Question

A sample size of n = 64 is drawn from a population whose standard deviation is o = 5.6.

Answer 1

Answer:

The margin of error for a 99% confidence interval for the population mean is 1.8025.

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 problem:

$\sigma = 5.6$

So

$M = 2.575*\frac{5.6}{\sqrt{64}} = 1.8025$

The margin of error for a 99% confidence interval for the population mean is 1.8025.