Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/o4/e/solution

At a true variance , . Therefore the power function is
The argument of decreases as increases. Since the chi-squared density is positive on the positive half-line, the power is continuous and strictly increasing, with
These features determine the requested rough sketch. The curve passes through the size at the null value and approaches one for very large variance.
Figure 1.
Power rises from zero through the test size to one as the variance increases
. Original illustrative power curve for n=4 and alpha=0.05, plotted against the ratio of the true variance to the null variance. The marked point has ratio one and power alpha. Details.

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