In: Statistics and Probability
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%?
[1] n total = 2044; n control group = 1,022; n case group = 1,022
[2] n total = 2044; n control group = 812; n case group = 1,232
[3] n total = 4088; n control group = 1,623; n case group = 2,465
[4] n total = 4088; n control group = 2,044; n case group = 2,044
sample proportion , p̂ =
0.397
sampling error , E = 0.03
Confidence Level , CL= 0.95
alpha = 1-CL = 0.05
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (
1.960 / 0.03 ) ² *
0.40 * ( 1 - 0.40 ) =
1021.8
so,Sample Size required=
1022
[1] n total = 2044; n control group = 1,022; n case group = 1,022