1. ONE WAY ANOVA
Dr. R. MUTHUKRISHNAVENI
SAIVA BHANU KSHATRIYA
COLLEGE, ARUPPUKOTTAI
2. ANOVA
ANOVA – Analysis of Variance is statistical techniques specially designed to test
whether the means of more than two quantitative populations are equal
It was developed by R.A. Fisher in 1920’s
Eg.
Production variability of same type of more than two plant(machinery)
Mark variability of more than two group of students in a common test
Yield variability of more than two plots of agricultural land
3. ASSUMPTIONS
Normality – Data are normally distributed
Homogeneity – variance within each group are equal
Independence of error –the error is independent for each group
4. TECHNIQUE OF ANOVA
For greater clarity the technique of ANOVA has been discussed
for separately for
One way classification
Two classification
5. ANOVA – ONE WAY
Set a hypothesis H0: µ1= µ2= µ3=…. µk or H1: µ1≠ µ2 ≠ µ3 ≠ …. µk
Calculate variance between the samples
Calculate variance within the samples
Calculate the ratio F =
𝑆1
2
𝑆2
2 or
𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
Compare the calculated value of F with Table value of F the degrees
of freedom V1 = c – 1 and V2 = n – c at a 5% 0r 1% level of
significance
1. Calculate the mean of each samples
2. Calculate the grand average(X double bar)
3. Take the difference between mean of various sample and the
grand average
4. Square these deviations and obtain the sum(of square
between the sample) and divided by c-1 or k-1
1. Calculate the mean of each sample
2. Take the deviation of various items in a sample from the
mean value
3. Square these deviations and obtain sum(of deviation within
the sample) divided by n-c or n-k
6. ONE WAY ANOVA - INFERENCE
Source of
Variation
SS(Sum of
Squares)
V(Degrees of
Freedom
MS(Mean
Square)
Variance Ratio
of F
Between
Samples
SSC V1 = c – 1 MSC =
𝑆𝑆𝐶
𝐶 −1 𝑀𝑆𝐶
𝑀𝑆𝐸Within Samples SSE V2 = n – c MSE =
𝑆𝑆𝐸
𝑛 − 𝑐
Total SST n – 1
CV < TV, the null hypothesis is accepted
CV > TV, the null hypothesis is rejected
7. ILLUSTRATION - 1
To assess the significance of possible variation in performance in a
certain test between the convent schools of a city, a common test was
given to a number of students taken at random from the senior fifth
class of each of the four schools concerned, the results are given
below. Make an analysis of variance of data
Schools
A B C D
8 12 18 13
10 11 12 9
12 9 16 12
8 14 6 16
7 4 8 15
8. SOLUTION
Let us take the hypothesis that there is no significance difference in
performance in a certain common test among convent schools of a
certain city H0: µA= µB= µC = µD
Calculation of sample means and Grand mean
Sample means
Sample A → XA =
8+10+12+8+7
5
=
45
5
= 9
Sample B → XB =
12+11+9+14+4
5
=
50
5
= 10
Sample C → XC =
18+12+16+6+8
5
=
60
5
= 12
Sample D → XD =
13+9+12+16+15
5
=
65
5
= 13
Grand Mean X=
𝑋 𝐴+𝑋 𝐵+𝑋 𝑐+𝑋 𝐷
4
=
9+10+12+13
4
=
44
4
= 11
9. Variance between samples
Sum of (Sample mean – Grand mean)2/C -1
Sample A
(XA – X)2
(9-11)2
Sample B
(XB – X)2
(10-11)2
Sample C
(XC – X)2
(12-11)2
Sample D
(XD – X)2
(13-11)2
4 1 1 4
4 1 1 4
4 1 1 4
4 1 1 4
4 1 1 4
20 5 5 20
Sum of square
between the
samples =
20+5+5+20 = 50
MSC =
𝑆𝑆𝐶
𝐶−1
=
50
3
=16.7
10. Variance within the samples
Sum of (individual Sample value – sample mean)2/n – c
XA (XA – XA)2 XB (XB – XB)2 XC (XC – XC)2 XD (XD – XD)2
8 1 12 4 18 36 13 0
10 1 11 1 12 0 9 16
12 9 9 1 16 16 12 1
8 1 14 16 6 36 16 9
7 4 4 36 8 16 15 4
16 58 104 30
Sum of square within the
samples = 16+58+104+30
= 208
MSE =
𝑆𝑆𝐸
𝑛 −𝑐
=
208
16
= 13
12. Inference
The table value for F for V1 = 3 and V2 = 16 at 5% level of
significance = 3.24 . The CV of F is less than the TV, Hence there is
no significance difference in performance in a certain common test
among convent schools of a certain city (the samples could have
come from the same universe)
Source of
Variation
SS(Sum of
Squares)
V(Degrees of
Freedom
MS(Mean
Square)
Variance Ratio
of F
Between
Samples
50 3 16.7
16.7
13.0
= 1.285
Within Samples 208 16 13.0
Total 258 19
