Hi Edita,

Here's a link that should provide some insight: https://books.google.com/books?id=KQ25WExx5usC&pg=PA678&dq=mann+whitney&hl=en&sa=X&ved=0ahUKEwig3vHBlK_aAhUixYMKHY89Bm0QuwUITTAG#v=onepage&q=mann%20whitney&f=false

However, I concur with the preceding responses. Your sample is sufficiently large enough to perform some form of parametric analysis.

Have a great day!

--Adrian

Hello Edita,

Beyond 20 samples you can use the normal distribution to calculate the exact p-value. The formula can be found at:

https://math.usask.ca/~laverty/S245/Tables/wmw.pdf

The exact value is more informative than comparison to a critical value. You report the exact p-value, then compare it to the usual 5% criterion. The calculations can be done in spreadsheet, including the p-value from the Z statistic.

=NORM.DIST(A2,0,1,0)

where A2 is the cell with the Z statistic

Hope that helps,

David S

In fact, with such large sample sizes, I'm not sure why you are bothering with a non-parametric test.

Hi Edita,

Here's a link that should provide some insight: https://books.google.com/books?id=KQ25WExx5usC&pg=PA678&dq=mann+whitney&hl=en&sa=X&ved=0ahUKEwig3vHBlK_aAhUixYMKHY89Bm0QuwUITTAG#v=onepage&q=mann%20whitney&f=false

However, I concur with the preceding responses. Your sample is sufficiently large enough to perform some form of parametric analysis.

Have a great day!

--Adrian

As your reference states that the MW test is about comparing means, you should probably check out discussion in this feed:

https://www.researchgate.net/post/Assumptions_for_Kruskal-Wallis_and_Mann-Whitney-U_violated_are_there_alternatives

Best

John-Kåre

Thank you all for the answers and have a great day!

Hi, you have to use parametric test than non-parametric, your sample is large enough

I agree with those above who say that you can use a parametric test. If you want an old reference that's still pretty good, look at W. J. Conover, Practical Nonparametric Statistics. Wiley, 1971, pp. 223-227.

Hello Edita,

When I wrote to you, I assumed you were a student instructed to do MW test. And so I sent you the large sample (parametric) formula from a citable source. Which is what the subsequent posts advocate, sometimes in somewhat different words. If you plan to go on in science, or you hold a research or teaching position, I urge you to look beyond ritual use of MW U-test.

It is too often used when there are better and more informative statistics.

Wishing you well in future endeavors,

David S

Hi again Edita. I looked again at your question and became a bit confused because you described n1 and n2 twice with different n's. Are the second cluster of groups drawn from the first? (if so, it may violate the independence assumption). Or are they follow-up values from the first groups (then you should rather have used the signed ranked test to examine related samples).

Best

John-Kåre

Hi Edita, if the normality test for your variables is clear enough to have reasonable evidence that their distribution is not normal as you said, the nonparametric test is the answer no matter if your sample size is small or large. Then for a large sample size the distribution of Mann-Whitney U, can be approximated to a normal distribution (Z), that it didn’t means that your variable has a normal distribution.

You can check this link:

http://users.sussex.ac.uk/~grahamh/RM1web/Wilcoxon%20Large%20N%202009.pdf

Best regards.

Dear Edita

Please refer to:

1. Landau, S. and Everitt, B. S.(2004). A Handbook of Statistical analyses using SPSS.

2.Howitt, D. and Cramer, D.(2008). Introduction to SPSS.

Regards,

Zuhair

Good question and informative answers thank you all!

Assumptions and formal statement of hypotheses

Although Mann and Whitney developed the Mann–Whitney U test under the assumption of continuous responses with the alternative hypothesis being that one distribution is stochastically greater than the other, there are many other ways to formulate the null and alternative hypotheses such that the Mann–Whitney U test will give a valid test.

A very general formulation is to assume that:

Under the general formulation, the test is only consistent (i.e., has power to reject that approaches 100% as the sample size approaches infinity) when the following occurs under H1:

Under more strict assumptions than the general formulation above, e.g., if the responses are assumed to be continuous and the alternative is restricted to a shift in location, i.e., F1(x) = F2(x + δ), we can interpret a significant Mann–Whitney U test as showing a difference in medians. Under this location shift assumption, we can also interpret the Mann–Whitney U test as assessing whether the Hodges–Lehmann estimate of the difference in central tendency between the two populations differs from zero. The Hodges–Lehmann estimate for this two-sample problem is the median of all possible differences between an observation in the first sample and an observation in the second sample.

The Mann–Whitney U test / Wilcoxon rank-sum test is not the same as the Wilcoxon signed-rank test, although both are nonparametric and involve summation of ranks. The Mann–Whitney U test is applied to independent samples. The Wilcoxon signed-rank test is applied to matched or dependent samples.

Calculations

The test involves the calculation of a statistic, usually called U, whose distribution under the null hypothesis is known. In the case of small samples, the distribution is tabulated, but for sample sizes above ~20, approximation using the normal distribution is fairly good. Some books tabulate statistics equivalent to U, such as the sum of ranks in one of the samples, rather than U itself.

The Mann–Whitney U test is included in most modern statistical packages. It is also easily calculated by hand, especially for small samples. There are two ways of doing this.

Method one:

For comparing two small sets of observations, a direct method is quick, and gives insight into the meaning of the U statistic, which corresponds to the number of wins out of all pairwise contests (see the tortoise and hare example under Examples below). For each observation in one set, count the number of times this first value wins over any observations in the other set (the other value loses if this first is larger). Count 0.5 for any ties. The sum of wins and ties is U for the first set. U for the other set is the converse.

Method two:

For larger samples:

U 1 = R 1 − n 1 ( n 1 + 1 ) 2 {\displaystyle U_{1}=R_{1}-{n_{1}(n_{1}+1) \over 2}\,\!} 📷 where n1 is the sample size for sample 1, and R1 is the sum of the ranks in sample 1. Note that it doesn't matter which of the two samples is considered sample 1. An equally valid formula for U is U 2 = R 2 − n 2 ( n 2 + 1 ) 2 {\displaystyle U_{2}=R_{2}-{n_{2}(n_{2}+1) \over 2}\,\!} 📷 The smaller value of U1 and U2 is the one used when consulting significance tables. The sum of the two values is given by U 1 + U 2 = R 1 − n 1 ( n 1 + 1 ) 2 + R 2 − n 2 ( n 2 + 1 ) 2 . {\displaystyle U_{1}+U_{2}=R_{1}-{n_{1}(n_{1}+1) \over 2}+R_{2}-{n_{2}(n_{2}+1) \over 2}.\,\!} 📷 Knowing that R1 + R2 = N(N + 1)/2 and N = n1 + n2, and doing some algebra, we find that the sum is U1 + U2 = n1n2.

Properties

The maximum value of U is the product of the sample sizes for the two samples. In such a case, the "other" U would be 0.

Examples

Illustration of calculation methods

Suppose that Aesop is dissatisfied with his classic experiment in which one tortoise was found to beat one hare in a race, and decides to carry out a significance test to discover whether the results could be extended to tortoises and hares in general. He collects a sample of 6 tortoises and 6 hares, and makes them all run his race at once. The order in which they reach the finishing post (their rank order, from first to last crossing the finish line) is as follows, writing T for a tortoise and H for a hare:

T H H H H H T T T T T H

What is the value of U?

• Using the direct method, we take each tortoise in turn, and count the number of hares it beats, getting 6, 1, 1, 1, 1, 1, which means that U = 11. Alternatively, we could take each hare in turn, and count the number of tortoises it beats. In this case, we get 5, 5, 5, 5, 5, 0, so U = 25. Note that the sum of these two values for U = 36, which is 6×6.
• Using the indirect method:

rank the animals by the time they take to complete the course, so give the first animal home rank 12, the second rank 11, and so forth. the sum of the ranks achieved by the tortoises is 12 + 6 + 5 + 4 + 3 + 2 = 32. Therefore U = 32 − (6×7)/2 = 32 − 21 = 11 (same as method one). the sum of the ranks achieved by the hares is 11 + 10 + 9 + 8 + 7 + 1 = 46, leading to U = 46 − 21 = 25.

Illustration of object of test

A second example race illustrates the point that the Mann–Whitney U test does not test for inequality of medians, but rather for difference of distributions. Consider another hare and tortoise race, with 19 participants of each species, in which the outcomes are as follows, from first to last past the finishing post:

H H H H H H H H H T T T T T T T T T T H H H H H H H H H H T T T T T T T T T

If we simply compared medians, we would conclude that the median time for tortoises is less than the median time for hares, because the median tortoise here (in bold) comes in at position 19, and thus actually beats the median hare (in bold), which comes in at position 20. However, the value of U is 100 (using the quick method of calculation described above, we see that each of 10 tortoises beats each of 10 hares, so U = 10×10). Consulting tables, or using the approximation below, we find that this U value gives significant evidence that hares tend to have lower completion times than tortoises (p < 0.05, two-tailed). Obviously these are extreme distributions that would be spotted easily, but in larger samples something similar could happen without it being so apparent. Notice that the problem here is not that the two distributions of ranks have different variances; they are mirror images of each other, so their variances are the same, but they have very different skewness.

Normal approximation and tie correction

For large samples, U is approximately normally distributed. In that case, the standardized value

z = U − m U σ U , {\displaystyle z={\frac {U-m_{U}}{\sigma _{U}}},\,} 📷

where mU and σU are the mean and standard deviation of U, is approximately a standard normal deviate whose significance can be checked in tables of the normal distribution. mU and σU are given by

m U = n 1 n 2 2 , {\displaystyle m_{U}={\frac {n_{1}n_{2}}{2}},\,} 📷 and σ U = n 1 n 2 ( n 1 + n 2 + 1 ) 12 . {\displaystyle \sigma _{U}={\sqrt {n_{1}n_{2}(n_{1}+n_{2}+1) \over 12}}.\,} 📷

The formula for the standard deviation is more complicated in the presence of tied ranks; the full formula is given in the text books referenced below.

If there are ties in ranks, σ should be corrected as follows:

σ corr = n 1 n 2 12 ( ( n + 1 ) − ∑ i = 1 k t i 3 − t i n ( n − 1 ) ) {\displaystyle \sigma _{\text{corr}}={\sqrt {{n_{1}n_{2} \over 12}\left((n+1)-\sum _{i=1}^{k}{{t_{i}}^{3}-t_{i} \over n(n-1)}\right)}}\,} 📷

where n = n1 + n2, ti is the number of subjects sharing rank i, and k is the number of (distinct) ranks.

If the number of ties is small (and especially if there are no large tie bands) ties can be ignored when doing calculations by hand. The computer statistical packages will use the correctly adjusted formula as a matter of routine.

Note that since U1 + U2 = n1n2, the mean n1n2/2 used in the normal approximation is the mean of the two values of U. Therefore, the absolute value of the z statistic calculated will be same whichever value of U is used.

Effect sizes

It is a widely recommended practice for scientists to report an effect size for an inferential test.[5][6]

Common language effect size

One method of reporting the effect size for the Mann–Whitney U test is with the common language effect size. As a sample statistic, the common language effect size is computed by forming all possible pairs between the two groups, then finding the proportion of pairs that support a hypothesis.[8] To illustrate, in a study with a sample of ten hares and ten tortoises, the total number of pairs is ten times ten or 100 pairs of hares and tortoises. Suppose the results show that the hare ran faster than the tortoise in 90 of the 100 sample pairs; in that case, the sample common language effect size is 90%. This sample value is an unbiased estimator of the population value, so the sample suggests that the best estimate of the common language effect size in the population is 90%.[9]

Rank-biserial correlation

A second method of reporting the effect size for the Mann–Whitney U test is with a measure of rank correlation known as the rank-biserial correlation. Edward Cureton introduced and named the measure.[10] Like other correlational measures, the rank-biserial correlation can range from minus one to plus one, with a value of zero indicating no relationship.

There is a simple difference formula to compute the rank-biserial correlation from the common language effect size: the correlation is the difference between the proportion of pairs favorable to the hypothesis (f) minus the proportion that is unfavorable (u). The value of f is the common language effect size. This simple difference formula is as follows:[7]

r = f − u {\displaystyle r=f-u} 📷

Stated a

nother way, the correlation is the difference between the common language effect size and its complement. For example, consider the example where hares run faster than tortoises in 90 of 100 pairs. The common language effect size is 90%, so the rank-biserial correlation is 90% minus 10%, and the rank-biserial r = 0.80.

There is a formula to compute the rank-biserial from the Mann–Whitney U and the sample sizes of each group:[11]

r = 1 − 2 U n 1 n 2 {\displaystyle r=1-{2U \over n_{1}n_{2}}\,} 📷

This formula is useful when the data are not available, but when there is a published report, because U and the sample sizes are routinely reported. Using the example above with 90 pairs that favor the hares and 10 pairs that favor the tortoise, U is the smaller of the two, so U = 10. This formula then gives r = 1 – (2×10) / (10×10) = 0.80, which is the same result as with the simple difference formula above.

Hello Edita,

The short answer to your question is, probably no text has tables of critical values for the Mann-Whitney U test with n values of the sizes you mention. There are three reasons:

1. The critical value is specific to a particular n1, n2 value set (as is the case for df in the F-distribution).

2. The work needed to derive critical values for U with n-values of 150+ is computationally quite intensive. For n1 = n2 = 150, you would have to evaluate 9.376 x 10^88 combinations of ranks (assuming no ties) to find the target percentile point for U representing the desired alpha level. That's a value higher than the estimated number of atoms in the known universe.

3. The accuracy of the "large sample z-approximation" used by most statistical software is excellent with n values over 100. So, just use that. Note, however, this is not the same as declaring that you're using a parametric test; you're simply using the standard normal distribution to estimate the chance-level probability of a result as extreme or more extreme as what you obtain with the M-W test.

Good luck with your work!

I recommend John-Kåre's answer

Best Regards

Hai Edita,

Please refer page 398 and 399 in

"Montgomery, D.C., Runger, G.C.: ‘Applied statistics and probability for engineers’ (John Wiley & Sons Inc., NY, USA, 2003, 3rd edn.)".