Affirmative Action-free GRE Scores

Steve Sailer publishes a table of graduate school admissions tests results, broken out by race. Steve explains:

To make all the numbers comprehensible, I’ve converted them to show where the mean for each race would fall in percentile terms relative to the distribution of scores among non-Hispanic white Americans. Most of us have some sense of what the distribution of talent is among whites—political correctness doesn’t demand we avert our eyes when it comes to whites—so I’ll use whites as benchmarks.

Here is is table:

Mean Score as Percentile of White Distribution

Test	Degree	Year	White	Black	Asian	Tot Hisp	Mex-Am
GMAT	M.B.A.	2008	50%	13%	55%	27%	24%
GRE-Verbal	Ph.D./M.A.	2007	50%	18%	47%	29%	28%
GRE-Quant		2007	50%	14%	66%	29%	28%
LSAT	J.D.	2006	50%	12%	47%	19%	29%
MCAT-Verbal	M.D.	2007	50%	10%	36%/td>	19%	21%
MCAT-Phys Sci		2007	50%	14%	61%	24%	25%
MCAT-Biol Sci		2007	50%	10%	54%	24%	25%
DAT	D.D.S.	2005	50%	16%	60%	27%	NA

Steve has his own commentary on what these numbers mean. What I want to examine is the median black score as a percentile of the white median. Looking at the GRE scores, for instance, we see that only 16% (cumulative estimate) of blacks scored above the white mean. This is extraordinary when you consider that the test is presumably taken exclusively by college graduates, among which the black representation presumably reflects their -1SD mean cognitive ability.

But what should the black percentile be? Our initial guess might be: 50%. After all, whatever affirmative action might be present in college admissions, blacks have presumably met the same graduation standards, correct?

Actually, no. First, there is abundant evidence that black graduates are overrepresented in less-demanding fields of study: elementary ed, say, rather than engineering. But second, even aside from academic specialty, there is additional reason to expect that black GRE scores would be lower: the distribution of test takers is not normally distributed; rather, they are distributed along the upper tail of a normal distribution.

Consider a hypothetical. Let's assume that college graduates taking the GRE possess a minimum +1SD academic ability above the white mean. This strikes me as reasonable. Granted, many more students than this attend college, but the dropout rate is high, and will will assume at present that the just-got-by students don't aspire to professional school. Thus, in the graph below, if the blue line represents the distribution of academic ability among the white population (μ = 0, σ = 1), the shaded area represents the distribution of white GRE test-takers.

It is from the shaded area only that the median white score (in SD's) must be calculated. For the math geeks, given a cutoff score of t_c = 1, we are trying to find the median score t_med:

For the bit-heads, the Matlab code looks like this:

t_c = 1;

t = t_c:.001:(t_c + 2);

t_med = ...

max(t((1 - normcdf(t))./(1-normcdf(t_c))>.5))

Given my assumptions, t_med turns out to equal around 1.4 SD. Thus, assuming no selection bias or other means of skewing the results, the black percentile score would be calculated by comparing the percentage of the black distribution (μ = -1, σ = 1) that exceeds the median white ability t_med = 1.4 to the percentage of the black population that exceeded the cutoff for taking the test, t_c = 1:

Again with the Matlab:

percentile = ...

(1 - normcdf(t_med,-1,1))/(1 - normcdf(t_c,-1,1))

This percentile score turns out to be about 35%. In other words, lower than 50%, but not as low as the measured value of 16%. How can we account for the difference?

Steve provides some additional data on the number of test takers among the different ethnic groups as compared to their representation in the population. It turns out that blacks have only half the representation among test takers (e.g. 47% for the GRE) as their percentage of the population would predict, which sounds low . . . until you consider that the black share of those with +1SD academic ability is only 14% of their percentage of the population.

Well, I did make up the +1SD standard. Is there an ability cutoff whereby black representation among test takers would be 50% of its representation in the population? For those of you keeping score at home, we are trying to find:

t = -1:.001:0;

t_c = ...

max(x((1 - normcdf(t,-1,1))./(1-normcdf(t))>.5))

t = t_c:.001:0;

t_med = ...

max(t((1 - normcdf(t))./(1-normcdf(t_c))>.5));

percentile = ...

(1 - normcdf(t_med,-1,1))/(1 - normcdf(t_c,-1,1))

The solution to the standard t_c to this equation turns out to be -0.68 SD. Not only is this number absurd (equivalent to pretending that an IQ of 90 is sufficient to graduate from college), but it still only gives a percentile score of 25%. In other words, it's closer to the measured value of 16%, but still not close enough. In fact, the only standard by which only 16% of black test takers would exceed the white median would be: no standard. In other words, if we were to administer the GRE to the entire population (or some random sample thereof), the white median would be zero (in a standard distribution), and 16% of blacks would exceed it. Bear in mind what we are measuring: as the standard goes up, the black representation among those that meet it decreases; however, among the blacks that do meet the standard, the percentile score compared to the white median improves.

In order to account for the black percentile score, it is necessary for us to drop the assumption of a single standard. Clearly, the GRE and other graduate school exams dip much further in the pool of black college graduates than they do in the pool of white college graduates. How much more? Well, let's return to our assumption of a t_c of +1SD, but apply it only to the white students. What would the standard applied to black test takers (herewith t_cb) be in order to lower the percentile score to 16%?

t_c = 1;

t = t_c:.001:(t_c + 2);

t_med = ...

max(t((1 - normcdf(t))./(1-normcdf(t_c))>.5))

t = 0:.001:t_c;

t_cb = ...

max(t((1 - normcdf(t_med,-1,1))./(1 - normcdf(t,-1,1))<.16))

The value thus calculated for t_cb is 0.64. This is equivalent to an IQ cutoff for black college graduates of 110 versus an IQ cutoff for white college graduates of 115. In conclusion, while we can partially account for the black percentile score on the GRE by improving our model of the distribution of test-takers, we cannot account for all of it. It is therefore fairly evident that Steve's analysis is almost certainly correct: the GRE and other graduate school exams dips much further into the pool of black college graduates than it does in the pool of white college graduates. Why this is, and indeed why there is a deeper pool from which they can draw, is a question that merits more examination.