In a survey of 7200 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.

Margin of Error=M. E = ± 0.0113

Explanation:

Margin of Error = M. E=?

Probability that watched network news programs = p = 0.4

α = 95%

Margin of Error = M. E = zₐ/₂√p (1-p) / n

Margin of Error=M. E = ±1.96 √0.4 (1-0.4) / 7200

Margin of Error=M. E = ±1.96√0.24/7200

Margin of Error=M. E - ±1.96 * 0.005773

Margin of Error=M. E = ±0.0113

The Margin of Error is the estimate of how much error is possible as a result of random sampling.