Question

​If, based on a sample size of 900​, a political candidate finds that in582 people would vote for him in a​ two-person race, what is the 90​% confidence interval for his expected proportion of the​ vote? Would he be confident of winning based on this​ poll?

Answer 1

The 90% confidence interval for the political candidate's expected proportion of the vote, based on a sample size of 900 with 582 people indicating they would vote for him in a two-person race, is approximately 60.8% to 65.3%.

However, winning confidence cannot be solely determined based on this poll, as the outcome depends on various factors beyond the confidence interval.

Step-by-step explanation:

In order to calculate the 90% confidence interval for the expected proportion of the vote, we can use the formula:

`\[ \text{Confidence Interval} = \hat{p} \pm z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]`

where
`\(\hat{p}\)` is the sample proportion (582/900), n is the sample size (900), and z is the Z-score corresponding to the desired confidence level (for 90%,
`\(z \approx 1.645\))`.

Substituting the values:

`\[ \text{Confidence Interval} = (582)/(900) \pm 1.645 * \sqrt{((582)/(900) * \left(1-(582)/(900)\right))/(900)} \]`

After calculations, the confidence interval is approximately 60.8% to 65.3%. This means that we are 90% confident that the true proportion of people voting for the candidate lies within this range.

It's important to note that confidence intervals provide a range of plausible values, but winning an election depends on factors such as the opponent's popularity, undecided voters, and potential changes in public opinion.

Therefore, while the candidate may have a favorable confidence interval, it does not guarantee victory, and additional considerations are needed for a comprehensive assessment of electoral prospects.