Wilcoxon Signed Rank Test

Wilcoxon signed rank test is a non-parametric test used to test or compare two related samples. Wilcoxon signed rank test, which is also known as Wilcoxon matched pair test is a non-parametrc equivalent of the paired t-test. The test is based on the magnitude of the difference between the pairs of observations.

The test is performed differently for small and large samples. Groups with size less than 30 are considered as small samples.

Null and Alternalte Hypothesis: Null hypothesis for Wilcoxon test states that there are no difference between the two samples and the alternate hypothesis states that there are some differences.

Example: Consider the following sets of scores from two different groups of people for a certain objects O1...O8.  Consider level of significane as 0.05.

Objects	Score I	Score II
O1 O2 O3 O4 O5 O6 O7 O8	82 69 73 43 58 56 76 85	63 42 74 37 51 43 80 82

Step 1: Calculate the difference between the two scores.

Objects	Score I	Score II	d
O1 O2 O3 O4 O5 O6 O7 O8	82 69 73 43 58 56 76 85	63 42 74 37 51 43 80 82	19 27 -1 6 7 13 -4 3

Step 2: Take the absolute value of the difference d.

Objects	Score I	Score II	d	| d |
O1 O2 O3 O4 O5 O6 O7 O8	82 69 73 43 58 56 76 85	63 42 74 37 51 43 80 82	19 27 -1 6 7 13 -4 3	19 27 1 6 7 13 4 3

Step 3: Rank the difference and record the sign of each difference.

Objects	Score I	Score II	d	| d |	Rank	Sign
O1 O2 O3 O4 O5 O6 O7 O8	82 69 73 43 58 56 76 85	63 42 74 37 51 43 80 82	19 27 -1 6 7 13 -4 3	19 27 1 6 7 13 4 3	7 8 1 4 5 6 3 2	+ + - + + + - +

Note:

1.       If the difference d between two score is 0, then we have to ignore the value while ranking the difference.

2.       If there are some 0 difference, then the total number of the sample will be decreased. For example, in this example if there is one object with same score, then the total number of samples to consider will be 7.

3.       If there are some differences with same value, first rank the whole entries in increasing order then take the average of the ranks of those entries.

For example, consider the following set of data.

Objects	Score I	Score II	d	| d |	Rank	Re-Rank
O1 O2 O3 O4 O5	82 69 73 69 58	63 42 74 42 51	19 27 -1 27 7	19 27 1 27 7	3 4 1 5 2	3 4.5 1 4.5 2

Step 4: Add the rank with same sign and consider the minimum of the two for getting the statistics T.

In our example,

T+ = 7+8+4+5+6+2 = 32;

T- = 1+3 = 4

T = min (T+, T-) = min (32, 4) = 4.

Step 5: Find out the critical value with level of significance as 0.05 from the Table G.

N	Level of significance for one-tailed test
	.025	.01	.005
	Level of significance for two-tailed test
	.05	.02	.01
6	0	—	—
7	2	0	—
8	4	2	0
9	6	3	2
10	8	5	3
11	11	7	5
12	14	10	7
13	17	13	10
14	21	16	13
15	25	20	16
16	30	24	20
17	35	28	23
18	40	33	28
19	46	38	32
20	52	43	38
21	59	49	43
22	66	56	49
23	73	62	55
24	81	69	61
25	89	77	68

From the above table, the critical value is 4.

Step 6: Compare the statistics value and critical value and take decision.

Rule: Reject null hypothesis if statistics is greater than or equal to critical value.

Here,

Statistics T = 4,

Critical Value = 4.

So, according to rule reject null hypothesis if T >= CV.

So, we are rejecting the null hypothesis because the two sample scores are not equal.