Answer :
Answer:
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. = [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, s = sample standard deviation = [tex]\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }[/tex] = 5.063
[tex]\sigma[/tex] = 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, [tex]\sigma[/tex] is ;
P(11.59 < [tex]\chi^{2}__2_1[/tex] < 32.67) = 0.90 {As the critical value of chi at 21 degrees
of freedom are 11.59 & 32.67}
P(11.59 < [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex] < 32.67) = 0.90
P( [tex]\frac{ 11.59}{(n-1) \times s^{2}}[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{ 32.67}{(n-1) \times s^{2}}[/tex] ) = 0.90
P( [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ) = 0.90
90% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] , [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ]
= [ [tex]\frac{21 \times 5.063^{2} }{32.67 }[/tex] , [tex]\frac{21 \times 5.063^{2} }{11.59 }[/tex] ]
= [16.48 , 46.45]
90% confidence interval for [tex]\sigma[/tex] = [[tex]\sqrt{16.48}[/tex] , [tex]\sqrt{46.45}[/tex] ]
= [4.06 , 6.82]
Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].