91影视

The sample size is 153(n).

The confidence level is 99%.

The actual critical values are ${蠂}_L^2=110.84$and${蠂}_R^2=200.657$

The z-score is obtained from the standard normal table.

The significance level is 0.01, corresponding to a 99% level of confidence.

The critical value is expressed as follows.

$$\begin{gathered} P\left(Z>z_{\frac{伪}{2}}\right)=\frac{伪}{2} \\ P\left(Z>z_{\frac{0.01}{2}}\right)=\frac{0.01}{2} \\ P\left(Z<z_{0.005}\right)=0.995 \end{gathered}$$

By using technology, the critical value is computed as ${\mathit{z}}_{\frac{\mathbf{0}\mathbf{.}\mathbf{01}}{\mathbf{2}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{575829303}$.

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =153-1 \\ =152 \end{gathered}$$

Use the given formula for calculation.

${\mathit{蠂}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left[卤z_{\frac{伪}{2}}+\sqrt{2k-1}\right]$

Substitute the value in the formula as shown below.

$$\begin{gathered} {蠂}_L^2=\frac{1}{2}{\left[-\text{2.575829303}+\sqrt{2\left(152\right)-1}\right]}^2 \\ =109.9803 \\ {蠂}_R^2=\frac{1}{2}{\left[\text{2.575829303}+\sqrt{2\left(152\right)-1}\right]}^2 \\ =199.6546 \end{gathered}$$

Therefore, the values are 109.980 and 199.655, respectively, which are close to the actual critical values ${蠂}_L^2=110.84$and ${蠂}_R^2=200.657$.