Question

You are planning a survey of starting salaries for recent computer science majors. In a recent survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,467. Grade: 5 4 3 2 1 Communication strong ←− − − − − − −→ weak ×1 If the standard deviation is known to be $12,000, what sample size do you need to have a margin of error equal to $1000, with 95% confidence

Answer 1

A sample size of 554 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 known to be $12,000

This means that
`\sigma = 12000`

What sample size do you need to have a margin of error equal to $1000, with 95% confidence?

This is n for which M = 1000. So

`M = z(\sigma)/(√(n))`

`1000 = 1.96(12000)/(√(n))`

`1000√(n) = 1.96*12`

Dividing both sides by 1000:

`√(n) = 1.96*12`

`(√(n))^2 = (1.96*12)^2`

`n = 553.2`

Rounding up:

A sample size of 554 is needed.