Testing the hypothesis equality of two variances
The test for equality of two population variances is based upon the variances in two independently chosen random samples drawn from two normal populations
Beneath the null hypothesis σ_{1}^{2} = σ_{2}^{2}
F = (S^{2}_{1}/ σ_{1}^{2})/ (S^{2}_{2}/ σ_{2}^{2}) here under the H_{0 }: it follows that
F = S^{2}_{1}/ S^{2}_{2} which is the test statistic.
This follows F - distribution along with V_{1} and V_{2} degrees of freedom. The larger sample variance is placed in the numerator and the smaller one in the denominator
If the computed value of F exceeds the table value of F, we reject the null hypothesis that is the alternate hypothesis is accepted
Illustration
In one sample of observations the sum of the squares of the deviations of the sample values from sample mean was 120 and in another sample of 12 observations it was 314. Test where the difference is significant at 5 percent level of significance
Solution
Given that n1 = 10, n2 = 12, Σ(x_{1} - x¯_{1} )^{2} = 120
Σ(x_{2} - x¯_{2} )^{2} = 314
Assume that take the null hypothesis that the two samples are drawn from the similar normal population of equal variance
H_{0} : s_{1}^{2 }= s_{2}^{2}
H_{1}: s_{1}^{2 }≠ s_{2}^{2}
Applying F test that is
F = S^{2}_{1}/ S^{2}_{2}
=
= (120/9)/(314/11)
= 13.33/28.55
As the numerator should be greater than denominator
F = 28.55/13.33 = 2.1
The table value of F at 5 percent level of significance for V_{1} = 9 and V_{2} = 11. Because the calculated value of F is less than the table value, we accept the hypothesis. The samples may have been drawn from the two populations having the similar variances.