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CHAPTER 14
Design Of Experiments With Several Factors
LEARNING OBJECTIVES
• Design and conduct engineering experiments involving several factors
• Analyze and interpret main effects and interactions
• How the ANOVA is used to analyze the data
• Use the two-level series of factorial designs
• Design and conduct two-level fractional factorial designs
Factorial Experimental Design• Performed in all engineering disciplines
– Learn about how systems and processes work• Focus on experiments that include two or
more factors• Experimental trials are performed at all
combinations of factor levels• Single-factor experiments can be extended
to the factorial experiments• ANOVA as the primary tools
FACTORIAL EXPERIMENTS• Factorial experimental design should be used• Mean that in each complete trial all possible
combinations of the levels• Two factors A and B with a levels of factor A and b
levels of factor B
Effect of Factor• Effect of a factor• Called a main effect• Consider the following
data
• Two factors, A and B, each at two levels
• Main effect of factor A• Using the data
A=(40+60)/2–(20+30)/2= 25• Changing factor A from
the low level to the high level causes an average response increase of 25
• Main effect of B B=(30+60)/2–(20+40)/2= 15
Factor B
Factor A Blow
Bhigh
Alow 20 30
Ahigh 40 60
Interaction Effect
• Difference in response is not the same at all levels
• When this occurs, there is an interaction between the factors
• Consider the following data
• At low level of factor B, A effect
A=40–20=20• At high level of factor
B, A effect A=10–30=-20• There is interaction
between A and B• Knowledge of the AB
interaction is more useful
Factor B
Factor A Blow
Bhigh
Alow 20 30
Ahigh 40 10
TWO-FACTOR FACTORIAL EXPERIMENTS
• Involves only two factors • a levels of factor A and
b levels of factor B• The format of wo-factor
factorial
• Experiment has n replicates
• Observation is denoted by yijk
• abn observations would be run in a random order
• Two-factor factorial is a completely randomized design
Linear Statistical Model
• Described by the linear statistical model
• Where µ is the overall mean effect i is the effect of the i th level of factor A
• βj is the effect of the jth level of factor B
• (β)ij is the effect of interaction between A and B
• εijk is a random error component
Page 511 Eq 14-1
Testing The Hypotheses
• Interested in testing the hypotheses• Analysis of variance (ANOVA) will be used to
test these hypotheses• Test procedure is sometimes called the two-way
analysis of variance• A and B are fixed factors• Chosen by the experimenter• Test hypotheses about the main factor effects of
A and B and the AB • Need some symbols
Notation• Let yi.. denote the total of
the observations taken at the ith level of factor A
• y.j. denote the total of the observations taken at the jth level of factor B
• yij. denote the total of the observations in the ij th
• y… denote the grand total of all the observations
Pg 511
• Define y... yi.., y.j., yij., and as the row, column, cell, and grand averages
Hypotheses
• Hypotheses that we will test– No main effect of factor A
– No main effect of factor B
– No interaction
0 oneleast at :
0...:
1
21
j
bo
H
H
0)( oneleast at :
0)(...)()(:
1
1211
ij
abo
H
H
0 oneleast at :
0...:
1
21
i
ao
H
H
Pg 512 Eq 14-2
Total Variability• ANOVA tests these hypotheses by decomposing
the total variability into component parts• Total variability is measured by the total sum of
squares of the observations
• Sum of squares decomposition
2
1 1 1
...)( yySSa
i
b
j
n
kijkT
Pg 512
Decomposition of SST
• SST is partitioned
– Sum of squares for the row factor A (SSA)
– Sum of squares for the column factor B (SSB)
– Sum of squares for the interaction between A and B (SSAB)
– an error sum of squares (SSE)
• abn - 1 total degrees of freedom• A and B have a - 1 and b - 1 degrees of freedom• AB has (a - 1)(b - 1) degrees of freedom• ab(n-1) degrees of freedom for error
Mean Squares• If we divide each of the sum of squares by the
corresponding number of degrees of freedom– Obtain the mean squares for A, B, the interaction,
and error
1b
SSMS B
B
)1(
nab
SSMS E
E
1a
SSMS A
A
)1)(1(
ba
SSMS AB
AB
Pg 513
Test Statistics• Test that the row factor effects are all equal to zero
– F-distribution with a -1 and ab(n - 1) d.o.f– Null hypothesis is rejected if fo > fα,a-1,ab(n-1)
• Test the hypothesis that all the column factor effects are equal to zero
– F-distribution with b -1 and ab(n - 1) d.o.f– Null hypothesis is rejected if fα,b-1,ab(n-1)
• Test that all interaction effects are zero
– F-distribution with a -1 and ab(n - 1) d.o.f– Null hypothesis is rejected if fo > fo > fα,(a-1)(b-1),ab(n-1)
E
AO MS
MSF
E
BO MS
MSF
E
ABO MS
MSF
ANOVA Table
• ANOVA table
Interaction or Main Effects?
• Conduct the test for interaction first
• Interpretation of the tests on the main effects
• When interaction is significant– Main effects of the factors may not have much
practical interpretative value
MINITAB Output
• Shows some of the output from the Minitab• Upper portion gives factor name and level information
• Lower portion presents the analysis of variance
Pg 517 Table 14- 7
MINITAB Output
• Shows some of the output from the Minitab• Upper portion gives factor name and level information
• Lower portion presents the analysis of variance
Example• An engineer who suspects that the surface finish of metal
parts is influenced by the type of paint used and the drying time. He selected three drying times—20, 25, and 30 minutes—and used two types of paint. Three parts are tested with each combination of paint type and drying time. The data are as follows:
• State and test the appropriate hypotheses using the analysis of variance with α=0.05
Analysis of Variance• ANOVA Table• Test statistic for the factors are and 0.07• f0=1.90<f0.05,1,12 =4.75 and f0=0.07<f0.05,2,12 =3.89• Main effects do not affect surface finish• f0=5.03>f0.05,2,12= 3.89• Indication of interaction between these factors• Last column shows the P-value• P-values for the main effects >0.05, while the P-value for the
interaction <0.05 Source DF SS MS F PPaint 1 355.6 355.6 1.90 0.193Drying 2 27.4 13.7 0.07 0.930paint*drying 2 1878.8 939.4 5.03 0.026Error 12 2242.7 186.9Total 17 4504.4
More Than Two Factors
• Involve more than two factors• a levels of factor A, b levels of factor B, c
levels of factor C, and so on• abc n total observations• Three main effects, three two-factor
interactions, a three-factor interaction, and an error term
• Must be at least two replicates (n - 2) to compute an error sum of squares
ANOVA Table
Example• The percentage of hardwood concentration in raw pulp, the
freeness, and the cooking time of the pulp are being investigated for their effects on the strength of paper
• The data from a three-factor factorial experiment are shown in the following table
• Analyze the data using the analysis of variance assuming that all factors are fixed. Use α=0.05.
Solution• ANOVA tableSource DF SS MS F PHardwood 2 8.37 4.18 7.64 0.003Cooking time 1 17.36 17.36 31.66 0.000freeness 2 21.85 10.92 19.92 0.000hardwood*cookingt 2 3.20 1.60 2.92 0.075hardwood*freeness 4 6.51 1.62 2.97 0.042cookingt*freeness 2 1.05 0.52 0.96 0.399Error 22 12.06 0 .5484Total 35 70.42 • All main factors are significant• Interaction of hardwood*freeness is also significant
2k FACTORIAL DESIGNS
• Experiments involving several factors are widely employed in research work
• k factors each at only two levels• Levels may be quantitative or they may be
qualitative• A complete replicate of such a design
requires 2 x 2 x …x 2 = 2k observations• 2k design is particularly useful• Provides the smallest number of runs
22 Design• Simplest type of 2k design is the 22
• Think of these levels as the low and high levels of the factor• Represented geometrically as a square with the 22=4 runs
• Denote the levels of the factors A and B by the signs - and +• Called the geometric notation for the design
Main and Interaction Effects• Effects of interest are A
and B and AB• Let (1), a, b, and ab
represent the totals of all n observations
• Main effect of A
• Main effect of B
• AB interaction
• Quantities in brackets are called contrasts
• A Contrast
Contrast A=a+ab-b-(1)
How to set up contrasts• Make a table where the rows are treatment
combinations ((1), a, b, ab, etc.) and the columns are factorial effects (A, B, AB, etc.)
• For each treatment combination, write a plus sign for that factorial effect if it's high and a minus if it's low
• To get interaction effects, multiply the signs together like arithmetic– if two signs are the same, their product is a plus sign,
and if not, their product is a minus sign
Signs for Effects
• A table of plus and minus signs can be used to determine the sign on each treatment
• Column headings are the main effects A and B, the AB interaction, and I
• Row headings are the treatment combinations
• Note that the signs in the AB column
• Multiply the signs in the appropriate column by the treatment combinations
• Note that the signs in the AB column
• Multiply the signs in the appropriate column by the treatment combinations
Analysis of Variance• Effect estimates and the sums of squares for A, B,
and the AB interaction• Sums of squares
• Analysis of variance– SST with 4n - 1 d.of.
– SSE with 4(n -1) d.o.f
Pg 526 Eq 14-14
2k Design for k>3 Factors• Methods presented in the
previous section can be easily extended to more than two factors
• Consider k=3 factors, each at two levels
• 23 factorial design and it has eight runs
• Geometrically, the design is a cube with the eight runs
• Lists the eight runs with each row representing one of the runs
Developing the Signs• Denote factors with capital letters as usual: A, B, C, etc. • Denote the "high" level by its associated letter (a, b, c, etc.)
and the "low" level as (1)• Make a table where the rows are (1), a, b, ab, etc. and the
columns are A, B, AB, etc.• “+” for the factorial effect if it's high and “-” if it's low• Get interaction effects, multiply the signs together• AB column are the products of the A and B column signs • If two signs are the same, their product is a plus sign
Calculating the Effects• Letters (1), a, b, ab, c, ac, bc, and abc • Main and interaction effects
A = 1/4n [a+ab+ac+abc-(1)-b-c-bc]
B = 1/4n [b+ab+bc+abc-(1)-a-c-ac]
C = 1/4n [c+ac+bc+abc-(1)-a-b-ab]
AB = 1/4n [ abc-bc+ab-b-ac+c-a+(1)]
AC = 1/4n [(1)-a+b-ab-c+ac-bc+abc]
BC = 1/4n [(1)+a-b-ab-c-ac+bc+abc]
ABC= 1/4n[abc-bc-ac+c-ab+b+a-(1)]
• Quantities in brackets are contrasts
Effect Estimates and SS
• Effect estimates are computed from
• Sum of squares for any effect
Example• An engineer is interested in the effect of cutting speed (A),
metal hardness (B), and cutting angle (C) on the life of a cutting tool
• Two levels of each factor are chosen, and two replicates of a 23 factorial design are run
• The tool life data (in hours) are shown in the following table:
• Analyze the data from this experiment.
Calculations
• Main effects are estimated
A=1/4n [ a+ab+ac+abc-(1)-b-c-bc]
=1/8(325+435)+(552+472)+(406+377)+(392+419)-(221+311)- 354+348)-(440+453) (605+550)=146
• Sum of squares for A
SSSPEED =(146)2/16 = 1332
• Easy to verify that the other effects
Source dof SS MS F P
Speed 1 1332 1322 0.49 0.502
Hardness 1 28932 28932 10.42 0.010
Angle 1 20592 20592 7.56 0.023
Speed,
Hardness
1 506 506 0.19 0.677
Speed,
Angle
1 56882 56882 20.87 0.000
Hardness,
Angle
1 2352 2352 0.86 0.377
Error 9 24530 2726
Total 15 134588
Solution• ANOVA tableSource DF SS MS F P
Speed 1 1332 1332 0.49 0.502
Hardness 1 28392 28392 10.42 0.010
Angle 1 20592 20592 7.56 0.023
speed*hardness 1 506 506 0.19 0.677
speed*angle 1 56882 56882 20.87 0.000
hardness*angle 1 2352 2352 0.86 0.377
Error 9 24530 2726
Total 15 134588
Next Agenda
• Methods and applications of nonparametric statistics