Question

The National Center for Education Statistics surveyed 4400 college graduates about the lengths of time required to earn their bachelor's degrees. The mean was 5.15 years and the standard deviation was 1.68 years. Based on the above information, find the margin of error for a 98% confidence level. Round your answer to the nearest hundredth.

Answer 1

Answer:

The margin of error for a 98% confidence level is of 0.06 years.

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:

$\alpha = \frac{1-0.98}{2} = 0.01$

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 $1-0.01 = 0.99$, so $z = 2.327$

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:

$\sigma = 1.68, n = 4400$

So

$M = 2.327*\frac{1.68}{\sqrt{4400}}$

$M = 0.06$

The margin of error for a 98% confidence level is of 0.06 years.