Analysis of Fractional Factorial Experiments

15.7 Analysis of Fractional Factorial Experiments

The difficulty of making formal significance tests using data from fractional factorial experiments lies in the determination of the proper error term. Unless there are

15.7 Analysis of Fractional Factorial Experiments 633 data available from prior experiments, the error must come from a pooling of

contrasts representing effects that are presumed to be negligible.

Sums of squares for individual effects are found by using essentially the same procedures given for the complete factorial. We can form a contrast in the treat- ment combinations by constructing the table of positive and negative signs. For

example, for a half-replicate of a 2 3 factorial experiment with ABC the defining contrast, one possible set of treatment combinations, along with the appropriate algebraic sign for each contrast used in computing effects and the sums of squares for the various effects, is presented in Table 15.13.

Table 15.13: Signs for Contrasts in a Half-Replicate of a 2 3 Factorial Experiment

Treatment

Factorial Effect

Combination

A B C AB AC BC ABC

+ + Note that in Table 15.13 the A and BC contrasts are identical, illustrating the

abc

aliasing. Also, B ≡ AC and C ≡ AB. In this situation, we have three orthogonal contrasts representing the 3 degrees of freedom available. If two observations were obtained for each of the four treatment combinations, we would then have an estimate of the error variance with 4 degrees of freedom. Assuming the interaction effects to be negligible, we could test all the main effects for significance.

An example effect and corresponding sum of squares is

a − b − c + abc

(a − b − c + abc) 2

In general, the single-degree-of-freedom sum of squares for any effect in a 2 −p fraction of a 2 k factorial experiment (p < k) is obtained by squaring contrasts in the treatment totals selected and dividing by 2 k−p n, where n is the number of replications of these treatment combinations.

Example 15.6: Suppose that we wish to use a half-replicate to study the effects of five factors, each at two levels, on some response, and it is known that whatever the effect of each factor, it will be constant for each level of the other factors. In other words, there are no interactions. Let the defining contrast be ABCDE, causing main effects to be aliased with four-factor interactions. The pooling of contrasts involving interactions provides 15 − 5 = 10 degrees of freedom for error. Perform an analysis of variance on the data in Table 15.14, testing all main effects for significance at the 0.05 level.

Solution : The sums of squares and effects for the main effects are (11.3 − 15.6 − · · · − 14.7 + 13.2) 2 (−17.5) 2

SSA =

634 Chapter 15 2 k Factorial Experiments and Fractions

Table 15.14: Data for Example 15.6 Treatment Response Treatment Response

a 11.3 bcd

b 15.6 abe

c 12.7 ace

d 10.4 ade

E= 8.9 8 = 1.11. All other calculations and tests of significance are summarized in Table 15.15.

The tests indicate that factor A has a significant negative effect on the response, whereas factor B has a significant positive effect. Factors C, D, and E are not significant at the 0.05 level.

Exercises

15.18 List the aliases for the various effects in a (c) Show the analysis-of-variance table (sources of vari- 2 5 factorial experiment when the defining contrast is

ation and degrees of freedom) for testing all uncon- ACDE.

founded main effects, assuming that all interaction

15.19 (a) Obtain a 1 fraction of a 2 2 4 factorial design

effects are negligible.

using BCD as the defining contrast. 15.20 Construct a 1 fraction of a 2 6 factorial design (b) Divide the 1

2 fraction into 2 blocks of 4 units each using ABCD and BDEF as the defining contrasts. by confounding ABC.

Show what effects are aliased with the six main effects.

Exercises 635

Table 15.15: Analysis of Variance for the Data of a Half-Replicate of a 2 5 Factorial Experiment

Source of

Sum of

Degrees of

Main effect:

15.21 (a) Using the defining contrasts ABCE and A B C D Response ABDF , obtain a 1 4 6 fraction of a 2 design.

−1 6.6 (b) Show the analysis-of-variance table (sources of vari-

1 6.9 ation and degrees of freedom) for all appropriate

1 7.9 tests, assuming that E and F do not interact and

−1 6.1 all three-factor and higher interactions are negligi-

1 1 −1 10.4 15.22 Seven factors are varied at two levels in an ex-

periment involving only 16 trials. A 1 fraction of a

2 7 factorial experiment is used, with the defining con- 8 15.24 In an experiment conducted at the Department

trasts being ACD, BEF , and CEG. The data are as of Mechanical Engineering and analyzed by the Statis- follows:

tics Consulting Center at Virginia Tech, a sensor de- Treat.

tects an electrical charge each time a turbine blade Comb.

Treat.

makes one rotation. The sensor then measures the am- (1)

31.1 plitude of the electrical current. Six factors are rpm A,

32.0 temperature B, gap between blades C, gap between abce

ad 28.7 cdg

32.8 blade and casing D, location of input E, and location cdef

35.3 of detection F . A 1 fraction of a 2 4 6 factorial experi- acef

33.1 beg

33.6 adef g

32.4 ment is used, with defining contrasts being ABCE and bcde

33.7 ef g

35.3 BCDF . The data are as follows: abdf

34.2 abdeg

32.5 bcdf g

35.6 A B C D E F Response

35.1 −1 −1 −1 −1 −1 −1 3.89 Perform an analysis of variance on all seven main ef-

bf 27.8 abcf g

1 −1 10.46 fects, assuming that interactions are negligible. Use a

1 1 25.98 0.05 level of significance.

1 1 −1 −1 −1 1 39.88 15.23 An experiment is conducted so that an en-

1 1 61.88 gineer can gain insight into the influence of sealing

1 −1 −1 1 3.22 temperature A, cooling bar temperature B, percent

1 1 −1 −1 −1 8.94 polyethylene additive C, and pressure D on the seal

1 −1 20.29 strength (in grams per inch) of a bread-wrapper stock.

1 −1 1 32.07 A 1 fraction of a 2 4 factorial experiment is used, with

1 1 1 50.76 the defining contrast being ABCD. The data are pre- 2

1 1 −1 2.80 sented here. Perform an analysis of variance on main

1 −1 −1 8.15 effects only. Use α = 0.05.

636 Chapter 15 2 k Factorial Experiments and Fractions Perform an analysis of variance on main effects and aliases.

two-factor interactions, assuming that all three-factor and higher interactions are negligible. Use α = 0.05.

15.27 There are six factors and only eight design points can be used. Construct a 2 6−3 by beginning 15.25 In the study Durability of Rubber to Steel Ad- 3 with a 2 and use D = AB, E = −AC, and F = BC hesively Bonded Joints, conducted at the Department as the generators. of Environmental Science and Mechanics and analyzed by the Statistics Consulting Center at Virginia Tech,

15.28 Consider Exercise 15.27. Construct another an experimenter measured the number of breakdowns 2 6−3 that is different from the design chosen in Ex- in an adhesive seal. It was postulated that concentra- ercise 15.27. tion of seawater A, temperature B, pH C, voltage D, and stress E influence the breakdown of an adhesive

15.29 For Exercise 15.27, give all aliases for the six

seal. A 1 2 fraction of a 2 5 factorial experiment was main effects.

used, with the defining contrast being ABCDE. The data are as follows:

15.30 In Myers, Montgomery, and Anderson-Cook (2009), an application is discussed in which an engi-

neer is concerned with the effects on the cracking of a −1

A B C D E Response

titanium alloy. The three factors are A, temperature; 1 −1

B, titanium content; and C, amount of grain refiner. −1

The following table gives a portion of the design and 1 1 −1

the response, crack length induced in the sample of the −1

A B C −1 Response −1

(a) What is the defining relation? 1 −1

(b) Give aliases for all three main effects assuming that −1

two-factor interactions may be real.

(c) Assuming that interactions are negligible, which Perform an analysis of variance on main effects and two

main factor is most important? factor interactions AD, AE, BD, BE, assuming that (d) At what level would you suggest the factor named all three-factor and higher interactions are negligible.

in (c) be for final production, high or low? Use α = 0.05. (e) At what levels would you suggest the other factors

15.26 Consider a 2 5−1 design with factors A, B, C,

be for final production?

D, and E. Construct the design by beginning with a (f) What hazards lie in the recommendations you 2 4 and use E = ABCD as the generator. Show all

made in (d) and (e)? Be thorough in your answer.