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In order to convert the new GRE to IQ equivalents, we must first know the means and standard deviations of the Americans who take the GRE.  Commenter George kindly provided that information:

The ETS publishes the “GRE Worldwide Test Taker Report” periodically. The report for test takers between 2013-2016…shows a US mean/SD of 152.7/7.6 for Verbal and 150.2/7.8 for Quantitative

Next, we’d like to know the mean and SD of the composite score (V+Q).

The mean can be determined simply by adding the mean V and mean Q, which gives 302.9.  To get the SD of the composite, we must know the correlation between these subscales.  Among the subset of people who took the old GRE after also taking the SAT, the correlation was 0.56.  If we assume the correlation is the same for all GRE takers, and also for new GRE takers, then we can apply the following formula to get the SD of new GRE V + Q composite:

formula

r is the correlation between the two tests that make up the composite and σ is the standard deviation of the two tests.

So let’s get out our calculators:

Composite SD = SQUARE ROOT OF: 7.6(7.6) + 7.8(7.8) + 2(0.56)(7.6)(7.8)

Composite SD = SQUARE ROOT OF: 57.76 + 60.84 + 66.39

Composite SD = SQUARE ROOT OF: 184.99

Composite SD = 13.6

Now that we know the mean and SD for the verbal, quantitative and composite scores we can convert them to the IQ scale (where the U.S. mean and SD are defined as 100 and 15 respectively).  The problem is U.S. GRE takers are an academic elite, and thus have a different IQ distribution from the general U.S. population.  How different?  I’m no longer comfortable answering that without doing a bit more research, so we’ll save that for part 2.

 

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