Question

You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $60,038. 5 If you assume that the standard deviation is $4300, what sample size do you need to have a margin of error equal to $500 with 95% confidence

Answer 1

Answer:

A sample size of 285 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

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation is $4300

This means that $\sigma = 4300$

What sample size do you need to have a margin of error equal to $500 with 95% confidence?

A sample size of n is needed. n is found when M = 500. So

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

$500 = 1.96\frac{4300}{\sqrt{n}}$

$500\sqrt{n} = 1.96*4300$

$\sqrt{n} = \frac{1.96*4300}{500}$

$n = 284.1$

Rounding up

A sample size of 285 is needed.