Statistical
Model of
&
Analysis
Lecture - 24
The
Yijk is the yield of ith level of factor A and jth level of factor B in kth block.
μ is the overall effect.
(
∑(
that's
A factorial experiment is not considered as an experimental design because of the fact that the
basic designs namely, CRD, RCBD, and LSD are used to carry out the factorial experiments. For this purpose of analysis of variance, the basic sums of squares are computed in a usual manner, with addition that the treatment sum of squares is further partitioned into component parts of main effects and interaction effects.
Analysis
Let Yijk
Add the observations in each cell denoted by
Where:
The data for blocks can be arranged as:
The following term is called total variation:
ANOVA Table:
|
S.V |
d.f |
S
S |
M
S |
F |
|
Block |
r - 1 |
SS
Block |
MS Block |
|
|
Treatment |
t - 1 |
SST |
MST |
F |
|
|
||||
|
A |
a
- 1 |
SSA |
MSA |
FA |
|
B |
b
- 1 |
SSB |
MSB |
FB |
|
AB |
(a
– 1) (b – 1) |
SSAB |
MSAB |
FAB |
|
Error |
(r
– 1)(ab- 1) |
SSE |
MSE |
|
|
Total |
rab - 1 |
SS
Total |
|
|
Where a = 2 & b = 2
Conventional procedure
of ANOVA to compute various sums of squares:
Error will be
equivalent to Interaction sum of squares in table – II:
SSAB = SST - SSA - SSB
Where SST in table - II
Alternative procedure
to compute sums of squares:
The simplest form of 2^2 factorial experiment is given as follows:
|
SV |
d.f |
SS |
MS |
F |
|
|
A |
1
|
SSA |
MSA |
FA |
|
|
B |
1
|
SSB |
MSB |
FB |
|
|
AB |
1
|
SSAB |
MSAB |
FAB |
|
|
Error |
4
(r – 1) |
SSE |
MSE |
|
|
|
Total |
4 r - 1 |
SS
Total |
|
|
|
The interpretation of factorial experiment depends on
interaction effect(s). if the interaction effect is significant then there is
no real mean whether the main effect is significant or not.
Example
- 1:
A 2^2
Write the statistical model and perform ANOVA and test the
significance of varieties and manures?
Method – I:
Solution: The given 2^2
The effect of four treatment combinations (T1 = a0b0, T2 = a0b1, T3 = a1b0, and T4 = a1b1) is considered as treatment and setup
hypothesis as:
i. H0 :The effect of four treatments is non significant. Vs. H0: The effect of four treatments is significant.
a. H01: The effect of factor A is non significant Vs. H11: The effect of factor A is significant.
b. H02: The effect of factor B is non significant Vs. H12: The effect of factor A is non significant.
c. H03: There interaction b/w factor A & B is zero. Vs. H13: There interaction effect is not zero
ii. The significance level; α = 0.05
iii. The test statistic:
v.
Computation:
ANOVA Table:
|
SV |
d.f |
S.S |
MS |
F |
|
Block |
2 |
18.50 |
9.25 |
|
|
Treatment |
3 |
38.25 |
12.75 |
4.36 |
|
A |
1 |
36.75 |
|
12.58 |
|
B |
1 |
0.75 |
|
0.25 |
|
AB |
1 |
0.75 |
|
0.25 |
|
Error |
1 |
17.50 |
|
|
|
Total |
11 |
74.25 |
|
|
The calculated F value falling in the
acceptance region, it is concluded that treatments (factors combination) are
insignificant at
- Read more:Solved Questions
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
No comments:
Post a Comment