Question 11.APP.5: The Normal Case (continued) Testing Further Hypotheses About......
The Normal Case (continued) Testing Further Hypotheses About the Mean In reference to Example 11, T(x) = x and Q(θ) is strictly increasing. Therefore, for testing H_{0}\colon\theta\leq\theta_{1}\ \mathrm{or}\ \theta\geq\theta_{2} against H_{A}\colon\theta_{1}\,\lt \,\theta\,\lt \,\theta_{2} at level of significance α, the test to be employed is the one given by (11) and (12)–(13),
\varphi(x_{1},\dots,x_{n})= \begin{cases} 1 & {\mathrm{if~}}C_1\lt V(x_1,\dots,x_n)\lt C_2\\ \gamma_1 & {\mathrm{if~}}V(x_1,\dots,x_n)=C_1 \\\gamma_2 & {\mathrm{if~}}V(x_1,\dots,x_n)=C_2 \\ 0 & {\mathrm{otherwise,}} \end{cases} \qquad(11)
E_{\theta_{2}}\varphi(X_{1}{,\,\dots\,,\,X_n})= P_{\theta_{2}}[C_{1}\lt V(X_{1},\,\dots\,,X_{n})\lt C_{2}] +\gamma_{1}P_{\theta_{2}}[V(X_{1},\dots,X_{n})=C_{1}] +\gamma_{2}P_{\theta_{2}}[V(X_{1},\ldots,X_{n})=C_{2}]=\alpha.\qquad(13)
which here becomes (\gamma_{1}=\gamma_{2}=0);
\varphi(x_{1},\dots,x_{n})= \begin{cases} 1 & {\mathrm{if~}} C_1\leq\textstyle\sum_{i=1}^{n}x_{i}\leq C_2\\ 0 & {\mathrm{otherwise,}} \end{cases} \qquad(35) \\ E_{\theta_{1}}\varphi(X_{1},\dots, X_{n})=P_{\theta_{1}}{\Biggl(}C_{1}\leq\sum\limits_{i=1}^{n}X_{i}\leq C_{2}{\Biggr)}=\alpha,(36)
{ E}_{\theta_{2}}\varphi(X_{1},\dots,X_{n})=P_{\theta_{2}}\!\left(C_{1}\leq\sum\limits_{i=1}^{n}X_{i}\leq C_{2}\right)=\alpha,and \textstyle\sum_{i=1}^{n}X_{i}\sim N(n\theta_{i},n\sigma^{2}),\,i=1,2.
For the purpose of utilizing the Normal tables, (36) are rewritten thus:
The power of the test is calculated as follows:
\pi_{\varphi}(\theta)=\Phi\!\left(\frac{C_{2}-n\theta}{\sigma\!\sqrt{n}}\right)-\Phi\!\left(\frac{C_{1}-n\theta}{\sigma\!\sqrt{n}}\right),\quad(\theta_{1}\lt \theta\lt \theta_2).\qquad\qquad(38)~~~N u m e r i c a l\:Ex amp l e In reference to Example 4, suppose n=25,\theta_{1}= 1,\theta_{2}=3 ,~\mathrm {and}~\alpha=0.01. For simplicity, let us take σ = 1.
Here n\theta_{1}=2{5},\,n\theta_{2}=7{5}, and (37) yields:
\Phi\biggl({\frac{C_{2}-25}{5}}\biggr)-\Phi\biggl({\frac{C_{1}-25}{5}}\biggr)=\Phi\biggl({\frac{C_{2}-75}{5}}\biggr)-\Phi\biggl({\frac{C_{1}-75}{5}}\biggr)=0.01.\ \ \ (39)
Placing the four quantities {\frac{C_{1}-75}{5}},\;{\frac{C_{2}-75}{5}},\;{\frac{C_{1}-25}{5}},\;\mathrm{and}\;{\frac{C_{2}-25}{5}} under the N(0, 1) curve, we observe that relation (39) obtains only for:
\frac{C_{1}-25}{5}=-\frac{C_{2}-75}{5}\quad\mathrm{and}\quad\frac{C_{2}-25}{5}=-\frac{C_{1}-75}{5},
which imply that C_{1}+C_{2}=100.~{\mathrm{Setting}}\,C_{1}=C, we have then that C_{2}=100-C, and (39) gives:
\Phi\bigg(\frac{75-C}{5}\bigg)-\Phi\bigg(\frac{C-25}{5}\bigg)=0.01.\quad\quad\quad\quad\quad\quad\quad(40)
From the Normal tables, we find that (40) is closely satisfied for C = 36.5. So C_{1}\,=\,{3}{6.5}\,{\mathrm{and~hence}}\,\,C_{2}\,=\,6{3}{\cdot{{5}}}, and the test rejects the hypothesis \textstyle H_{0} whenever \textstyle\sum_{i=1}^{25}x_{i} is between 36.5 and 63.5 and accepts it otherwise.
The power of the test, calculated through (38), is, for example, for θ = 2.5 and θ = 2:
\pi_{\varphi}(1.5)=\pi_{\varphi}(2.5)=0.57926~~~\mathrm{and}~~~\beta_{\varphi}(2)=0.99307.