Question

Find the margin of error for the given values of c, o, and n.

Answer 1

Using the z-distribution, it is found that the margin of error is of 0.059.

We are given the standard deviation for the population, hence the z-distribution is used.

What is the margin of error for a z-confidence interval?

It is given by:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• z is the critical value.

• $\sigma$ is the population standard deviation.

• n is the sample size.

In this problem, the parameters are:

• $\sigma = 0.25, n = 49$.

• Confidence level of 0.90, hence, using a z-distribution calculator, the critical value is z = 1.645.

Then:

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.645\frac{0.25}{\sqrt{49}}$

$M = 0.059$

The margin of error is of 0.059.

You can learn more about the z-distribution at brainly.com/question/12517818