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

At least 1022 new moms should be recruited.

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (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{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.397*0.603)/(n)}`

`0.03√(n) = 1.96√(0.397*0.603)`

`√(n) = (1.96√(0.397*0.603))/(0.03)`

`(√(n))^(2) = ((1.96√(0.397*0.603))/(0.03))^(2)`

`n = 1021.8`

Rounding up

At least 1022 new moms should be recruited.