The problem does not state that a most powerful test is to be used, but assume that is the case. The Neyman-Pearson Lemma states that for a (used here for alpha) in the null hypothesis, reject Ho if the ratio of the likelihood function (here equal to the density function due to a sample of size one) at the null to the likelihood function at the alternative is small.
The numerator is .5(10000^.5)(10000+x)^-1.5 and the denominator is a(10000^a)(10000+x)^(-a-1). The ratio is
..5(1/a)(10000^(.5-a))((10000+x)^(a-.5)). Taking logs gives
log(.5) - log(a) + (.5-a)[log(10000) - log(10000+x)] < k as the rejection region. Move the first two terms to the right and then divide by .5-a (which must be positive since a is in the alternative hypothesis) to get
log(10000) - log(10000+x) < k. Then move the first term to the right, change the sign, and exponentiate to get
10000 + x > k or x > k as the rejection region.
For x = 9,600,000, the p-value is Pr(X > 9,600,000 given a = .5) = 0.032 and so reject at .05, but not at .02.
You don't really need to go through all this work. For a sample of size 1 the only reasonable rejection regions are x < k and x > k. The alternative is a small alpha which implies large values of x and so x > k is the rejection region to use.
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Stuart Klugman, FSA, PhD
Principal Financial Group Professor of Actuarial Science
Drake University
2507 University Avenue
Des Moines, IA 50311 USA
ph: 515-271-4097
e-mail: Stuart.Klugman@drake.edu
Drake Act. Sci. web site: www.drake.edu/cbpa/acts