Question

a simple random sample of 25 observations is dervied from a normally distributed population with a population standard deviation of 8.2. compute the margin of error with 80% confidence.

Answer 1

The margin of error with 80% confidence interval is 2.099

Given that,

A straightforward random sample of 25 observations is taken from a population that is normally distributed and has a standard deviation of 8.2.

To find : The margin of error with 80% confidence

Margin of Error : The margin of error is a statistic that describes how much random sampling error there is in survey results. One should have less faith that a poll's findings would accurately reflect those of a population census the higher the margin of error.

Here we have

n=25,σ=8.2

For 80% confidence interval, the critical value of z is 1.28. So requried margin of error is

ME=$Z_{c}\cdot \frac{\sigma}{\sqrt{n}}$=1.28$\cdot \frac{8.2}{\sqrt{25}}$=2.0992

Thus, with 80% confidence, the error margin is 2.099.

To learn more about margin of error click here:

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