The principal of a school surveyed 200 students on the time they spent on the Internet after school . The mean and the standard deviation of the sample data are 2.4 hours and 0.8 hours , respectively . What is the margin of error for 95 % confidence interval ?

11%

Explanation:

Sample size, n = 200

Mean, m = 2.4

Standard deviation, s = 0.8

Z critical at 95% = 1.96

Margin of Error = Zcritical * sqrt(s^2 / n)

Margin of Error = 1.96 * sqrt(0.0032)

Margin of Error = 1.96 * 0.0565685

Margin of Error = 0.1108743

Hence, margin of error at 95% confidence interval is (0.1108743 * 100%) = 11.08% = 11%

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