Respuesta :
Answer:
80% confidence interval for the population variance = (0.07 , 0.55).
Step-by-step explanation:
We are given that the following sample of heights was taken from 5 air filters off the assembly line;
3.8, 4.3, 3.8, 4.5, 3.6
So, firstly the pivotal quantity for 80% 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, s = sample standard deviation
[tex]\sigma[/tex] = population standard deviation
n = sample of rods = 5
Also, [tex]s^{2} = \frac{\sum (X-\bar X)^{2} }{n-1}[/tex] , where X = individual data value
[tex]\bar X[/tex] = mean of data values = 4
[tex]s^{2}[/tex] = 0.145
So, 80% confidence interval for population variance, [tex]\sigma^{2}[/tex] is;
P(1.064 < [tex]\chi^{2} __4[/tex] < 7.779) = 0.80 {As the table of [tex]\chi^{2}[/tex] at 4 degree of freedom
gives critical values of 1.064 & 7.779}
P(1.064 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 7.779) = 0.80
P( [tex]\frac{ 1.064}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{ 7.779}{(n-1)s^{2} }[/tex] ) = 0.80
P( [tex]\frac{ (n-1)s^{2}}{7.779 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{ (n-1)s^{2}}{1.064 }[/tex] ) = 0.80
80% confidence interval for [tex]\sigma^{2}[/tex] = ( [tex]\frac{ (n-1)s^{2}}{7.779 }[/tex] , [tex]\frac{ (n-1)s^{2}}{1.064 }[/tex] )
= ( [tex]\frac{ (5-1)\times 0.145}{7.779 }[/tex] , [tex]\frac{ (5-1)\times 0.145}{1.064 }[/tex] )
= (0.07 , 0.55)
Therefore, 80% confidence interval for the population variance for all air filters that come off the assembly line is (0.07 , 0.55).
