Question

The airfare prices (in dollars) for a one-way ticket from Atlanta to Chicago was chosen by Newsweek in 2001 and is listed below. Calculate a 90% confidence interval for the population cost of a ticket. 87 90 94 96 98 99 101 101 102 103 104 105 105 107 108 111

Answer 1

90% confidence interval for the population cost of a ticket = [97.85 , 103.55]

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. =
`(xbar -\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`xbar` = Sample mean = Sum of all above values ÷ Total values

=
`(87 +90 +94 +96+ 98 +99+ 101+ 101+ 102+ 103+ 104+ 105+ 105+ 107+ 108 +111)/(16)` = 100.7

s = Sample standard deviation =
`\sqrt{(\sum (x-xbar)^(2) )/(n-1)}` = 6.5

n = sample size = 16

So, 90% confidence interval for the population cost of a ticket is given by;

P(-1.753 <
`t_1_5` < 1.753) = 0.90

P(-1.753 <
`(xbar -\mu)/((s)/(√(n) ) )` < 1.753) = 0.90

P(-1.753 *
`(s)/(√(n) )` <
`xbar-\mu` < 1.753 *

P(-xbar - 1.753 *
`(s)/(√(n) )` <
`-\mu` < -xbar + 1.753 *

P(xbar - 1.753 *
`(s)/(√(n) )` <
`\mu` < xbar + 1.753 *

So, 90% confidence interval for
`\mu` = [xbar - 1.753 *
`(s)/(√(n) )` , xbar - 1.753 *

=
`[100.7 - 1.753*(6.5)/(√(16) ) , 100.7 + 1.753*(6.5)/(√(16) ) ]`

= [97.85 , 103.55]

Therefore, confidence interval for the population cost of a ticket is [97.85 , 103.55] .