A random sample of 40 students were asked how much money they spent on entertainment in the last week. They spent an average of $28, with a standard deviation of $18. What is the 90% confidence interval for mean amount spent on entertainment?

In between $23.3 and $32.7

Explanation:

standard deviation = $18

sample mean = $28

sample size = 40 students

Significance level = 1 - confidence interval = 1 - 0.9 = 0.1

Using the confidence interval calculator

90% Confidence Interval: $28 ± $4.68

($23.3 to $32.7)

With 90% confidence the population mean is between 23.3 and 32.7 based on 40 samples."