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?

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. =  ~

where,  = sample variance = = 9409

              n = sample of boxes = 15

             = population variance

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

So, 95% confidence interval for the population variance, is ;

P(5.629 < < 26.12) = 0.95  {As the critical value of chi-square at 14

                                         degree of freedom are 5.629 & 26.12}  

P(5.629 < < 26.12) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                                                  = [ , ]

                                                  = [5043.11 , 23401.31]

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

oof tough question

step-by-step explanation:

sorry i can't ya : c gl tho