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 81,788 bottles with a sample standard deviation of 4,524 bottles. Construct a 90% confidence interval to estimate the average beer consumption per cruise (upper and lower limit)

Answer 1

Solution :

Given that,

= 81788

s = 4524

n = 19

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

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,18 = 1.734

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

= 1.734 * (4524 / 19)

= 1799.68

Margin of error = 1799.68

The 90% confidence interval estimate of the population mean is,

- E < < + E

81788 - 1799.68 < < 81788 +1799.68

79988.32 < < 83587.68

Lower limit = 79988.32

Upper limit =  83587.68