Chapter 8.4, Problem 23E

To determine

The closeness of the confidence interval based on the accurate normal approximation and the exact one

Explanation of Solution

Let’s calculate the confidence interval using exact one.

Given,

  $\begin{array}{l}n=101\\ s=40\\ \alpha =0.05\end{array}$

Using chi square table,

  $\begin{array}{l}{\chi }_{\frac{\alpha }{2}}^2=129.5611\\ {\chi }_{1-\frac{\alpha }{2}}^2=74.2219\end{array}$

It is known that confidence interval for $\sigma$ is $\sqrt{\frac{(n-1)s^2}{{\chi }_{\frac{\alpha }{2}}^2}}<\sigma <\sqrt{\frac{(n-1)s^2}{{\chi }_{1-\frac{\alpha }{2}}^2}}$

Now, putting all the values in the above formula

  $\begin{array}{l}\sqrt{\frac{(n-1)s^2}{{\chi }_{\frac{\alpha }{2}}^2}}<\sigma <\sqrt{\frac{(n-1)s^2}{{\chi }_{1-\frac{\alpha }{2}}^2}}\\ \Rightarrow \sqrt{\frac{160000}{129.5611}}<\sigma <\sqrt{\frac{160000}{74.2219}}\\ \Rightarrow \sqrt{1234.9385}<\sigma <\sqrt{2155.697}\\ \Rightarrow 35.141<\sigma <46.4294\end{array}$

Confidence interval for $\sigma$ using is $35.141<\sigma <46.4294$

Let’s find confidence interval using accurate normal approximation

Given,

  $\begin{array}{l}n=101\\ s=40\\ \alpha =0.05\end{array}$

Using accurate normal approximation,

  $\begin{array}{l}{\chi }_{\alpha }^2\approx 0.5{\left(z_{\alpha }+\sqrt{2k-1}\right)}^2\\ {\chi }_{0.975}^2\approx 0.5{\left(\text{-1}\text{.960}+\sqrt{199}\right)}^2\\ {\chi }_{0.975}^2\approx 73.7715\\ {\chi }_{0.025}^2\approx 0.5{\left(\text{1}\text{.960}+\sqrt{199}\right)}^2\\ {\chi }_{0.025}^2\approx 129.070\end{array}$

It is known that confidence interval for $\sigma$ is $\sqrt{\frac{(n-1)s^2}{{\chi }_{\frac{\alpha }{2}}^2}}<\sigma <\sqrt{\frac{(n-1)s^2}{{\chi }_{1-\frac{\alpha }{2}}^2}}$

Now, putting all the values in the above formula

  $\begin{array}{l}\sqrt{\frac{(n-1)s^2}{{\chi }_{\frac{\alpha }{2}}^2}}<\sigma <\sqrt{\frac{(n-1)s^2}{{\chi }_{1-\frac{\alpha }{2}}^2}}\\ \Rightarrow \sqrt{\frac{160000}{129.070}}<\sigma <\sqrt{\frac{160000}{73.7715}}\\ \Rightarrow \sqrt{1239.63}<\sigma <\sqrt{2168.8592}\\ \Rightarrow 35.208<\sigma <46.571\end{array}$

Confidence interval for $\sigma$ is $35.208<\sigma <46.571$

Hence, confidence interval using exact one is $(35.141,46.4294)$

Confidence interval using accurate normal approximation is $(35.208,46.571)$

It can be clearly seen that lower bound differ by $0.1$ and upper bound differ by $0.1$ . Hence we can say that they are roughly equal.