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

A sample size of 285 is needed.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))`

`500 = 1.96(4300)/(√(n))`

`500√(n) = 1.96*4300`

`√(n) = (1.96*4300)/(500)`

`n = 284.1`

Rounding up

A sample size of 285 is needed.