What is the margin of error for a sample statistic for a population with a standard deviation of 7.75? enter your answer in the box.


Margin error of a sample mean is given by 
ME=Z_{critical} \cdot  \frac{\sigma}{ \sqrt{n}}
where σ is the standard deviation, Z_{critical} is the critical value and n is the sample mean.

So beside the given standard deviation, we're supposed to know the sample size and critical value to answer this question. 

Step-by-step explanation:

Given that there is a sample statistic for a population.  Population std deviation sigma is given as 5.75.

Margin of error = z critical value * std error

For standard error we must know the sample size n.

Std error = \frac{\sigma}{\sqrt{n} } =\frac{5.75}{\sqrt{n} }

Margin of error = Z critical * std error =1.96*\frac{5.75}{\sqrt{n} } for 95%

=2.58*\frac{5.75}{\sqrt{n} } for 99%

Z critical = 1.96 for 95% and 2.58 for 99%

Hence margin of error =

Margin of error =z-score value for chosen confidence level×population standard deviation

Assuming a 95% confidence level , then


standard deviation=5.75

Margin of error= 1.96×5.75=11.27

Do you know the answer?

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