Question

5. A cruise company would like to estimate the average beer consumption to plan its beer inventory levels on future​ seven-day cruises.​ (The ship certainly​ doesn't want to run out of beer in the middle of the​ ocean!) The average beer consumption over 12 randomly selected​ seven-day cruises was 81,845 bottles with a sample standard deviation of 4,528. complete parts a and b below.

a. Construct a 90​% confidence interval to estimate the average beer consumption per cruise.

lower limit of ___ bottles to an upper limit of ___ bottles

b. . What assumptions need to be made about this​ population?

6. A national air traffic control system handled an average of 47,574 flights during 28 randomly selected days in a recent year. The standard deviation for this sample is 6,421 flights per day. Complete parts a through c below.

a. Construct a 99​% confidence interval to estimate the average number of flights per day handled by the system.

The 99​% confidence interval to estimate the average number of flights per day handled by the system is from a lower limit of ___ to an upper limit of ___

b. Suppose an airline company claimed that the national air traffic control system handles an average of​ 50,000 flights per day. Do the results from this sample validate the airline​ company's claim?

c. What assumptions need to be made about this​ population?

Answer 1

5)a) Degrees of freedom = 12 - 1 = 11

At 90% confidence level, the critical value is t^* = 1.796

The 90% confidence interval is

+/- t^* * s/

= 81845 +/- 1.796 * 4528/

= 81845 +/- 2347.589

= 79497.411, 84192.589

Lower limit = 79497.411

Upper limit = 84192.589

b) Assume that the population is normally distributed.

6) a) Degrees of freedom = 28 - 1 = 27

At 99% confidence level, the critical value is t^* = 2.771

The 99% confidence interval is

+/- t^* * s/

= 47574 +/- 2.771 * 6421/

= 47574 +/- 3362.48

= 44211.52, 50936.48

Lower limit = 44211.52

Upper limit = 50936.48

b) Yes, the results from this sample validates the airline company's claim . Because the confidence interval contains 50,000.

c) Assume that the population is normally distributed.