Question

In a sample of 198 observations, there were 80 positive outcomes. Find the margin of error for the 95% confidence interval used to estimate the population proportion.

Answer 1

Using the z-distribution, the margin of error for the 95% confidence interval used to estimate the population proportion is 0.0683 = 6.83%.

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 given by:

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

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.

For this problem, the parameters are given as follows:

$n = 198, \pi = \frac{80}{198} = 0.404$

Hence the margin of error is:

$M = 1.96\sqrt{\frac{0.404(0.596)}{198}} = 0.0683$

The margin of error for the 95% confidence interval used to estimate the population proportion is 0.0683 = 6.83%.

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

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