I have collected data from two populations, M (males) and F (females) through a Likert scale of their agreement to a statement X

The data is the following for females F

enter image description here

And for M males

enter image description here

As you can see it is from strongly disagree to strongly agree. For analysis this was converted to a scale from 1 to 5 and a Mann-Whitney U test was done to compare the distribution of both populations' answers.

  1. Could you tell me whether I have explained this adequately in the 'analysis' part of my paper and if I have reported the results in an appropriate format? Also is using the mean (+/- SD) OK for comparing the two groups' distribution qualitatively as I have done?

Analysis: "Likert-scale data was treated as ordinal (1-5) and subsequently analysed using the Mann-Whitney U-test when appropriate"

Results: "There was no significant difference between females’ opinion (mean Likert score: 3.06 ± 1.095) and males’ opinion (mean Likert score: 3.00 ± 1.113 ) of the importance of being asked x (U = 5813, z = 0.587, p = .5552)."

  1. Are the results correct? I haven't used any stats software, just an online calculator (as I have no skills in R or even SPSS). Is anyone able to check?
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    $\begingroup$ The Mann-Whitney test is not about comparing sample means. $\endgroup$
    – Nick Cox
    Jun 18, 2020 at 16:48
  • $\begingroup$ I know this and understand why you pointed this out, however I am not implying the mann whitney u test does this. What I am attempting to do is compare the spread using mann whitney U test which shows no difference. I then give additional info about the data (in a very vague qualitative way) by pointing out the two means? What would be wrong with this? Or is finesse your concern? Do you have an alternative suggestion? Many thanks $\endgroup$ Jun 18, 2020 at 18:02
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    $\begingroup$ It doesn't much matter what you want to imply if a reader is likely to infer that you think, or are telling them, that a Mann-Whitney test is about means, because you don't say otherwise. Why ask Is this correct? when you know it isn't? A better formulation of Mann-Whitney is as as assessment of how likely it is that, comparing values from two different groups, that values from one group are higher than those in another. For this point of view see e.g. stata-journal.com/sjpdf.html?articlenum=st0007 $\endgroup$
    – Nick Cox
    Jun 18, 2020 at 18:16
  • 2
    $\begingroup$ No; what you can do is keep discussions of mean and SD and the Mann-Whitney test completely separate. $\endgroup$
    – Nick Cox
    Jun 18, 2020 at 18:32
  • 1
    $\begingroup$ I don't post on that site but it's possible that someone posted a link to something I wrote. Either way, happy to be thought constructive! $\endgroup$
    – Nick Cox
    Jun 19, 2020 at 5:27

4 Answers 4


As for the check with SPSS or R, suitable R code could be the following. Unfortunately I can only tell you a way via Wilcoxon W, not Mann-Whitney U. The tests are equivalent, though:

f <- c(rep(1,21), rep(2,17), rep(3, 82), rep(4,34), rep(5,18))
m <- c(rep(1,7), rep(2,15), rep(3,28), rep(4,13), rep(5,8))
wilcox.exact(f, m)

The result would be

> wilcox.exact(f, m)

    Asymptotic Wilcoxon rank sum test

data:  f and m
W = 6399, p-value = 0.5343
alternative hypothesis: true mu is not equal to 0

Where you could cite R in the literature as

R Core Team (2020). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL https://www.R-project.org/.

and the package exactRankTests as

Torsten Hothorn and Kurt Hornik (2019). exactRankTests: Exact Distributions for Rank and Permutation Tests. R package version 0.8-31. https://CRAN.R-project.org/package=exactRankTests

As for the rest of the description, that depends a lot on personal taste, faculty etc. I for one would be careful to call something that has been measured by only one Likert-type item as a Likert scale. Also you seem to use Likert scale data and Likert score somewhat identical. Why two different words then? Apparently, you have interviewed 243 persons. Does it seem appropriate to use that many digits for standard deviation and p-value?

So the calculation is about right, detail in the wording has to do with personal taste.

  • $\begingroup$ Thanks! Appreciate the help. I think I get a bit paranoid with the wording and there probably isn't a strict way of doing it as much as I imagine. You raise a good point re scale/score I think I need to go over what wording should be used. And the p-value and SD sigh idk how that got in there! $\endgroup$ Jun 18, 2020 at 18:05
  • $\begingroup$ Also my likert item qualifies as a scale (Neutral middles value, 1st and last value are opposite etc) so what would be wrong in using it as a scale? $\endgroup$ Jun 18, 2020 at 18:29
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    $\begingroup$ I know people who believe that a Likert scale is the sum of a number (n>1) of Likert type items. I know others who talk about single-item-scales. Somehow taking Likert values as pseudo-metric seems to be less implausible if its a sum and can take more then 5 different values. I do not hold an opinion, just telling you to reconsider in which of these worlds you live. Naming something Likert scale is a difficult matter, cf. john-uebersax.com/stat/likert.htm $\endgroup$
    – Bernhard
    Jun 18, 2020 at 20:34
  • $\begingroup$ Oh right thanks for the link was an interesting read. And I recall reading something similar before starting my paper actually. hmmmm $\endgroup$ Jun 18, 2020 at 21:28
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    $\begingroup$ On the vexed question of means for ordinal scales see e.g. stats.stackexchange.com/questions/67551/… An executive summary of the spectrum of opinion might run from "quite wrong in principle" to "may work well in practice". $\endgroup$
    – Nick Cox
    Jun 19, 2020 at 5:30

I have no disagreement with @Bernhard's Answer (+1), but I will give my own comments on this using R, especially because you have not up-voted or accepted the answer, and you still seem puzzled in some of your comments.

The Likert scores and summaries are as follows:

wom = rep(1:5, c(21,17,92,34,18))

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00    3.00    3.00    3.06    4.00    5.00 

men = rep(1:5, c(7,15,28,15,8))

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.000   2.000   3.000   3.027   4.000   5.000 

The two sample medians are 3.0, so I think it is better just to say that, than to try to give confidence intervals. Giving confidence intervals for means seems undesirable because

  • The methods for making those confidence intervals seem to be based on an assumption that data are from a continuous normal distribution, while they are actually ordinal categorical data.

  • Also, I agree with the objection that CIs for means (besides being pointless) might confuse your readers, making them wonder what those CIs have to do with your nonparametric test (which is nothing at all).

Boxplots leave little doubt that the medians for men and women are both $3.$

boxplot(men, wom, col="skyblue2", pch=20)

enter image description here

I agree that a 2-sample Wilcoxon rank sum test does not find a difference between the two samples of Likert scores.

wilcox.test(men, wom)

        Wilcoxon rank sum test 
      with continuity correction

data:  men and wom
W = 6829, p-value = 0.711
alternative hypothesis: 
  true location shift is not equal to 0

Data summaries and box plots seem to show a few more low (disagree) scores among women than among men. However, a chi-squared test of homogeneity of Likert scores for men and women does not reject the null hypothesis of homogeneity.

TAB = rbind(c(21,17,92,34,18),
            c( 7,15,28,15, 8))
     [,1] [,2] [,3] [,4] [,5]
[1,]   21   17   92   34   18
[2,]    7   15   28   15    8


        Pearson's Chi-squared test

data:  TAB
X-squared = 7.1942, df = 4, p-value = 0.126

I think it may be sufficient to say that both Men and Women have median 3 Likert scores and that a Wilcoxon rank sum test (equivalent to Mann-Whitney) finds no significant difference in locations, with P-value 0.71. If you feel you need to say more, then perhaps mention the P-value 0.13 for the chi-squared test of homogeneity.

Finally, I think it is worth mention somewhere the exact numbers of men and of women in the study (and if not obvious from the context, the reason for such different numbers).

  • 1
    $\begingroup$ No disagreement in both directions and (+1). However, if you include the test of homogeneity as well IMHO your Material&Methods part should clearly indicate, which of the two is the primary test of concern (cannot find a good translation of "primärer Endpunkt") of the study and which one is posted just as additional information. One of them investigates, whether females or males score higher then the other gender and the other tests for different answering patterns. $\endgroup$
    – Bernhard
    Jun 19, 2020 at 6:06
  • $\begingroup$ Agree about making the main objective clear. (Context. in my mind anyhow, was trying to promote chi-sq results as a substitute for CIs.) // Just read google-translate version of my Answer in German. Often do that on the theory that a coherent google translation means my writing is coherent--never a sure bet at this time of night. And it freshens up my German, never good and disused for 40 yrs. $\endgroup$
    – BruceET
    Jun 19, 2020 at 6:23
  • $\begingroup$ As far as I am concerned, it is 8:30 am SCNR $\endgroup$
    – Bernhard
    Jun 19, 2020 at 6:25
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    $\begingroup$ I have taken the liberty to edit in "using R", which is no doubt utterly obvious to many readers but helpful to anyone else who may be in doubt. $\endgroup$
    – Nick Cox
    Jun 19, 2020 at 7:11
  • 1
    $\begingroup$ I too am happy to use a chi-square test here, but add nevertheless that it ignores the ordering of the grades. $\endgroup$
    – Nick Cox
    Jun 19, 2020 at 7:45

This is partly a comment on @Bruce ET's helpful answer, but the graph here won't fit in a comment -- and inviting or expecting readers to enter the data and draw it for themselves is unrealistic.

The box plot isn't wrong, as box plots go, and makes the point that the medians are the same for males and females. But box plot conventions make the display overstate the difference between males and females in distribution.

Also, the box plot does precisely what is implied to be wrong about calculating means, treat the grades or ratings Strongly agree to Strongly disagree as equally spaced points on a measured scale, here 1 2 3 4 5. This is important because the box plot display hinges on calculation of median and quartiles and (specifically here) uses 1.5 IQR in deciding where whiskers stop and whether data points are shown beyond the ends of whiskers.

Indeed, experience on Cross Validated and elsewhere shows that box plots for graded or ordinal data like these -- more generally, for data with many ties -- are often puzzling. They can even provoke suspicions that something is wrong. (Usually the software is put in question, not the reader of the graph.) These example threads understate the puzzlement box plots can cause.

Boxplot interpretation: is it correct that a boxplot is missing a whisker?

Help needed with my box plot

A plain bar chart explains why and how the box plot muddies the picture. Bar lengths here are proportional to percents given gender, but the annotation shows absolute counts too. Indeed, my bar chart also shows grades equally spaced, but nothing depends on that conventional spacing.

enter image description here

For males, the distribution is such that median and lower quartile agree at 3. So, the interquartile range is just 1: this is clear from the graph, as it is the height of the box. So, the lowest value 1 qualifies for separate display: it is 2 below the lower quartile, and so more than 1.5 IQR away from the lower quartile, which is the most common convention for separate display of low values and that used by R in this case. (I don't join the poor practice of shouting "outlier" here.)

For females small differences between the distributions make the lower quartile emerge as 2, and the lowest value 1 is not selected for separate display.

The box plot does not, and cannot, tell you much about the relative frequency of grades of 1, which are not much different for males and females, or about the relative frequency of any other grade for that matter.


I would say that your presentation of the Mann-Whitney U test is slightly sloppy, although in practice it hardly matters. Intuitively, you are in the right direction, but it wouldn't hurt to be more correct.

Mixing of concepts

"There was no significant difference between females’ opinion (mean Likert score: 3.06 ± 1.095) and males’ opinion (mean Likert score: 3.00 ± 1.113 ) of the importance of being asked x (U = 5813, z = 0.587, p = .5552)."

This sentence might be confusing because it is combining three concepts. It is talking about:

  • General differences between distributions:

    "There was no significant difference between females’ opinion ... and males’ opinion"

    For this, if you just want to test whether there are any differences, you might better use a chi-squared test.

  • Means of distributions and their error estimates:

    (mean Likert score: 3.06 ± 1.095) ... (mean Likert score: 3.00 ± 1.113 )

    You write scores with confidence intervals or with expressions for the error. For these types of statistics, to compare significance, one would expect something like a t-statistic, instead of a U-statistic.

  • A U-statistic:

    (U = 5813, z = 0.587, p = .5552)

    The U-statistic (and related z-score) is a test for equivalence of distributions, but it is only sensitive to a specific type of alternative hypothesis. The Mann-Whitney test is only sensitive for the alternative P(X>Y). A chi-squared test relates to all possible differences between the distributions and might be more intuitive when you wish to express whether the opinions differ.

    Use the Mann Whitney test when you wish to specifically test the idea that one variable is larger (higher order) than the other (personally I would not do this when you only have 5 categories and variations might be occurring in more than just differences in order).

Why Mann-Whitney test is not presented appropriately

  • One aspect is that the Mann-Whitney U test is not a test for differences between means.

    This, Mann-Whitney U test is used to test differences in means, is a bit implied when you mix those three concepts (described above) in the same sentence.

    The Mann-Whitney U test relates to the question of stochastic dominance $P(X>Y) \neq 0.5$ and not to the question of different means.

    On the one hand you can have different means but no stochastic dominance. On the other hand you can have stochastic dominance but not different means. They are different things.

    In practice they may coincide: for instance, if you envision the same distribution with only a shift in the location then you get that the difference in means will coincide with a difference in stochastic dominance. But in your case I would not use that assumption with 5 points.

  • You are comparing the means of a Likert scale by converting the categories into a scalar number. This might seem right since both the 'Likert scale' and the 'number system/scale' have an order. However, something that is not equal between them is a concept of scale or distance.

    This does not mean that you can not compare the means. The resulting 'mean' of that scale is a number which you can compare for different groups. However, you need to be careful in the interpretation (the same would be true if you are dealing with actual scalars).

    Comparing means becomes tricky when distributions differentiate on more aspects than just a shift in the mean. If the distributions are different in more ways than just a shift then the differences in the mean are dependent on the scale.

    The difference in the mean will not be invariant for a change in the scale.

    For instance take your distributions:

              SD    D     N     A     SA
     men      7     15    28    13    8
     women    21    17    82    34    18

    If you assign the values $1,2,3,4,5$ to those categories then you will get averages $$3.064 = \bar{X}_{women} > \bar{X}_{men} = 3.000$$ but if you assign values $e^2,e^4,e^6,e^8,e^{10}$ to the categories (or anything else that increases the weight of the fifth category) then $$3092 = \bar{X}_{women} < \bar{X}_{men} = 3199$$

About the Mann-Whitney U test

Intuitively you can consider the Mann-Whitney U test as comparing something like an empirical joint distribution (the numbers in the cells are the product of the numbers in the margins, e.g. the upper left number $147 = 7 \times 21$):

$$\begin{array}{cc | cccccccc} &&\text{SD} &\text{D}&\text{N}&\text{A}&\text{SA}\\ & &7 & 15& 28 & 13 & 8\\ \hline \text{SD}&21& \color{gray}{147} & \color{blue}{315} & \color{blue}{588} & \color{blue}{273} & \color{blue}{168}\\ \text{D}&17& \color{red}{119} & \color{gray}{255} & \color{blue}{476} & \color{blue}{221} & \color{blue}{136} \\ \text{N}&82& \color{red}{547} & \color{red}{1230} & \color{gray}{2296} & \color{blue}{1066} & \color{blue}{656}\\ \text{A}&34& \color{red}{238} & \color{red}{510} & \color{red}{952} & \color{gray}{442} & \color{blue}{272} \\ \text{SA}&18 & \color{red}{126} & \color{red}{270}& \color{red}{504} & \color{red}{234} &\color{gray}{144} \\ \end{array}$$

And the question is: Do I get more observations in the upper right corner (men more often higher than women, blue) or in the lower left corner (women more often higher than men, red)?

This table relates to the probability that two random men and women from your sample will be equal (gray) or different, men>women (blue) or men<women (red).

You get the following score if you compare how often the men's score is higher than the women's. $$\color{blue}{315+588+273+168+476+221+136+1066+656+272}+\frac{1}{2}\color{gray}{(147+255+2296+442+144)} = 5813$$

You get the following score if you compare how often the women score higher than the men. $$\color{red}{119+574+1230+238+510+952+126+270+504+234}+\frac{1}{2}\color{gray}{(147+255+2296+442+144)} = 6399$$

The distribution of these scores can be imagined by considering randomly ordering the two categories. This is what Mann and Whitney did and they showed that the distribution of the U score is approximately a normal distribution.

Graphical representation

It may help to plot the percentages of the results.


You can see that for women and men you have more or less similar frequencies in the 'strongly disagree' and in the 'agree' and 'strongly agree' categories. It is in the categories 'disagree' and 'neutral' that you see that men are relatively more often in the disagree category and less often in the neutral category (or from the other perspective women less often in the disagree category and more often in the neutral category).

These differences are not very large. We can also see this based on a chi-squared test for the equivalence of the two distributions ($\chi^2 = 5.9037, df = 4, p = 0.2065$). But it might be interesting for further investigation to see whether men are often less nuanced (less often 'N') in comparison to women, and in place of that more often slightly negative (more often 'D').

  • $\begingroup$ A nuance is to plot not cumulative probabilities, but midpoints for each bin, So if your cumulative probabilities are $P_1, P_2, P_3, P_4, P_5 = 100$ plot instead $P_1/2, (P_1 + P_2)/2, (P_2 + P_3)/2, (P_3 + P_4)/2, (P_4 + P_5)/2$. Otherwise all curves end at 100, which is tautologically correct but doesn't help. More at stata-journal.com/sjpdf.html?articlenum=gr0004 pp.203-207. $\endgroup$
    – Nick Cox
    Jun 19, 2020 at 16:51
  • $\begingroup$ With regard to your first point "General differences between distributions": What would be a better way to word the results of the Mann Whitney test then? "There was no significant difference in the distribution of the answers for men and women...". With regard to your point on converting Likert scale into numerical:the assumption I am making in doing that is that the difference between each category is the same (i.e. 1), with this assumption in mind, I fail to see how this would be problematic? $\endgroup$ Jun 19, 2020 at 17:09
  • $\begingroup$ @NickCox I agree with you that in some sense the midpoints might be better, but in the end we will have anyway only 4 independent degrees of freedom. I am personally not so much bothered by the last points equal to 100. $\endgroup$ Jun 19, 2020 at 17:28
  • 1
    $\begingroup$ @chaudryshahidiqbal with a Mann Whitney U test you compare whether one distribution is or is not stochastically greater than the other. With a t-test you compare the means. And with a chi-squared test you test any difference. If you assume that the change occurs only by means of a shift in the value than stochastic order and difference in means tell the same story. But you should keep in mind that this is not always true (and it becomes troubling when you have multiple changes, or not just a shift) $\endgroup$ Jun 19, 2020 at 17:48
  • 2
    $\begingroup$ @ChaudryShahidIqbal it depends on what question you wish to answer. All these tests are ways to express whether the distributions are different, but they pick up in different ways the differences. The chi-squared test is more general, that is both an advantage (able to pick up more different type of differences) and a disadvantage (less powerful for specific differences). $\endgroup$ Jun 19, 2020 at 19:08

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