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Mathematics 11 Online
OpenStudy (anonymous):

Assume that a sample is used to estimate a population mean mu. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation. 95% confidence; n = 21; x-bar = 0.16; s = 0.16

OpenStudy (anonymous):

\[E=(1.96(0.16))/\sqrt{21}\] = 0.068

OpenStudy (toxicsugar22):

hey meth.party can u help me after her

OpenStudy (anonymous):

I may have done this wrong, Are you looking for the margin of error of the mean or that of the proportion?

OpenStudy (anonymous):

margin of error

OpenStudy (anonymous):

The error of what exactly?

OpenStudy (anonymous):

of population mean mu

OpenStudy (anonymous):

If thats what is sought then what I did above should be correct. What course is this?

OpenStudy (anonymous):

statistics thank you for all of your help so far

OpenStudy (anonymous):

No problem!

OpenStudy (anonymous):

how about this one... Find the minimum sample size. You want to be 95% confident that the sample variance is within 20% of the population variance.

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