Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to have a population standard deviation of $3,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is (a) [1 pt] $500

Answer 1

Answer:

A sample of 139 should be taken.

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

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.

Population standard deviation of $3,000.

This means that $\sigma = 3000$

How large a sample should be taken if the desired margin of error is $500?

A sample of n is needed.

n is found when M = 500. So

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

$500 = 1.96\frac{3000}{\sqrt{n}}$

$500\sqrt{n} = 1.96*3000$

Dividing both sides by 500

$\sqrt{n} = 1.96*6$

$(\sqrt{n})^2 = (1.96*6)^2$

$n = 138.3$

Rounding up:

A sample of 139 should be taken.