First, we put both samples together and organize it in ascending order, which is shown in the table below:
| Sample | Value |
| 1 | 23 |
| 2 | 33 |
| 2 | 37 |
| 1 | 38 |
| 1 | 41 |
| 1 | 45 |
| 2 | 48 |
| 2 | 52 |
| 1 | 57 |
| 1 | 69 |
| 2 | 74 |
| 2 | 82 |
| 1 | 95 |
| 2 | 103 |
Now, that the values that are in ascending order are assigned ranks to them, taking care of assigning the average rank to values with rank ties
| Sample | Value | Rank | Rank (Adjusted for ties) |
| 1 | 23 | 1 | 1 |
| 2 | 33 | 2 | 2 |
| 2 | 37 | 3 | 3 |
| 1 | 38 | 4 | 4 |
| 1 | 41 | 5 | 5 |
| 1 | 45 | 6 | 6 |
| 2 | 48 | 7 | 7 |
| 2 | 52 | 8 | 8 |
| 1 | 57 | 9 | 9 |
| 1 | 69 | 10 | 10 |
| 2 | 74 | 11 | 11 |
| 2 | 82 | 12 | 12 |
| 1 | 95 | 13 | 13 |
| 2 | 103 | 14 | 14 |
The sum of ranks for sample 1 is:
R1=1+4+5+6+9+10+13=48
and the sum of ranks of sample 2 is:
R2=2+3+7+8+11+12+14=57
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H0: Median (Difference) = 0 i.e Median delay time for airport 1 = Median delay time for airport 2
Ha: Median (Difference)
0 i.e Median delay time for airport 1
Median
delay time for airport 2
Test statistic W : 20
p-value : 0.620047
null hypothesis μ0=μ0= 0.0
alternative hypothesis: two sided,
Since p value=0.620047 > 0.05. there is not enough evidence to claim that the population median of differences is different than 0, at the 0.05 significance level.

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