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 18 randomly selected​ seven-day was 81,903 bottles with a sample standard deviation of 4,585 bottles. Complete parts a and b below.

A. Construct a 95% confidence interval to estimate the average beer consumption per cruise.

The 95% 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​ Student's t-distribution.

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

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

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 81903

sample standard deviation = s = 4585

sample size = n = 18

Degrees of freedom = df = n - 1 = 18 - 1 = 17

A) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,17 = 2.110

Margin of error = E = t_/2,df * (s /n)

= 2.110 * (4585 / 18)

Margin of error = E = 2280

The 95% confidence interval estimate of the population mean is,

  ± E

= 81903 ± 2280

= ( 79623, 84183 )

The 95% confidence interval to estimate the average beer consumption per cruise is from a lower limit of 79623 bottles to an upper limit of 84183 bottles.

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