A stockbroker at critical securities reported that the ...
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 [pic] and [pic] be the variances of the return of oil and utility stocks respectively.
Here we want to test the null hypothesis H0: [pic] = [pic] against the alternative hypothesis H1: [pic] > [pic].
Let [pic]and [pic] be the sample standard deviation of the return of oil and utility stocks respectively.
The test statistic for testing H0 is
[pic], the F distribution with [pic] degrees of freedom.
From the given data,
[pic]= 10, [pic]= 8, [pic] = 3.9, [pic] = 3.5
Thus the calculated value of [pic] = 1.2416
Since the significance level is 0.05, using the F distribution with [pic]= (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 [pic]= (9, 7) degrees of freedom, we get the P-value as
P-value = P[ F > 1.2416] = 0. 3964
Since the level of significance[pic]= 0.05, we reject the null hypothesis [pic] 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
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