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Lecture

22 factorials.pdf

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Department
Statistics
Course
STAT368
Professor
Douglas Wiens
Semester
Winter

Description
125 2 17. 2 factorials • We’ll 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 Total Label 28 25 27 80 = (1) + 36 32 32 100 = + 18 19 23 60 = + + 31 30 29 90 = 126 • Notation (1) = sum of obs’ns at low levels of both factors, = sum of obs’ns with A high and B low, = sum of obs’ns with B high and A low, = sum of obs’ns 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 dene 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 ˆ = average at high A - average at low A + (1) + = 2 2 + (1) = 2 This is the LSE. Reason: We know that the LSE of 2is ˆ = average at high overall average 2 and that of is 1 ˆ 1= average at low overall average so that ˆ = ˆ2 ˆ1 = average at high A - averageat low A. 128 • Often the ‘hats’ are omitted (as in the text). Sim- ilarly, + (1) = 2 = di erence between e ect of A at high B, and e ect of A at low B (1) = 2 2 = + (1) 2 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 isn’t much interaction e ect. To conrm this we would do an ANOVA. 129 > A B I I
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