Question

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.

Answer 1

Answer:

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.