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.
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.
2 answers
288.24
0 answers