Question

The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12 minutes. and a standard deviation of 3 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 28 customers visiting her store this week resulted in a mean of 10.5 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12 minutes? Use α = .02.

(a) Find z. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion. Reject the null hypothesis, there is not significant evidence that the mean checkout time is less than 12 minutes. Reject the null hypothesis, there is significant evidence that the mean checkout time is less than 12 minutes. Fail to reject the null hypothesis, there is significant evidence that the mean checkout time is less than 12 minutes. Fail to reject the null hypothesis, there is not significant evidence that the mean checkout time is less than 12 minutes.

You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

Given data are

Population mean u=12 minutes.

and population  standard deviation= σ =3 minutes

Sample n=28

Sample mean x= 10.5

H0:null hypothesis :- u=12

Ha: alternative hypothesis: - u < 12

α =level of significance=0.02

a) Z = (x-u)/( σ /√ n)

  Z=(10.5-12) / (3/√ 28)

=-1.5/ 0.5669

= -2.6458= - 2.65

b) p (z < -2.65) = 0.0040 in left tail from z table

C) Pcal< 0.02 ( i.e 0.004 < 0.02)

So, Reject the null hypothesis, there is significant evidence that the mean checkout time is less than 12 minutes.