Analysis of the difference in the Florida vote counts

Robert Weiss

UCLA Department of Biostatistics

School of Public Health
Los Angeles, CA 90095-1772

robweiss at ucla.edu

Summary. Neither a t-test nor a Bayesian analysis suggests that the Florida results are well enough determined to be certain that Bush should be declared the winner.

There have been two vote counts in Florida, one giving the lead of 1784 to Bush, and the second compiled by the Associated Press, giving a lead of 327 votes to Bush. I analyzed this as a sample of size n=2 from a normal population whose mean M is the true difference between Bush’s vote total and Gore’s vote total.

The question of interest is whether, based on this data, we can be sure that M is indeed positive. If there is reasonable doubt that M is positive, then Mr. Gore should not yet concede the election and Mr. Bush should not yet claim victory.

One method to analyze this data is to use a t-test to test the null hypothesis
H₀: M <= 0
versus
H_A: M > 0.
If we do this, we get x-bar = 1055.5, and s = 1030.255, and a t-statistic of 1.0245. The p-value is .246, not allowing us to reject the null hypothesis in favor of the alternative. Note that if we reverse the hypotheses, we can’t reject that null hypothesis either. Basically this data isn’t strong enough to determine whether M is greater than zero or not.

A second method of analyzing this data is to run a Bayesian model with a flat or near flat prior and calculate the posterior probability that M < 0, given the data. This calculation gives a posterior probability of approximately .19 that M might be negative, again suggesting that the result is not well enough determined to identify for certain that Bush is the winner.

This last Bayesian calculation was done in a program called WinBUGS (http://www.mrc-bsu.cam.ac.uk/bugs/welcome.shtml). The WinBUGS program gives all assumptions used and is given below.

Data, program/model and initial values.

list(x=c(1784, 327), n=2)

model{

for(i in 1:n) {

x[i] ~ dnorm(m,tau) }

sigma <- 1/sqrt(tau)

m ~ dnorm(0,.0000000001)

msign<-step(m)

tau ~dgamma(.00001,.00001)

}

list(m=0,tau=.000001)

The parameter msign gives the probability that M is positive. One minus this calculation gives the probability that M is negative. An MCMC sample of size 100000 was used. The prior sets a mean of zero and a standard deviation of 100000 votes for the difference M.