Asked by Rachel

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.

Answers

Answered by MathGuru
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.
Answered by pmc
288.24
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions