Question

Suppose we want to formulate a 95% confidence interval for the population proportion and we would like our margin of error to be not greater than 0.04. We have no history with this characteristic, so we have no idea as to what the proportion might be. Conservatively, what is the minimum sample size we should anticipate using? (NOTE: Round up to the nearest whole number)

Answer 1

Answer:

The minimum sample size we should anticipate using is 601.

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:

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

We have no history with this characteristic, so we have no idea as to what the proportion might be.

This means that we use $\pi = 0.5$, which is when the largest sample size will be needed.

95% confidence level

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

What is the minimum sample size we should anticipate using?

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up, 601

The minimum sample size we should anticipate using is 601.