Respuesta :
Answer:
99% confidence interval for the population standard deviation = (0.17 , 0.75).
Step-by-step explanation:
We are given that the following sample of lengths was taken from 9 rods off the assembly line;
13.6, 13.8, 14.1, 13.6, 13.3, 13.5, 13.9, 13.3, 14.1
So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;
P.Q. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, s = sample standard deviation
[tex]\sigma[/tex] = population standard deviation
n = sample of rods = 9
Also, [tex]s^{2} = \frac{\sum (X-\bar X)^{2} }{n-1}[/tex] , where X = individual data value
[tex]\bar X[/tex] = mean of data values = 13.7
[tex]s^{2}[/tex] = 0.094
So, 99% confidence interval for population standard deviation, is;
P(1.344 < [tex]\chi^{2} __8[/tex] < 21.95) = 0.99 {As the table of [tex]\chi^{2}[/tex] at 8 degree of freedom
gives critical values of 1.344 & 21.95}
P(1.344 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 21.95) = 0.99
P( [tex]\frac{ 1.344}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{ 21.95}{(n-1)s^{2} }[/tex] ) = 0.99
P( [tex]\frac{ (n-1)s^{2}}{21.95 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{ (n-1)s^{2}}{1.344 }[/tex] ) = 0.99
99% confidence interval for [tex]\sigma^{2}[/tex] = ( [tex]\frac{ (n-1)s^{2}}{21.95 }[/tex] , [tex]\frac{ (n-1)s^{2}}{1.344 }[/tex] )
= ( [tex]\frac{ (9-1)\times 0.094}{21.95 }[/tex] , [tex]\frac{ (9-1)\times 0.094}{1.344 }[/tex] )
= (0.03 , 0.56)
99% confidence interval for [tex]\sigma[/tex] = ( [tex]\sqrt{0.03}[/tex] , [tex]\sqrt{0.56}[/tex] )
= (0.17 , 0.75)
Therefore, 99% confidence interval for the population standard deviation for all rods that come off the assembly line is (0.17 , 0.75).
