Question

3. Zero Artificial Agro Limited furnished the following information from their farmhouse. Set up an analysis of variance table for the following two-way design results: Per acre production data of Wheat A Varieties of seeds Varieties of Fertilizers W a) Find the differences concerning varieties of seeds are insignificant at 5% level of significance. b) Also obtain a separate independent measure of inherent variations. (6 Marks)

Answer 1

Solution:

ANOVA table can be set up for the given problem as shown below:

THE ANOVA TABLE

Source of variation SS d.f                         MS             F-ratio        5% F-limit(or the tables values)

Between columns 8 (3-1)=2                     8/2 = 4        4/1 = 4       F(2,6) = 5.14

(i.e., between varieties

of seeds)

Between rows 18 (4-1) = 3                   18/3 = 6        6/1 = 6       F(3,6) = 4.76

(i.e., between varieties

of fertilizers)

Residual or error                         6 (3-1)*(4-1) = 6              6/6 = 1

Total                                            32 (3*4 )-1 = 11

             From the given ANOVA table, we can find that the differences concerning varieties of seeds are insignificant at 5% level as the calculated F-ratio of 4 is less than the table value of 5.14,

but the variety differences concerning fertilizers are significant as the calculated F-ratio of 6 is more than its table value of 4.76.

ANOVA technique in context of two-way design when repeated values are there:

Separate independent measure of inherent or smallest variations

          For this measure , we can calculate the sum of squares and degrees of freedom in the same way as the sum of squares for variance within samples in the case of one-way ANOVA.

Total SS,

            SS between columns and SS between rows can also be worked out as stated above. We then find left-over sums of squares and left-over degrees of freedom which are used for what is known as ‘interaction variation’ (Interaction is the measure of inter relationship among the two different classifications).

After making all these computations, ANOVA table can be set up for drawing inferences

  We can illustrate the same with an example.

Step (i) T = 60, n= 12, therefore Correction factor = (T)2/n = 60*60/12 = 300 Step (ii) Total SS = ( 36+25+25+49+25+16+9+9+9+64+49+16) – (60*60/12) = 332-300 = 32 Step (iii)        SS between columns treatment = [24*24/4 + 20*20/4 + 16*16/4 ]– [60*60/12]                                                                                 = 144+100+64-300                                                                                = 8 Step (iv)   SS between rows treatment =[16*16/3 + 16*16/3 + 9*9/3 + 19*19/3] –[60*60/12]                                                                       = 85.33 + 85.33 + 27.00 + 120.33 – 300 = 317.99 - 300 = 17.99                                                                       = 18 Step (v)     SS residual or error   = Total SS – (SS between columns + SS between rows)                                                          = 32 – (8+18)                                                          = 6