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 $21.2$ 21.2, and the variance is known to be $34.81$ 34.81. How large of a sample would be required in order to estimate the mean per capita income at the 80%80% level of confidence with an error of at most $0.52$ 0.52? Round your answer up to the next integer.

Answer 1

Answer:

We need a sample size of at least 170.

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:

$\alpha = \frac{1-0.8}{2} = 0.1$

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 $1-0.1 = 0.9$, so $z = 1.28$

Now, find M as such

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

In which $\sigma$ is the standard deviation of the population(square root of the variance) and n is the size of the sample.

How large of a sample would be required in order to estimate the mean per capita income at the 80%80% level of confidence with an error of at most $0.52$ 0.52?

A sample size of at least n, in which n is found when $M = 0.52, \sigma = \sqrt{34.81} = 5.9$

So

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

$0.58 = 1.28*\frac{5.9}{\sqrt{n}}$

$0.58\sqrt{n} = 1.28*5.9$

$\sqrt{n} = \frac{1.28*5.9}{0.58}$

$(\sqrt{n})^{2} = (\frac{1.28*5.9}{0.58})^{2}$

$n = 169.5$

Rounding up

We need a sample size of at least 170.