In Exercises 1–3, use the data listed below. The values are departure delay times (minutes) for American Airlines flights from New York to Los Angeles. Negative values correspond to flights that departed early.

Flight 1(min)	-2	-1	-2	2	-2	0	-2	-3
Flight 19 (min)	19	-4	-5	-1	-4	73	0	1
Flight 21(min)	18	60	142	-1	-11	-1	47	13

Departure Delay Times Use a nonparametric test to test the claim that the three samples are from populations with the same median departure delay time.

Threesamples show the departure delay times (in minutes) for three flights.

To test if the samples come from populations with equal medians, you can use theKruskal-Wallis test,a non-parametric test.

The null hypothesis for the test is as follows:

The three samples come from populations with the same median.

The alternative hypothesis for the test is as follows:

The three samples do not come from populations with the same median.

The test is two-tailed.

The ranks of the observations from the three samples are obtainedusing the following steps:

• Combine the three samples and mark each observation with the name/number of the sample from which it is obtained.
• The smallest observation is given rank one;the next smallest is given rank two, and so on until the largest value is reached.
• If two observations have the same rank, they are given the mean of the ranks.

The following table shows the ranks:

The sum of the ranks corresponding to Flight 1 is computed as follows:

\(\begin{array}{c}{R_1} = 17.5 + 13.5 + .... + 20\\ = 122\end{array}\)

The sum of the ranks corresponding to Flight 19 is computed as follows:

\(\begin{array}{c}{R_2} = 5 + 21.5 + .... + 9\\ = 106\end{array}\)

The sum of the ranks corresponding to Flight 21 is computed as follows:

\(\begin{array}{c}{R_3} = 6 + 3 + ... + 7\\ = 72\end{array}\)

Let n be the sample size.

Let N be the total number of observations

Here,\({n_1} = {n_2} = {n_3} = 8\)

and

\(\begin{array}{c}N = 8 + 8 + 8\\ = 24.\end{array}\)

The value of the test statistic is computed below:

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{24\left( {25} \right)}}\left( {\frac{{{{122}^2}}}{8} + \frac{{{{106}^2}}}{8} + \frac{{{{72}^2}}}{8}} \right) - 3\left( {25} \right)\\ = 3.26\end{array}\)

Let k be the number of samples.

The degrees of freedom are computed as follows:

\(\begin{array}{c}df = k - 1\\ = 3 - 1\\ = 2\end{array}\)

The critical value of chi-square for \(\alpha = 0.05\)with 2 degrees of freedom is 5.991.

Since the test statistic value is less than the critical value, the decision fails to reject the null hypothesis.

It can be concluded that there is no difference in the medians of the three samples.