Day 25 – Wilcoxon test – Practice Data

Our dataset:

 

Participants Condition 1 Condition 2
1 3 5
2 4 5
3 5 2
4 1 5
5 5 4
6 2 5
7 3 5
8 4 4
9 1 5
10 3 5

Our hypothesis: participants will learn a list better with music (Condition 2) than without music (Condition 1).

So we not only predict diferences among the means in the two conditions. We also predict that the differences will be in a way such that Condition 2  will have higher mean scores than  Condition 1.

From the above data we need to calculate:

  • Means in Condition 1
  • Means in Condition 2
  • Differences Column
  • Rank of Differences Column
  • Signed ranks (positive)
  • Signed ranks (negative)

Now we fill out the rest of columns with the calculations. Reminder: the Signed Ranks column get their sign from the D column and their number from the Rank of Differences column.

 

Participants Condition 1 Condition 2 d Ranks of d Signed ranks (plus) Signed ranks (negative)
1 3 5 -2 5 -5
2 4 5 -1 2 -2
3 3 2 +1 2 +2
4 1 5 -4 8.5 -8.5
5 5 4 +1 2 +2
6 2 5 -3 7 -7
7 3 5 -2 5 -5
8 4 4 0 (tie)
9 1 5 -4 8.5 -8.5
10 3 5 -2 5 -5
Means 2.9 4.5 Rank Total: -4 Rank Total: -41

The smaller Rank Total value (ignoring the sign) is the value of W. We can see in the table that W is 4.

Rationale of Wilcoxon test

The Wilcoxon test aims to compare the number of positive and negative values from the Signed Ranks columns. In other words, it aims to compare the negative and positive differences in scores of each participant. The idea is that if the differences between conditions are random (which is what the null hypothesis claims), the number of positive and negative differences (i.e. the two Rank Total values) should be roughly equal. But, if the numbers of positive and negative differences show some significant differences, the null hypothesis could be rejected.

We have seen that the is a  difference between the two Rank Totals. Now we need to see if the difference is statistically significant. So we check the Wilcoxon Table.

The Wilcoxon Table

The Wilcoxon Table enables you to check whether given your one/two tailed hypothesis, your W value and your sample size, the probability that the differences found between conditions were likely to occur by chance. Reminder:  the significance levels range from 5% to 1%.

Our values are: W = 4, N = 9 (it was 10 but one participant does not count because he got the same scores in the two conditions) and hypothesis = one-tailed.

We need to go to the N column and stop at the row where N = 9, following the 9 there will be several columns with different values, starting from the lowest value, we need to stop at the column that gives true when asked whether W is lower than it. In our case, that value is 6. When looking at the one-tailed significance levels, our W is lower than the number that gives a p value of 0.025. That means that our p-value is lower than 0.025. In other words, the probability that the differences found between conditions can occur due to chance is less than 2.5% which is below 5%, this enables us to claim that the differences are statistically significant and thus we can reject the null hypothesis.

Reminder: always check that the differences are in the direction predicted by the one-tailed hypothesis. In our case, the data shows that participants scored higher in Condition 2 than they did in Condition 1 and the differences are statistically significant (p < 0.05).

 

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