Question

Marge conducted a survey by asking 350 citizens whether they frequent the city public parks. of the citizens surveyed, 240 responded favorably. what is the approximate margin of error for each confidence level in this situation?

Answer 1

Using the z-distribution, considering the standard 95% confidence level, the margin of error is of 0.0486 = 4.86%.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

The margin of error is:

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

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.

The sample size and the estimate are given by:

$n = 350, \pi = \frac{240}{350} = 0.6857$

Hence the margin of error is:

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

$M = 1.96\sqrt{\frac{0.6857(0.3143)}{350}}$

M = 0.0486 = 4.86%.

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

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