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A stockbroker at critical securities reported that the mean rate of return on a sample of 10 oil stocks was 12.6 percent with a standard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 with a standard deviation of 3.5 percent . At .05 significance level can we conclude that there is more variation in the oil stocks? Let 12 and 22 be the variances of the return of oil and utility stocks respectively. Here we want to test the null hypothesis H0: 12 = 22 against the alternative hypothesis H1: 12 > 22 . Let s1 and s2 be the sample standard deviation of the return of oil and utility stocks respectively. The test statistic for testing H0 is F s12 s22 F( n1 1,n2 1) , the F distribution with (n1 1, n2 1) degrees of freedom. From the given data, n1 = 10, n2 = 8, s1 = 3.9, s2 = 3.5 Thus the calculated value of F 3.92 = 1.2416 3.52 Since the significance level is 0.05, using the F distribution with (n2 1, n1 1) = (9, 7) degrees of freedom, we get the critical value = 3.667 Thus the critical region of the test is F > 3.667 That is we reject the null hypothesis if F > 3.667. Here, F = 1.2416 < 3.667 So we fail to reject the null hypothesis Ho. Thus at .05 significance level we cannot conclude that there is more variation in the oil stocks P-value approach Using the F distribution with (n2 1, n1 1) = (9, 7) degrees of freedom, we get the P-value as P-value = P[ F > 1.2416] = 0. 3964 Since the level of significance = 0.05, we reject the null hypothesis H 0 if the P-value < 0.05. Here the P-value = 0. 3964 > 0.05 So we do not reject the null hypothesis Ho. Thus at .05 significance level we cannot conclude that there is more variation in the oil stocks