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

A sample of 1699 would be required.

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 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(\sigma)/(√(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(\sigma)/(√(n))`

`0.39 = 1.96(8.2)/(√(n))`

`0.39√(n) = 1.96*8.2`

`√(n) = (1.96*8.2)/(0.39)`

`(√(n))^2 = ((1.96*8.2)/(0.39))^2`

`n = 1698.3`

Rounding up:

A sample of 1699 would be required.