A population has a known standard deviation of 1.27 and a sample space contains 85 values find the margin of error needed to create a 99% confidence interval estimate of the mean of the population

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

In this question:

So

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent