Question

Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean.

Flight Arrivals Listed below are arrival delays (minutes) of randomly selected American Airlines flights from New York (JFK) to Los Angeles (LAX). Negative numbers correspond to flights that arrived before the scheduled arrival time. Use a 95% confidence interval. How good is the on-time performance?

-5 -32 -13 -9 -19 49 -30 -23 14 -21 -32 11

Short answer

The 95% confidence interval is given by $-24.3\min <\mu <6.0\min$.

As the confidence interval expresses that most flights arrive before the scheduled time, the on-time performance is good for the flights of American Airlines from New York to Los Angeles.

Step-by-step solution

Given information

The values for different arriving times of flights are given.

Sample size $n=12$.

Calculate the mean

Let X represent the arrival times of different flights.

The sample mean is given as follows.

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{\left(-5\right)+\left(-32\right)+...+\left(11\right)}{12} \\ =-9.1667 \end{gathered}$$

Therefore, the mean value is -9.1667 min.

Calculate the standard deviation

The standard deviation is given as follows.

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(-5-\left(-9.1667\right)\right)}^2+{\left(-32-\left(-9.1667\right)\right)}^2+...+\left(11-\left(-9.1667\right)\right)}{12-1}} \\ =23.8626 \end{gathered}$$

Therefore, the standard deviation is 23.8626 min.

Calculate the critical value

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

The critical value is computed by using the significance level of 0.05 and the degree of freedom 11 from the standard normal table.

The level of significance is given as follows.

$$\begin{gathered} \alpha =\left(100-95\right)\% \\ =0.05 \end{gathered}$$

From the t-table, the critical value is $t_{\frac{\alpha }{2}}=2.201$.

Calculate the margin of error

The margin of error is as follows.

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ =2.201\times \frac{23.8626}{\sqrt{12}} \\ =15.1616 \end{gathered}$$

Therefore, the margin of error is 15.1616 min.

Calculate the confidence interval

The confidence interval for the mean is given as

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E \\ -9.1667-15.1616<\mu <-9.1667+15.1616 \\ -24.328<\mu <5.995 \end{gathered}$$

The lower limit of the 95% confidence interval is -24.3 min.

The upper limit of the 95% confidence interval is 6.00 min.

As the confidence interval includes more negative measures, it implies that most flights arrive before the scheduled time. It indicates a good on-time schedule.