Question

In a recent survey by the National Association of Colleges and Employers, the average starting salary for college graduate with a computer and information sciences degree was reported to be $62,194.40 You are planning to do a survey of starting salaries for recent computer science majors from your university. Using an estimated standard deviation of $11,605, what sample size do you need to have a margin of error equal to $5000 with 95% confidence

Answer 1

Answer:

A sample size of 21 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 $1 - \alpha$.

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

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

Using an estimated standard deviation of $11,605

This means that $\sigma = 11605$

What sample size do you need to have a margin of error equal to $5000 with 95% confidence

A sample size of n is needed. n is found when M = 5000. So

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

$5000 = 1.96\frac{11605}{\sqrt{n}}$

$5000\sqrt{n} = 1.96*11605$

$\sqrt{n} = \frac{1.96*11605}{5000}$

$(\sqrt{n})^2 = (\frac{1.96*11605}{5000})^2$

$n = 20.69$

Rounding up,

A sample size of 21 is needed.