Randomized Block Design/ Two-Way Classification (R.B.D.)

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Let measurements of n treatments are distributed to m blocks where,

yij be the observation of i-th treatment for j-th block

For this, the following layout shall be used,

Blocks:

B1   B2——-Bj———-Bm Means
Treatment 1

Treatment 2

y11   y12——-y1j———-y1m

Y21   y22——-y2j———-y2m

.1
Treatment i Yi1   yi2——-yij———-yim .2
Treatment n Yn1yn2——-ynj———-ynm .j
Means .1.2——-.j———-.m .m

 

Consider this model as,

yij = µ + αi + βj + eij  where, i = 1, 2, …….. n and j = 1, 2, …….. m

Here,

µ = Grand Mean

αi = i-thTreatment Effects ()

βj = j-thTreatment Effects ()

eij = Random Errors, these are identically distributed {N (0, σ2)

 

Hypothesis:

To get the statistic hypothesis, it can be written as-

H01 = α2 = ………. = αn = 0

Or,

H01 = β2 = ………. = βn = 0

 

H1 :At least one value of these effects should not be zero

 

Constructing the ANOVA Table:

 

Source of Variation Degree of Freedom Sum of

Squares

Mean

Square

F
Treatments n – 1 SS (Tr) MS (Tr)

=

FTr =
Blocks m – 1 SS (Bl) MS (Bl)

=

FBl =
Error (n – 1) (m – 1) SSE MSE

=

Total nm – 1 SST

 

Where,

SSE       = Error Sum of Squares

SST       = SS (Tr) + SSE

It can be calculated as,

SST       = Total Sum of Square

=

SS(Tr)   =

SS(Bl)   =

c          =

 

Where,

Ti.         = Total of m Observations for i-th Treatment

T.j         = Total of n Observations for j-thBlock

T..         = Grand Total of all ‘nm’ Observations

So,

SSE       = SST – SS (Tr) – SS (Bl)

Set up the value of α

It’s Conclusion:

Reject H0:        αi = 0               i           if FTr> Fα.n – 1. (n – 1) (m – 1)

RejectH0:         βj= 0                j           if FBL> Fα.m – 1. (n – 1) (m – 1)

 

Example No. 3

When four students appeared for three different IQ test, the following table has been obtained. The scores are tabulated as-

Students
1 2 3 4
Test-I

Test-II

Test-III

75

83

86

73

72

61

59

56

53

68

69

70

 

By using the method of two-way analysis of variance, state the reasons whether these tests are equivalent or not. Assume level of significance = 0.01

Solution:

  1. As there are three equivalent tests,

H01 = α2 = α3 = 0

H1i≠ 0 for at least one value of i

Since students’ IQs are equivalent

H01 = β2 = β3 = β4 = 1340 = 0

H1j≠ 0 for at least one value of j

 

  1. Given:α = 0.01, n = 3 and m = 4

Ftab(Tr) = F0.01, 2, 6= 10.92

Ftab(Bl) = F0.01, 3, 6 = 9.78

 

  • Computations:

T1. = 75 + 73 + 59 + 68= 275

T2. = 83 + 72 + 56 + 69 = 280

T3. = 86 + 61 + 53 + 70 = 270

T.1 = 75 + 83 +86= 244

T.2 = 73 + 72 + 61 = 206

T.3 = 59 + 56 + 53 = 168

T.4 = 68 + 69 + 70 = 207

Thus,

T.. = 825

Now,

c = (825)2/ 12 = 56718.75

So,

SST = 57855 – 56718.75 = 1136.25

SS (Tr) = 226925/ 4 – 56718.75

= 12.5

SS (Bl)  = 173045/ 3 – 56718.75

= 962.92

Hence,

SSE = SST – SS (Tr) – SS (Bl)

= 160.83

 

ANOVA Table:

Sources of

Variation

Degrees of Freedom Sum of Squares Mean Square F
Treatment 2 12.5 12.5/ 2 = 6.25 6.25/ 26.81 = 0.23
Blocks 3 962.92 962.92/ 3 = 320.97 320.97/ 26.81 = 11.97
Error 6 160.83 160.83/ 6 = 26.81
Total 11 1136.25

 

  1. Conclusion:

Here,

Fcal (Tr)<Ftab(Tr)

Hence,

H01 = α2 = α3 = 0 is accepted for all the three tests as equivalent

 

And,

Fcal (Bl) >Ftab (Bl)

Hence,

H01 = β 2 = β 3 = β4 = 0 is rejected for all the three tests as students’ IQs are different.

 

Note:

The R.B.D is greatly accepted above the C.R.D. because of its greater flexibility in solving different types of experimental works. This is because there are no restrictions on numbers of treatments along with numbers of replicates. But the fact is that- for the large number of treatments or for the wide variable blocks, the R.B.D. is not suitable.

 

Problems:

  1. Three samples are tabulated below. To obtain the hypothesis, consider the population means are equal and level of significance as 5%:
A B C
8

10

7

14

11

7

5

10

9

9

12

9

13

12

14

 

  1. From the group of three main brands of certain powder, 120 sales are examined. Allocate these brands of powder considering the table below:
Brands Groups
A B C D
I

II

III

0

5

18

4

8

19

8

13

11

15

6

13

 

  • Fours batches of electric bulbs with 1000 hrs. of lives are tabulated below:
Batches
A

B

C

D

1.6

1.58

1.46

1.51

1.61

1.64

1.55

1.52

1.65

1.64

1.6

1.53

1.68

1.7

1.62

1.57

1.7

1.75

1.64

1.6

1.72

 

1.66

1.68

1.8

 

1.74

 

 

 

1.82

 

 

Show that the significance test is accepted by using the analysis of variance method.

  1. After conducting five tests in Computer Science paper, the following table has been tabulated with their test scores.
Batch-I Batch-II Batch-III
86

77

81

84

69

79

78

82

83

68

88

83

86

89

82

 

The test was conducted in three batches. Take the level of significance as 5% for the analysis of variance.

  1. Construct a two-way ANOVA table using the following data-
Prices of Field Treatment
A B C D
P

Q

R

45

43

39

40

41

39

38

45

41

37

38

41

 

If required, you can shift the origin to 40.

  1. In a factory, four different workers are hired for working on five different machines. After taking proper observations, some data has been made. You have to calculate in two different ways whether all men are similar in production or not along with the efficiency if each machines. The following has been given-
  2. Sum of Squares for Variance between machines = 35.2
  3. Sum of Square for Variance between workmen = 53.8
  4. Sum of Squares for Total Variance = 174.2

 

Answers:

  1. Reject H0 : µ1 = µ2 = µ3, Fcal = 4

 

  1. No such difference in the brands

 

  1. Reject H0 as the means are different

 

  1. ANOVA Table:
Sources of

Variation

Degrees of Freedom Sum of Squares Mean Square F
Between

Treatment

3 22.917 7.639 1.0
Between

Fields

2 8.167 4.083 0.534
Error 6 45.833 7.639
Total 11 76.917

 

  1. Workers’ productivity is similar to mean productivity and it is same for all the f machines.

Here,

F (Workers) = 2.53

F (Machines) = 1.24

 

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