Bendixen's Mathematical Challenge

Friday, November 9, 2012
Pollster Sergio Bendixen has put out a press release claiming that the overall Hispanic precinct vote tally in Miami-Dade County (Obama 51-Romney 49) confirms his exit polling data of Cuban-American votes (Romney 52-Obama 48).

But actually, it proves his math is wrong.

Using Bendixen's own numbers, if non-Cuban Hispanics, which account for 25% of Miami-Dade Hispanics, voted for Obama 82-18, then for the overall vote tally to reach 51-49, the Cuban-American vote, which account for 75% of the county's Hispanics, would have to be at least 60-40 Romney-Obama.

This works inversely also.  If the Cuban-American vote were Romney 52-Obama 48, then to reach the overall 51-49 tally, the non-Cuban Hispanic vote would be in the mid-50s, which is well below the national Hispanic average.

And if you aren't lazy and actually correlate the census numbers, percentages, per precinct, and don' try to confuse observers by focusing on non-Cuban precincts in Kendall, Doral, etc., then this number is likely to be even higher.

Here's the math:

51 = 0.25*82 + 0.75*x

Solve for x:

x = (51 – 0.25*82)/0.75
= 40.7

You can do the same for y, the proportion of Cuban-Americans who voted for Romney:

y = (49 – 0.25*18)/0.75
= 59.3

Or, alternatively,

y = 100 – x
= 59.3

Note you can also do scenarios, for example, if 50% Cuban-Americans voted for Obama, what percent of non-Cubans could have voted for him keeping the 51 percent total unchanged? Assume Cuban-American vote was 50% and let z be the non-Cuban proportion. Solve for z:

z = (51 – 0.75*50)/.25
= 54.0

That is a much lower number than the national average for Hispanics.

Elementary, my dear Bendixen.