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 23 randomly selected​ seven-day cruises was 81,648 bottles with a sample standard deviation of 4,596 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 BLANK bottles to an upper limit of BLANK 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 follows the normal probability distribution.

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

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 81648

sample standard deviation = s = 4596

sample size = n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 95% confidence level

= 1 - 95%

=1 - 0.95 = 0.05

/2 = 0.025

t/2,df = t_0.025,22 = 2.074

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

= 2.074 * (4596 / 23)

Margin of error = E = 1987.6

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

- E < < + E

81648 - 1987.6 < < 81648 + 1987.6

79660.4 < < 83635.6

79661 < < 83636

( 79661 , 83636 )

The 95​% confidence interval to estimate the average beer consumption per cruise is from a lower limit of 79661 bottles to an upper limit of  83636 bottles.

b.

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