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

  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

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

  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) .

  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

  − 15.6 − · · · − 14.7 + 13.2)

  = −17.5) = 19.14,

  SSA =

  5 −1

  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

  A= 17.5 − 8 = −2.19,

  SSB = −11.3 + 15.6 − · · · − 14.7 + 13.2) 5 −1 =

  SSC = −11.3 − 15.6 + · · · + 14.7 + 13.2) 5 −1 =

  SSD = −11.3 − 15.6 − · · · + 14.7 + 13.2)

  D= −7.7 8 = −0.96,

  SSE = −11.3 − 15.6 − · · · + 14.7 + 13.2)

  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 2 fraction of a 2 4 factorial design

  effects are negligible.

  using BCD as the defining contrast.

  15.20 Construct a 1 fraction of a 2 4 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

  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 fraction of a 2 6 design.

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

  −1

  ation and degrees of freedom) for all appropriate

  1 −1 6.1

  tests, assuming that E and F do not interact and

  −1

  −1

  all three-factor and higher interactions are negligi-

  15.22 Seven factors are varied at two levels in an ex- 1 1 1 periment involving only 16 trials. A 7.3

  fraction of a

  2 factorial experiment is used, with the defining con- 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

  makes one rotation. The sensor then measures the am-

  31.6 acg

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

  ad 28.7 cdg

  32.0 temperature B, gap between blades C, gap between

  abce

  33.1 beg

  32.8 blade and casing D, location of input E, and location of detection F . A 1 fraction of a 2 cdef 6 33.6 adef g 35.3 4 factorial experi-

  acef

  33.7 ef g

  32.4 ment is used, with defining contrasts being ABCE and

  bcde

  34.2 abdeg

  35.3 BCDF . The data are as follows:

  abdf

  32.5 bcdf g

  35.6 A B C D E F Response

  bf 27.8 abcf g

  35.1 −1 −1 −1 −1 −1 −1

  Perform an analysis of variance on all seven main ef-

  fects, assuming that interactions are negligible. Use a

  −1

  0.05 level of significance.

  1 −1 39.88

  −1 −1

  15.23 An experiment is conducted so that an en-

  gineer can gain insight into the influence of sealing

  1 −1 3.22

  temperature A, cooling bar temperature B, percent

  −1

  1 −1 8.94

  polyethylene additive C, and pressure D on the seal

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

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

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

  sented here. Perform an analysis of variance on main

  −1 −1

  effects only. Use α = 0.05.

  1 1 16.80 1 −1

  −1

  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- with a 2 3 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

  2 an experimenter measured the number of breakdowns 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 fraction of a 2 5 factorial experiment was 2 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-

  A B C D E Response

  neer is concerned with the effects on the cracking of a

  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

  the response, crack length induced in the sample of the

  −1 Response 1 764 A B C

  (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.