Arbitron Media Research Inc conducted a study of the iPod li

Arbitron Media Research Inc. 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 men was 26 minutes per day. The standard deviation of the sample of the 10 men studied was 11 minutes per day. The mean listening time for the 12 women studied was also 26 minutes, but the standard deviation of the sample was 15 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women? (Round your answer to 3 decimal places.)

Arbitron Media Research Inc. 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 men was 26 minutes per day. The standard deviation of the sample of the 10 men studied was 11 minutes per day. The mean listening time for the 12 women studied was also 26 minutes, but the standard deviation of the sample was 15 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women? (Round your answer to 3 decimal places.)

Arbitron Media Research Inc. 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 men was 26 minutes per day. The standard deviation of the sample of the 10 men studied was 11 minutes per day. The mean listening time for the 12 women studied was also 26 minutes, but the standard deviation of the sample was 15 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women? (Round your answer to 3 decimal places.) The test statistic is . Decision (Click to select)Do not rejectReject

Solution

Men: Mean = 26 minutes, n = 10, s = 11.

Women: Mean = 26 minutes, n = 12, s = 15

Test statistic = F = 15^2/11^2 = 1.8595

DF in numerator = 14, DF in denominator = 10.

The critical value in the right tail can be obtained from Excel, by using the formula: =F.INV.RT(0.05,14,10), which gives a critical value of 2.864728.

The critical value in the left tail can be obtained from Excel by using the formula: =F.INV(0.05,14,10), which gives a critical value of 0.384297.

Since the test statistic of 1.860 is neither greater than the critical value in the right tail nor less than the critical value in the left tail, we fail to reject Ho.

Answer: Critical value = 1.860, Decision: Do not reject Ho.