ANOVA SNEHA.docx

1
GOVERNMENT NAGARJUNA P.G.
COLLEGE OF SCIENCE RAIPUR (C.G.)
M.Sc. MICROBIOLOGY
SEMESTER - 2
2022-23
WRITE UP ON
ANALYSIS OF VARIENCE {ANOVA}
SUBMITTED BY :- SUBMITTED
TO:-
SNEHA AGRAWAL DEPT.OF MICROBIOLOGY

2
INDEX
• INTRODUCTION
• HISTORY
• ASSUMPTIONS
• TYPES
• CALCULATION OF ONE WAY ANOVA
• CALCULATION OF TWO WAY ANOVA
• REFERENCE

3
ANOVA
INTRODUCTION
Test of significance for ‘t’ test is used to compare the mean values of
two samples. The difference in the means of more than two samples
or population cannot compared. The statistical technique used to
compare the means of variation of more than two samples for
population is called analysis of variance (ANOVA). ANOVA is based
on two types of variation:
a) Variations existing within the sample
b) Variations existing between the samples.
HISTORY
The concept of ANOVA was introduced by R.A.Fisher.
DEFINITIONS
ANOVA - analysis of variance is the synthesis of several ideas. It is
used for multiple purpose. As a consequence it is difficult to define
ANOVA concisely or. Precisely. ANOVA is a statistical test used to
analyse whether or not the means of several groups are equal.

4
F-value / F – ratio –the ratio between two variations is an indication
of sample difference. Therefore, ANOVA helps in estimating whether
more variations exist among the group means or within the groups.
The ratio of these two variation is denoted by ‘F’ . The F- value or F-
ratio is the indication of difference between the samples.
HYPOTHESIS
It is a definite statement about the population parameters.
1. NULL HYPOTHESIS :- It states that no association exists
between the two cross- tabulated variables in the population,
and therefore the variables are statistically independent.
For Instance, if we compare 2 methods method A and B for its
superiority and if the assumptions is that both methods are
equally good, the this assumptions are said to be null
hypothesis.
• H₀=μ₁=,μ₂=,μ₃=….=μĸ
• Where μ=group mean ,K=no. of group
2. ALTERNATIVE HYPOTHESIS :- It says that the 2 variables
are related in the populations. If we assume that from 2
methods A is superior than B then assumptions is called as
alternative hypothesis.

5
H₀:μ₁≠μ₂≠μ₃≠…≠μκ
SUM OF SQUARES (SS)
Sum of square is computed as the sum of squares of deviation of
the value for mean of the sample. It means
SS = 𝜮(𝒙 − 𝒙′)𝟐
Or
SS= 𝜮𝒙𝟐
−
[𝜮𝒙]𝟐
𝒏
Here, SS= represent sum of squares
x = represents mean of sample
x = represents value of variable x
n = represents total number of observation
MEAN SQUARE (MS)
Mean square (MS) is the mean of all the sum of squares. It is
obtained by dividing SS with appropriate degree of freedom(df)
MS = 𝑺𝒖𝒎 𝒐𝒇 𝒔𝒒𝒖𝒂𝒓𝒆/ 𝒅𝒆𝒈𝒓𝒆𝒆 𝒐𝒇 𝒇𝒓𝒆𝒆𝒅𝒐𝒎 =
𝑺𝑺
𝒅𝒇
DEGREE OF FREEDOM
The degree of freedom is equal to the sum of the individual degrees of
freedom for each sample since each sample has degree of freedom

6
equal to one less than sample size and there are k- sample, the total
degrees of freedom is k less than the total sample size
df = N – k
F -TEST STATISTIC
It is the ratio , after scaling by the degree of freedom.
𝐹 =
𝑀𝑆𝐵
𝑀𝑆𝑊
ASSUMPTIONS
1. All ANOVA require random sampling
2. The items for analysis of variance should be independent to each
other or mutually exclusive.
3. Equality or homogeneity of variances in a group of sample is
important.
4. The residual components should be normally distributed.
5. The variable of interest for each population or group has a normal
probability distribution.

7
PRINCIPLE OF ANOVA
The basic principle of ANOVA is to test for difference among the
means of the population by examining the amount of variation within
each of these samples, relative to amount of variation between the
samples. In terms of variation within the given population, it is
assumed that the values of (F) differ from the mean of this
population only because of random effects i.e. there are influencers
on (F) which are unexplainable, whereas in examining differences
between the mean of F population and grand means is attributable to
what is called a “specific factor” or what is technically described as
treatment effect.
Thus, while using ANOVA, we assume that each of the sample is
drawn from a normal population and that each of this population has
the same variance. We also assume that all factors other than the one
or more being tested are effectively controlled.This, in other words
means that we assume the absence of many factors that might affect
our conclusions concerning that factor to be studied.

8
TYPES OF ANOVA
Observation for analysis of variance are classified according to one
factor analysis or two factor criteria into two groups:-
1.ONE WAY ANALYSIS OF VARIANCE
Simplest type of analysis of variance is known as one way analysis of
variance. In one way ANOVA only one source of variation (or factor)
is investigated. But investigations are carried out in three or more
samples simultaneously. ANOVA is used to test the null hypothesis
that three or more treatments are equally effective.
Examples:- to compare the effectiveness of two or more drugs in
controlling or curing a particular disease.
COMPUTATION OF ONE-WAY ANALYSIS OF
VARIANCE (ANOVA)
For the computation of ANOVA the variance is estimated at two
different levels.
A) POPULATION VARIANCE BETWEEN THE GROUPS.
The variation due to the interaction between the samples, denoted
SSB for sum of squares between groups. If the sample means are
close to each other (and therefore the grand means) this will be

9
small. There are k samples involved with one data volume for each
sample (the sample mean) so there are k – 1 degrees of freedom.
The variance due to the interaction between the samples, denoted
MSB for mean square between groups. This is the between group
variation divided by its degree of freedom.
B)POPULATION VARIANCE WITHIN THE GROUP
The variation due to differences within the individual samples,
denoted SSW for sum of squares within groups. Each sample is
considered independently, snow interaction between samples is
involved. The degree of freedom is equal to the sum of individual
degrees of freedom for each sample. Since each sample has degree
of freedom equal to one less than their sample size, and there are k
samples, the total degrees of freedom is k less than the total sample
size
df= N – k
The variance due to the difference within individual samples,
denoted MSW for mean square within groups. This is the within
group variation divided by its degree of freedom.

10
PROCEDURE FOR CALCULATION OF ONE WAY ANOVA
i. Obtain the mean of each sample.
ii. Take out the average mean of the sample means.
iii. Calculate sum of squares for variance between the sample ( or
SS between)
iv. Obtain variance for mean square (MS) between samples.
v. Calculate sum of squares for variance within samples (or SS
within).
vi. Obtain the various or means square (MS) within samples.
vii. Find sum of squares or deviation for total variance
viii. Finally, find F- rrati.
All the calculation for determining variance between
and within the samples or group can be represented in a tabular
form. This table is called ANOVA table.

11
TABLE :- ANOVA TABLE: FOR ONE WAY CALCULATION.
S.No. Source of
variation
Sum of
squares
(SS)
Degree of
freedom
(df)
Mean of
squares
(MS)
Test
statistic
1 Between SSB(SB²) K – 1
𝑴𝑺𝑩
=
𝑺𝑺𝑩
𝒌 − 𝟏 𝑭 =
𝑴𝑺𝑾
𝑴𝑺𝑾
2 Within SSW(SW²) N- k
𝑴𝑺𝑾
=
𝑺𝑺𝑾
𝑵 − 𝒌
Examples :1
Q) The yield of 3 varieties of wheat A,B,C in 4 separated fields is
show in following table. Test of significance of different in the yields
of 3 varieties of wheat.

12
S.No.
Wheat
variety
Fields
Field 1 Field 2 Field 3 Field 4
1 A 12 18 14 16
2 B 19 17 15 13
3 C 14 16 18 20
STEP 1:- CALCULATION OF SAMPLE MEANOF EACH OF THE
SAMPLE
S.No. Fields Wheat variety
A B C
1 1 12 19 14
2 2 18 17 16
3 3 14 15 18
4 4 16 13 20
Σx1= 60 Σx2 = 64 Σx3 = 68
STEP2 :- 𝑥1 = 60/4 = 15
x2 = 64/4 = 16
x3 = 68/4 = 17
STEP :- 3 GRAND SIMPLE MEAN

13
X = 𝑥1 + x2 + X3 /3
= 15+16+17/3
=16
STEP 4 :- CALCULATE VARIANCE OF SUM OF SQUARE
BETWEEN THE GROUP.(SSB)
= Σni (X1 – x)2
= n1(X1 – x)2
+ n2 (X2 – x)2
+ n3 (X3 – x)2
=
4(15-16) 2
+ 4(16-16) 2
+ 4(17-16) 2
= 4+ 0+ 4
= 8
Here, n = total number of observation.
STEP 5:- DEGREE OF FREEDOM
Degree of freedom = k – 1
= 4 – 1 = 3
= 3-1 = 2
MSB = SSB / (k – 1)
= 8 / 2 = 4

14
STEP 6:- CALCULATE THE VARIANCE OF SUM OF SQUARE
WITHIN THE GROUP (SSW).
SSW = (x1–x”)²
TABLE :- CALCULATION OF SUM OF SQUARE WITHIN
SAMPLES.
Wheat varietysample 1 Wheat variety
sample 2
Wheat variety
sample 3
X1 (x1 – x1)² x2 (x2 – x2)² x3 (x3 – x3)²
12 (12-15)²= 9 19 (19-16)²=9 14 (14-17)²=9
18 (18 -15)²=9 17 (17-16)²=1 16 (16-17)²=1
14 (14-15)²=1 15 (15-16)²=1 18 (18-17)²=1
16 (16-15)²=1 1 (13-16)²=9 20 (20-17)²=9
Total 20 20 20
SSW = Σ(x1-x1)² + Σ(x2-x2)² + Σ(x3-x3)²
= 20 + 20 + 20
= 60
d.f. = N- k

15
= 12-3
= 9
MEAN SQUARE = SSW + (N-k)
= 60/9
= 6.7
STEP :- 7 ANOVA TABLE: ONE WAY ANOVA TABLE
S.No. Source
of
variation
Sum of
square
Degree
of
freedom
Mean
square
Test
statistic
1 Between
sample
8 2 4 F = 6.7/4
= 1.57
2 Within
sample
60 9 6.7
TOTAL 68 (SST) N-1=11

16
CONCLUSION
Here, F(calculated) is < F(tabulated) at 0.05 (2.l9)
i.e. F(calculated) = 1.67 < 4.26
The value of F is less than 4.26 at 5% level with degree of freedom
being V1 = 2 and V2 = 9 and hence could have arisen due to chance.
There is no significant difference in the yield of three wheat
varieties.
Hence, this analysis supports the the null hypothesis of no difference
in sample mean.
2.TWO WAY ANALYSIS OF VARIANCE
An extension to the one-way analysis of variance. There are two
independent variables. There are three sets of hypothesis with the two
way ANOVA. The first null hypothesis is that there is no interaction
between the two factors. The second null hypothesis is that the
population means of the first factor equal. The third null hypothesis
is that the population means of the second factor are equal.

17
Example:- productivity of several nearby fields influenced by seed
varieties and types of manure used.
COMPUTATION OF TWO WAY ANALYSIS OF VARIANCE
( ANOVA).
The two-way analysis of variance is an extension to the the one-way
analysis of variance. There are two independent variables (hence the
name two way). Two way ANOVA is used to study the impact of two
different factors on a specific variable of the population such as
effect of different varieties of seeds and different types of fertilizers
on crop yield in several nearby field.
ASSUMPTIONS OF TWO WAY ANOVA
i. The two factors under study are independent i.e. they do not
interact.
ii. Variable under study has normal distribution.
iii. Selection of variable and population is at random.
PROCEDURE OF CALCULATION OF TWO WAY ANOVA

18
The total variance for sum of squares of variance in two way
ANOVA includes three parts:-
a. Sum of squares between the column = SSC
b. Sum of square between the rows = SSR
c. Sum of square for the residuals = SSE
Therefore, sum of square of total variance= SST
i. SST = SSC + SSR + SSE , and
T = Σx1+ Σx2 + … + Σxk
ii. Correlation factor (CF) =
𝑻𝟐
𝑵
iii. SSC =
𝜮(𝜮𝜲𝒄)𝟐
𝒏𝒄
−
𝑻
𝑵
Here, nc= number of units in each column
iv. SSR =
𝜮(𝜮𝜲𝒓)𝟐
𝒏𝒓
−
𝑻𝟐
𝑵
v. SST = Σx²1+ Σx²2 + … + Σx²k−
𝑻𝟐
𝑵
Here, nr= number of units in each row
vi. SSE = SST- (SSC + SSR)
vii. Total number of degree of freedom
(N- 1) = Cr – 1
d.f between column = C-1
d.f between row. = r- 1
d.f between residuals = (C-1)(r-1)

19
TABLE :- ANOVA TABLE: FOR TWO WAY ANOVA
CALCULATION.
S.No.
Source of
variation
Sum of
squares
SS
Degree
of
freedom
d.f
Mean square
MS
Total
Statistic
1
Between
column
SSC C-1
MSS=SSC+C-
1
𝐹 =
𝑀𝑆𝐶
𝑀𝑆𝐸
2
Between
Rows
SSR r-1
MSR=SSR+
r-1
𝐹 =
𝑀𝑆𝑅
𝑀𝑆𝐸
3 Residual SSE
(C-1)(r-
1)
MSE=SSE
Total SST N-1
Example 1
Q. 1) In the the following table the crop yield of three varieties of
crop from 4 different fields is shown, test whether (i) the mean yield
of these varieties are equal and so also, (ii) the quality of field mean.

20
Variety of crops Field
I II III IV
A 9 5 9 7
B 7 9 5 5
C 7 8 4 9
CALCULATION
STEP 1:- Substract 5 from each value and arrange the
data into columns and rows showings fields or blocks
and varieties that represent factors.
Variety of
crop
Fields
I II III IV Total of
rows
A 4 0 4 2 R1= 10
B 2 4 0 0 R2= 6
C 2 3 -1 4 R3= 8
Total of
columns
C1 =8 C2= 7 C3=3 C4=6 R=24

21
Here, T = 24
N = 12
Correlation factor (F) =
𝑇2
𝑁
=
(24)2
12
=
576
12
= 48
STEP -2 SSC = SUM IF SQUARE BETWEEN COLUMNS (
VARIETY OF CROP)
=
(𝛴𝐶1)2
𝑛𝑐
+
(𝛴𝐶2)2
𝑛𝑐
+
(𝛴𝐶3)2
𝑛𝑐
+
(𝛴𝐶4)2
𝑛𝑐
−
(𝑇)2
𝑁
=
(8)2
3
+
(7)2
3
+
(3)2
3
+
(6)2
3
−
(24)2
12
= 21.3 + 16.3 + 3 + 12 − 48
= 52.6 − 48
= 4.6
d.f. = (C-1) = 4-1
= 3
STEP :- 3 SSR = SUM OF SQUARE BETWEEN ROWS

22
SSR =
(𝛴𝑅1)2
𝑛𝑟
+
(𝛴𝑅2)2
𝑛𝑟
+
(𝛴𝑅3)2
𝑛𝑟
−
(𝑇)2
𝑁
=
(10)2
4
+
(6)2
4
+
(8)2
4
−
(𝑇)2
𝑁
= (25)2
+ (9)2
+ (8)2 𝑇2
𝑁
= 50 − 48 = 2
Degree of freedom = (r – 1)
= 3 – 1
= 2
STEP :- 4 SST = TOTAL SUM OF SQUARE
= (16 + 0 +16 + 4 + 4 +16 + 0 +0 + 4 + 9 + 1 + 16) – 48
= 86 – 48
= 38
Degree of freedom = N – 1
= 12 – 1
= 11
STEP 5 :- SSE = SST – SSC + SSR
= 38 – 4.6 + 2
= 31.4
Degree of freedom = (C – 1) (r – 1)

23
= (4-1) (3-1)
= 3 × 2
= 6
CONCLUSION
F(tabulated) is compared with F(calculated) at 5% for (C-1),
(r-1)(C-1) i.e. F(3,6) and for (r-1), (r-1)(C-1) i.e. F(2,6).
F(calculated) for F(3,6) = 0.29
SOURCE OF
VARIATION
SUM OF
SQUARES
DEGREE
OF
FREEDOM
MEAN
SQUARE
(MS)
VARIANC
E RATIO
BETWEEN
THE
COLUMN
(VARIETIES
)
2 2 2/2 = 1.00 F=
5.23/1.00
= 5.23
BETWEEN
THE ROW
(FIELDS)
4.6 3 4.6/3 = 1.53 F=5.23/
1.53 = 3.42
RESIDUAL 31.4 6 31.4/6 = 5.23
TOTAL 38.0 11

24
F(calculated) for F (2,6) = 0.19
And, F(tabulated) for F(3,6)= 4.76
F(tabulated) for F(2,6)= 5.14
F (calculated) < F( tabulated)
Hence, F(calculated) is less than F(tabulated), so the null hypothesis
is accepted.
REFERENCE
 BIOSTATISTICS BOOK BY VEER BALA RASTOGI
 BIOSTATISTICS BOOK BY P.N. ARORA & P.K. MALHAN

ANOVA SNEHA.docx

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to ANOVA SNEHA.docx

Similar to ANOVA SNEHA.docx (20)

More from SNEHA AGRAWAL GUPTA

More from SNEHA AGRAWAL GUPTA (20)

Recently uploaded

Recently uploaded (20)

ANOVA SNEHA.docx