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Solved examples on Analysis of Variance (ANOVA)
One Way ANOVA
If there are three samples ๐
1
2
and ๐
3
having values as follows:
๐
๐
๐
11
21
31
12
22
32
13
23
33
To test the null hypothesis ๐ป 0
1
2
3
, we compute the following values.
๐๐
๐ ๐
Total Sum
๐๐
2
๐ ๐
2
Total sum of squares of
variation
๐ 1 ๐
2
1
๐ 2 ๐
2
2
๐ 3 ๐
2
3
2
๐
๐
Sum of square of variation
between the columns
Sum of squares of
variation within the
columns
ANOVA TABLE
Source of
variation
Sum of
squares
Degrees of freedom Mean squares ๐ญ
Between Samples ๐๐๐ถ
1
1 ๐น
๐๐๐
Within samples
(Errors)
2
2
We use the following table to find the table value of ๐น at ๐ 1
2
degrees of freedom.
If ๐น
๐๐๐
- 05
1
2
), then we accept the null hypothesis (i.e., ๐
1
2
3
) at 5% LOS.
Otherwise, we reject the null hypothesis.
Challenge 0 2 : To test the significance of variation of retail prices of a commodity in three
principal cities: Mumbai, Kolkata and Delhi, four shops were chosen at random
from each city and prices observed in rupees were as follows:
Do the data indicate that the prices of the commodity in the three cities are
significantly different.
Solution: Null Hypothesis ๐ป 0
1
2
3
Totalโ 58 48 38
i. ๐ = โ โ ๐ฅ
๐ ๐ ๐๐
ii. ๐๐๐ =
๐๐
2
๐ ๐
๐
2
๐
144
2
12
iii. ๐๐๐ถ =
(โ ๐ฅ
๐ 1 ๐
)
2
๐
1
(โ ๐ฅ
๐ 2 ๐
)
2
๐
2
(โ ๐ฅ
๐ 3 ๐
)
2
๐
3
๐
2
๐
58
2
4
48
2
4
38
2
4
iv. ๐๐๐ธ = ๐๐๐ โ ๐๐๐ถ = 136 โ 50 = 86.
ANOVA TABLE
Source of
variation
Sum of
squares
Degrees of freedom Mean squares ๐ญ
Between Samples ๐๐๐ถ = 50
1
1
๐๐๐
Within samples
(Errors)
2
= 3 ร 3 = 9
2
Now, ๐น
- 05
( 2 , 9 ) = 4. 26. Here ๐น
๐๐๐
- 05
We accept the null hypothesis. Hence, the prices of the commodity in the three
cities are not significantly different.
More problems on One Way ANOVA.
- Three samples each of size 5 were drawn and the following data was observed. Test the
hypothesis that the population mean are equal.
Challenge 0 1: A farmer applies three types of fertilisers on 4 separate plots. The figures on yield
per acre are tabulated as following.
Find out if the plots are materially different and if the fertilisers make any
material difference.
Solution: Null hypothesis ๐ป 0
: a) Plots do not differ materially
b) The fertilisers do not differ materially.
i. ๐ = โ โ ๐ฅ
๐ ๐ ๐๐
ii. ๐๐๐ =
๐๐
2
๐ ๐
๐
2
๐
84
2
12
iii. ๐๐๐ถ =
(โ ๐ฅ ๐ 1 ๐
)
2
๐
1
(โ ๐ฅ ๐ 2 ๐
)
2
๐
2
(โ ๐ฅ ๐ 3 ๐
)
2
๐
3
(โ ๐ฅ ๐ 4 ๐
)
2
๐
4
๐
2
๐
21
2
3
15
2
3
24
2
3
24
2
3
iv. ๐๐๐
=
(โ ๐ฅ
๐ ๐ 1
)
2
๐
1
(โ ๐ฅ
2 ๐ 2
)
2
๐
2
(โ ๐ฅ
๐ ๐ 3
)
2
๐
3
๐
2
๐
24
2
4
28
2
4
32
2
4
v. ๐๐๐ธ = ๐๐๐ โ ๐๐๐ถ โ ๐๐๐
= 36 โ 18 โ 8 = 10.
ANOVA TABLE
Source of
variation
Sum of
squares
Degrees of freedom Mean squares ๐ญ
Between Samples ๐๐๐ถ = 18
1
1
๐
๐
Between the rows ๐๐๐
= 8
2
2
Within samples
(Errors)
2
3
Now, ๐น
- 05
= 4. 76 and ๐น
- 05
Here ๐น
๐
- 05
( 3 , 6 ) and ๐น
๐
- 05
( 2 , 6 ). Hence we accept the null hypothesis.
Plots
Fertilisers
๐จ ๐ฉ ๐ช ๐ซ Total
More problems:
- Set up a two-way ANOVA table for the following per hectare yield for 4 varieties of
yield of rice on 3 plots.
Yield
Plot
