Answer to Question #251267 in Statistics and Probability for Melca

Question

Find the 95% confidence interval for the variance and standard deviation for the sugar content in ice cream (in mg) if a sample of nine servings has a variance of 36.

Answer 1

The critical values for $\alpha = 0.05$ and $df =n-1=9-1= 8$ degrees of freedom are $\chi^2_L=\chi^2_{1-\alpha/2,n-1}=2.1797$ and $\chi^2_L=\chi^2_{\alpha/2,n-1}=17.5345.$

The corresponding confidence interval is computed as shown below:

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

$=\bigg(\dfrac{8(36)}{17.5345},\dfrac{8(36)}{2.1797}\bigg)$

$=(16.4248, 132.1283)$

Now that we have the limits for the confidence interval, the limits for the 95% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

$CI(Standard\ Deviation)=\bigg(\sqrt{\dfrac{288}{17.5345}}, \sqrt{\dfrac{288}{2.1797}}\bigg)$

$=(4.0527,11.4947)$

Therefore, based on the data provided, the 95% confidence interval for the population variance is $16.4248 < \sigma^2 < 132.1283,$ and the 95% confidence interval for the population standard deviation is $4.0527 < \sigma < 11.4947.$