}\left( 
\boldsymbol{\beta }^{\ast }\right) =1-\phi _{n}\left( \frac{1}{\sigma \left( 
\boldsymbol{\beta }^{\ast }\right) }\left( \frac{\chi _{r,\alpha }^{2}}{\sqrt{n}}-\sqrt{n}l_{\boldsymbol{\beta }^{\ast }}(\boldsymbol{\beta }^{\ast
})\right) \right)  \label{D}$$where $\phi _{n}\left( x\right) $ tends uniformly to the standard normal distribution $\phi \left( x\right) $ and $\sigma \left( \boldsymbol{\beta }^{\ast }\right) $ is given by $$\sigma \left( \boldsymbol{\beta }^{\ast }\right) ^{2}=\left. \frac{\partial
l_{\boldsymbol{\beta }}(\boldsymbol{\beta }^{\ast })}{\partial \boldsymbol{\beta }^{T}}\right\vert _{\boldsymbol{\beta }=\boldsymbol{\beta }^{\ast }}\widehat{\mathbf{Q}}_{n,\lambda }\left( \boldsymbol{\beta }^{\ast }\right)
\left. \frac{\partial l_{\boldsymbol{\beta }}(\boldsymbol{\beta }^{\ast })}{\partial \boldsymbol{\beta }}\right\vert _{\boldsymbol{\beta }=\boldsymbol{\beta }^{\ast }}.$$

It is clear that $$\lim_{n\rightarrow \infty }Po_{W_{n}\left( \widehat{\boldsymbol{\beta }}_{\lambda ,Q}\right) }\left( \boldsymbol{\beta }^{\ast }\right) =1$$for all $\alpha \in \left( 0,1\right) $. Therefore, the Wald-type test statistics are consistent in the sense of Fraser.

Based on the previous theorem we can obtain the sample size necessary to get a fix power $Po_{W_{n}\left( \widehat{\boldsymbol{\beta }}_{\lambda
,Q}\right) }\left( \boldsymbol{\beta }^{\ast }\right) =\pi _{0}.$ By (\[D\]) we must solve the equation $$1-\pi _{0}=\phi \left( \frac{1}{\sigma \left( \boldsymbol{\beta }^{\ast
}\right) }\left( \frac{\chi _{r,\alpha }^{2}}{\sqrt{n}}-\sqrt{n}l_{\boldsymbol{\beta }^{\ast }}(\boldsymbol{\beta }^{\ast })\right) \right)$$and we get that $n=\left[ n^{\ast }\right] +1$ with $$n^{\ast }=\frac{A+B+\sqrt{A(A+2B)}}{2l_{\boldsymbol{\beta }^{\ast }}(\boldsymbol{\beta }^{\ast })^{2}}$$ being $A=\sigma ^{2}\left( \boldsymbol{\beta }^{\ast }\right) \left( \phi
^{-1}\left( 1-\pi _{0}\right) \right) ^{2}\text{ and }B=\frac{1}{2}\chi
_{r,\alpha }^{2}l_{\boldsymbol{\beta }^{\ast }}(\boldsymbol{\beta }^{\ast }).$

We may 