Design Of Experiment DOE
• Reference Book: Douglas C. Montgomery
• Regents’ Professor of Industrial Engineering and Statistics
• ASU Foundation Professor of Engineering
•
Chapter 4
Arizona State University
1
Chapter 4
Experiments with Blocking Factors
Chapter 4
2
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
3
2. The Randomized Complete
Blocking Design (RCBD)
• If the nuisance variable is known and controllable, we use
blocking
• If the nuisance factor is known and uncontrollable,
sometimes we can use the analysis of covariance to
remove the effect of the nuisance factor from the analysis
• If the nuisance factor is unknown and uncontrollable (a
“lurking” variable), we hope that randomization
balances out its impact across the experiment
• Sometimes several sources of variability are combined in
a block, so the block becomes an aggregate variable
Chapter 4
4
Principle of analysis as Blocking Design
Chapter 4
5
3. 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 overall mean, 𝑖 the effect of the 𝑖𝑡ℎ treatment, 𝑗
the effect of 𝑗𝑡ℎ block, 𝑖𝑗 the usual NID (0,1) random
error term
• The relevant (fixed effects) hypotheses are
H 0 : 1  2 
Chapter 4
 a where i  (1/ b) j 1 (    i   j )    i
b
6
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
7
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
8
𝑀𝑆Blocks
𝑀𝑆𝐸
Manual computing:
Chapter 4
9
4. Example: Vascular grafts
A medical device manufacturer produces vascular grafts (artificial veins).
These grafts are produced by extruding billets of polytetrafluoroethylene
(PTFE) resin combined with a lubricant into tubes.
Frequently, some of the tubes in a production run contain some defects
called “flicks “.
The product developer suspects that the extrusion pressure affects the
occurrence of flicks and therefore intends to conduct an experiment to
investigate this hypothesis. However, the resin is manufactured by an
external supplier and is delivered to the medical device manufacturer in
batches. The engineer also suspects that there may be significant batch-tobatch variation
Therefore, the product developer decides to investigate the effect of four
different levels of extrusion pressure on flicks using a randomized
complete block design considering batches of resin as blocks.
The order in which the extrusion pressures are tested within each block is
random. The response variable is yield, or the percentage of tubes in
the production run that did not contain any flicks.
Chapter 4
10
Problem presentation
Inputs
Extrusion Pressure
(EP)
Output
Vascular grafts
Resin (RE)
Yield (percentage of tubes
not contain any flicks)
EP : factor of treatment
RE: Blocking factor
H0_1: Test of Hypothesis EP not affects Flicks
H0_2: Test of Hypothesis RE not affects Flicks
Chapter 4
11
Results of Experiments
• 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
12
SST = 480.31
SSTreatment = 178.17 so MSTreatment = 178.17/3 = 59.39
SSBlocks = 192.25 so MSBlocks =192.25/5= 38.45
SSE = SST- SSTreatment- SSBlocks= 109.89 so MSE=109.89/15 = 7.33
F0.05; 3; 15 = 3.29, F0.05; 5; 15 = 2.90
FTreatment = MSTreatment/MSE = 59.39/7.33 = 8.11
FBlocks = MSBlocks/MSE =38.45/7.33= 5.24
H0_1: Test of Hypothesis EP not affects Flicks
H0_2: Test of Hypothesis RE not affects Flicks
H0_1 and H 0_2 are rejected because FTreatment > 3.29 and FBlocks > 2.90
Chapter 4
13
ANOVA with Blocking
P-Value
Because the P-value (very small) is less than 0.05, we would
still reject the null hypothesis and conclude that extrusion
pressure significantly affects the mean yield.
Because the P-value (very small) is less than 0.05, we would
still reject the null hypothesis and conclude that the noise
caused by different batches of the resin could inflate the
experimental error.
Chapter 4
15
ANOVA without Blocking
Note that the mean square for error has more than doubled,
increasing from 7.33 in the RCBD to 15.11. All of the
variability due to blocks is now in the error term.
Chapter 4
16
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
According to the table , we can conclude that
-
The difference between 1 and 2 (8500 and 8700 for psi), 2 and 3, 3 and 4 is
not significant  𝜇1 = 𝜇2 , 𝜇2 = 𝜇3 , and 𝜇3 = 𝜇4 .
-
The difference between 1 and 3, 1 and 4, 2 and 4 is significant  𝜇1 ≠ 𝜇3 ,
𝜇1 ≠ 𝜇4 , and 𝜇2 ≠ 𝜇4
Chapter 4
17
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
Multiple comparisons:
1) Replace the number of replicate (n) by the
number of blocks (b).
2) Also, replace the number of error degrees of
freedom from [a(n-1)] to [(a-1)(b-1)]
𝑳𝑺𝑫 = 𝒕𝜶/𝟐,(𝒂−𝟏)(𝒃−𝟏) 𝟐𝑴𝑺𝑬 /𝒃
𝑳𝑺𝑫 = 𝟐. 𝟏𝟑𝟏 𝟐 ∗ 𝟕. 𝟑𝟑/𝟔
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
Chapter 4
18
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
Chapter 4
19
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
Chapter 4
𝜇1 ≠ 𝜇3
20
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
Chapter 4
𝜇1 ≠ 𝜇3
𝜇1 ≠ 𝜇4
21
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
𝜇2 = 𝜇3
Chapter 4
𝜇1 ≠ 𝜇3
𝜇1 ≠ 𝜇4
22
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
𝜇2 = 𝜇3
𝜇1 ≠ 𝜇3
𝜇1 ≠ 𝜇4
𝜇2 ≠ 𝜇4
Chapter 4
23
Multiple Comparisons for the Vascular Graft
Example – Which Pressure is Different?
𝑳𝑺𝑫 = 𝟑. 𝟑𝟑𝟏
That means:
𝜇1 = 𝜇2
𝜇2 = 𝜇3
𝜇1 ≠ 𝜇3
𝜇1 ≠ 𝜇4
𝜇3 = 𝜇4
𝜇2 ≠ 𝜇4
Chapter 4
24
Model Adequacy Checking
Residual Analysis for the Vascular Graft Example
Chapter 4
25
According to this figures, we can confirm that the model
is adequate
Chapter 4
26
5. The Latin Square Design
• These designs are used to simultaneously control
(or eliminate) two sources of nuisance
variability; that is, it systematically allows
blocking in two directions.
• 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
30
6. Application
Suppose that an experimenter is studying the effects of five
different formulations of a rocket propellant used in aircrew
escape systems on the observed burning rate.
Each formulation is mixed from a batch of raw material that
is only large enough for five formulations to be tested.
Furthermore, the formulations are prepared by several
operators, and there may be substantial differences in the
skills and experience of the operators. Thus, it would seem
that there are two nuisance factors to be “averaged out” in the
design: batches of raw material and operators.
The appropriate design for this problem consists of testing
each formulation exactly once in each batch of raw material
and for each formulation to be prepared exactly once by each
of five operators.
Chapter 4
31
• This is a 5  5 Latin square design
• Table 4.9, the rows and columns actually represent two
restrictions on randomization.
Chapter 4
32
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 overall mean, αi the effect of the ith row, j
the effect of jth treatment, k the effect of the kth
column, and ijk the usual NID (0,1) random error
term.
Chapter 4
33
The degrees of freedom for the sums of squares in
are as follows:
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, which is distributed as
under
the null hypothesis.
Chapter 4
34
The statistical analysis (ANOVA) is much like the analysis
for the RCBD.
Chapter 4
35
Chapter 4
36
Chapter 4
37
1.59
3.51
F0.05; 4; 12 = 3.26
• H01: Test of Hypothesis Formulation not affects burning rate
• H02: Test of Hypothesis Batches of raw material not affects the experiments
output.
• H03: Test of Hypothesis Operators not affects the experiments output.
• H01 and H 03 are rejected because FFormulation> 3.26 and FOperators > 3.26  There is
a significant difference in the mean burning rate generated by the different rocket
propellant formulations. There is also an indication that differences between
operators exist.
• H02 is accepted because FBatches of raw material < 3.26  There is no strong evidence
of a difference between batches of raw material.
Chapter 4
38
Application
A consumer products company relies on
direct mail 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 considered as blocks. The number of
responses to each mailing is as follows.
(a) Analyze the data from this
experiment.
Solution
a)
Chapter 4
40
Application
A consumer products company relies on
direct mail 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 considered as blocks. The number of
responses to each mailing is as follows.
(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.
Solution
b)
Based on the Fisher LSD = 52.05
𝜇1 ≠ 𝜇2
𝜇 2 ≠ 𝜇3
Chapter 4
𝜇1 = 𝜇3
42
Application
A consumer products company relies on
direct mail 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 considered as blocks. The number of
responses to each mailing is as follows.
(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.
Solution
c)
Chapter 4
44
Solution
c)
Chapter 4
45
Solution
c)
We can conclude that there is a concern with normality as
well as inequality of variance are presented
Chapter 4
46