Question

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi Vin, given in increasing order)] 70.0 71.9 72.6 73.1 73.3 73.5 75.5 75.7 75.9 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.0 (a) Calculate a 99% CI for the standard deviation of the fracture toughness distribution. (Give answer accurate to 2 decimal places.) Is this interval valid whatever the nature of the distribution? Explain. Yes, the interval is always valid. The distribution needs to be skewed to the left. The distribution needs to be skewed to the right The distribution needs to be approximately normal. (b) Calculate a 90% CI for the standard deviation of the fracture toughness distribution. (Give answer accurate to 2 decimal places.)

Answer 1

a) Given n=22, s= 4.89 (based on the data)

$\alpha$ =0.01, chisquare with 0.005, df=n-1=21 is 8.03 (from chisquare table)
chisquare with 0.995, df=n-1=21 is 41.4 (from chisquare table)

So 99% CI is
(√(n-1)*s/√41.4 , √(n-1)*s/√8.03)

= (√21)*4.89/√41.4, √21*4.89/√8.03)

= (3.48 , 7.91)

b) $\alpha$ =0.01, chisquare with 0.05, df=n-1=21 is 11.59 (from chisquare table)
chisquare with 0.95, df=n-1=21 is 32.67 (from chisquare table)

So 90% CI is
(√(n-1)*s/√32.67 , √(n-1)*s/√11.59)

= (√21*4.89/√32.67, √21*5.16/√11.59)

= (3.92 , 6.58)