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What's the best way to solve this ? 24 former students were randomly surveyed. The average and standard deviation of monthly salaries of these students

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24 former students were randomly surveyed. The average and standard deviation of monthly salaries of these students were $3,399 and $569, respectively. What is the margin of error of the 99% confidence interval for the true mean monthly salary of all former students? Assume the population is normally distributed. Round your answer to two decimal places (the hundredths place). Question 20 4 pts

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