Question

alculate the margin of error for a sampling distribution that has a size of 36 for a 95% confidence level. The population standard deviation is known to be 12.

Answer 1

For a sampling distribution that has a sample size of 36 for a 95% confidence level and the population standard deviation is known to be 12, then the margin of error will be 3.98.

Given that the sample size is 36, confidence level is 95% and the population standard deviation be 12.

We are required to find the margin of error of the sampling distribution.

Margin of error is the difference between real values and the calculated values of the sampling distribution.

Margin of error=z*σ/$\sqrt{n}$

,in which σ is population standard deviation and n is the sample size.

z value of 95% confidence level is 1.96.

Margin of error=1.96*12/$\sqrt{36}$

=1.96*12/6

=1.96*2

=3.98

Hence for a sampling distribution that has a sample size of 36 for a 95% confidence level and the population standard deviation is known to be 12, then the margin of error will be 3.98.

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