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
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
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](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.19)
(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](https://tex.z-dn.net/?f=M%20%3D%200.03)
Then
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.03 = 1.96\sqrt{\frac{0.19*0.81}{n}}](https://tex.z-dn.net/?f=0.03%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.19%2A0.81%7D%7Bn%7D%7D)
![0.03\sqrt{n} = 1.96\sqrt{0.19*0.81}](https://tex.z-dn.net/?f=0.03%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.19%2A0.81%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.19*0.81}}{0.03}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.19%2A0.81%7D%7D%7B0.03%7D)
![(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.19*0.81}}{0.03})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%5Csqrt%7B0.19%2A0.81%7D%7D%7B0.03%7D%29%5E%7B2%7D)
![n = 656.91](https://tex.z-dn.net/?f=n%20%3D%20656.91)
Rounding up to the nearest whole number.
A sample size of 657 is needed.