1 The Wilcoxon rank-sum test

Sometimes called the Mann-Whitney test, the Wilcoxon rank-sum test may be applied to continuous distributions which have the same shape and spread but may have different means. If we take the distributions as $X_1$ with mean ${\mu }_1$ and $X_2$ with mean ${\mu }_2$ then the Wilcoxon rank-sum test may be used to test the null hypothesis

$H_0:{\mu }_1={\mu }_2$

Against the alternatives

$\begin{array}{c}\hfill H_1:{\mu }_1\ne {\mu }_2\hfill \\ \hfill \hfill \\ \hfill H_1:{\mu }_1>{\mu }_2\hfill \\ \hfill \hfill \\ \hfill H_1:{\mu }_1<{\mu }_2\hfill \\ \hfill \hfill \\ \hfill \hfill \end{array}$

Now assume that a random sample of size $n_1$ is taken from population $X_1$ and a random sample of size $n_2$ is taken from population $X_2$ . As with the Wilcoxon signed-rank test, the theory is demanding but the application is straightforward. The test procedure is as follows:

1. Arrange all of the $n_1+n_2$ sample members in ascending order and assign ranks to them. Equal ranks are dealt with in the usual way.
2. Find the sum of the ranks assigned to members of the smaller of the two samples and call this $S_1$ .
3. Find the sum of the ranks assigned to members of the larger of the two samples and call this $S_2$ . Normally, this is not done directly. It may be shown that

   $S_2=\frac{\left(n_1+n_2\right)\left(n_1+n_2+1\right)}{2}-S_1$

   and it is usual to use this relationship to find $S_2$ rather than do the direct calculation to save both time and effort.

4. When testing $H_0:{\mu }_1={\mu }_2$ against $H_1:{\mu }_1\ne {\mu }_2$ , Tables 2 and 3 given at the end of this Workbook may be used directly to test at both the 5% and 1% levels of significance. Rejection of the null hypothesis occurs when either rank sum is less than the tabulated critical value.
5. In the case of one-tailed tests the same tables may be used but with these tables the levels of significance are restricted to 2.5% (from the 5% table) and 0.5% (from the 1% table). Examples given here will normally use a two-tailed test and the 5% level of significance.
6. The tables gives critical values for sample sizes $n\le 25$ . For $n>25$ we use a normal distribution as an approximation to the distribution of the rank sum.

Example 6

The properties of a new alloy for potential use in aircraft wing construction are being investigated. If the new alloy is to replace the one in current use, it must be established that the mean axial twisting resistance of the two alloys does not differ significantly. 10 samples of each alloy are tested and the mean axial twisting resistance is measure. The results are given in the table below.

Use the Wilcoxon rank-sum test to decide, at the 5% level of significance, whether there is evidence of a significant difference in the mean axial twisting resistance of the two alloys.

Solution

Denoting the mean axial twisting resistance of the current alloy by ${\mu }_1$ and the mean axial twisting resistance of the new alloy by ${\mu }_2$ , we will test the hypothesis

$H_0:{\mu }_1={\mu }_2$

against the alternative

$H_1:\mu \ne {\mu }_2.$

Note that in the following table the use of - $c$ and - $n$ to denote current and new alloys is simply a device to enable use to distinguish between the two samples for the purposes of calculation.

Note that a spreadsheet such as Excel will sort quickly and accurately when this notation is used.

We now calculate the sum of the ranks assigned to the current (- $c$ ) alloy. Note that in this case the choice of which sum to calculate is arbitrary since both samples are the same size. We have

$S_C=\left(1+4+5+6+7+9+12+15+16+20\right)=95$

The sum $S_N$ of the ranks assigned to the new alloy is calculated as follows:

$S_N=\frac{\left(10+10\right)\left(10+10+1\right)}{2}-S_C=\frac{20\times 21}{2}-95=115$

From Table 2, the critical value at the 5% level of significance corresponding to two samples each of size 10 is 78. As neither rank sum is less than (or equal to) this value we conclude that on the basis of the available evidence we cannot reject the null hypothesis at the 5% level of significance.

Now do this Task.

Task!

A motorcycle engineer is investigating the resistance to stretching of two alloy steels for potential use in chains. The engineer wishes to establish in the first instance whether there is any difference in the mean resistance to stretch of the two alloys. 10 samples of one alloy and 12 samples of the second alloy are tested under the same conditions and the actual stretch is measured. All samples are the same length. The results are given in the table below.

Use the Wilcoxon rank-sum test to decide, at the 5% level of significance, whether there is evidence of a significant difference in the mean resistance to stretching of the two alloys.

Answer

1. It can be shown that in cases where the underlying distribution is normal, the $t$ -test is preferable to the Wilcoxon rank-sum test.
2. In cases where the underlying distribution in non-normal and the conditions for the $t$ -test cannot reasonably be met, it may well be preferable to use the Wilcoxon rank-sum test.
3. In cases where the underlying distribution is symmetric but non-normal and exhibits substantially larger tails then the normal distribution, it is often preferable to use the Wilcoxon rank-sum test since the mean of such distributions is often unstable.

Example 7

A civil engineer is investigating the compressive strength of a new type of insulating block for potential use in the building of new houses.

The engineer wishes to establish whether there is any difference in the mean compressive strengths of the blocks in current usage and the proposed new block.

Ten samples of the current block and 14 samples of the new block are tested under the same conditions and their compressive strength in pounds per square inch (psi) is measured. All samples are of the standard size used in the building industry.

The results are given in the table below.

Use the Wilcoxon rank-sum test to decide, at the 5% level of significance, whether there is evidence of a significant difference in the mean compressive strengths of the two types of insulating blocks.

Solution

Denoting the mean compressive strength of the current blocks by ${\mu }_1$ and the mean compressive strength of the new blocks by ${\mu }_2$ , we will test the hypothesis

$H_0:{\mu }_1={\mu }_2$ against the alternative $H_1:{\mu }_1\ne {\mu }_2.$

Note that the use of - $c$ and - $n$ to denote the current and new blocks is simply to device to enable us to distinguish between the two samples for the purposes of calculation.

We now calculate the sum $S_C$ of the ranks assigned to the blocks in current usage since this is the smallest sample. We have:

$S_C=\left(1+6+10+11+12+14+16+17+18+19\right)=124$

The sum $S_N$ of the ranks assigned to the new type of block is calculated as follows:

$S_N=\frac{\left(10+14\right)\left(10+14+1\right)}{2}-S_C=\frac{24\times 25}{2}-124=176$

From Table 2, the critical value at the 5% level of significance corresponding to samples of sizes 10 and 14 is 91. As neither rank sum is less than (or equal to) this value we conclude that on the basis of the available evidence we cannot reject the null hypothesis at the 5% level of significance.

Task!

The breaking strengths of cables made with two different compounds are compared. Standard lengths of ten samples using compound A and twelve using compound B are tested. The breaking strengths in newtons are as follows.

Use a Wilcoxon rank-sum test to test the null hypothesis that the mean breaking strengths for the two compounds are the same against the two-sided alternative. Use the 5% level of significance.

Answer

Exercises

1. The lifetimes of plastic clips with two different designs are compared by subjecting clips to continuous flexing until they break. Twelve of each design are tested. The lifetimes in hours are as follows.

   Design A				Design B
   36.1	16.6	24.6	38.5	62.5	28.2	19.9	33.9
   15.6	28.3	16.0	44.7	13.3	39.4	19.3	23.7
   14.3	10.8	0.7	6.5	12.7	122.0	168.0	55.0

   Use a Wilcoxon rank-sum test to test the null hypothesis that the mean lifetimes are equal for the two designs against the alternative that they are not. Use the 5% level of significance. Comment on any assumptions which are necessary.

2. An experiment is conducted to test whether the installation of cavity-wall insulation reduces the amount of energy consumed in houses. Out of twenty otherwise similar houses on a housing estate, ten are selected at random for insulation. The total energy consumption over a winter is measured for each house. The data, in mwh, are as follows.

   Without insulation					With insulation
   12.6	11.8	12.8	11.4	14.4	10.8	9.9	9.5	10.0	10.4
   12.3	11.5	13.2	11.0	11.8	10.7	11.8	7.5	11.8	10.1

   Use a Wilcoxon rank-sum test to test the null hypothesis that the insulation has no effect against the alternative that it reduces energy consumption. Use the 1% level of significance.

Answer

1.2 Critical values for the Wilcoxon signed-rank test

Table 1

For $n>25$ the rank sum has an approximately normal distribution with mean $M=\frac{1}{4}n\left(n+1\right)$ and standard deviation $s=\sqrt{n\left(n+1\right)\left(2n+1\right)∕24}$ .

1.3 Critical Values for the Wilcoxon Rank-Sum Test (5% Two-tail Values)

Table 2

1.4 Critical Values for the Wilcoxon Rank-Sum Test (1% Two-tail Values)

Table 3