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

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

y_ij 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,

y_ij = µ + α_i + β_j + e_ij  where, i = 1, 2, …….. n and j = 1, 2, …….. m

Here,

µ = Grand Mean

α_i = i-thTreatment Effects ()

β_j = j-thTreatment Effects ()

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

 

Hypothesis:

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

H₀ :α₁ = α₂ = ………. = α_n = 0

Or,

H₀ :β₁ = β₂ = ………. = β_n = 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,

T_i.         = 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:        α_i = 0               i           if F_Tr> F_{α.n – 1. (n – 1) (m – 1)}

RejectH₀:         β_j= 0                j           if F_BL> F_{α.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,

H₀ :α₁ = α₂ = α₃ = 0

H₁ :α_i≠ 0 for at least one value of i

Since students’ IQs are equivalent

H₀ :β₁ = β₂ = β₃ = β₄ = 1340 = 0

H₁ :β_j≠ 0 for at least one value of j

 

1. Given:α = 0.01, n = 3 and m = 4

F_tab(Tr) = F_{0.01, 2, 6}= 10.92

F_tab(Bl) = F_{0.01, 3, 6} = 9.78

 

• Computations:

T_1. = 75 + 73 + 59 + 68= 275

T_2. = 83 + 72 + 56 + 69 = 280

T_3. = 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,

F_cal (Tr)<F_tab(Tr)

Hence,

H₀ :α₁ = α₂ = α₃ = 0 is accepted for all the three tests as equivalent

 

And,

F_cal (Bl) >F_tab (Bl)

Hence,

H₀ :β₁ = 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, F_cal = 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

 

Links of Previous Main Topic:-

• Introduction to statistics
• Knowledge of central tendency or location
• Definition of dispersion
• Moments
• Bivariate distribution
• Theorem of total probability addition theorem
• Random variable
• Binomial distribution
• What is sampling
• Estimation
• Statistical hypothesis and related terms
• Analysis of variance introduction
• Completely randomized design c r d one way classification

Links of Next Statistics Topics:-

• Definition of stochastic process
• Introduction operations research
• Introduction and mathematical formulation in transportation problems
• Introduction and mathematical formulation
• Queuing theory introduction
• Inventory control introduction
• Simulation introduction
• Time calculations in network
• Introduction of game theory