Question

5. 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 Calculate a 99% CI for the variance σ 2 , and the standard deviation σ of the coating layer thickness distribution

Answer 1

99% CI for the variance
`\sigma^(2)` , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

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

where,
`s^(2)` = sample variance =
`(\sum (X-\bar X)^(2) )/(n-1)` = 67.43

n = sample of observations = 8

`\sigma^(2)` = population variance

Here for constructing 99% confidence interval we have used chi-square test statistics.

So, 99% confidence interval for the population variance,
`\sigma^(2)` is ;

P(0.9893 <
`\chi^(2)_7` < 20.28) = 0.99 {As the critical value of chi-square at 7

degree of freedom are 0.9893 & 20.28}

P(0.9893 <
`((n-1)s^(2) )/(\sigma^(2) )` < 20.28) = 0.99

P(
`(0.9893 )/((n-1)s^(2) )` <
`(1)/(\sigma^(2) )` <
`(20.28 )/((n-1)s^(2) )` ) = 0.99

P(
`((n-1)s^(2) )/(20.28 )` <
`\sigma^(2)` <
`((n-1)s^(2) )/(0.9893 )` ) = 0.99

99% confidence interval for
`\sigma^(2)` = [
`((n-1)s^(2) )/(20.28 )` ,
`((n-1)s^(2) )/(0.9893 )` ]

= [
`(7* 67.43 )/(20.28 )` ,
`(7* 67.43 )/(0.9893 )` ]

= [23.275 , 477.115]

99% confidence interval for
`\sigma` = [
`√(23.275)` ,
`√(477.115)` ]

= [4.824 , 21.843]

Therefore, 99% CI for the variance
`\sigma^(2)` , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.