Yates Algorithm for 2^k Factorial Experiment lecture - 27

 

Yates Algorithm for 2^k Factorial Experiment 

lecture - 27

History

Frank Yates (12 May 1902 – June 17, 1994) was a pioneer in the field of statistics throughout the 20th century. He served as President of the British Computer Society from 1960 to 1961. Frank Yates was the eldest son of a seed merchant and botanist Percy Yates. In 1931, Frank Yates started his career as mathematics teacher, teaching with the secondary level students in Africa. Frank Yates was appointed as assistant statistician at Rothamsted Experimental Station by R.A Fisher and in 1933, he became head of statistics when R. A Fisher left the Rothamsted Experimental Station and joined University College London. Frank Yates contributed to the theory of analysis of variance while working on experimental design at Rothamsted Experimental Station, where he also developed Yates’s algorithm and Balance incomplete block design.

Yates Algorithm

Frank Yates developed a unique structure to generate the main effects as well as their interaction effects of a factorial experiment. To compute the main and their interaction effects in a factorial experiment, the Yates algorithm is a considerably more convenient method than the contrast method, which can time consuming and laborious.

Frank Yates Algorithm for 2^k Factorial Experiment

The Frank Yates proposed an algorithm to compute the sum of squares for various effects in 2^k factorial experiment. The Frank Yates algorithm is outlined below:

1.      Step - 1:  Write the treatment combination in standard order in the beginning of the table.

                      e.g. in 2^2 factorial experiment 1, a, b, 

                      e.g. in 2^3 factorial experiment  1, a, b, ab, c, ac, bc, abc.

1.      Step - 2: Yield column is obtained from the total of each treatment enclosed [].

 In 2 ^2 factorial experiment the treatment combination total is represented by; [1],  [a]. [b], [ab]

In 2^3 factorial experiment [1], [a], [b], [ab], [c], [ac], [bc], [abc]

Step - 3: The column – I, column – II, …. , is obtained from yield column as:

a.  Upper half is obtained by adding the yield values in pairs

b.  Lower half is obtained by taking the difference in pairs (lower minus upper.

Step - 4: The other columns are obtained in the same discussed in step – 3.

1.    Step - 5: The k – columns will be developed for 2^k factorial experiment. (Number of columns is depend on the number of factors).

In 2 ^2 factorial experiment two columns (column- I, column - II) will be develop. 

In 2 ^3 factorial experiment, column – I, column – II and column - III will be develop and so on.

1.      Step - 6: The sum of squares effect column is obtained as:

The sketch of 2^2 Factorial Experiment

Treatment Combination

Total

Column – I

Column – II

Effects

SS. Effects

1

[1]

[a] + [1]

[ab] + [b] +[a] + [1]

Grand

 

a

[a]

[ab] + [b]

[ab] - [b] + [a] - [1]

[A]

[A]^2 / 2^2x r

b

[b]

[a] - [1]

[ab] + [b] - ([a] + [1])

[B]

[B]^2 / 2^2x r

ab

[ab]

[ab] - [b]

[ab] - [b] –([a] - [1])

[AB]

[AB]^2 / 2^2x r

Example: 

A 2^2 factorial experiment i.e. with 2 varieties (factors) and 2 manures (level) was carried out in a randomized complete block design with 3 replications. The yield given in the following table:


Test the significance of varieties and manures using Yates Algorithm.

Solution: 
i. Hypotheses Statement:

    H01: Factor A has no significant effect    Vs.    H11: Factor A has significant effect 

    H02: Factor B has no significant effect    Vs.   H12: Factor B has significant effect 

   H03: The interaction effect of  AB has no significant effect   
     
                                               Vs.  
    H13: The interaction effect of  AB has  significant effect .

ii. The significance level;, α = 0.05
iii
iii. Test Statistic: Frank Yates Algorithm
iv. Reject H01, H02, H03, if F >=  5.12
v. Computation:

Treatment Combination

Yield

Column – I

Column – II

SS Effects

1

15

39

81

 

a

24

42

21

36.75

b

15

9

3

0.75

ab

27

12

3

0.75
  

vi. Remarks:
     The interaction effect of AB is insignificant and it is concluded that the varieties levels and  manures levels has not significant effect on the yield. 

Alternative Way

Some times it is desired to compute the block and treatment effects. so, add the conventional design approach to Yates method. An alternate solution is provided for the aforementioned situation.

i.  Statement of Hypotheses

     H0 : T1 = T2 = T3 = T4   Vs.  H1 : T1 ≠  T2 ≠  T3 ≠ T4

    H01: Factor A has no significant effect    Vs.    H11: Factor A has significant effect 

    H02: Factor B has no significant effect    Vs.   H12: Factor B has significant effect 

   H03: The interaction effect of  AB has no significant effect   
     
                                               Vs.  
   H13: The interaction effect of  AB has  significant effect .

ii. The significance level;, α = 0.05
iii
iii. Test Statistic: Frank Yates Algorithm
iv.  Reject H0, if F > = 4.07
      Reject H01, H02, H03, if F >=  5.12
v. Computation:

Yates – Method:

Treatment Combination

Yield

Column – I

Column – II

SS Effects

1

15

39

81

 

a

24

42

21

36.75

b

15

9

3

0.75

ab

27

12

3

0.75

SST = SSA + SSB + SSAB

SST = 36.75 + 0.75 + 0.75

SST = 38. 25

SSE = SSTotal – SST – SSBlock

SSE = 74.25 – 38.25 – 18.50

SSE = 17. 50

ANOVA Table

SV

df

SS

MS

F

F tab

Block

2

18.50

9.25

3.16

 

Treatment

3

38.25

12.75

4.36

4.07

A

1

36.75

36.75

12.58

5.12

B

1

0.75

0.75

0.25

5.12

AB

1

0.75

0.75

0.25

5.12

Error

6

17.50

2.92

 

 

Total

11

74.25

 

 

 

vi. Remarks:
     The interaction effects of AB is insignificant

Example: The following table gives treatment combination of 2 factors:

Replicate

a

b

c

abc

1

ab

ac

bc

1

1.9

1.6

2.1

3.8

1.3

3.2

2.8

3.2

2

3.0

2.7

3.0

4.8

2.2

4.1

3.9

4.1

3

4.0

3.8

2.1

5.9

4.2

5.2

5.1

5.0

4

1.9

1.5

2.0

3.9

1.1

3.4

3.0

3.0

Write statistical model for the above stated experiment and test the significance of factors effects and interaction effects.

Solution: The statistical model of the above experiment (2^3 factorial experiment) is:

Yijkl = μ + ρi + αj + βk +γl + (αβ)jk + (αγ)jl + (βγ)kl + (αβγ)jkl + ϵijkl

i =1, 2, 3, 4

j, k, l = 0, 1

i. We setup our hypotheses as:

H01: The interaction effects of ABC is insignificant  

                             Vs. 

H11: The interaction effects of ABC is insignificant 

H02: The interaction effects of AB is insignificant  

                             Vs. 

H12: The interaction effects of AB is insignificant 

H03: The interaction effects of AC is insignificant  

                             Vs. 

H13: The interaction effects of AC is insignificant 

H04: The interaction effects of BC is insignificant  

                             Vs. 

H14: The interaction effects of BC is insignificant 

ii. The significance level;  α = 0.05

iii. Test statistic: Frank Yates Algorithm

iv. Reject all null hypotheses, if F => 4.32

v. Calculation

Treatment Combination

Yield

Column I

Column II

Column III

S.S effect

F0.05(1, (r-1) 2^3-1)

(1)

8.8

19.6

45.1

103.4

 

a

10.8

25.5

57.7

17

9.03125

4.32

b

9.6

24

8.3

15.6

7.60

4.32

ab

15.9

33.7

8.7

1.8

0.101

4.32

c

9.2

2

5.9

12.6

4.96

4.32

ac

14.8

6.3

9.7

0.4

0.005

4.32

bc

15.3

5.6

4.3

3.8

0.45

4.32

abc

18.4

3.1

- 2.5

- 6.8

1.48

4.32

vi. Remarks:

    The interaction effect of ABC, AB, AC, and BC are insignificant.

Alternative Way

i. We setup our hypotheses as:

H01: The interaction effects of ABC is insignificant  

                             Vs. 

H11: The interaction effects of ABC is insignificant 

H02: The interaction effects of AB is insignificant  

                             Vs. 

H12: The interaction effects of AB is insignificant 

H03: The interaction effects of AC is insignificant  

                             Vs. 

H13: The interaction effects of AC is insignificant 

H04: The interaction effects of BC is insignificant  

                             Vs. 

H14: The interaction effects of BC is insignificant 

ii. The significance level;  α = 0.05

iii. Test statistic: 

iv. Reject all null hypotheses, if F => 4.32

v. Calculation

SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC

SST = 9.03 + 7.60 + 4.96  + 0.101 + 0.005 + 0.45 + 1.48

SST = 23.26

SSE = SSTotal - SST - SSBlock

SSE = 46.775 - 23.26 - 20.62

SSE = 2 . 90

 ANOVA Table

vi. Remarks
     The high order interaction is significant.

Example: The data of the following table are form  factorial experiment. Partition the treatment sum of squares into main effects and interaction counter parts. Interpret your results.

Solution: The data look like from homogeneous experimental units. So, we are using CRD  factorial experiment. The statistical model for this experiment is given by;

Yijk = μ + Aj + Bk + (AB)jk + ϵijk      i= 1, 2, 3, 4, 5. j, k = 0, 1

i. we setup our hypotheses as:

H0 : T1 = T2 =T3 =T4   Vs. H0 : T1 ≠  T2  ≠  T3  ≠  T4

H01 : aj = 0 Vs. H11 : aj ≠ 0

H02 : bk = 0 Vs. H12 : bk ≠ 0

H03 : (ab)jk = 0 Vs. H13 : (ab)jk ≠ 0

ii. The significance level;  α = 0.05

iii. The test statistic:

iv. Reject H0, if F => F0.05 (3, 16) = 3.24
     Reject H0i, if F => F0.05 (1, 16) = 4.49  i = 1, 2, 3.
v.  Computation:

Now partitioned the treatments sum of squares into component parts by using Yates method:

Treatment Combination

 

Yield

Column I

Column II

S.S

44

107

323

 

63

216

-         7

2.45

121

19

109

594.05

95

-         26

-         45

101.25

ANOVA Table

SV

d.f

SS

MS

F

Treatment

3

697.75

232.58

21.535

A

1

2.45

2.45

0.227

B

1

594.05

594.05

55.005

AB

1

101.25

101.25

9.375

Error

16

172.80

10.80

 

Total

19

870.55

 

 

vi. Remarks:

    H0 is significant

   H02 and H03 are significant.

Example: 

Eight treatments have been applied to a certain crop and their effects have been reflected in the following table with 2 replications:

Give a method of analysis and interpret your result.

Solution:

Now using Yates method

Treatment Combination

Yield

Column I

Column II

Column III

S.S

(1)

27

48

102

176

 

a

21

54

74

0

0

b

25

42

-          2

-          4

1

ab

29

32

2

4

1

c

19

             -    6

6

            - 28

49

ac

23

4

-          10

4

1

bc

17

4

10

    -   16

16

abc

15

   - 2

                   - 6

                 - 16

16

In this experiment the error sum of squares is zero; we combine the interaction sum of squares of little importance. So, their pooled sum of squares may be used as SSE.

MSE = ab + ac 

MSE = 1 + 1 = 2

ANOVA Table:

SV

d.f

SS

MS

F

F tab

Block

1

4

 

 

Treatment

7

84

12

6.00

3.73

A

1

0

0

0

     5.32

B

1

1

1

0.50

5.32

C

1

49

49

24.50

5.32

AB

1

1

1

0.25

5.32

AC

1

1

1

0.25

5.32

BC

1

16

16

8

5.32

ABC

1

16

16

8

5.32

Error

8

 

2

 

 

Total

15

88

 

 

 

The treatment combinations ABC and BC are significant.


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