F - test
z
Dr. R. MUTHUKRISHNAVENI
SAIVA BHANU KSHATRIYA COLLEGE,
ARUPPUKOTTAI
z
F – test
 The F-test(Variance Ratio Test) is developed by
Statistician R.A. Fisher
 The object of the test is find out whether the two
random independent samples variance differ
significantly or not
 Formula for F-test , F =
𝑆12
𝑆22
; 𝑆12 > 𝑆22
 Table value for degrees of freedom(v1,v2) at 5% or 1
% level of significance
z
F- Test
𝑆12
𝑆22

F=

𝑆12

𝑆22

Degrees of freedom v1 = 𝑛1 − 1; v2= 𝑛2 − 1
= Estimate of Variance of sample1 =
(𝑋1 − 𝑋1 )2
𝑛1 −1
= Estimate of Variance of sample1 =
(𝑋2 − 𝑋2 )2
𝑛2 −1
z
Assumptions

Normality – Data are normally distributed

Homogeneity – variance within each group are equal

Independence of error –the error is independent for each group
z
Illustration - 1

In a sample of 8 observations, the sum of squared deviations of
items from the mean was 84.4. In another samples of 10
observations, the value was found to be 102.6. Test whether the
difference is significant at 5% level.
V1 = 7 & V2 = 9 degrees of
freedom at 5% level of
significance F0.05 = 3.29
z
Solution

Let us take hypothesis that the difference in the variance of two
samples is not significant H0:𝜎12 = 𝜎22

n1= 8 (𝑋1 − 𝑋1 )2 = 84.4 n2 = 10

𝑆12

𝑆22

=
(𝑋1 − 𝑋1 )2 84.4 84.4
=
=
= 12.06
8−1
7
𝑛1 −1
=
(𝑋2 − 𝑋2 )2
𝑛1 −1
F=
𝑆12
𝑆22
=
12.6
11.4
=
102.6
10−1
= 1.06
=
102.6
9
= 11.4
(𝑋2 − 𝑋2 )2 = 102.6
z
Solution - continue

Calculated value of F = 1.06

Table value of F = 3.29 (v1 = 7
and v2 = 9 at 5% level of
significance)

Inference

CV < TV, hence we accept the
null hypothesis and conclude
that the difference in the
variance of two samples is not
significant at 5%
z
Illustration 2

The following data present the yields in Kilogram of common 10
subdivision of equal area of two agricultural plots
Plot
A
620
570
650
600
630
580
570
600
600
580
Plot
B
560
590
560
570
580
570
600
550
570
550

Test whether two samples taken from two random populations
have same variance.(5% point of F for V1=9 and V2=9 is 3.18)
z
Solution

Let us take the null hypothesis that the samples come from
populations having the same variance H0:𝜎12 = 𝜎22 applying Ftest

F=

𝑆12
𝑆12
𝑆22
=
(𝑋1 − 𝑋1 )2
𝑛1 −1
;
𝑆22
=
(𝑋2 − 𝑋2 )2
𝑛1 −1
z
Calculation of sample variance
𝑋1
(𝑋1 − 𝑋1 ) (𝑋1 − 𝑋1 )2
𝑋2
(𝑋2 − 𝑋2 ) (𝑋2 − 𝑋2 )2
620
20
400
560
-10
100
570
-30
900
590
20
400
X1=
650
50
2500
560
-10
100
=
600
O
0
570
0
0
630
30
900
580
10
100
X2=
580
-20
400
570
0
0
=
570
-30
900
600
30
900
600
0
0
550
-20
400
600
0
0
570
0
0
580
-20
400
550
-20
400
6000
0
6400
5700
0
2400
𝑋1
𝑁
6000
10
60
𝑋2
𝑁
5700
10
57
𝑜𝑟 𝑆𝑢𝑚 𝑜𝑟 𝑇𝑜𝑡𝑎𝑙
z
Solution - Continue



𝑆12
=
𝑆22
F=
(𝑋1 − 𝑋1 )2
𝑛1 −1
=
(𝑋2 − 𝑋2 )2
𝑛1 −1
𝑆12
𝑆22
=
711.11
266.27
=
6400
10−1
=
2400
10 −1
= 2.67
= 711.11
= 266.67
z
Solution - continue

Calculated value of F = 2. 67

Table value of F = 3.18 (v1 = 9
and v2 = 9 at 5% level of
significance)

Inference

CV < TV, hence we accept the
null hypothesis and conclude
that the samples come from
populations having the same
variance at 5%