Suppose that a random sample of 25 retail merchants from all of the 5,000 merchants in a large city yielded a mean advertising expense (x¯) for the past year of $1,250. If the annual advertising expenditures are known to be normally distributed and the standard deviation of the population (σ) is $750, what formula would you use to determine the 95% confidence interval for the true mean advertising expense?
F = s12 ÷ s22
x¯ - t(s ÷ √n) < µ < x¯ + t(s ÷ √n)
z = (x - µ) ÷ σ
x¯ - z(σ ÷ √n) < µ < x¯ + z(σ ÷ √n)