Answer:
Critical F value = 4.9503
Explanation:
Given that:
The sample size of the numerator = 7
The sample size of the denominator = 6
The degree of freedom for the numerator df = n -1
The degree of freedom for the numerator df = 7 - 1
The degree of freedom for the numerator df = 6
The degree of freedom for the denominator df = n - 1
The degree of freedom for the denominator df = 6 - 1
The degree of freedom for the denominator df = 5
The assume that the test is two tailed and using a level of significance of ∝ = 0.10
The significance level for the two tailed test = 0.10/2 = 0.05
From the standard normal F table at the level of significance of 0.05
Critical F value = 4.9503