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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

Row

R–1

Column

C–1

Interaction

(R–1)(C–1)

Within cells

RC (n-1)

Total

nRC – 1

Mean
Square

Fvalue

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

t1
t2

V2

V3

26 14
41 16
28 29
92 31
51 35
96 36
97 28
22 76

41 82
26 86
19 45
59 37
39 114
104 92
130 87
122 64

36 87
39 99
59 126
27 104
42 133
92 124
156 68
144 142
Hypothesis
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.
1.
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
Freedom

Row

R–1

Column

C–1

Interaction

(R–1)(C–1)

Within cells

N-RC

Mean
Square

Fvalue
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

R1

R2

7 6
6 2
4 3

23 22
14 26
9
18

C2
8
12
16
24

17
19
21
22

11 26
15 14
26 13
31

C3
16
17

10

9
27
31
42

13
14

16
17
18
20
Summary of the Data
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
Computation for Harmonic Mean:

Means of each cell and other
Computations of the data

C1
C2
C3
R1 4.67 17.38 14
R2 18.67 19.43 22.5
TOTAL 23.34 36.81 36.5

Total
36.05
60.6
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 

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ANOVA 2-WAY Classification

  • 2. This test is designed for more than two groups of objects studies to see if each group is affected by two different experimental conditions.
  • 3. This test is used when the number of observation in the subclasses are equal.
  • 4. Formulas for Two Way ANOVA For Equal and Proportionate Entries in the Subclasses Source of Variation Sum of Squares Degrees of Freedom Row R–1 Column C–1 Interaction (R–1)(C–1) Within cells RC (n-1) Total nRC – 1 Mean Square Fvalue
  • 5.
  • 6. 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.
  • 7. Types of Fertilizer Varieties of Rice V1 t1 t2 V2 V3 26 14 41 16 28 29 92 31 51 35 96 36 97 28 22 76 41 82 26 86 19 45 59 37 39 114 104 92 130 87 122 64 36 87 39 99 59 126 27 104 42 133 92 124 156 68 144 142
  • 8. Hypothesis 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. 1.
  • 9. 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
  • 10. Sum of Squares Computations 
  • 11. 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
  • 12. 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.
  • 13. 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.
  • 14.
  • 15. Formulas for Two Way ANOVA For Unequal Frequency in the Subclasses Source of Variation Sum of Squares Degrees of Freedom Row R–1 Column C–1 Interaction (R–1)(C–1) Within cells N-RC Mean Square Fvalue
  • 16. 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.
  • 17. C1 R1 R2 7 6 6 2 4 3 23 22 14 26 9 18 C2 8 12 16 24 17 19 21 22 11 26 15 14 26 13 31 C3 16 17 10 9 27 31 42 13 14 16 17 18 20
  • 18. Summary of the Data 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
  • 20. Means of each cell and other Computations of the data C1 C2 C3 R1 4.67 17.38 14 R2 18.67 19.43 22.5 TOTAL 23.34 36.81 36.5 Total 36.05 60.6 96.65
  • 21. Sum of Squares Computations 
  • 22. 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.