Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to have a population standard deviation of $3,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is (a) [1 pt] $500

Answer 1

A sample of 139 should be taken.

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.

Population standard deviation of $3,000.

This means that
`\sigma = 3000`

How large a sample should be taken if the desired margin of error is $500?

A sample of n is needed.

n is found when M = 500. So

`M = z(\sigma)/(√(n))`

`500 = 1.96(3000)/(√(n))`

`500√(n) = 1.96*3000`

Dividing both sides by 500

`√(n) = 1.96*6`

`(√(n))^2 = (1.96*6)^2`

`n = 138.3`

Rounding up:

A sample of 139 should be taken.