In an election poll, 56 % of 400 people voters choose a certain candidate. Find margin of sampling error?

Generally, we refer to a margin of error with a certain confidence. For a poll of n people that results in choice one p percent of the time and choice two (100-p) percent of the time one standard deviation (or standard error) is given by sqrt(p (100-p) / n). In this case, that would be

std = sqrt(56 * 44 / 400) = 2.48%

The canonical choice is to choose the 95% confidence interval. For a 95% confidence interval, the margin of error is

1.96 * std = 4.9% (approx)

For a 99% confidence interval, the margin of error is 2.58 * std.

what "following statement"?

step-by-step explanation: