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

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

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(156.5)/(7) = 22.35`

95% Confidence interval:

`\bar{x} \pm z_(critical)(\sigma)/(√(n))`

Putting the values, we get,

`z_(critical)\text{ at}~\alpha_(0.05) = 1.96`

`22.35 \pm 1.96((2.5)/(√(7)) ) = 22.35 \pm 1.85 = (20.5,24.2)`