SOLUTION: **Please help** 12-3 Arbitron Media Research Incorporated conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time.

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Question 1202979: **Please help** 12-3
Arbitron Media Research Incorporated conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for a sample of 8 men was 44 minutes per day. The standard deviation was 9 minutes per day. The mean listening time for a sample of 9 women was also 44 minutes, but the standard deviation of the sample was 24 minutes. At the 0.02 significance level, can we conclude that there is a difference in the variation in the listening times for men and women?
b. State the decision rule. (Round your answer to 2 decimal places.)
c. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
d. Compute the p-value. (Round your answer to 4 decimal places.)

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
with the f-test, the sample with the higher value of variance is the numerator and the other is the denominator.
you have:

n1 = 8
dof1 = 7
m1 = 44
sd1 = 9
v1 = 9^2

n2 = 9
dof2 = 8
m2 = 44
sd2 = 24
v2 = 24^2

sample 2 has the greatest variatiion, so it's the numerator.
the test f-value is 24^2 / 9^2 = 7.11.
the critical f-value is 8/7 = 5.32733457 at .02 level of significance.

the test f-value is greater than the critical f-value so the conclusion is that the variances are not equal.

here's a refeence.

http://www.statistics4u.com/fundstat_eng/cc_test_2sample_ftest.html

here's a critical f-stat calcjulator.

https://www.danielsoper.com/statcalc/calculator.aspx?id=4