USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

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

Let the measurements pertaining to n treatments distributed over m blocks be available.

Let yij denotes the observation of the i-th treatment over the j-th block.

We shall use the following layout :

Blocks

Consider the model

yij  = µ +αi + βj + eij,               i = 1, 2, ... , n

j = 1, 2, ... , m

 

where,                         µ=       Grand mean

α=        Effect of the i-th treatment with ∑αj = 0

βj = Effect of the j-th block with  ∑ βj = 0

eij = Random errors which are identically distributed as N ( 0, σ2)

 

Hypothesis :

H0 :      α1 = α 2 = · · · =         α n       = 0

or         H0:       β1= β2 = ...      =          βn        = 0

H1 : At least one of the effects is not zero.

 

Thus we construct the following

ANOVA TABLE :

Ti· = Sum of m observations for the i-th treatment

T.j = Sum of n observations for the j-th block

T.. = Grand total of all the observations

SSE = SST - SS (Tr) - SS (Bl)

 

Set up α

Conclusions:

 

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

Reject H0              : 13; = 0          j      if         FBl > Fα. m -1. (n -1) (m -1)

 

 

Example 3. Three different types of I.Q. test were conducted to four students and the following are the scores which they detained.

Perform a two way analysis of variance to test at the level of significance a = 0.01 whether it is reasonable to treat the three tests as equivalent.

 

Solution. 1 . H0 : Three tests are equivalent (a1 = a2 = a3 = 0)

H1 : At least one a; not zero.

 

Also    H0 : β1= β2 = β3 = β4 =  1340 = 0 (Student's I.Q. are equivalent),

H1 : At least one βi ≠ 0

 

2. α = 0.01, Here n = 3, m = 4

 

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

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

 

3. 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

T•• = 825

 

c = (825)2 / 12= 56718.75

 

 

 

 

SST     =

SS (Tr) =

SS (Bl) =

SSE =

57855 - 56718.75 = 1136.25

226925 / 4 - 56718.75 = 12.5

173045 / 3 - 56718.75 = 962.92

SST - SS (Tr) - SS (B/) = 160.83

 

ANOVATABLE

4. Conclusions.

Since Fcal (Tr)  <  Ftab (Tr)

 

=> H0 : a.1 = a.2 = a.3 = 0 is accepted.

=> All three .tests are equivalent.

 

But since Fcal (Bl) > Ftab (Bl)

=> H0 : β1= β2 = β3 = β4 = 0 is rejected.

=> I.Q. of the students is not same.

Note. The R.B.D is more accurate than C.R.D. for most types of experimental work. It has greater flexibility, i.e., no restrictions are placed on the number of treatments or the number of replicates. However, R.B.D is not suitable to the problems with large number of treatments or to the wide variable blocks.

 

PROBLEMS

1. Three samples below have been obtained from normal populations with equal variances. Test the hypothesis at 5% level of significance that the population means are equal.

2. There arc three main brands of a certain powder. A set of its 120 sales is examined and found to be allocated among four groups (A, 8, C, D) and brands (I, II and III) as shown below :

3. The following table shows the lives (in 1000 hrs) of four batches of electric bulbs:

Perform an analys1s of variance of these data and show that a significance test does not reject their homogeneity.

 

4. Five tests were conducted in Computer Science papers to candidates in three batches selected based on their strengths in Mathematics, English and Numerical computing respectively. The following average test scores were obtained :

Carry out an analysis of variance. Test the hypothesis whether the differences among the means obtained for the three batches are significant at 5% level of significance.

 

5. Set up a two-way ANOVA table for the data given below :

(You can shift the origin to 40)

 

6. In a certain factory, production can be accomplished by four different workers on five different types of machines. A sample study in context of two way design without repeated values is being made with two fold objectives of examining whether the four workers differ with respect to mean productivity and whether the mean productivity is the same for the five machines. The researcher involved in this study reports while analyzing the gathered data as under:

(i)              Sum of squares for variance between machines = 35.2

(ii)            Sum of squares for variance between workmen = 53.8

(iii)          Sum of squares for total variance = 174.2

 

ANSWERS

 

1. Reject H0 : µ1 = µ 2 = µ3, Fcal = 4

 

2. No significant difference in brands

 

4. Reject H0, means are different

 

5.

6. Workers do not differ w.r.t. mean productivity and mean productivity is the same for 5 different machines.

 

F (machines) = 1.24, F (Workers) = 2.53.