Question

The following sample of lengths was taken from 9 rods off the assembly line. Construct the 99% confidence interval for the population standard deviation for all rods that come off the assembly line. Round your answers to two decimal places.

13.6,13.8,14.1,13.6,13.3,13.5,13.9,13.3,14.1

Answer 1

99% confidence interval for the population standard deviation = (0.17 , 0.75).

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

where, s = sample standard deviation

`\sigma` = population standard deviation

n = sample of rods = 9

Also,
`s^(2) = (\sum (X-\bar X)^(2) )/(n-1)` , where X = individual data value

`\bar X` = mean of data values = 13.7

`s^(2)` = 0.094

So, 99% confidence interval for population standard deviation, is;

P(1.344 <
`\chi^(2) __8` < 21.95) = 0.99 {As the table of
`\chi^(2)` at 8 degree of freedom

gives critical values of 1.344 & 21.95}

P(1.344 <
`((n-1)s^(2) )/(\sigma^(2) )` < 21.95) = 0.99

P(
`( 1.344)/((n-1)s^(2) )` <
`(1 )/(\sigma^(2) )` <
`( 21.95)/((n-1)s^(2) )` ) = 0.99

P(
`( (n-1)s^(2))/(21.95 )` <
`\sigma^(2)` <
`( (n-1)s^(2))/(1.344 )` ) = 0.99

99% confidence interval for
`\sigma^(2)` = (
`( (n-1)s^(2))/(21.95 )` ,
`( (n-1)s^(2))/(1.344 )` )

= (
`( (9-1)* 0.094)/(21.95 )` ,
`( (9-1)* 0.094)/(1.344 )` )

= (0.03 , 0.56)

99% confidence interval for
`\sigma` = (
`√(0.03)` ,
`√(0.56)` )

= (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).