Respuesta :
Answer:
80% confidence interval for the average age of people who attend U of O football games is [42.795 , 44.604].
Step-by-step explanation:
We are given that a sample of 100 attendees and find the average age to be 43.7 years old with a standard deviation of 7 years.
So, the pivotal quantity for 80% confidence interval for the population average start up cost is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\mu[/tex] = sample average age = 43.7 years old
[tex]\sigma[/tex] = sample standard deviation = 7 years
n = sample of attendees = 100
[tex]\mu[/tex] = population average age of people
So, 80% confidence interval for the average age of people, [tex]\mu[/tex] is ;
P(-1.2915 < [tex]t_9_9[/tex] < 1.2915) = 0.80
P(-1.2915 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.2915) = 0.80
P( [tex]-1.2915 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]1.2915 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.80
P( [tex]\bar X -1.2915 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +1.2915 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.80
80% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -1.2915 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +1.2915 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]43.7 -1.2915 \times {\frac{7}{\sqrt{100} }[/tex] , [tex]43.7 +1.2915 \times {\frac{7}{\sqrt{100} }[/tex] ]
= [42.795 , 44.604]
Therefore, 80% confidence interval for the population average age of people who attend U of O football games is [42.795 , 44.604].
