Question

In a survey of 5100 randomly selected T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.

Answer 1

Margin of error = 0.01344

Explanation:

Margin of error = critical value × standard deviation.

critical value for 95% confidence interval = 1.96

Standard deviation = √[(p)(q)/n]

p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100

Standard deviation = √[(0.6×0.4)/5100] = 0.00686

Margin of error = 1.96 × 0.00686 = 0.0134