Answer in Statistics and Probability for Melca #251267
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.”
