A bank located in a commercial district of a city has developed an improved process for serving customers during the noon to 1:00 PM peak lunch period. The waiting time (as defined as the time the customer enters the line until he or she is served) of all customers during this hour is recorded over a period of 1 week. A random sample of 15 customers is selected, and the results are as follows:

4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79

(a) Set up a 95% confidence interval for estimating the average waiting time during peak hours.
(b) At the 0.05 level of significance, is there evidence that the average waiting time is less than 5 minutes?
(c) Find the p-value for part(b).
(d) What assumption must hold in order to perform the test in part(b)?

Step-by-step explanation:

Given that a bank located in a commercial district of a city has developed an improved process for serving customers during the noon to 1:00 PM peak lunch period. The waiting time (as defined as the time the customer enters the line until he or she is served) of all customers during this hour is recorded over a period of 1 week.

A random sample of 15 customers is selected, and the results are as follows:

4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79

a) Mean difference= -0.713

Std error = 0.4229

df = 14

Margin of error = 0.9082

Confidence interval = (4.287-0.9082, 4.287+0.9082)

= (3.3788,5.1352)

b) H0: Mean = 5

Ha: Mean <5

Mean diference = 4.287-5 =-0.713

t=-1.686

c) p value = 0.0569

Since p >0.05 we accept null hypothesis.  The average time is less than 5 minutes cannot be accepted.

d)

Assumptions are:

The data is normally distributed

Scale of measurement should be interval or ratio

A randomized sample from a defined population

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)