A car salesman claims that the variance of prices on convertibles is higher than the variance of prices on station wagons. The standard deviation of the list price on 16 convertibles is $6800 and the standard deviation on 24 station wagons is $3900. What should the test value be?
A) 1.74
B) 3.04
C) 2.25
D) 1.53

Question

Quizplus · Accepted Answer

We can use the F-test to compare the variances of the two groups. 
The test value can be calculated as F = s1^2/s2^2, where s1 and s2 are the sample standard deviations of the two groups. 
In this case, F = (6800)^2/(3900)^2 = 3.038. 
The degrees of freedom for the numerator is n1-1 = 15 and the degrees of freedom for the denominator is n2-1 = 23. 
Looking at an F distribution table with 15 and 23 degrees of freedom, we can see that the critical value for a 0.05 level of significance is 2.24. 
Since our calculated F-value (3.038) is greater than the critical value (2.24), we can reject the null hypothesis and conclude that the variance of prices on convertibles is indeed higher than the variance of prices on station wagons. 
Therefore, the test value is 3.04, which corresponds to choice B.

A Car Salesman Claims That the Variance of Prices on Convertibles