Mann-Whitney U Test (go to the calculator)

The Mann-Whitney U test, also called the Wilcoxon rank-sum test, is a non-parametric test. It checks continuous or ordinal data for a significant difference between two independent groups. The test merges the data from the two groups. Then, it sorts the data by the value. Unlike the t-test that compares the groups' averages, the rank test compares the entire distributions.
When the two groups' distributions have a similar shape, the test will also compare the median of each group.
For symmetrical distribution, the median is the average.

When to use?

• Not normal, the data is not normally distributes and the sample size is less than 30.
• Ordinal data, but not interval scaled. You know the order but not the differences between the values.
  for example - unhappy, neutral, happy
• Outliers the test is more robust to outliers than the T-test

What does a higher rank mean?

Higher rank, R_i, for group i (lower U_i), says that the probability to get higher value from this group is higher.
Following an example:

R_B > R_C, μ_B < μ_C.
The rank of group B, 108, is greater than the rank of group C, 63. But the mean of group B, 20, is less than the mean of group C, 65.67.
There is a high probability that a random value from group B will be greater than random value from group C, but if you repeat this process 20 times, most likely you will accumulate a greater total from group C.

Med_A = Med_B, R_B > R_A.
The median of group A, 20, equals to the median of group B, 20. But the rank of group B, 101.5, is greater than the rank of group A, 69.5. There is a high probability that the random value from group B will be greater than the random value from group A.

Assumptions

• Independent observations.
• Ordinal / Continuous - the compared data consist of ordinal data or continuous data.
• Similar Shape - If the test is required to compare the medians, both groups should have a similar shape. Otherwise, the test can compare only the ranks.

U Calculation

• Merge the data from the two groups to one group.
• Sort the data from low value to high value.
• Rank the merged list, the lower value gets rank 1 , the second rank 2, etc.
  When having ties group, identical value for several observations, the rank will be the average of the ranks for the entire group.
• Calculate the ranks
  R_1i - the rank of the i member in group 1.
  R_2i - the rank of the i member in group 2.
  n₁ the number of observations in group 1.
  n₂ the number of observations in group 2.
  $$R_1=\sum_^>$$ $$R_2=\sum_^>$$
• Calculate Ui
  $$U_1=n_1n2+\frac- R_1.\\ U_2=n_1n2+\frac- R_2.\\ (U_1+U_2=n_1n_2)$$ Since the distribution is symmetrical, usually U is the minimum between U₁ and U₂. $$U=min(U_1,U_2).$$ It is good for the two tails test, but for the one tail test, it will always assume the following H₁: the sample with larger values is bigger than the sample with the smaller values.
  In this tool, the statistics is U₂, in this way, we can calculate the left-tailed or the right-tailed like any other test.

Calculation method

Exact

Calculated the distribution, p(U1 or to get higher value from Group₂. It calculates the U for all the combinations:
$$p(U = u) = \frac$$ The method requires high calculation power, the calculation duration grows exponentially to the total sample size (n₁ + n₂).
The calculation is accurate only when the data does not have ties. The tool uses pre-calculated data to save performance time.

Corrected normal approximation

To get more accurate results, the tool uses the ties corrections, and optionally use the continuity correction and. ties is a group of observations with the same value. $$ Z = \frac < U_2 - \mu_u + C_> $$ $$ \mu= \frac $$ $$ \sigma^2= \frac (1 - C_)$$

Ties correction

$$n = n_1 + n_2
\\ C_ = \sum_^>$$ t - group number of ties.
f_t - number of values in group t.

Continuity correction

When using continuous distribution to calculate discrete data it is better to use continuity correction.
P(X < a) =>P(X < a - 0.5)
P(X > a) => P(X > a + 0.5)

As a result:
Right tail, or two tails with positive Z, (U₂ > μ) , C_continuity = -0.5 .
Left tail, or two tails with negative Z, (U₂ < μ) , C_continuity = 0.5 .
When we don't correct the data, C_continuity = 0.

Choose the method

The tool will use the exact method or the normal approximation per the method definition:

Automatic

This is the recommended method!.
When n is small, n1 ≤ 20 and n2 ≤ 20, and the data doesn't have ties the tool will use the exact value from the pre-calculated data. Otherwise, the tool will use the normal approximation.

Exact

Using the 'Exact' method force the tool to use the exact method even when having ties.
When n is small, n1 ≤ 20 and n2 ≤ 20, the tool will use the exact value from tables. Otherwise, the tool will use the normal approximation.

Z approximation

The tool uses the normal approximation.

Effect Size

The common language effect size is the probability that a random value from Group₁ is greater than random value from Group₂.
$$f=\frac\\ r=\frac>$$

Example

The following example checks the number of questions answered correctly by two independent groups. One group completed training before performed the test and the other group didn't do the training. Following the test results. The sample sizes: n1=8, n2=10. The significant level (α) is 0.05.

Indirect method

• Merge the lists of the two groups to one list.
• Sort by the value, the smallest value first.
  

$$U_1 = n_1n2 + \frac - R_1\\ U_1 = 8*10+\frac - 54.5 = 61.5 \\ U_2=n_1n2+\frac - R_2 \\ U_2 =8*10+\frac - 116.5 = 18.5 \\ (U_1+U_2=61.5 + 18.5 = 80, n_1n_2=8*10 = 80)$$ U = min(61.5 , 18.5) = 18.5

Direct method

• Merge the lists of the two groups to one list.
• Sort by the value, the smallest value first.

Exact calculation

The Mann-Whitney U distribution is discrete. We can calculate the exact cumulative probability, but we can not have critical value for the exact significance level. Hence we calculate the critical value that get the maximum significance level which is not greater than the required significance level (α).
When the statistic equal the critical value, you should reject the null assumption.
Since the data in this example contains ties, the exact calculation is not accurate. Therefore you should use the normal approximation with continuous correctiom.

Statistical tables

Two-tailed (H₀: Group a = Group b)

• Critical Value
  P(X≤17)=0.02171. 0.02171P(X≤18)=0.02726. 0.02726>0.025.
  The left critical value is 17, and the left edge of the region of acceptance is 18.

P(X>63)=1 - P(X≤62)=0.02171. 0.02171P(X>62)=1 - P(X≤61)=0.02726. 0.02726>0.025.
The right critical value is 63, and the right edge of the region of acceptance is 62.
You may calculate as following right=n₁n₂-left. (8*10-17=63)

Left tail (H₀: Group a ≥ Group b)

• Critical Value
  P(X≤20)=0.02171. 0.0416P(X≤21)=0.0506. 0.02726>0.05.
  The left critical value is 20, and the left edge of the region of acceptance is 21.

Right tail (H₀: Group a < Group b)

• Critical Value
  P(X>60)=1 - P(X≤59)=0.04157. 0.04157P(X>59)=1 - P(X≤58)=0.0506. 0.0506>0.05.
  The right critical value is 60, and the right edge of the region of acceptance is 59.

Corrected normal approximation

• $$group_1: [13,13,13], \quad f_1=3.\\ group_2: [17,17], \quad f_2=2.\\ group_3: [24,24], \quad f_3=2.$$ There are 3 tie groups (t=3):
  $$n=n_1+n_2=8+10=18. \\ C_ = \sum_^>\\ C_ = >\\ C_ =\frac=\frac=0.00619$$
• $$ \mu_u= \frac =\frac =40 $$ $$ \sigma_u^2= \frac (1 - C_)\\ \sigma_u^2= \frac <8*10(8 + 10 + 1)>(1 - 0.00619)\\ \sigma_u^2= 125.8826, \sigma_u = 11.22$$ Since the data is discrete and U2 < μ , C_continuity = 0.5. $$ Z = \frac < U_2 - \mu_u + C_>= \frac < 18.5 - 40 + 0.5>= -1.872$$
• P( z ≤ Z) = P( z ≤ -1.872) = 0.0306

Two tailed (H₀: Group a = Group b)

• P-value = 2 * 0.0306 = 0.0612
• Since 0.0612 > 0.05, accept H₀.

Left tail (H₀: Group a ≥ Group b)

• P-value = P( z ≤ -1.872) = 0.0306
• Since 0.0306 < 0.05, reject H₀.

Right tail (H₀: Group a < Group b)

• P-value = P( z ≥ -1.872) = 1 - P( z ≤ -1.872) = 1 - 0.0306 = 0.9694
• Since 0.9604 > 0.05, accept H₀.

Exact Example2

Two-tailed (H₀: Group a = Group b)

α / 2 = 0.05. $$P(X \leq 0)=\frac=\frac=0.029\\ P(X \leq 1)=\frac=\frac=0.057$$ The left critical value - the first value that gets a cumulative probability greater than 0.05 is x=1 $$P(X \geq 12)=\frac=\frac=0.029\\ P(X \geq 11)=\frac=\frac=0.057$$ The right critical value - the first value that gets cumulative probability which is greater than 0.05 is for x=11. Since the distribution is symmetrical you may get the same result as following: 12 - 1 = 11.

Left tail (H₀: Group a ≥ Group b)

$$P(X \leq 1)=\frac=0.057\\ P(X \leq 2)=\frac=0.114$$ Critical value - the first value that gets cumulative probability which is greater than 0.1 is for x=2.

Right tail (H₀: Group a < Group b)

$$P(X \geq 11)=\frac=0.057\\ P(X \geq 10)=\frac=0.114$$ Critical value - the first value that gets cumulative probability which is greater than 0.1 is for x=10.