Question

In the past, 19% of all homes with a stay-at-home parent had the father as the stay-at-home parent. An independent research firm has been charged with conducting a sample survey to obtain more current information. (a) What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03? Use a 95% confidence level. (Round your answer up to the nearest whole number.)

Answer 1

Answer:

A sample size of 657 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:

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

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 $Z = 1.96$.

In the past, 19% of all homes with a stay-at-home parent had the father as the stay-at-home parent.

This means that $\pi = 0.19$

(a) What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03?

A sample size of n is needed.

n is found when $M = 0.03$

Then

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

$0.03 = 1.96\sqrt{\frac{0.19*0.81}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.19*0.81}$

$\sqrt{n} = \frac{1.96\sqrt{0.19*0.81}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.19*0.81}}{0.03})^{2}$

$n = 656.91$

Rounding up to the nearest whole number.

A sample size of 657 is needed.