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

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

answer: c

step-by-step explanation