Answer:
At least 1022 new moms should be recruited.
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]
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].
How many individuals should the researcher recruit to participate in the study if the researcher wants to be 95% confident that the margin of error is no more than 3.0%?
At least n new moms should be recruited.
n is found when [tex]M = 0.03, \pi = 0.397[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.397*0.603}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.397*0.603}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.397*0.603}}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.397*0.603}}{0.03})^{2}[/tex]
[tex]n = 1021.8[/tex]
Rounding up
At least 1022 new moms should be recruited.
