An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $22.1, 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 98% level of confidence with an error of at most $0.36

A sample size of at least 2809 would be required in order to estimate the mean per capita income at the 98% level of confidence with an error of at most $0.36

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 and n is the size of the sample.

Sample size with margin of error of at most $0.36

Sample size of at least n, in which n is found when . So

A sample size of at least 2809 would be required in order to estimate the mean per capita income at the 98% level of confidence with an error of at most $0.36

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hope this ^_^

step-by-step explanation: