Question

A survey was given to a random sample of 45 residents of a town to determine
whether they support a new plan to raise taxes in order to increase education
spending. of those surveyed, 9 respondents said they were in favor of the plan. at the
95% confidence level, what is the margin of error for this survey expressed as a
proportion to the nearest thousandth?.

Answer 1

Using the z-distribution, it is found that the margin of error for the 95% confidence interval is of 0.117.

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 sample size and the estimate are given as follows:

$n = 45, \pi = \frac{9}{45} = 0.2$

Hence the margin of error is given by:

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

$M = 1.96\sqrt{\frac{0.2(0.8)}{45}}$

M = 0.117.

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

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