Question

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 21 randomly selected​ seven-day cruises was 81785 bottles with a sample standard deviation of 4,544 bottles.

Complete parts a and b below.

a. Construct a 80​% confidence interval to estimate the average beer consumption per cruise.

The 80% confidence interval to estimate the average beer consumption per cruise is from a lower limit of __bottles to an upper limit of ___bottles. ​(Round to the nearest whole​ numbers.)

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

A.

The only assumption needed is that the population follows the normal probability distribution.

B.

The only assumption needed is that the population distribution is skewed to one side.

C.

The only assumption needed is that the population follows the​ Student's t-distribution.

D.

The only assumption needed is that the population size is larger than 30.

Answer 1

sample mean, xbar = 81785
sample standard deviation, s = 4544
sample size, n = 21
degrees of freedom, df = n - 1 = 20

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, tc = t(α/2, df) = 1.325

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (81785 - 1.325 * 4544/sqrt(21) , 81785 + 1.325 * 4544/sqrt(21))
CI = (80471 , 83099)

The 80% confidence interval to estimate the average beer consumption per cruise is from a lower limit of 80471 bottles to an upper limit of 83099 bottles.

A. The only assumption needed is that the population follows the normal probability distribution.