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How to Convert Norm Referenced Test Scores

Updated on October 26, 2013

Norm Referenced Scales

Norm Referenced scales are an expression of raw test data in terms of a statistically useful scale with a set mean and a set standard deviation. Prominent versions of such scales are CEEB scores, stanines, sten scores, T scores, Deviation IQ scores, and NCE scores.

The information needed to convert a score expressed on one of these scales into an expression of the equivalent score on any other scale is merely the mean and the standard deviation of the scale to which you want to convert.

The CEEB has mean of 500 and S.D. of 100.

The Deviation IQ has a mean 100 and a S.D. of 15.

T scores have a mean 50 and a S.D. of 10.

Stanines (standard nines) have a mean of 5 and a S.D. of 2.

Sten Scores have a mean of 4.5 and a S.D. of 2.

NCE (Normal Curve Equivalent) scores have a mean of 50 and a S.D. of 21.06.

Finding Z and Scale Conversion

To convert between the different norm referenced scores it's all about first finding Z which is the number of standard deviations (expressed as a whole number and decimal) that a score occupies either above or below the mean,

So take the measured score on a test and subtract the mean score of all test takers and then divide by the standard deviation of that test, or Z=(score-mean)/S.D. This gives you a kind of constant that you can use to find the scores for other tests with different means and different standard deviations.

So once you have Z, plug it into this formula for whatever other norm referenced scale you want to convert to. New test score=Z(S.D.') + mean', where S.D.' is the Standard deviation of the new scale and mean' is the mean of the new scale.

So for example) If someone scored 115 on the WAIS IQ test and you want to convert it to a T score, first find their Z

For the WAIS IQ test the mean=100 and S.D.=15, they scored 115 so

Z=(score-mean)/S.D. or,

Z=(115-100)/15 or,

Z=1

Now that we have the Z value we can convert to any norm referenced test for which we know the mean and the standard deviation. For T scores mean=50, S.D.=10.

So,

T=S.D.'(Z)+mean' where S.D.' is just the S.D. of the new norm referenced test, and mean' is the mean of the new norm referenced test. So,

T=10(1)+50 or

T=60

You could also convert it to a CEED (S.D.=100, mean=500) or a stanine (S.D.=2, mean =5) or an NCE (S.D.=21.06, mean=50). Using the same formula and it would look like this;

CEED=100(z)+500 or,

CEED=100(1)+500 or,

CEED=600

For Stanines;

Stanine=2(Z)+5 or,

Stanine=2(1)+5 or,

Stanine=7

And for NCE Scores

NCE=21.06(Z)+50 or,

NCE=21.06(1)+50 or,

NCE=71.06

So if we find the Z score for one type of norm referenced test, we can convert the score for that test to a score for any other norm referenced scale.

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