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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 ?

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Answer:

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%

The mean and the standard deviation of the sample data are 2.4 hours and 0.8 hours . Therefore, the margin of error for 95 % confidence interval is equal to 11%.

What is the term Margin of Error about?

A margin of error helps to provide the information about the difference between the real population value and expected results.

Given Information:

Sample size, n = 200

Mean, m = 2.4

Standard deviation, s = 0.8

Z critical at 95% = 1.96

Solution:

Margin of Error = Z critical * 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%

Learn more about margin of error, refer to the link:

https://brainly.com/question/13990500

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