<p>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?</p><p>2 2 Let s1 and s 2 be the variances of the return of oil and utility stocks respectively. 2 2 Here we want to test the null hypothesis H0: s1 = s 2 against the alternative hypothesis 2 2 H1: s1 > s 2 .</p><p>Let s1 and s2 be the sample standard deviation of the return of oil and utility stocks respectively.</p><p>The test statistic for testing H0 is s2 F= 1 : F (n- 1, n - 1) 2 (n1- 1, n 2 - 1) , the F distribution with 1 2 degrees of freedom. s2 From the given data, n1 = 10, n2 = 8, s1 = 3.9, s2 = 3.5</p><p>3.92 Thus the calculated value of F = = 1.2416 3.52</p><p>Since the significance level is 0.05, using the F distribution with (n2- 1, n 1 - 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><p>P-value approach</p><p>Using the F distribution with (n2- 1, n 1 - 1) = (9, 7) degrees of freedom, we get the P-value as P-value = P[ F > 1.2416] = 0. 3964</p><p>Since the level of significancea = 0.05, we reject the null hypothesis H0 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</p>