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Mann–Whitney U Test:
An Extended Presentation
Introduction
Independent Groups Design
Analysis Using the Mann–Whitney U Test
Determining the Probability of U if Chance Alone Is Responsible
Using Tables C.1–C.4
Summary of Protein–IQ Data Analysis
Tied Ranks
Practical Considerations in Using the Mann–Whitney U Test
Summary
Important Terms
Questions and Problems
Notes
Web Connection
INTRODUCTION
In this presentation, we shall consider hypothesis testing using an independent
groups design. We have chosen the Mann–Whitney U test as the inference test to
use when first considering this design because it has a probability distribution
that is easy to understand. Unlike the sign test, however, the Mann–Whitney U
test is a powerful test and therefore has practical utility. Once you understand
how the Mann–Whitney U test works, you will have covered two probability distributions (the binomial distribution and the distribution of U). With this background, you will be ready to grasp the importance of probability distributions to
hypothesis testing. We treat this topic formally in Chapter 12.
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Mann–Whitney U Test: An Extended Presentation
INDEPENDENT GROUPS DESIGN
There are essentially two basic experimental designs used in studying behavior.
We met the first when discussing the sign test. This design is called the repeated
or replicated measures design. The simplest form of the design uses two conditions: an experimental and a control condition. The essential feature of the design
is that there are paired scores between conditions, and difference scores are analyzed to determine whether chance alone can reasonably explain them.
The other type of design is called the independent groups design. In this design, subjects are randomly selected from the population and then randomly divided into two or more groups. The most basic experiment has only two groups.
There is no basis for pairing of subjects, and each subject is tested only once. All
of the subjects in one of the groups (it doesn’t matter which group) are run in the
experimental condition. These subjects are referred to as the experimental group.
All of the subjects in the other group receive the control condition. Subjects in
this condition constitute the control group. In analyzing the data, there is no basis for pairing scores between the conditions. Rather, a comparison is made
between the scores of each group to determine whether chance alone is a reasonable explanation of the differences between the group scores.
example
Effect of High-Protein Diet on
Intellectual Development
To illustrate this design, let’s consider an example. Suppose you are a developmental
psychologist with special competence in nutrition. Based on previous research and theory, you believe that a high-protein diet eaten during early childhood is important for
intellectual development. The diet in the geographical area where you live is low in
protein. You believe the low-protein diet eaten during the first few years of childhood
is detrimental to intellectual development. If you are correct, a high-protein diet should
result in higher intelligence. You decide to investigate this directional alternative hypothesis, and you choose the independent groups design for your experiment.
Alpha is set at the beginning of the experiment to 0.051 tail. Six children are randomly chosen from the 1-year-old children living in your city. Note that in an actual experiment the sample size would be much larger. We have limited it to six in this example for clarity in probability determinations. The six children are then randomly divided
into two groups of three each. One group is fed the usual low-protein diet for 3 years,
whereas the other group receives a diet high in protein for the same duration. We shall
call the low-protein group the control group and the high-protein group the experimental group. At the end of the 3 years, each child is given an IQ test. The scores are
given in Table 1.
What can we conclude from these data? Most of the scores in the experimental
group are higher than the scores in the control group. The mean IQ of the experimental group (XE 100) is higher than the mean IQ of the control group (XC 90). These
results are consistent with the hypothesis that a high-protein diet increases IQ. Can we
therefore conclude that the high-protein diet was responsible for the higher IQ scores?
Not necessarily.
Suppose that, instead of giving the experimental group a high-protein diet, we
gave them the same diet as the control group. Isn’t it possible that, just due to chance
alone, we would get scores as high as or even higher than those obtained in the original experiment? The answer is yes. Thus, as with the repeated measures design, we need
to evaluate chance before we can conclude for H1. The null hypothesis for the inde-
Analysis Using the Mann–Whitney U Test
3
t a b l e 1 Data from the protein–IQ
experiment
Control Group,
Low Protein
C
1
Experimental Group,
High Protein
E
2
84
194
88
101
98
105
XC 90
XE 100
pendent groups design is the same as for the repeated measures design. Since the alternative hypothesis is directional for this experiment, the null hypothesis specifies that
a high-protein diet during infancy does not increase intellectual development compared to a low-protein diet. As with the repeated measures design, we always evaluate
H0 first and then indirectly decide about H1. Of course, we evaluate H0 by determining
whether chance is a reasonable explanation of the results.
In the independent groups design, we do not analyze difference scores as in the repeated measures design. Instead, we analyze the two groups of scores as separate samples. The control group scores can be considered a random sample taken from a theoretical population of IQ scores that would have resulted after giving all the 1-year-old
children living in the city the low-protein diet for 3 years. The scores in the experimental group can be considered a random sample from a population of IQ scores that
would have resulted had the same children been fed the high-protein rather than the
low-protein diet. If chance is the correct explanation of the results, then the high- and
low-protein diets have the same effect on IQ, and the two theoretical populations
would have identical distributions.
To evaluate chance, we must determine the probability of getting the obtained results or
results even more extreme if the two groups of sample scores are random samples from
the two populations having identical scores.
As with the replicated measures design, in assessing H0, we assume chance to be true
and calculate the previously mentioned probability. If the obtained probability is equal
to or lower than the alpha level, we reject H0 and conclude in favor of H1. If the obtained probability is higher than alpha, we retain H0. Whichever way we conclude, we
run the same risks as discussed previously.
ANALYSIS USING THE MANN–WHITNEY U TEST
The Mann–Whitney U test analyzes the separation between the two sets of
sample scores and allows us to determine the probability of getting the obtained
separation or even greater separation if both sets of sample scores are random
samples from identical populations. Although separation between the two samples is not a quantity you are used to dealing with, it should be intuitively clear
that the greater the separation between the two sets of scores, the more reasonable it is that they are not random samples from the same or identical populations. Conversely, the more overlap between the two sets of scores, the more
reasonable chance becomes.
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Mann–Whitney U Test: An Extended Presentation
IQ:
Low-protein diet
High-protein diet
µ = 90
σ = 10
µ = 170
σ = 10
f i g u r e 1 Population scores resulting from eating a low-protein or a high-protein diet
To illustrate, let’s assume that if we gave the low-protein diet to all of the 1year-old children in your geographical area, the resulting set of IQ scores would
be normally distributed with 90 and 10. Next, suppose the high-protein
diet has a very large effect on IQ, such that if we gave it to the same children, we
would find 170 and 10. The two IQ populations are shown in Figure 1.
Since there is negligible overlap between them, random sampling from these
populations would produce very little overlap of scores in the two samples. For
instance, if we were randomly sampling three scores from each population, it is
quite probable that all of the scores in the high-protein sample would be higher
than the three scores in the low-protein sample. Table 2 presents the original
three scores for the low-protein sample and, for the high-protein sample, three
randomly selected scores from the high-protein population. The scores have been
rank-ordered to show more clearly the separation between the groups. In this example, there is complete separation. All of the control group scores are lower
than the experimental group scores.
From Figure 1, we can readily see that, as the effect of high protein on IQ decreases, the high-protein population slides to the left, and its overlap with the
low-protein population increases. Finally, when there is no difference in effect on
IQ between the two diets, the populations superimpose on one another. Overlap
is complete. In this case, random sampling from identical populations of scores
accounts for any separation between the scores in the sample experimental and
control groups. It should be apparent that, when the two populations have identical distributions, the overlap between sample group scores is likely to be greater
t a b l e 2 Hypothetical data with no overlap
Control Group
Low Protein
C
Experimental Group
High Protein
E
84
163
88
172
98
175
C1
C2
C3
E1
E2
E3
84
88
98
163
172
175
Analysis Using the Mann–Whitney U Test
5
than when there is a great separation between populations. Thus, the degree of
separation in sample group scores is a measure of how reasonable chance is as an
explanation. The more separated the sample group scores, the less reasonable it
is to conclude that chance is responsible for the separation. Conversely, the less
separated the sample group scores, the more reasonable it is to conclude in favor
of chance.
Calculation of Separation (Uobt and Uobt )
The degree of separation between the two samples can be calculated in two
ways: (1) by counting the total number of C scores that are lower than E scores
or by counting the total number of E scores that are lower than C scores an
(2) by using two equations. Whichever method is used, there are always two numbers that result. The smaller of the two numbers is arbitrarily called Uobt and the
larger Uobt. Thus,
Uobt Smaller of the two numbers indicating the degree of separation
between the two samples
Uobt Larger of the two numbers indicating the degree of separation
between the two samples
The subscript “obt” stands for obtained, and Uobt means the U value calculated
from the experimental data. Note that Uobt and Uobt indicate the same degree of
separation. Let’s first see how to calculate Uobt and Uobt by counting Es and Cs,
and then by using the equations.
Determining Uobt and Uobt by Counting Es and Cs
We’ll begin with the example where there was complete separation between the
samples. The scores are in Table 3.
t a b l e 3 Hypothetical data with no overlap
repeated
Control Group
C
Experimental Group
E
84
163
88
172
98
175
There are two steps involved in calculating U and U by this method.
STEP 1:
Combine the scores from both groups and rank-order them:
84
C1
STEP 2:
88
C2
98
C3
163
E1
172
E2
175
E3
Count the number of E scores that are lower than C scores or the number of C scores that are lower than E scores:
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Mann–Whitney U Test: An Extended Presentation
Es Lower Than Cs
Cs Lower Than Es
Es C1 0
Cs E1 3
Es C2 0
Cs E2 3
Es C3 0
Cs E3 3
Total Es Cs 0
Total Cs Es 9
To count the total number of E scores that are lower than C scores, count the
number of E scores that are lower than each C score and sum these values. This
determination is shown in the table. There are no E scores lower than the C1
score, no E scores lower than C2, and no E scores lower than C3. Their sum equals
zero. Thus, there are zero E scores that are lower than C scores. The same procedure is followed in counting the number of C scores that are lower than E scores.
In all, there are nine C scores that are lower than E scores: three lower than E1,
three lower than E2, and three lower than E3. Since U is always the lower of these
two totals and U the highest,
Uobt 0
Uobt 9
This is the greatest degree of separation possible for three C scores and three
E scores.
Now let’s calculate the U value for the data of the original experiment. The
original obtained IQ scores are as shown again in Table 4.
t a b l e 4 Data from original experiment
Control Group
C
Experimental Group
E
84
94
88
101
98
105
STEP 1:
Combine the scores and rank-order them:
C1
84
C2
88
E1
94
C3
98
E2
101
E3
105
We can see that these scores are not as separate as in the previous example. There is some overlap (E1 is lower than C3).
STEP 2:
Count the total number of E scores that are lower than C scores or the
number of C scores that are lower than E scores:
Es Lower Than Cs
Cs Lower Than Es
Es C1 0
Cs E1 2
Es C2 0
Cs E2 3
Es C3 1
Cs E3 3
Total Es Cs 1
Total Cs Es 8
Analysis Using the Mann–Whitney U Test
7
Thus, for the original data,
Uobt 1
Uobt 8
Calculating Uobt and Uobt from Equations
Next, let’s illustrate how to calculate U and U from equations. The equations are
as follows:
Uobt n1n2 n1(n1 1)
R1
2
Uobt n1n2 n2(n2 1)
R2
2
where
general equation for finding Uobt or Uobt
general equation for finding Uobt or Uobt
n1 number of scores in group 1
n2 number of scores in group 2
R1 sum of ranks for scores in group 1
R2 sum of ranks for scores in group 2
According to this method, we identify one of the samples as group 1 and the
other as group 2. Then we just go ahead and solve the equations. It doesn’t matter which sample is labeled 1 and which is labeled 2. We still obtain the same two
numbers from the equations. What does change with labeling is which equation
yields the high number and which yields the low number. Since this depends on
which group is labeled 1 and which 2, these equations are both written in terms
of U. In an actual analysis, the equation that yields the lower number is the U
equation; the one that yields the higher number is the U equation.
Let’s try this method on the previous problem. The scores are repeated in
Table 5 for your convenience. We have labeled the control group as 1 and the experimental group as 2.
t a b l e 5 Data from original experiment repeated
Control Group
1
Experimental Group
2
84
94
88
101
98
105
This method of calculating U and U involves three steps:
STEP 1:
STEP 2:
Combine the scores, rank-order them, and assign each a rank score using
1 for the lowest score:
Original score
84
88
94
98
101
105
Rank
1
2
3
4
5
6
Sum the ranks for each group; that is, determine R1 and R2. To find R1, we
add the ranked scores for group 1, and to find R2, we add the ranked
scores for group 2, as shown here:
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Mann–Whitney U Test: An Extended Presentation
Control Group
1
Original
score
Experimental Group
2
Original
score
Rank
84
1
194
3
88
2
101
5
98
4
R1 7
105
6
R2 14
n1 3
STEP 3:
Rank
n2 3
Solve the equations for U and U:
Uobt n1n2 n1(n1 1)
R1
2
Uobt n1n2 n2(n2 1)
R2
2
3(4)
14
2
9 6 14 1
3(4)
7
2
9678
3(3) 3(3) Thus,
Uobt 1
Uobt 8
Again, we point out that both equations are written in terms of U rather than U.
In this case, the equation with R2 yielded U, and the one with R1 gave U, because
the group we labeled 1 had the lower sum of ranks (R1 R2). If we had labeled
the experimental group as 1 and the control group as 2, the equation with R2
would have yielded U, and the one with R1 would have given U. Therefore, we
can’t label the equations generally as U and U. We assign U and U after doing
the calculations. Note that the answers we obtained for U and U are the same as
those we calculated by counting Es and Cs. The equation solution can be used under all circumstances. The counting Es and Cs solution cannot be used when
there are tied scores between samples. After all, which score would we put first
when doing the rank ordering, the tied E score or the C score? Since the equation solution is more general, we shall use it for the remainder of the chapter.
Generalizations
Some generalizations are now in order:
1. The Mann–Whitney U test analyzes the separation between the control
and experimental group scores. For every experiment, there will be two
numbers that indicate the degree of separation between the groups. They
are U and U. Both numbers indicate the same degree of separation for
that set of data. Thus, in the first example, U 0 and U 9. Both 0 and
9 indicate the same degree of separation for the data in that example.
2. A U value of 0 represents the greatest degree of separation possible. It
means there is no overlap between the groups. This is true regardless of
the number of scores in each group. In the second example, there was not
Determining the Probability of U if Chance Alone Is Responsible
9
complete separation between groups. The E1 score intruded into the C
scores, and the C3 score intruded into the E scores. In this case, the Uobt
value was 1. Thus, as U gets larger, the scores become more mixed.
3. The sum of Uobt and Uobt must equal n1n2. Thus,
Uobt Uobt n1n2
In the two examples,
0 9 n1n2
9 3(3)
99
and
1 8 n1n2
9 3(3)
99
Since both Uobt and Uobt indicate the same degree of separation, it is only
necessary to evaluate one of them. We shall always evaluate Uobt because
we know its smallest possible value is always 0 irrespective of n1 and n2.
However, as a computation check, it is a good idea to calculate both of
them and see that the calculated values satisfy the relationship
Uobt Uobt n1n2
4. When using the Mann–Whitney U test to evaluate H0, we must determine
the following:
a. The degree of separation between the control and experimental group
scores. We do this by calculating Uobt or Uobt. For the data of our protein–IQ experiment, Uobt 1 and Uobt 8. Since both Uobt and Uobt indicate the same degree of separation, it is only necessary to calculate
one of them. We shall always work with Uobt.
b. The probability of getting a U value equal to or less than Uobt assuming
the sample scores are random samples from populations with identical
distributions. This value is one-tailed or two-tailed depending on H1. In
this experiment, we need to determine p(Uobt 1)1 tail.
DETERMINING THE PROBABILITY OF U
IF CHANCE ALONE IS RESPONSIBLE
If chance alone is responsible, then the separation between the groups is due to
random sampling from populations with identical distributions. Suppose that we
have two populations with identical distributions and that we randomly sample
three scores from each population. Let’s arbitrarily assign the three scores from
one population to the experimental group and the other three to the control
group. If we rank-order the scores, it turns out there are 20 different orders possible. These are listed in column 2 of Table 6. In order 1 (CCCEEE), there is complete separation between the E and C scores, with the C scores being lower than
the E scores. Order 2 (CCECEE) is one step less separated. As we proceed from
order 1 to order 10, the scores become more and more mixed. Beginning with order 11, the group scores begin to separate again. However, now the E scores are
10
Mann–Whitney U Test: An Extended Presentation
t a b l e 6 Generation of the probability distribution of U and U for n1 n2 3
Order
No.
(1)
Order
(2)
No. of
Es Lower
Than Cs
(3)
No. of
Cs Lower
Than Es
(4)
U
(5)
U
(6)
p(U)
(7)
1
CCCEEE
0
9
0
9
0.05
2
CCECEE
1
8
1
8
0.05
3
CCEECE
2
7
2
7
4
CECCEE
2
7
2
7
f
0.10
0.15
5
CCEEEC
3
6
3
6
6
CECECE
3
6
3
6¶
7
ECCCEE
3
6
3
6
8
CECEEC
4
5
4
5
9
CEECCE
4
5
4
5¶
10
ECCECE
4
5
4
5
11
CEECEC
5
4
4
5
12
ECCEEC
5
4
4
5¶
13
ECECCE
5
4
4
5
14
CEEECC
6
3
3
6
15
ECECEC
6
3
3
6¶
16
EECCCE
6
3
3
6
0.15
0.15
0.15
17
ECEECC
7
2
2
7
18
EECCEC
7
2
2
7
19
EECECC
8
1
1
8
0.05
20
EEECCC
9
0
0
9
0.05
f
0.10
generally lower than the C scores. Finally, in order 20, the scores are again completely separated, with all of the E scores being lower than all of the C scores.
In columns 5 and 6 of Table 6, we have listed the U and U values that correspond to each order. Note that U can take on a value of 0, 1, 2, 3, or 4, whereas
U can be 9, 8, 7, 6, or 5. Thus, no matter what the absolute magnitudes of the
three scores in the control group and the three scores in the experimental group,
these are the only possible values for U and U.
Now, let’s determine the probability of getting each U value. In calculating
these values, we shall consider each half of the distribution separately. For the
half where C scores are lower than E scores,
p(U 0) p(order 1) p(CCCEEE)
Number of favorable events
1
0.05
Total number of possible events
20
1
p(U 1) p(order 2) p(CCECEE) 0.05
20
p(U 2) p(order 3 or order 4) 0.05 0.05 0.10
p(U 3) p(order 5, 6, or 7) 0.05 0.05 0.05 0.15
p(U 4) p(order 8, 9, or 10) 0.05 0.05 0.05 0.15
11
Determing the Probability of U if Chance Alone Is Responsible
Cs < Es
(5, 6, 7)
0.15
(8, 9, 10)
(14, 15, 16)
(11, 12, 13)
Probability
(3, 4)
(a) n1 = n2 = 3
(17, 18)
0.10
0.05
(1)
(2)
0
1
Es < Cs
(19)
(20)
1
0
0
2
3
4
4
3
2
U
Cs < Es
0.08
Es < Cs
Probability
0.06
0.04
0.02
(b) n1 = n2 = 5
0
0
2
4
6
8
10
12 12
U
10
8
6
4
2
0
f i g u r e 2 Two probability distributions for U
For the half of the distribution where E scores are lower than C scores,
p(U 4) p(order 11, 12, or 13) 0.05 0.05 0.05 0.15
p(U 3) p(order 14, 15, or 16) 0.05 0.05 0.05 0.15
p(U 2) p(order 17 or 18) 0.05 0.05 0.10
p(U 1) p(order 19) 0.05
p(U 02 p(order 202 0.05
These values have been entered in column 7 of Table 6.*
The probability distribution of U for n1 and n2 3 has been graphed in Figure
2(a). The distribution is symmetrical and tails off from the center in two directions.
The direction, of course, depends on whether the C scores are higher or lower than
the E scores. The probability distribution of U for n1 and n2 5 is shown in Figure
2(b). Note that it too is symmetrical and has a shape very close to the normal
curve. From these two examples, we would like to make the generalizations that
(1) all probability distributions of U are symmetrical regardless of the size of n1
and n2, and (2) as n1 and n2 increase, the distribution approaches normality.
*
See Note 1 for an alternate method of calculating these probabilities.
12
Mann–Whitney U Test: An Extended Presentation
Knowing the probability distribution of U for n1 and n2 3 when chance
alone is operating, we are now in a position to evaluate H0 for our experiment.
Since H1 is directional (i.e., it predicts the Es to be higher than the Cs), we shall
use a one-tailed evaluation. Our obtained U value was Uobt 1.
p(U 1)1 tail p(U 0)1 tail p(U 1)1 tail
p(order 1) p(order 2)
0.05 0.05 0.10
Since 0.10 is greater than alpha ( 0.05), we retain H0. Random sampling from
populations with identical distributions is a reasonable explanation of the data.
The experiment therefore fails to establish that a high-protein diet increases IQ
relative to a low-protein diet. In concluding this way, we may be making a Type
II error (). The probability of making a Type II error depends, of course, on the
power of the experiment. Everything we said about power in Chapter 11 applies
to the Mann–Whitney U test. Actual calculation of power for this test is complicated and beyond the scope of this presentation. However, with only three subjects in each group, you can guess that the power is low for all but the strongest
effects. Increasing n1 and n2 will, of course, increase power.
USING TABLES C.1–C.4
In evaluating the protein experiment data, we developed the probability distribution of U for n1 n2 3. That distribution was the basis for our decision to retain H0. Fortunately, we do not have to derive a probability distribution of U each
time we do an experiment. The probability distributions of U with different n1
and n2 have already been worked out. Since it would take too many pages to give
the p(U) for each n1 and n2, summary tables have been derived. These are presented in Tables C.1–C.4 in Appendix D. For each cell, there are two entries. The
upper entry is the highest value of U for various n1 and n2 combinations that will
allow rejection of H0. The lower entry is the lowest value of U that will allow rejection of H0. Since both U and U measure the same degree of separation, we
shall only consider the values of U. Each table is for a different alpha level. Let’s
see how to use these tables.
Practice Problem 1
Suppose n1 5 and n2 7, 0.011 tail, and Uobt 2. Can we reject H0?
SOLUTION
STEP 1:
STEP 2:
Find the appropriate U table. Since 0.011 tail, Table C.2 applies.
Locate the U value at the intersection of n1 and n2. Since n1 5 and
n2 7,
U3
Summary of Protein–IQ Data Analysis
STEP 3:
13
This value is the highest value of U that will allow rejection of H0 at
the alpha level heading the table. Thus, a U value of 3 or less will allow rejection of H0 for an alpha level of 0.011 tail. This value has been
derived by determining the probability distribution of U for n1 5
and n2 7. A U value of 3 or less occurs due to chance less than 1
time in 100.
Conclude: Since Uobt 3, we reject H0.
Let’s do another problem.
Practice Problem 2
Suppose n1 8 and n2 10. Alpha has been set at 0.052 tail. Uobt 15. Can
we reject H0?
SOLUTION
STEP 1:
STEP 2:
Find the appropriate U table. Since 0.052 tail, Table C.3 applies.
Locate the U value at the intersection of n1 and n2. Since n1 8 and
n2 10,
U 17
STEP 3:
Conclude: Reject H0 because Uobt 17.
Now let’s evaluate the protein and IQ data. In this experiment, n1 n2 3.
Alpha has been set at 0.051 tail, and Uobt 1. Can we reject H0?
SOLUTION
STEP 1:
STEP 2:
Find the appropriate U table. Since 0.051 tail, Table C.4 applies.
Locate the U value at the intersection of n1 and n2. Since n1 n2 3,
U0
STEP 3:
Conclude: Retain H0 because Uobt 0.
SUMMARY OF PROTEIN–IQ DATA ANALYSIS
In explaining the use of an independent groups design and the analysis of the
data with the Mann–Whitney U test, we necessarily digressed at various stages in
the analysis. We would now like to present the total analysis without digressions.
14
Mann–Whitney U Test: An Extended Presentation
example
The Protein–IQ Experiment Revisited
You believe a high-protein diet during infancy will increase intellectual functioning in
children in your geographical area. You decide to investigate this hypothesis using the
independent groups design. Alpha is set at the beginning of the experiment to 0.051 tail.
Six children are randomly sampled from all children 1 year old living in your geographical area and then further randomly divided into two groups of three children
each. The control group is fed a low-protein diet for 3 years, and the experimental
group gets a high-protein diet for the same time period. At the end of the 3 years, each
child is given an IQ test. The results are in Table 7.
t a b l e 7 Data from protein–IQ experiment repeated
Control Group 1
Experimental Group 2
84
194
88
101
98
105
1.
2.
3.
4.
5.
What is the directional alternative hypothesis?
What is the null hypothesis?
What can we conclude? Use 0.051 tail.
What type error may be involved?
To what population does this conclusion apply?
SOLUTION
1. Directional alternative hypothesis: The directional alternative hypothesis states that
a high-protein diet eaten during infancy will increase intellectual functioning relative to a low-protein diet.
2. Null hypothesis: The null hypothesis states that a high-protein as compared to a lowprotein diet eaten during infancy does not increase intellectual functioning.
3. Conclusion, using 0.051 tail:
STEP 1: Calculate Uobt for the data:
a. Combine the scores, rank-order them, and assign them each a rank
score, using 1 for the lowest score:
Original score
84
88
94
98
101
105
Rank
1
2
3
4
5
6
b. Sum the ranks for each group; that is, determine R1 and R2:
Control Group
1
Original
score
Experimental Group
2
Rank
Original
score
Rank
84
1
194
3
88
2
101
5
4
105
98
6
R1 7
R2 14
n1 3
n2 3
Summary of Protein–IQ Data Analysis
15
c. Solve the equations for Uobt and Uobt:
Uobt n1n2 3(3) n1(n1 1)
2
3(4)
2
R1
7
9678
Uobt n1n2 3(3) n2(n2 1)
2
3(4)
2
R2
14
9 6 14 1
Thus,
Uobt 1
Uobt 8
STEP 2: Evaluate Uobt. With 0.051
tail, Table C.4 is appropriate. With n1 n2 3, the U value needed to reject H0 is U 0. Since Uobt 0, we retain H0.
The null hypothesis is a reasonable explanation of the data. Since we have
not rejected the null hypothesis, we cannot accept H1.
4. Type error: We may have made a Type II error. This high-protein diet may be responsible for the generally higher IQ scores in the experimental group, but due to
relatively low power, we have not been able to reject H0.
5. Population: This conclusion applies to the 1-year-old children living in your geographical area at the time of the experiment.
Practice Problem 3
Let’s increase the N in the previous experiment to increase power and analyze the results. The problem is the same as before except n1 9 and n2 8. The following results are obtained:
Control Group
1
Experimental Group
2
102
110
104
115
105
117
107
122
108
125
111
130
113
135
118
140
120
a. What is the directional alternative hypothesis?
b. What is the null hypothesis?
(continued)
c. What can we conclude? Use 0.051 tail.
d. What type error may be involved?
e. To what population does this conclusion apply?
SOLUTION
a. Directional alternative hypothesis: same as before.
b. Null hypothesis: same as before.
c. Conclusion, using 0.051 tail:
STEP 1:
Calculate Uobt for the data:
a. Combine the scores, rank-order them, and assign each a rank
score, using 1 for the lowest score:
Original score
102
104
105
107
108
110
111
113
115
1
2
3
4
5
6
7
8
9
Original score
117
118
120
122
125
130
135
140
Rank
10
11
12
13
14
15
16
17
Rank
b. Sum the ranks for each group; that is, determine R1 and R2:
Control Group
1
Original
score
Experimental Group
2
Original
score
Rank
Rank
102
1
110
6
104
2
115
9
105
3
117
10
107
4
122
13
108
5
125
14
111
7
130
15
113
8
135
16
118
11
140
17
R2 100
120
12
n2 8
R1 53
n1 9
c. Solve the equations for Uobt and Uobt:
Uobt n1n2 n1(n1 1)
R1
2
9(10)
53
2
72 45 53
64
9(8) 16
Uobt n1n2 n2(n2 1)
R2
2
8(9)
100
2
72 36 100
8
9(8) Tied Ranks
17
Therefore,
Uobt 8
Uobt 64
STEP 2:
Evaluate Uobt. With 0.051 tail, Table C.4 is appropriate. With
n1 9 and n2 8, the U value needed to reject H0 is U 18. Since
Uobt 18, we reject H0. Increasing the power of the experiment
allowed H0 to be rejected. We can now accept H1; that is, a highprotein diet eaten during infancy increases intellectual functioning relative to a low-protein diet.
d. Type error: We may have made a Type I error, rejecting H0
when it is true.
e. Population: same as before.
TIED RANKS
Thus far, none of the problems involved two or more scores of the same value.
When this occurs, Uobt and Uobt are still computed with the equations we’ve just
been using. However, ranking the scores is a little more complicated because of
the ties. We’ve already shown how to do the ranking when tied scores are involved when we discussed the Spearman rho correlation coefficient (p. 118). To
review, tied scores are handled by assigning them the average of the tied ranks.
For example, consider the following two sets of scores:
Group 1
Group 2
12
11
14
12
15
16
17
17
18
17
20
To rank-order the combined scores, we proceed as follows. First, the scores are
arranged in ascending order. Thus,
Raw score
Rank
11
12
12
14
15
16
17
17
17
18
20
1
2.5
2.5
4
5
6
8
8
8
10
11
Next, we assign each raw score its rank beginning with 1 for the lowest score. This
has been shown previously. Note that the two raw scores of 12 are tied at the
ranks of 2 and 3. They are assigned the average of these tied ranks. Thus, they
each get a rank of 2.5 [(2 3)2 2.5]. Since we have used up the ranks of 2 and
3, the next score gets a rank of 4. The raw scores of 17 are tied at the ranks of 7,
18
Mann–Whitney U Test: An Extended Presentation
8, and 9. Therefore, they receive the rank of 8, which is the average of 7, 8, and 9
[(7 8 9)3 8]. Note that the next rank is 10 (not 9) because we’ve already
used ranks 7, 8, and 9 in computing the average. If the ranking is done correctly,
unless there are tied ranks at the end, the last raw score should have a rank equal
to N. In this case, N 11 and so does the rank of the last score. Once the ranks
have been assigned, Uobt and Uobt are calculated in the usual way.
PRACTICAL CONSIDERATIONS IN USING
THE MANN–WHITNEY U TEST
The Mann–Whitney U test is used with an independent groups design. It does not
depend on the shape of the population of scores. It is therefore appropriate for
testing the hypothesis that both sets of scores are random samples from identical
populations regardless of the shape of the populations. It does, however, require
that the data be at least ordinal. You must be able to rank-order the scores. This
is a fairly powerful, practical test that is often used when the data are only ordinal. It is also used as a substitute for Student’s t test when the assumptions of that
test are not met. However, since it only uses the ordinal property of the scores, it
is not as powerful as the t test, which uses the interval property of the scores.
SUMMARY
In this presentation, we have discussed the Mann–
Whitney U test and the independent groups design.
In the independent groups design, subjects are randomly selected from the population and randomly
assigned to conditions. In the most basic form of the
design, there are only two conditions: the experimental and control conditions. Each subject serves in
only one condition, and there is no basis for pairing
of scores. The scores of each sample are analyzed
separately. If chance alone is responsible for differences between sample scores, both sets of sample
scores can be considered random samples from the
same population of scores.
The Mann–Whitney U test analyzes the degree
of separation between the samples. The less the
separation, the more reasonable chance is as the underlying explanation. For any analysis, there are two
values indicating the degree of separation. They both
indicate the same degree of separation. The lower
value is arbitrarily called Uobt, and the higher value is
called Uobt. The lower the Uobt value, the greater the
separation. The higher the Uobt value, the greater the
separation. Tables C.1–C.4 give all the values of Uobt
and Uobt that allow rejection of the null hypothesis
for various combinations of n1 and n2. As with the repeated measures design, the null hypothesis is evaluated directly. If it is rejected, H1 is accepted. If not, we
can’t accept H1. With the Mann–Whitney U test, we
calculate the Uobt or Uobt value of the data. Since
they both indicate the same degree of separation, we
just used the Uobt value. If Uobt is equal to or less than
the tabled U value for rejecting the null hypothesis,
we reject H0. If not, we retain H0. The Mann–Whitney U test is appropriate for an independent groups
design where the data are at least ordinal in scaling.
It is a powerful test, often used in place of Student’s
t test when the data do not meet the assumptions of
the t test.
IMPORTANT TERMS
Control group (p. 2)
Degree of separation (p. 5)
Experimental group (p. 2)
Independent groups design (p. 2)
Mann–Whitney U test (p. 1)
Population of scores (p. 4)
Rank order (p. 5)
Tied ranks (p. 17)
Uobt (p. 5)
Uobt (p. 5)
Questions and Problems
QUESTIONS AND PROBLEMS
1. Briefly define or explain each of the terms in the
“Important Terms” section.
2. Briefly contrast the repeated measures and independent groups design.
3. What are the practical considerations in using
the Mann–Whitney U test?
4. Briefly describe the process of hypothesis testing
using the Mann–Whitney U test.
5. In an independent groups design, if the alternative
hypothesis is nondirectional, what does the null
hypothesis say with respect to the two samples?
6. An independent groups experiment is conducted
to see whether treatment A differs from treatment B. Eight subjects are randomly assigned to
treatment A and seven to treatment B. The following data are collected:
Treatment A
b. What is the null hypothesis?
c. What is your conclusion? Use 0.052 tail.
d. What error may you be making by concluding
as you did in part c?
e. To what population do the results apply?
Theology
Professors
Psychology
Professors
36
13
42
25
22
40
48
29
31
10
35
26
47
43
38
17
12
Treatment B
32
30
14
27
35
8
32
34
25
40
16
19
26
32
28
21
9
23
a. What is the alternative hypothesis? Use a
nondirectional hypothesis.
b. What is the null hypothesis?
c. What is your conclusion? Use 0.052 tail.
7. A social scientist believes that university theology professors are more conservative in political
orientation than their colleagues in psychology.
A random sample of 8 professors from the theology department and 12 professors from the psychology department at a local university are
given a 50-point questionnaire that measures the
degree of political conservatism. The following
scores were obtained. Higher scores indicate
greater conservatism.
a. What is the alternative hypothesis? In this
case, assume a nondirectional hypothesis is
appropriate because there are insufficient theoretical and empirical bases to warrant a directional hypothesis.
19
8. Someone has told you that men are better in abstract reasoning than women. You are skeptical,
so you decide to test this idea using a nondirectional hypothesis. You randomly select eight
adult men and eight adult women living in your
hometown and administer an abstract reasoning
test. A higher score reflects better abstract reasoning abilities. You obtain the following scores:
Men
Women
70
81
86
80
60
50
92
95
82
93
65
85
74
90
94
75
a. What is the alternative hypothesis? Assume a
nondirectional hypothesis is appropriate.
b. What is the null hypothesis?
c. Using 0.052 tail, what do you conclude?
20
Mann–Whitney U Test: An Extended Presentation
d. What error may you have made by concluding
as you did in part c?
e. To what population do your results apply?
9. An ornithologist thinks that injections of folliclestimulating hormone (FSH) increase the singing
rate of his captive male cotingas (birds). To test
this hypothesis, he randomly selects 20 singing
cotingas and divides them into two groups of 10
birds each. The first group receives injections of
FSH and the second injections of saline solution,
as a control for the trauma of receiving an injection. He then records the singing rate (in songs per
hour) for both groups. The results are given in the
following table. Note that two of the FSH birds escaped during injection and were not replaced.
Saline
FSH
17
10
31
29
14
37
12
41
29
16
23
45
7
34
19
57
28
3
a. What is the alternative hypothesis? Use a directional alternative hypothesis.
b. What is the null hypothesis?
c. Using 0.051 tail, what is your conclusion?
10. A psychologist is interested in determining
whether left-handed and right-handed people differ in spatial ability. She randomly selects 10 lefthanders and 10 right-handers from the students
enrolled in the university where she works and administers a test that measures spatial ability. The
scores are in the following table (a higher score indicates better spatial ability). Note that one of the
subjects did not show up for the testing.
a. What is the alternative hypothesis? Use a
nondirectional hypothesis.
b. What is the null hypothesis?
c. Using 0.052 tail, what do you conclude?
Left-Handers
Right-Handers
87
47
94
68
56
92
74
73
98
71
83
82
92
55
84
61
76
75
85
11. A university counselor believes that hypnosis is
more effective than the standard treatment given
to students who have high test anxiety. To test his
belief, he randomly divides 22 students with high
test anxiety into two groups. One of the groups
receives the hypnosis treatment and the other
group the standard treatment. When the treatments are concluded, each student is given a test
anxiety questionnaire. High scores on the questionnaire indicate high anxiety. The following are
the results:
Hypnosis
Treatment
Standard
Treatment
20
42
21
35
33
30
40
53
24
57
43
26
48
37
31
30
22
51
44
62
30
59
a. What is the alternative hypothesis? Assume
there is sufficient basis for a directional hypothesis.
b. What is the null hypothesis?
c. Using 0.051 tail, what do you conclude?
Web Connection
21
NOTES
1. These probabilities could also have been calculated from multiplication and addition rules. For
example, to determine p(U 0) such that all the
C scores are lower than E scores, we need to determine p(CCCEEE). Assume we have a bag containing three C scores and three E scores and we
are randomly sampling one score at a time without replacement. What is the probability we shall
get the order CCCEEE? The solution is as follows. By using the multiplication rule,
p(U 0) p(CCCEEE)
36 . 25 . 14 . 33 . 22 . 11 0.05
Note that each order yields the same probability
value (0.05). Thus, to find the other probabilities,
all we need do is multiply 0.05 by the number of
orders yielding the particular U value. For example, find p(U 2) with Cs lower than Es, since there
are two orders that yield U 2 in this direction:
P(U 2) 2(0.05) 0.10
WEB CONNECTION
Go to http://psychology.wadsworth.com/courses/
statistics/ and test your knowledge of this chapter by
taking the online quiz. You can also find additional
fully solved practice problems there.