Respuesta :
Answer:
90% confidence interval for the population cost of a ticket = [97.85 , 103.55]
Step-by-step explanation:
We are given below the airfare prices (in dollars) for a one-way ticket from Atlanta to Chicago that was chosen by Newsweek in 2001 ;
87, 90, 94, 96, 98, 99, 101, 101, 102, 103, 104, 105, 105, 107, 108, 111
We have to calculate a 90% confidence interval for the population cost of a ticket.
The Pivotal quantity is given by;
P.Q. = [tex]\frac{xbar -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]xbar[/tex] = Sample mean = Sum of all above values ÷ Total values
= [tex]\frac{87 +90 +94 +96+ 98 +99+ 101+ 101+ 102+ 103+ 104+ 105+ 105+ 107+ 108 +111}{16}[/tex] = 100.7
s = Sample standard deviation = [tex]\sqrt{\frac{\sum (x-xbar)^{2} }{n-1}}[/tex] = 6.5
n = sample size = 16
So, 90% confidence interval for the population cost of a ticket is given by;
P(-1.753 < [tex]t_1_5[/tex] < 1.753) = 0.90
P(-1.753 < [tex]\frac{xbar -\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.753) = 0.90
P(-1.753 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]xbar-\mu[/tex] < 1.753 *
P(-xbar - 1.753 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]-\mu[/tex] < -xbar + 1.753 *
P(xbar - 1.753 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < xbar + 1.753 *
So, 90% confidence interval for [tex]\mu[/tex] = [xbar - 1.753 * [tex]\frac{s}{\sqrt{n} }[/tex] , xbar - 1.753 *
= [tex][100.7 - 1.753*\frac{6.5}{\sqrt{16} } , 100.7 + 1.753*\frac{6.5}{\sqrt{16} } ][/tex]
= [97.85 , 103.55]
Therefore, confidence interval for the population cost of a ticket is [97.85 , 103.55] .
