Question

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independent research firm has been charged with conducting a sample survey to obtain more current information.
Required:
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 (round up to the next whole number).
b. Repeat part (a) using a 99% confidence level (round up to the next whole number).

Answer 1

Answer:

a)

$n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$, in which z is related to the confidence level.

b) A sample size of 991 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:

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

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent

This means that $\pi = 0.16$

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 (round up to the next whole number).

This is n for which $M = 0.03$. So

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

$0.03 = z\sqrt{\frac{0.16*0.84}{n}}$

$0.03\sqrt{n} = z\sqrt{0.16*0.84}$

$\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}$

$(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$

$n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$, in which z is related to the confidence level.

Question b:

99% confidence level,

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

$n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$

$n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2$

$n = 990.2$

Rounding up

A sample size of 991 is needed.