Experimental Design Problem Set II

Experimental Experimental Design Lab Exercise III ASAAD, Al-Ahmadgaid B. July 6, 2012

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

strength strength of a particular particular type of cloth. cloth. Because Because there might be varivariability from one bolt to another, the chemist decides to use a randomized 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 experiment (use α = 0.05) and draw appropriate conclusions.

Bolt Chemical

1

2

3

4

5

1

73

6 68 8

7 74 4

7 71 1

6 67 7

2

73

6 67 7

7 75 5

7 72 2

7 70 0

3

75

6 68 8

7 78 8

7 73 3

6 68 8

4

73

7 71 1

7 75 5

7 75 5

6 69 9

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

in Problem 4-1 and compare them to an appropriatety scaled t distribution. What conclusions would you draw from this display? 4-9. Assuming Assuming that chemical chemical types and bolts are fixed, estimate the

model parameters τ i and β i in Problem 4-1. 4-11. Suppose Suppose that the obersvation obersvation for chemical chemical type 2 and bolt 3 is

missing in Problem 4-1. Analyz Analyzee the problem problem by estimat estimating ing the missing value. Perform the exact analysis and compare the results.

experimen expe rimenta tal l design design lab lab exerci exercise se iii

Manual Computation and Graphical Illustration 4-1. A chemist wishes wishes to test the effect effect of four chemical chemical agents on the

strength strength of a particular particular type of cloth. cloth. Because Because there might be varivariability from one bolt to another, the chemist decides to use a randomized 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 experiment (use α = 0.05) and draw appropriate conclusions.

Bolt Chemical

1

2

3

4

5

1

73

6 68 8

7 74 4

7 71 1

6 67 7

2

73

6 67 7

7 75 5

7 72 2

7 70 0

3

75

6 68 8

7 78 8

7 73 3

6 68 8

4

73

7 71 1

7 75 5

7 75 5

6 69 9

Solution:

• Hypotheses: Hypotheses: Treatment: – H 0 : τ 1 = τ 2 = τ 3 = τ 4 = 0 – H 1 : τ i = 0 at least one i



Block: – H 0 : β1 = β2 =

· · · = β5 = 0

– H 1 : β j = 0 at least one j



• Level Level of significanc significance: e: α = 0.05 • Test Statisti Statistic: c: FT =

MS Treatments MS E

FB =

MS Blocks MS E

• Rejection Rejection Region: Region: FT > 3.4903

FB

>

3.2592

2


• Computatio Computations: ns:

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

a

SS T =

b

∑ ∑ yij2 −



a

b

∑ ∑ yij i=1 j=1

2

= (73 + 68 + = 103153

=

2

· · · + 75

2

+ 69 )

−

14352 20

− 102961.25

= 191.75

SS Treatments

2

ba

i =1 i=1 2



1 a 2 yi. b i∑ =1

−



a

b

∑ ∑ yij i=1 j=1



(1) 2

ba

(3532 + 3572 + 3622 + 3632 ) 5 = 102974 102961.25 =

2

− 1435 20

−

= 12.95

SS Blocks

=

1 b 2 y. j j a j∑ =1

−



a

b

∑ ∑ yij i=1 j=1



(2) 2

ba

(2942 + 2742 + 3022 + 2912 + 2742 ) 4 = 103118.25 102961.25 =

2

− 1435 20

−

= 157 SS E

= SS T =

− SSTreatments − SSBlocks 191.75 − 12.95 − 157

= 21.8

(3)

3


ANOVA Table Source of Variation

SS

DF

MS

F

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

Treatments

• Decision: Decision: Since Since the FT is less than the tabulated value F0.05,3,12 = 3.4903 3.4903.. Then, Then, the null hypothe hypothesis sis of treatm treatment ent is not rejected rejected at α = 0.05. Moreove Moreoverr, since 21.602 of FB is greater than than F0.05,4,12 = 3.2592. 3.2592. Then, the null hypothesis of block is rejected. rejected. • Conclusio Conclusion: n: Hence, the four chemical chemical agents tested by the chemist chemist on the strength of a particular type of cloth is not significant, which which means it has no effect. Furthermore, Furthermore, the bolt of cloths 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) (bolt),, since since the chemist chemist wishes wishes only to test test the effect effect of four four chemical agents on the strength of a particular type of cloth. 4-3. Plot the mean tensile tensile strengths observed observed for each chemical type

in Problem 4-1 and compare them to an appropriatety scaled t distribution. 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 Assuming that chemical chemical types and bolts are fixed, estimate the

model parameters τ i and β i in Problem 4-1. Solution: If both treatments and blocks are fixed, we may estimate the parameters parameters in the RCBD model by least squares. Recall Recall that the

4


Figure 1: Tensile strength averages from the Chemical experiment in relation to a t distribution with a scale factor

Scaled t Distribution 0.4

qq q q q q q

MS E = a

q

q

q

q

0.3

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

0.1

q

q

q

q

q

q

q

q

0.0

q qqqq qqqqqqqqqqq

−4

q q q qq qqq

q

q q q q

−2

1.82 = 0.675 4

q

q

) x (0.2 P

 

q

q

5

q q q q

0

q q qq qq qqq

qqqqqqqqqqqq qqq

2

4

x values with intercept of average of the Chemicals

linear statistical model is yij = µ + τ i + β j + ij

 

i = 1,2,...,a j = 1 , 2 , . . . , b

Applying the rules in section 3-9.2 (Montgomery, D. C., Design and Analysis of Experiments. Experiments. Fifth Fifth Edition) for finding finding the normal equaequations for an experimental model, we obtain

The long equation array of µ’s, τ ’s, ’s, and β’s can actually be simplified, using the

µ

                                                                   

τ1 + 5τ τ2 + 5τ τ3 + 5τ τ4 + 4 β β 1 + 4 β β 2 + : 20µ + 5τ

4 β β 3 + 4 β β 4 + 4 β β 5 = 1435

τ 1

τ1 + β β 1 + β β 2 + β β 3 + β β 4 + β β 5 = 353 : 5µ + 5τ

τ 2

: 5µ + 5τ τ2 + β β 1 + β β 2 + β β 3 + β β 4 + β β 5 = 357

τ 3

τ3 + β β 1 + β β 2 + β β 3 + β β 4 + β β 5 = 362 : 5µ + 5τ

τ 4

τ4 + β β 1 + β β 2 + β β 3 + β β 4 + β β 5 = 363 : 5µ + 5τ

β 1

τ1 + τ τ2 + τ τ3 + τ τ4 + 4 β β 1 = 294 : 4µ + τ

β 2

τ1 + τ τ2 + τ τ3 + τ τ4 + 4 β β 2 = 274 : 4µ + τ

β 3

τ1 + τ τ2 + τ τ3 + τ τ4 + 4 β β 3 = 302 : 4µ + τ

β 4

τ1 + τ τ2 + τ τ3 + τ τ4 + 4 β β 4 = 291 : 4µ + τ

β 5

τ1 + τ τ2 + τ τ3 + τ τ4 + 4 β β 5 = 274 : 4µ + τ

following solutions below, µ

=

y..

(4)

τ τi

=

yi .

i = 1 , 2 , . . . , a (5)

β j

=

 

− y.. y. j j − y..

j = 1 , 2 , . . . , b (6)

Since the usual constraints are a

b



τi = 0 ∑ β β j = 0 ∑ τ



i= 1

(7)

j =1

Which simplifies the long array of equations into ab µ

=

y..

τi bµ + bτ

=

yi .

i = 1,2,..., a

β aµ + a β

=

yij

j = 1,2,..., b

    


Using the equations 4, 5, and 6, and applying the constraints in equation 7, we obtain µ

=

τ τ2

=

τ τ4

=

β β 2

=

β β 3

=

β β 4

=

β β 5

=

      

1435 ; 20 357 5 363 5 274 4 302 4 291 4 274 4

23 − 1435 =− 20 20 1435 7 362 1435 − 20 = − 20 ; τ τ3 = 5 − 20 = 13 20 17 294 1435 7 5 β 1 = = ; β = − 1435 − 20 20 4 20 4 5 τ τ1 =



353 5

− 1435 = 20 − − −

   −    −



13 5 65 = 4 5 20 1435 15 5 75 = = 20 4 5 20 1435 20 = 20 20 1435 13 5 65 = = 20 4 5 20

=

35 20

−

−

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

missing in Problem 4-1. Analyz Analyzee the problem problem by estimat estimating ing the missing value. Perform the exact analysis and compare the results. Solution: Bolt Chemical

1

2

3

4

5

1

73

6 68 8

7 74 4

7 71 1

6 67 7

2

73

67

x

72

70

3

75

6 68 8

7 78 8

7 73 3

6 68 8

4

73

7 71 1

7 75 5

7 75 5

6 69 9

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

Bolt

Treatment

Chemical

1

2

3

4

5

Totals

1

73

6 68 8

7 74 4

7 71 1

6 67 7

353

2

73

67

x

72

70

282

3

75

6 68 8

7 78 8

7 73 3

6 68 8

362

4

73

7 71 1

7 75 5

7 75 5

6 69 9

363

Blocks Total

294

274

227

291

274

y.. = 1360

6


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 usual anal analysi ysiss of variance may now be performed using x = y23 and reducing the error and total degrees of freedom by 1. The new computat computation ion 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

b

∑ ∑ yij2 −



a

∑ ∑ yij i =1 j=1

= 103190.56



2

ba

i=1 i =1

= (732 + 682 +

b

2

· · · + 752 + 692 ) − 1435.25 20

− 102997.13

= 193.43 SS 

Treatments

[ y . j j

− ( a − 1) x ] − t(t − 1) [227 − 3(75.25)] 2 − 12.95

= SS Treatments =

(9)

= 12.95 = 12.82

12

− 0.1302 (10)

7


 SS Blocks

1 b 2 y. j a j∑ =1

=

−



a

b

∑ ∑ yij i=1 j=1



2

ba

(2942 + 2742 + 302.252 + 2912 + 2742 ) 4 = 103156.02 102997.13 =

2

− 1435.25 20

−

= 158.89 SS E

(11)

 = SS T =

 − SSTreatments − SSBlocks 193.43 − 12.82 − 158.89

= 21.72

(12)


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.

8


Latin Square Problem 1. Shown Shown below below the yield yield (ton (ton per 1/4-ha.p 1/4-ha.plot lots) s) of sugar sugar cane in a

Latin square experiments comparing five (5) fertilizer levels.

Where: A=no fertilizer

Rows

C=10 tons manure/ha

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 Analyze the data completely and interpret interpret your results. results. b. Obtain the treatment means, treatment effects, standard deviation of a treatment mean and treatment mean difference, and the CV of the experiment. c. Obtain the efficiency efficiency of this design with respect to CRD and with respect to RCB i. if columns columns were were used as blocks; blocks; ii. if rows were were used as blocks and interpret interpret your results. results. Solution: i. Hypotheses: Hypotheses: H 0 : The five fertilizer fertilizerss have have equal effects effects on the yields of sugar cane. H 1 : At least least one of the fertiliz fertilizers ers has an effect effect on the yields yields of sugar cane. ii. Level Level of Significanc Significance: e: α = 0.05 iii. Test Statistics: Statistics: F=

MS Treatments MS E

iv. iv. Rejection Rejection Region: Region: Reject the null hypothesi hypothesiss if, F > Fα, p−1,( p−1)( p−2)

tha that is F

>

( F0.05,4,12 = 3.2592)

E=30 tons manure/ha B=complete inorganic fertilizer D=20 tons manure/ha

9

experimental experimental design lab exercise iii

v. Computation:


Rows

yi..

Columns

C=10 tons manure/ha E=30 tons manure/ha

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 ..k

100

99

103

97

103

502

y. j. j.

A=80

B=101

C=93

D=113

E=115

SS T =

2

∑ ∑ ∑ y i

j

k

ijk

= (142 + 222 + = 10318

−

B=complete inorganic fertilizer D=20 tons manure/ha

y2 ... N

) · · · + 172 + 192 ) − (502 25

2

− 10080.16

= 237.84

(13)

p

SS Rows

=

1 y2 i.. ∑ p i=1

2

− yN ...

1 (992 + 972 + 992 + 1092 + 982 ) 5 = 10099.2 10080.16

=

) − (502 25

2

−

= 19.04

(14)

p

SSColumns

=

1 y2 ..k ∑ p k=1 ..k

−

y2

...

N

1 (1002 + 992 + 1032 + 972 + 1032 ) 5 = 10085.6 10080.16

=

) − (502 25

2

−

= 5.44

(15)

p

SS Treatments

=

1 y2 . j. j. p j∑ =1

2

− yN ...

1 (802 + 1012 + 932 + 1132 + 1152 ) 5 = 10248.8 10080.16

=

) − (502 25

2

−

= 168.64

(16)

10


SS Error

= SS T =

− 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: Decision: Thus, the null hypothesi hypothesiss is rejected since since 11.302 is greater than 3.2592. v. Conclusion: Hence, the five fertilizers are significantly significantly different, which means that they do have an effect on the yield of sugar cane. vii. Multiple Comparison Test: Test: Solution: Using Using the Least Significan Significance ce Differenc Difference, e, the critical critical value value is, LSD = t α2 ,N − p





2 MS E = 2.086 n

2(3.73) = 2.548 5

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

= 7

∗

(18)

= 4.4

(19)

=

(20)

∗ 2.8 ∗

= 0.4 = 6.6 = 4

(21)

∗

(22)

∗

(23)

= 2.4

(24)

11


− y. A. A. = 20.2 − 16 y.B. − y.C. = 20.2 − 18.6 y.C. − y. A. A. = 18.6 − 16 y.B.

= 4.2

(25)

∗

= 1.6

(26)

= 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 Obtain treatment treatment means, means, y. j. j.

A=80

B=101

C=93

D=113

E=115

y. j. j.

y. A. A. =16

y.B. =20.2

y.C. =18.6

y.D. =22.6

y.E. =23

b. Treatment Effects µ

=

τ τ1

=

τ τ2

=

τ τ3

=

τ τ4

=

τ τ5

=

     

502 = 20.08 25 y. A. A. y... = 16

(28)

− − 20.08 = −4.08 y.B. − y... = 20.2 − 20.08 = 0.12 y.C. − y... = 18.6 − 20.08 = −1.48 y.D. − y... = 22.6 − 20.08 = 2.52 y.E. − y... = 23 − 20.08 = 2.92

(29) (30) (31) (32) (33)

c. Standard Standard Deviation Deviation of a Treatmen Treatmentt mean S y... =

  MS E = n

3.73 = 0.8637 5

d. Mean Differen Difference ce The ascending ascending order of the means y.E. = 23 y.D. = 22.6 y.B. = 20.2 y.C. = 18.6 y. A. A. = 16

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

= 7

(34)

= 4.4

(35)

= 2.8

(36)

= 0.4

(37)

12


− y. A. A. = 22.6 − 16 y.D. − y.C. = 22.6 − 18.6 y.D. − y.B. = 22.6 − 20.2 y.B. − y. A. A. = 20.2 − 16 y.B. − y.C. = 20.2 − 18.6 y.C. − y. A. A. = 18.6 − 16 y.D.

= 6 .6

(38)

= 4

(39)

= 2 .4

(40)

= 4 .2

(41)

= 1 .6

(42)

= 2 .6

(43)

e. Coefficient of Variation: Variation: CV =

√ MS

E

y...

=

√3.73 20.08

= 0.0962

× 100 = 9.62%

c. Obtain Obtain the efficiency efficiency of this design with respect respect to CRD and with respect respect to RCBD i. if columns columns were were used as blocks; blocks; ii. if rows were were used as blocks and interpret interpret your results. results. Solution: Completely Randomized Design Data Layout: Fertilizers

Yield

Total

Means

A

14

15 15

16 16

17 17

18 18

80

16

B

19

25

20

18

19

101

20 .2

C

21

16 16

20 20

18 18

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 SS T SS Treatments SS Error

5022 = 10080.16 25 = 237.84

=

= 168.64

(44) (45) (46)

= SS T =

− SSTreatments 237.84 − 168.64 = 69.2

(47)

13



SS

DF

MS

F

168.64

4

42 .16

F1 =12.185

Error

69.2

20

3.46

Total

237.84

24

Treatments

i. Decisi Decision: on: Thus, Thus, the null hypothe hypothesis sis is reject rejected ed since 12.185 is greater than 2.87, for F0.05,4,20 . ii. Conclusion: Hence, the five fertilizers are significantly significantly different, different, implying that they do have an effect on the yield of sugar cane. iii. Multiple Multiple Comparis Comparison on Solution: Using Using the Tukey Honestly Significant Significant Differenc Difference, e, the critical value is,



MS E = q(5,20) T α = q(a, f ) n



3.46 = 4.23(0.8319) = 3.519 5

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

= 7

(48)

∗

= 4.4

∗

(49)

= 2.8

(50)

= 0.4

(51)

= 6.6 = 4

∗

(52)

∗

(53)

= 2.4

(54)

= 4.2

∗

(55)

= 1.6

(56)

= 2.6

(57)

The starred values indicates pairs of mean that are significantly different. iv. iv. Relative Efficiency

14


RE

  LS CRD

=

f =

= =

RE

=

MS R + MSC + ( a 1) MS E ( f ) ( a + 1) MSC ( f 1 + 1)( f 2 + 3) ( f 2 + 1)( f 1 + 3) (12 + 1)(20 + 3) = 0.949 (20 + 1)(12 + 3) 4.76 + 1.36 + 4(3.73) (0.949) 6(1.36) 2.447

−

(58)

Randomized Complete Block Design Data Layout if columns were used as blocks:

Fert Fertiilize lizers rs

Block locks( s(Co Collumn umns)

Treat reatme ment nt Tota otals

Mea 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 SS T SS Treatments

SS Blocks(Columns)

5022 = 10080.16 25 = 237.84

=

= 168.64

=

1 b 2 y. j a j∑ =1

−



a

(59) (60)

b

∑ ∑ yij i=1 j=1



(61) 2

ba

(1002 + 992 + 1032 + 972 + 1032 ) 5 = 10085.6 10080.16 =

− 502 25

2

−

= 5.44

(62)

15


= SS T

SS Error

− SSTreatments − SSBlocks 237.84 − 168.64 − 5.44 = 63.76

=

(63)


SS

DF

MS

F

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

Treatments

v. Decision: Thus, the null hypothesis is rejected since 10.58 is greater than 3.0069. vi. Conclusion Conclusion:: Hence, Hence, the five fertilizers fertilizers are significant significantly ly different, different, which means that they do have an effect on the yield of sugar cane when columns were used as blocks. vii. Multiple Multiple Comparis Comparison on Using the Least Significance Difference, the critical value is, LSD = t α ,15 2





2 MS E = 2.131 n

2(3.985) = 2.69 5

Thus, any pair of treatment averages differ by more than 2.69 would imply that the correspondi corresponding ng pair of population population means are signifisignificantly different. The differences in averages are,

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

= 7

∗

(64)

= 4.4

(65)

=

(66)

∗ 2.8 ∗

= 0.4 = 6.6 = 4

(67)

∗

(68)

∗

(69)

= 2.4

(70)

= 4.2 = 1.6

∗

(71) (72)

16


y.C.

− y. A. A. = 18.6 − 16

= 2.6

(73)

The starred values indicates pairs of mean that are significantly different.

viii. Relative Efficiency Efficiency of RCBD RCBD with columns as as blocks

RE

  LS RCBD

MS R + ( a 1) MS E ( f ) ( a) MS E ( f 1 + 1)( f 2 + 3) ( f 2 + 1)( f 1 + 3) (12 + 1)(20 + 3) = 0.926 (16 + 1)(16 + 3) 4.76 + 4(3.73) (0.926) 5(3.73) 0.977

−

=

f =

= RE

= =

(74)

Randomized Complete Block Design Data Layout if rows were used as blocks:

Fertilizers

Blocks (Rows)

Treatment Totals

Mea 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 SS T SS Treatments

5022 = 10080.16 25 = 237.84

=

= 168.64

(75) (76) (77)

17


1 b 2 y. j j a j∑ =1

=

SS Blocks(Rows)

−



a

b

∑ ∑ yij i=1 j=1



2

ba

(1002 + 992 + 1032 + 972 + 1032 ) 5 = 10085.6 10080.16 =

2

− 502 25

−

= 5.44

(78)

= SS T

SS Error

− SSTreatments − SSBlocks 237.84 − 168.64 − 5.44 = 63.76

=

(79)


SS

DF

MS

F

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

Treatments

ix. Decision: Thus, the null hypothesis is rejected since 10.58 is greater than 3.0069. x. Conclusion Conclusion:: Hence, Hence, the five fertilizers fertilizers are significantly significantly different different,, which means that they do have an effect on the yield of sugar cane. xi. Multiple Multiple Comparison: Comparison: Using Using Least Significa Significance nce Differenc Difference, e, the process is just the same with (ii.). xii. Relative Efficiency Efficiency of RCBD with rows rows as blocks

RE

  LS RCBD

=

f =

= RE

= =

MSC + ( a 1) MS E ( f ) ( a) MS E ( f 1 + 1)( f 2 + 3) ( f 2 + 1)( f 1 + 3) (12 + 1)(20 + 3) = 0.926 (16 + 1)(16 + 3) 1.36 + 4(3.73) (0.926) 5(3.73) 0.808

−

(80)

18


Graeco-Latin Square Problems 4-22. The yield of a chemical process was measured measured using five five batches

of raw materials, five acid concentrations, five standing times (A, B, C, D, E). and five catalyst concentrations (α, β, γ, δ, ). The GraecoLatin Latin square square that that follo follows ws was used. Analyz Analyzee the data from this experiment (use α = 0.05) and draw conclusions. Acid Concentration Batch

1

2

3

4

5

1

Aα=26

B β=16

Cγ=19

Dδ=16

E=13

2

Bγ=18

Cδ=21

D=18

Eα=11

A β=21

3

C=20

Dα=12

E β=16

Aγ=25

Bδ=13

4

D β=15

Eγ=15

Aδ=22

B=14

Cα=17

5

Eδ=10

A=24

Bα=17

C β=17

Dγ=14

Solution: Acid Concentration Batch

1

2

3

4

5

Totals

1

Aα=26

B β=16

Cγ=19

Dδ=16

E=13

90

2

Bγ=18

Cδ=21

D=18

Eα=11

A β=21

89

3

C=20

Dα=12

E β=16

Aγ=25

Bδ=13

86

4

D β=15

Eγ=15

Aδ=22

B=14

Cα=17

83

5

Eδ=10

A=24

Bα=17

C β=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,  = 89. Computation of Sum of Squares: G.. 4302 CF = 2 = = 7396 25 a

+ 172 + 142 ) - 7396 = 436 Total SS or TSS = (26 2 + 162 + (892 + 882 + 922 + 832 + 782 ) Acid SS or ASS = - 7396 = 24.4 5 (902 + 892 + 862 + 832 + 822 ) Batch SS or BSS = - 7396 = 10 5 (1182 + 782 + 942 + 752 + 652 ) Times SS or TrSS = - 7396 = 342 5

···

19


(832 + 852 + 912 + 822 + 892 ) - 7396 = 12 5 Error SS or SSE = TSS - BSS - ASS - CSS - TrSS

Catalyst SS or CSS =

= 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 critical 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 standing times are significantly significantly different. different. b. The five batches of raw materials have have no significant difference. c. The five acid concentrations concentrations have have no significant difference. difference. d. The five catalyst concentrations concentrations have have no significant difference. difference. xiii. Multiple comparison for five five standing times, times, Using Tukey Tukey Honestly Significant Difference, the critical value is obtain,



MS E = q(5, 8) T α = q( a, f ) n





5.95 = 4.89 5

5.95 = 5.33 5

Treatment Means: y A = 23.6, yC = 18.8, yB = 15.6, y D = 15, yE = 13 Thus, any pair of treatment averages differ by more than 5.33 would imply that the correspondi corresponding ng pair of population population means are signifisignifi-

20


cantly different. The differences in averages are,

− yE = 23.6 − 13 y A − yD = 23.6 − 15 y A − y B = 23.6 − 15.6 y A − yC = 23.6 − 18.8 yC − yE = 18.8 − 13 yC − yD = 18.8 − 15 yC − y B = 18.8 − 15.6 yB − yE = 15.6 − 13 yB − yD = 15.6 − 15 y D − yE = 15 − 13 y A

= 10.6 = =

∗ 8.6 ∗ 8∗

= 4.8 = 5.8

∗

= 3.8 = 3.2 = 2.6 = 0.6 = 2

The starred values indicates pairs of mean that are significantly different.

21


22

Computation Computation using SPSS Software 3-1. The tensile strength of portland cement is being being studied. Four dif-

ferent mixing techniques can be used economically. economically. The following data have been collected: Mixing Mixing Technique echnique

Tensile ensile Strength Strength (lb/ lb /in 2 )

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 above table table is entered entered in SPSS in this manner.

Step 2 Figure 3: The second step after inputting your your data, go to analyze compare means one - way anova.

⇒

⇒


23

Step 3

Figure 4: Next, enter the variable yield to the dependent (yield d ep list ep en en de de nt nt l is is t) treatment factor and to (treatment factor). After that you can click the post hoc .. for choosing the test for multiple comparison.

⇒

⇒

Table 1: The output of the performed steps. steps. In multiple comparison comparison table, the test test performed performed was was Scheffé. Scheffé. You can check it in the post hoc .. section of the Step 3

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

strength strength of a particular particular type of cloth. cloth. Because Because there might be varivariability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order


24

to each bolt. The resulting tensile strength follow. Analyze the data from this experiment experiment (use α = 0.05) and draw appropriate conclusions. Bolt Chemical

1

2

3

4

5

1

73

6 68 8

7 74 4

7 71 1

6 67 7

2

73

6 67 7

7 75 5

7 72 2

7 70 0

3

75

6 68 8

7 78 8

7 73 3

6 68 8

4

73

7 71 1

7 75 5

7 75 5

6 69 9

Solution Step 1 Figure 5: The above above table table is entered entered in SPSS in this manner.

Step 2 Figure 6: The second step after inputting your your data, data, go to analyze general linear model

⇒ ⇒univariate


25

Step 3 Figure 7: Next, enter the varidependen ndent t list able yield to the the depe (yield dependent list ) and treat ment and block to fixed factor ( s ) (treatment and block fixed fixed factor ( s )). Afte Afterr that you you can click click the for choosi choosing ng the the test test for post hoc .. for multiple comparison.

⇒

⇒

Step 4 Figure 8: Before Before clicki clicking ng the ok button, go first to the model (seen on Step the univariate: model win3). In th dow, click (custom) then put the treat ment and block to the model box, as shown shown in the figure. figure. Then, Then, change change the type to main effects and unchecked unchecked the include intercept in model , before the clicking the continue button.

output of the performed performed test. Refer to Manual Computation and

Graphical Illustration Section item 4-1 for the interpretation. 4-11. Suppose Suppose that the obersvation obersvation for chemical chemical type 2 and bolt 3 is

missing in Problem 4-1. Analyz Analyzee the problem problem by estimat estimating ing the


26

missing value. Perform the exact analysis and compare the results. Solution: Bolt Chemical

1

2

3

4

5

1

73

6 68 8

7 74 4

7 71 1

6 67 7

2

73

67

x

72

70

3

75

6 68 8

7 78 8

7 73 3

6 68 8

4

73

7 71 1

7 75 5

7 75 5

6 69 9

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 This is the output of the test performed.

Refer to Manual Computation and Graphical Illustration Section item 4-11 for the interpretation. 1. Shown Shown below below the yield yield (ton (ton per 1/4-ha.p 1/4-ha.plot lots) s) of sugar sugar cane in a

Latin square experiments comparing five (5) fertilizer levels.


Row

C=10 tons manure/ha

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 Analyze the data completely and interpret interpret your results. results.

E=30 tons manure/ha B=complete inorganic fertilizer D=20 tons manure/ha


27

Solution: Step 1 Figure 9: Enter Enter the data to SPSS SPSS in this this manner.

Step 2 Figure 10: The second step after inputting your our data, go to

⇒ ⇒

analyze g e n e r a l model univariate

l ine ar


28

Step 3 Figure 11: Next, ente nter the varidependen ndent t list able yield to the the depe (yield dependent list ) and treat ment , row and column to fixed fac tor ( s ) ( row and column fixed fac Afterr that you you can click click the tor ( s )). Afte post hoc .. for for choosi choosing ng the the test test for multiple comparison.

⇒

⇒

Step 4 Figure 12: Before clicking clicking the ok button, go first to the model (seen on Step 3). In the univariate: model window, click custom then put the treatment , col umn , and row to the model box, as shown shown in the figure. figure. Then, Then, change change the type to main effects and unchecked unchecked the include intercept in model , before the clicking the continue button.

Figure 13: The output of the performed steps


29

Figure 14: The output generated using post hoc ..- lsd Method

Refer to the Manual Computation and Graphical Illustration Section for the interpretation of these outputs.

Experimental Design Problem Set II

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