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.

Step-by-step explanation:

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

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;

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

answer is 67%

step-by-step explanation:

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step-by-step explanation: