Respuesta :
Answer:
99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.
Step-by-step explanation:
We are given that the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for Wire electrical-discharge machining (WEDM) : 21, 16, 29, 35, 42, 24, 24, 25
Firstly, the pivotal quantity for 99% confidence interval for the population variance is given by;
P.Q. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]
where, [tex]s^{2}[/tex] = sample variance = [tex]\frac{\sum (X-\bar X)^{2} }{n-1}[/tex] = 67.43
n = sample of observations = 8
[tex]\sigma^{2}[/tex] = population variance
Here for constructing 99% confidence interval we have used chi-square test statistics.
So, 99% confidence interval for the population variance, [tex]\sigma^{2}[/tex] is ;
P(0.9893 < [tex]\chi^{2}_7[/tex] < 20.28) = 0.99 {As the critical value of chi-square at 7
degree of freedom are 0.9893 & 20.28}
P(0.9893 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 20.28) = 0.99
P( [tex]\frac{0.9893 }{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{20.28 }{(n-1)s^{2} }[/tex] ) = 0.99
P( [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ) = 0.99
99% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] , [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ]
= [ [tex]\frac{7\times 67.43 }{20.28 }[/tex] , [tex]\frac{7\times 67.43 }{0.9893 }[/tex] ]
= [23.275 , 477.115]
99% confidence interval for [tex]\sigma[/tex] = [ [tex]\sqrt{23.275}[/tex] , [tex]\sqrt{477.115}[/tex] ]
= [4.824 , 21.843]
Therefore, 99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.
