Respuesta :
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. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]
where, [tex]s^{2}[/tex] = sample variance = [tex]97^{2}[/tex] = 9409
n = sample of boxes = 15
[tex]\sigma^{2}[/tex] = population variance
Here for constructing 95% confidence interval we have used chi-square test statistics.
So, 95% confidence interval for the population variance, [tex]\sigma^{2}[/tex] is ;
P(5.629 < [tex]\chi^{2}__1_4[/tex] < 26.12) = 0.95 {As the critical value of chi-square at 14
degree of freedom are 5.629 & 26.12}
P(5.629 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 26.12) = 0.95
P( [tex]\frac{5.629 }{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{26.12 }{(n-1)s^{2} }[/tex] ) = 0.95
P( [tex]\frac{(n-1)s^{2} }{26.12 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1)s^{2} }{5.629 }[/tex] ) = 0.95
95% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2} }{26.12 }[/tex] , [tex]\frac{(n-1)s^{2} }{5.629 }[/tex] ]
= [ [tex]\frac{14 \times 9409 }{26.12 }[/tex] , [tex]\frac{14 \times 9409 }{5.629 }[/tex] ]
= [5043.11 , 23401.31]
Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].
