CS代考 Math 558 Lecture #22

Math 558 Lecture #22

The empirical model
The calculations done in lecture 21 can be used to determine an empirical model of a response using a 2k factorial design when we have small number of levels and the experimental design is without replications. In this case the number of runs is equal to the number of parameters in the model.

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If we include only first order variables, we have one more experimental run than there are variables in our empirical model, the summary provides some information on the significance of the model’s parameters; however, with just one degree of freedom this information is not really reliable. While including interactions we are left with 0 degrees of freedom. In order to study the statistical significance of the factors of interest and their interaction we need to have more replicates of our experiments. The example discussed in the following slides will give you an idea.

Analysis of 2 × 2 Factorial Designs
This example considers the effect of the the concentration of a reactant and the amount of a catalyst on the conversion (yield) in a chemical process. The objective of the experiment is to investigate the effect of these two factors on the conversion. In other words the objective is to optimize the amount of both factors to maximize the yield. Let us denote the reactant concentration by the factor A with the two levels of interest be 15 and 25 percent. Similarly represent the catalyst by factor B, with the levels, 1 and 2 pounds. There are three replications of the experiments. This means that all the treatments (factor combinations) are replicated 3 times. The experimental runs are performed in the random order so this is a completely randomized experiment. The data obtained are given in the next slide.

Analysis of 2 × 2 Factorial Designs
Factor Replicate Combinations I II −−2825
+−3632 −+1819 ++3130
III Total 27 80 32 100 23 60 29 90

Analysis of 2 × 2 Factorial Designs
We have already seen that the simple effect of A at low levels of B is
(a) − (1) and at high levels of B is (ab) − (b) when the experiment is
performed once. When there are n replicates, the simple effects are the
average of the effects for each replicate. That is (a)−(1) for low level of n
B and (ab)−(b) . So the main effect of A for n replicates is n
A = 1 [(ab)−(b)+(a)−(1)] 2n
B = 1 [(ab)−(a)+(b)−(1)] 2n
AB = 1 [(ab)−(b)−(a)−(1)] 2n

Analysis of 2 × 2 Factorial Designs
A = 1 (90+100−60−80) = 8.33 2(3)
1 (90−100+60−80) = −5.00 2(3)
AB = 1 2(3)
(90−100−60+80) = 1.67

Analysis of 2 × 2 Factorial Designs
In experiments using factorial designs, it is always important to examine the magnitude and direction of the factor effects to determine which variables are likely to be important. The analysis of variance or t-tests can generally be used to confirm this interpretation .

Calculations for sum of squares for A, B, and AB
From the expression giving the main effects of the factors we can write the contrasts as
ContrastA = (ab) − (b) + (a) − (1)
ContrastB = (ab) + (b) − (a) − (1) ContrastAB = (ab) − (b) − (a) + (1)
This contrasts can also be termed as the total effects. From the SS formulas for contrast discussed earlier
SSA = [(ab)−(b)+(a)−(1)]2 4n
SSB = [(ab)+(b)−(a)−(1)]2 4n
SSAB = [(ab) − (b) − (a) + (1)]2 4n

Calculations for sum of squares for A, B, and AB
The Total Sum of Squares is
SST = ∑∑∑y2ijk − …
For our example
222 y2 i=1 j=1 k=1 4n
SST = SSTr + SSE
SST = SSA +SSB +SSAB +SSE
SSA = 4 × 3 = 208.33
SSB = 4×3 =75.00
SSSAB = 4 × 3 = 8.33
SST = 9398.00 − 9075.00 = 323.00
SSE = 31.34 9/14

Statistical Analysis
Source of variation A
B AB Error Total
sum of squares 208.33 75.00 8.33 31.24 323.00
Degrees of freedom 1
Square F P-value 208.33 53.15 0.001
75.00 19.13 0.0024 1 8.33 2.13 0.1826

The empirical model
yˆ = 27.5 + 0.5 × 8.33×1 + 0.5 × (−5.00)x2

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