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Genrish500
16 days ago
6

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.
Computers and Technology
1 answer:
Amiraneli [921]16 days ago
6 0

Answer:

Margin of Error=M.E= ± 0.0113

Explanation:

Margin of Error= M.E=?

The probability of viewers watching network news programs = p = 0.4

α= 95%

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

By substituting values: M.E= ±1.96 √0.4(1-0.4)/7200

Now it gives us Margin of Error= ±1.96√0.24/7200

Thus, the Margin of Error equals M.E= ±1.96* 0.005773

Finally, the Margin of Error is ±0.0113, indicating the potential error from random sampling.

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