Question

Flight Arrivals. Listed below are the arrival delay times (min) of randomly selected American Airlines flights that departed from JFK in New York bound for LAX in Los Angeles. Negative values correspond to flights that arrived early and ahead of the scheduled arrival time. Use these values for Exercises 1–4.

-30 -23 14 -21 -32 11 -23 28 103 -19 -5 -46

Confidence Interval Construct a 95% confidence interval estimate of the mean arrival delay time for the population of all American Airlines flights from JFK to LAX.

Short answer

The 95% confidence interval estimate of the population’s mean arrival delay time is equal to (-28.9 minutes, 21.8 minutes).

Step-by-step solution

Given information

A sample of arrival delay times (in minutes) is given for a set of American Airlines flights from New York to LA.

Formula of the confidence interval

The formula for computing the confidence interval estimate of the population mean arrival delay time is

\(CI = \left( {\bar x - E,\bar x + E} \right)\)

Where,

\(E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\).

Sample mean and sample standard deviation

The sample mean is computed as shown below.

\(\begin{array}{c}\bar x = \frac{{\left( { - 30} \right) + \left( { - 23} \right) + ...... + \left( { - 46} \right)}}{{12}}\\ = - 3.6\end{array}\)

Thus, \(\bar x = - 3.6\)minutes.

The sample standard deviation of the arrival delay times is computed below.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 30 - 3.6} \right)}^2} + {{\left( { - 23 - 3.6} \right)}^2} + .... + {{\left( { - 46 - 3.6} \right)}^2}}}{{12 - 1}}} \\ = 39.9\end{array}\)

Thus, s=39.9 minutes.

Sample size and t-value

The sample size is n=12.

The confidence level is equal to 95%. Thus, the level of significance is equal to 0.05.

The degree of freedom is equal to

\(\begin{array}{c}df = n - 1\\ = 12 - 1\\ = 11\end{array}\).

By referring to the t-table, the value of \({t_{\frac{\alpha }{2}}}\) with \(\alpha = 0.05\) and the degree of freedom equal to 11 is equal to 2.201.

Calculation

The margin of error becomes

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\\ = 2.201 \times \frac{{39.9}}{{\sqrt {12} }}\\ = 25.3514\end{array}\).

The 95% confidence interval becomes

\(\begin{array}{c}CI = \left( {\bar x - E,\bar x + E} \right)\\ = \left( { - 3.6 - 25.3514, - 3.6 + 25.3514} \right)\\ = \left( { - 28.9,21.8} \right)\end{array}\).

Therefore, the 95% confidence interval estimate of the population’s mean arrival delay time is equal to (-28.9 minutes, 21.8 minutes).