experimental design problem set ii
DESCRIPTION
A problem set about the Randomized Complete Block Design, Latin-Square, Graeco-Latin SquareTRANSCRIPT
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
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
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
experimental design lab exercise iii 3
• 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)
experimental design lab exercise iii 4
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
experimental design lab exercise iii 5
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0.4
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−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
experimental design lab exercise iii 6
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
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.
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
experimental design lab exercise iii 7
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)
experimental design lab exercise iii 8
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.
experimental design lab exercise iii 9
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)
experimental design lab exercise iii 10
v. Computation: 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 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)
experimental design lab exercise iii 11
SSError = SST − SSTreatments − SSRows − SSColumns
= 237.84− 168.64− 19.04− 5.44
= 44.72 (17)
ANOVA Table
Source of Variation SS DF MS F
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)
experimental design lab exercise iii 12
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)
experimental design lab exercise iii 13
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)
experimental design lab exercise iii 14
ANOVA Table
Source of Variation SS DF MS F
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
experimental design lab exercise iii 15
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)
SSTreatments = 168.64 (61)
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)
experimental design lab exercise iii 16
SSError = SST − SSTreatments − SSBlocks
= 237.84− 168.64− 5.44 = 63.76 (63)
ANOVA Table
Source of Variation SS DF MS F
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)
experimental design lab exercise iii 17
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)
SSTreatments = 168.64 (77)
experimental design lab exercise iii 18
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
Source of Variation SS DF MS F
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)
experimental design lab exercise iii 19
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
experimental design lab exercise iii 20
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-
experimental design lab exercise iii 21
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.
experimental design lab exercise iii 22
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.
experimental design lab exercise iii 23
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
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
experimental design lab exercise iii 24
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
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
experimental design lab exercise iii 25
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.
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
experimental design lab exercise iii 26
missing value. Perform the exact analysis and compare the results.
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
B=complete inorganic fertilizer
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.
experimental design lab exercise iii 27
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
experimental design lab exercise iii 28
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
experimental design lab exercise iii 29
Figure 14: The output generated usingpost hoc..-lsd Method
Refer to the Manual Computation and Graphical Illustration Section for
the interpretation of these outputs.