Answer:
The sample size must be approximately 1067 to get a margin of error of 0.03.
Step-by-step explanation:
We are given the following in the question:
Margin of error = 0.03
Significance level = 5%
Margin of error =
[tex]z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
Since, no prior proportions are given, we take
[tex]\hat{p} = 0.5[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]0.03 = 1.96\sqrt{\dfrac{0.5(1-0.5)}{n}}\\\\\sqrt{n} = \dfrac{1.96\times 0.5}{0.03}\\\\\\sqrt{n} = 32.67\\n = 1067.3289\approx 1067[/tex]
Thus, the sample size must be approximately 1067 to get a margin of error of 0.03.
