A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.Question 1. Find the margin of error zsqrt(((p^^\^)((1-p^^\^)))/n), abbreviated ME, for this poll if we want 90% confidence in our estimate of the proportion of adults who believe in ghosts.ME=(Round to 3 decimal places.)Question 2. Find the margin of error needed to be 99% confident.ME= (Round to 3 decimal places.)

Answer:Margin of error at 90% is 0.024Margin of error at 99% is 0.037Step-by-step explanation:Sample size = 1076A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.So, No. of adults said they believe ghosts = [tex]\frac{64}{100} \times 1076=688.64\sim 688[/tex]So, x = 688n = 1076[tex]\widehat{p} = \frac{x}{n}[/tex][tex]\widehat{p} = \frac{688}{1076}[/tex][tex]\widehat{p} = 0.639[/tex][tex]ME=z \times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}[/tex]z at 90% confidence is 1.64[tex]ME=1.64 \times \sqrt{\frac{0.639(1-0.639)}{1076}}[/tex][tex]ME=0.024[/tex]So, margin of error at 90% is 0.024 Find the margin of error needed to be 99% confident.z at 99% confidence is 2.58[tex]ME=2.58 \times \sqrt{\frac{0.639(1-0.639)}{1076}}[/tex][tex]ME=0.024[/tex]So, margin of error at 99% is 0.037