Question

Construct a 99% confidence interval for the population standard deviation σ if a sample of size 8 has standard deviation 

Answer 1

We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size:

$$s=7.5, s^2=7.5^2=56.25, n=8.$$

The critical values for $\alpha=0.01$ and $df=n-1=7$ degrees of freedom are $\chi_L^2=\chi^2_{1-\alpha/2, n-1}=0.9893$  and$\chi_U^2=\chi^2_{\alpha/2, n-1}=20.2777.$

The corresponding confidence interval is computed as shown below:

$$CI(Variance)=(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}})$$

$$=(\dfrac{(8-1)56.25}{20.2777}, \dfrac{(8-1)56.25}{0.9893})$$

$$\approx(19.4178, 398.0265)$$

$$CI(Standard \ Deviation)=(\sqrt{19.4178}, \sqrt{398.0265})$$

$$\approx(4.4066, 19.9506)$$