The research is about the tea consuming persons in Malaysia. This survey decide that the effect of tea consumption in our economic growth.This is decided in survey report. It relates to many SDGs like reduction of poverty, women empowerment, and others etc.
https://sdgs.un.org/goals
the data needs to be collected according to one of the 17 sdgs mentioned here
MST2034 / November 2022
1. Preparation of Experimental Design
Decide on a unique research topic which satisfies a 2k factorial design with at least four
replications related to your personal interest that fit into Sustainable Development Goals
(SDGs). In this section, you are required to give an overview on your research topic, such as
how you decide on the components of the experimental design, how you control the data
collection and etc. Append your dataset on the last page of your report.
(10 marks)
2. Model Analysis
Conduct relevant model analysis on a self-determined conventional significance level. You
should consider blocking on a confounding system in your model analysis.
(10 marks)
3. Modified Model Analysis
Based on section 2 Model Analysis, consider reasonable adjustment towards the original model
and perform relevant model analysis again on a same/ different self-determined conventional
significance level on the modified model.
(10 marks)
4. Conclusion and Reporting
Conclude on your findings. You are required to discuss how this research topic could help in
achieving SDGs.
(10 marks)
Marking Rubric for
Unacceptable
1–2
Irrelevant
preparation.
1. Preparation of Experimental Design
Poor
Fair
Good
3–4
5–6
7–8
Unorganized
Missing certain
Some components
with many
parts of
are unclear.
mistakes.
explanation.
2. Model Analysis
3. Modified Model Analysis
Unacceptable
Poor
Fair
Good
1–2
3–4
5–6
7–8
Irrelevant
Analysis
Analysis
Analysis
data analysis performed with
performed with
performed without
many mistakes.
some minor
reasoning.
mistakes
4. Conclusion and Reporting
Unacceptable
Poor
Fair
Good
1–2
3–4
5–6
7–8
Irrelevant
Unorganized
Reasonable report
Clear and concise
conclusion
report with many
with some minor
report.
mistakes
mistakes
–END OF PAPER-2
Excellent
9 – 10
Clear research topic
relates to SDGs. All
components are
defined correctly.
Excellent
9 – 10
Excellent flow of
data analysis with
arguments.
Excellent
9 – 10
Clear, concise,
comprehensive and
with convincing
analysis.
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
MST2034 DESIGN OF EXPERIMENTS
(TUTORIAL 3) (ANSWERS)
(Chapter 3) (Factorial Experiments)
Question 1
i = 1, 2,…, 4
yijk = + i + j + ( )ij + ijk , j = 1, 2,…, 4
k = 1, 2
Source of
Variation
Temperature
Copper
Content
TC
Error
Total
df
SS
MS
F0
3
3
94.8437 31.6146
599.5937 199.8646
3.9364 > 3.24
24.8858 > 3.24
9
16
31
112.5313
128.5001
1.5569 < 2.54
12.5035
8.03126
Hence, the analysis from ANOVA suggested that we could reject null hypotheses for factors A and
B. However, we do not have sufficient evidence to conclude that there is existence of interaction
between the factors.
Question 2
i = 1, 2
yijk = + i + j + ( )ij + ijk , j = 1, 2,3, 4
k = 1, 2
Source of
Variation
Drill speed (D)
Feed rate (F)
DF
Error
Total
df
SS
MS
F0
1
3
3
8
15
0.1444
0.0925
0.0417
0.0234
0.302
0.1444
0.03083
0.0139
0.02925
49.3675 > 5.32
10.5402 > 4.07
4.7521 > 4.07
Hence, the analysis from ANOVA suggested that we can reject all the null hypotheses. Hence, we
conclude that there is any difference in drill speed and feed rate in affecting the thrust force. There
is existence of interaction too.
MST2034 Design of Experiments
BinDS, BAS
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
Question 3
Source of
Variation
Operator
Machine
OM
Error
Total
df
SS
MS
F0
2
3
6
12
23
160.3333
12.4583
44.6667
45.5
262.9583
80.1667
4.1528
7.4444
3.7917
21.1426 > 3.89
1.09523 < 3.49
1.9633 < 3.00
Hence, the ANOVA indicates that operators influenced the breaking strength of a synthetic fiber
significantly. However, no evidence shows that production machines affecting the braking strength
and existence of interaction between the two factors.
Question 4
i = 1, 2,3, 4
yijk = + i + j + k + ( )ij + ijk j = 1, 2,3, 4
k = 1, 2
Source of
Variation
Blocks
df
SS
MS
F value
1
2.5312
2.5312
0.3014
Temperature (T)
3
y2
1
2
y
−
i abn = 94.8437
bn i
31.6146
3.7646 > 3.29
Copper content
(C)
3
y2
1
y2j − = 599.5937
an j
abn
199.8646
23.7993 > 3.29
TC
9
y2
1
2
y
−
ij abn − SSA − SSB = 112.5313
n i
12.5035
1.4889 < 2.59
Error
Total
15
31
125.9689
8.3979
2
y
y − abn = 935.4688
2
ijk
i
j
k
With block effects introduced, the error sum of squares has been reduced, however, block effect does
not seem to contribute a lot to the error reducing effect in this experiment, the treatments,
temperature and copper content still affect the warping of copper plates. The interaction effect is
again does not exist.
MST2034 Design of Experiments
BinDS, BAS
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
Question 5
i = 1, 2,3
yijk = + i + j + k + ( )ij + ijk j = 1, 2,3, 4
k = 1, 2
Source of Variation
Blocks
df
1
SS
2.04163
MS
2.04163
F value
0.5144
Operator (O)
Machine (M)
OM
Error
Total
2
3
6
11
23
160.3333
12.4583
44.6667
43.6584
262.9583
80.1667
4.1528
7.4444
3.9689
20.1987 > 3.98
1.04633 < 3.59
1.8757 < 3.09
With block effects introduced, the error sum of squares has been reduced, however, it does not
change the conclusion which we have concluded earlier. Only operators affect the breaking strength
of a synthetic fiber. Again, there is no sufficient evidence to show the existence of the interaction
effect.
Question 6
i = 1, 2,3
j = 1, 2
yijkl = + i + j + k + ( )ij + ( )ik + ( ) jk + ( )ijk + ijkl
k = 1, 2,3
l = 1, 2,3
Source
Cycle time (A)
Temperature (B)
Operator (C)
AB
AC
BC
ABC
Error
Total
df
2
1
2
2
4
2
4
36
53
SS
426.3333
44.4629
279
68.9260
349.6667
8.0307
39.01273
123.401
1338.8333
MS
213.1667
44.4629
139.5
34.463
87.4167
4.01535
9.7532
3.4278
F value
62.1876 > 3.23
13.1012 > 4.08
40.6967 > 3.23
10.0540 > 3.23
25.5023 > 2.61
1.1714 < 3.23
2.8453 > 2.61
From the ANOVA table, we see that the cycle time, temperature, operators significantly affect the
fill volume. All interaction effects are tested to be exist in this experiment affecting the dyeing of
cotton-synthetic cloth, except for the interaction effect between temperature and operators.
MST2034 Design of Experiments
BinDS, BAS
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
Question 7
(a)
yijk = + i + j + ( )ij + k + ijk , i = 1, 2,3; j = 1, 2,3; k = 1, 2
= overall mean
i = effect of the ith level of the row factor (Temperature)
j = effect of the jth level of the column factor (Pressure)
( )ij = effect of interaction between i and j (interaction effect)
k = effect of the kth block
ijk = random error component
(b)
ANOVA Table
Source
Day (block)
Temperature (A)
Pressure (B)
Temp : Pressure
Error
Total
(c)
df
1
2
2
4
8
17
Sum of Squares
13.004956
99.854
5.508
4.452
4.2504
127.0694
Mean Square
13.004956
49.927
2.754
1.113
0.5313
F Value
93.9807
5.1838
2.0952
H 0 : ( )ij = 0
H1 : At least one ( )ij 0 .
Test statistic: F0 = 2.0952 < F0.05,4,8 = 3.84
We do not reject H0 and conclude that there is no statistically significant in (temperature x
pressure) interaction.
(d)
t0.025, N −a =15 = 2.131 ;
LSD = ( 2.131)
( 0.5312555) ( 3(12) + 3(12) ) = 0.8968
y1 − y2 = 0.2167 LSD
y1 − y3 = 1.05 LSD *
y2 − y3 = 1.2667 LSD *
Pressure of 250 & 270, 260 & 270 are significantly different.
MST2034 Design of Experiments
BinDS, BAS
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
MST2034 DESIGNS OF EXPERIMENTS
(TUTORIAL 3)
(Chapter 3) (Factorial Experiments)
Question 1
Johnson describe an experiment to investigate warping of copper plates. The two factors studied
were the temperature and the copper content of the plates. The response variable was a measure of
the amount of warping. The data were as follows. Is there any indication that either factor affects
the amount of warping? Is there any interaction between the factors? Use = 0.05 .
Temperature (oC)
50
75
100
125
Copper Content (%)
40
60
80
100
17, 20 16, 21 24, 22 28, 27
12, 19 18, 13 17, 12 27, 31
16, 12 18, 21 25, 23 30, 23
21, 17 23, 21 23, 22 29, 31
Question 2
A mechanical engineer is studying the thrust force developed by a drill press. He suspects that the
drilling speed and the feed rate of the material are the most important factors. He selects four feed
rates and uses a high and low drill speed chosen to represent the extreme operating conditions. He
obtains the following results. Analyze the data and draw conclusions. Use = 0.05 .
Drill Speed
125
200
0.015
2.70
2.78
2.83
2.86
Feed rate
0.030 0.045
2.45
2.60
2.49
2.74
2.85
2.86
2.80
2.87
0.060
2.75
2.86
2.94
2.88
Question 3
The factors that influence the breaking strength of a synthetic fiber are being studied. Four
production machines and three operators are chosen and a factorial experiment is run using fiber
from the same production batch. The results are as follows. Use = 0.05 .
Operator
1
2
3
1
109
110
110
112
116
114
Machine
2
3
110 108
115 109
110 111
111 109
112 114
115 119
4
110
108
114
112
120
117
Analyze the data and draw conclusions.
MST2034 Design of Experiments
BinDS, BAS
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
Question 4
Referring to Question 1 again, analyze the data assuming that the replicates are blocks. Conclude
whether there are any changes on the conclusions drawn earlier.
Question 5
Referring to Question 3 again, analyze the data assuming that the replicates are blocks. Conclude
whether there are any changes on the conclusions drawn earlier.
Question 6
The quality control department of a fabric finishing plant is studying the effect of several factors on
the dyeing of cotton-synthetic cloth used to manufacture men’s shirts. Three operators, three cycle
times and two temperatures were selected and three small specimens of cloth were dyed under each
set of conditions. The finished cloth was compared to a standard and a numerical score was
assigned. The results are as follows. Analyze the data and draw conclusions.
Temperature
350oC
Operator
Operator
1
2
3
1
2
3
23 30 31 24 38 34
24 28 32 23 36 36
25 26 29 28 35 39
36 34 33 37 34 34
35 38 34 39 38 36
36 39 35 35 36 31
28 35 26 26 36 28
24 35 27 29 37 26
27 34 25 25 34 24
300oC
Cycle Time
40
50
60
Question 7
The yield of a chemical process is being studied. The two factors of interest are temperature and
pressure. Three levels of each factor are selected; however, only 9 runs can be made in one day.
The experimenter runs a complete replicate of the design on each day. The data are shown in the
following table. Assuming that the days are blocks.
Day 1 Pressure
Day 2 Pressure
Temperature 250
260 270 250 260
270
Low
86.3 84.0 85.8 86.1 85.2
87.3
Medium
88.5 87.3 89.0 89.4 89.9
90.3
High
89.1 90.2 91.3 91.7 93.2
93.7
(a)
(b)
(c)
(d)
Write a linear statistical model for the experiment, explain the terms.
Compute the analysis of variance for the data.
Test the null hypothesis of no Temperature x Pressure interaction.
Use LSD test to determine which levels of the pressure factor are significantly different.
MST2034 Design of Experiments
BinDS, BAS
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
CHAPTER 3
FACTORIAL EXPERIMENTS
3.1)
Introduction
Many experiments involve the study of effects of two or more factors. In general, factorial
designs are most efficient for this type of experiment. By a factorial design, we mean that
in each complete trial or replication of the experiment all possible combinations of the
levels of the factors are investigated. For example, if there are a levels of factor A and b
levels of factor B, each replicate contains all ab treatment combinations. When factors are
arranged in a factorial design, they are often said to be crossed.
The effect of a factor is defined to be the change in response produced by a change
in the level of the factor. This is frequently called a main effect because it refers to the
primary factors of interest in the experiment. For example, consider the simple experiment
in the figure below. This is a two-factor factorial experiment with both design factors at
two levels.
We have called these levels “low” and “high” and denoted them “ –” and “+”, respectively.
The main effect of factor A in this two-level design can be thought of as the difference
between the average response at the low level of A and the average response at the high
level of A. Numerically, this is A = 40+252 − 20+230 = 21 .
That is, increasing factor A from the low level to the high level causes an average response
increase of 21 units. Similarly, the main effect B is B = 30+252 − 20+2 40 = 11 .
If the factors appear at more than two levels, the above procedure must be modified because
there are other ways to define the effect of a factor. In some experiments, we may find that
the difference in response between the levels of one factor is not the same at all levels of
the other factors. When this occurs, there is an interaction between the factors. For
example, consider the two-factor factorial experiment shown in figure below.
At the low level of factor B (or B − ), the A effect is A = 50 − 20 = 30 and at the high level
of factor B (or B + ), the A effect is A = 12 − 40 = −28 .
School of Mathematical Sciences (SMS)
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MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
Because the effect of A depends on the level chosen for factor B, we see that there is
interaction between A and B. The magnitude of the interaction effect is the average
−28 −30
difference in these two A effects or AB = ( 2 ) = −29 . Clearly, the interaction is large in
this experiment. These ideas may be illustrated graphically, shown below. Two-factor
interaction graphs such as these are frequently very useful in interpreting significant
interactions and in reporting results to nonstatistically trained personnel. However, they
should not be utilized as the sole technique of data analysis because their interpretation is
subjective and their appearance is often misleading.
Left-hand figure: Note that the B − and B + lines are parallel, indicating a lack of interaction
between factors A and B.
Right-hand figure: Here, we see that the B − and B + lines are not parallel, indicating an
interaction between factors A and B.
3.2)
The Two-Factor Factorial Design
The simplest types of factorial designs involve only two factors or sets of treatments. There
are a levels of factor A and b levels of factor B and these are arranged in a factorial design;
that is, each replicate of the experiment contains all ab treatment combinations. In general,
there are n replicates.
Example 1 (The Battery Design Experiment)
As an example of a factorial design involving two factors, an engineer is designing a battery
for use in a device that will subjected to some extreme variations in temperature. The only
design parameter that he can select at this point is the plate material for the battery and he
has three possible choices. When the device is manufactured and is shipped to the field,
the engineer has no control over the temperature extremes that the device will encounter
and he knows from experience that temperature will probably affect the effective battery
life. However, temperature can be controlled in the product development laboratory for
the purposes of a test.
The engineer decides to test all three plate materials at three temperature levels –
15, 70 and 125oF – because these temperature levels are consistent with the product enduse environment. Because there are two factors at three levels, this design is sometimes
called a 32 factorial design. Four batteries are tested at each combination of plate material
and temperature and all 36 tests are run in random order. The experiment and the resulting
observed battery life data are given below.
School of Mathematical Sciences (SMS)
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August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
Material Type
1
2
3
Temperature
15oF
70oF
130 155
34
40
74
180
80
75
150 188 136 122
159 126 106 115
138 110 174 120
168 160 150 139
125oF
20
70
82
58
25
70
58
45
96 104
82
60
In this problem the engineer wants to answer the following questions:
1. What effects do material type and temperature have on the life of the battery?
2. Is there a choice of material that would give uniformly long life regardless of
temperature?
This last question is particularly important. It may be possible to find a material alternative
that is not greatly affected by temperature. If this so, the engineer can make the battery
robust to temperature variation in the field. This is an example of using statistical
experimental design for robust product design, a very important engineering problem.
This design is a specific example of the general case of a two-factor factorial.
3.3)
Statistical Analysis of Two-Factor Factorial Design
To pass to the general case, let yijk be the observed response when factor A is at the ith
level ( i = 1, 2,..., a ) and factor B is at the jth level ( j = 1, 2,..., b ) for the kth replicate
( k = 1, 2,..., n ) . In general, a two-factor factorial experiment will appear as follows. The
order in which the abn observations are taken is selected at random so that this design is a
completely randomized design.
Factor B
1
2
b
1
y1b1 , y1b 2 ,
y111 , y112 ,
y121 , y122 ,
2
Factor A
a
, y11n
, y12 n
, y1bn
y211 , y212 ,
y221 , y222 ,
y2b1 , y2b 2 ,
, y21n
, y22 n
, y2bn
ya11 , ya12 ,
ya 21 , ya 22 ,
yab1 , yab 2 ,
, ya1n
, ya 2 n
, yabn
The observations in a factorial experiment can be described by a model. There are
several ways to write the model for a factorial experiment. The effects model is
School of Mathematical Sciences (SMS)
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August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
i = 1, 2,..., a
yijk = + i + j + ( )ij + ijk , j = 1, 2,..., b
k = 1, 2,..., n
where is the overall mean effect, i is the effect of ith level of the row factor A, j is
the effect of the jth level of the column factor B, ( )ij is the interaction between i and
j , ijk is a random error component. Both factors are assumed to be fixed and the
treatment effects are defined as deviations from the overall mean, so
a
b
= 0 ; = 0 ;
i =1
i
j =1
j
Similarly, the interaction effects are fixed and are defined such that
a
b
( ) = ( ) = 0 .
i =1
ij
j =1
ij
Because there are n replicates of the experiment, there are abn total observations.
Another possible model for a factorial experiment is the means model
i = 1, 2,..., a
yijk = ij + ijk , j = 1, 2,..., b
k = 1, 2,..., n
where the mean of the ijth cell is ij = + i + j + ( )ij .
In the two-factor factorial, both row and column factors, A and B are of equal
interest. Specifically, we are interested in testing hypotheses about the equality of row
treatment effects, say
H 0 : 1 = 2 = ... = a = 0
H1 : at least one i 0
and the equality of column treatment effects, say
H 0 : 1 = 2 = = b = 0
H1 : at least one j 0 .
We are also interested in determining whether row and column treatments interact. Thus,
we also wish to test
H 0 : ( )ij = 0 for all i, j
H1 : at least one ( )ij 0 .
We now discuss how these hypotheses are tested using a two-factor analysis of variance.
School of Mathematical Sciences (SMS)
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August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
Let yi denote the total of all observations under the ith level of factor A, y j denote the
total of all observations under the jth level of factor B, yij denote the total of all
observations in the ijth cell and y denote the grand total of all the observations. Define
yi , y j , yij and y as the corresponding row, column, cell and grand averages. Express
mathematically,
b
n
yi
, i = 1, 2,..., a
bn
j =1 k =1
a
n
y
y j = yijk ; y j = j , j = 1, 2,..., b
an
i =1 k =1
n
y
yij = yijk ; yij = ij , i = 1, 2,..., a
n j = 1, 2,..., b
k =1
b
n
y
y = yijk ; y = .
abn
j =1 k =1
yi = yijk ; yi =
The total corrected sum of squares may be written as
( y − y ) = ( y − y ) + ( y − y − y + y ) + ( y − y )
a
b
n
i =1 j =1 k =1
2
ijk
...
a
b
n
i =1 j =1 k =1
i
ij
i
j
ijk
2
ij
= bn ( yi.. − y... ) + an ( y. j . − y... ) + n ( yij . − yi.. − y. j . + y... )
a
b
2
i =1
j =1
+ ( yijk − yij. )
a
2
b
n
a
b
2
i =1 j =1
2
i =1 j =1 k =1
because the six cross products on the right-hand side are zero. Notice that the total sum of
squares has been partitioned into a sum of squares due to “rows”, or factor A, (SSA); a sum
of squares due to “columns”, or factor B, (SSB); a sum of squares due to the interaction
between A and B, (SSAB); and a sum of squares due to error, (SSE). This is the fundamental
ANOVA equation for the two-factor factorial. Symbolically,
SSTO = SSA + SSB + SSAB+SSE
The number of degrees of freedom associated with each sum of squares is
Effect
A
B
AB interaction
Degrees of Freedom
a–1
b–1
( a − 1)( b − 1)
Error
ab ( n − 1)
Total
abn − 1
We may justify this allocation of the abn − 1 total degrees of freedom to the sum of squares
as follows: The main effects A and B have a and b levels, respectively; therefore, they have
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MST2034 Design of Experiments
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B Sc (Hons) in Actuarial Studies
a – 1 and b – 1 degrees of freedom as shown. The interaction degrees of freedom are
simply the number of degrees of freedom for cells (which is ab − 1 ) minus the number of
degrees of freedom for the two main effects A and B; that is
ab − 1 − ( a − 1) − ( b − 1) = ( a − 1)( b − 1) . Within each of the ab cells, there are n – 1 degrees
of freedom between the n replicates; thus, there are ab ( n − 1) degrees of freedom for error.
If there are column treatment effects or interaction present, then the corresponding
mean squares will be larger than MSE. Therefore, to test the significance of both main
effects and their interaction, simply divide the corresponding mean square by the error
mean square. Large values of this ratio imply that the data do not support the null
hypothesis. Computationally, manual computing of the sum of squares is shown below.
a
b
n
SSTO = yijk2 −
i =1 j =1 k =1
y2
;
abn
SSA =
1 a 2 y2
yi − abn ;
bn i =1
SSB =
y2
1 b 2
y
−
j abn .
an j =1
It is convenient to obtain the SSAB in two stages. First, we compute the sum of squares
between ab cell totals, which is called the sum of squares due to “subtotals”:
SSSubtotals =
1 a b 2 y2
yij − abn .
n i =1 j =1
This sum of squares also contains SSA and SSB. Therefore, the second step is to compute
SSAB as SSAB = SSSubtotals − SSA − SSB .
We may compute SSE by subtraction as
SSE = SSTO − SSSubtotals .
SSE = SSTO − SSAB − SSA − SSB
The ANOVA Table
Source
Factor A
a −1
SS
SSA
MSA
F0
MSA
MSE
Factor B
b −1
SSB
MSB
MSE
SSAB
MSAB
MSAB
MSE
Error
( a − 1)( b − 1)
ab ( n − 1)
SSB
b−1
AB Interaction
SSE
MSE
Total
abn − 1
SSTO
School of Mathematical Sciences (SMS)
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August 2022 Semester
or
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
Example 2
Perform ANOVA for Example 1.
Temperature
150F
700F
Material Type
1
130 155
74
180
150 188
159 126
138 110
168 160
144.833
2
3
y. j.
34
40
80
75
136 122
106 115
174 120
150 139
107.583
1
2
3
yi 2.
yi1.
yi1.
134.75
155.75
144.00
b
20
70
82
58
25
70
58
45
96 104
82
60
64.167
83.167
108.333
125.083
1250F
yi 2.
n
2
ijk
yi 3.
yi 3.
57.25
119.75
145.75
y2
2
2
SSTO = y − = (130 ) + (155) +
abn
i =1 j =1 k =1
a
yi..
Temperature
700F
150F
Material Type
1250F
57.5
49.5
85.5
+ ( 60 )
( 3799 ) = 77646.97 .
−
2
2
36
( 3799 ) = 10683.72 .
y2
1 a
1
2
2
2
SSMaterial = yi2 − =
( 998) + (1300 ) + (1501) −
bn i =1
abn ( 3)( 4 )
36
2
( 3799 ) = 39118.72 .
y2
1 b 2
1
2
2
2
SSTemperature =
y j − =
(1738) + (1291) + ( 770 ) −
an j =1
abn 3 ( 4 )
36
2
SSInteractions =
y2
1 a b 2
y
−
ij abn − SSMaterial − SSTemperature
n i =1 j =1
( 3799 ) − 10683.72 − 39118.72 = 9613.78
+ ( 342 ) −
36
and
SSE = SSTO − SSMaterial − SSTemperature − SS Interaction
.
= 77646.97 − 10683.72 − 39118.72 − 9613.78 = 18230.75
2
1
2
2
= ( 539 ) + ( 229 ) +
4
Source
Temperature
Material Type
Temperature X MT
Error
Total
School of Mathematical Sciences (SMS)
2
Df
2
2
4
27
35
SS
39118.72
10683.72
9613.78
28230.75
77646.97
page 7
MS
19559.36
5341.86
2403.44
675.21
F0
28.97
7.91
3.56
August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
B Sc (Hons) in Actuarial Studies
Because F0.05,4,27 = 2.73 , we conclude that there is a significant interaction between
material types and temperature. Furthermore, F0.05,2,27 = 3.35 , so the main effects of
material type and temperature are also significant.
- - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - To assist in interpreting the results of this experiment, it is helpful to construct a
graph of the average responses at each treatment combination. The graph is shown below.
The significant interaction is indicated by the lack of parallelism of the lines. In general,
longer life is attained at low temperature, regardless of material type. Changing from low
to intermediate temperature, battery life with material type 3 may actually increase,
whereas it decreases for type 1 and 2. From intermediate to high temperature, battery life
decreases for material types 2 and 3 and is essential unchanged for type 1. Material type 3
seems to give the best results if we want less loss of effective life as the temperature
changes.
3.3.1
Multiple Comparisons
When the ANOVA indicates that row or column means differ. It is usually of interest to
make comparisons between the individual row or column means to discover the specific
differences. Referring to Example 1, it indicates that interaction effect is significant. When
interaction is significant, comparisons between the means of one factor (e.g. A) may be
obscured by the AB interaction. One approach to this situation is to fix factor B at a specific
level and apply Tukey’s test to the means of factor A at the level. To illustrate, let us refer
to the following example.
Example 3
From Example 1, we are interested in detecting differences among the means of three
material types. Because the interaction is significant, we make this comparison at just one
level of temperature, say level 2 (70 o F). We assume that the best estimate of the error
variance is the MSE from the ANOVA table, utilizing the assumption that the experimental
error variance is the same over all treatment combinations.
The three material type averages at 70 oF arranged in ascending order are
y12 = 57.25 (material type 1)
y22 = 119.75 (material type 2)
y32 = 145.75 (material type 3)
and
T0.05 = q0.05 ( 3, 27 )
School of Mathematical Sciences (SMS)
MSE
n
= ( 3.50 )
page 8
675.21
4
= 45.47
August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
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where we obtained q0.05 ( 3, 27 ) 3.50 by interpolation. The pairwise comparisons yield
3 vs 1: 145.75 − 57.25 = 88.50 T0.05 = 45.47
3 vs 2: 145.75 − 119.75 = 26.00 T0.05 = 45.47
2 vs 1: 119.75 − 57.25 = 62.50 T0.05 = 45.47
This analysis indicates that at the temperature level 70 oF, the mean battery life is the same
for material types 2 and 3 and that the mean battery life for material type 1 is significantly
lower in comparison to both types 2 and 3.
3.4)
The General Factorial Design
The results for the two-factor factorial design may be extended to the general case where
there are a levels of factor A, b levels of factor B, c levels of factor C and so on, arranged
in a factorial experiment. In general, there will be abc…n total observations if there are n
replicates to determine a sum of squares due to error if all possible interactions are included
in the model.
If all factors in the experiment are fixed, we may easily formulate and test
hypotheses about the main effects and interactions using the ANOVA. For a fixed effects
model, test statistics for each main effect and interaction by the mean square error. All of
these F tests will be upper-tail, one-tail tests. The number of degrees of freedom for any
main effect is the number of levels of the factor minus one and the number of degrees of
freedom for an interaction is the product of the number of degrees of freedom associated
with the individual components of the interaction. For example, consider the three-factor
analysis of variance model:
i = 1, 2,..., a
j = 1, 2,..., b
yijkl = + i + j + k + ( )ij + ( )ik + ( ) jk + ( )ijk + ijkl
k = 1, 2,..., c
l = 1, 2,..., n
Assuming that A, B, C are fixed, the ANOVA table is shown below. The F tests on main
effects and interactions follow directly from the expected mean squares.
Source
A
a −1
df
SS
SSA
MS
MSA
B
b −1
SSB
MSB
C
c −1
SSC
MSC
AB
( a − 1)( b − 1)
SSAB
MSAB
School of Mathematical Sciences (SMS)
page 9
F0
MSA
MSE
MSB
MSE
MSC
MSE
MSAB
MSE
August 2022 Semester
MST2034 Design of Experiments
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AC
( a − 1)( c − 1)
SSAC
MSAC
BC
( b − 1)( c − 1)
SSBC
MSBC
ABC
( a − 1)( b − 1)( c − 1)
SSABC
MSABC
Error
abc ( n − 1)
SSE
MSE
Total
abcn − 1
SSTO
MSAC
MSE
MSBC
MSE
MSABC
MSE
Manual computing formulas for the sum of squares are as follows.
a
b
c
n
y2
2
SSTO = yijkl
− ;
abcn
i =1 j =1 k =1 l =1
y2
1 a 2
yi − ;
bcn i =1
abcn
b
y2
1
2
SSB =
y
−
j abcn ;
acn j =1
SSA =
SSC =
y2
1 c 2
y
−
k abcn ;
abn k =1
To compute the two-factor interaction sums of squares, the total for the A B , A C and
B C cells are needed. It is frequently helpful to collapse the original data table into three
two-way tables to compute these quantities. The sums of squares are found from
SSAB =
y2
1 a b 2
y
−
ij abcn − SSA − SSB = SSSubtotals( AB) − SSA − SSB ;
cn i =1 j =1
y2
1 a b 2
y
−
ik abcn − SSA − SSC = SSSubtotals( AC ) − SSA − SSC ;
bn i =1 k =1
y2
1 b c 2
SSBC =
y
−
jk abcn − SSB − SSC = SSSubtotals( BC ) − SSB − SSC .
an j =1 k =1
SSAC =
Note that the sums of squares for the two-factor subtotals are found from the totals at each
two-way table. The three-factor interaction sum of squares is computed from the threeway cell totals yijk as
SSABC =
y2
1 a b c 2
y
−
ijk abcn − SSA − SSB − SSC − SSAB − SSAC − SSBC .
n i =1 j =1 k =1
= SSSubtotals( ABC ) − SSA − SSB − SSC − SSAB − SSAC − SSBC
The error sum of squares may be found by subtracting the sum of squares for each main
effect and interaction from the total sum of squares or by SSE = SSTO − SSSubtotals( ABC ) .
School of Mathematical Sciences (SMS)
page 10
August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
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Example 4 The Soft Drink Bottling Problem
A soft drink bottler is interested in obtaining more uniform fill heights in the bottles
produced by his manufacturing process. The filling machine theoretically fills each bottle
to the correct target height, but in practice, there is variation around this target and the
bottler would like to understand the sources of this variability better and eventually reduce
it.
The process engineer can control three variables during the filling process: the
percent carbonation (A), the operating pressure in the filler (B) and the bottles produced
per minute or the line speed (C). The pressure and speed are easy to control but the percent
carbonation is more difficult to control during actual manufacturing because it varies with
product temperature. However, for purposes of an experiment, the engineer can control
carbonation at three levels: 10, 12 and 14%. She chooses two levels for pressure (25 and
30 psi) and two levels for line speed (200 and 250 bpm). She decided to run 2 replicates
of a factorial design in these three factors, with all 24 runs taken in random order. The
response variable observed is the average deviation from the target fill height observed in
a production run of bottles at each set of conditions. The data that resulted from this
experiment is given below. Positive deviations are fill heights above the target, whereas
negative deviations are fill heights below the target.
Operating Pressure (B)
25 psi
30 psi
Line Speed (C) Line Speed (C)
200
250
200
250
−1
−1
−3
1
0
0
−1
1
0
2
2
6
1
1
3
5
5
7
7
10
4
6
9
11
6
15
20
34
% of Carbonation (A)
10
12
14
B C Totals, y jk
21
y. j ..
B
25
−5
4
22
−4
20
59
75 = y....
54
A C Totals, yi.k .
C
A
200
250
−5
10
1
12
6
14
14
25
34
A B Totals, yij ..
A
10
12
14
yi
30
1
16
37
The total corrected sum of squares is found as
( 75)
y2
SSTO = y − = 571 −
= 336.625
abcn
24
i =1 j =1 k =1 l =1
and the sum of squares for the main effects are
a
b
c
2
n
2
ijkl
School of Mathematical Sciences (SMS)
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August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
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( 75)
y2
1 a 2
1
2
2
2
SSCarbonation =
yi −
= ( −4 ) + ( 20 ) + ( 59 ) −
= 252.75 ;
bcn i =1
abcn 8
24
2
( 75)
1
2
2
y2
1 b 2
= ( 21) + ( 54 ) −
= 45.375 ; and
SSPressure =
y
−
j
24
acn j =1
abcn 12
2
( 75)
1
y2
1 c 2
2
2
= ( 26 ) + ( 49 ) −
= 22.042 ;
y
−
SSSpeed =
k
24
abn k =1
abcn 12
2
To calculate the sums of squares for the two-factor interactions, we must find the two-way
cell totals. Hence, the sums of squares are
y2
1 a b
SSAB = yij2 − − SSA − SSB
cn i =1 j =1
abcn
;
2
75 )
(
1
2
2
2
= ( −5 ) + (1) + + ( 37 ) −
− 252.75 − 45.375 = 5.25
4
24
The carbonation-speed or AC interaction uses the A C cell totals yik shown below.
SSAC =
y2
1 a b 2
y
−
ik abcn − SSA − SSC
bn i =1 k =1
1
2
2
= ( −5) + (1) +
4
( 75)
;
2
+ ( 34 ) −
− 252.75 − 22.042 = 0.583
24
2
The pressure-speed or BC interaction is found from the B C cell totals y jk shown
below.
y2
1 b c 2
SSBC =
y jk − abcn − SSB − SSC
an j =1 k =1
=
( 75)
;
2
1 2
2
2
2
− 45.375 − 22.042 = 1.042
( 6 ) + (15) + ( 20 ) + ( 34 ) −
6
24
The three-factor interaction sum of squares is found from the A B C cell totals yijk
as follows.
y2
1 a b c 2
SSABC = yijk
−
− SSA − SSB − SSC − SSAB − SSAC − SSBC
n i =1 j =1 k =1
abcn
1
2
2
= ( −4 ) + ( −1) +
2
= 1.083.
( 75)
+ ( 21) −
− 252.75 − 45.375 − 22.042 − 5.25 − 0.583 − 1.042
24
2
2
y2
1 a b c 2
y
−
ijk abcn = 328.125 , we have
n i =1 j =1 k =1
SSE = SSTO − SSSubtotals( ABC ) = 336.625 − 328.125 = 8.5 .
Finally, noting that SSSubtotals( ABC ) =
School of Mathematical Sciences (SMS)
page 12
August 2022 Semester
MST2034 Design of Experiments
Source
% of carbonation (A)
Pressure (B)
Line Speed (C)
AB
AC
BC
ABC
Error
Total
B Sc (Hons) in Industrial Statistics
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df
2
1
1
2
2
1
2
12
23
SS
252.75
45.375
22.042
5.25
0.583
1.042
1.083
8.500
336.625
MS
126.375
45.375
22.042
2.625
0.292
1.042
0.542
0.708
F value
178.412
64.059
31.118
3.706
0.412
1.471
0.765
From the ANOVA table, we see that the percentage of carbonation, operating pressure and
line speed significantly affect the fill volume. The carbonation-pressure interaction F ratio
indicating some interaction between these factors.
3.5)
Blocking in a Factorial Design
We have discussed factorial designs in the context of a completely randomized experiment.
Sometimes, it is not feasible or practical to completely randomize all of the runs in a
factorial. For example, the presence of a nuisance factor may require that the experiment
be run in blocks. We discussed the basic concepts of blocking in the context of a singlefactor experiment in previous chapter. We now show how blocking can be incorporated in
a factorial.
Consider a factorial experiment with two factors (A and B) and n replicates. The
linear statistical model for this design is
i = 1, 2,..., a
yijk = + i + j + ( )ij + ijk j = 1, 2,..., b
k = 1, 2,..., n
where i , j and ( )ij represent the effects factor A, B and the AB interaction,
respectively.
Now suppose that to run this experiment a particular raw material is required. This
raw material is available in batches that are not large enough to allow all abn treatment
combinations to be run from the same batch. However, if a batch contains enough material
for ab observations, then an alternative design is to run each of the n replicates using a
separate batch of raw material. Consequently, the batches of raw material represent a
randomization restriction or a block and a single replicate of a complete factorial
experiment is run within each block. The effects model for this new design is
i = 1, 2,..., a
yijk = + i + j + ( )ij + k + ijk j = 1, 2,..., b
k = 1, 2,..., n
where k is the effect of the kth block. Of course, within a block the order in which the
treatment combinations are run is completely randomized.
School of Mathematical Sciences (SMS)
page 13
August 2022 Semester
MST2034 Design of Experiments
B Sc (Hons) in Industrial Statistics
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ANOVA Table for a Two-Factor Factorial in a Randomized Complete Block
Source of
df
SS
F value
Variation
Blocks
n −1
y2
1
2
y − abn
k
ab k
Factor A
a −1
y2
1
yi2 −
bn i
abn
MSA
MSE
Factor B
b −1
y2
1
2
y
−
j abn
an j
MSB
MSE
AB
( a − 1)( b − 1)
MSAB
MSE
Error
( ab − 1)( n − 1)
( abn − 1)
y2
1
yij2 − − SSA − SSB
n i
abn
Total
Subtraction
yijk2 −
i
j
k
y2
abn
The model assumes that interaction between blocks and treatments is negligible.
This was assumed previously in the analysis of randomized block designs. If these
interactions do exist, they cannot be separated from the error component. In fact, the error
2
2
2
term in this model really consists of the ( )ik , ( ) jk and ( )ijk interactions. The
ANOVA outlined a layout which closely resembles that of a factorial design, with the error
sum of squares reduced by the sum of squares for blocks. Computationally, we find the
sum of squares for blocks as the sum of squares between the n block totals yk .
Example 5
An engineer is studying methods for improving the ability to detect targets on a radar scope.
Two factors she considers to be important are the amount of background noise or “ground
clutter”, on the scope and the type of filter placed over the screen. An experiment is
designed using three levels of ground clutter and two filter types. We will consider these
as fixed-type factors. The experiment is performed by randomly selecting a treatment
combination (ground clutter level and filter type) and then introducing a signal representing
the target into the scope. The intensity of this target is increased until the operator observes
it. This intensity level at detection is then measured as the response variable. Because
operator availability, it is convenient to select an operator and keep him or her at the scope
until all the necessary runs have been made. Furthermore, operators differ in their skill and
ability to use the scope. Consequently, it seems logical to use the operators as blocks.
Once an operator is chosen, the order in which the six treatment combinations are run is
randomly determined. Thus, we have a 3 2 factorial experiment run in a randomized
complete block. The data are shown below.
School of Mathematical Sciences (SMS)
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August 2022 Semester
MST2034 Design of Experiments
Operators (blocks)
Filter Type
Low
Medium
High
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1
2
2
1
2
Ground clutter
90
86
96
84
102 87 106 90
114 93 112 91
1
3
4
1
2
1
2
100
105
108
92
97
95
92
96
98
81
80
83
The linear model for this experiment is
i = 1, 2,..., a
yijk = + i + j + ( )ij + k + ijk j = 1, 2,..., b
k = 1, 2,..., n
where i represents the ground clutter effect, j represents the filter type effect ( )ij is
the interaction, k is the block effect and ijk is the NID ( 0, 2 ) error component. The
sum of squares for ground clutter, filter type and their interaction are computed in the usual
manner.
The sum of squares due to blocks is found from the operator totals yk as follows.
SSBlocks =
y2
1 n 2
y
−
k abn
ab k =1
( 2278)
1
2
2
2
2
=
572 ) + ( 579 ) + ( 597 ) + ( 530 ) −
= 402.17
(
( 3)( 2 )( 4 )
( 3) 2
2
ANOVA Table
Source
Ground clutter (G)
Filter type (F)
GF
Blocks
Error
Total
Df
2
1
2
3
15
23
SS
MS
F0
335.58 157.79 15.13
1066.67 1066.67 96.19
77.08
38.54
3.48
402.17 134.06
166.33
11.09
2047.83
The complete ANOVA indicates that all effects are tested by dividing their mean squares
by the mean square error. Both ground clutter level and filter type are significant at the 1
percent level, whereas their interaction is significant only at the 10 percent level. Thus, we
conclude that both ground clutter level and the type of scope filter used affect the operator’s
ability to detect the target and there is some evidence of mild interaction between these
factors.
School of Mathematical Sciences (SMS)
page 15
August 2022 Semester
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