Question

Heine Brothers’, a coffee shop chain from Louisville, Kentucky, wants to determine if the amount of time customers spend in the “Germantown” coffee shop is different from the “Crescent Hill” coffee shop. An employee at each location times the amount of time 20 customers spends in each coffee shop. For the “Germantown” location the sample mean is 56.70 minutes with a sample standard deviation of 13.48 minutes. For the “Crescent Hill” location, the sample mean is 77.25 minutes with a standard deviation of 37.06 minutes.

• (a) Determine a 90% confidence interval for the difference in the mean time customers spend in each location.
• (b) Determine if there is evidence that the average time customers spend in each location is different (use ). What is the p-value or level of significance of this test?

Answer 1

Part a)

Confidence interval :-

t(α/2, DF) = t(0.1 /2, 23 ) = 1.714

DF = 23


Lower Limit =
Lower Limit = -35.6641
Upper Limit =
Upper Limit = -5.4359
90% Confidence interval is ( -35.6641 , -5.4359 )

Part b)

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 ≠ µ2

Test Statistic :-


t = -2.3305

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, DF)

DF = 23

Critical value t(α/2, DF) = t(0.1 /2, 23 ) = 1.714
| t | > t(α/2, DF) = 2.3305 > 1.714
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 2.3305 ) = 0.0289
Reject null hypothesis if P value < α = 0.1 level of significance
P - value = 0.0289 < 0.1 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence that the average time customers spend in each location is different.