Using data from the first 55 unique individuals to take the sequence addition test, a cumulative frequency table was made:

This allowed me to convert raw scores on the test percentile equivalents and these percentiles were then assigned normalized Z scores (the Z scores they would equate to in a perfectly normal curve) relative to the test taking population, not the general population.

RawPercentile (%)Z-Score (relative to the test-taking population NOT the general (U.S.) population
01.8−2.08
14.5−1.69
210.9−1.23
320.9−0.81
429.0−0.55
536.0−0.36
644.0−0.15
759.0+0.23
883.0+0.95
996.0+1.75
1099.0+2.33

The next question is how self-selected is the test taking population. On a scale where Americans average IQ 100 (SD = 15), those test takers who reported Wechsler IQs had a mean of 127 (SD = 24) suggesting they are much brighter and much more variable than the general U.S. population. There is no reason to think the test takers who reported Wechsler IQs are brighter than the test takers in general; in fact the latter scored higher on the sequence addition test (mean 5.65 SD = 2.51 vs mean 4.2 SD = 3.19) though given the small sample size of those with self-reported Wechsler scores (n=5) it’s not statistically significant.

Now when Ron Hoeflin normed the Mega Test, he used equipercentile equating, meaning that even though the two tests are imperfectly correlated, he assumed that the percentile distribution of the two tests would perfectly match. So if the highest SAT score among his Mega takers was one in a million, then the highest Mega score among the Mega takers was one in a million, even if they were not the same person.

Hoeflin’s approach makes sense IF one assumes self-selection for taking the Mega Test correlates about as well with Mega scores as it does with SAT scores, however I highly doubt my test correlates about as well with self-selection as the Wechsler does, and the reason is a comprehensive test like the Wechsler is going to be much more reliable and valid than my brief test is, and thus correlate better with most external criteria. So assuming my sample averages +1.8 SD on the Wechsler, I’d expect them to average (0.58)(1.8 SD) = 1.04 on sequence addition. Why 0.58? Because before contamination, that’s the correlation between WAIS-R full-scale IQ and Digit Span (the subtest most similar to sequence addition). Also, if my sample has a Wechsler SD that is 160% the U.S. SD, on this test they should be 160(0.58) = 93% of the U.S. SD.

Armed with these statistics, I transformed the normalized Z scores calculated with reference to the test taking population to Z scores with reference to the U.S. population.

U.S. population Z score = (test taker Z score)(0.93) + 1.04

Raw score on sequence additionEstimated Z score relative to the U.S. population
0−0.8944
1−0.5317
2−0.1039
3+0.2867
4+0.5285
5+0.7052
6+0.9005
7+1.2539
8+1.9235
9+2.6675
10+3.2069

I deliberately did not convert these scores to IQ equivalents because people shouldn’t get the impression that a simple test like this measures overall IQ but if I had, I would have done so by multiplying them by 15 and adding 100.