There is a follow-up article by Kolata in today's Times (Sunday).
There was blogospheric clamor for the full distribution of the number of sexual partners of men and women; the original report from the CDC, it turns out, doesn't have that, but groups people into four groups -- those who have had 0 or 1 sexual partners, 2 to 6, 7 to 14, and 15 or more. This strikes me as insufficient resolution. In particular, zero is much different than one, as any virgin could tell you. Two seems a lot different than six, as well; two sexual partners could be someone who had sex with their spouse and one other person, whereas six sexual partners in a lifetime, although not a lot, can't have such a simple story behind it.
In any case, I'll reproduce the table. (The numbers are percentages.)
| Partners | 0-1 | 2-6 | 7-14 | 15+ |
| Men | 16.6 | 33.8 | 20.7 | 28.9 |
| Women | 25.0 | 44.3 | 21.3 | 9.4 |
The claimed medians for the number of sexual partners for men and women are seven and four, respectively. But 50.4% of men have had six or less sexual partners, according to this data. My earlier claim that the data might support a two-peaked distribution for women seems unlikely, but can't be ruled out at this resolution. (But I've seen enough other distributions that would explain the difference in medians that I don't really believe my own theory any more.) You can't extract means from this data -- and in fact at this resolution, it's theoretically possible that all the men could have 0, 2, 7, or 15 sex partners, for a mean of 6.46, and all the women 1, 6, 14, or [large number] sex partners, for a mean of at least 7.3 (if [large number] is in fact 15), making the female mean actually higher than the male mean. It would be simple (though I won't do it) to tweak the numbers so that the two means came out exactly equal.
Kolata (who has a master's in math, according to Wikipedia), however, claims that the data is inconsistent, in that there's no way to make the means equal: "I got between 40 percent and 75 percent more male than female partners depending on how you guess the average on each interval." I wonder what she tried. Sure, I'm just showing that it's possible the means are equal, not that it's likely. But someone with mathematical training should know better.
3 comments:
Wow, seriously? They use that rough a scale in the actual study? No wonder nobody takes social scientists seriously.
I commented on your original post -- the explanation of two different peaks in the pattern of partners could apply exactly equally well to either men or women: the situations are completely symmetrical. The only difference is that one sex would have a larger peak in the group with a greater number of partners, whereas the other sex would have to have a greater peak in the group with fewer partners, in order to make the medians different. (I think you'd need more men in the promiscuous category, whereas you'd need more women in the less promiscuous category, to make it work out.) It could even be a combination of the two effects.
I would actually tend to suppose that the male distribution would be more like this and the female distribution more uniform, based very loosely on the evolutionary concept of differential investment and the differnce in mating patterns by sex that follow from it.
Whoops -- "uniform" was a bad word choice there. I didn't mean uniform in the technical sense. I should have said that I would guess the male distribution would be the one more likely to be divided into two separate peaks.
But, actually, an even simpler explanation is simply that the distribution is more skewed, and that's closer to my guess here: that the male distribution of sex partners per capita is more strongly negatively skewed, leading to a larger median than mean, and the female distribution is less skewed, not skewed, or skewed in the opposite direction. But that's a wild guess.
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