Answer:
[tex]p \± ME[/tex] [tex]\to[/tex] [tex]71.2\% \± 7.9\%[/tex]
Step-by-step explanation:
Given
[tex](63.3\%,79.1\%)[/tex]
Required
Express in form of: [tex]p \± ME[/tex]
First, calculate the midpiont (p)
[tex]p = \frac{63.3\% + 79.1\%}{2}[/tex]
[tex]p = \frac{142.4\%}{2}[/tex]
[tex]p = 71.2\%[/tex]
The margin of error (ME) is:
[tex]ME =79.1\% - 71.2\%[/tex]
[tex]ME =7.9\%[/tex]
So, we have:
[tex]p \± ME[/tex] [tex]\to[/tex] [tex]71.2\% \± 7.9\%[/tex]
