Question 5 4 pts
DINING OUT
It is reported that the average American eats out at a restaurant 5.9 times per week. A random sample of 40
Americans showed that the average amount of times they ate out at a restaurant per week was 5.1 times
with a standard deviation of 1.7. Is there enough evidence at the 5% level that the average American eats
out less than 5.9 times per week?
Find the 95% Confidence Interval and explain:
5.9 - (2.02)(0.2688) - 5.36 5.9 +(2.02)(0.2688) = 6.44 We are 95% confident the average amount of times American
eat out at a restaurant per week is between 5.36 and 6,44
5.1 (2.02/0.2688) - 4.56 5.1 + (2.02)(0.2688) - 5.64 We are 95% confident the avenge amount of times Americans
eat out at a restaurant per week is between 4.56 and 5.64
5.9 - (0.95)(0.2688) - 5.65 5.9 + (0.95)(0.2688) - 6.16 We are 95% confident the average amount of times Americans
eat out at a restaurant per week is between 5.65 and 6.16.
o 5.1 - (0.95X0.2688) - 4.85% 5.1 + (0.95)(0.2688) - 5.36% We are 95% confident the average amount of times
Americans eat out at a restaurant per week is between 4.853 and 5.36%