Industrial Engineering Chapter 14 Homework Source Main Effects Residual Error Total Seqss

Source

DF

SeqSS

AdjSS

AdjMS

F

P

Main Effects

5

0.1944

0.0389

3.3

0.051

A

1

0.0946

8.02

0.018

Reserve Supplemental Exercises Chapter 14 Problem 6

An article in Biotechnology Progress (December 2002, Vol. 18(6), pp. 1170–1175) presented a

73

2−

fractional factorial to evaluate factors promoting astaxanthin production. The data are

shown in the following table.

Runs

A

B

C

D

E

F

G

Weight Content (mg/g)

Cellular Content (pg/cell)

1

-1

1

-1

4.2

10.8

2

1

-1

1

4.4

24.9

3

-1

1

-1

1

-1

1

7.8

27.3

4

1

-1

1

-1

14.9

36.3

5

-1

1

-1

1

25.3

112.6

6

1

-1

1

-1

1

-1

26.7

159.3

7

-1

1

-1

1

-1

23.9

145.2

8

1

21.9

243.2

9

1

-1

1

24.3

72.1

10

-1

1

-1

20.5

112.2

11

1

-1

1

-1

1

-1

10.8

22.5

12

-1

1

-1

1

20.8

149.7

13

1

-1

1

-1

13.5

140.1

14

-1

1

-1

1

-1

1

10.3

47.3

15

1

-1

1

-1

1

23.0

153.2

16

-1

12.1

35.2

The factors and levels are shown in the following table.

Factor

-1

+1

A

Nitrogen concentration (mM)

4.06

0

B

Phosphorus concentration (mM)

0.21

0

C

Photon flux density

( )

22

Em s



−−

100

500

B

1

0.0298

2.52

0.143

D

1

0.0410

3.48

0.092

E

1

0.0116

0.98

0.345

Residual Error

10

0.1179

0.01179

Total

15

0.3123

D

Magnesium concentration (mM)

1

0

E

Acetate concentration (mM)

0

15

F

Ferrous concentration (mM)

0

0.45

G

NaCl concentration (mM)

OHM

25

OHM: Optimal Haematococcus Medium

(a) Choose the correct complete defining relation and generators for the design.

(b) Estimate the main effects.

For Weight content:

For Cellular content:

(c) Plot the effect estimates on normal probability and interpret the results. Use

0.05



=

.

(d) Which factors (including two-ways interactions) are significant for the cellular content?

SOLUTION

(a)

Alias Structure (up to order 3):

I

A + B*C*G + B*E*F + C*D*F + D*E*G

B + A*C*G + A*E*F + C*D*E + D*F*G

Aliases:

A = BEF = BCG = CDF = DEG = ABCDE = ABDFG = ACEFG

B = AEF = ACG = CDE = DFG = ABCDF = ABDEG = BCEFG

AE = BF = DG = ABCD = ACFG = BCEG = CDEF = ABDEFG

(b)

For weight content estimated effects and coefficients are:

Term

Effect

Coef

Constant

16.525

A

1.825

0.912

B

1.225

0.612

C

10.5

5.25

D

E

1.55

0.775

F

G

A*B

A*C

A*D

0.75

0.375

A*E

6.125

3.063

A*F

A*G

0.525

0.262

B*D

A*B*D

For cellural content:

Term

Effect

Coef

Constant

93.24

A

26.41

13.21

B

19.44

9.72

C

67.71

33.86

D

E

62.46

31.23

F

9.44

4.72

G

21.09

10.54

A*B

13.51

6.76

A*C

A*D

16.66

8.33

A*E

72.54

36.27

A*F

A*G

12.71

6.36

B*D

15.54

7.77

A*B*D

3.41

1.71

(c)

C, F, and AE interaction looks significant for weight content. So the important effects are photon

Reserve Supplemental Exercises Chapter 14 Problem 7

An article in European Food Research and Technology [“Factorial Design Optimisation of Grape

(Vitis vinifera) Seed Polyphenol Extraction” (2009, Vol. 229(5), pp. 731–742)] used a central

composite design to study the effects of basic factors (time, ethanol, and pH) on the extractability

of polyphenolic phytochemicals from grape seeds. Total polyphenol (TP in mg gallic acid

equivalents/100 g dry weight) from three types of grape seeds (Savatiano, Moschofilero, and

Agiorgitiko) were recorded. The data follow.

Run

Ethanol(%)

pH

Time(h)

TP

Moschofilero

TP

Savatiano

TP

Agiorgitiko

1

40

2

1

13,320

13,127

8,673

2

40

2

5

13,596

8,925

4,370

3

40

6

1

10,714

12,047

8,049

4

40

6

5

10,730

11,299

5,315

5

60

2

1

12,149

9,700

9,384

6

60

2

5

10,910

7,107

8,290

7

60

6

1

11,620

8,755

7,905

8

60

6

5

9,757

9,792

9,347

9

40

4

3

13,593

9,748

7,253

10

60

4

3

13,459

8,727

8,390

11

50

2

3

11,980

7,164

7,611

12

50

6

3

10,338

5,928

7,292

13

50

4

1

13,992

12,200

8,305

14

50

4

5

13,450

10,552

8,380

15

50

4

3

11,745

9,284

8,792

16

50

4

3

12,267

9,084

8,302

(a) Build a second-order response surface model for each seed type and compare the models.

Which factors are significant for total polyphenol from each seeds type? Use

0.05



=

.

(b) Which interaction and quadratic terms are significant for total polyphenol from each seeds

type? Use

0.05



=

.

SOLUTION

(a)

Response Surface analysis for TP Moschofilero:

Estimated Regression Coefficients for TP Moschofilero

Term

Coef

SE

Coef

T

P

Constant

12776.9

312.4

40.896

0.0

ethanol(%)

-405.8

208.7

-1.945

0.10

-879.6

208.7

-4.215

0.006

time(h)*time(h)

558.7

406.4

1.375

0.218

Analysis of Variance for TP Moschofilero

Source

DF

Seq SS

Adj SS

Adj MS

F

P

Regression

9

24800962

2755662

6.33

0.018

Linear

3

10507288

3502429

8.04

0.016

ethanol(%)

1

1646736

3.78

PH

1

7736962

17.77

0.006

time(h)

1

1123590

2.58

0.159

Square

3

10960575

3653525

8.39

0.014

ethanol(%)*ethanol(%)

1

361616

348646

0.405

PH*PH

1

9776161

10580757

24.3

0.003

time(h)*time(h)

1

822797

1.89

0.218

Interaction

3

3333099

1111033

2.55

0.152

ethanol(%)*PH

1

1795512

4.12

0.089

ethanol(%)*time(h)

1

1439904

3.31

0.119

PH*time(h)

1

0.22

0.653

Residual Error

6

2612927

435488

Lack-of-Fit

5

2476685

495337

3.64

0.378

Pure Error

1

136242

Total

15

27413889

Response Surface analysis for TP Savatiano:

Term

Coef

SE

Coef

T

P

Constant

8824.69

333.2

26.488

0

ethanol(%)

-1106.5

222.5

-4.972

0.003

179.8

222.5

0.808

0.45

time(h)

-815.4

222.5

-3.664

0.011

ethanol(%)*ethanol(%)

592.47

433.4

1.367

0.221

PH*PH

–

2099.03

433.4

-4.843

0.003

time(h)*time(h)

2730.97

433.4

6.301

0.001

ethanol(%)*PH

55.75

248.8

0.224

0.83

ethanol(%)*time(h)

424.25

248.8

1.705

0.139

Analysis of Variance for TP Savatiano

Source

DF

Seq SS

Adj SS

Adj MS

F

P

Regression

9

54006900

6000767

12.12

0.003

Linear

3

19215475

19215474

6405158

12.93

0.005

ethanol(%)

1

12243423

24.72

0.003

PH

1

323280

0.65

0.45

time(h)

1

6648772

13.43

0.011

Square

3

27053775

9017925

18.21

0.002

ethanol(%)*ethanol(%)

1

2952824

925404

1.87

0.221

PH*PH

1

4438496

11615675

23.46

0.003

ethanol(%)*PH

473.7

233.3

2.031

0.089

ethanol(%)*time(h)

-424.2

233.3

-1.818

0.119

PH*time(h)

-110.5

233.3

-0.474

0.653

Response Surface analysis for TP Agiorgitiko:

Estimated Regression Coefficients for TP Agiorgitiko

Term

Coef

SE

Coef

T

P

Constant

8231.1

289.8

28.4

0

ethanol(%)

193.6

4.988

0.002

193.6

–

0.217

0.835

time(h)

193.6

–

3.417

0.014

ethanol(%)*ethanol(%)

–

251.66

377

–

0.667

0.529

PH*PH

621.66

377

1.649

time(h)*time(h)

269.34

0.714

0.502

ethanol(%)*PH

216.4

–

0.429

0.683

ethanol(%)*time(h)

923.12

216.4

4.265

0.005

PH*time(h)

513.12

216.4

2.371

0.055

–

Analysis of Variance for TP Agiorgitiko

Source

DF

Seq SS

Adj SS

Adj MS

F

P

Regression

9

24391655

2710184

7.23

0.013

Linear

3

13715973

4571991

12.2

0.006

ethanol(%)

1

9323834

24.88

0.002

PH

1

17640

0.05

0.835

time(h)

1

4374500

11.67

0.014

Square

3

1683019

0.308

ethanol(%)*ethanol(%)

1

0.45

0.529

PH*PH

1

1018836

2.72

0.15

time(h)*time(h)

1

0.51

0.502

Interaction

3

8992662

2997554

8

0.016

ethanol(%)*PH

1

69006

0.18

0.683

ethanol(%)*time(h)

1

6817278

18.19

0.005

PH*time(h)

1

2106378

5.62

0.055

Residual Error

6

2248567

Lack-of-Fit

5

2128517

3.55

0.382

Pure Error

1

Total

26640222

0.05



=

Factor is significant if

P value



−

. Fron model, we find:

pH is significant for TP Moschofilero.

ethanol(%)*time(h)

1439904

PH*time(h)

6272882

Residual Error

2971311

Lack-of-Fit

2951311

Pure Error

20000

Total

15

56978211

(b)

For Moschofilero, the second-order term for pH (PH*PH) is significant.

Reserve Supplemental Exercises Chapter 14 Problem 8

This

3

2

factorial design was used to study the effects of basic factors (time, ethanol, and pH) on

the extractability of polyphenolic phytochemicals from grape seeds. Total polyphenol from three

types of grape seeds (Savatiano, Moschofilero, and Agiorgitiko) were recorded. Assume that the

experiment was conducted in blocks based on seed type.

Run

Ethanol

(%)

pH

Time

(h)

TP

Moschofilero

TP

Savatiano

TP

Agiorgitiko

1

40

2

1

13,320

13,127

8,673

2

40

2

5

13,596

8,925

4,370

3

40

6

1

10,714

12,047

8,049

4

40

6

5

10,730

11,299

5,315

5

60

2

1

12,149

9,700

9,384

6

60

2

5

10,910

7,107

8,290

7

60

6

1

11,620

8,755

7,905

8

60

6

5

9,757

9,792

9,347

(a) Analyze the factorial effects. Use

0.1



=

. Which effects are important?

(b) Develop a regression model to predict the response in terms of the actual factor levels. Use

the significant terms and blocks identified in (a).

SOLUTION

(a)

Estimated Effects and Coefficients for TP Extract (coded units)

Term

Effect

Coef

SE

Coef

T

P

Constant

9786.7

359.2

27.25

0

Block 1

1812.8

507.9

3.57

0.003

Block 2

307.3

507.9

0.555

ethanol(%)

-454.1

359.2

0.537

-351.7

-175.9

359.2

0.632

time(h)

-666.9

359.2

0.085

ethanol(%)*PH

291.1

145.5

359.2

0.41

0.691

ethanol(%)*time(h)

615.4

307.7

359.2

0.86

0.406

PH*time(h)

858.8

429.4

359.2

0.252

ethanol(%)*PH*time(h)

359.2

0.09

0.929

ethanol(%)

1

1237150

0.4

0.537

PH

1

742368

0.24

0.632

time(h)

1

10673334

3.45

0.085

2-Way Interactions

3

7205512

2401837

0.78

0.527

(b) We can develop a linear model, based on Time only. Blocks are important so coefficients for

blocks are included. For actual factor levels we use estimates for uncoded units:

Term

Coef

Constant

16838.5

Block1

1812.79

Block2

307.292

ethanol(%)

-88.231

PH

time(h)

–

1369.79

ethanol(%)*PH

4.8427

ethanol(%)*time(h)

12.1396

PH*time(h)

66.771

ethanol(%)*PH*time(h)

0.81146

Reserve Supplemental Exercises Chapter 14 Problem 9

Factorial design was used to study the effects of basic factors (time, ethanol, and pH) on the

extractability of polyphenolic phytochemicals from grape seeds. Total polyphenol from three

types of grape seeds (Savatiano, Moschofilero, and Agiorgitiko) were recorded. Assume that the

experiment was conducted in blocks based on seed type.

Run

Ethanol(%)

pH

Time(h)

TP

Moschofilero

TP

Savatiano

TP

Agiorgitiko

1

40

2

1

13,320

13,127

8,673

2

40

2

5

13,596

8,925

4,370

3

40

6

1

10,714

12,047

8,049

4

40

6

5

10,730

11,299

5,315

5

60

2

1

12,149

9,700

9,384

6

60

2

5

10,910

7,107

8,290

7

60

6

1

11,620

8,755

7,905

8

60

6

5

9,757

9,792

9,347

ethanol(%)*PH

1

508377

0.16

0.691

ethanol(%)*time(h)

1

2272426

0.73

0.406

PH*time(h)

1

4424709

1.43

0.252

3-Way Interactions

1

25285

0.01

0.929

ethanol(%)*PH*time(h)

1

25285

0.01

0.929

Residual Error

43343958

3095997

Total

126230769

Graphically study the assumption of no interactions between blocks and treatments. Suppose that

time is the only significant factor.

Select the correct answer and explanation.

SOLUTION

The means of extract versus Time and block are plotted below.

Reserve Supplemental Exercises Chapter 14 Problem 10

An article in Journal of Applied Microbiology (2012, Vol. 113(1), pp. 36-43) described a

fractional factorial design with three center points to study six factors (yeast extract, peptone,

inoculum concentration, agitation,

temperature, and pH) for protease activity. Response units were proteolytic activity per ml

(U/ml). The data follow.

Run

Yeast

(g/L)

Peptone

(g/L)

Inoculum

(vol.%)

Agitation

(rpm)

Temperature

(°C)

pH

Protease

Activity

(U/ML)

1

5

2

1

0

34

6

29.81

2

10

2

1

0

34

8

15.52

3

5

4

1

0

34

8

18.23

4

10

4

1

0

34

6

27.25

5

2

3

0

34

8

25.81

6

10

2

3

0

34

6

37.21

7

5

4

3

0

34

6

22.75

8

10

4

3

0

34

8

8.01

9

5

2

1

100

34

8

26.01

10

2

1

100

34

6

37.21

11

5

4

1

100

34

6

26.73

12

10

4

1

100

34

8

7.4

13

5

2

3

100

34

6

36.97

14

10

2

3

100

34

8

16.24

15

5

4

3

100

34

8

18.15

16

10

4

3

100

34

6

25.3

17

5

2

1

0

40

8

32.6

18

10

2

1

0

40

6

47.33

19

5

4

1

0

40

6

37.33

20

10

4

1

0

40

8

12.32

21

5

2

3

0

40

6

37.7

22

10

2

3

0

40

8

19.01

23

5

4

3

0

40

8

36.24

24

10

4

3

0

40

6

35.88

25

5

2

1

100

40

6

39.94

26

10

2

1

100

40

8

20.34

27

5

4

1

100

40

8

38.19

28

10

4

1

100

40

6

32.72

29

5

2

3

100

40

8

52.85

30

10

2

3

100

40

6

54.61

31

5

4

3

100

40

6

41.09

32

10

4

3

100

40

8

21.34

33

7.5

3

2

50

37

7

25.1

34

7.5

3

2

50

37

7

25.5

35

7.5

3

2

50

37

7

25.3

(a) What is the generator or generators for this design?

(b) What is the resolution of this design?

(c) Choose the factors and interactions, which cause the most effect in Protease activity (up to

five terms).

(d) Construct reduced model for the effects chosen in Part C. What is the estimated coefficients

in terms of actual factor levels?

SOLUTION

(a) The generator is F=ABCDE.

pH

1

1269.7

31742.55

0

2-Way Interactions

15

1049.74

69.98

1749.57

0.001

Yeast*Peptone

1

36.7

917.53

0.001

Yeast*Inoculum

1

0.85

21.21

0.044

Yeast*Agitation

1

22.5

562.38

0.002

Yeast*Temperature

1

55.31

1382.72

0.001

Yeast*pH

1

732.39

18309.8

0

Peptone*Agitation

1

21.57

539.15

0.002

Peptone*Temperature

1

14.7

367.54

0.003

Peptone*pH

1

16.86

421.59

0.002

Inoculum*Agitation

1

40.03

1000.72

0.001

Inoculum*Temperature

1

39.76

994.02

0.001

Inoculum*pH

1

5.99

149.86

0.007

Agitation*Temperature

1

34.55

863.72

0.001

Agitation*pH

1

5.67

141.75

0.007

Temperature*pH

1

6.26

156.42

0.006

3-Way Interactions

10

196.84

19.68

492.11

0.002

Yeast*Peptone*Inoculum

1

30.79

769.79

0.001

Yeast*Peptone*Agitation

1

6.31

157.75

0.006

Yeast*Peptone*Temperature

1

16.98

424.5

0.002

Yeast*Peptone*pH

1

58.08

1451.93

0.001

Yeast*Inoculum*Agitation

1

2.24

56.05

0.017

Yeast*Inoculum*Temperature

1

0.247

Yeast*Inoculum*pH

1

5.11

127.8

0.008

Yeast*Agitation*Temperature

1

0.01

0.634

Yeast*Agitation*pH

1

0.111

Yeast*Temperature*pH

1

76.91

1922.78

0.001

Curvature

1

44.22

1105.56

0.001

Residual Error

2

0.08

0.04

Pure Error

2

0.08

0.04

Source

SeqSS

AdjSS

AdjMS

F

P

Main Effects

6

534.83

13370.84

0

Yeast

1

329.67

8241.68

0

Peptone

1

451.73

11293.17

0

Inoculum

1

50.58

1264.42

0.001

Agitation

1

84.79

2119.82

0

Temperature

1

1022.54

25563.43

0

(d)

Factorial fit with chosen terms:

Term

Effect

Coef

SE Coef

T

P

Constant

29.315

0.8515

34.43

0

Yeast(g/L)

-6.419

-3.21

0.8515

-3.77

0.001

Peptone(g/L)

-7.514

-3.757

0.8515

-4.41

0

Temperature

11.306

0.8515

0

pH

-6.299

0.8515

0

Yeast(g/L)*pH

-9.568

-4.784

0.8515

-5.62

0

Analysis of variance:

Source

DF

Seq SS

Adj SS

Adj MS

F

P

Main Effects

4

768.41

33.12

0

Yeast(g/L)

1

329.67

14.21

0.001

Peptone(g/L)

1

451.73

19.47

0

Temperature

1

44.07

0

pH

1

1269.7

54.73

0

2-Way

Interactions

1

732.39

31.57

0

Yeast(g/L)*pH

1

732.39

31.57

0

Curvature

1

0.178

Residual Error

649.64

Lack of Fit

324.07

0.135

Pure Error

325.57

Total

For actual factor levels we use estimated coefficients in uncoded units:

Term

Coef

Constant

–

75.87

Yeast(g/L)

12.11

Peptone(g/L)

Temperature

pH

Yeast(g/L)*pH

Reserve Supplemental Exercises Chapter 14 Problem 11

Consider an unreplicated 2k factorial, and suppose that one of the treatment combinations is

missing. One logical approach to this problem is to estimate the missing value with a number

that makes the highest order interaction estimate zero. Apply this technique to the design

following, assuming that ab is missing.

A(Gap)

B(Pressure)

C(Flow)

D(Power)

Etch

Rate

(Å/min)

-1

550

1

-1

669

-1

1

-1

604

1

-1

missed

ab

-1

1

-1

633

1

-1

1

-1

642

-1

1

-1

601

1

-1

635

-1

1

1037

1

-1

1

749

-1

1

-1

1

1052

1

-1

1

868

-1

1

1075

1

-1

1

860

-1

1

1063

1

729

Estimate the missing value.

SOLUTION

Highest order interaction is ABCD. The table shows the contrast constants for 24 design.

A

B

AB

C

AC

BC

ABC

D

AD

BD

ABD

CD

ACD

BCD

ABCD

b

–

+

–

+

–

+

–

+

–

+

–

+

–

ab

+

–

+

–

+

–

+

–

+

–

+

–

ac

+

–

+

–

+

–

+

bc

–

+

–

+

–

+

–

+

–

+

–

+

–

+

abc

+

–

d

–

+

–

+

–

+

–

+

–

+

–

ad

+

–

+

–

+

bd

–

+

–

+

–

+

–

+

–

+

–

+

abd

+

–

+

–

+

–

+

–

+

–

+

acd

+

–

+

–

+

–

+

–

bcd

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

abcd

+

According to Contrast Table:

ABCD = 0:

–

+

–

+

–

+

–

+

–

+

a

+

–

+

–

+

–

Reserve Supplemental Exercises Chapter 14 Problem 12

What blocking scheme would you recommend if it were necessary to run a

4

2

design in four

blocks of four runs each? Select effects to be confounded with blocks.

SOLUTION

2

42=

, so we select 2 effects to be confounded with blocks, their general interaction will be also

Reserve Supplemental Exercises Chapter 14 Problem 13

Consider a

2

design in two blocks with AB confounded with blocks.

Which of the following proofs that

AB blocks

SS SS=

?

SOLUTION

A

B

AB

block

(1)

–

+

1

a

+

–

2

b

–

+

–

2

+

1

Reserve Supplemental Exercises Chapter 14 Problem 14

Consider a

3

2

design. Suppose that the largest number of runs that can be made in one block is

four, but you can afford to perform a total of 32 observations.

(a) Suggest a blocking scheme wich provides some information on all interactions.

(b) Determine the degrees of freedom for each source.

SOLUTION

(a)

A different interaction effect can be confounded in each replicate as follows.

Replicate1

Replicate2

Replicate3

Replicate4

ABC confounded

AB confounded

BC confounded

AC confounded

block1

block2

block3

block4

block5

block6

block7

block8

(1)

a

(1)

a

(1)

b

(1)

a

c

b

a

c

b

c

(b)

Source of

Variation

Degrees of

freedom

Replicates

3

Blocks with

replicates

AB (from

replicates 1, 3,

and 4)

1

replicates 1, 2,

and 3)

1

BC (from

replicates 1, 2,

and 4)

1

ABC (from

replicates 2, 3,

and 4)

1

Error (by

subtraction)

17

Total

31

AC (from

Reserve Supplemental Exercises Chapter 14 Problem 15

Construct a

51

2−

design. Suppose that it is necessary to run this design in two blocks of eight

runs each.

Choose the blocking scheme, that confounds a two-factor interaction (and its aliased three-factor

interaction) with blocks.

SOLUTION

A

B

C

D

E =

ABCD

AB =

CDE

block

–

+

1

4

A

1

B

1

C

1

+

–

+

1

–

+

–

+

1

+

–

+

–

2

–

+

–

+

–

2

+

–

+

–

+

1

–

+

1

+

–

+

–

2

–

+

–

2

+

1

Reserve Supplemental Exercises Chapter 14 Problem 16

Construct

72

2IV

−

design in four blocks of eight runs each.

(a) Which minimum number of two-factor interactions can be confounded in blocks?

(b) Choose the effects to confound with blocks to minimize the number of confounded two-

factor interactions.

SOLUTION

The generators are F = ABCD and G = ABDE. The complete defining relation is I = ABCDF =

Reserve Supplemental Exercises Chapter 14 Problem 17

Consider a

73

2IV

−

design with the generators E = ABC, F = BCD, and G = ACD.

Choose block generator using which this design can be confounded in two blocks of eight runs

each without losing information on any of the two-factor interactions.

SOLUTION

The generators are E = ABC, F = BCD, and G = ACD.

Reserve Supplemental Exercises Chapter 14 Problem 18

+

–

2

–

+

–

2

+

–

+

1

–

+

–

+

1

+

–

+

–

+

–

2

–

+

–

+

–

2

Set up a

74

2III

−

design using D = AB, E = AC, F = BC, and G = ABC as the design generators.

Ignore all interactions above two-factor interactions.

(a) Verify that each main effect is aliased with three two-factor interactions. Choose the correct

aliases for A.

(b) Construct a second

74

2III

−

design with generators D = –AB, E = –AC, F = –BC, and G = ABC

is run. What are the aliases of the D in this design?

(c) What factors may be estimated if the two sets of factor effect estimates in (a) and (b) are

combined?

SOLUTION

(a)

A

B

C

D = AB

E = AC

F = BC

G = ABC

–

+

–

+

–

+

The alias structure follows (including only one- and two-factor effects).

A = BD = CE = FG

B = AD = CF = EG

(b) The aliases (up to two-factor effects) are:

A = -BD = -CE = –FG

B = -AD = -CF = –EG

Reserve Supplemental Exercises Chapter 14 Problem 19

Consider the square root of the sum of squares for curvature and divide by the square root of

mean square error.

(a) What is the sum of squares for curvature?

(b) When the square root of the sum of squares for curvature is divided by the square root of

mean squared error, what is the resulting statistic?

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

(c) Can this statistic be used to conduct a t-test for curvature that is equivalent to the F test in the

ANOVA?

SOLUTION

(a) A coefficient for curvature is defined to be

FC

yy−

FC

(b)

When the square root of the sum of squares for curvature is divided by the square root of mean

squared error, the resulting statistic is

(c) If this t-statistic is significant,

FC

yy−

is large, meaning curvature is significant. This statistic

is compared to a t distribution with the degrees of freedom associated with the estimate of σ.