Chapter 4 Experimental Design

8/6/2019 Chapter 4 Experimental Design

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Experiment 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.



Experiment design

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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)



Experiment design

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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.



Experiment design

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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



Experiment design

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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.



Experiment design

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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



Experiment design

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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.



Experiment design

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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



Experiment design

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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.



Experiment design

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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



Experiment design

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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



Experiment design

13

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.



Experiment design

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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



Experiment design

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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



Experiment design

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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



Experiment design

17

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



Experiment design

18

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.



Experiment design

19

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.



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.

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Chapter 4 Experimental Design