ANALYSIS OF
VARIANCE
(F-RATIO TEST)
TWO WAY
CLASSIFICATION
This test is designed for more than
two groups of objects studies to see if
each group is affected by two different
experimental conditions.
This test is used when the
number of observation in the
subclasses are equal.
Formulas for Two Way ANOVA
For Equal and Proportionate Entries in the
Subclasses
Source of
Variation
Sum of Squares Degrees
of
Freedom
Mean
Square
F-
value
Row
R – 1
Column
C – 1
Interaction
(R–1)(C–1)
Within cells
RC (n-1)
Total
nRC – 1
Formula for Mean Square
Example
An agricultural experiment was
conducted to compare the yields of
three varieties of rice applied by
two types of fertilizer. The
following table represents the yield
in grams using eight plots.
Types of
Fertilizer
Varieties of Rice
V1 V2 V3
t1
26 14 41 82 36 87
41 16 26 86 39 99
28 29 19 45 59 126
92 31 59
37
27 104
t2
51 35 39 114 42 133
96 36 104 92 92 124
97 28 130 87 156 68
22 76 122 64 144
142
Hypothesis
1. There is no difference in the yields of the
three varieties of rice.
2. The two types of fertilizer does not
significantly affect the yields meaning
the yield is not dependent of the type of
fertilizer used.
3. There is no significant interaction
between the variety of rice and the types
of fertilizer used.
Summary of the Data
(Sum of all entries in each cell)
Types of
Fertilizer
Varieties of Rice Total
V1 V2 V3
t1 277 395 577 1249
t2 441 752 901 2094
Total 718 1147 1478 3343
Sum of Squares Computations

Source of
Variation
Sum of
Squares
Degree of
Freedom
Mean
Square
F-Value
Rows 14, 875.52 1 14,875.52 14.64
Columns 18,150.04 2 9,075.02 8.93
Interaction 1,332.04 2 666.02 0.656
Within Cells 42,667.38 42 1,015.89
TOTAL 77,024.98 47
Interpretations
1.For the different varieties of rice, we have Fc=8.93 with 2 df
associated with the numerator and 42 df with the denominator.
The values required for significance at 5% and 1% levels are 3.22
and 5.15, respectively. We conclude that the different varieties of
rice differ significantly in their yields.
2.For the different types of fertilizer, we have Fr=14.64 with 1 df
associated with the numerator and 42 df with the denominator.
The value required for the significance at 5% and 1% levels are
4.072 & 7.287 respectively. We conclude therefore that the
different types of fertilizer affect significantly the yields of rice.
3. For significant interaction, we have Frc=0.656 which is lower
than the table value. Therefore hypothesis number three is
accepted.
This method is applied in two way
ANOVA where the number of observations in
the subclasses or cell frequency is unequal.
The data is to be adjusted by the method of
unweight mean. This method is in effect the
analysis of variance applied to the means of
the subclasses. The sum of the squares for
rows, columns, and interaction are then
adjusted using the harmonic mean.

Formulas for Two Way ANOVA
For Unequal Frequency in the Subclasses
Source of
Variation
Sum of Squares
Degrees
of
Freedo
m
Mean
Square
F-
value
Row R – 1
Column C – 1
Interaction (R–1)(C–1)
Within cells N-RC
Example
The following table shows of
factitious data for a two
way
classification experiment with two
levels of one factor and three
levels of the other factor.
C1 C2 C3
R1
7
6
6
2
8
12
17
19
16
17
13
14
4 3 16 21 10
24 22
R2
23 22 11 26 9 16
14 26 15 14 27 17
9 18 26 13 31 18
31 42 20
C1 C2 C3
R1
N=6
T=28
N=8
T=139
N=5
T=70
R2
N=6
T=112
N=7
T=136
N=8
T=180
Summary of the Data
Computation for Harmonic Mean:

Means of each cell and other
Computations of the data
C1 C2 C3 Total
R1 4.67 17.38 14 36.05
R2 18.67 19.43 22.5 60.6
TOTAL 23.34 36.81 36.5 96.65
Sum of Squares Computations

Source of
Variation
Sum of
Squares
Degree of
Freedom
Mean
Square
F-Value
Rows 650.92 1 650.92 13.97
Columns 383.1 2 191.55 4.11
Interaction 231.85 2 115.93 2.49
Within Cells 1,584.25 34 46.60
Interpretation: In this factitious data, the row effect
is significant at .01 level of significance, the column
effect is also significant at .05 level while the
interaction effect is not significant.
THANK YOU


Analysis of Variance (F-ratio Test) Two Way Classification

  • 1.
  • 2.
    This test isdesigned for more than two groups of objects studies to see if each group is affected by two different experimental conditions.
  • 3.
    This test isused when the number of observation in the subclasses are equal.
  • 4.
    Formulas for TwoWay ANOVA For Equal and Proportionate Entries in the Subclasses Source of Variation Sum of Squares Degrees of Freedom Mean Square F- value Row R – 1 Column C – 1 Interaction (R–1)(C–1) Within cells RC (n-1) Total nRC – 1
  • 5.
  • 6.
    Example An agricultural experimentwas conducted to compare the yields of three varieties of rice applied by two types of fertilizer. The following table represents the yield in grams using eight plots.
  • 7.
    Types of Fertilizer Varieties ofRice V1 V2 V3 t1 26 14 41 82 36 87 41 16 26 86 39 99 28 29 19 45 59 126 92 31 59 37 27 104 t2 51 35 39 114 42 133 96 36 104 92 92 124 97 28 130 87 156 68 22 76 122 64 144 142
  • 8.
    Hypothesis 1. There isno difference in the yields of the three varieties of rice. 2. The two types of fertilizer does not significantly affect the yields meaning the yield is not dependent of the type of fertilizer used. 3. There is no significant interaction between the variety of rice and the types of fertilizer used.
  • 9.
    Summary of theData (Sum of all entries in each cell) Types of Fertilizer Varieties of Rice Total V1 V2 V3 t1 277 395 577 1249 t2 441 752 901 2094 Total 718 1147 1478 3343
  • 10.
    Sum of SquaresComputations 
  • 11.
    Source of Variation Sum of Squares Degreeof Freedom Mean Square F-Value Rows 14, 875.52 1 14,875.52 14.64 Columns 18,150.04 2 9,075.02 8.93 Interaction 1,332.04 2 666.02 0.656 Within Cells 42,667.38 42 1,015.89 TOTAL 77,024.98 47
  • 12.
    Interpretations 1.For the differentvarieties of rice, we have Fc=8.93 with 2 df associated with the numerator and 42 df with the denominator. The values required for significance at 5% and 1% levels are 3.22 and 5.15, respectively. We conclude that the different varieties of rice differ significantly in their yields. 2.For the different types of fertilizer, we have Fr=14.64 with 1 df associated with the numerator and 42 df with the denominator. The value required for the significance at 5% and 1% levels are 4.072 & 7.287 respectively. We conclude therefore that the different types of fertilizer affect significantly the yields of rice. 3. For significant interaction, we have Frc=0.656 which is lower than the table value. Therefore hypothesis number three is accepted.
  • 13.
    This method isapplied in two way ANOVA where the number of observations in the subclasses or cell frequency is unequal. The data is to be adjusted by the method of unweight mean. This method is in effect the analysis of variance applied to the means of the subclasses. The sum of the squares for rows, columns, and interaction are then adjusted using the harmonic mean.
  • 14.
  • 15.
    Formulas for TwoWay ANOVA For Unequal Frequency in the Subclasses Source of Variation Sum of Squares Degrees of Freedo m Mean Square F- value Row R – 1 Column C – 1 Interaction (R–1)(C–1) Within cells N-RC
  • 16.
    Example The following tableshows of factitious data for a two way classification experiment with two levels of one factor and three levels of the other factor.
  • 17.
    C1 C2 C3 R1 7 6 6 2 8 12 17 19 16 17 13 14 43 16 21 10 24 22 R2 23 22 11 26 9 16 14 26 15 14 27 17 9 18 26 13 31 18 31 42 20
  • 18.
  • 19.
  • 20.
    Means of eachcell and other Computations of the data C1 C2 C3 Total R1 4.67 17.38 14 36.05 R2 18.67 19.43 22.5 60.6 TOTAL 23.34 36.81 36.5 96.65
  • 21.
    Sum of SquaresComputations 
  • 22.
    Source of Variation Sum of Squares Degreeof Freedom Mean Square F-Value Rows 650.92 1 650.92 13.97 Columns 383.1 2 191.55 4.11 Interaction 231.85 2 115.93 2.49 Within Cells 1,584.25 34 46.60 Interpretation: In this factitious data, the row effect is significant at .01 level of significance, the column effect is also significant at .05 level while the interaction effect is not significant.
  • 23.