The Mitt Romney campaign must surely be concerned about the lack of support for it in the black community, with some surveys putting it at an astounding zero percent. In response, they have set up a Black Leadership Council, no doubt to bring more African-Americans into the fold.
I quickly scanned the people who are on this council and was shocked to find that Alan Keyes was not there. How could Romney leave him out? Keyes ran unsuccessfully as a Republican for the US senate in Maryland in 1988, 1992, and in Illinois in 2004 (when he lost to Barack Obama) and for the Republican nomination for president in 1996, 2000, and 2008.
Keyes’s main claim to fame is as the source of the data that fixes the ‘crazification factor’ in America, the number of people who will vote for any crazy person, however crazy, at 27%. He is a birther with a proven ability to attract the crazy. Assuming that the percentage of crazy in the black voting bloc is the same as in the general population, Romney could improve his chances by aligning himself with Keyes.
After all, he can’t do worse than zero percent.
Romney probably didn’t invite Keyes because Keyes has left the Republican Party.
I also notice that Ohio’s Ken Blackwell isn’t on the list. Blackwell seemed to do a decent job as Secretary of State, but then after he lost to Strickland he seemed to let the crazy loose quite a bit.
Sure, if by “decent job” you mean the conflict of interest* and voter disenfranchisement that led to him being named in at least 16 lawsuits.
http://en.wikipedia.org/wiki/Ken_Blackwell#Involvement_in_the_2004_U.S._Presidential_Election
* Chief Elections Official of Ohio and honorary co-chair of the “Committee to re-elect George W. Bush” during the 2004 election. Simultaneously.
Wow! Then it really is strange that Blackwell didn’t end up on that panel!
I’m curious about the margin of error for that zero percent black support for Romney.
The article says that the survey polled 1000 registered voters, with a margin of error of 3.1 percent.
I assume (?) that the margin of error is relative to the overall population—so there’s only a small (1 in 3?) chance that more than 3.1 percent of registered voters are blacks who support Romney, right?
That would be a much bigger percentage of black people, though.
IIRC, only something like 1 in 7 American adults is black, so that 3.1 percent of the population would be about 21.7 percent of blacks—more than a fifth of blacks could support Romney with some significant probability.
Also IIRC, margins of error are usually reported with about 2/3 probability—two out of three times, the actual value will be within the margin, and one out of three times it won’t.
I guess that this would be a one-tailed distribution, because you can’t get lower than zero, so that would give a one in six chance that over a fifth of blacks support Romney, right?
Now that’s a shocking result. I’d have put the probability of more than a fifth of blacks supporting Romney much lower than that!
I hope someone will correct me if my math is substantially off, but it seems to me that the poll just doesn’t say much about black support for Romney that we couldn’t already guess—e.g., that it’s likely to be somewhere in the single digits.
It doesn’t tell us that it’s likely to be less than one percent. It’s fairly likely to be several percent for all this poll can tell us, right?
What it does seem to say is that while 27 percent of Americans may be crazy enough for a wackaloon like Keyes, probably less than a fifth of blacks are crazy enough to vote for a wackaloon like Romney.
I guess blacks are relatively sane when it comes to Romney. 🙂
Ugh, please ignore my previous post.
Wikipedia says that a sample size of 1000 gives you a three percent margin of error with 95 percent confidence, not 67 percent (1 SE?) as I thought. Doh. (I guess I need a stats refresher.)
That changes things a lot.
(Mano, feel free to delete both of these posts.)