Mega Test Norms

Maintained by Darryl Miyaguchi
Last updated: November 1, 1997

Table T1 shows raw scores on the Mega Test and the corresponding I.Q. scores, using 16 I.Q. points per standard deviation. These I.Q. assignments are the results of Ronald Hoeflin's sixth norming of the Mega Test.

¹ Does not accept scores from unsupervised tests such as the Mega as qualifying for admission.
² Accepts Mega Test scores as qualifying for admission.
³ No longer accepts the Mega Test as qualifying for admission (it used to, up until 1992). Its own self-administered tests are acceptable, though.
⁴ Currently (11/97) debating whether or not it will continue to accept the Mega Test. It doesn't appear likely that it will.

The Sixth Norming of the Mega Test⁴
by Ronald K. Hoeflin

The chief impetus behind this new norming of the Mega Test was my acquisition of data from the Educational Testing Service showing combined verbal plus mathematical aptitude SAT scores (on a scale from 400 to 1600) for the years 1985, 1986, 1987, and 1988, supplementing the data I already had for 1984, upon which my fifth norming was entirely based. I had hoped that with data on over 5 million SAT test subjects I would be enabled to refine my norms for the upper end of the Mega Test scale, in particular permitting me to pinpoint the one-in-a-million level more accurately. Unfortunately, this goal could not be achieved by means of this extra data since the number of SAT scores reported to me by Mega Test participants, 222, remains inadequate. I did succeed, however, in finding a striking new approach to extrapolating the Mega Test scale to the one-in-a-million level and beyond.

I began by calculating that there were almost precisely one-third as many SAT participants from 1984 to 1988 as there were 18-year-olds, namely about 5 million vs. 15 million. I assumed that close to 100% of 18-year-olds in the top few percent in ability would attempt the SAT, and that whatever shortfall there might be would be roughly balanced by the extra foreign participants. I then found the percentile equivalents of standard deviations (sigmas) ranging from 1.25 to 4.25 above the mean at intervals of 0.25 sigmas, using standard statistical tables for the normal (Gaussian) distribution curve, since my aim was to map Mega Test raw scores into this curve. I then made a factor-of-3 shift in these percentiles to allow for the above-average ability of SAT participants. These adjusted percentiles were then converted into SAT scores for each year at each sigma level using the data supplied by the Educational Testing Service. After averaging these scores for all five years, I equated the resulting SAT averages with Mega Test raw scores at each sigma level by ranking all the reported SAT scores from 1 to 222 and by ranking all the Mega Test raw scores achieved by those reporting SAT scores likewise from 1 to 222 and equating scores of equal rank. These results are reported in tables T2 and T3 below.

I then examined the data I had compiled in my fourth norming, in which I had used scores reported on five other tests: the AGCT (Army General Classification Test), CTTM (California Test of Mental Maturity), LAIT (Langdon Adult Intelligence Test), S-B (Stanford-Binet), and the WAIS (Wechsler Adult Intelligence Scale). In the fourth norming I had found the equivalent Mega Test scores for each of these tests at each sigma level from 1.25 to 4.50 at intervals of 0.25 and then averaged these figures. The resulting graph had a noticeable dip in it between 3.50 and 4.50 sigmas. This dip can be largely eliminated, however, by weighting the results for each test by the number of persons who had reported scores for each test. These weighted averages differ from the SAT-based results reported in table T3 by less than one Mega Test raw score point at each of the twelve sigma levels from 1.25 to 4.00, overall, the SAT-based results averaging just one-sixth of a point higher than the weighted averages from the other five tests. But at 4.25 sigmas the results differ by 2.4 Mega Test raw score points, which suggests that the data from these tests is becoming too unreliable to be trusted at any higher levels. I averaged the SAT-based and non-SAT-based results and report the outcome in table T5 below.

⁴ The norming date was not specified, but I believe it was performed sometime in 1989 (DTM).

Graph G1. depicts the foregoing calculations and extrapolations

Discussion: Six times as many participants scored above the 99th percentile as would have been expected to on the basis of the number who scored above the 90th percentile divided by 10; 3.8 times as many scored above the 99.9th percentile as would have been expected to on the basis of the number who scored above the 99th percentile divided by 10; and so forth. Graph G1 suggests that the last three ratios for table T6 should be approximately 2.2, 2.1, and 2.0. Multiplying these numbers by the expected number of participants yields the number that ought to be observed above these levels, from which the Mega Test score can be determined.

⁵ As of the date of this norming, Ron still believed that "Eric Hart's" score of 47 was a result of a first attempt at the Mega. Marilyn vos Savant's first-attempt score is shown as the sole 46 (DTM).

Graph G2. Equivalencies between Mega Test raw scores and standard deviations above the mean (with IQs)

In order to extrapolate to the 99.9999 percentile and beyond, I determined the equivalent sigma scores for the 90, 99, 99.9, 99.99, and 99.999 percentiles from standard statistical tables for the normal distribution curve. I then equated these percentiles with raw scores on the Mega Test by interpolating between (or, in the case of the 99.999 percentile, extrapolating slightly beyond) the results given in table T5. I then determined how many Mega Test participants had scored above each of these raw scores (and hence above each of these percentiles) using the distribution of Mega Test raw scores shown in Table T7. I obtained fractional results by assuming, for example, that the 96 people who scored 24 right were spread evenly over the interval from 23.5 to 24.5. By comparing one-tenth of the number who exceeded each of these percentiles with the number who actually exceeded the next higher percentile, I found that 6.0 times as many people exceeded the 99th percentile as would have been expected to by dividing the number who exceeded the 90th percentile by 10, and the corresponding figures for the 99.9, 99.99, and 99.999 percentiles were 3.8, 2.8, and 2.4, respectively. Graphing these data points, one finds that they are leveling off fairly quickly. I estimated that the next three figures would probably be about 2.2, 2.1, and 2.0. Since 54.8 people exceeded the 99.999 percentile, I estimated that 2.2 times (54.8/10) = 12.06 people ought to have exceeded the 99.9999 percentile, that 2.1 times (12.06/10) = 2.54 people ought to have exceeded the 99.99999 percentile, and that 2.0 times (2.54/10) = 0.50 people ought to have exceeded the 99.999999 percentile. Using the table of actual score distribution in table T7, I found that 12.06 people exceeded a raw score of 42.6, that 2.54 people exceeded a raw score of 45.2, and that 0.5 people exceeded a raw score of 47.0. Extrapolating from these results, the ceiling of the test⁶ would appear to be roughly the one-in-300,000,000 level or 99.9999997 percentile for a perfect score of 48. The results described in this paragraph are compiled in table T6, graph G1, and the associated discussion.

In Graph G2 I graphed the results from tables T5 and T6, the table T5 results appearing as black dots and the table T6 data appearing as small circles with unfilled-in interiors. I then drew a line through these sixteen data points, using a straightedge for the middle section and French curves for the lower and upper curved sections.

Finally, in table T1 I compiled my final norms, based on graph G2. A uniform scaling of 0.075 sigmas per raw score point was used for the straight-line section from a raw score of 8 to 38. Since I use 16 I.Q. points per standard deviation, this works out to 1.2 I.Q. points per raw score point. The percentiles were then, of course, derived using a standard statistical table for a normal curve, which can be found in any statistics book, since for example 3 standard deviations above the mean or 148 I.Q. is always equivalent to the 99.87 percentile for a normal curve. The right-hand column lists nine high-IQ societies next to their respective minimum qualifying scores, of which all but Mensa and Intertel accept the Mega Test for admission purposes⁷.

⁶ Ron no longer assigns a specific IQ level for a Mega raw score of 48 right; thus the ceiling is IQ 190+ rather than IQ 193.
⁷ Table T1 has been updated to reflect existing high-IQ societies and the new ceiling (DTM).

Graph G3 below shows the Mega Test score distribution as tabulated in table T7. The median score of all 3920 participants is 15 (IQ = 139).

Graph G3. Mega Test Score Distribution as of the Sixth Norming (1989)

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