Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 73%. You would like to be 95% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required

Answer 1

Answer:

A sample size of 474 is required.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

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

Based on previous evidence, you believe the population proportion is approximately 73%.

This means that $\pi = 0.73$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

How large of a sample size is required?

A sample size of n is required, and n is found when M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.73*0.27}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.73*0.27}$

$\sqrt{n} = \frac{1.96\sqrt{0.73*0.27}}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.73*0.27}}{0.04})^2$

$n = 473.24$

Rounding up:

A sample size of 474 is required.