Steps for working out the Wilcoxon Test in Excel
How to carry out a Wilcoxon test
STEP ONE:
Find the differences between each pair of scores (i.e. Condition A – Condition B).
|Participant |Condition A |Condition B |Difference (d) |
|1 |25 |32 | |
|2 |29 |30 | |
|3 |10 |8 | |
|4 |31 |32 | |
|5 |27 |20 | |
|6 |24 |32 | |
|7 |26 |27 | |
|8 |29 |30 | |
|9 |30 |32 | |
|10 |32 |32 | |
|11 |20 |30 | |
|12 |5 |32 | |
STEP TWO:
Rank the differences, ignoring any “0” differences and ignoring the sign of the difference
To rank the differences:
Ignoring the sign of the difference (whether its positive or negative), the lowest difference is – 1, of which there are 4. So, we add up the ranks they would take e.g., _ + _ + _ + _ = __, and then divide this by the number of ranks, so __ / _ = __.
The next lowest rank is 2 (there are both positive and negative differences here, but ignore the signs). So, _ + _ = __. The ranks assigned would therefore be __ / _ = __.
|Participant |Condition A |Condition B |Difference (d) |Rank |
|1 |25 |32 | | |
|2 |29 |30 | | |
|3 |10 |8 | | |
|4 |31 |32 | | |
|5 |27 |20 | | |
|6 |24 |32 | | |
|7 |26 |27 | | |
|8 |29 |30 | | |
|9 |30 |32 | | |
|10 |32 |32 | | |
|11 |20 |30 | | |
|12 |5 |32 | | |
STEP THREE:
Add together the ranks belonging to scores with a positive sign: __ + __ = __
STEP FOUR:
Add together the ranks belonging to scores with a negative sign:
__ + __ + __ + __ + __ + __ + __ + __ + __ = __
STEP FIVE:
Whichever of these sums is the smaller, is our value of W. So, W = __.
STEP SIX:
N is the number of differences (omitting “0” differences). We have 12 – 1 = 11 differences. [Important to note that these are not the same as degrees of freedom. We only calculate N here, using N – 1, as we have 1 difference which equals zero]
STEP SEVEN:
Use the table of critical Wilcoxon values on Graham’s website. With an N of 11, what is the critical value for a two-tailed test at the 0.05 significance level?
The critical value = __. With the Wilcoxon test, an obtained W is significant if it is LESS than the critical value. So, in this case observed W = __ and critical W = __.
Our obtained value of __ is ____ than __, and so we can conclude that there is not a difference between Condition A and Condition B.
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