Question

Finding Critical Values In constructing confidence intervals for $\sigma$or ${\sigma }^2$, Table A-4 can be used to find the critical values ${\chi }_L^2$and ${\chi }_R^2$only for select values of n up to 101, so the number of degrees of freedom is 100 or smaller. For larger numbers of degrees of freedom, we can approximate ${\chi }_L^2$and${\chi }_R^2$ by using,

${\chi }^2=\frac{1}{2}{\left[\pm z_{\frac{\alpha }{2}}+\sqrt{2k-1}\right]}^2$

where k is the number of degrees of freedom and $z_{\frac{\alpha }{2}}$is the critical z score described in Section 7-1. Use this approximation to find the critical values ${\chi }_L^2$and ${\chi }_R^2$for Exercise 8 “Heights of Men,” where the sample size is 153 and the confidence level is 99%. How do the results compare to the actual critical values of ${\chi }_L^2$= 110.846 and ${\chi }_R^2$= 200.657?

Short answer

The calculated values are as follows.

$$\begin{gathered} {\chi }_L^2=109.980 \\ {\chi }_R^2=199.655 \end{gathered}$$

Both values are quite close to the actual critical values.

Step-by-step solution

Given information

The sample size is 153(n).

The confidence level is 99%.

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

Calculate the critical z-score 

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.

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

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

Compute the degree of freedom

The degree of freedom is computed as follows.

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

Compute the critical values 

Use the given formula for calculation.

${\mathit{\chi }}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left[\pm z_{\frac{\alpha }{2}}+\sqrt{2k-1}\right]$

Substitute the value in the formula as shown below.

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

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