Experimental Design Problem Set II

Experimental Design Lab Exercise IIIASAAD, Al-Ahmadgaid B.July 6, 2012 al in the cloud:

website: www.alstat.weebly.comblog: www.alstatr.blogspot.comemail: [email protected]

4-1. A chemist wishes to test the effect of four chemical agents on the

strength of a particular type of cloth. Because there might be vari-

ability from one bolt to another, the chemist decides to use a ran-

domized block design, with the bolts of cloth considered as blocks.

She selects five bolts and applies all four chemicals in random order

to each bolt. The resulting tensile strength follow. Analyze the data

from this experiment (use α = 0.05) and draw appropriate conclu-

sions.

Bolt

Chemical 1 2 3 4 5

1 73 68 74 71 67

2 73 67 75 72 70

3 75 68 78 73 68

4 73 71 75 75 69

4-3. Plot the mean tensile strengths observed for each chemical type

in Problem 4-1 and compare them to an appropriatety scaled t dis-

tribution. What conclusions would you draw from this display?

4-9. Assuming that chemical types and bolts are fixed, estimate the

model parameters τi and βi in Problem 4-1.

4-11. Suppose that the obersvation for chemical type 2 and bolt 3 is

missing in Problem 4-1. Analyze the problem by estimating the

missing value. Perform the exact analysis and compare the results.

experimental design lab exercise iii 2

Manual Computation and Graphical Illustration








sions.

Bolt

Chemical 1 2 3 4 5

1 73 68 74 71 67

2 73 67 75 72 70

3 75 68 78 73 68

4 73 71 75 75 69

Solution:

• Hypotheses:

Treatment:

– H0 : τ1 = τ2 = τ3 = τ4 = 0

– H1 : τi 6= 0 at least one i

Block:

– H0 : β1 = β2 = · · · = β5 = 0

– H1 : β j 6= 0 at least one j

• Level of significance: α = 0.05

• Test Statistic:

FT =MSTreatments

MSEFB =

MSBlocksMSE

• Rejection Region:

FT > 3.4903 FB > 3.2592


• Computations:

Bolt Treatment Treatment

Chemical 1 2 3 4 5 Totals Means

1 73 68 74 71 67 353 70.6

2 73 67 75 72 70 357 71.4

3 75 68 78 73 68 362 72.4

4 73 71 75 75 69 363 72.6

Blocks Total 294 274 302 291 274 y.. = 1435 y.. = 71.75

SST =a

∑i=1

b

∑i=1

y2ij −

(a

∑i=1

b

∑j=1

yij

)2

ba

= (732 + 682 + · · ·+ 752 + 692)− 14352

20= 103153− 102961.25

= 191.75 (1)

SSTreatments =1b

a

∑i=1

y2i. −

(a

∑i=1

b

∑j=1

yij

)2

ba

=(3532 + 3572 + 3622 + 3632)

5− 14352

20= 102974− 102961.25

= 12.95 (2)

SSBlocks =1a

b

∑j=1

y2.j −

(a

∑i=1

b

∑j=1

yij

)2

ba

=(2942 + 2742 + 3022 + 2912 + 2742)

4− 14352

20= 103118.25− 102961.25

= 157

SSE = SST − SSTreatments − SSBlocks

= 191.75− 12.95− 157

= 21.8 (3)


ANOVA Table

Source of Variation SS DF MS F

Treatments 12.95 3 4.317 FT=2.376

Blocks 157 4 39.25 FB=21.602

Error 21.8 12 1.817

Total 191.75 19

• Decision: Since the FT is less than the tabulated value F0.05,3,12 =

3.4903. Then, the null hypothesis of treatment is not rejected

at α = 0.05. Moreover, since 21.602 of FB is greater than than

F0.05,4,12 = 3.2592. Then, the null hypothesis of block is rejected.

• Conclusion: Hence, the four chemical agents tested by the chemist

on the strength of a particular type of cloth is not significant,

which means it has no effect. Furthermore, the bolt of cloths has

a significant effect on the strength of a particular type of cloth.

There will be no multiple comparison test to happen in block

(bolt), since the chemist wishes only to test the effect of four

chemical agents on the strength of a particular type of cloth.

4-3. Plot the mean tensile strengths observed for each chemical type

in Problem 4-1 and compare them to an appropriatety scaled t dis-

tribution. What conclusions would you draw from this display?

Solution: Refer to Figure 1.

Conclusion: there is no obvious difference between the means. This

is the same conclusion given by the analysis of variance.

4-9. Assuming that chemical types and bolts are fixed, estimate the

model parameters τi and βi in Problem 4-1.

Solution: If both treatments and blocks are fixed, we may estimate

the parameters in the RCBD model by least squares. Recall that the


0.0

0.1

0.2

0.3

0.4

●●●●●●●●●●●●●●●●●●●●●

●●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●●●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●●

●●

●●●●●●●●●●●●●●●●●●●●

−4 −2 0 2 4x values with intercept of average of the Chemicals

P(x

)

Scaled t DistributionFigure 1: Tensile strength averages fromthe Chemical experiment in relation to at distribution with a scale factor√

MSE

a=

√1.82

4= 0.675

linear statistical model is

yij = µ + τi + β j + εij

i = 1, 2, . . . , a

j = 1, 2, . . . , b

Applying the rules in section 3-9.2 (Montgomery, D. C., Design and

Analysis of Experiments. Fifth Edition) for finding the normal equa-

tions for an experimental model, we obtain The long equation array of µ’s, τ’s, and

β’s can actually be simplified, using the

following solutions below,

µ̂ = y.. (4)

τ̂i = yi .− y.. i = 1, 2, . . . , a (5)

βj = y.j − y.. j = 1, 2, . . . , b (6)

Since the usual constraints are

a

∑i=1

τ̂i = 0b

∑j=1

β̂ j = 0 (7)

Which simplifies the long array of equa-

tions into

abµ̂ = y..

bµ̂ + bτ̂i = yi. i = 1, 2, . . . , a

aµ̂ + aβ̂ = yij j = 1, 2, . . . , b

µ : 20µ̂ + 5τ̂1 + 5τ̂2 + 5τ̂3 + 5τ̂4 + 4β̂1 + 4β̂2 +

4β̂3 + 4β̂4 + 4β̂5 = 1435

τ1 : 5µ̂ + 5τ̂1 + β̂1 + β̂2 + β̂3 + β̂4 + β̂5 = 353

τ2 : 5µ̂ + 5τ̂2 + β̂1 + β̂2 + β̂3 + β̂4 + β̂5 = 357

τ3 : 5µ̂ + 5τ̂3 + β̂1 + β̂2 + β̂3 + β̂4 + β̂5 = 362

τ4 : 5µ̂ + 5τ̂4 + β̂1 + β̂2 + β̂3 + β̂4 + β̂5 = 363

β1 : 4µ̂ + τ̂1 + τ̂2 + τ̂3 + τ̂4 + 4β̂1 = 294

β2 : 4µ̂ + τ̂1 + τ̂2 + τ̂3 + τ̂4 + 4β̂2 = 274

β3 : 4µ̂ + τ̂1 + τ̂2 + τ̂3 + τ̂4 + 4β̂3 = 302

β4 : 4µ̂ + τ̂1 + τ̂2 + τ̂3 + τ̂4 + 4β̂4 = 291

β5 : 4µ̂ + τ̂1 + τ̂2 + τ̂3 + τ̂4 + 4β̂5 = 274


Using the equations 4, 5, and 6, and applying the constraints in

equation 7, we obtain

µ̂ =143520

; τ̂1 =353

5− 1435

20= −23

20

τ̂2 =357

5− 1435

20= − 7

20; τ̂3 =

3625− 1435

20=

1320

τ̂4 =363

5− 1435

20=

1720

; β̂1 =294

4− 1435

20=

74

(55

)=

3520

β̂2 =274

4− 1435

20= −13

4

(55

)= −65

20

β̂3 =3024− 1435

20=

154

(55

)=

7520

β̂4 =291

4− 1435

20=

2020

β̂5 =274

4− 1435

20= −13

4

(55

)= −65

20




Solution:

Bolt

Chemical 1 2 3 4 5

1 73 68 74 71 67

2 73 67 x 72 70

3 75 68 78 73 68

4 73 71 75 75 69

let us solve first the value of x in the table,

Bolt Treatment

Chemical 1 2 3 4 5 Totals

1 73 68 74 71 67 353

2 73 67 x 72 70 282

3 75 68 78 73 68 362

4 73 71 75 75 69 363

Blocks Total 294 274 227 291 274 y.. = 1360


x =a(y′ i.) + b(y′ .j)− y′ ..

(a− 1)(b− 1)

=4(282) + 5(227)− 1360

12= 75.25 (8)

Thus, the estimated value of x is 75.25. The usual analysis of

variance may now be performed using x = y23 and reducing the

error and total degrees of freedom by 1. The new computation of

sum of squares are shown below using the estimated value,

Bolt Treatment

Chemical 1 2 3 4 5 Totals

1 73 68 74 71 67 353

2 73 67 75.25 72 70 357.25

3 75 68 78 73 68 362

4 73 71 75 75 69 363

Blocks Total 294 274 302.25 291 274 y.. = 1435.25

SS′T =a

∑i=1

b

∑i=1

y2ij −

(a

∑i=1

b

∑j=1

yij

)2

ba

= (732 + 682 + · · ·+ 752 + 692)− 1435.252

20= 103190.56− 102997.13

= 193.43 (9)

SS′Treatments = SSTreatments −[y′ .j − (a− 1)x]

t(t− 1)

= 12.95− [227− 3(75.25)]2

12= 12.95− 0.1302

= 12.82 (10)


SS′Blocks =1a

b

∑j=1

y2.j −

(a

∑i=1

b

∑j=1

yij

)2

ba

=(2942 + 2742 + 302.252 + 2912 + 2742)

4− 1435.252

20= 103156.02− 102997.13

= 158.89 (11)

SS′E = SS′T − SS′Treatments − SS′Blocks

= 193.43− 12.82− 158.89

= 21.72 (12)

ANOVA Table

Source of Variation SS Degrees of Freedom MS F

Treatments 12.82 3 4.27 F1=2.17

Blocks 158.89 4 39.723 F2=20.16

Error 21.72 11 1.97

Total 193.43 18

The results for both ANOVA’s are very close, but with the estimated

value of x and an adjustment of sum of squares of treatment, the

FComputed now becomes smaller, which means getting far from the

rejection region.


Latin Square Problem

1. Shown below the yield (ton per 1/4-ha.plots) of sugar cane in a

Latin square experiments comparing five (5) fertilizer levels. Where:

A=no fertilizer

C=10 tons manure/ha

E=30 tons manure/ha

B=complete inorganic fertilizer

D=20 tons manure/ha

Rows Columns

1 2 3 4 5

1 14(A) 22(E) 20(B) 18(C) 25(D)

2 19(B) 21(D) 16(A) 23(E) 18(C)

3 23(D) 15(A) 20(C) 18(B) 23(E)

4 21(C) 25(B) 24(E) 21(D) 18(A)

5 23(E) 16(C) 23(D) 17(A) 19(B)

a. Analyze the data completely and interpret your results.

b. Obtain the treatment means, treatment effects, standard devia-

tion of a treatment mean and treatment mean difference, and the

CV of the experiment.

c. Obtain the efficiency of this design with respect to CRD and with

respect to RCB

i. if columns were used as blocks;

ii. if rows were used as blocks and interpret your results.

Solution:

i. Hypotheses:

H0: The five fertilizers have equal effects on the yields of sugar

cane.

H1: At least one of the fertilizers has an effect on the yields of

sugar cane.

ii. Level of Significance: α = 0.05

iii. Test Statistics:

F =MSTreatments

MSE

iv. Rejection Region: Reject the null hypothesis if,

F > Fα,p−1,(p−1)(p−2) that is F > (F0.05,4,12 = 3.2592)


v. Computation: Where:

A=no fertilizer

C=10 tons manure/ha

E=30 tons manure/ha


D=20 tons manure/ha

Rows Columns yi..

1 2 3 4 5

1 14(A) 22(E) 20(B) 18(C) 25(D) 99

2 19(B) 21(D) 16(A) 23(E) 18(C) 97

3 23(D) 15(A) 20(C) 18(B) 23(E) 99

4 21(C) 25(B) 24(E) 21(D) 18(A) 109

5 23(E) 16(C) 23(D) 17(A) 19(B) 98

y..k 100 99 103 97 103 502

y.j. A=80 B=101 C=93 D=113 E=115

SST = ∑i

∑j

∑k

y2ijk −

y2...

N

= (142 + 222 + · · ·+ 172 + 192)− (502)2

25= 10318− 10080.16

= 237.84 (13)

SSRows =1p

p

∑i=1

y2i.. −

y2...

N

=15(992 + 972 + 992 + 1092 + 982)− (502)2

25= 10099.2− 10080.16

= 19.04 (14)

SSColumns =1p

p

∑k=1

y2..k −

y2...

N

=15(1002 + 992 + 1032 + 972 + 1032)− (502)2

25= 10085.6− 10080.16

= 5.44 (15)

SSTreatments =1p

p

∑j=1

y2.j. −

y2...

N

=15(802 + 1012 + 932 + 1132 + 1152)− (502)2

25= 10248.8− 10080.16

= 168.64 (16)


SSError = SST − SSTreatments − SSRows − SSColumns

= 237.84− 168.64− 19.04− 5.44

= 44.72 (17)

ANOVA Table


Treatments 168.64 4 42.16 F1=11.303

Rows 19.04 4 4.76

Columns 5.44 4 1.36

Error 44.72 12 3.73

Total 237.84 24

vi. Decision: Thus, the null hypothesis is rejected since 11.302 is

greater than 3.2592.

v. Conclusion: Hence, the five fertilizers are significantly different,

which means that they do have an effect on the yield of sugar

cane.

vii. Multiple Comparison Test:

Solution: Using the Least Significance Difference, the critical value

is,

LSD = t α2 ,N−p

√2MSE

n= 2.086

√2(3.73)

5= 2.548

Thus, any pair of treatment averages differ by more than 2.548

would imply that the corresponding pair of population means

are significantly different. The differences in averages are,

y.E. − y.A. = 23− 16 = 7 ∗ (18)

y.E. − y.C. = 23− 18.6 = 4.4 ∗ (19)

y.E. − y.B. = 23− 20.2 = 2.8 ∗ (20)

y.E. − y.D. = 23− 22.6 = 0.4 (21)

y.D. − y.A. = 22.6− 16 = 6.6 ∗ (22)

y.D. − y.C. = 22.6− 18.6 = 4 ∗ (23)

y.D. − y.B. = 22.6− 20.2 = 2.4 (24)


y.B. − y.A. = 20.2− 16 = 4.2 ∗ (25)

y.B. − y.C. = 20.2− 18.6 = 1.6 (26)

y.C. − y.A. = 18.6− 16 = 2.6∗ (27)

The starred values indicates pairs of mean that are significantly

different.

b. Obtain the treatment means, treatment effects, standard deviation

of a treatment mean and treatment mean difference, and the CV of

the experiment. Solution:

a. Obtain treatment means,

y.j. A=80 B=101 C=93 D=113 E=115

y.j. y.A.=16 y.B.=20.2 y.C.=18.6 y.D.=22.6 y.E.=23

b. Treatment Effects

µ̂ =50225

= 20.08 (28)

τ̂1 = y.A. − y... = 16− 20.08 = −4.08 (29)

τ̂2 = y.B. − y... = 20.2− 20.08 = 0.12 (30)

τ̂3 = y.C. − y... = 18.6− 20.08 = −1.48 (31)

τ̂4 = y.D. − y... = 22.6− 20.08 = 2.52 (32)

τ̂5 = y.E. − y... = 23− 20.08 = 2.92 (33)

c. Standard Deviation of a Treatment mean

Sy...=

√MSE

n=

√3.73

5= 0.8637

d. Mean Difference The ascending order of the means

y.E. = 23 y.D. = 22.6 y.B. = 20.2 y.C. = 18.6 y.A. = 16

y.E. − y.A. = 23− 16 = 7 (34)

y.E. − y.C. = 23− 18.6 = 4.4 (35)

y.E. − y.B. = 23− 20.2 = 2.8 (36)

y.E. − y.D. = 23− 22.6 = 0.4 (37)


y.D. − y.A. = 22.6− 16 = 6.6 (38)

y.D. − y.C. = 22.6− 18.6 = 4 (39)

y.D. − y.B. = 22.6− 20.2 = 2.4 (40)

y.B. − y.A. = 20.2− 16 = 4.2 (41)

y.B. − y.C. = 20.2− 18.6 = 1.6 (42)

y.C. − y.A. = 18.6− 16 = 2.6 (43)

e. Coefficient of Variation:

CV =

√MSEy...

=

√3.73

20.08= 0.0962× 100 = 9.62%

c. Obtain the efficiency of this design with respect to CRD and with

respect to RCBD

i. if columns were used as blocks;

ii. if rows were used as blocks and interpret your results.

Solution:

Completely Randomized Design Data Layout:

Fertilizers Yield Total Means

A 14 15 16 17 18 80 16

B 19 25 20 18 19 101 20.2

C 21 16 20 18 18 93 18.6

D 23 21 23 21 25 113 22.6

E 23 22 24 23 23 115 23

502 20.08

CF =5022

25= 10080.16 (44)

SST = 237.84 (45)

SSTreatments = 168.64 (46)

SSError = SST − SSTreatments

= 237.84− 168.64 = 69.2 (47)


ANOVA Table


Treatments 168.64 4 42.16 F1=12.185

Error 69.2 20 3.46

Total 237.84 24

i. Decision: Thus, the null hypothesis is rejected since 12.185 is

greater than 2.87, for F0.05,4,20.

ii. Conclusion: Hence, the five fertilizers are significantly different,

implying that they do have an effect on the yield of sugar cane.

iii. Multiple Comparison

Solution: Using the Tukey Honestly Significant Difference, the

critical value is,

Tα = q(a, f )

√MSE

n= q(5, 20)

√3.46

5= 4.23(0.8319) = 3.519

Thus, any pair of treatment averages differ by more than 3.519

would imply that the corresponding pair of population means

are significantly different. The differences in averages are,

y.E. − y.A. = 23− 16 = 7 ∗ (48)

y.E. − y.C. = 23− 18.6 = 4.4 ∗ (49)

y.E. − y.B. = 23− 20.2 = 2.8 (50)

y.E. − y.D. = 23− 22.6 = 0.4 (51)

y.D. − y.A. = 22.6− 16 = 6.6 ∗ (52)

y.D. − y.C. = 22.6− 18.6 = 4 ∗ (53)

y.D. − y.B. = 22.6− 20.2 = 2.4 (54)

y.B. − y.A. = 20.2− 16 = 4.2 ∗ (55)

y.B. − y.C. = 20.2− 18.6 = 1.6 (56)

y.C. − y.A. = 18.6− 16 = 2.6 (57)

The starred values indicates pairs of mean that are significantly

different.

iv. Relative Efficiency


RE(

LSCRD

)=

MSR + MSC + (a− 1)MSE(a + 1)MSC

( f )

f =( f1 + 1)( f2 + 3)( f2 + 1)( f1 + 3)

=(12 + 1)(20 + 3)(20 + 1)(12 + 3)

= 0.949

RE =4.76 + 1.36 + 4(3.73)

6(1.36)(0.949)

= 2.447 (58)

Randomized Complete Block Design Data Layout if columns were

used as blocks:

Fertilizers Blocks(Columns) Treatment Totals Means

1 2 3 4 5

A 14 15 16 17 18 80 16

B 19 25 20 18 19 101 20.2

C 21 16 20 18 18 93 18.6

D 23 21 23 21 25 113 22.6

E 23 22 24 23 23 115 23

Block Totals 100 99 103 97 103 502 20.08

CF =5022

25= 10080.16 (59)

SST = 237.84 (60)


SSBlocks(Columns) =1a

b

∑j=1

y2.j −

(a

∑i=1

b

∑j=1

yij

)2

ba

=(1002 + 992 + 1032 + 972 + 1032)

5− 5022

25= 10085.6− 10080.16

= 5.44 (62)


SSError = SST − SSTreatments − SSBlocks

= 237.84− 168.64− 5.44 = 63.76 (63)

ANOVA Table


Treatments 168.64 4 42.16 F1=10.58

Blocks (Columns) 5.44 4 1.36

Error 63.76 16 3.985

Total 237.84 24

v. Decision: Thus, the null hypothesis is rejected since 10.58 is greater

than 3.0069.

vi. Conclusion: Hence, the five fertilizers are significantly different,

which means that they do have an effect on the yield of sugar cane

when columns were used as blocks.

vii. Multiple Comparison

Using the Least Significance Difference, the critical value is,

LSD = t α2 ,15

√2MSE

n= 2.131

√2(3.985)

5= 2.69

Thus, any pair of treatment averages differ by more than 2.69 would

imply that the corresponding pair of population means are signifi-

cantly different. The differences in averages are,

y.E. − y.A. = 23− 16 = 7 ∗ (64)

y.E. − y.C. = 23− 18.6 = 4.4 ∗ (65)

y.E. − y.B. = 23− 20.2 = 2.8 ∗ (66)

y.E. − y.D. = 23− 22.6 = 0.4 (67)

y.D. − y.A. = 22.6− 16 = 6.6 ∗ (68)

y.D. − y.C. = 22.6− 18.6 = 4 ∗ (69)

y.D. − y.B. = 22.6− 20.2 = 2.4 (70)

y.B. − y.A. = 20.2− 16 = 4.2 ∗ (71)

y.B. − y.C. = 20.2− 18.6 = 1.6 (72)


y.C. − y.A. = 18.6− 16 = 2.6 (73)

The starred values indicates pairs of mean that are significantly dif-

ferent.

viii. Relative Efficiency of RCBD with columns as blocks

RE(

LSRCBD

)=

MSR + (a− 1)MSE(a)MSE

( f )

f =( f1 + 1)( f2 + 3)( f2 + 1)( f1 + 3)

=(12 + 1)(20 + 3)(16 + 1)(16 + 3)

= 0.926

RE =4.76 + 4(3.73)

5(3.73)(0.926)

= 0.977 (74)

Randomized Complete Block Design Data Layout if rows were used

as blocks:

Fertilizers Blocks (Rows) Treatment Totals Means

1 2 3 4 5

A 14 15 16 17 18 80 16

B 19 25 20 18 19 101 20.2

C 21 16 20 18 18 93 18.6

D 23 21 23 21 25 113 22.6

E 23 22 24 23 23 115 23

Block Totals 100 99 103 97 103 502 20.08

CF =5022

25= 10080.16 (75)

SST = 237.84 (76)



SSBlocks(Rows) =1a

b

∑j=1

y2.j −

(a

∑i=1

b

∑j=1

yij

)2

ba

=(1002 + 992 + 1032 + 972 + 1032)

5− 5022

25= 10085.6− 10080.16

= 5.44 (78)

SSError = SST − SSTreatments − SSBlocks

= 237.84− 168.64− 5.44 = 63.76 (79)

ANOVA Table


Treatments 168.64 4 42.16 F1=10.58

Blocks (Columns) 5.44 4 1.36

Error 63.76 16 3.985

Total 237.84 24

ix. Decision: Thus, the null hypothesis is rejected since 10.58 is greater

than 3.0069.

x. Conclusion: Hence, the five fertilizers are significantly different,

which means that they do have an effect on the yield of sugar cane.

xi. Multiple Comparison: Using Least Significance Difference, the pro-

cess is just the same with (ii.).

xii. Relative Efficiency of RCBD with rows as blocks

RE(

LSRCBD

)=

MSC + (a− 1)MSE(a)MSE

( f )

f =( f1 + 1)( f2 + 3)( f2 + 1)( f1 + 3)

=(12 + 1)(20 + 3)(16 + 1)(16 + 3)

= 0.926

RE =1.36 + 4(3.73)

5(3.73)(0.926)

= 0.808 (80)


Graeco-Latin Square Problems

4-22. The yield of a chemical process was measured using five batches

of raw materials, five acid concentrations, five standing times (A, B,

C, D, E). and five catalyst concentrations (α, β, γ, δ, ε). The Graeco-

Latin square that follows was used. Analyze the data from this

experiment (use α = 0.05) and draw conclusions.

Acid Concentration

Batch 1 2 3 4 5

1 Aα=26 Bβ=16 Cγ=19 Dδ=16 Eε=13

2 Bγ=18 Cδ=21 Dε=18 Eα=11 Aβ=21

3 Cε=20 Dα=12 Eβ=16 Aγ=25 Bδ=13

4 Dβ=15 Eγ=15 Aδ=22 Bε=14 Cα=17

5 Eδ=10 Aε=24 Bα=17 Cβ=17 Dγ=14

Solution:

Acid Concentration

Batch 1 2 3 4 5 Totals

1 Aα=26 Bβ=16 Cγ=19 Dδ=16 Eε=13 90

2 Bγ=18 Cδ=21 Dε=18 Eα=11 Aβ=21 89

3 Cε=20 Dα=12 Eβ=16 Aγ=25 Bδ=13 86

4 Dβ=15 Eγ=15 Aδ=22 Bε=14 Cα=17 83

5 Eδ=10 Aε=24 Bα=17 Cβ=17 Dγ=14 82

Totals 89 88 92 83 78 430

Treatment (Times) Totals: A = 118, B = 78, C = 94, D = 75, E = 65.

Catalyst Totals: α = 83, β = 85, γ = 91, δ = 82, ε = 89.

Computation of Sum of Squares:

CF =G..

a2 =4302

25= 7396

Total SS or TSS = (262 + 162 + · · ·+ 172 + 142) - 7396 = 436

Acid SS or ASS =(892 + 882 + 922 + 832 + 782)

5- 7396 = 24.4

Batch SS or BSS =(902 + 892 + 862 + 832 + 822)

5- 7396 = 10

Times SS or TrSS =(1182 + 782 + 942 + 752 + 652)

5- 7396 = 342


Catalyst SS or CSS =(832 + 852 + 912 + 822 + 892)

5- 7396 = 12

Error SS or SSE = TSS - BSS - ASS - CSS - TrSS

= 436− 10− 24.4− 342− 12 = 47.6

ANOVA table for 5×5 Graeco-Latin Square (p=5)

SV DF SS MS F

Times p-1=4 342 85.5 14.37

Batch p-1=4 10 2.5 0.42

Acid p-1=4 24.4 6.1 1.025

Catalyst p-1=4 12 3 0.504

Error (p-1)(p-3)=8 47.6 5.95

Total 24 436

Decision: All FComputed of each Source Variation is less than the crit-

ical value, Fα,4,8 = 3.8379, except for the treatments which is 14.37.

And thus, the following decision is obtain,

a. The five standing times are significantly different.

b. The five batches of raw materials have no significant difference.

c. The five acid concentrations have no significant difference.

d. The five catalyst concentrations have no significant difference.

xiii. Multiple comparison for five standing times,

Using Tukey Honestly Significant Difference, the critical value is ob-

tain,

Tα = q(a, f )

√MSE

n= q(5, 8)

√5.95

5= 4.89

√5.95

5= 5.33

Treatment Means:

yA = 23.6, yC = 18.8, yB = 15.6, yD = 15, yE = 13

Thus, any pair of treatment averages differ by more than 5.33 would

imply that the corresponding pair of population means are signifi-


cantly different. The differences in averages are,

yA − yE = 23.6− 13 = 10.6 ∗

yA − yD = 23.6− 15 = 8.6 ∗

yA − yB = 23.6− 15.6 = 8 ∗

yA − yC = 23.6− 18.8 = 4.8

yC − yE = 18.8− 13 = 5.8 ∗

yC − yD = 18.8− 15 = 3.8

yC − yB = 18.8− 15.6 = 3.2

yB − yE = 15.6− 13 = 2.6

yB − yD = 15.6− 15 = 0.6

yD − yE = 15− 13 = 2

The starred values indicates pairs of mean that are significantly dif-

ferent.


Computation using SPSS Software

3-1. The tensile strength of portland cement is being studied. Four dif-

ferent mixing techniques can be used economically. The following

data have been collected:

Mixing Technique Tensile Strength (lb/in2)

1 3129 3000 2865 2890

2 3200 3300 2975 3150

3 2800 2900 2985 3050

4 2600 2700 2600 2765

Steps:

Step 1

Figure 2: The above table is entered inSPSS in this manner.

Step 2

Figure 3: The second step after inputtingyour data, go to analyze⇒compare

means⇒one-way anova.


Step 3

Figure 4: Next, enter the vari-able yield to the dependent

list (yield⇒dependent list)and treatment to factor

(treatment⇒factor). After thatyou can click the post hoc.. for choos-ing the test for multiple comparison.

Table 1: The output of the performedsteps. In multiple comparison table, thetest performed was Scheffé. You cancheck it in the post hoc.. section of theStep 3









sions.

Bolt

Chemical 1 2 3 4 5

1 73 68 74 71 67

2 73 67 75 72 70

3 75 68 78 73 68

4 73 71 75 75 69

Solution

Step 1

Figure 5: The above table is entered inSPSS in this manner.

Step 2

Figure 6: The second step after inputtingyour data, go to analyze⇒general

linear model⇒univariate


Step 3

Figure 7: Next, enter the vari-able yield to the dependent list

(yield⇒dependent list) and treat-ment and block to fixed factor(s)(treatment and block⇒fixed fac-tor(s)). After that you can click thepost hoc.. for choosing the test formultiple comparison.

Step 4

Figure 8: Before clicking the ok but-ton, go first to the model (seen on Step3). In the univariate: model win-dow, click (custom) then put the treat-ment and block to the model box, asshown in the figure. Then, change thetype to main effects and uncheckedthe include intercept in model, be-fore the clicking the continue button.

output of the performed test. Refer to Manual Computation and

Graphical Illustration Section item 4-1 for the interpretation.





Solution:

Bolt

Chemical 1 2 3 4 5

1 73 68 74 71 67

2 73 67 x 72 70

3 75 68 78 73 68

4 73 71 75 75 69

Solution: For missing value, just replace x to 75.25 as computed in

the Manual Computation and Graphical Illustration Section. After

that, perform the above steps in 4-1 of this section.

Table 2: This is the output of the testperformed.

Refer to Manual Computation and Graphical Illustration Section

item 4-11 for the interpretation.

1. Shown below the yield (ton per 1/4-ha.plots) of sugar cane in a

Latin square experiments comparing five (5) fertilizer levels. Where:

A=no fertilizer

C=10 tons manure/ha

E=30 tons manure/ha


D=20 tons manure/ha

Row Columns

1 2 3 4 5

1 14(A) 22(E) 20(B) 18(C) 25(D)

2 19(B) 21(D) 16(A) 23(E) 18(C)

3 23(D) 15(A) 20(C) 18(B) 23(E)

4 21(C) 25(B) 24(E) 21(D) 18(A)

5 23(E) 16(C) 23(D) 17(A) 19(B)

a. Analyze the data completely and interpret your results.


Solution:

Step 1

Figure 9: Enter the data to SPSS in thismanner.

Step 2

Figure 10: The second step af-ter inputting your data, go toanalyze⇒general linear

model⇒univariate


Step 3

Figure 11: Next, enter the vari-able yield to the dependent list

(yield⇒dependent list) and treat-ment, row and column to fixed fac-tor(s) (row and column⇒fixed fac-tor(s)). After that you can click thepost hoc.. for choosing the test formultiple comparison.

Step 4

Figure 12: Before clicking the ok button,go first to the model (seen on Step 3). Inthe univariate: model window, clickcustom then put the treatment, col-umn, and row to the model box, asshown in the figure. Then, change thetype to main effects and uncheckedthe include intercept in model, be-fore the clicking the continue button.

Figure 13: The output of the performedsteps


Figure 14: The output generated usingpost hoc..-lsd Method

Refer to the Manual Computation and Graphical Illustration Section for

the interpretation of these outputs.

Documents

Experimental Design Problem Set II