Question 11.APP.4: The Normal Case (continued) Testing Hypotheses About the Var......
The Normal Case (continued) Testing Hypotheses About the Variance Refer to Example 12 , where T(x)=(x-\mu)^{2}\,(\mu known) and Q is strictly increasing. Then, for testing the hypothesis H_{0}\colon\sigma^{2}\geq\sigma_{0}^{2} against the alternative H_{A}\!:\!\sigma^{2}\lt \sigma_{0}^{2}(\mathrm{or}\,\theta\geq\theta_{0} against \theta\,\lt \theta_{0}\,\mathrm{with}\,\theta=\sigma^{2}\,\mathrm{and}\,{\theta_{0}=\sigma_{0}^{2}}) at level of significance α, the appropriate test is given by (8) and (9)
\varphi(x_{1},\dots,x_{n}) = \begin{cases} 1 & \mathrm{if~}~~~V(x_{1},\ldots,x_{n})\lt C \\ \gamma & {\mathrm{if~}}~~~V(x_{1},\ldots,x_{n})=C \\ 0 & {\mathrm{if~}}~~~V(x_{1},\ldots,x_{n})\gt C, \end{cases}\qquad(8)
E_{\theta_{0}}\varphi(X_{1},\,\dots ,\,{X_n})= P_{\theta_{0}}[V(X_{1},\,\dots,\,{X_n})\lt C] +\gamma P_{\theta_{0}}[V(X_{1},\dots,\,{X_{n}})= C]=\alpha.\qquad\qquad(9)
(with \gamma = 0), and it is here:
or
\varphi(x_{1},\dots,x_{n})= \begin{cases} 1 & {\mathrm{if~}}\textstyle\sum_{i=1}^{n}(\frac {x_{i}-\mu}{\sigma _0})^2\lt \chi_{n;1-\alpha}^2\\\\ 0 & {\mathrm{otherwise,~}} \end{cases} \qquad(32^\prime)because
\alpha=E_{\sigma_{0}^{2}}\varphi(X_{1},\ldots,X_{n})={{P}}_{\sigma_{0}^{2}}\left[\sum\limits_{i=1}^{n}(X_{i}-\mu)^{2}\lt C\right] =P_{\sigma_{0}^{2}}\left[\sum\limits_{i=1}^{n}\left({\frac{X_{i}-\mu}{\sigma_{0}}}\right)^{2}\lt {\frac{C}{\sigma_{0}^{2}}}\right],so that {\frac{C}{\sigma_{0}^{2}}}=\chi_{n;1-\alpha}^{2} and therefore
C=\sigma_{0}^{2}\chi_{n;1-\alpha}^{2};\qquad\qquad\qquad\qquad\qquad\qquad(33)this is so, because \textstyle\sum_{i=1}^{n}({\frac{X_{i}-\mu}{\sigma_{0}}})^{2}\sim\,\chi_{n}^{2}.
By slightly abusing the notation and denoting by \chi_{n}^{2} also a r.v. which has the \chi_{n}^{2} distribution, the power of the test is given by:
because, on account of (33):
\pi_{\varphi}(\sigma^{2})=P_{\sigma^{2}}\Biggl[\sum\limits_{i=1}^{n}(X_{i}-\mu)^{2}\lt C\Biggl]=P_{\sigma^{2}}\Biggl[\sum\limits_{i=1}^{n}\Biggl(\frac{X_{i}-\mu}{\sigma}\Biggr)^{2}\lt \frac{C}{\sigma^{2}}\Biggr] =P_{\sigma^{2}}\bigg[\sum\limits_{i=1}^{n}\bigg(\frac{X_{i}-\mu}{\sigma}\bigg)^{2}\lt \frac{\sigma_{0}^{2}}{\sigma^{2}}\chi_{n;1-\alpha}^{2}\bigg]=P{\bigg(}\chi_{n}^{2}\lt \frac{\sigma_{0}^{2}}{\sigma^{2}}\chi_{n;1-\alpha}^{2}\bigg),since \textstyle\sum_{i=1}^{n}({\frac{X_{i}-\mu}{\sigma}})^{2}\sim\,\chi_{n}^{2}.
~~~N u m e{r i c a l}\,Ex a m p l e~~~ Suppose n=40,\sigma_{0}=2,\,\mathrm{and}\,\alpha\,=\,0.02{5}. For simplicity, take μ = 0.
Here \chi_{n;1-\alpha}^{2}~=~\chi_{40;0.975}^{2}~=~24.433~\mathrm{and}~C~={4}~\times~24.433~=97.732. Thus, by means of (32), the hypothesis is rejected if \textstyle\sum_{i=1}^{40}x_{i}^{2}\,\lt 97.732, and it is accepted otherwise.
For σ = 1.25, for example, the power of the test is, by means of (34),
\pi_{\varphi}(\sigma^{2})=\pi_{\varphi}(2.25)=P\biggl(\chi_{40}^{2}\lt {\frac{97.732}{1.5625}}\biggr)
=P{\left(\chi_{40}^{2}\lt 62.548\right)}=0.986
(by linear interpolation).
If we suppose that the observed value of \textstyle\sum_{i=1}^{40}x_{i}^{2} is 82.828, then the P-value is
P_{4}\left(\sum\limits_{i=1}^{40}X_{i}^{2}\lt 82.828\right)=P_{4}\left[\sum\limits_{i=1}^{40}\left(\frac{X_{i}}{2}\right)^{2}\lt 20.707\right]=0.05,
which indicates strong rejection.
