Title: Gamma
Course: UGQuants (SSPS) > Section: Association for Nominal and Ordinal Variables

Gamma

Gamma, also referred to as Goodman and Kruskal’s Gamma, is a measure of association for ordinal data.

In a table we can think of cells which contain concordant and discordant pairs of values. Concordance exists when a case which takes a higher value on one variable also takes a higher value on the other variable. A pair of values is discordant when a case which has a higher value on one variable takes a lower value on the other variable.

An example will help to clarify this. Research suggests that male candidates with a full head of hair are more successful in elections than those who are bald.  We might ask candidates to rank the state of their hair (full head of hair, partially bald, totally bald) and how they fared in their last campaign (very well, well, not well at all).  We would have concordance if those with full heads of hair fared very well and the bald candidates fared poorly, and discordance if all the bald candidates fared very well and those with full heads of hair fared poorly.

Gamma computes the ratio of the number of concordant pairs minus the number of discordant pairs to the total of all pairs. If each of our variables took only two values, the formula for Gamma will be exactly the same as that for Yule's Q insert link to page ( = (ad-bc/ ad+ bc). All the cases in the cells a and d (top left and bottom right) of our table will be concordant, and those in cells b and c (top right and bottom left) will be discordant. When there are more than two values on either variable, the formula and calculations for Gamma become more complex but the principle remains exactly the same.

To calculate Gamma it is useful to start with a cross tabulation table but we should always arrange the table in decreasing order from left to right and from top to bottom.

The formula for Gamma is as follows:

where N_a is the number of agreements or concordant pairs and N_i is the number of inversions or discordant pairs.

Let us pretend for the moment that we have two variables that are perfectly and positively related: as one increases, so too does the other. Let us assume, for example, that the larger the town a person lives in, the higher their perception of the prevalence of crime. This is a perfect and positive relationship. A perfect inversion would suggest a perfect negative relationship: as the town one lives in gets larger, the perception of crime decreases.

Let us assume that we have asked 80 people from different sized towns about their perceptions of crime and ended with the following results

Town size and perception of crime

Perception of crime | Town size | Total
Big | Medium | Small
High | 15 | 8 | 7 | 30
Medium | 10 | 10 | 8 | 28
Low | 4 | 7 | 11 | 22
Total | 29 | 25 | 26 | 80

To calculate Gamma we must follow seven basic steps:Note we have arranged the table in decreasing order from left to right and from top to bottom so that the highest category (big, high) of both the IV and DV are in the top left and the lowest value of both is in the bottom right.

1 | Check to make sure that the cross table is ordered correctly so that the ordinal values for the IV decrease from left to right and the ordinal values for the DV decrease from top to bottom
2 | Identify N_a (the number of agreements) assuming a perfect positive relationship.

If we have a perfect positive relationship - as town size increases so too does perception of crime - we’d expect to focus in the cell at the top left (big city, lots of crime). But what about those other positive relationships such as small city, small perception of crime. For gamma we concentrate on those cells diagonally below and to the right of the first cell.  In our example, this includes 10 (medium, medium), 7 (medium, low), 8 (medium, small), and 11 (small, small). We add these together  10+7+8+11 = 36.

and then multiply them by the cell we were originally examining: (15 (big, high)  36 * 15 = 540

If we are interested in a positive relationship we wouldn’t just be interested in the top left cell. We’d also be interested in those living in a medium city with a high perception of crime (8) and would multiple this by the cells diagonally below and to the right …  (8+11)*8 =152… as well as those living in a big city with a medium perception of crime (7+11) x 19… and those living in a medium sized city with a medium perception of crime  11 *10 = 110.

Of course those living in a small city with a low perception of crime are also of interest to us but there are no cells diagonally below and to the right of it, so for now we ignore this bottom right cell.

3
We add together all these agreements

N_a = 540+152+180+110 =982.
4 | Identify N_i, the number of inversions assuming a perfect inverse relationship (as city size decreases, perception of crime increases).

We do this in a similar process to that pursued above except we start in the upper RIGHT hand cell of the table, identify the appropriate cells and then examine the cells diagonally below and to the LEFT.

Small city, high crime =(10+10+7+4) * 7 = 217

Medium city, high crime = (10+4) * 8 = 112

Small city, medium crime (7+4) * 8 = 88

Medium city, medium crime 10 * 4 = 40

We add together all of these inversions  Ni = 217+112+88+40 = 457

5 | Now we can insert each of these into the formula:

There is, then, a moderate, positive correlation between the size of the town you live in and your perception of crime.  If the answer had been -.36 there would be a moderate negative correlation.

Gamma is also useful because it can tell us whether there is a statistically significant relationship between two variables. This means that it is not only a measure of association but can be adapted to become a test of significance.  To determine whether gamma is significant we use a z test. The null hypothesis is that the two variables (IV and DV) are not associated (in other words, that gamma = 0).

Our number must exceed the critical value in order to reject the null hypothesis. If we use the numbers from the previous example we get 1.65

We must consult a critical value table. We know that the z score for the .05 level of significance is 1.96 and we know that the z score for the .01 level of significance is 2.58, so our number must be greater than either of these numbers in order for us to reject the null.  Because our calculated figure is smaller than both of these critical values we fail to reject the null and conclude that there is not a significant relationship between the size of the town one lives in and one’s perception of crime.