Flight Departure Delays Listed below are departure delay times (minutes) for American Airlines flights from New York to Los Angeles. Negative values correspond to flights that departed early. Use a 0.05 significance level to test the claim that the different flights have the same mean departure delay time. What notable feature of the data can be identified by visually examining the data?

Flight 1	-2	-1	-2	2	-2	0	-2	-3
Flight 19	19	-4	-5	-1	-4	73	0	1
Flight21	18	60	142	-1	-11	-1	47	13

Data are given on the departure delay of the three different flights.

Let\({\mu _1},{\mu _2},{\mu _3}\)be the population mean departure delay times for three flights.

The null hypothesis to test the difference in the departure delay times of the 3 samples is as follows:

\(\begin{aligned}{l}{H_0}:{\mu _1} = {\mu _2} = {\mu _3}\\{H_1}:{\rm{at}}\,{\rm{least}}\;{\rm{one}}\;{\rm{mean}}\;{\rm{is}}\;{\rm{different}}\end{aligned}\)

If the computed F-statistic is greater than the critical value, the null hypothesis is rejected at a 0.05 level of significance.

If the computed F-statistic is less than the critical value, the null hypothesis fails to reject at a 5% level of significance.

Let n denote the sample sizes.

As the 3 samples are of equal size, which is 8(n).

Let\({\bar x_i}\)denote the sample means.

The 3 sample means are computed below:

\[\begin{aligned}{c}{{\bar x}_1} = \frac{{ - 2 - 1 + \ldots - 3}}{8}\\ = - 1.25\\{{\bar x}_2} = \frac{{19 - 4 + \ldots + 1}}{8}\\ = 9.875\\{{\bar x}_3} = \frac{{18 + 60 + \ldots + 13}}{8}\\ = 33.37\end{aligned}\]

Let\(s_{\bar x}^2\)denote the variance of the sample means:

The mean of the 3 sample means is equal to:

\(\begin{aligned}{c}\bar \bar x = \frac{{ - 1.25 + 9.875 + 33.37}}{3}\\ = 14.00\end{aligned}\)

The variance of the sample means is computed below:

\(\begin{aligned}{c}s_{\bar x}^2 = \frac{{\sum\limits_{i = 1}^3 {{{({{\bar x}_i} - \bar \bar x)}^2}} }}{{3 - 1}}\\ = \frac{{{{\left( { - 1.25 - 14} \right)}^2} + {{\left( {9.875 - 14} \right)}^2} + {{\left( {33.37 - 14} \right)}^2}}}{{3 - 1}}\\ = 312.39\end{aligned}\)

The variance between sample means is equal to:

\(\begin{aligned}{c}ns_{\bar x}^2 = 8\left( {312.39} \right)\\ = 2499.875\end{aligned}\)

Therefore, the variance between sample means is equal to 2499.875.

Now, the 3 sample variances denoted by\({s_i}^2\)are computed below:

\[\begin{aligned}{c}{s_1}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( { - 2 + 1.25} \right)}^2} + {{\left( { - 1 + 1.25} \right)}^2} + \ldots + {{\left( { - 3 + 1.25} \right)}^2}}}{{8 - 1}}\\ = 2.5\end{aligned}\]

\(\begin{aligned}{c}{s_2}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {19 - 9.875} \right)}^2} + {{\left( { - 4 - 9.875} \right)}^2} + \ldots + {{\left( {1 - 9.875} \right)}^2}}}{{8 - 1}}\\ = 709.8393\end{aligned}\)

\(\begin{aligned}{c}{s_3}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{3i}} - {{\bar x}_3})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {18 - 33.37} \right)}^2} + {{\left( {60 - 33.37} \right)}^2} + \ldots + {{\left( {13 - 33.37} \right)}^2}}}{{8 - 1}}\\ = 2525.411\end{aligned}\)

The variance within samples\(\left( {s_p^2} \right)\)is equal to:

\(\begin{aligned}{c}{s_p}^2 = \frac{{{s_1}^2 + {s_2}^2 + {s_3}^2}}{3}\\ = \frac{{2.5 + 709.8393 + 2525.411}}{3}\\ = 1079.25\end{aligned}\)

Therefore, the variance within samples is equal to 1079.25.

Now, the F-statistic is computed as shown below:

\(\begin{aligned}{c}F = \frac{{{\rm{variance}}\;{\rm{between}}\;{\rm{samples}}}}{{{\rm{variance}}\;{\rm{within}}\;{\rm{samples}}}}\\ = \frac{{n{s_{\bar x}}^2}}{{{s_p}^2}}\\ = \frac{{2499.10}}{{1079.25}}\\ = 2.316\end{aligned}\)

Therefore, the value of the F-statistic is equal to 2.316.

Let k be the number of samples.

It is known that the value of k is 3.

Thus, the degrees of freedom are calculated as shown below:

\(\begin{aligned}{c}df = \left( {k - 1,k\left( {n - 1} \right)} \right)\\ = \left( {3 - 1,3\left( {8 - 1} \right)} \right)\\ = \left( {2,21} \right)\end{aligned}\)

The critical value of F at a 0.05 level of significance with (2,21) degrees of freedom from the F-distribution table is equal to 3.4668.

It can be observed that:

\(\begin{aligned}{c}F < {F_{crit}}\\\left( {2.316} \right) < \left( {3.4668} \right)\end{aligned}\)

Since the value of the F-statistic is less than the critical value, the null hypothesis is failed to reject at 0.05 level of significance.

Therefore, there is not enough evidence to conclude that there is a significant difference in the departure delay time.

From the observations, the flight 21 has a large departure delay times as compared to other flights.