A car-rental company is interested in the amount of time its vehicles are out of operation for repair work. State all assumptions and find a 90% confidence interval for the mean number of days in a year that all vehicles in the company's fleet are out of operation if a random sample of 10 cars showed the following number of days that each had been inoperative.
10
16
18
20
23
17
15
9
5
12
Find a 90% confidence interval for the population mean.
Assumptions:
1. The sample of 10 cars is a random sample from the entire fleet of the car-rental company.
2. The number of days each car is out of operation is independent of each other.
3. The population of days that cars are out of operation is normally distributed.
Calculations:
Mean of the sample = (10 + 16 + 18 + 20 + 23 + 17 + 15 + 9 + 5 + 12) / 10 = 14.5 days
Standard deviation of the sample = √[Σ(xi - x̄)² / (n - 1)] = 5.084 days
Standard error = standard deviation / √n = 5.084 / √10 = 1.61 days
A 90% confidence interval for the population mean can be calculated using the t-distribution with n-1 degrees of freedom, where n is the sample size. Since n = 10, the degrees of freedom is 9.
Using a t-table or calculator, the t-value for a 90% confidence interval with 9 degrees of freedom is approximately 1.833.
Margin of error = t-value * standard error = 1.833 * 1.61 = 2.95 days
Therefore, the 90% confidence interval for the population mean is 14.5 ± 2.95 days, or (11.55, 17.45) days.