Chapter 4 Experimental Design
Transcript of Chapter 4 Experimental Design
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Chapter 4 Experimental design
4.1 Overview
A designed experiment is a test or series of tests in which purposeful changes are made to
the input variables of a process so that the resulted changes in the output variables(responses) may be observed and identified. Many of the earlier applications of
experimental design were in the agriculture and biological sciences. It has found broad
applications in many disciplines now.
Consider the tension bond strength of cement mortar which is an important characteristic
of ready-mixed cement. An engineer is interested in comparing the strength of a modifiedformulation in which polymer latex emulsions (referred to as a factor) have been added
during the mixing to the strength of unmodified formulation. The engineer will collect
observations on tested items selected these two formulations. The two different
formulations are referred to as treatments. A two-sample t-test can be used to testwhether the modified one on the average has larger tension bond strength. If we have
more than two formulations to compare the statistical techniques ‘ANOVA’ we discussed
earlier may be used to test the difference in the bond strength of different formulations.
Formally, this involves several steps in the experimental process.
1. Statement of the problem – objectives
2. Factors – choose the factors to be varied in the experiment, and the specific levels atwhich runs will be made. Need practical experience and theoretical understanding.
3. Selection of response variable – provides useful information about the process under
study.4. Choice of experimental design – sample size, replications, run order, blocking,
randomization. Keep design and analysis as simple as possible. Should consider costsand other resources.
5. Performing the experiment6. Data analysis – appropriate statistical methods, results and conclusions should be
objective rather than judgmental in nature.
7. Conclusions and recommendations – practical conclusions about results andrecommendations for action.
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4.2 Factorial experiments
Factorial experiments are employed to study simultaneously the effects of two or more
factors, each covers a pre-determined number of levels. By a factorial design we meanthat in each complete trial or replicate of the experiment all possible combinations of the
levels of the factors are investigated. Thus if there are a levels of factor A and b levels of factor B, then each replicate contains all ab treatment combinations. We shall have a two-
way or higher-way classification experiment.
The effect of a factor is defined as the change in response produced by a change in the
level of the factor. This is called a main effect because it refers to the primary factor
under study.
In some experiments, 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 effect between the factors.
Factorial designs are more efficient than one-factor-at-a-time experiments. A factorial
design is necessary when interactions are present.
4.2.1 Two-factor factorial design
Let these factors be A and B. Each observation will be denoted by ijt x . Formally, we
assume
( )ijt i j ij ijt x A B ABμ ε = + + + +
i refers to factor A (a treatments) j refers to factor B (b treatments)t refers to the replications (n replicates) Ai refers to the main effect of A B j refers to the main effect of B
( AB)ij refers to the interaction effect of A and B
ε ijt are independent N (0, σ 2)
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The treatment sums will be
a factor A treatment sums =..i
X , (n A = nb observations each)
b factor B treatment sums = . . j X , (n B = na observations each)
ab factor AB combined treatment sums = .ij X , (n observations each)
Total treatment = ... X , ( N = nab observations)
Treatment A SS = 2
.. ...
1
( )a
i
i
nb x x=
−∑ =2 2
.. ...
1
ai
i A
X X
n N =
−∑ = SS A
Treatment B SS = 2
. . ...
1
( )b
j
j
na x x=
−∑ =
2 2. . ...
1
b j
j B
X X
n N =
−∑ = SS B
Total SS = 2
...
1 1 1
( )a b n
ijt
i j t
x x= = =
−∑∑∑ = 22 ...
1 1 1
a b n
ijt
i j t
X x N = = =
−∑∑∑
When factors A and B interact the interaction variation is:
AB interaction SS = 2
. .. . . ...
1 1
( )a b
ij i j
i j
n x x x x= =
− − +∑∑ =
2 2. ...
1 1
a bij
i j
X X
n N = =
−∑∑ - SS A - SS B
The total SS is decomposed as
Total SS = Treatment A SS + Treatment B SS + AB Interaction SS + Error SS
Error SS is obtained by subtraction.
As before,
Error SS = 2 2 2 2 2 2 2
.. 1 ( 1)
1 1 1 1 1
( ) ~a b n a b
ijt i n ab n N ab
i j t i j
x x σ χ σ χ σ χ − − −
= = = = =
− = =∑∑∑ ∑∑
E(Error SS) = ( N - ab) σ 2
Here we do not require H 0. The error SS always has a 2 χ - distribution whether or not
H 0 is true.
Under H 0,
Treatment A SS = 2
.. ...
1
( )a
i
i
nb x x=
−∑ ~ σ 2
1a χ
−
Treatment B SS = 2
. . ...
1
( )b
j
j
na x x=
−∑ ~ σ 2
1b χ −
AB interaction SS = 2
. .. . . ...
1 1
( )a b
ij i j
i j
n x x x x= =
− − +∑∑ ~ σ 2
( 1)( 1)a b χ − −
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Mean squared SS
2s = Error SS / ( N-ab)
2
As = Factor A SS / (a-1)2
Bs = Factor B SS / (b-1)
2
ABs = Interaction SS / (a-1) (b-1)
Whether or not H 0 is true E( 2s ) = σ
2, thus 2
s is unbiased for σ 2.
2
2
As
s~ 1,a N abF
− −
2
2
As
s ~ 1,b N abF − − 2
2
ABs
s~
( 1)( 1),a b N abF
− − −
4.2.2 ANOVA Table
Source of
variation
Sum of squares Degree of
freedom
Mean
square
F-ratio
Treatments A
2 2
.. ...
1
ai
i A
X X
n N =
−∑ a-12
As
2
2
As
s
Treatments B
2 2. . ...
1
b j
j B
X X
n N =
−∑ b-12
Bs
2
2
As
s
AB Interaction
2 2. ...
1 1
a bij
i j
X X
n N = =
−∑ ∑ - SS A - SS B (a-1)(b-1)
2
ABs 2
2
ABs
s
Random
variation or
Error
By subtraction
By subtraction
= ab(n-1)
= N - ab
2s
Total variation2
2 ...
1 1 1
a b n
ijt
i j t
X x
N = = =
−∑∑∑ N -1
Note: When the interaction effect is not significant or when there are no replications (i.e.
n = 1), we may omit the row of interaction. In the former situation, the random SS will be
the original random SS plus the AB interaction SS and the d.f. will accordingly be
increased. In the latter, no interaction SS can be calculated due to no replications.
The randomized block design discussed earlier may be regarded as a two-factor (one
factor is plot and the other is fertilizer) experiment without replication. The interaction
effect cannot be separated from the error SS.
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Example
A department store conducted an experiment to study the sales of a product for different
package designs and different selling prices. There were three different package designs
at two different selling prices. (This is equivalent to a randomized block design with two
replications. Package is the block and price is the treatment.)
Package design Selling price (Higher) Selling price (Lower)
1 14
12
17
13
2 12
16
11
9
3 37
41
35
41
They want to test:
1. The hypothesis that there are no interactions for any combination of package
design and a selling price.
2. The hypothesis that the mean population amount of sales is the same for three
package designs.
3. The hypothesis that the mean population sales amount is the same for two
selling prices.
4. Locate where the difference are if the hypothesis is significant.
Totals: the values inside table are interaction totals .ij X
the row and column totals are factor totals ..i X , . . j X
the overall total is grand treatment total ... X
Package
design
Selling price
(Higher)
Selling price
(Lower). . j X
1 26 30 56
2 28 20 48
3 78 76 154
..i
X 132 126 258
a = 2, b = 3, n =2, N = 12,
Total SS = 142
+ 122
+ 172
+ 132
+ 372
+ 412
+ 352
+ 412
+ 122
+ 162
+ 112
+ 92
- 2582 /12
= 1809
Package SS = 562 /4 + 154
2 /4 + 48
2 /4 - 258
2 /12 = 1742
Price SS = 1322 /6 + 126
2 /6 + 48
2 /4 - 258
2 /12 = 3
Interaction SS = 262 /2 + 30
2 /2 + 78
2 /2 + 76
2 /2 + 28
2 /2 + 20
2 /2 - 258
2 /12 - 1742 – 3 = 18
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ANOVA Table
Source of
variation
Sum of
squares
Degree of
freedom
Mean
square
F-ratio F-critical
at 5%
Package 1742 2 871 113.6 5.143
Price 3 1 3 0.391 5.987
Package x Price 18 2 9 1.173 5.143
Error 46 6 7.67
Total1809 11
Conclusion
At 5% level, the difference in package design effect on the mean sales is significant,
whereas the difference in price effect and interaction effect of price and package design
are not significant.
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Below is computer output from EXCEL
Anova: Two-Factor With Replication
SUMMARY Selling price (Higher) Selling price (Lower) Total
Package design 1
Count 2 2 4
Sum 26 30 56
Average 13 15 14
Variance 2 8 4.666667
Package design 2
Count 2 2 4
Sum 78 76 154Average 39 38 38.5
Variance 8 18 9
Package design 3
Count 2 2 4
Sum 28 20 48
Average 14 10 12
Variance 8 2 8.666667
Total
Count 6 6
Sum 132 126Average 22 21
Variance 177.2 184
ANOVA
Source of Variation SS df MS F P-value F crit
Sample 1742 2 871 113.6087 1.7E-05 5.143249
Columns 3 1 3 0.391304 0.554646 5.987374
Interaction 18 2 9 1.173913 0.371307 5.143249
Within 46 6 7.666667
Total 1809 11
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4.3 Example Three factor design
An experiment is conducted to determine the thrust forces in drilling at different speeds,feeds and in different materials. Five speeds are used, three feeds and two materials with
two samples tested under each set of conditions. The order of the experiment is
completely randomized and the levels of all factors are fixed. Data in coded units are
obtained as follows:
Speed
Material Feed 100 220 475 715 870
B .004 122
110
108
85
108
60
66
50
80
60
.008 332
330
276
310
248
295
248
275
276
310.014 640
500
612
500
543
450
612
610
696
610
V .004 196
170
136
130
122
85
108
75
136
75
.008 386
365
333
330
318
330
472
350
499
390
.014 810
725
779
670
810
750
893
890
1820
890
4.3.1 Three-factor Interaction totals , Speed totals
Speed Row
totalMaterial Feed 100 220 475 715 870
B .004 232 193 168 116 140 849
.008 662 586 543 523 586 2900
.014 1140 1112 993 1222 1306 5773
V .004 366 266 207 183 211 1233
.008 751 663 648 822 889 3773
.014 1535 1449 1560 1783 2710 9037
Speed
total
4686 4269 4119 4649 5842 23565
The above is similar to a three-factorial design without replications.
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4.3.2 Two-factor Interaction totals and Factor totals
Speed x Material
Speed Material
totalMaterial 100 220 475 715 870
B 2034 1891 1704 1861 2032 9522
V 2652 2378 2415 2788 3810 14043
Speed
total4686 4269 4119 4649 5842 23565
Speed x Feed
Speed Row
totalFeed 100 220 475 715 870
.004 598 459 375 299 351 2081
.008 1413 1249 1191 1345 1475 6673
.014 2675 2561 2553 3005 4016 14810
Speed total 4686 4269 4119 4649 5842 23565
Material x Feed
Feed Material
totalMaterial .004 .008 .014
B 849 2900 5773 9522
V 1233 3773 9037 14043
Feed total 2081 6673 14810 23565
Total SS = 1222
+ 1102
+ 1082
+ … + 18202
+ 8902
- 235652
/ 60 = 5831591.25
Speed SS = 46862 /12 + 4269
2 /12 + 4119
2 /12 + 4649
2 /12 + 5842
2 /12
- 235652
/ 60 = 152453.17
Material SS = 95222 /30 + 14043
2 /30 - 23565
2/ 60 = 340657.35
Feed SS = 20822 /20 + 6673
2 /20 + 14810
2 /20 - 23565
2/ 60 = 4154833.9
SxM SS = 20342 /6 + 1891
2 /6 + … + 2788
2 /6 + 3810
2 /6
- 235652 / 60 - 152453.17 - 340657.35 = 88111.56
SxF SS = 5982 /4 + 459
2 /4 + … + 3005
2 /4 + 4016
2 /4
- 235652
/ 60 - 152453.17 - 4154833.9 = 255471.43
MxF SS = 8492 /10 + 2900
2 /10 + … + 3773
2 /10 + 9037
2 /10
- 235652
/ 60 - 340657.35 - 4154833.9 = 237506.7
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SxMxF SS = 2322 /2 + 193
2 /2 + … + 1783
2 /2 + 2710
2 /2
- 235652
/ 60 - 152453.17 - 340657.35 -4154833.9
- 88111.56 - 255471.43 - 237506.7 = 113084.64
Higher Interaction SS = sum of interaction total squares/replication number
- Total correction SS – All main factor SS and lower interaction SS
4.3.3 ANOVA Table
Source SS DF MS F Remark
Speed 152453.17 4 38113.3 2.336
Material 340657.35 1 340657.35 20.879 * at 0.001
Feed 4154833.9 2 2077416.95 127.326 * at 0.001
S x M 88111.56 4 22027.89 1.350
S x F 255471.43 8 31933.9 1.957
M x F 237506.7 2 118753.35 7.278 * at 0.01
S x M x F 113084.64 8 14135.58 0.866
Error 489472.5 30 16315.75
Total 5831591.25 59
Since the three factor interaction effect is not significant, we may assume no three factor
interaction so that its SS is combined with Error SS to form a new Error SS with 38
degrees of freedom.
ANOVA Table without three factor interaction
Source SS DF MS F
Speed 152453.17 4 38113.3 2.404
Material 340657.35 1 340657.35 21.483
Feed 4154833.9 2 2077416.95 131.011
S x M 88111.56 4 22027.89 1.389
S x F 255471.43 8 31933.9 2.0139
M x F 237506.7 2 118753.35 7.489
Error 602557.14 38 15856.77
Total 5831591.25 59
The conclusions are not affected. The latter is a more powerful test because it has larger
d.f. for the Error SS.
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4.3.4 Main effects
Look at the treatment means of Feed at different levels of the other factors.
Main effect of Feed at different levels of Speed
Speed
Feed 100 220 475 715 870
.004 149.5 114.75 93.75 74.75 87.75
.008 353.25 312.25 297.75 336.25 368.75
.014 668.75 640.25 638.25 751.25 1004
Main effect of Feed at different levels of Material
FeedMaterial .004 .008 .014
B 84.9 290 577.3
V 123.3 377.3 903.7
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Interaction Plot of Feed at Levels of Speed
50
150
250
350
450
550
650
750
850
950
1050
F1 F2 F3
Feed
R e s p o n s e
S1
S2
S3
S4
S5
Interaction Plot of Feed at Levels of Material
80
180
280
380
480
580
680
780
880
F1 F2 F3
Feed
R e s p o n s e
B
V
No
interaction
effect
means
lines have
the same
slope
Significant
interaction
effects
means lines
have
differentslopes
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Computer output from MINITAB
General Linear Model: Thrust versus Speed, Feed, Material
Factor Type Levels Values
Speed fixed 5 100, 220, 475, 715, 870
Feed fixed 3 0.004, 0.008, 0.014
Material fixed 2 B, V
Analysis of Variance for Thrust, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Speed 4 152453 152453 38113 2.34 0.078
Feed 2 4154834 4154834 2077417 127.33 0.000
Material 1 340657 340657 340657 20.88 0.000
Speed*Feed 8 255471 255471 31934 1.96 0.088Speed*Material 4 88112 88112 22028 1.35 0.275
Feed*Material 2 237507 237507 118753 7.28 0.003
Speed*Feed*Material 8 113085 113085 14136 0.87 0.555
Error 30 489473 489473 16316
Total 59 5831591
S = 127.733 R-Sq = 91.61% R-Sq(adj) = 83.49%
Unusual Observations for Thrust
Obs Thrust Fit SE Fit Residual St Resid
55 1820.00 1355.00 90.32 465.00 5.15 R60 890.00 1355.00 90.32 -465.00 -5.15 R
R denotes an observation with a large standardized residual.
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M e a n
o f T h r u s t
870715475220100
800
600
400
200
0.0140.0080.004
VB
800
600
400
200
Speed Feed
Material
Main Effect Plot
SpeedSpeed
1000
500
0
FeedFeed
M ater ia lM ater ia l
VB
0.0140.0080.004
1000
500
0
870715475220100
1000
500
0
Speed
475
715
870
100
220
Feed
0.014
0.004
0.008
Material
B
V
Interaction Plot
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4.4 Latin Square
In randomized block design we attempt to eliminate a source of error by applying each of
our treatments just once in each block. In a Latin square we try to eliminate two sorts of
undesired variability from our treatment comparisons. These are called ‘rows’ and
‘columns’. Again each treatment occurs just once within each row and just once within
each column.
A Latin square of side n or n x n Latin square is an arrangement of n letters, each
repeated n times, in a square array of side n in such a manner that each letter appears
exactly once in each row and in each column.
Examples of Latin squares:
4 x 4 5 x 5 5 x 5
A B D C A B C D E A B C D E
B C A D B C D E A C D E A B
C D B A C D E A B E A B C D
D A C B D E A B C B C D E A
E A B C D D E A B C
Model
ijt i j t ijt x R C T μ ε = + + + + , i, j, t = 1, 2, …, n.
Ri : ith row effect
C j : jth column effect
T t : t th treatment effect
Assumptions
i
i
R∑ = 0
j
j
C ∑ = 0
t
t T ∑ = 0
2independently ~ (0, )ijt
N ε σ
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4.3.1 Sum of squares
The total sum of squares can be decomposed into components attributable to different
effects.
2
...
1 1
( )n n
ijt
i j
x x= =
−∑∑ = 2
.. ...
1
( )n
i
i
n x x=
−∑ + 2
. . ...
1
( )n
j
j
n x x=
−∑ + 2
.. ...
1
( )n
t
t
n x x=
−∑
+ 2
.. . . .. ...
1 1
( 2 )n n
ijt i j t
i j
x x x x x= =
− − − +∑∑
Total SS = Row SS + Column SS + Treatment SS + Error SS
4.3.2 Computational formulae
2 N n=
Row total =..
1
n
i ijt
j
X x=
= ∑
Column total = . .
1
n
j ijt
i
X x=
= ∑
Treatment total =..
or 1
n
t ijt
i j
X x=
= ∑
Grand total = ...1 1
j n
ijt i j X x= =
=
∑∑
Total SS = 2
1 1ijt
n n
i j
x= =
∑∑ -2
... X
N d.f. = N - 1
Row SS =2
..
1
ni
i
X
n=
∑ -2
... X
N d.f. = n - 1
Column SS =
2
. .
1
n j
j
X
n=
∑ -2
... X
N d.f. = n - 1
Treatment SS =
2
..
1
n
t
t
X
n=∑ -
2
...
X
N d.f. = n – 1
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4.3.3 ANOVA Table
Source SS DF MS F
Treatments
2
..
1
nt
t
X
n=
∑ -2
... X
N n – 1
2
t s
2
t s / 2s to test
H 0: T t = 0
Rows
2
..
1
ni
i
X
n=
∑ -2
... X
N n – 1
2
Rs
2
Rs / 2s to test
H 0: Ri = 0
Columns
2
. .
1
n j
j
X
n=
∑ -2
... X
N n – 1
2
C s
2
C s / 2s to test
H 0: C j = 0
Error By subtraction (n – 1)( n – 2) 2s
Total2
1 1
ijt
n n
i j
x
= =
∑∑ -2
... X
N
N – 1
E( 2s ) = σ 2 whether of not H 0 are true. Thus 2s is unbiased for σ
2.
Example
An experimenter is studying the effect of 5 different formulations of an explosive mixturein the manufacture of dynamite on the observed explosive force. Each formulation is
mixed from a batch of raw materials that is only enough for five formulations to be
tested. Furthermore, the formulations are prepared by several operators. The appropriatedesign 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 fiveoperators. A Latin square design is resulted for dynamite formulation
Operator
Batch 1 2 3 4 5 Row total
1 A=24 B=20 C=19 D=24 E=24 111
2 B=17 C=24 D=30 E=27 A36 134
3 C=18 D=38 E=26 A=27 B21 130
4 D=26 E=31 A=26 B=23 C22 128
5 E=22 A=30 B=20 C=29 D31 132
Column
Total 107 143 121 130 134 635
A, B, C, D and E are the five formulations.Treatment totals are A = 143, B = 101, C = 112, D = 149, E = 130
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Operator SS = 1072 /5 + 1432 /5 + 1212 /5 + 1302 /5 + 1342 /5 - 6352 /5
Batch SS = 1112 /5 + 1342 /5 + 1302 /5 + 1282 /5 + 1322 /5 - 6352 /5
Formulation SS = 1432 /5 + 101
2 /5 + 112
2 /5 + 149
2 /5 + 130
2 /5 - 635
2 /5
Total SS = 242 + 172 + … + 312 - 6352 /5
ANOVA Table
Source SS DF MS F
Formulations 330 4 82.50 7.73
Batches 68 4 17.00
Operators 150 4 37.50Error 128 12 10.67
Total 676 24
Conclusion: Different formulations have different explosive forces.
Though we may test the batches and operator effects these usually are not of primeinterest.
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Practical Exercise
Question 1. (Training.xls)
The personnel manager of a large insurance company wished to evaluate the
effectiveness of four different sales-training programs designed for new employees. Agroup of 32 recently hired college graduates were randomly assigned to four programs so
that there were eight subjects in each program. At the end of the month-long training
period a standard exam was administered to the 32 subjects; the scores are given below:
Programs
A B C D
66 72 61 6374 51 60 6182 59 57 76
75 62 60 84
73 74 81 58
97 64 55 65
87 78 70 6978 63 71 80
Determine whether there is evidence of a difference in the four sales-training programs atthe 0.05 level of significance.
Question 2. (Cholesterol.xls)
The following are the cholesterol contents, in milligrams per package, which four
laboratories obtained for 6-ounce packages of three very similar diet foods:
Laboratory
Diet food
A
Diet food
B
Diet food
C
1 3.4 2.6 2.8
2 3.0 2.7 3.1
3 3.3 3.0 3.4
4 3.5 3.1 3.7
Test whether the cholesterol contents differ among the 3 diet foods at the 0.05 level of
significance.
8/6/2019 Chapter 4 Experimental Design
http://slidepdf.com/reader/full/chapter-4-experimental-design 20/20
Experiment design
20
Question 3. (Effluent.xls)
Ozonization as a secondary treatment for effluent, following absorption by ferrous
chloride, was studied for three reaction times and three pH levels. The study yielded thefollowing results for effluent decline.
Reaction time (min) pH level 7.0 pH level 9.0 pH level 10.5
20 2321
22
1618
15
1413
16
40 20
22
19
14
13
12
12
11
10
60 21
2019
13
1212
11
1312
a) Test for significant interaction.b) Test for significant differences among reaction times.
c) Test for significant differences among pH levels.