How to interpret Mann-Whitney's statistical significance if median is equal? Testing the difference between the observations of two groups by using Mann-Whitney Test has given the following output (from minitab):
                N  Median
positives  137  1.0000
negatives  892  1.0000
Point estimate for η1 - η2 is 0.0000
99.0 Percent CI for η1 - η2 is (0.0001,-0.0001)
W = 56899.5
Test of η1 = η2 vs η1 ≠ η2 is significant at 0.0000
The test is significant at 0.0000 (adjusted for ties)

When I was testing other pairs of groups, I concluded that a group tends to have larger values than another group based on their medians, if the test shows a statistically significant difference. However, the example above shows that I may not be interpreting the results correctly.
Here is some additional descriptive statistics for the example above:
Variable   |  Mean | StDev  | Minimum | Q1 |Median| Q3 | Maximum
positives  |  4.13 | 13.17  |  1.00   |1.00| 1.00 |1.00| 116.00
negatives  | 6.851 | 20.503 |  0.000  |1.00| 1.00 |5.00| 434.000



*

*What have I done wrong?

*How can I figure out the direction of the difference? Can I conclude that the negatives tends to have larger values based on their larger Q3, for example?
 A: The Mann-Whitney is not a test of medians. At best, the Mann-Whitney test can only be claimed to a be a test of differences in mean-rank between two populations' pooled ranking.
You can easily calculate medians empirically and perform a basic Wald test if you need a test of medians. 
The Mann-Whitney test happens to be a reasonably powerful test of medians only when the underlying distributions are symmetric, an assumption that is clearly violated in these data. However, if a distribution is symmetric, the median also happens to be the mean (when variance is finite). This means the Mann-Whitney and the t-test are testing the same hypothesis in symmetric distributions.
A: Mann-Whitney U test is a rank-sum test, hence it doesn't really care about distribution properties such as mean, media, etc, it only cares that one of your variables tends to have higher values than the other, hence the former has a higher sum of ranks. Nevertheless, if you look closely at this table: 
Variable   |  Mean | StDev  | Minimum | Q1 |Median| Q3 | Maximum
positives  |  4.13 | 13.17  |  1.00   |1.00| 1.00 |1.00| 116.00
negatives  | 6.851 | 20.503 |  0.000  |1.00| 1.00 |5.00| 434.000

you might notice that both variables have equal 25-percentiles, means and medians, while the seconds one has a higher 75-percentile. That supports the observation that the second distribution is likely to have a higher rank-sum. 
Edit
Inspired by AdamO's comment, I made a little research on the U-test. According to this published response in Arthritis and Rheumatism journal (impact-factor >7) the test can only compare two distributions of similar shape. This assumption is clearly violated.

The Mann-Whitney U test (2) and the Kruskal-Wallis test (3) are
  nonparametric methods designed to detect whether 2 or more samples
  come from the same distribution or to test whether medians between
  comparison groups are different, under the assumption that the shapes
  of the underlying distributions are the same. Thus, these
  nonparametric tests are commonly used to determine whether medians,
  not means, are different between comparison groups. Although these
  tests are often used to compare means when normality assumption is not
  violated, strictly speaking, interpreting the results of non-
  parametric tests for mean comparison is inaccurate. When the
  distribution of a variable is skewed (for example, as in the values
  for C-reactive protein that van der Helm-van Mil et al present in
  Table 2 of their article 1), only assertions on whether medians, and
  not means, were different between groups should be made using
  nonparametric methods.

Given that your sample is not small I would recommend you to try a permutation test to answer your question. Here is a good discussion on their limitations
