The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in increasing order)]. 68.6 71.9 72.6 73.1 73.3 73.5 75.5 75.7 75.8 76.1 76.2
76.2 77.0 77.9 78.1 79.6 79.8 79.9 80.1 82.2 83.7 93.4
Calculate a 90% CI for the standard deviation of the fracture toughness distribution. (Give answer accurate to 2 decimal places.)

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2,  76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

                             P.Q.  =    ~

where, s = sample standard deviation = = 5.063

             = population standard deviation

            n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, is ;

P(11.59 < < 32.67) = 0.90  {As the critical value of chi at 21 degrees  

                                                  of freedom are 11.59 & 32.67}  

P(11.59 < < 32.67) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                                     = [ , ]

                                     = [16.48 , 46.45]

90% confidence interval for = [ , ]

                                                 = [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

the answer is d. (1.2, -4.7) on edgn

hope this !