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Severe Discrepancy Determination by Formula

Kate obtains an IQ score of 90 and an achievement score of 74. Is this 16-point difference large enough to be considered a ‘significant difference’ between ability and achievement? Below is a table showing a statistical manipulation of the scores.

Using this approach to assessment, Kate should be considered to be functioning significantly below what was expected of her.

Above is an example of the use of a regression equation to determine a severe discrepancy. Regression is necessary because of the imperfect correlations between the ability and achievement measures. Just because someone has a low IQ score, we should not assume that they would also have a correspondingly low score on other measures. No one, I hope, would assume that just because a person had a very high IQ that the person would obviously be able to excel at gymnastics, or be able to hold their breath longer than ‘normal.’ Just because someone is good or bad at one thing does not guarantee that the person will also be good or bad at other things. The IQ test does not measure the same thing as an achievement test. Regressing the scores on the tests used also allows one to compare individuals to others with the same IQ. For example, if one child has an IQ of 120, while another an IQ of 80, the expectation of how each might be performing on an achievement test would be different. If the tests correlated at the .60 level, the first child would be expected to obtain an achievement score of 112 while the second child would be expected to achieve a score of 88.

Because most tests do not report scores in terms of Z, we can use a formula to calculated Z:

Z=

The simplest way to regress a particular score (either IQ or achievement) is to multiply the correlation between the two measures by the Z score you wish to regress. The formula would be:

Regressed score = (Correlation*Z )

The best way to determine the correlation between two measures is to look the correlation up in the manual of the test used. Unfortunately, not all manuals offer that information; the information is often based on absurdly small samples; and not all tests have been compared to each other. One way around this shortcoming is to estimate the correlation between the two tests. If we know the reliability of the ability test and the reliability of the achievement test, the correlation between the two tests can be estimated.

The equation for estimating the correlation between two tests is

**The values rxx and ryy are the internal consistency reliability coefficients for the aptitude and achievement tests used.

*** note that in the example at the beginning of this page, the correlation is listed as .47.

If the user knows from some source what the correlation really is, that should override the estimated correlation obtained from a formula. The formula provides the upper end of what is possible given the two reliabilities but those are estimates. In the example given, since  the correlation between the IQ and the achievement (.47) was known,  it was used. The example is something like using the PIQ versus a reading test (correlation .47) versus using the VIQ versus a reading test (correlation .65).

Use this link for Tables for the Reliability

Finally, the score must be turned back into a Standard Score with a mean of 100 and a standard deviation of 15. This is done by multiplying the "regressed Z" by the standard deviation and adding 100 to the result.

Predicted achievement score =

or

15*(Correlation*((Ability-100)/15))+100

This regressed score provides the examiner with a different point for comparison. Comparing the expected score with the actual score gives a better idea of the magnitude of the difference between the scores obtained. When Kate obtained an IQ score of 90 and an achievement score of 74 her 16-point difference was considered non-significant. However, she was expected (using the regression formula) to obtain an achievement score of 95, not 90. This increases the simple difference from 16 points to 21 points. We still don’t know if this new 21-point difference is large enough to be considered a ‘severe discrepancy.’ For that determination, a different formula is needed.

Magnitude of Difference required at .05 level =

1.96*15*(SQRT(1-(Correlation^2)))-1.65*((15*SQRT(1-(Correlation^2))*

SQRT(1-((RAbility)+(RAchievement*(Correlation^2))-(2*(Correlation^2)))/(1-(Correlation^2)))))

Thank heavens for computers!!

The first part of this monstrous equation:

1.96*15*(SQRT(1-(Correlation^2))

sets the level of our decision making to the 95% confidence (z=1.96) and then determines the Standard Error of Estimate for the obtained score. The second part of the equation:

-1.65*((15*SQRT(1-(Correlation^2))*SQRT(1-((RAbility)+(RAchievement*(Correlation^2)) -(2*(Correlation^2)))/(1-(Correlation^2))))

reduces the final cut-off score by subtracting the standard error of the relevant difference score. The end result is a statistically justifiable ‘severe discrepancy’ upon which to make clinical decisions.

When a psychologist doesn’t bother to struggle with the true identification problems, the result can be an over-identification of children as disabled when they are not. This increases the caseloads of the school and the individual staff, as well as making the whole label of "Learning Disability" into a meaningless catch-all. It must also be noted that even when a child does have a severe discrepancy between ability and achievement, this by itself does not constitute a diagnosis. It is a necessary, but insufficient factor, to be used in the determination.

For a wonderful explanation of this and other issues see:

Reynolds, C. R. Conceptual and technical problems in learning disability diagnosis, (Chapter 24) in Handbook of Psychological and Educational Assessment of Children: Intelligence and Achievement (Reynolds & Kamphaus) (1990) Guilford Press 

A copy of the template for determining severe discrepancy using this method is available.