Question

A researcher wants to estimate the impact prenatal care during pregnancy can have on anemia rates in pregnant mothers by conducting a retrospective case-control study on new moms. If the prevalence of anemia is approximately 39.7%, 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%?

Answer 1

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 $\pi$, 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:

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

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

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 $M = 0.03, \pi = 0.397$

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

$0.03 = 1.96\sqrt{\frac{0.397*0.603}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.397*0.603}$

$\sqrt{n} = \frac{1.96\sqrt{0.397*0.603}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.397*0.603}}{0.03})^{2}$

$n = 1021.8$

Rounding up

At least 1022 new moms should be recruited.