Question

a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97. Assume that the population is normally distributed. What is the 95% confidence interval for the population variance of the number of tissues per box?

Answer 1

95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

Explanation:

We are given that a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97.

Firstly, the pivotal quantity for 95% 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 =
`97^(2)` = 9409

n = sample of boxes = 15

`\sigma^(2)` = population variance

Here for constructing 95% confidence interval we have used chi-square test statistics.

So, 95% confidence interval for the population variance,
`\sigma^(2)` is ;

P(5.629 <
`\chi^(2)__1_4` < 26.12) = 0.95 {As the critical value of chi-square at 14

degree of freedom are 5.629 & 26.12}

P(5.629 <
`((n-1)s^(2) )/(\sigma^(2) )` < 26.12) = 0.95

P(
`(5.629 )/((n-1)s^(2) )` <
`(1)/(\sigma^(2) )` <
`(26.12 )/((n-1)s^(2) )` ) = 0.95

P(
`((n-1)s^(2) )/(26.12 )` <
`\sigma^(2)` <
`((n-1)s^(2) )/(5.629 )` ) = 0.95

95% confidence interval for
`\sigma^(2)` = [
`((n-1)s^(2) )/(26.12 )` ,
`((n-1)s^(2) )/(5.629 )` ]

= [
`(14 * 9409 )/(26.12 )` ,
`(14 * 9409 )/(5.629 )` ]

= [5043.11 , 23401.31]

Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].