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PaperclipPerfector  1yr ago (text post)   31988 thread views

Culture War Roundup for the week of July 24, 2023

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Assuming aptitude is normally distributed and doesn't contain any ceilings for some races, if you filter you applicants beforehand with a threshhold, race loses its predictive power. Of course this only works without concepts like affirmative action:

If the threshhold is 130 IQ, than the share of <140 IQ and >140 IQ will the same for blacks as for whites. The only difference willl be that whites will have a big positive multiplies for both sets.

Or phrased differently, P(130<IQ<135) / P(135<IQ) should be equal regardless of race

  • -13

This would be true for an exponential distrubution but it is false for a normal distribution

The notion that conditioning on a threshold can wash away differences in mean is wrong intuitively and its wrongness can be expressed formulaically and precisely, but I also did a quick simulation to confirm because why not.

Suppose we have a population with a mean of 100 (stand-in for white) and a second population with a mean of 85 (stand-in for black) for a given trait (e.g., IQ), both with standard deviations of 15. (1 - normcdf(140, 100, 15))/(1 - normcdf(130, 100, 15)) is about 16.8%. (1 - normcdf(140, 85, 15))/(1 - normcdf(130, 85, 15)) is about 9.1%. So whites have almost twice as much of those over 140 than blacks.

I generated 10 million hypothetical individuals for each population as a check, and got the same figures out to the digits shown in the percentages.

This is incorrect. Even with a cutoff, whites would have a higher average IQ.

With a standard distribution, group differences will be amplified among outliers.

About 5.9% of people who are above +2 deviations will also be above +3.

Only about 2.3% of people who are above +3 deviations will also be above +4.

I encourage you to calculate these numbers yourself: https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

I encourage you to calculate these numbers yourself:

Should've just done it in advance. Lesson learned, thanks for the correction

Suppose that Pop A has mean 100, sigma 15 and Pop B has mean 85, sigma 15.

The ratio shown above is not the same for the two populations. If the cutoff is 85 and the high mark is 100 instead, this is obvious without computing Z-scores.

Is that really true? Consider a threshold of 0 IQ (so 6 std dev below the mean). No-one (well, 1 in a billion) is excluded by this, but as the sets are the same, if there was predictive power before, there is after.

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