# DOE (Design of Experiments) - CQE Primer question

M

#### Mehtap

I have been trying to understand one question from CQE Primer, and the solution is given, but I can not understand. If simple explanation is given, greatly appreciated.

Question:An experiment has 3 factors with 2 levels and 1 factor with 3 levels. What is the minimum number of trials necessary if all interactions are ignored?

Solution:
When experimenting, 1 degree of freedom is required to compute the overall mean, and n-1 degrees of freedom are required for each factor, where n is the number of levels for the factor. The degrees of freedom required for interactions are the product of the interactions for the factors involved in the interactions. This experiment has 1 degree of freedom for the overall mean, 3 factors with 2 levels, which require 1 degree of freedom each, and 1 factor with 3 levels, which requires 2 degrees of freedom. Thus, 6 degrees of freedom are required, so a minimum of 6 trials are required.

#### Ajit Basrur

Staff member
Does any one have answer to Mehtap's question ?

C

#### CertifiedDataJunkie

Question:An experiment has 3 factors with 2 levels and 1 factor with 3 levels. What is the minimum number of trials necessary if all interactions are ignored?
The first part of the experiment is a 2^3 design -- three factors A, B, and C, each at two (2) levels + (high) and - (low).

From p. 301 of Montgomery's "Design and Analysis of Experiments, 4th Ed.," the setup for the mean (1) and main factors a, b, and c are as follows (recalling that this problem ignores interactions ab, ac, bc, and abc):

Run A B C Label A B C
1 - - - (1) 0 0 0
2 + - - a 1 0 0
3 - + - b 0 1 0
4 - - + c 0 0 1

Note: Above is easier to view in table format of 8 columns x 5 rows

response y = uo + beta1*x1 + beta2*x2 + beta3*x3 + error

For this part, you have four unknowns (uo, beta1, beta2, and beta3), hence the four runs are needed to provide the required information.

The second part of the experiment is a 3^1 design -- one factor D at three (3) levels +, 0, and -.

Again from Montgomery (p. 438), the addition of a third factor level allows the relationship between the response and the design factors to be MODELED AS A QUADRATIC.

Assuming that D is independent of A, B, and C (to again ignore interactions, including ad, bd, cd, abd, acd, bcd, and abcd),

response y = uo + beta4*x4 + beta44*x4^2 + error

Thus, the two additional runs, 5 and 6, are needed to characterize the two coefficients for this additional part of the response equation for the model.

I haven't looked at DOE stuff in a while, but I hope this helps explain the provided solution further.

G

#### George Weiss

I see a problem with the solution:
Given:
A minimum 1 degree of freedom is needed for each factor.
A degree of freedom = n-1.
n= number of trial(s).
So: Trials = degrees of freedom +1
But: 6 degrees of freedom = 6 trials in solution sounds like a goof.
Maybe:
5 DF = 6 Trials -or- 6DF = 7 trails
This is just an ignorant person's side question/oberservation
I am speaking from an uncertainty degrees of freedom view.

C

#### CertifiedDataJunkie

I see a problem with the solution:
A degree of freedom = n-1.
n= number of trial(s).
But, n is the number of levels for the factor (not the number of trials)

for 2 levels, + and -, n = 2 - 1 = 1

for 3 level, +, 0, and -, n = 3 - 1 = 2

Therefore, for the 3^1 design portion of the model, you needed 2 df while the 2^3 design required only 1 df.

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