Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 60%. You would like to be 98% confident that your estimate is within 2.5% of the true population proportion. How large of a sample size is required?

Answer 1

Answer:

A sample size of 2080 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

Based on previous evidence, you believe the population proportion is approximately 60%.

This means that $\pi = 0.6$

How large of a sample size is required?

We need a sample of n.

n is found when $M = 0.025$. So

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

$0.025 = 2.327\sqrt{\frac{0.6*0.4}{n}}$

$0.025\sqrt{n} = 2.327\sqrt{0.6*0.4}$

$\sqrt{n} = \frac{2.327\sqrt{0.6*0.4}}{0.025}$

$(\sqrt{n})^{2} = (\frac{2.327\sqrt{0.6*0.4}}{0.025})^{2}$

$n = 2079.3$

Rounding up

A sample size of 2080 is needed.