Question

Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.50. The last seven times John has taken a taxi from Logan to downtown Boston, the fares have been $22.10, $23.25, $21.35, $24.50, $21.90, $20.75, and $22.65.
What is a 95% confidence interval for the population mean taxi fare?

Answer 1

Answer:

95% Confidence interval for taxi fare:  ($20.5,$24.2)

Step-by-step explanation:

We are given the following data set: for fares:

$22.10, $23.25, $21.35, $24.50, $21.90, $20.75, and $22.65

Formula:

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{156.5}{7} = 22.35$

95% Confidence interval:

$\bar{x} \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$22.35 \pm 1.96(\frac{2.5}{\sqrt{7}} ) = 22.35 \pm 1.85 = (20.5,24.2)$