Florida: Slight advantage for Clinton in recent Mason-Dixon poll
Results of a new poll carried out by Mason-Dixon were circulated on October 3. The poll asked participants from Florida for whom they will vote: Hillary Clinton or Donald Trump.
Florida is traditionally a battleground state, where the candidates of both major parties have historically gained similar levels of support among voters. Therefore, the election outcome in that state is regarded critical in determining which party will win the majority of electoral votes.
Mason-Dixon poll results
Of those who answered the question, 46.0% said that they would vote for former New York Senator Hillary Clinton, while 42.0% said that they would give their vote to real estate developer Donald Trump.
This poll was conducted from September 27 to September 29, among a random sample of 820 likely voters. Considering the poll's margin of error of +/-3.5 percentage points, the results reflect a statistical tie.
Putting the results in context
Single polls should be interpreted with caution, as they may include substantial errors. Rather than relying on results from single polls, the best practice is to look at combined polls or, even better, the combined PollyVote forecast that draws upon forecasts from different methods, each of which draws upon different data.
In order to make the results comparable to forecasts from other methods, we translate them into shares of the two-party vote. This yields figures of 52.3% for Clinton and 47.7% for Trump.
Comparison to other polls
If we look at an average of Florida polls, Clinton's current two-party vote share is at 51.6%. This value is 0.7 percentage points lower than her corresponding numbers in the Mason-Dixon poll. This margin is within the poll's sampling error, which means that the poll is not an outlier.
The poll compared with PollyVote's forecast
The most recent PollyVote expects Clinton to gain 50.4% of the two-party vote in Florida. That is, the PollyVote forecast is 1.9 points below her polling numbers. Again, a look at the poll's sampling error reveals that this difference is insignificant.