Question

A group conducted a poll of 2071 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 49​% of the popular vote and candidate B would receive 48​% of the popular vote. The margin of error was reported to be 2​%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means.

Answer 1

Let $\hat{p_1}$ be the proportion of votes for Candidate A = 0.49

Let $\hat{p_2}$ be the proportion of votes for Candidate B = 0.48

The confidence interval has 2 values, a lower limit and the upper limit, which is given by ($\hat{p_1}$ - $\hat{p_2}$) $\pm$ ME

If both the signs are positive, it would mean that $\hat{p_1}$ is > $\hat{p_2}$ and they cannot be equal, and it is more likely that Candidate A wins the election.

If both the signs are negative, it would mean that $\hat{p_1}$ is < $\hat{p_2}$ and they cannot be equal, and it is more likely that Candidate B wins the election.

The confidence interval that we obtain is (0.49 - 0.48) $\pm$ 0.02

Lower Limit = 0.01 - 0.02 = -0.01

Upper Limit = 0.01 + 0.02 = 0.03

Here we have an interval where the signs are different. This means that it could go both ways, i.e either Candidate A wins or B wins, or even a deadlock is possible as the CI contains 0. Hence it is too close to call.