Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 32%. You would like to be 90% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required?

Answer 1

Answer:

We need a sample size of at least 369.

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:

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

For this problem, we have that:

$\pi = 0.32$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

How large of a sample size is required?

We need a sample size of at least n.

n is found when $M = 0.04$. So

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

$0.04 = 1.645\sqrt{\frac{0.32*0.68}{n}}$

$0.04\sqrt{n} = 1.645\sqrt{0.32*0.68}$

$\sqrt{n} = \frac{1.645\sqrt{0.32*0.68}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.32*0.68}}{0.04})^{2}$

$n = 368.02$

Rounding up

We need a sample size of at least 369.