Design of Engineering Experiments
– The Blocking Principle
• Text Reference, Chapter 4
• Blocking and nuisance factors
• The randomized complete block design or
the RCBD
• Extension of the ANOVA to the RCBD
• Other blocking scenarios…Latin square
designs
Chapter 4
1
The Blocking Principle
• Blocking is a technique for dealing with nuisance factors
• A nuisance factor is a factor that probably has some effect
on the response, but it’s of no interest to the
experimenter…however, the variability it transmits to the
response needs to be minimized
• Typical nuisance factors include batches of raw material,
operators, pieces of test equipment, time (shifts, days, etc.),
different experimental units
• Many industrial experiments involve blocking (or should)
• Failure to block is a common flaw in designing an
experiment (consequences?)
Chapter 4
2
The Hardness Testing Example
• Text reference, pg 139, 140
• We wish to determine whether 4 different tips produce
different (mean) hardness reading on a Rockwell hardness
tester
• Gauge & measurement systems capability studies are
frequent areas for applying DOE
• Assignment of the tips to an experimental unit; that is, a
test coupon
• Structure of a completely randomized experiment
• The test coupons are a source of nuisance variability
• Alternatively, the experimenter may want to test the tips
across coupons of various hardness levels
• The need for blocking
Chapter 4
3
The Hardness Testing Example
• Suppose that we use b = 4 blocks:
• Notice the two-way structure of the experiment
• Once again, we are interested in testing the equality of
treatment means, but now we have to remove the
variability associated with the nuisance factor (the blocks)
Chapter 4
4
Extension of the ANOVA to the RCBD
• Suppose that there are a treatments (factor levels)
and b blocks
• A statistical model (effects model) for the RCBD
is
i 1, 2,..., a
yij i j ij
j 1, 2,..., b
• The relevant (fixed effects) hypotheses are
H 0 : 1 2
Chapter 4
a where i (1/ b) j 1 ( i j ) i
b
5
Extension of the ANOVA to the RCBD
ANOVA partitioning of total variability:
a
b
a
b
2
(
y
y
)
ij .. [( yi. y.. ) ( y. j y.. )
i 1 j 1
i 1 j 1
( yij yi. y. j y.. )]2
a
b
i 1
j 1
b ( yi. y.. ) 2 a ( y. j y.. ) 2
a
b
( yij yi. y. j y.. ) 2
i 1 j 1
SST SSTreatments SS Blocks SS E
Chapter 4
6
Extension of the ANOVA to the RCBD
The degrees of freedom for the sums of squares in
SST SSTreatments SSBlocks SSE
are as follows:
ab 1 a 1 b 1 (a 1)(b 1)
Therefore, ratios of sums of squares to their degrees of
freedom result in mean squares and the ratio of the mean
square for treatments to the error mean square is an F
statistic that can be used to test the hypothesis of equal
treatment means
Chapter 4
7
ANOVA Display for the RCBD
Manual computing (ugh!)…see Equations (4-9) –
(4-12), page 124
Design-Expert analyzes the RCBD
Chapter 4
8
Manual computing:
Chapter 4
9
Chapter 4
10
Vascular Graft Example (pg. 126)
• To conduct this experiment as a RCBD, assign all 4
pressures to each of the 6 batches of resin
• Each batch of resin is called a “block”; that is, it’s a
more homogenous experimental unit on which to test
the extrusion pressures
Chapter 4
11
Chapter 4
12
Vascular Graft Example
Design-Expert Output
Chapter 4
13
Residual Analysis for the
Vascular Graft Example
Chapter 4
14
Residual Analysis for the
Vascular Graft Example
Chapter 4
15
Residual Analysis for the
Vascular Graft Example
• Basic residual plots indicate that normality,
constant variance assumptions are satisfied
• No obvious problems with randomization
• No patterns in the residuals vs. block
• Can also plot residuals versus the pressure
(residuals by factor)
• These plots provide more information about the
constant variance assumption, possible outliers
Chapter 4
16
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
Also see Figure 4.2
Chapter 4
17
Other Aspects of the RCBD
See Text, Section 4.1.3, pg. 132
• The RCBD utilizes an additive model – no
interaction between treatments and blocks
• Treatments and/or blocks as random effects
• Missing values
• What are the consequences of not blocking if we
should have?
• Sample sizing in the RCBD? The OC curve
approach can be used to determine the number of
blocks to run..see page 133
Chapter 4
18
Random Blocks and/or
Treatments
σ2β
Chapter 4
Design & Analysis of Experiments
8E 2012 Montgomery
19
Chapter 4
Design & Analysis of Experiments
8E 2012 Montgomery
20
Chapter 4
Design & Analysis of Experiments
8E 2012 Montgomery
21
The Latin Square Design
• Text reference, Section 4.2, pg. 158
• These designs are used to simultaneously control
(or eliminate) two sources of nuisance
variability
• A significant assumption is that the three factors
(treatments, nuisance factors) do not interact
• If this assumption is violated, the Latin square
design will not produce valid results
• Latin squares are not used as much as the RCBD
in industrial experimentation
Chapter 4
22
The Rocket Propellant Problem –
A Latin Square Design
•
•
•
•
This is a 5 5 Latin square design
Page 159 shows some other Latin squares
Table 4-12 (page 162) contains properties of Latin squares
Statistical analysis?
Chapter 4
23
Statistical Analysis of the
Latin Square Design
• The statistical (effects) model is
i 1, 2,..., p
yijk i j k ijk j 1, 2,..., p
k 1, 2,..., p
• The statistical analysis (ANOVA) is much like the
analysis for the RCBD.
• See the ANOVA table, page 160 (Table 4.10)
• The analysis for the rocket propellant example
follows
Chapter 4
24
Chapter 4
25
Chapter 4
26
Graeco-Latin Squares
=
=
In gen-
4.5 Problems
177
4.5 Problems
4.1.
The ANOVA from a randomized complete block
experiment output is shown below.
(4.51)
Р
Source
Treatment
Block
Error
Total
DF
SS
MS
F
4 1010.56 ? 29.84
?
? 64.765
?
20 169.33
?
29 1503.71
?
?
4.5. Plot the mean tensile strengths observed for each
chemical type in Problem 4.3 and compare them to an appro-
priately scaled + distribution. What conclusions would you
draw from this display?
4.6. Plot the average bacteria counts for each solution in
Problem 4.4 and compare them to a scaled distribution. What
conclusions can you draw?
Consider the hardness testing experiment described in m
Section 4.1. Suppose that the experiment was conducted as
described and that the following Rockwell C-scale data
(coded by subtracting 40 units) obtained:
4.7.
Coupon
(4.52)
Tip
1
2
3
4
(a) Fill in the blanks. You may give bounds on the P-value.
(b) How many blocks were used in this experiment?
(c) What conclusions can you draw?
4.2.
Consider the single-factor completely randomized sin-
gle factor experiment shown in Problem 3.4. Suppose that this
experiment had been conducted in a randomized complete
block design, and that the sum of squares for blocks was 80.00
Modify the ANOVA for this experiment to show the correct
analysis for the randomized complete block experiment.
4.3. A chemist wishes to test the effect of four chemical
agents on the strength of a particular type of cloth. Because
there might be variability from one bolt to another, the
chemist decides to use a randomized block design, with the
bolts of cloth considered as blocks. She selects five bolts and
applies all four chemicals in random order to each bolt. The
resulting tensile strengths follow. Analyze the data from this
experiment (use a = 0.05) and draw appropriate conclusions.
1
2
3
4
9.3
9.4
9.2
9.7
9.4
9.3
9.4
9.6
9.6
9.8
9.5
10.0
10.0
9.9
9.7
10.2
4.53a)
1.53b)
Bolt
Chemical
1
2
3
4
5
(a) Analyze the data from this experiment.
(b) Use the Fisher LSD method to make comparisons
among the four tips to determine specifically which
tips differ in mean hardness readings.
(c) Analyze the residuals from this experiment.
4.8. A consumer products company relies on direct mail m
marketing pieces as a major component of its advertising
campaigns. The company has three different designs for a
new brochure and wants to evaluate their effectiveness, as
there are substantial differences in costs between the three
designs. The company decides to test the three designs by
mailing 5000 samples of each to potential customers in four
different regions of the country. Since there are known
regional differences in the customer base, regions are consid-
ered as blocks. The number of responses to each mailing is as
follows.
25 we
adjusted)
ation
1
2
67
70
73
73
75
73
68
67
68
71
74
75
78
75
71
72
73
75
3
4
68
69
m 4.4. Three different washing solutions are being compared
to study their effectiveness in retarding bacteria growth in
5-gallon milk containers. The analysis is done in a laboratory,
and only three trials can be run on any day. Because days could
represent a potential source of variability, the experimenter
decides to use a randomized block design. Observations are
taken for four days, and the data are shown here. Analyze the
data from this experiment (use a = 0.05) and draw conclusions,
Region
Design
NE
NW
SE
SW
com-
e also
cause
1
2
3
250
400
275
350
525
340
219
390
200
375
580
310
Days
mate
Solution
1
2
3
4
22
1
2
3
13
16
5
18
17
1
24
4
(a) Analyze the data from this experiment.
(b) Use the Fisher LSD method to make comparisons
among the three designs to determine specifically
which designs differ in the mean response rate.
(c) Analyze the residuals from this experiment.
39
44
22
180
Chapter 4 Randomized Blocks, Latin Squares, and Related Designs
4.22. The effect of five different ingredients (A, B, C, D, E)
on the reaction time of a chemical process is being studied.
Each batch of new material is only large enough to permit five
runs to be made. Furthermore, each run requires approximately
1 hours, so only five runs can be made in one day. The exper-
imenter decides to run the experiment as a Latin square so that
day and batch effects may be systematically controlled. She
obtains the data that follow. Analyze the data from this exper-
iment (use a = 0.05) and draw conclusions.
Consider the randomized complete block design in
4.27.
Problem 4.11. Assume that the software projects that
were used as blocks are random. Estimate the block variance
component.
4.28. Consider the gene expression experiment in Problem
4.12. Assume that the subjects used in this experiment are rah.
dom. Estimate the block variance component.
4.29. Suppose that in Problem 4.20 the observation from
batch 3 on day 4 is missing. Estimate the missing value and
perform the analysis using the value.
4.30. Consider a p X p Latin square with rows (a),
columns (B2), and treatments (T)) fixed. Obtain least squares
estimates of the model parameters ai, Bk, and Tje
4.31. Derive the missing value formula (Equation 4.27) for
the Latin square design.
4.32. Designs involving several Latin squares. (See
Cochran and Cox (1957), John (1971).] The p X p Latin square
contains only p observations for each treatment. To obtain more
replications the experimenter may use several squares, say n. It
is immaterial whether the squares used are the same or differ-
ent. The appropriate model is
Day
Batch
1
2
3
4
5
1
2
3
4
5
A = 8
C = 11
B = 4
D= 6
E=4
B= 7
E = 2
A = 9
C = 8
D = 2
D = 1
A = 7
C = 10
E = 6
B = 3
C = 7
D = 3
E= 1
B= 6
A= 8
E = 3
B = 8
D = 5
A = 10
C = 8
= 1,2, ...,
u + Ph + ich)
Ij = 1,2, ...,
+
T;
+ Bun)
k = 1,2,...,
+ (TP)in + Eijkh
h = 1,2,...,n
Vijkh =
4.23. An industrial engineer is investigating the effect of
four assembly methods (A, B, C, D) on the assembly time for
a color television component. Four operators are selected for
the study. Furthermore, the engineer knows that each assem-
bly method produces such fatigue that the time required for
the last assembly may be greater than the time required for the
first, regardless of the method. That is, a trend develops in the
required assembly time. To account for this source of variabil-
ity, the engineer uses the Latin square design shown below.
Analyze the data from this experiment (a = 0.05) and draw
appropriate conclusions.
Operator
Order of
Assembly
1
2
4
1
2
3
4
C = 10
B = 7
A = 5
D= 10
D = 14
C = 18
B = 10
A = 10
A = 7
D = 11
C = 11
B = 12
B = 8
A = 8
D= 9
C = 14
where
Уijkh
is the observation on treatment j in row i and col-
umn k of the hth square. Note that aich and Bu are the row
and column effects in the hth square, Pn is the effect of the hth
square, and (TP)jn is the interaction between treatments and
squares.
(a) Set up the normal equations for this model, and solve
for estimates of the model parameters. Assume that
appropriate side conditions on the parameters are
Efn = 0, ,âich = Q and ExBxn) = 0 for each h, 9,9 =
0, E,(TP);A = 0 for each h, and ECP);n = 0 for
each j.
(b) Write down the analysis of variance table for this
design.
4.33. Discuss how the operating characteristics curves in
the Appendix may be used with the Latin square design.
4.34. Suppose that in Problem 4.22 the data taken on day 3
were incorrectly analyzed and had to be discarded. Develop
an appropriate analysis for the remaining data.
4.35. The yield of a chemical process was measured using
five batches of raw material, five acid concentrations, five
standing times (A, B, C, D, E), and five catalyst concentrations
(a, b, 7, 8, €). The Graeco-Latin square that follows was used.
Analyze the data from this experiment (use a = 0.05) and
draw conclusions.
4.24. Consider the randomized complete block design in
Problem 4.4. Assume that the days are random. Estimate the
block variance component.
4.25. Consider the randomized complete block design in
Problem 4.7. Assume that the coupons are random. Estimate
the block variance component.
4.26. Consider the randomized complete block design in
Problem 4.9. Assume that the trucks are random. Estimate the
block variance component.
4.5 Problems
181
Acid Concentration
e
Batch
1
2
3
time constraint, he must use an incomplete block design. He
runs the balanced design with the five blocks that follow.
Analyze the data from this experiment (use a = 0.05) and
draw conclusions.
1
2
3
4
5
Aa = 26
By = 18
Ce = 20
DB = 15
ES = 10
BB = 16
C8 = 21
Da = 12
Ey = 15
A6 = 24
Cy = 19
De = 18
EB = 16
A8 = 22
Ba = 17
Car
Additive
1
2
3
4
5
14
17
14
1
2.
3
4
13
13
12
12
10
9
Acid Concentration
5
14
12
13
11
Batch
4
11
12
بر سر و
8
T
e
1
2
3
4
5
D8 = 16
Ea = 11
Ay = 25
Be = 14
CB = 17
Ee = 13
AB = 21
B8 = 13
Ca = 17
Dy= 14
e
t
4.41. Construct a set of orthogonal contrasts for the data in m
Problem 4.33. Compute the sum of squares for each contrast.
4.42. Seven different hardwood concentrations are being m
studied to determine their effect on the strength of the paper
produced. However, the pilot plant can only produce three
runs each day. As days may differ, the analyst uses the bal-
anced incomplete block design that follows. Analyze the data
from this experiment (use a = 0.05) and draw conclusions.
-
4.36. Suppose that in Problem 4.23 the engineer suspects
that the workplaces used by the four operators may represent
an additional source of variation. A fourth factor, workplace
(a, b, 7, 8) may be introduced and another experiment con-
ducted, yielding the Graeco-Latin square that follows.
Analyze the data from this experiment (use a = 0.05) and
draw conclusions.
Days
Hardwood
Concentration (%)
1
2
3
4
2
4
114
126
Operator
v
h
d
120
137
Ord of
Assembly
1
2
3
4
117
129
141
149
150
6
8
10
12
14
145
1
2
3
e
at
120
CB = 11 By = 10 D8 = 14 Aa = 8
Ba = 8 C8 = 12 Ay = 10 DB = 12
AS = 9 Da = 11 BB = 7
Cy = 15
Dy = 9 AB = 8 Ca = 18 B8 = 6
136
e
4
Days
or
Hardwood
Concentration (%)
5
6
7
is
120
117
2
4
119
134
n
6
8
5
р
4.37. Construct a 5 x 5 hypersquare for studying the
effects of five factors. Exhibit the analysis of variance table
for this design.
4.38. Consider the data in Problems 4.23 and 4.36.
Suppressing the Greek letters in problem 4.36, analyze the
data using the method developed in Problem 4.32.
4.39. Consider the randomized block design with one miss-
ing value in Problem 4.19. Analyze this data by using the
exact analysis of the missing value problem discussed in
Section 4.1.4. Compare your results to the approximate analy-
sis of these data given from Problem 4.19.
m 4.40. An engineer is studying the mileage performance
characteristics of five types of gasoline additives. In the road
test he wishes to use cars as blocks; however, because of a
10
12
14
143
118
123
130
127
g
we
as
4.
d
4.43. Analyze the data in Example 4.5 using the general L
regression significance test.
4.44. Prove that k(9-19/(a) is the adjusted sum of squares 0
for treatments in a BIBD.
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