Question

52% of 1000 registered voters intend to vote for Steven Collins for mayor. Based on a 95% confidence interval, can we assume that Steven Collins will win the election if all votes are cast for either him or his opponent, Eric Camden?
A. Cannot be determined as samples cannot reliably be used to estimate population parameters
B. No, since the confidence interval includes values less than 50%, his opponent may get more votes
C. Yes, 52% is a majority of the votes
D. No, sampling variation causes all confidence interval conclusions to be questionable.

Answer 1

Using the z-distribution, it is found that the correct option regarding the question in the problem is given by:

B. No, since the confidence interval includes values less than 50%, his opponent may get more votes.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

In this problem, the parameters are given as follows:

$\pi = 0.52, n = 1000$

Hence, the bounds are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 1.96\sqrt{\frac{0.52(0.48)}{1000}} = 0.489$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 1.96\sqrt{\frac{0.52(0.48)}{1000}} = 0.551$

The lower bound is below 0.5, hence option B is correct.

More can be learned about the z-distribution at brainly.com/question/25890103

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