Answer
$[58.5,63.5]$
Work Step by Step
To find the margin of error for the survey, we apply $M=\pm \frac{1}{\sqrt n}=\pm\frac{1}{\sqrt 1600}\approx0.025=2.5\%$
The interval that is likely to contain the exact percent of all U.S. adults who have purchased a product online is: $[58.5,63.5]$
