Question

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

Answer 1

Answer:

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

In this question:

$\sigma = 1.27, n = 85$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$M = 2.575*\frac{1.27}{\sqrt{85}}$

$M = 0.3547$

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