Question

A researcher wishes to estimate with 95% confidence, the proportion of the people who own a home computer. A previous study shows that 40% of the interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary?

Answer 1

Answer:

The minimum sample size necessary is 2305.

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:

$\pi = 0.4$

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 $Z = 1.96$.

The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary?

We need a sample size of n.

n is found when M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.4*0.6}{n}}$

$0.02\sqrt{n} = 1.96*\sqrt{0.4*0.6}$

$\sqrt{n} = \frac{1.96*\sqrt{0.4*0.6}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.4*0.6}}{0.02})^{2}$

$n = 2304.96$

Rounding up

The minimum sample size necessary is 2305.