Question

A researcher is interested in estimating the proportion of low birth weight (LBW) infants delivered at less than 6.0 lbs in a population of low income mothers. Suppose the researcher is estimating based on a national prevalence for LBW published by the US Census bureau indicating that 16% of infants are born categorized as LBW. Calculate the margin of error for a 95% confident sample size and estimating the need to enroll 36 patients.

Answer 1

Solution :

Given that,

n = 36

Point estimate = = 16% = 0.16

1 - = 1 - 0.16 = 0.84

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

E = 1.96 (((0.16 * 0.84) / 36)

E = 0.120