Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $30.8, and the standard deviation is known to be $8.2. How large of a sample would be required in order to estimate the mean per capita income at the 95% level of confidence with an error of at most $0.39

Answer 1

Answer:

A sample of 1699 would be required.

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 Z-table as such z has a p-value 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

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation is known to be $8.2.

This means that $\sigma = 8.2$

How large of a sample would be required in order to estimate the mean per capita income at the 95% level of confidence with an error of at most $0.39?

This is n for which M = 0.39. So

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

$0.39 = 1.96\frac{8.2}{\sqrt{n}}$

$0.39\sqrt{n} = 1.96*8.2$

$\sqrt{n} = \frac{1.96*8.2}{0.39}$

$(\sqrt{n})^2 = (\frac{1.96*8.2}{0.39})^2$

$n = 1698.3$

Rounding up:

A sample of 1699 would be required.