Question

A politician estimates that 61% of his constituents will vote for him in the coming election. How many constituents are required for a random sample to obtain a margin of error of at most 0. 03 with 95% confidence

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that 1016 constituents are required.

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.

The margin of error is given by:

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

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.

The estimate is of $\pi = 0.61$, while the margin of error is of M = 0.03, hence solving for n we find the minimum sample size.

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

$0.03 = 1.96\sqrt{\frac{0.61(0.39)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.61(0.39)}$

$\sqrt{n} = \frac{1.96\sqrt{0.61(0.39)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.61(0.39)}}{0.03}\right)^2$

$n = 1015.5$

Rounding up, 1016 constituents are required.

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