Given Information:
Population size = n = 300
standard deviation = σ = 15
confidence level = 99%
Required Information:
Margin of error = ?
Explanation:
The margin of error is a measure of the effectiveness of your survey.
Margin of error = z*(σ/√n)
z-score corresponding to 99% confidence level = 2.58
Margin of error = ±2.58*(15/√300)
Margin of error = ±2.58*(0.866)
Margin of error = ±2.23
Hence the correct option is C ±2.23