Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 98% confident that you estimate is within 2% of the true population proportion. How large of a sample size is required?

Answer 1

Using the z-distribution, it is found that a sample size of 3,385 is required.

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 98% confidence level, hence$\alpha = 0.98$, z is the value of Z that has a p-value of $\frac{1+0.98}{2} = 0.99$, so the critical value is z = 2.327.

We have no prior estimate, hence $\pi = 0.5$ is used, which is when the largest sample size is needed. To find the sample size, we solve the margin of error expression for n when M = 0.02, hence:

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

$0.02 = 2.327\sqrt{\frac{0.5(0.5)}{n}}$

$0.02\sqrt{n} = 2.327 \times 0.5$

$\sqrt{n} = \left(\frac{2.327 \times 0.5}{0.02}\right)$

$(\sqrt{n})^2 = \left(\frac{2.327 \times 0.5}{0.02}\right)^2$

n = 3,385.

A sample size of 3,385 is required.

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

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