Answer :
Answer:
95% confidence interval for the population variance = (1.42 , 2.62).
Step-by-step explanation:
We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.
So, 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 = 1.88
[tex]\sigma^{2}[/tex] = population variance
n = sample of windshields = 83
So, 95% confidence interval for population variance, [tex]\sigma^{2}[/tex] is;
P(58.85 < [tex]\chi^{2} __8_2[/tex] < 108.9) = 0.95 {As the table of [tex]\chi^{2}[/tex] at 82 degree of freedom
gives critical values of 58.85 & 108.9}
P(58.85 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 108.9) = 0.95
P( [tex]\frac{ 58.85}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{108.9}{(n-1)s^{2} }[/tex] ) = 0.95
P( [tex]\frac{ (n-1)s^{2}}{108.9 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{ (n-1)s^{2}}{58.85 }[/tex] ) = 0.95
95% confidence interval for [tex]\sigma^{2}[/tex] = ( [tex]\frac{ (n-1)s^{2}}{108.9 }[/tex] , [tex]\frac{ (n-1)s^{2}}{58.85 }[/tex] )
= ( [tex]\frac{ (83-1)\times 1.88}{108.9 }[/tex] , [tex]\frac{ (83-1)\times 1.88}{58.85 }[/tex] )
= (1.42 , 2.62)
Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).