Question

Use the given data to find the minimum sample size required to estimate the population proportion.
Margin of​ error: 0.008; confidence​ level: 98%; ModifyingAbove p with caret and ModifyingAbove q with caret unknown

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that the minimum sample size is of 21,153.

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 of:

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

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.

The margin of error is of M = 0.008, while there is no estimate of the proportion, hence $\pi = 0.5$, and solving for n we find the sample size needed.

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

$0.008 = 2.327\sqrt{\frac{0.5(0.5)}{n}}$

$0.008\sqrt{n} = 2.327(0.5)$

$\sqrt{n} = \frac{2.327(0.5)}{0.008}$

$(\sqrt{n})^2 = \left(\frac{2.327(0.5)}{0.008}\right)^2$

$n = 21152.1$

Rounding up, the minimum sample size is of 21,153.

More can be learned about the z-distribution at https://brainly.com/question/14398287