Question

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 (x̄) 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

x̄ - t(s ÷ √n) < µ < x̄ + t(s ÷ √n)

z = (x - µ) ÷ σ

x̄ - z(σ ÷ √n) < µ < x̄ + z(σ ÷ √n)

Answer 1

As the distribution is normal and the population standard deviation is known, the most appropriate sampling distribution is the Z-distribution. The formula to be used to determine the 95% confidence interval for the true mean advertising expense is given as

Substitute the values in the above interval.

956 < < 1544

The 95% confidence interval for the true mean advertising expense is 956 < < 1544.