Question 11.8: On the basis of a random sample of size 1 from the p.d.f. f ......
On the basis of a random sample of size 1 from the p.d.f. f(x;\theta)=1+\theta^{2}({\frac{1}{2}}- x),\;0\lt x\lt 1,\;-1\leq\theta\leq1 :
(i) Use the Neyman–Pearson Fundamental Lemma to derive the MP test for testing the hypothesis H_{0}\colon\theta\ =\ 0 (i.e., the p.d.f. is U(0, 1)) against the alternative H_{A}\colon\theta=\theta_{1} at level of significance α.
(ii) Investigate whether or not the test derived in part (i) is a UMP test for testing H_{0}\colon\theta=0 against the alternative H_{A}^{\prime}\colon\theta\neq0.
(iii) Determine the test in part (i) for α = 0.05.
(iv) Determine the power of the test in part (i).
First, the function given is a p.d.f., because it is nonnegative and \textstyle\int_{0}^{1}[1+\theta^{2}(\frac{1}{2}-x)]d x=1+\theta^{2}(\frac{1}{2}-\frac{1}{2})=1. Next:
(i) \textstyle H_{0} is rejected whenever 1+\theta_{1}^{2}({\frac{1}{2}}-x)\;\gt \;C^{*}, or x < C, where {\textstyle C=\frac{1}{2}}-(C^{*}-1)/\theta_{1}^{2}, and C is determined by P_{0}(X\lt C)=\alpha, so that C = α, since X ∼ U(0, 1) under \textstyle H_{0}. Thus, \textstyle H_{0}. is rejected when x < α.
(ii) Observe that the test is independent of \theta_{1}, and since it is MP against each fixed \theta_{1}, it follows that it is UMP for testing \textstyle H_{0} gainst H_{A}^{\prime}\colon\theta\neq0.
(iii) For α = 0.05, the test in part (i) rejects \textstyle H_{0} whenever x < 0.05.
(iv) For θ ≠ 0, the power of the test is:
\pi(\theta)=P_{\theta}(X\lt \alpha)=\int_{0}^{\alpha}\!\left[1+\theta^{2}\left({\frac{1}{2}}-x\right)\right]d x={\frac{1}{2}}\alpha(1-\alpha)\theta^{2}+\alpha.
Thus, e.g., \pi(\pm1)=\textstyle{\frac{1}{2}}\alpha(1-\alpha)+\alpha,\ \pi(\pm{\textstyle{\frac{1}{2}}})=\textstyle{\frac{1}{8}}\alpha(1-\alpha)+\alpha, which for α = 0.05 become: \pi({\pm}1)\simeq0.074,\pi(\pm\frac {1}{2}){\simeq}\,0.056.