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

We need a sample size of at least 170.

Explanation:

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

`\alpha = (1-0.8)/(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*(\sigma)/(√(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 = √(34.81) = 5.9`

So

`M = z*(\sigma)/(√(n))`

`0.58 = 1.28*(5.9)/(√(n))`

`0.58√(n) = 1.28*5.9`

`√(n) = (1.28*5.9)/(0.58)`

`(√(n))^(2) = ((1.28*5.9)/(0.58))^(2)`

`n = 169.5`

Rounding up

We need a sample size of at least 170.