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A test was performed to estimate the level of distress of a sample of male patients...

A test was performed to estimate the level of distress of a sample of male patients and a sample of patients shortly before performing the same surgery. The sizes of the samples and the variances calculated from the obtained scores

They are the following:

Men: n = 13, s2 = 150

Women: n = 21, s2 = 275

Do these data provide sufficient evidence to indicate that, in the populations represented, the scores obtained by women are more variable than those obtained by men? Note: the probability of not rejecting true H0 equal to 0.90 and the probability of rejecting false H0 equal to 0.95.

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Answer #1

Solution: Women n, = 21 S1 275 Men n2- 13 s2 150 Null Hypothesis Alternative Hypothesis: HaUT > Critical Value: Degrees of freedom for numerator-dh Degrees of freedom for numerator-df2 21-1 20 13-1 = 12 Given that the probability of not rejecting true Ho is equal to 0.90 Hence, probability of rejecting true Ho is equal to 1 0.90 0.10 Hence, level of significance a 0.10 From F-distribution table, the critical value of ‘F, having area in the right tail equal to α = 0.10 and for df1 = 20 and df 12, is given by: 2.060 Hence, critical value-Fcritical- - 2.060Test statistic: S 275 1.833 s 150 Decision Since the value of test statistic (F = 1.833) is lower than the critical value of 2.060, we fail to reject the null hypothesis. Conclusion: Since the null hypothesis cannot be rejected, data does NOT provide sufficient evidence to indicate that, in the populations represented, the scores obtained by women are more variable than those obtained by men

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