Question

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.)

Answer 1

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

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. =
`((n-1) * s^(2) )/(\sigma^(2) )` ~
`\chi^(2) __n_-_1`

where, s = sample standard deviation =
`\sqrt{(\sum (X - \bar X^(2)) )/(n-1) }` = 5.063

`\sigma` = 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,
`\sigma` is ;

P(11.59 <
`\chi^(2)__2_1` < 32.67) = 0.90 {As the critical value of chi at 21 degrees

of freedom are 11.59 & 32.67}

P(11.59 <
`((n-1) * s^(2) )/(\sigma^(2) )` < 32.67) = 0.90

P(
`( 11.59)/((n-1) * s^(2))` <
`(1)/(\sigma^(2) )` <
`( 32.67)/((n-1) * s^(2))` ) = 0.90

P(
`((n-1) * s^(2) )/(32.67 )` <
`\sigma^(2)` <
`((n-1) * s^(2) )/(11.59 )` ) = 0.90

90% confidence interval for
`\sigma^(2)` = [
`((n-1) * s^(2) )/(32.67 )` ,
`((n-1) * s^(2) )/(11.59 )` ]

= [
`(21 * 5.063^(2) )/(32.67 )` ,
`(21 * 5.063^(2) )/(11.59 )` ]

= [16.48 , 46.45]

90% confidence interval for
`\sigma` = [
`√(16.48)` ,
`√(46.45)` ]

= [4.06 , 6.82]

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