Research Design: Factorial Experiment-122799

  1. The major purpose was to focus on the survival and the growth of the mice. Hence, there were efforts made as per the animal distribution based on the below table:



Description Level





A Sprouts —- Free
B Frequency of Weighing One per day One per 3 days
C Nest Box No Yes
D Remove young No Yes
E Male Yes No
F Exercise wheel Locked Free
G Food 80 % Free
H water 80 % Free



The generators 28-4 which fractional design were:




The 7 strings of estimable confounded two factor interactions:









The different blocks are works on allowing to prevent the uncontrolled features. The duration chose k-p-q =8-4-1=3.

There is a generalise degree of 8-1=7 degree of freedom for blocks. The block contrasts the chosen factor:





The two factor







The alias relationship for 2k-pfractional factorial designs with k<=15 and n<=64.

Hence, X in the above figure can have a resolution with 14 factor design

Or design with 15 factor.




The four block 22=4. For the pattern of k-p-q =6-2-2=2 the block defining contrasts. The generators are based on the E=ABC and F=ABD

The defining relation is I=ABCE=ABDF=CDEF

The block management as per the contrasts BDE and ACDE. The generalised interaction BDE(ACDE)=ABC which accounts for the three degree of freedom for block.

The aliasing is confounded





The choice of generators are based on the better finding would be ACD and AB. There are many choices for different blocks which are based on aliasing the different structures.




Standard Yates order

Workable YO




Estimated Effects





A – B – BCD + ACD



– C +ABC + ABD – D




To get the workable Yates Order, one uses one or both of the dead letters to the chosen treatments from the Standard Yates order.  Thus to get the b of block 2 we multiply 1 by the dead letter b, we leave a alone, to get acd we multiply ac by the dead letter d and to get bcd we multiply c by the dead letters b and d.  The estimated effects need to be filled.  They are associated with the effects listed in the standard Yates order, so, 1 row is aliased with AB, ABCD, and CD.  The A row is aliased with B, BCD, and ACD (must include A).




  No of letters in the Effect    
No letters treatment combination has in common with he Effect Even Odd  
Even +  
Odd +  


To complete the table,  for the a row, the effect B has an odd number of letters and the treatment A has an even (0) number of letters in common with it.  So B gets a – sign, BCD is an effect with an odd number of letters and it has an even number of letters in common with the treatment A, so BCD gets a – sign, A has an odd number of letters in common with itself and so receives a + sign.   ACD has an odd number of letters but it has an odd number in common with the treatment, so its sign is +.  For the c row, the treatment is BCD, the aliases are C, ABC, ABD, and D, all odd.  BCD has an odd number in common with C, so the sign on C is -, it has an even in common with ABC and with ABD, so the signs for them are + and it has an odd number in common with D, so the sign is -.



So, we have run the following experiments, possibly with replication and gotten our average yields:

Treatment Average Yield
b 3
a 2
bcd 9
acd 5

One now applies the Yates algorithm to the workable Yates order on the treatments to obtain the effects.


Applying the Yates algorithm to the table above we would have


Step 1                                                               Step 2                                   Effect Estimated


a + b                                                                  a + b + bcd + acd                               —

bcd + acd                                                         a – b + bcd – acd               (A – B – BCD – ACD)2

a – b                                                                  a + b –bcd – acd                (-C + ABC + ABD – D)2

bcd – acd                                                         a – b – bcd + acd         (AC – BC + AD – BD)2


Unfortunately, we have designed very badly, because, with this design we cannot estimate any of the principal effects without assuming that one or two of them insignificant, but then why include them in the experiments?


It remains to see how to evaluate error when the designs are replicated.  Recall, we can consider a  experiment as if it were a one-way ANOVA with 4 factor levels.  In that case, we might summarize the data from the experiments as follows:



1 a b ab
Y11 Y12 Y13 Y14
Y21 : : :
Y31 : : :
: : : :
Yn1 Yn2 Yn3 Yn4

The Yates algorithm then yields an orthogonal contrast matrix


A = (-1 + a – b + ab)/2

B = (-1 – a + b + ab)/2

AB = (1 – a – b + ab)/2


So, we get an Orthogonal matrix  which we can use to evaluate the error for each of the effects being calculated.  The MSW term will have 4n – 4 df and the component of the error represented by each of the effects is .  The Fcalcs are  which must be evaluated in the FDIST(1,4n -4).