I have a rating system which can take 5 possible outcomes, which are coded to:

0 : No problem
1 : Minor problem
2 : Mild problem
3 : Moderate problem
4 : Severe problem

This scale relates to severity of individual's medical conditions and I have 12 separate items (each with the same 5 possible outcomes) I would like to aggregate the 12 to make a combined score.

The problem I have is when I combine the scales, for example using a sum. If individual 'A' scores a 2 (mild problem) across all 12 items they would be assigned a aggregated score of 24. However if individual 'B' scores a 4 (severe problem) in 2 of the 12 items and 1 (minor problem) in the rest they would be assigned a aggregated score of 18.

So using this crude example individual 'A' would be ranked higher than individual 'B' but 'B' requires the more urgent attention. A mean would also yield similar results.

Is there any analysis that can be applied to this to favour the more severe cases without skewing results.


As you point out, these weights are too close from each others. More than a change of analysis, I would suggest to keep the sum (or mean) but change the weights. I would suggest three ways depending to the degree of time you dispose and rigour you want :

  • Simplest one : use weights that makes more sense to you like a multiplicative scale (0,1,2,4,8).

  • Simple one : Ask an expert questions like "how many Moderate problems worth a Severe problem", "how many Mild Moderate problems worth a Severe problem" etc... and build a scale which respect these constraints.

  • More heavy one : Use individual profiles (may be fake if necessary), ask one or several experts to rate how severe is the case presented, run a regression model where your predictors are the number of mild,minor,moderate and severe problems that the patient have to find the weights predicting best the overall severity rating given by the expert.

  • $\begingroup$ Thanks for the response - will attempt for the third option, being the most robust. $\endgroup$ – sharkey32 Jun 17 '15 at 10:19

A simple match solution is to apply exponentiation to your values (with two, or three, or more - even not integer - as exponent), and then go for sum or mean. This leads to an 'auto' higher weight for mild to severe problems.

$$ v = \frac{\sum_1^n x^a}{ n} , \quad a> 1$$

With your example values ($n=12$), $a=3$ exponent leads to $v_A= 8$ and $v_B=11.5$, that fits your 'urgence' factor.

  • 1
    $\begingroup$ Note that these are sometimes known as generalised means en.wikipedia.org/wiki/Generalized_mean if you take the ath root (so, the cube root in @Tomaso example) $\endgroup$ – mdewey Apr 29 '16 at 14:38

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