Answer:
Test value = 3.49
Step-by-step explanation:
We are given the following in the question:
[tex]S_1 = \$ 23, n_1 = 12\\S_2 = \$ 43, n_2 = 8[/tex]
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
[tex]H_{0}: \sigma_1^2 = \sigma_2^2\\H_A: \sigma_1^2 \neq \sigma_2^2[/tex]
We use F-test to determine the difference in variation between the two samples.
Formula:
[tex]F_{stat} = \displaystyle\frac{S_2^2}{S_1^2}\\\\(As ~S_2 > S_1)[/tex]
Putting all the values, we have
[tex]F_{stat} = \displaystyle\frac{1849}{529} = 3.49[/tex]
Now, [tex]F_{critical} \text{ at 0.05 level of significance, degree of freedom(8-1,12-1)} = 3.01[/tex]
Comparison:
[tex]F_{stat} > F_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, there is variation in advertising cost of salon and the advertising cost of nails.