Question

An economist wants to estimate the mean income for the first year of work for college graduates who have had the profound wisdom to take a statistics course. How many such incomes must be found if we want to be 95% confident that the sample mean is within $500 of the true population mean? Assume that a previous study has revealed that for such incomes, σ = $6250

Answer 1

We need at least 601 incomes.

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.

How many such incomes must be found if we want to be 95% confident that the sample mean is within $500 of the true population mean?

We have to find n, for which
`M = 500, \sigma = 6250`. So

`M = z*(\sigma)/(√(n))`

`500 = 1.96*(6250)/(√(n))`

`500√(n) = 1.96*6250`

`√(n) = (1.96*6250)/(500)`

`(√(n))^(2) = ((1.96*6250)/(500))^(2)`

`n = 600.25`

Rounding up

We need at least 601 incomes.