Completely Randomized Design Analysis along with LSD test
Problem Statement
In order to find out the yielding abilities of 5 varities of sesamum an experiment was conducted in the green house using a CRD with 4 pots per variety. The data for Yield (in g) is given below.
|
Varieties |
||||
|
V1 |
V2 |
V3 |
V4 |
V5 |
|
10 |
12 |
20 |
30 |
17 |
|
8 |
15 |
21 |
32 |
19 |
|
12 |
14 |
25 |
34 |
18 |
|
11 |
13 |
24 |
29 |
20 |
F value (0.05,4,15) = 3.05 t value (0.05,15) = 2.13
F value (0.01,4,15) = 4.89 t value (0.01,15) = 2.94
Work out the CRD analysis and perform LSD test if required. Interpret the results.
Solution:
Step 1: Calculate Correction Factor
Grand Total = Sum of all observation = 10 + 8 +12 + . . . + 18 + 20 = 384
𝐶𝐹 = 𝐺𝑇2/r*t
= 3842/4 ∗ 5
= 7372.8
Step 2: Calculate Total SS
Total SS = Sum of Square of all observations – CF
= 102 + 82 + 122+ . . . + 182 + 202 − 7372.8 = 1127.20
Step 3: Calculate Treatment SS
|
V1 = 41 |
V2 = 54 |
V3 = 90 |
V4 = 125 |
V5 = 74 |
Treatment SS = (1/r)* [ 𝑇2 + 𝑇2 + 𝑇2 + 𝑇2 + 𝑇2] − 𝐶𝐹
= (1/4) * [ 412 + 542 + 902 + 1252 + 742 ] − 7372.8
= 1076.7
Step 4: Calculate Error SS
Error SS = Total SS – Treatment SS
= 1127.20 - 1076.7 = 50.50
Step 5: Calculate degree of freedom
Treatment DF = t – 1 = 5 - 1 = 4
Total DF = r*t – 1 = 5*4 – 1 = 19
Error DF = Total DF – Treatment DF = 19 – 4 = 15
Step 6: Calculate Mean Square for all sources
Treatment MS = Treatment SS / Treatment DF
= 1076.7 / 4
= 269.175
Error MS = Error SS / Error DF
= 50.50 / 15
= 3.37
Step 7: Calculate F value
Calculated F = Treatment MS / Error MS
= 269.175 / 3.37
= 79.95
Step 8: Prepare ANOVA
|
Source of Variation |
Degrees of Freedom |
Sum of Squares |
Mean Square |
Calculated F value |
|
Treatment |
4 |
1076.70 |
269.18 |
79.95 |
|
Error |
15 |
50.50 |
3.37 |
- |
|
Total |
19 |
1127.20 |
- |
- |
The null hypothesis is 
which mean that mean performance of all the treatments is same.
In order to check this null hypothesis we are required to compare the Calculate F value (79.95) with Table F value (3.05). As the Calculated F value is greater than Table F value our results are significant at 5 % level of significance. Further we can test for 1 % level by comparing Calculate F value (79.95) with Table F value (4.89), so result is also significant at 1 % level of significance, and we reject our Null hypothesis.
This mean that mean performance of all the treatment is not same. This raises the question that which treatment gives a better Yield. In order to get this answer, we need to perform LSD test.
LSD TEST (FOR MEAN COMPARISON)
Step 1: Calculate CD (critical difference) value
Step 2: Calculate the treatment means and arrange them is order
The treatment mean is obtained by dividing treatment total with number of replications
|
V4 = 31.25 |
V3 = 22.50 |
V5 = 18.50 |
V2 = 13.50 |
V1 = 10.25 |
Step 3: Allocate same letters to treatment that are at par
V4 mean – CD = 31.25 – 3.82 = 27.43. The means ranging from 31.25 to 27.43 are at par with V4 but none of the means are in this range. So, V4a.
V3 mean – CD = 22.50 – 3.82 = 18.68. The means ranging from 22.50 to 18.68 are at par with V3 and share letter b but none of the mean are in this range. So, V3b.
V5 mean – CD = 18.50 – 3.82 = 14.68. The means ranging from 18.50 to 14.68 are at par with V5 and share letter c but none of the mean are in this range. So, V5c.
V2 mean – CD = 13.50 – 3.82 = 9.68. The means ranging from 13.50 to 9.68 are at par with V2 and share letter d and V1 (10.25) is in this range. So, V2d and V1d .
Calculation of CV
Conclusion
The ANOVA results reveal that the treatment component is significant at 1 % level of significance. The LSD test reveals that highest yield was observed for treatment V4 and none of the treatments was at par with it.
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Quiz on Completely Randomized Design
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