Day 37 – Friedman test (related) – FD13Ror (DVCD)

When do we use it? When there are:

a) Differences between conditions

b) One variable

c) Three conditions (or more)

d) Related-design

We will code it like this: FD13Ror. The “or” comes from ordinal data and means that the Friedman test is non-parametric (i.e. uses ordinal data). Since we are using ordinal data, we will rank the scores like we did in the Wilcoxon test. However, since the Friedman test uses more than 2 conditions, we cannot compute a difference column which renders Wilcoxon-style ranking useless. Since we cannot rank vertically, we will rank horizontally, across a participant’s conditions. See the raw data below:

Participants Condition 1 Condition 2 Condition 3
1 1.5 1.5 4
2 3 5 7
3 1 7 4
4 2 6 7

And see the computed data below:

Participants Condition 1 Ranks Condition 2 Ranks Condition 3 Ranks
1 1 1.5 1 1.5 4 3
2 3 1 5 2 7 3
3 1 1 7 3 4 2
4 2 1 6 2 7 3

Since we have 3 conditions, the ranking has 3 positions where 1 represents the lowest score and 3 the highest score. And the tied scores are computed like it was done in the Wilcoxon test.

————————————————————————

See the practice data below:

Hypothesis: children would rate some pictures higher (prettier) than others.

Since we are not stating which picture will be the most/lowest rated, our hypothesis is two-tailed.

Note: Friedman can only tackle two-tailed hypothesis.

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

From the above table, we calculate:

  • Ranks row for every participant in every condition
  • Means for every condition
  • Ranks total for every condition
Participants Condition 1 Condition 1 Ranks Condition 2 Condition 2 Ranks Condition 3 Condition 3 Ranks
1 2 1 5 3 4 2
2 1 1 5 3 3 2
3 3 1 5 2.5 5 2.5
4 3 2 5 3 2 1
5 2 1 3 2 5 3
6 1 1 4 2.5 4 2.5
7 5 3 3 2 2 1
8 1 1 4 3 3 2
Rank Totals 11 21 16
Means 2.25 4.25 3.5

We see that on average,  there are some wide differences in the rating scores. But we need to check whether or not these differences are statistically significant. So we will use the Friedman test.

Rationale of the Friedman test (related)

The Friedman test aims to find out whether the rank totals are different for the conditions. If the differences between the rank totals are due to random factors (as stated by the null hypothesis), the differences across the conditions would be relatively small. In that case, the null hypothesis could be rejected.

Friedman test

N = Number of participants

C = Number of conditions

∑T² = Sum of Squared Rank Totals (for all conditions)

The Friedman Test Table

The Friedman Critical Values Table (I have not found any straightforward table so I will upload one myself soon) enables you to check whether given your one/two tailed hypothesis, your value and your sample size, the probability that the differences found between conditions were likely to occur by chance.

We open up the table and we check against our N (8) and Friedman value (6.25) values. Our value has to be equal or larger than the values in the table. The critical value with our sample size is 0.047. Our value is larger than 0.047 so the probability that the differences found between conditions can occur due to chance is less than 5%, this enables us to claim that the differences are statistically significant and thus we can reject the null hypothesis.

In our case, taking a look at the means, the data shows that the picture in Condition 2 was the highest rated, followed by the picture in Condition 3 and the picture in Condition 2, we know that the differences are statistically significant (p < 0.05).

Remainder: the Friedman test can only be used with two-tailed hypotheses.

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One thought on “Day 37 – Friedman test (related) – FD13Ror (DVCD)

  1. […] data). Since we are using ordinal data, we would ideally rank the scores like we did in the Friedman test. However, since the Friedman test is unrelated, we cannot do horizontal ranking, instead, we will […]

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