The analysis of variance(ANOVA) is used to
determine whether there is any statistical
significant difference between the means of
three or more independent(unrelated) group.
One wayANOVA uses one independent
variable while two-wayANOVA uses two
independent variable.

 Developed by R.A Fisher in 1920.
 Also known as F-test , which is based on F-
distribution( G.W. Snedecor ).
 It compares the means between the groups
and determine whether any of those means
are statistically significantly different from
each other
 It determines whether all groups are taken
from common population or not.

 ANOVA is a ratio between “mean sum of squares between
(MSSB) and “mean sum of squares within (MSSW) ”.
F=MSSB/MSSW =Between variance/within variance
 The variation among the observations of each specific group
is called its internal variation and the totality of the internal
variation is called variability within groups.
 The totality of variation from one group to another i.e
variation to group is called variability between group.

 within
 group
between group
GROU
PA
19
25
32
58
59
94
Group
B
14
27
39
51
66
70
Group
C
20
22
33
50
52
55

 It test the null hypothesis
Ho :there is no significant difference between the
means of all groups.(all groups are same)
Ho: µ1= µ2= µ3=……= µk
Where µ=group mean k=number of groups
 It test the alternative hypothesis
HA : there are at least two group means that are
statistically significantly different from each other
Ho: µ1 ≠ µ2 ≠ µ3 ≠ …… ≠ µk

1. Random selection
2. Normal distribution
3. Homogeneity of variance
4. Additivity of variance

Source of
variance
Degree of
freedom(df)
Sum of
square(ss)
Mean of sum
of
square(MSS)
F Calculated F
Tabulated
at 5% and
1%
Between
(factors)
k-1 SSB MSSB
=SSB/(k-1)
Within
(error)
N-k SSW MSSW
=SSW/(N-K)
MSSB/MSSW
Total N-1 TSS

1. Correction factor(c)
2. Total sum of square(SST)
3. Between sum of square(SSB)
4. Within sum of squares(SSW)
5. Preparation of analysis of variation table
6. Degree of freedom(df)
7. Variance estimates/mean sum of
squares(MSS)
8. F-ratio
9. interpretation

An experiment is conducted to study the effectiveness of three
methods lecture ,questions-answers and library method . In
each group four students assigned randomly . The obtained
scores are given in the following table . Is there significant
difference among three methods of teaching?
Lecture Questions-answers library
x1 x2 x3
4 9 2
5 10 4
1 9 2
2 6 2

1. Correction Factor(c):-
c = (∑X)2/N
=T2/N
=(56)2/12
C=261.33
2.Total sum of square(SST):-
SST= ∑(Xi)2-c
=41+298+28-261.33
=110.67

3. Between sum of squares(SSB):-
SSB={(∑X1) 2+(∑X2) 2+(∑X3)2/n}-c
=(144+1156+100)/4-261.33
=88.67
4.Within sum of squares(SSW):-
SSW=SST-SSB
=110.67-88.07
=22.00

5.
Source of
variance
Degree of
freedom(df)
Sum of
square(ss)
Mean of sum
of
square(MSS)
F Calculated F
Tabulated
at 5% and
1%
Between
(factors)
k-1 SSB MSSB
=SSB/(k-1)
Within
(error)
N-k SSW MSSW
=SSW/(N-K)
MSSB/MSSW
Total N-1 TSS

Source of
variance
Degree of
freedom(df)
Sum of
square(ss)
Mean of sum
of
square(MSS)
F Calculated F
Tabulated
at 5% and
1%
Between
(factors)
2 88.67 88.67/2
=43.33
Within
(error)
9 22.00 22/9
=2.44
43.33/2.44
=18.16
Total 11 110.67

9.Interpretation of F-ratio
F(2,9)=18.16
F -> F ratio
2 -> df for variance between groups
9 -> df for variance within groups
18.16 -> value of F ratio
 The obtained F value is greater than the table value at 0.01
level of significance.The null hypothesis is rejected at both
the level of significance.
 Hence,it can be interpreted that there is higher significant
difference among these three methods of teaching
F value with df (2,9)
At 0.05=4,46
At 0.01=8.02

 When you have collected data about one
categorial independent variable and one
quantitative dependent variable.The
independent variable should have at least
three levels(i.e at least three different groups
or categories)
 ANOVA tells you if the dependent variable
changes according to the level of the
independent variable.

 It overcomes the limitation of T-test(not
applicable on more than two group).
 An ANOVA controlTYPE I Error remains at
5% .
 ANOVA evaluates both between and within
variance and calculates its ratio

 It will tell you that at least two groups were
different from each other. But it will not tell
you which groups were different.

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