Question

A car rental company is interested in the amount of time its vehicles are out of operation for repair work. A random sample of 21 cars showed that, over the past year, the numbers of days that each of these cars had been inoperative were as follows: 15, 11, 19, 24, 6, 18, 20, 15, 18, 12, 14, 19, 5, 21, 12, 11, 8, 10, 7, 2, 14 Assuming that the population is normally distributed, construct a 95% confidence interval for the mean number of days that vehicles in the company's fleet are out of operation.

Answer 1

(10.891; 15.869)

Explanation:

Given the data::

15, 11, 19, 24, 6, 18, 20, 15, 18, 12, 14, 19, 5, 21, 12, 11, 8, 10, 7, 2, 14

The confidence interval is obtained using the relation :

Mean ± margin of error

Margin of Error = Zcritical * s/sqrt(n)

Using a calculator, the mean value of the data and standard deviation above is :

Mean = Σx / n ; n = sample size

Mean, xbar = 281 / 21

Mean = 13.38

Standard deviation , s = sqrt[Σ(X - xbar)²/ n-1]

Standard deviation, s = 5.82

Margin of Error = Zcritical * s/sqrt(n)

Zcritical at 95% confidence level = 1.96

Margin of Error = 1.96 * 5.82/sqrt(21) = 2.489

Confidence interval :

Mean ± margin of error

13.38 ± 2.489

Lower boundary = (13.38 - 2.489) = 10.891

Upper boundary = (13.38 + 2.489) = 15.869

(10.891; 15.869)