Question

part b: What assumptions need to be made about this population ?

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 oceanl) The average beer consumption over 19 randomly selected seven-day cruises was 81,911 bottles with a sample standard deviation of 4,513 bottles. Complete parts a and b below a. Construct a 95% confidence interval to estimate the average beer consumption per cruise. bottles to an upper limit of bottles. The 95% confidence interval to estimate the average boer consumption per cruise is from a lower limit of (Round to the nearest whole numbers.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 81911

sample standard deviation = s = 4513

sample size = n = 19

Degrees of freedom = df = n - 1 = 18

a)

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,19 = 2.093

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.093 * ( 4513/ $\sqrt$ 19)

= 2166.99

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

81911 - 2166.99 < $\mu$ < 81911 + 2166.99

79744 < $\mu$ < 84078

( 79744 , 84078)

b)

The 95% confidence interval to estimate the average beer consumption per cruise is from a lower limit of 79744 bottles to an limit of 84078 bottles.