In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 90% confidence in our estimate of the percent of T.V. viewers who watch network news programs
To find the margin of error for this survey, we can use the formula:
Margin of Error = Z * (sqrt(p * (1 - p) / n))
Where:
- Z is the Z-score corresponding to the desired confidence level. For a 90% confidence level, the Z-score is approximately 1.645.
- p is the estimated proportion of the population that possesses the characteristic of interest. In this case, it is the proportion of TV viewers who watch network news programs, which is 40% or 0.40.
- n is the sample size, which is 5100.
Let's calculate the margin of error step by step:
Step 1: Calculate p * (1 - p):
p * (1 - p) = 0.40 * (1 - 0.40) = 0.40 * 0.60 = 0.24
Step 2: Calculate sqrt(p * (1 - p)):
sqrt(0.24) ≈ 0.4899
Step 3: Calculate Z * (sqrt(p * (1 - p) / n)):
1.645 * (0.4899 / sqrt(5100)) ≈ 0.0181
Therefore, the margin of error for this survey, with a 90% confidence level, is approximately 0.0181 or 1.81%.