Question

In a survey of 300 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:

0.05543

Step-by-step explanation:

The formula for calculating the margin of error is expressed as;

$M.E = z * \sqrt{\frac{p*(1-p)}{n} }$ where;

z is the z-score at 95% confidence = 1.96 (This is gotten from z-table)

p is the percentage probability of those that watched network news

p = 40% = 0.4

n is the sample size = 300

Substituting this values into the formula will give;

$M.E = 1.96*\sqrt{\frac{0.4(1-0.4)}{300} }\\ \\M.E = 1.96*\sqrt{\frac{0.4(0.6)}{300} }\\\\\\M.E = 1.96*\sqrt{\frac{0.24}{300} }\\\\\\M.E = 1.96*\sqrt{0.0008}\\\\\\M.E = 1.96*0.02828\\\\M.E \approx 0.05543$

Hence, 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 is approximately 0.05543