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.

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

he spend 5.1 minutes chatting with his friends.

step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5.1

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42