Question

A Gallup poll reported that 58% of Americans say they are satisfied with the costs of healthcare. If the margin of error was 1.5%, approximately what was the sample size?

Answer 1

we can use the following formula for the margin of error:

`ME=1.96\sqrt[]{(p(1-p))/(n)}`

where p is the proportion and n the sample size

p = 58% = 0.58

ME = 1.5% = 0.015

therefore:

`0.015=1.96\sqrt[]{(0.58(1-0.58))/(n)}`

now, we need to solve for n

`\begin{gathered} (0.015)/(1.96)=\sqrt[]{(0.58(1-0.58))/(n)} \\ (0.58(1-0.58))/(n)=((0.015)/(1.96))^2 \\ 0.58(1-0.58)=n\cdot((0.015)/(1.96))^2 \\ 0.58\cdot\: 0.42=n\cdot0.00765^2 \\ n=(0.58\cdot0.42)/(0.00765^2)=4159.17226 \end{gathered}`