The document describes the Mann-Whitney U test, a non-parametric test used as an alternative to the unpaired t-test. It tests whether two independent samples come from the same population by comparing their medians. The test assumes the two distributions are similar in shape but not necessarily normal. To perform it, the data is ranked and a test statistic U is calculated and compared to a critical value to determine if the null hypothesis that the two populations are equal can be rejected. Advantages are that it can handle skewed data and data sets of different sizes, while disadvantages include a lengthy calculation process and inability to explain why a difference exists.

1. MANN- WHITNEY U TEST
Mrs. D. Melba Sahaya Sweety RN,RM
PhD Nursing , MSc Nursing (Pediatric Nursing),BSc Nursing
Associate Professor
Department of Pediatric Nursing
Enam Nursing College, Savar,
Bangladesh.
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2. INTRODUCTION
 The Mann-Whitney U test is a non-parametric test that can be used in
place of an unpaired t-test. It is used to test the null hypothesis that two
samples come from the same population (i.e. have the same median) or,
alternatively,
 The Mann-Whitney test is a test of both location and shape. Given two
independent samples, it tests whether one variable tends to have values higher
than the other. Although it is a non-parametric test it does assume that the
two distributions are similar in shape.
 Thus the hypotheses of Mann-Whitney U Test results in:
• The null hypothesis (H0) is that the two populations are equal.
• The alternative hypothesis (H1) is that the two populations are not equal.
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3. ASSUMPTIONS
• The assumptions for the Mann-Whitney U Test include:
1. Ordinal or Continuous : The variable you’re analyzing is ordinal or continuous.
Examples of ordinal variables include Likert items (e.g., a 5-point scale from
“strongly disagree” to “strongly agree”). Examples of continuous variables include
height (measured in inches), weight (measured in pounds), or exam scores (measured
from 0 to 100).
2. Skewed Distribution : free to use a Mann Whitney u test when the variable is
skewed rather than normally distributed.
3. Random Sample :The data should be two randomly selected independent samples,
meaning the groups have no relationship to each other.
4. Enough Data : The sample size (or data set size) should be greater than 5 in each
group
5. Similar Shape Between Groups : While the data in both groups are not assumed to
be Normal, the data are assumed to be similar in shape across the two groups.
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4. STEPS OF U TEST
1. State the hypotheses. In most cases, a Mann-Whitney U test is performed as a
two-sided test. The null and alternative hypotheses are written as:
• H0: The two populations are equal
• Ha: The two populations are not equal
2. Determine a significance level to use for the hypothesis.
• Decide on a significance level. Common choices are .01, .05, and .1.
3. Find the test statistic.
To assign rank to each observation of the two samples
• We combine the two samples into a single sample
• Arrange their values in ascending order of magnitude
• Then ranks are assigned to the combined series data 4

5. STEPS OF U TEST
• The test statistic is denoted as U and is the smaller of U1 and U2, as defined below:
• U1 = n1n2 + n1(n1+1)/2 – R1
• U2 = n1n2 + n2(n2+1)/2 – R2
• where n1 and n2 are the sample sizes for sample 1 and 2 respectively, and R1 and
R2 are the sum of the ranks for sample 1 and 2 respectively.
• U = min(U1 U2)
4. Reject or fail to reject the null hypothesis.
• Using the test statistic, determine whether to reject or fail to reject the null hypothesis
based on the significance level and critical value found in the Mann-Whitney U
Table.
5. Interpret the results.
• Interpret the results of the test in the context of the question being asked.
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6. MANN-WHITNEY
U TEST
Test the hypothesis that the median HDL cholesterol levels in adult population of city A and
city B are the same. Using the following observation and the Mann-Whitney test at 5%level of
significance by the following data
City A 42 20 51 39 57 60 23
City B 30 42 25 29 35
Solution:
Hypothesis:
Null Hypothesis H0: The two population are same (n1 = n2)
Alternative Hypothesis H1: The two population are not same (n1 ≠ n2)
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7. MANN-WHITNEY
U TEST
n1= 7, n2= 5
U1 = n1n2 + n1(n1+1)/2 – R1
2
Cholesterol
level
Rank City
20 1 A
23 2 A
25 3 B
29 4 B
30 5 B
35 6 B
39 7 A
42 8.5 A
42 8.5 B
51 10 A
57 11 A
60 12 A
R1 = Sum of ranks of A city
R1 = 1+2+7+8.5+10+11+12
R1 = 51.5
R2 = Sum of ranks of B city
R2 = 3+4+5+6+8.5
R2 = 26.5
- 51.5
7 x 5 +
U1 = 7(7+1)
U1 = 35+ 28 – 51.5
U1 = 63 – 51.5
U1 = 11.5
U2 = n1n2 + n2(n2+1)/2 – R2
U2 = 7 x 5 + 5(5+1)
2
- 26.5
U2 = 35+ 15 – 26.5
U2 = 50 – 26.5
U2 = 23.5
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8. MANN-WHITNEY
U TEST
U = min(U1 U2)
U = 11.5
The critical value of U for n1= 7, n2= 5 at 5% significance for two tail is 5
Since the U(cal) > U(tab)
Hence We reject null hypothesis
Therefore we may conclude that the median HDL cholesterol levels in the
adult population of the cities A and B are not equal
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9. ADVANTAGES
• States whether the difference is significant or occurred by
chance
• Shows the median between 2 sets of data
• You can use data sets of different sizes
• Good with dealing with skewed data so data doesn't need to be
normally distributed
DISADVANTAGES
• Lengthy calculation - prone to human error
• Does not explain why there is a difference
• More appropriate when the data sets are independent of each other
• More appropriate when both sets of data have the same shape distribution
• Become less accurate when the sample size is below 5 or above 20
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10. Slide Title
Product A
• Feature 1
• Feature 2
• Feature 3
Product B
• Feature 1
• Feature 2
• Feature 3
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