[tex]\pm6.479[/tex]
1) Considering that we can find out the Margin of Error (MOE) by using this formula
[tex]M=\frac{z\sqrt[]{p\cdot(1-p)}}{\sqrt[]{n}}[/tex]
2) Then we can plug into that the given data. Note that z =2.807 for a confidence level of 99.5%, as well as "p" stands for proportion, n the sample size. So z=2.807, p=0.86, n=226
[tex]\begin{gathered} MOE=\frac{z\sqrt[]{p\cdot(1-p)}}{\sqrt[]{n}} \\ MOE=\frac{2.807\sqrt[]{0.86\cdot(1-0.86)}}{\sqrt[]{226}} \\ \text{MOE}=\frac{0.974}{15.033}\times100=6.479\% \end{gathered}[/tex]
3) Hence, the answer is:
[tex]\pm6.479\%[/tex]
