Reply #51: Re-doing the math [View All]

43 percent of 122.6 million voters is 52.6 million, which is 2.14 million more than Bush's 50.46 million in 2000. Even if the actual weight is 42.6 percent, we are left with 52.1 million, which is still a 103 percent turnout.

If we instead use an actual 97% turnout of Bush voters, we now see that the maximum proportion of Bush-2000 voters among the 2004 electorate is 40.0 percent. This means that the exit poll had a weighing error of at least 6.5 percent (2.6 divided by 40.0). This level of error is simply unacceptable. And it means that the entire 'fixed' exit poll data is useless until it is again re-weighted.

And, if we multiply out the partiles, we have a final result of Kerry with 50.4% of the popular vote and Bush, 49.2%. The reported popular vote margin of 2.46 percent falls well outside of the expected range (-0.8 to +3.2). In fact, if 1% = MOE at a 95% level, then the reported popular vote has a T-score of 6.00 and a probability of 2.06 x 10 to the -9th. Or about a 1 in 500 million chance of occurring.

Can some statistician help me out? Is there any way that Bush could actually have won the popular vote by 2.46 percent?

Even using the outside edge of the margin of error, i.e. a Bush margin of victory of 0.8 percent, would mean that just over 1,000,000 votes were switched over from Kerry to Bush or that just over 2,000,000 votes were fraudulently added to Bush's total nationwide (or, more likely, some combination of the two types of fraud).