Respuesta :
Answer:
a)
[tex]n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2[/tex], 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 [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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent
This means that [tex]\pi = 0.16[/tex]
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 [tex]M = 0.03[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = z\sqrt{\frac{0.16*0.84}{n}}[/tex]
[tex]0.03\sqrt{n} = z\sqrt{0.16*0.84}[/tex]
[tex]\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2[/tex]
[tex]n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2[/tex], in which z is related to the confidence level.
Question b:
99% confidence level,
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
[tex]n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2[/tex]
[tex]n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2[/tex]
[tex]n = 990.2[/tex]
Rounding up
A sample size of 991 is needed.
