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.

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.

the answer is -1 for this question

step-by-step explanation: