Question

Find the margin of error if p = 71% and n = 100. round to four decimal places.

Answer 1

Using the z-distribution, it is found that the margin of error for the 95% confidence interval is of 0.0889.

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 and the sample size are given as follows:

$\pi = 0.71, n = 100$

Hence, the margin of error is:

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

$M = 1.96\sqrt{\frac{0.71(0.29)}{100}}$

M = 0.0889.

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

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