A Gallup poll reported that 58% of Americans say they are satisfied with the costs of healthcare. If the margin of error was 5%, approximately what was the sample size?
A) 10
B) 200
C) 400
D) 1600

Question

Quizplus · Accepted Answer

The sample size can be calculated using the formula for the margin of error in a proportion: $ME = Z \sqrt{\frac{p(1-p)}{n}}$, where ME is the margin of error, Z is the Z-score for the desired confidence level (typically 1.96 for 95% confidence), p is the proportion (0.58 in this case), and n is the sample size. Rearranging to solve for n gives $n = \frac{Z^2 p(1-p)}{ME^2}$. Plugging in the values (with ME = 0.05 and assuming a 95% confidence level so Z = 1.96), we get a sample size close to 400.

A Gallup Poll Reported That 58% of Americans Say They