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

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. = $\frac{xbar -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

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

           = $\frac{87 +90 +94 +96+ 98 +99+ 101+ 101+ 102+ 103+ 104+ 105+ 105+ 107+ 108 +111}{16}$ = 100.7

            s = Sample standard deviation = $\sqrt{\frac{\sum (x-xbar)^{2} }{n-1}}$ = 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 < $\frac{xbar -\mu}{\frac{s}{\sqrt{n} } }$ < 1.753) = 0.90

  P(-1.753 * $\frac{s}{\sqrt{n} }$ < $xbar-\mu$ < 1.753 *

  P(-xbar - 1.753 * $\frac{s}{\sqrt{n} }$ < $-\mu$ < -xbar + 1.753 *

  P(xbar - 1.753 * $\frac{s}{\sqrt{n} }$ < $\mu$ < xbar + 1.753 *

So, 90% confidence interval for $\mu$ = [xbar - 1.753 * $\frac{s}{\sqrt{n} }$ , xbar - 1.753 *

                                                        = $[100.7 - 1.753*\frac{6.5}{\sqrt{16} } , 100.7 + 1.753*\frac{6.5}{\sqrt{16} } ]$

                                                        = [97.85 , 103.55]

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