In a survey of 1300 T.V.viewers,30% 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.
A)2.86%
B)1.87%
C)2.49%
D)3.27%

Question

Quizplus · Accepted Answer

To find the margin of error, we need to use the formula:

margin of error = z*sqrt(p*(1-p)/n)

where:
- z is the z-score for the confidence level (95% confidence level has a z-score of 1.96)
- p is the estimated proportion (30% or 0.3)
- n is the sample size (1300)

So, plugging in the values, we get:

margin of error = 1.96*sqrt(0.3*(1-0.3)/1300) ≈ 0.025 or 2.49%

Therefore, the best choice is C) 2.49%.

In a Survey of 1300 T