The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi vin, given in increasing order)]. 69.5 71.9 72.6 73.1 73.3 73.5 75.5 75.7 76.0 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.6 79.9 80.1 82.2 83.7 94.5 a) Calculate a 99% CI for the standard deviation of the fracture toughness distribution. Is this interval valid whatever the nature of the distribution? b) Calculate a 90% CI for the standard deviation of the fracture toughness distribution. Solution a. Given n=22, s= 5.16 (based on the data) a=0.01, chisquare with 0.005, df=n-1=21 is 8.03 (check chisquare table) chisquare with 0.995, df=n-1=21 is 41.4 (check chisquare table) So 99% CI is ((n-1)*s/41.4 , (n-1)*s/8.03) --> (sqrt(21)*5.16/sqrt(41.4), sqrt(21)*5.16/sqrt(8.03)) --> (3.675016,8.344524).