RANDOMIZED BLOCK DESIGN :

This is a technique of ANOVA which is an extension of the equality of means of two populations with matched (paired) samples to more than two populations. In this technique the subjects which are similar in everything except the characteristic of interest are arranged into blocks. For instance suppose we take three independent samples of n1 n2 and n3 students to administer three different teaching methods. We can do an ANOVA to test the equality of effectiveness of the three methods (treatments). The results (response) (as measured by the performance of the students on a common test) might be blurred by the differences in the students' other characteristics (other than the method of teaching) such as GPA, previous courses, age, sex etc. One can apply randomized block design concept to control as many of these (confounding) factors as possible. This may be done by selecting several blocks of three students who are very similar in their other characteristics --GPA, age etc. Each of the three students belonging to a block can then be randomly assigned to one of the teaching methods (treatment groups). This way, most of the variation observed in the performance of the students can more directly be attributed to the treatment, rather than being masked by other factors.

In some cases the blocks might consist of the same entity which is subjected to different treatments. In this extreme case the confounding factors can be controlled even more tightly—all of the observed differences can then be attributed to the treatment.  Here is an example: There are three appraisers (A, B, C) of real estate property and we wish to test the null hypothesis that appraisals of any property by these three do not significantly differ from one another (in other words,  none of the appraisers consistently over or under appraise properties compared to the others).

Formally we represent the null and the alternative hypotheses as:

Ho : m A= mB = m C
H1: At least one m is different

The sample consists of appraisals of properties 1 through 5 by appraisers A, B and C as given in the table below:
 
PROPERTY
A
B
C
1
90
93
92
2
94
96
88
3
91
92
84
4
85
88
83
5
88
90
87

Each appraiser (treatment) appraises each of five properties. The three appraisals for each property is a block (same entity subjected to three treatments). The test of the hypothesis is done on the basis of the partitioning of the total variation in the 15 sample observations into a) due to appraisers (treatment); b) due to property (block) and c) due to random error. If we find the variation due to appraisers to be significantly large compared to variation due to error we would then be inclined to reject the null hypothesis that appraisers appraise about equally.

To illustrate these, let us assume j is the index of the appraisers (treatments) (j = A,B,C); i is the index of properties (blocks) (i = 1,2,3,4,5); Xij the observation of the ith block of jth treatment; Xi and Xj the means of the ith block and jth treatment respectively and X the grand mean. Let us also use k to denote the number of treatments and b the number of blocks (here 3 and 5 respectively).

The total variation in all b * k = 15 observations as measured by SST (the total sum of squares) can be broken down to three sources: Treatments (SSTR); Blocks (SSB) and error (SSE) so that

SST = SSTR + SSB + SSE.

Calculation of Sums of Squares:

  1. Total Sum of Squares: (SST)

    2. SST = å å (Xij - X)2 with b * k – 1 degrees of freedom
    In the example the grand mean = X = 89.4.  Thus
    SST = (90 - 89.4)2 + (93 - 89.4)2 + (92 - 89.4)2 + (94 - 89.4)2 + (96 - 89.4)2 . . . .
    = 195.6 with 5 * 3 – 1 = 14 degrees of freedom.
     
  2. Treatment Sum of Squares (SSTR)

    3. SSTR = b * å (Xj - X )2 with k - 1 degrees of freedom
    In the example XA = 89.6 ; XB = 91.8; XC = 86.8. Thus
    SSTR = 5 * [(89.6 - 89.4)2 + (91.8 - 89.4)2 + (86.8 - 89.4)2] = 62.8 with 3 - 1 = 2 degrees of freedom
     
  3. Block Sum of Squares (SSB)

    4. SSB = k * å (Xi - X )2 with b - 1 degrees of freedom
    In the example X1 = 91.67; X2 = 92.67; X3 = 89; X4= 85.33 and X5= 88.33.
    Thus SSB = 3 * [(91.67 - 89.4)2 + ( 92.67 - 89.4)2 + (89 - 89.4)2 + (85.33 -89.4)2 + (88.33 -89.4)2] = 100.93. with 5 - 1 = 4 degrees of freedom
     
  4. Error Sum of Squares (SSE)
SSE = SST - SSTR - SSB with (k-1) * (b-1) degrees of freedom
Thus in the example
SSE = 195.6 - 62.8 - 100.93 = 31.87. with (3-1) (5-1) = 8 degrees of freedom. Note that just like the sum of the squares are additive e.g., SSTR and SSB and SSE add up to SST, so are the degrees of freedom: treatment df +  block df + error df add up to total df.

As before the F statistic is computed as the ratio of mean squares :
 
Source
Sum of Squares
df
Mean Squares
F-Stat
F-Crit (5%)
p-value
Treatment
62.8
2
31.4
7.882
 4.4589 .0128
Block
100.93
4
25.23
6.335
 3.8378 .0134
Error
31.87
8
3.98
      
Total
195.6
14
13.97
      

The F statistic for Treatment is MSTR/MSE = 31.4/3.98 while (optionally) the F statistic for blocks is MSB/MSE = 25.23/3.98. These may be compared to the critical F values to test the hypotheses. We reject the null hypothesis that the appraisers’ appraisals are on the average the same since the computed F value = 7.882 exceeds the Critical F value (%5) with 2 and 8 degrees of freedom ( 4.4589).  Excel function "=FINV(0.05,2,8)" returns 4.4589.
Although we were not specifically interested in this, yet the analysis also shows that there are significant differences in the values of the 5 properties. (F block = 6.335 > F Critical = 3.8378). The p-values lead to the same conclusions.

Supplementary Problem:

A marketing research firm has been asked to compare the percentage increase in sales of three brands of low cholesterol margarine. Six supermarkets are selected from the city to serve as blocks and account for variation in sales that result from demographic and socio-economic differences among the customers. The percentage increases in sales for the three brands (A, B and C) over the last three months in the six supermarkets are shown below:
 
 
BRAND
Supermarket
A
B
C
1
4.2
2.8
3.4
2
9.5
8.2
7.8
3
8.2
6.3
5.2
4
2.4
1.8
2.1
5
9.8
8.9
9.8
6
6.5
6.4
6.8

  1. Use the randomized block method of ANOVA to test the hypothesis that the three brands experienced similar growth rates and that there were no differences in the increased use of margarine across the socio-economic groups represented by the supermarkets. Use a significance level of .05.
  2. Repeat the test using Excel's ANOVA