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 $28, and the standard deviation is known to be $5.5. How large of a sample would be required in order to estimate the mean per capita income at the 90% level of confidence with an error of at most $0.42? Round your answer up to the next integer.

In order to estimate the mean per capita income at the 90% level of confidence with an error of at most $0.42. The sample size would be 85How to find the size of sample when the error of at most $0.42Information form the questionthe mean income is $28the standard deviation is known to be $5.5the mean per capita income at the 90% level of confidencean error of at most $0.42What is the formula for the sample size?This is given as from margin of error:This is the term refers to the minimum error that is to be allowed Margin of error, E = Zc * √(SD²/n)n = sample sizen = SD x (Zc/E)²SD = standard deviation Zc = would be the critical value at 90 percent level of confidence = 1.65plugging in the valuesn = 5.5 x (1.65/0.42)²n = 0.25 x 15.4337n = 84.8854n = 85Read more on margin of error here:https://brainly.com/question/13897400#SPJ1