Q:

A business major wants to determine whether the variation in advertising costs of hair salons is different from the variation in advertising costs of nail salons. He surveys several businesses and finds the standard deviation in monthly advertising costs is $23 for 12 hair salons, and $43 for 8 nail salons. What is the test value for this hypothesis test? Test value:

Accepted Solution

A:
Answer:Test value = 3.49Step-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 = 0.05First, 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.