In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout time, with a standard deviation of 3.1 minutes. A sample of 72 customers who used a cashier averaged 6.1 minutes with a standard deviation of 2.8 minutes.

(a) Can you conclude that the mean checkout time is less for people who use the self-service lane?
(b) Can you conclude that if everyone used the self-service lane, that the mean checkout time would decrease? Consider the number of items checked out when formulating your answer.

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AlonsoDehner AlonsoDehner · Answer 1 · 2019-08-02T16:27:13+02:00

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 = [tex]\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2} } \\=0.5187[/tex]

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