F-test is named after the more prominent analyst R.A. Fisher. F-test is utilized to test whether the two autonomous appraisals of populace change contrast altogether or whether the two examples may be viewed as drawn from the typical populace having the same difference. For doing the test, we calculate F-statistic is defined as:

Formula

F=Larger estimate of population variancesmaller estimate of population variance=S12S22 where S12>S22F=Larger estimate of population variancesmaller estimate of population variance=S12S22 where S12>S22

Procedure

Its testing procedure is as follows:

1.     Set up null hypothesis that the two population variance are equal. i.e. H0:σ12=σ22H0:σ12=σ22

2.     The variances of the random samples are calculated by using formula:

S21=∑(X1−X¯1)2n1−1, S22=∑(X2−X¯2)2n2−1S12=∑(X1−X¯1)2n1−1, S22=∑(X2−X¯2)2n2−1

3.     The variance ratio F is computed as:

F=S12S22 where S12>S22F=S12S22 where S12>S22

4.     The degrees of freedom are computed. The degrees of freedom of the larger estimate of the population variance are denoted by v1 and the smaller estimate by v2. That is,

1.     v1v1 = degrees of freedom for sample having larger variance = n1−1n1−1

2.     v2v2 = degrees of freedom for sample having smaller variance = n2−1n2−1

5.     Then from the F-table given at the end of the book, the value of FF is found for v1v1 and v2v2 with 5% level of significance.

6.     Then we compare the calculated value of FF with the table value of F.05F.05for v1v1 and v2v2 degrees of freedom. If the calculated value of FF exceeds the table value of FF, we reject the null hypothesis and conclude that the difference between the two variances is significant. On the other hand, if the calculated value of FF is less than the table value, the null hypothesis is accepted and concludes that both the samples illustrate the applications of F-test.

Example

Problem Statement:

In a sample of 8 observations, the entirety of squared deviations of things from the mean was 94.5. In another specimen of 10 perceptions, the worth was observed to be 101.7 Test whether the distinction is huge at 5% level. (You are given that at 5% level of centrality, the basic estimation of FF for v1v1= 7 and v2v2 = 9, F.05F.05 is 3.29).

Solution:

Let us take the hypothesis that the difference in the variances of the two samples is not significant i.e. H0:σ12=σ22H0:σ12=σ22

We are given the following:

n1=8,∑(X1−X¯1)2=94.5,n2=10,∑(X2−X¯2)2=101.7,S21=∑(X1−X¯1)2n1−1=94.58−1=94.57=13.5,S22=∑(X2−X¯2)2n2−1=101.710−1=101.79=11.3n1=8,∑(X1−X¯1)2=94.5,n2=10,∑(X2−X¯2)2=101.7,S12=∑(X1−X¯1)2n1−1=94.58−1=94.57=13.5,S22=∑(X2−X¯2)2n2−1=101.710−1=101.79=11.3

Applying F-Test

F=S12S22=13.511.3=1.195F=S12S22=13.511.3=1.195

For v1v1 = 8-1 = 7, v2v2 = 10-1 = 9 and F.05F.05 = 3.29. The Calculated value of FFis less than the table value. Hence, we accept the null hypothesis and conclude that the difference in the variances of two samples is not significant at 5% level.