The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi in. , given in increasing order)]: 65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8 Calculate a 99% CI for the standard deviation of the fracture toughness distribution. (Round your answers to one decimal place.)

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Step-by-step explanation:

Data given:

65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8

The sample mean would be given by:

We can calculate the sample deviation with this formula:

And we got:

s=5.437 represent the sample standard deviation

represent the sample mean

n=22 the sample size

Confidence=99% or 0.99

The confidence interval for the population variance is given by:

The degrees of freedom given by:

The Confidence is 0.99 or 99%, the value of significance is and , and the critical values are:

And the confidence interval would be:

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8