This document discusses strategies for designing factorial experiments with multiple factors. It explains that factorial experiments involve studying the effect of varying levels of factors on a response variable. The optimal design strategy depends on whether the circumstances are unusual or normal. For normal circumstances where there is some noise and factors influence each other, a fractional factorial or full factorial design is typically best. The document provides details on analyzing the data from factorial experiments to determine if factor effects and interactions are significant. It includes examples of calculating main effects and interactions from 2-level factorial data.

1. Factorial Experiments
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2. Strategies for Experimentation with Multiple FactorsStrategies for Experimentation with Multiple Factors
•Good Strategy for Data Collection (Design)
⇒ Analysis and interpretation of data is easy
•Sophisticated Data Analysis cannot remedy problems caused
by a bad Data Collection Strategy

3. Circumstances Dictate Optimal Data Collection Strategy
Unusual circumstances:
•Response is a complicated function of the factors
•The effect or optimum level of one factor is not
influenced by the levels of other factors
Normal circumstances
•There is little ‘noise’ and experiments are quite repeatable
•Response is a smooth function of factors, linear or
Curved but no sharp kinks or inflection points
•The effect or optimum level of one factor is
influenced by the levels of other factors
•There is ‘noise’ that makes repeat experiments vary

4. •This is a Good Strategy for the Unusual Circumstances
•Methodically Vary One Factor while holding all others
constant
•Repeat for each of the other factors in turn (this may take
quite a few experiments
Typical Strategy by one untrained is: The One-at-a-time Method

5. •Simple and lends itself easily to graphical display

6. •It’s a poor strategy under normal circumstances

7.  In general, a factorial experiment involves
several variables.
 One variable is the response variable, which
is sometimes called the outcome variable or
the dependent variable.
 The other variables are called factors.
 The question addressed in a factorial
experiment is whether varying the levels of the
factors produces a difference in the mean of
the response variable.

8.  If there is just a single factor, then we say that
it is a one-factor experiment.
 The different values of the factor are called the
levels of the factor and can also be called
treatments.
 The objects upon which measurements are
made are called experimental units.
 The units assigned to a given treatment are
called replicates.

10.  The data from the experiment thus consists of
several random samples, each from a different
population.
 The population means are called treatment
means.
 The questions of interest concern the treatment
means – whether they are all equal, and if not,
which ones are different, how big the differences
are, and so on.
 To make a formal determination as to whether the
treatment means differ, a hypothesis test is
needed.

11.  We have I samples, each from a different
treatment.
 The treatment means are denoted µ1,…, µI.
 The sample sizes are denoted J1,…, JI.
 The total number in all the samples combined is
denoted by N, N = J1+…+ JI.
 The hypothesis that we wish to test is
H0: µ1=…= µI versus H1:
two or more of the µi are different
 If there were only two samples, we might use the
two-sample t test to test for the null hypothesis.
 Since there are more than two samples, we use a
method known as one-way analysis of
variance (ANOVA).

12.  Since there are several samples, we use a
double subscript to denote the observations.
 Specifically, we let Xij denote the jth
observation in the ith sample.
 The sample mean of the ith sample:
 The sample grand mean:
i
J
j ij
i
J
X
X
i
∑ =
=
1
.
N
XJ
N
X
X
I
i ii
I
i
J
j ij
i
∑∑ ∑ == =
== 1 .1 1
..

13. Question: For the data in Table 1, find I, J1,
…,JI, N, X23, , and.3X
TABLE 1
Flux Sample Values Mean SD
A 250, 264, 256, 260, 239 253.8 9.7570
B 263, 254, 267, 265, 267 263.2 5.4037
C 257, 279, 269, 273, 277 271.0 8.7178
D 253, 258, 262, 264, 273 262.0 7.4498
..X

14. Answer: There are four samples, so I = 4. Each
sample contains five observations, so J1= J2= J3= J4=
5. The total number of observations is N = 20. The
quantity X23is the third observation in the second
sample, which is 267. The quantity is the
sample mean of the third sample. This value is
presented in the table and is 271.0. We can use the
equation on a previous slide:
.3X
5.262
20
)0.262(5)0.271(5)2.263(5)8.253(5
.. =
+++
=X

15.  The variation of the sample means around the sample grand
mean is measured by a quantity called the treatment sum
of squares (SSTr), which is given by
 Note that each squared distance is multiplied by the sample
size corresponding to its sample mean, so that the means for
the larger samples count more.
 SSTr provides an indication of how different the treatment
means are from each other.
 If SSTr is large, then the sample means are spread widely, and
it is reasonable to conclude that the treatment means differ
and to reject H0.
 If SSTr is small, then the sample means are all close to the
sample grand mean and therefore to each other, so it is
plausible that the treatment means are equal.
( ) 2
..1 .
2
1
..SSTr XNXJXXJ
I
i ii
I
i ii −=−= ∑∑ ==

16.  In order to determine, whether SSTr is large enough to
reject H0, we compare it to another sum of squares,
called the error sum of squares (SSE).
 SSE measures the variation in the individual sample
points around their respective sample means.
 This variation is measured by summing the squares of
the distances from each point to its own sample
mean.
 SSE is given by
∑ ∑= =
−=
I
i
J
j iij
i
XX
1 1
2
. )(SSE

17.  The term that is squared in the formula for SSE is
called a residual.
 Therefore, SSE is the sum of the squared residuals.
 SSE depends only on the distances of the sample
points from their own means and is not affected by the
location of the treatment means relative to one
another.
 So, SSE measures only the underlying random
variation in the process being studied.
 An easier computational formula is:
∑ ∑∑= ==
−=
I
i
I
i ii
J
j ij XJX
i
1 1
2
.1
2
SSE

18. For the data in Table 1, compute SSTr and SSE.

19. To test H0: µ1=…= µI versus H1:
two or more of the µi are different
1. Compute SSTr.
2. Compute SSE.
3. Compute MSTr = SSTr/(I – 1) and MSE = SSE/(N – I).
4. Compute the test statistic: F = MSTr / MSE.
5. Find the P-value by consulting the F table (Table A.7
in Appendix A) with I – 1 and N – I degrees of
freedom.
Note: The total sum of squares, SST = SSTr + SSE.

20. For the data in Table 1, compute MSTr, MSE, and
F. Find the P-value for testing the null hypothesis
that all the means are equal. What do you
conclude?

22.  Three separation methods were compared in a certain
chemical process to study their effects on yield. Three
runs were made with each method, and the yields, in
percent of a theoretical maximum, are as follows:
 Method A: 84.6 83.3 85.1
 Method B: 87.3 85.9 88.2
 Method C: 87.2 86.0 86.3
a. Construct an ANOVA table. You may give a range for
the P-value.
b. Can you conclude that there are differences among
the mean yields?

23. A level 100(1 - α)% confidence interval for µi is
given by
i
Ni
J
tX
MSE
2/,1. α−±

24. Find a 95% confidence interval for the mean
hardness of welds produced with flux A.

25.  The fifth column is the one titled “F”. This gives the test statistic that we just
discussed.
 The column “P” presents the P-value for the F test.
 Below the ANOVA table, the value “S” is the pooled estimate of the error standard
deviation, σ.
 Sample means and standard deviations are presented for each treatment group, as
will as a graphic that illustrates a 95% CI for each treatment mean.
One-way ANOVA: A, B, C, D
Source DF SS MS F P
Factor 3 743.40 247.800 3.87 0.029
Error 16 1023.60 63.975
Total 19 1767.00
S = 7.998 R-Sq = 42.07% R-Sq(adj) = 31.21%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ----+---------+---------+---------+-----
A 5 253.80 9.76 (-------*------)
B 5 263.20 5.40 (------*-------)
C 5 271.00 8.72 (-------*-------)
D 5 262.00 7.45 (-------*-------)
----+---------+---------+---------+-----
250 260 270 280
Pooled StDev = 8.00

26. Present Goal
 
0% Knowledge 100%
Objective:
No. ofFactors:
Model:
Information:
Designs:
Screening
6 - 20
Continuous &/or
Discrete
Linear
Identify important
variables;
Crude predictions
of effects
Fractional-Factorial
or
Plackett-Burman
Constrained
Optimization
3 - 8
Continuous &/or
Discrete
Linear
+ Cross-products
(interactions)
Good predictions of
effects and
interactions
2-Level Factorial
(+ Center Points)
Unconstrained
Optimization
3 - 6
Continuous only
Linear
+ Cross-products
+ Quadratics
Good predictions
of effects,
interactions, and
curvature.
Central Composite
or
Box-Behnken
Extrapolation
or Optimization
1 - 5
Mechanistic Model
Estimate parameters
in theoretical model
Special
(computer
generated)
Constrained
Optimization
3 - 8
Continuous &/or
Discrete
Linear
+ Cross-products
(interactions)
Good predictions of
effects and
interactions
2-Level Factorial
(+ Center Points)
Constrained
Optimization
3-8
Continuous &/or
Discrete
Linear
+ Cross-products
(interactions)
Good predictions of
effects and
interactions
2-Level Factorial
(+ Center Points)

27. Why start in the middle?Why start in the middle?
2-Level Factorial
Designs
Screening
Designs
Optimization
Designs
1.
2.
3.
Fractionate
Augm
ent
2-Level Factorial
Designs

28. •This is costly when the underlying relationship may be smooth
Run Ambient Warm-up Measured
Number Temperature Time Voltage
1 20.0 C 0.5 min ---
2 25.0 C 0.5 min ---
3 30.0 C 0.5 min ---
4 35.0 C 0.5 min ---
5 20.0 C 3.0 min ---
6 25.0 C 3.0 min ---
7 30.0 C 3.0 min ---
8 35.0 C 3.0 min ---
9 20.0 C 5.5 min ---
10 25.0 C 5.5 min ---
11 30.0 C 5.5 min ---
12 35.0 C 5.5 min ---
Table 8.1 -- Full 4×3 Factorial Design to Check the Accuracy of a Voltmeter
°C
°C
°C
°C
°C
°C
°C
°C
°C
°C
°C
°C
°C

29. Figure 8.2 -- Two-Level Factorial Design for Two Factors (22 Factorial Design)Figure 8.2 -- Two-Level Factorial Design for Two Factors (22 Factorial Design)

30. Figure 8.3 -- Two-Level Factorial Design for Three Factors (23 Factorial Design)

31. Figure 8.4 -- Estimating the Main Effect of Factor 1 (X1) for the 22
Factorial Design
22
Factorial – Main Effects
2/)]()[(
)(
)(
34121
3421
1221
YYYYX
YYXX
YYXX
−+−=
−=
−=
ofEffectAverage
highatofEffect
lowatofEffect
Write the equation for the main effect of X2:

32. Figure 8.5 -- Estimating the Main Effect of Factor 2 (X2) for the 22 Factorial Design
2/)]()[(
)(
)(
24132
2412
1312
YYYYX
YYXX
YYXX
−+−=
−=
−=
ofEffectAverage
highatofEffect
lowatofEffect

33. •36
Figure 8.6 -- Estimating the Main Effect of Factor 1 (X1) for the 23
Factorial Design

34. ( a ) D e te r m i n e t h e e f f e c ts o f X 1 a n d X 2 f o r E x a m p l e 1 .
( b ) D e te r m in e t h e e f f e c t s o f X 1 a n d X 2 fo r E x a m p l e 2 .
Group Problem
X2
X1X1
X2
32 11
56 35
11
56
11
14
Example 1 Example 2

35. Interaction Effects

36. Figure 8.8 -- Calculation of the Interaction for a 22 Factorial Design

37. 2/)(2/)( 324121 YYYYXX +−+=− ninteractio

38. Group Problem
(c) Determine the interaction of X1 and X2 in Example 2
X2
X1X1
X2
32 11
56 35
11
56
11
14
Example 1 Example 2

39. X1 X2 Y
-1 -1 80
1 -1 82
-1 1 86
1 1 88
From before:
Main X1 effect: (Y2+Y4-Y1-Y3)/2
Main X2 effect: (Y4+Y3-Y1-Y2)/2
X1X2 interaction: (Y4+Y1-Y2-Y3)/2

40. Table 8.2 -- The Factorial Pattern of Experiments
The General Pattern
For 2-Level Factorial
Experiments

45.  Objectives:
 Which main effects and interactions affect average process yield?
 Which main effects and interactions affect variability in process yield?
 What is the optimal process condition?
 Process Parameters:
 Temperature (T) 80 – 120 C
 Pressure (P) 50 - 70 psig
 Reaction time (R) 5 - 10 min

46. → Take 3 replicates to measure variability; randomize run
order
Run Mean T P R T*P T*R P*R T*P*R Yield 1 Yield 2 Yield 3
1 1 -1 -1 -1 1 1 1 -1 61.4 58.6 57.1
2 1 1 -1 -1 -1 -1 1 1 75.6 77.6 75.8
3 1 -1 1 -1 -1 1 -1 1 27.5 34.0 25.1
4 1 1 1 -1 1 -1 -1 -1 51.4 48.5 54.4
5 1 -1 -1 1 1 -1 -1 1 24.8 20.7 15.4
6 1 1 -1 1 -1 1 -1 -1 43.6 44.3 37.0
7 1 -1 1 1 -1 -1 1 -1 45.2 49.5 50.3
8 1 1 1 1 1 1 1 1 70.5 74.0 74.7

47. •51
→ Calculate average yield for each run, as well as variances:
Run Mean T P R T*P T*R P*R T*P*R
Ave
Yield Variance ln(s)
1 1 -1 -1 -1 1 1 1 -1 59.0 4.9 0.79
2 1 1 -1 -1 -1 -1 1 1 76.3 1.2 0.09
3 1 -1 1 -1 -1 1 -1 1 28.9 21.5 1.53
4 1 1 1 -1 1 -1 -1 -1 51.4 8.6 1.08
5 1 -1 -1 1 1 -1 -1 1 20.3 22.2 1.55
6 1 1 -1 1 -1 1 -1 -1 41.6 16.3 1.39
7 1 -1 1 1 -1 -1 1 -1 48.3 7.5 1.01
8 1 1 1 1 1 1 1 1 73.1 5.0 0.81
Sumprd 399 85.9 4.4 -32.3 8.6 6.2 114.5 -1.9
Effect 49.9 21.5 1.1 -8.1 2.2 1.6 28.6 -0.5

48. •52
→ When we calculate the significance of an effect, we are
comparing two averages. For example, for Temp we compare the
average YlargeT and the average YsmallT So an Effect is just the difference
between the two averages
→ When we did that before, we calculated t-value and compared to
t*
→ The noise was given by:
The ratio of these two values was t:
→ Here, the difference between the two averages is the effect
(experiments with high T – experiments with low T).
→How do we estimate sp(std dev of individual measurements)
21
p
n
1
n
1
s +
21
p
n
1
n
1
s
Effect
t
+
=

49. ∑
∑
=
=
= k
1i
i
k
1i
2
ii
2
p
df
sdf
s
∑=
==
k
1i
p 16dfdf
For our example: s2
1 = (4.9) , s2
2 = (1.2), … , s2
8 = (5.0)
Each estimate has degrees of freedom (3 repeats at each
factor-level setting):
df1 = 3 - 1 = 2
df2 = 3 - 1 = 2
df8 = 3 - 1 = 2
sp
2
= [2(4.9) + 2(1.2) + … + 2(5.0)]/16 = _________

50. •54
 We just learned how to calculate sp, above.
 n1 and n2 = number of experiments that went into avg 1
(high T) and avg 2 (low T), respectively.
 Half of the total number of experiments in our factorial design
(nF) went into the high T average and the other half into the low
T average. So:
 Rearranging:
 We call the denominator, sE
FF
p
E
n
2
1
1
n
2
1
1
s
Effect
t
+
=
F
p
E
n
s
2
Effect
t =

51. Run Mean T P R T*P T*R P*R T*P*R AveYield Variance ln(s)
1 1 -1 -1 -1 1 1 1 -1 59.0 4.9 0.79
2 1 1 -1 -1 -1 -1 1 1 76.3 1.2 0.09
3 1 -1 1 -1 -1 1 -1 1 28.9 21.5 1.53
4 1 1 1 -1 1 -1 -1 -1 51.4 8.6 1.08
5 1 -1 -1 1 1 -1 -1 1 20.3 22.2 1.55
6 1 1 -1 1 -1 1 -1 -1 41.6 16.3 1.39
7 1 -1 1 1 -1 -1 1 -1 48.3 7.5 1.01
8 1 1 1 1 1 1 1 1 73.1 5.0 0.81
Sumprd 399 85.9 4.4 -32.3 8.6 6.2 114.5 -1.9 sp
2
= 10.90
Effect 49.9 21.5 1.1 -8.1 2.2 1.6 28.6 -0.5 sp
= 3.30
t 15.9 0.8 -6.0 1.6 1.2 21.2 -0.3 sE
= 1.35
→ Now that we have signal-to-noise ratios for each effect and interaction, we need to compare them to some
‘hurdle’ value (t*) to determine which ones are statistically significant.
→ For 95% confidence, using the number of degrees of freedom in sp (v = 16) we find:
t* = 2.120
→ Which signal-to-noise ratios are above our hurdle value?

52. •56
Standardized Effect
ABC
B
AC
AB
C
A
BC
20151050
2.12
A Temperature
B Pressure
C Time
Factor Name
Pareto Chart of the Standardized Effects

53. •57
1
-1
1
-1
1-1
Time
Pressure
Temperature
73.0633
41.626720.3000
48.3425
51.4100
76.313359.0267
28.8700
Cube Plot (data means) for Yield

54. •58
Temperat ure
Pressure
Time
1-1 1-1
60
45
30
60
45
30
-1
1
Temperature
-1
1
Pressure
I nteraction Plot (data means) for Yield

55. → If we repeat this analysis for a second response, ln(s), we can determine which
factors influence variability of the process:
Run Mean T P R T*P T*R P*R T*P*R
Ave
Yield Variance ln(s)
1 1 -1 -1 -1 1 1 1 -1 59.0 4.9 0.79
2 1 1 -1 -1 -1 -1 1 1 76.3 1.2 0.09
3 1 -1 1 -1 -1 1 -1 1 28.9 21.5 1.53
4 1 1 1 -1 1 -1 -1 -1 51.4 8.6 1.08
5 1 -1 -1 1 1 -1 -1 1 20.3 22.2 1.55
6 1 1 -1 1 -1 1 -1 -1 41.6 16.3 1.39
7 1 -1 1 1 -1 -1 1 -1 48.3 7.5 1.01
8 1 1 1 1 1 1 1 1 73.1 5.0 0.81
Sumprd 8.3 -1.5 0.6 1.3 0.2 0.8 -2.9 -0.3
Effect 1.0 -0.4 0.2 0.3 0.1 0.2 -0.7 -0.1
→ Why did we choose ln(s) as our response and not variance?

56. Half-Normal
(i) Effect |Effect| q Z-score
T*P 1 0.05 0.05 0.54 0.09
T*P*R 2 -0.08 0.08 0.61 0.27
P 3 0.15 0.15 0.68 0.46
T*R 4 0.20 0.20 0.75 0.67
R 5 0.32 0.32 0.82 0.92
T 6 -0.38 0.38 0.89 1.24
P*R 7 -0.71 0.71 0.96 1.80

57. Half-Normal Plot of Effects
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80
|Effect|
Z-score
P*R

58. •A replicated full factorial design in three factors was run to estimate the effects of the three
factors on the strength of bread wrapper seals. The data (avg of 2 reps) and effects/interactions
are given above. If the pooled standard deviation, sp = 50.0, which effects and interactions are
statistically significant? (Note: s was estimated with 8 degrees of freedom from the replicates
at each corner.)
Group Problem:
Run Mean X1 X2 X3 X1X2 X1X3 X2X3 X1X2X3 Avg Y FACTORS
1 1 -1 -1 -1 1 1 1 -1 380 X1
= Seal Temperature
2 1 1 -1 -1 -1 -1 1 1 536 -1 = 225 F
3 1 -1 1 -1 -1 1 -1 1 636 1 = 285 F
4 1 1 1 -1 1 -1 -1 -1 616 X2
= Polyethylene Content
5 1 -1 -1 1 1 -1 -1 1 284
-1 = 0.008
1 = 0.0146 1 1 -1 1 -1 1 -1 -1 488
7 1 -1 1 1 -1 -1 1 -1 472 X3
= Sealing Time
8 1 1 1 1 1 1 1 1 652 -1 = 2.5 sec
SumPrd 4064 520 688 -272 -200 248 16 152 1 = 3.5 sec
Effect 508 130 172 -68 -50 62 4 38

59. 0bYˆ =
kk2211 XbXbXb ++++ 
Constant term
Linear terms
k1-k1)k-(k31132112 XXbXXbXXb ++++ 
Cross-Product (Interaction) terms
Where:
Xi is the coded value of Factor i, (-1, 0, +1)
b0 = mean of the factorial data
bi = Effect of Factor i/2
bij = Interactionij /2

60. Run Mean X1 X2 X3 X1X2 X1X3 X2X3 X1X2X3 Avg Y
1 1 -1 -1 -1 1 1 1 -1 692.5
2 1 1 -1 -1 -1 -1 1 1 635.5
3 1 -1 1 -1 -1 1 -1 1 692.5
4 1 1 1 -1 1 -1 -1 -1 632.0
5 1 -1 -1 1 1 -1 -1 1 663.0
6 1 1 -1 1 -1 1 -1 -1 679.5
7 1 -1 1 1 -1 -1 1 -1 693.5
8 1 1 1 1 1 1 1 1 660.0
Sumproduct 5348.5 -134.5 7.5 43.5 -53.5 100.5 14.5 -46.5
Effect 668.6 -33.63 1.875 10.88 -13.38 25.13 3.625 -11.63
t E = -3.721 0.208 1.204 -1.48 2.781 0.401 -1.287 t*(8)=2.306
sE = 9.0355
306.2*
=t

61.  Example (Circuit Voltage)
3131
3131
XX6.21X4.5X8.61668.6Voltage
XX
2
25.1
X
2
10.9
X
2
33.6
668.6Voltage
++−=
++−=
NOTE:
Terms that were not significant were dropped.
Exception -- Factors that have significant interactions
always have their linear term retained.

62.  Group Problem: Write the appropriate model
for the previous group problem

63. termsorderhigher-
termproductcross-
termslinear-
termconstant-
...
...
...
ˆ
321123
1,121132112
2211
0
++
++++
++++
=
−−
XXXb
XXbXXbXXb
XbXbXb
bY
kkkk
kk
•Uncoded Equation Form
YY ofvaluepredicted=ˆ






=
=
Center-ValueHigh
Center-ValueFactor
factorforvaluecoded iXi





 +
=
2
ValueFactorLowValueFactorHigh
Center

64. Additional Tools for Design and Analysis of Two
Level Factorials
Additional Tools for Design and Analysis of Two
Level Factorials
•How many replicates are needed for a specific level of precision?
•Is curvature present in the model, or is it linear?
•How do we account for background variables that may change over
the course of our experiment?
•What if we cannot completely randomize our experiment?

65. 2
8






=
δ
s
nF
•Number of Replicates Needed for Precision
δ = size of a practical difference, i.e.,
Ho : Ei = 0 Ha: Ei ≠0
•To guarantee β = P(II) = 0.05 with significance level α = 0.05
•When Ei = δ
k
F rn 2×= Round up to the nearest multiple of 2k
2
2.7






=
δ
σ
Fn

66. Size of
practical
difference
δ = σ
64
δ = 2σ 16
δ = 4σ 4
Fn
Largest
Practical?
2
8






=
δ
s
nF

67. Fn
s8
=δ
•Often the equation is used “backwards”
in other words if we have budget for nF experiments
how big of an effect can we detect?

68. 24.10)]05.0/()02.0(8[ 2
==Fn
4
2 05.0=δ 02.0=s
•Example 1 Purification process for crystalline product
Y = Purity
X1 = Temperature of dissolution
X2 = type liquid added at second step
X3 = stirring time
X4 = steam distillation time
16224.10 4
=< 1=r
04.016/)02.0(8/8 === Fnsδ

69. Group Problem:
The following questions refer to an experimental program you
are planning. The individual experiments have a response
standard deviation of s = 15. The smallest main effect or
interaction that you would not like to overlook is δ = 25.
(a) How many experiments do you need to run if you plan to study
three factors (k = 3)?
(b) How many experiments do you need to run if you plan to study
four factors (k = 4 )?
(c) If you can only run 16 experiments (nF = 16), what is the
resulting δ? Does your answer depend on k?

70. •9.4 Testing for Curvature in the Model
Figure 9.5--Curvature in Two-Level factorial Design with One Factor

71. - 0 +
- -0 0+ +
Center
Point
Center
Point
Center
Point
A. One Factor
B. Two Factors C. Three Factors
X
X
X X
X
X
1
1
2 2
3
Figure 9.4--Center Points in two Level FactorialDesigns

72. y
X X
12
Average of Corners
Center Point Average
Curvature
Figure 9.6--Curvature in Two-Level Factorial Design with two Factors

73. FactorialCenter YYC −=






+=
FC
pC
nn
ss
112
•Curvature Effect
•Difference of two averages
•Yardstick for measuring significance of the curvature effect
CC Ycenter,at theresponseaveragetheinnsobservatioofnumberthen =
FF Ypoints,factorialat theresponseaveragein thensobservatioofnumberthen =

74. CC Ycenter,at theresponseaveragetheinnsobservatioofnumberthen =
C
CC
stC
or
sCt
**
/
=
=
FF Ypoints,factorialat theresponseaveragein thensobservatioofnumberthen =

75. •Example 22
Chemical Reaction Experiment






=
9
1
113-Temp
X 





=
45.
2
.55-Ratio
X
Table 9.1--Yields from Chemical Experiment and Summary Statistics
Run
Temp
X1
Aq/Org
X2 Observed Yields (Y)
Means
¯Y
Variances
s2
df
1 - - 68.5 68.3 68.4 0.02 1
2 + - 72 72 72.0 0.0 1
3 - + 73 70 71.5 4.5 1
4 + + 72.4 73 72.7 0.18 1
5 0 0 74 75 74.125 0.395833 4
74 73.5
Means
¯Y
68.4
3

76. Temperature104 122
.1
1.0
Aqueous/Organic
68.4 72.0
71.5 72.7
74.125
Figure 9.7--Average Yields Chemical Reaction Experiment
•Temperature•104 •122

77. 9171.08411.0,8411.0
31111
)3958.0(3)18.0(1)0.0(1)02.0(12
===
++++
×+×+×+×
= pp ss
975.215.71125.74
4
7.725.710.724.68
125.74 =−=
+++
−=
•C = Center Point Average - Factorial Average
5616.0
8
1
4
1
)9171.0(
11
=





+=+=
Fc
PC
nn
ss
297.55616.0/975.2Curvature/ === CC st

78. Group Problem:
You ran a twenty run design which had 16 factorial points (average
= 86.3) and four center points (with an average = 87.2).
(a) What is the value of Curvature?
(b) If the pooled standard deviation of the individual responses
was calculated to be 0.6, is curvature statistically significant?

79. •9.5 Blocked Factorial Designs
•Blocks are different levels of known “lurking variables”
•We want to know if “Effects” (+) (-) are the same regardless of
•which block or level of lurking variable we are in
•We want to know if optimum is the same regardless of
•which block or level of lurking variable we are in

80. •Purposes for Blocking
•Prevent Known Lurking Variables from Biasing or
Confusing Effects
•Broaden the Basis for Conclusions
•Increase Precision of Treatment Effects

81. Unblocked Design Blocked Design
Temp
X1
Time
X2
Run
Order Day Yield
Temp
X1
Time
X2
Block
Day
Run
Order Yield
- - 5 2 Y1 - - 1 1 Y1
- - 7 2 Y2 + - 1 3 Y2
+ - 4 1 Y3 - + 1 2 Y3
+ - 1 1 Y4 + + 1 4 Y4
- + 6 2 Y5 - - 2 2 Y5
- + 3 1 Y6 + - 2 3 Y6
+ + 2 1 Y7 - + 2 1 Y7
+ + 8 2 Y8 + + 2 4 Y8
•-,- •+,- •-,+ •+, •-,- •+,- •-,+ •+,
•Day 1
•Day 2

82. •Calculating SE when you don’t have replicates
s Y Y n2 2
1= − −∑ ( ) / ( )
s Y n2 2
= −∑ ( ) /µ
s B I mE i
2 2
0= −∑ ( ) /
s B I mE i= ∑ 2
/
•Note that in our design, when we add “Day” as a blocking variable, we go from a 22
design with replication to a 23
design with no replication. How do we determine se
in
the blocking design?
•Block effects should not interact with other variables (otherwise we could not
generalize our conclusions about a consistent effect over all blocks)
•Thus, interaction terms that include the blocking variable are a measure of noise.
•se
will be the standard deviation of blocking interactions around the mean of zero.
•Note here we know the mean of the block interactions (BIs) – we assume them to be
zero.
•Thus, te
= effect/sE

83. •A Two Block Example
F ig u r e 9 .8 - - G r o w t h C h a m b e r s f o r P e s t ic id e D e g r a d a t io n E x p e r im e n t
( S ig n s o n C o n ta in e r s R e p r e s e n t L e v e l o f M o i s tu r e )

84. Coded Levels Uncoded Levels
Random
Soil
Container
Half
Life
Y
ln(half
life)
Computed
Effects
Run
No. X1 X2 Block Temp %Moist
Soil
Type
1 - - 1 50N
F 15% SL 2 886.3 6.787 5.0156=Mean
2 + - 1 80N
F 15% SL 3 187.9 5.236 -1.166=X1
3 - + 1 50N
F 25% SL 1 229.1 5.434 -0.630=X2
4 + + 1 80N
F 25% SL 4 129.8 4.866 0.0813=X1× X2
5 - - 2 50N
F 15% SC 3 167.5 5.121 -1.130=B
6 + - 2 80N
F 15% SC 4 65.2 4.178 -0.106=X1×B
7 - + 2 50N
F 25% SC 1 156.3 5.052 0.232=X2×B
8 + + 2 80N
F 25% SC 2 31.5 3.451 -.410=X1× X2×B
s B I mE i= =
− + + −
∑ 2
2 2 2
0 1 0 6 0 2 3 2 0 4 1 0
3
/
( . ) ( . ) ( . )
= =
0 2 3 3
3
0 2 7 9
.
.

89. Algal
Blocks BI BI BI BI BI BI Block BI BI BI BI BI BI BI BI BI BI BI BI BI BI BI Population
Blocks Mean A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE x 100
1 1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 -1 31.2
1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 39.1
1 1 -1 1 -1 -1 -1 -1 1 1 1 -1 -1 -1 1 1 1 1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 -1 1 -1 1 41.2
1 1 1 1 -1 -1 -1 1 -1 -1 -1 -1 -1 -1 1 1 1 -1 -1 -1 1 1 1 1 1 1 -1 1 1 1 -1 -1 -1 37.6
1 1 -1 -1 1 -1 -1 1 -1 1 1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 1 1 -1 1 -1 -1 1 -1 -1 1 47.9
1 1 1 -1 1 -1 -1 -1 1 -1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 1 1 -1 1 1 1 -1 1 -1 -1 48.1
1 1 -1 1 1 -1 -1 -1 -1 1 1 1 -1 -1 -1 -1 1 -1 1 1 1 1 -1 -1 -1 1 1 1 1 -1 -1 1 -1 46.5
1 1 1 1 1 -1 -1 1 1 -1 -1 1 -1 -1 -1 -1 1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 -1 1 1 1 1 45.1
2 1 -1 -1 -1 1 -1 1 1 -1 1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 1 1 -1 1 -1 -1 -1 1 44.8
2 1 1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 1 1 -1 1 1 -1 -1 24.2
2 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 1 -1 1 -1 1 1 -1 1 -1 1 -1 43.4
2 1 1 1 -1 1 -1 1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 1 1 25.1
2 1 -1 -1 1 1 -1 1 -1 -1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 1 1 -1 1 -1 -1 1 1 -1 63.9
2 1 1 -1 1 1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 1 1 -1 1 1 58.3
2 1 -1 1 1 1 -1 -1 -1 -1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 -1 -1 1 1 1 -1 1 65.7
2 1 1 1 1 1 -1 1 1 1 -1 1 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 -1 -1 76.8
3 1 -1 -1 -1 -1 1 1 1 1 -1 1 1 -1 1 -1 -1 -1 -1 1 -1 1 1 -1 1 1 1 1 -1 -1 -1 -1 1 57.6
3 1 1 -1 -1 -1 1 -1 -1 -1 1 1 1 -1 1 -1 -1 1 1 -1 1 -1 -1 -1 1 1 1 -1 1 1 1 -1 -1 30.9
3 1 -1 1 -1 -1 1 -1 1 1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 1 -1 -1 1 -1 1 1 -1 1 -1 28
3 1 1 1 -1 -1 1 1 -1 -1 1 -1 -1 1 1 -1 -1 -1 -1 1 1 -1 -1 1 -1 -1 1 1 -1 -1 1 1 1 20.1
3 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1 65.6
3 1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 1 63.1
3 1 -1 1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 -1 -1 1 -1 1 1 -1 1 73.6
3 1 1 1 1 -1 1 1 1 -1 1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 -1 81.4
4 1 -1 -1 -1 1 1 1 1 -1 -1 1 -1 -1 -1 -1 1 -1 1 1 1 1 -1 1 1 -1 -1 -1 -1 1 1 1 -1 32.6
4 1 1 -1 -1 1 1 -1 -1 1 1 1 -1 -1 -1 -1 1 1 -1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1 32.3
4 1 -1 1 -1 1 1 -1 1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 -1 1 -1 1 1 -1 1 -1 1 48.1
4 1 1 1 -1 1 1 1 -1 1 1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 31.2
4 1 -1 -1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 1 67.9
4 1 1 -1 1 1 1 -1 1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 64.8
4 1 -1 1 1 1 1 -1 -1 -1 -1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 -1 68
4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 79.9
Sumprd 1584 -68 39 449 70 106 33 105 -16 -7 75 59 -8 78 118 -61 47 1 22 36 27 57 -32 60 28 -99 45 -1 -63 -64 -100 36
Effect 50 -4 2 28 4 7 2 7 -1 0 5 4 -1 5 7 -4 3 0 1 2 2 4 -2 4 2 -6 3 0 -4 -4 -6 2
t -1.2 0.7 8.0 1.2 1.9 0.6 1.9 -0.3 -0.1 1.3 1.1 -0.1 1.4 2.1 -1.1 0.8 0.0 0.4 0.6 0.5 1.0 -0.6 1.1 0.5 -1.8 0.8 0.0 -1.1 -1.1 -1.8 0.6
se 4 Blocks BI BI BI BI BI BI Block BI BI BI BI BI BI BI BI BI BI BI BI BI BI BI
t*.95,15 2.080

90. Term
St andardized Effect
AB
B
ABC
A
BC
AC
C
9876543210
2.080
A A
B B
C C
Factor Name
Pareto Chart of the Standardized Effects
(response is Algal Population x 100, Alpha = .05)

91. Estimated Effects and Coefficients for Algal Population x 100 (coded
units)
Term Effect Coef SE Coef T P
Constant 49.500 1.871 26.46 0.000
A -4.250 -2.125 1.871 -1.14 0.267
B 2.462 1.231 1.871 0.66 0.517
C 28.075 14.038 1.871 7.50 0.000
A*B 2.088 1.044 1.871 0.56 0.582
A*C 6.550 3.275 1.871 1.75 0.093
B*C 4.712 2.356 1.871 1.26 0.220
A*B*C 2.963 1.481 1.871 0.79 0.436
S = 10.5815 R-Sq = 72.61% R-Sq(adj) = 64.62%
Analysis of Variance for Algal Population x 100 (coded units)
Source DF Seq SS Adj SS Adj MS F P
Main Effects 3 6498.66 6498.66 2166.22 19.35 0.000
2-Way Interactions 3 555.74 555.74 185.25 1.65 0.203
3-Way Interactions 1 70.21 70.21 70.21 0.63 0.436
Residual Error 24 2687.23 2687.23 111.97
Pure Error 24 2687.23 2687.23 111.97
Total 31 9811.84

92. Term
St andardized Effect
AB
B
ABC
A
BC
AC
C
876543210
2.064
A A
B B
C C
Factor Name
Pareto Chart of the Standardized Effects
(response is Algal Population x 100, Alpha = .05)

93. •Group Problem

94. •Group Problem (cont.)

95. R u n
N o .
T e m p
X 1
M o is tu r e
X 2 B lo c k X 3 X 4
R a n d o m S o il
C o n ta in e r
H a lf
L if e
1 - - 1 - - 2 886.3
2 + - 1 - - 3 187.9
3 - + 1 - - 1 229.1
4 + + 1 - - 4 129.8
5 - - 2 + - 3 167.5
6 + - 2 + - 4 65.2
7 - + 2 + - 1 156.3
8 + + 2 + - 2 31.5
9 - - 3 - + 1 253.9
10 + - 3 - + 3 65.83
11 - + 3 - + 4 188.3
12 + + 3 - + 2 54.7
13 - - 4 + + 2 240.7
14 + - 4 + + 4 75.5
15 - + 4 + + 3 166.2
16 + + 4 + + 1 45.2

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