NBC/WSJ/MaristNBC/WSJ published the results of a new poll. In this poll, interviewees from Colorado were asked for whom they will vote: Republican Donald Trump or Democrat Hillary Clinton.
Historically, Colorado has been a purple state, in which no single party has had overwhelming support to secure its electoral college votes. Hence, predictions here are of particular interest.
NBC/WSJ/MaristNBC/WSJ poll results
The results show that 43.0% of participants indicated that they would cast a ballot for former New York Senator Hillary Clinton, whereas 35.0% would vote for businessman Donald Trump.
The poll was conducted from July 5 to July 11. A total of 794 registered voters responded. Given the poll's error margin of +/-3.5 percentage points, the spread in voter support is statistically significant.
Putting the results in context
In general, however, don't have too much faith in the results of single polls, as they may incorporate large errors. Instead of trusting the results from single polls, the evidence-based approach is to rely on combined polls or, even better, a combined forecast that includes different methods and data.
In order to make the results comparable to benchmark forecasts, you can convert them into two-party vote shares. This procedure yields values of 55.1% for Clinton and 44.9% for Trump. To compare: Only 44.9% was obtained by Clinton in the NBC/WSJ/MaristNBC/WSJ poll on August 7, for Trump this number was only 0.0%.
Results compared to other polls
Clinton currently achieves 55.9% of the two-party vote in an average of recent polls in Colorado. This value is 0.8 percentage points higher than her corresponding numbers in the NBC/WSJ/MaristNBC/WSJ poll. This deviation is within the poll's margin of error, which suggests that the poll is not an outlier.
The poll in comparison with PollyVote's forecast
The current PollyVote forecasts Clinton to gain 53.4% of the two-party vote in Colorado. That is, Polly's prediction is 1.7 points below her polling numbers. Again, a look at the poll's margin of error shows that this deviation is negligible.