A medical researcher wishes to estimate what proportion of babies born at a particular hospital are born by caesarean section. In a random sample of 64 births at the hospital, 26% were Caesarean sections. Find the margin of error for the 95% confidence interval.

Question

Quizplus · Accepted Answer

The margin of error for a proportion is calculated using the formula $ME = Z \times \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score for the confidence level (1.96 for 95%), $p$ is the sample proportion (0.26), and $n$ is the sample size (64). Plugging in the values: $ME = 1.96 \times \sqrt{\frac{0.26(1-0.26)}{64}} = 0.1097$.

A Medical Researcher Wishes to Estimate What Proportion of Babies