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

A sample of 1699 would be required.

Step-by-step explanation:

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

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which 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

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

Rounding up:

A sample of 1699 would be required.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

maybe showing the graph would

step-by-step explanation:

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