Answer:
Step-by-step explanation:
I sample :
Sample size n1 = 60
Sample average = x bar = 5.2 minutes
STd deviation sample = s1 = 3.1 minutes
II sample:
n2 = 72
y bar = 6.1
s2 = 2.8 minutes
Mean difference = -0.9
Std error for difference =
t statistic = mean diff/se = -1.735
p value = 0.0848
Since p >0.05 we accept null hypothesis that there is no difference between the two averages
Conclusion:
a) We cannot conclude that the mean checkout time is less for people who use the self-service lane
b) NO, we cannot conclude that if everyone used the self-service lane, that the mean checkout time would decrease
