I have three datasets as follows:

  • Dataset A can have a maximum score of 30. (Individual scores can be from 1 to 30)
  • Dataset B can have a maximum score of 45. (Individual scores can be from 1 to 45)
  • Dataset C can have a maximum score of 25. (Individual scores can be from 1 to 25)

The three datasets contribute equally to a final score, hence it is 33.33% each. i.e. (33.33% for dataset A + 33.33% for dataset B + 33.33% for dataset C) = 100% (final score)

I have the following scores in each dataset:

  • Dataset A: 22
  • Dataset B: 15
  • Dataset C: 10

Is there a quick formula that I can use to work out the final score, based on the above datapoints?

(This is not a homework question.)

  • $\begingroup$ $(22+15+10)/3$ would work but you might want to scale the scores (e.g. divide by the number of points possible) because the 45 point scale will contribute more weight than it should purely because there are more points possible $\endgroup$
    – Macro
    Jan 1 '12 at 9:29
  • $\begingroup$ Thanks. How can I write the new formula (i.e. the one that takes into account the differences in the maximum scores possible) $\endgroup$ Jan 1 '12 at 12:22
  • $\begingroup$ When you say "The three datasets contribute equally to a final score" do you mean that having the highest possible score on test B and lowest possible score on test C is as "good" as the highest possible score on test C and lowest possible score on test B? If not, then you could just add the three scores together to get something in the range 3 to 100. $\endgroup$
    – Henry
    Jan 1 '12 at 23:31

There are different options for rescaling your summed scale scores (or scaled scores):

  • Express every score on a 0-100 point scale, with higher scores reflecting higher locations on the latent trait each scale purports to assess;
  • Standardize scores ($T$- or $z$-score) such that scores are deviations from the mean, expressed in standard deviation (SD) units. For $T$-scores, the mean and SD that are considered are 50 and 10, respectively.

(Percentile-based or normalized scores are also common options. Note that for $T$-scores, we usually rely on the empirical mean and SD of a larger population that responded to all items previously (e.g., during large-scale field study) and which might be considered as a "reference population". Of course, more complex methods exist in the case of grading or equating raw scoring gathered throughout different measurement instruments.)

The use of a common scale makes more sense with sum scores (it doesn't matter much if you consider the mean instead of the sum, which is what social scientists generally prefer compared to psychologists), as @Macro pointed out.

Simple formulae exist in this case (this is just a rescaling problem), but the general idea can be summarized as follows:

Scaled score = [(Raw score - Min response category score) / 
                 Range of possible response category scores] 
                 * 100

to get scores on a 100-point scale. If some items (or response categories) are negatively worded, you will need to reverse-score them first. Subtract the resulting score from 100 to get scores expressed in the reverse direction.

Once each scale score (A, B, C) has been expressed on a common scale, you can use the arithmetic (unweighted) mean to compute your final score.

  • $\begingroup$ The formula is illustrative (thanks a lot!) and I am able to follow it but I am still having difficulty working out how to use the arithmetic mean in the last paragraph. (This is the second part of the above problem to work out the final score based on one-third contribution of A+B+C). Suppose the means for raw scores in Dataset A is 10, for Dataset B is 15 and for Dataset C is 20, how do I use them. I will be most grateful for an example. $\endgroup$ Jan 3 '12 at 6:30
  • 2
    $\begingroup$ @Adhesh If each three scores are expressed on a common scale (say, 0-100 point), then $\bar x = (x_A+ x_B+ x_C)/3$ is what you are after (each scale contributes equally to the total score). With individual data like $(22,15,10)$, the final score would be $1/3\times(21/30+14/45+9/10)=0.637$ which you can conveniently express on any scale you want. (On a 0-45 point scale, it amounts to 28.7; on a 0-100 point, it is 63.7, etc.) $\endgroup$
    – chl
    Jan 3 '12 at 7:46

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