Respuesta :
Answer:
A sample size of 79 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]\pi = 0.21[/tex]
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
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 0.09?
A sample size of n is needed.
n is found when M = 0.09. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}[/tex]
[tex]0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}[/tex]
[tex]n = 78.68[/tex]
Rounding up to the nearest whole number.
A sample size of 79 is needed.
