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kirill [66]
3 years ago
11

A researcher wants to estimate the impact prenatal care during pregnancy can have on anemia rates in pregnant mothers by conduct

ing 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%?
Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

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.

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