Question 11.APP.1: The Binomial Case In reference to Example 9 with n = 1, we h......
The Binomial Case In reference to Example 9 with \textstyle n=1 , we have T(x)=x where x=0,1, and then in Theorem 2, V(x_{1},\ldots,x_{n})= \textstyle\sum_{i=1}^{n}\,x_{i}\,\,\,\mathrm{and}\,\,{V}(X_{1},\dots,X_{n})\ =\ \textstyle\sum_{i=1}^{n}\,X_{i}\ \sim\ B(n,\theta). Since Q(\theta)~=~\log(\frac{\theta}{1-\theta}) is strictly increasing, consider relations (5) and (6), which, for testing the hypothesis H_{0}\colon\theta\leq{{\theta}}_{0} against the alternative H_{A}\colon\theta\gt \theta_{0}, become here:
\varphi(x_{1},\dots,x_{n}) = \begin{cases} 1 & \mathrm{if~}~~~\textstyle\sum_{i=1}^{n} x_i\gt C \\ \gamma & {\mathrm{if~}}~~~\textstyle\sum_{i=1}^{n} x_i=C \\ 0 & {\mathrm{if~}}~~~\textstyle\sum_{i=1}^{n} x_i\lt C, \end{cases}\qquad(19) \\ E_{\theta_{0}}\varphi(X_{1},\dots, X_{n})=P_{\theta_{0}}(X\gt C)+\gamma P_{\theta_{0}}(X=C)=\alpha,\quad X\sim B(n,\theta_{0}).\quad(20)Relation (20) is rewritten below to allow the usage of the Binomial tables for the determination of C{\mathrm{~and~}}\gamma; namely,
P_{\theta_{0}}(X\leq C)-\gamma P_{\theta_{0}}(X=C)=1-\alpha,\quad X\sim B(n,\theta_{0}).\qquad\qquad(21)The power of the test is:
\pi_{\varphi}(\theta)=\ P_{\theta}(X\gt C)+\gamma P_{\theta}(X=C)=1-P_{\theta}(X\leq C)+\gamma P_{\theta}(X=C), (\theta\gt \theta_{0}),\quad X\sim B(n,\theta).\qquad\qquad(22)~~~~N u m e{r i c a l\:E x a m p l e} Referto Example 2 and suppose that n=25,\theta_{0}= 100~p\%= 0.125,{\mathrm{and}}~\alpha=0.05.
Here, for \theta=\theta_{0},\,X\sim B(25,~0.125), and (21) becomes
P_{0.125}(X\leq C)-\gamma P_{0.125}(X=C)=0.95.
From the Binomial tables, the value of C which renders P_{0.125}(X\leq C) just above 0.95 is 6 and P_{0.125}(X\leq6)=0.9703. Also, P_{0.125}(X=6)=0.9703-0.9169=0.0534, so that {{\gamma}={\frac{\ 0.9703-0.95}{0.0534}}={\frac{0.0203}{0.0534}}\simeq 0.38.} Thus, the test in (19) is:
\varphi(x_{1},\dots,x_{n}) = \begin{cases} 1 & \mathrm{if~}~~~x\gt 6 \\ 0.38 & {\mathrm{if~}}~~~x =6 \\ 0 & {\mathrm{if~}}~~~x\lt 6. \end{cases}
Reject outright the hypothesis that 100~p\%=12.5\% if the number of listeners among the sample of 25 is 7 or more, reject the hypothesis with probability 0.38 if this number is 6, and accept the hypothesis if this number is 5 or smaller.
The power of the test is calculated to be as follows, by relation (22):
\pi_{\varphi}(0.1875)\simeq1-0.8261+0.38\times0.1489\simeq0.230,
\pi_{\varphi}(0.25)\simeq1-0.5611+0.38\times0.1828\simeq0.508,
\pi_{\varphi}(0.375)\simeq1-0.1156+0.38\times0.0652\simeq0.909.
If we suppose that the observed value of X is 7, then the P-value is:
1-P_{0.125}(X\leq7)+0.38P_{0.125}(X=7)=1-0.9910+0.38\times0.0207\simeq0.017,
so that the result is statistically significant.
Next, for testing the hypothesis H_0:\theta\geq\theta_{0}. against the alternative H_{A}\colon\theta\lt \theta_{0}, relations (8) and (9) become:
\varphi(x_{1},\dots,x_{n}) = \begin{cases} 1 & \mathrm{if~}~~~\textstyle\sum_{i=1}^{n}x_{i}\lt C \\ \gamma & {\mathrm{if~}}~~~\textstyle\sum_{i=1}^{n}x_{i} =C \\ 0 & {\mathrm{if~}}~~~\textstyle\sum_{i=1}^{n}x_{i}\gt C, \end{cases}\qquad (23)
and
~~~P_{\theta_{0}}(X\leq C-1)+\gamma P_{\theta_{0}}(X=C)=\alpha,\quad X=\sum\limits_{i=1}^{n}X_{i}\sim B(n,\theta_{0}).\qquad(24)
The power of the test is:
\pi_{\varphi}(\theta)=P_{\theta}(X\leq C-1)+\gamma P_{\theta}(X=C)\quad(\theta\lt \theta_{0}),\quad X\sim B(n,\theta).\qquad\qquad (25)
~~N u m e{r i c a l}~E x a m p l e\quad Refer to Example 1 and suppose that n\;=\;25,\; \theta_{0}=0.0625,\,\mathrm{and}\,\alpha=0.05.
Here, under \theta_{0},X\sim B(25,\ 0.0625) and (24) becomes
P_{0.0625}(X\leq C-1)+\gamma P_{0.0625}(X=C)=0.05,
so that C=0,{\mathrm{and}}\,\gamma\,P_{0.0625}(X=0)=0.1992\,\gamma=0.05. It follows that \gamma\simeq0.251. Therefore the hypothesis is rejected with probability 0.251,\,\mathrm{if}\,x=0, and is accepted otherwise.