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 19 randomly selected​ seven-day cruises was 81,838 bottles with a sample standard deviation of 4,568 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.

Answer 1

Solution :

Given that,

= 81838

s =4568

n = Degrees of freedom = df = n - 1 = 19- 1 = 18

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,18 =  2.101

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

= 2.101* ( 4568/ 19)

=2201.7872

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

- E < < + E

81838 - 2201.7872< < 81838+ 2201.7872

79636.2128 < < 84039.7872

(lower limit =79636.2128 , upper limit= 84039.7872 )