum of obsns with both high.
E ects model. Use a more suggestive notation:
= + + +( ) + ( = 1 2 = 1 )
E.g. 1 = main e ect of low level of A2 =
main e ect of high level of A. But sin1e+
2= 0, we have 1 = 2.
We dene the main e ect of Factor A to be
2 1 127
What is the LSE of ? Since is the e ect of
changing factor A from high to low, we expect
= average at high A - average at low A
+ (1) +
This is the LSE.
Reason: We know that the LSE of 2is
= average at high overall average
and that of is
1= average at low overall average
= 2 1
= average at high A - averageat low A. 128
Often the hats are omitted (as in the text). Sim-
= di erence between e ect of A at high B,
and e ect of A at low B
= + (1)
With (1) = 80 = 100 = 60 = 90 we nd
= 8 33
= 5 0
= 1 67
It appears that increasing the level of A results in
an increase in yield; that the opposite is true of
B, and that there isnt much interaction e ect.
To conrm this we would do an ANOVA. 129
> A B I I