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.

We need a sample size of at least 170.

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 Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which 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

So

Rounding up

We need a sample size of at least 170.

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

gdp is the best phrase!