Past exam of the options II course of the University of Oxford / 2013 / O4 / e

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- Solution e

Solution

At a true variance

σ^{2} > 0

,

S / σ^{2} \sim χ_{n}^{2}

. Therefore the power function is

w (σ^{2}) = 1 - F_{n} (\frac{σ _{0}^{2}}{σ ^{2}} F_{n}^{- 1} (1 - α)) .

The argument of

F_{n}

decreases as

σ^{2}

increases. Since the chi-squared density is positive on the positive half-line, the power is continuous and strictly increasing, with

w (0^{+}) = 0, w (σ_{0}^{2}) = α, w (\infty) = 1.

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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