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

A sample of 217 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`.

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*(\sigma)/(√(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*(\sigma)/(√(n))`

`500 = 1.96*(3750)/(√(n))`

`500√(n) = 1.96*3750`

`√(n) = (1.96*3750)/(500)`

`(√(n))^(2) = ((1.96*3750)/(500))^(2)`

`n = 216.1`

Rounding up

A sample of 217 is needed.