CNN-ORC published the results of a new poll. In this poll, participants from Ohio were asked for whom they will vote: Democrat candidate Hillary Clinton or Republican candidate Donald Trump.
Ohio is traditionally a purple state, where the GOP and the Democrats have often achieved similar levels of voter support. This is the reason why the election outcome here is regarded important in determining the overall result of the presidential election.
CNN-ORC poll results
According to the results, 46.0% of interviewees will give their vote to former Secretary of State Hillary Clinton, while 50.0% plan to vote for businessman Donald Trump.
The poll was conducted from September 7 to September 12 with 769 likely voters. The error margin is +/-3.5 points, which means that the levels of voter support for Trump and Clinton do not differ significantly.
Putting the results in context
Single polls should be treated with caution, because they may include large errors. Rather than trusting the results from single polls, we recommend to consult combined polls or, even better, a combined forecast that incorporates forecasts from different methods, each of which incorporates different data.
To make the results comparable to forecasts from benchmark methods, we translate them into two-party vote shares. This procedure results in values of 47.9% for Clinton and 52.1% for Trump. In the latest CNN-ORC poll on March 6 Clinton received 53.8%, while Trump received only 46.2%.
Results in comparison to other polls
Trump currently runs at 50.6% of the major two-party vote according to an average of recent polls in Ohio. This value is 1.5 percentage points lower than his corresponding numbers in the CNN-ORC poll. This difference is within the poll's error margin, which means that the poll is not an outlier.
The poll compared with PollyVote's prediction
The most recent PollyVote foresees Trump to gain 48.7% of the two-party vote in Ohio. Hence, the PollyVote forecast is 3.4 points below his polling numbers. Again, a look at the poll's margin of error suggests that this deviation is insignificant.