Question

In a survey of 5100 TV 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 TV viewers who watch network news programs. Hint, find the CI using 1-PropZInt, then add and subtract from the proportion to find the margin of error.

Answer 1

Here is one way you can find the margin of error since the problem is using proportions:

Margin of error = (z-value)(√pq/n)
Note: q = 1 - p

Using your data:
Margin of error = (1.96)[√(.40)(.60)/5100)]

Finish the calculation.

Answer 2

288.24