2 Last year the owner of River Rouge Winery implemented a cu

2.         Last year the owner of River Rouge Winery implemented a customer loyalty program. She wants to estimate the mean number of bottles of wine purchased per loyal customer under this program. A sample of 12 customers showed an average of 24 bottles per month with a sample standard deviation of 6 bottles per month.

Given in the problem:

n = 12

= 24

s = 6

Confidence Interval = 95%

Degrees of freedom = n -1, so ___-1 = _____ critical value of __ = ____________

What is the relevant formula when is unknown?

What is the value of the population mean?

                b. What is the best estimate of the population mean?

                c. Do we use the z or the t distribution? ______________ Why? Explain your reasoning: ______________________________________________________________________________________________________________________________________________________________________________________

d. Develop the 95% Confidence Interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 24 bottles of wine per loyal customer?

Solution

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=24
Standard deviation( sd )=6
Sample Size(n)=12
Confidence Interval = [ 24 ± t a/2 ( 6/ Sqrt ( 12) ) ]
= [ 24 - 2.201 * (1.73205) , 24 + 2.201 * (1.73205) ]
= [ 20.18776,27.81224 ]

a)
Degrees of freedom = n -1, so 12-1 = 11 critical value of =2.201
b)Mean(x)=24
c) t distribution
d) [ 20.18776,27.81224 ]
e) Yes, as it lies in the interval