Question

Annual starting salaries for college graduates is unknown. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. If the population standard deviation is $3,750 how large should the sample be if margin of error is $500

Answer 1

Answer:

A sample of 217 is needed.

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 $1-\alpha$.

So it 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.

If the population standard deviation is $3,750 how large should the sample be if margin of error is $500

We have that $\sigma = 3750$.

We need a sample of n, and n is found when $M = 500$. So

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

$500 = 1.96*\frac{3750}{\sqrt{n}}$

$500\sqrt{n} = 1.96*3750$

$\sqrt{n} = \frac{1.96*3750}{500}$

$(\sqrt{n})^{2} = (\frac{1.96*3750}{500})^{2}$

$n = 216.1$

Rounding up

A sample of 217 is needed.