The amount of lateral expansion mils was determined for a sa

The amount of lateral expansion (mils) was determined for a sample of n = 10 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.88 mils. Assuming normality, derive a 95% CI for sigma 2 and for sigma. (Round your answers to two decimal places.) CI for sigma2 ( , ) mils^2 CI for sigma( , ) mils You may need to use the appropriate table in the Appendix of Tables to answer this question.

Solution

Given a=0.05, chisquare with 0.025,df=n-1=9 is 2.7 (check chisquare table)
chisquare with 0.975, df=9 is 19.02 (check chisquare table)

So 95% CI for ^2 is

((n-1)*s^2/19.02, (n-1)*s^2/2.7)

--> (9*2.87^2/19.02, 9*2.87^2/2.7)

--> (3.897587, 27.45633)