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

Answer:

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

Step-by-step 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. = $\frac{(n-1)s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

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 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 26.12) = 0.95

P( $\frac{5.629 }{(n-1)s^{2} }$ < $\frac{1}{\sigma^{2} }$ < $\frac{26.12 }{(n-1)s^{2} }$ ) = 0.95

P( $\frac{(n-1)s^{2} }{26.12 }$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{5.629 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2} }{26.12 }$ , $\frac{(n-1)s^{2} }{5.629 }$ ]

                                                  = [ $\frac{14 \times 9409 }{26.12 }$ , $\frac{14 \times 9409 }{5.629 }$ ]

                                                  = [5043.11 , 23401.31]

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