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

Answer:

We need at least 601 incomes.

Step-by-step explanation:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

$500 = 1.96*\frac{6250}{\sqrt{n}}$

$500\sqrt{n} = 1.96*6250$

$\sqrt{n} = \frac{1.96*6250}{500}$

$(\sqrt{n})^{2} = (\frac{1.96*6250}{500})^{2}$

$n = 600.25$

Rounding up

We need at least 601 incomes.