Following aggregation, each score is the mean of a set of z-scores – as such, for each indicator, the mean across Europe for all respondents is 0 (though the standard deviation is not necessarily 1 as these scores are the means of true z-scores, not z-scores themselves). This makes comparison easy. If an individual has a positive mean for positive functioning for example, then one immediately knows that their functioning is better than the European average. If their score for vitality is lower than their score for positive emotions, then one can say that, relative to Europe, low vitality is more of an issue for them than lack of positive emotions.

These aggregated standardised scores are the most appropriate for analysis purposes. However, for presentation outside the academic sphere, they are somewhat clumsy. Most people are not used to dealing with z-scores. Furthermore, whilst z-scores tell you very clearly how well a particular score compares to the mean, they tell you nothing about how well they compare to the theoretical minimum or maximum 1 for a given indicator. For one indicator , the theoretical maximum could be 1.2, for another it may be 4.2. To resolve this issue we looked for a transformation metric that maps the z-scores for each indicator onto 0–10 scales (t-scores), where a ‘0’ is the minimum for that indicator, ‘10’ is the maximum, and ‘5’ is the mean for Europe.

Before reading on, take a sheet of squared paper and try to imagine what an appropriate transformation metric would look like, using the autonomy score. The x-axis is for the z-scores which range from, in this case -2.57 to 1.52. The y-axis is for the desired t-scores, from 0-10. On the sheet, plot the three points that our transformation metric must pass through. For example, we know that, when the z-score is -2.57, the t-score should be 0. What are the other two points we know? Look at Figure 2.4

As you can see quickly, the three points (0, 5 and 10) do not sit on a straight line. This means that no linear transformation (i.e., of the form y = mx + c) is possible. Note that, if we weren’t interested in the middle point (i.e. a z-score of 0 did not need to equate to a t-score of 5), then we would not have this problem. A straight line could be drawn between the point (-2.57,0) and (1.52,10). Any idea what the formula would be?

The answer is found on the next page.

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