The amount of lateral expansion mils was determined for a sa

The amount of lateral expansion (mils) was determined for a sample of n=9 pulse-power gas metal arc welds used in LNG ship containment tanks, with resulting standard deviation s=2.81 mils. Assuming normality, derive

a) A 95% confidence interval for sigma^2

b) A 95% confidence interval for sigma.

Solution

a)

As              
              
df = n - 1 =    8          
alpha = (1 - confidence level)/2 =    0.025          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    17.53454614          
chi^2(alpha/2) =    2.179730747          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    3.602534077          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    28.98009311          
              
Thus, the confidence interval for the variance is              
              
(   3.602534077   ,   28.98009311   ) [ANSWER]

b)

              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   1.898034267   ,   5.383316181   ) [ANSWER]