Question

A bank with a branch located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1 P.M. lunch period. The bank has the business objective of reducing the waiting time (defined as the number of minutes that elapse from when the customer enters the line until he or she reaches the teller window) to increase customer satisfaction. A random sample of 15 customers is selected and waiting times(in minutes) are collected and stored in the Bank 1 column. Another branch, located in a residential area, is also concerned with the noon-to-1 p.m. lunch period. A random sample of 15 customers is selected and waiting times  (in minutes) are collected and stored in the Bank 2 column.

a.Assuming that the population variances from both banks are notequal, is there evidence of a difference in the mean waiting time between the two branches? (Use α = 0.05.)

b. Determine the p-value in (a) and interpret its meaning.

c.Construct and interpret a 95% confidence interval estimate of the difference between the population means in the two branches.

Observation	Bank1	Bank2
1	4.21	9.66
2	5.55	5.9
3	3.02	8.02
4	5.13	5.79
5	4.77	8.73
6	2.34	3.82
7	3.54	8.01
8	3.20	8.35
9	4.50	10.49
10	6.10	6.68
11	0.38	5.64
12	5.12	4.08
13	6.46	6.17
14	6.19	9.91
15	3.79	5.47

Thank you!

Answer 1

SolutioA:

H0:al, is are no  differences in the mean waiting time between the two branches

Ha: there is a difference in the mean waiting time between the two branches.

alpha=0.05

Run t test in R to get the test statistic and p value i:

Rcode is:

bank1 <- c(4.21,5.55,3.02,5.13,4.77,2.34,3.54,3.2,4.5,6.1,0.38,5.12,6.46,6.19,3.79)
bank2 <- c(9.66,5.9,8.02,5.79,8.73,3.82,8.01,8.35,10.49,6.68,5.64,4.08,6.17,9.91,5.47)
t.test(bank1,bank2)

Output:

Welch Two Sample t-test

data: bank1 and bank2
t = -4.1343, df = 26.529, p-value = 0.0003187
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.232685 -1.423315
sample estimates:
mean of x mean of y
4.286667 7.114667

Intrepretation:

t=t = -4.1343

p=0.0003187

p<0.05

Reject null hypothesis.

Accept alternative hypothesis.

Conclusion:

There is sufficient statistical evidence at 5% level of significance to conclude that  there is a difference in the mean waiting time between the two branches.

b. Determine the p-value in (a) and interpret its meaning.

how we got this value is

df=26.529

t=-4.1343

In excel type

=T.DIST.2T(4.1343;26.529)

p=0.000329

that is 0.000329*100 = 0.0329 in these cases only fail to reject H0 and rest 100-0.0329 =99.9671 cases reject Ho.

the test statistic is as more extreme than the observed value of the test statistic.

Solution1c:

95% confidence interval estimate of the difference between the population means in the two branches. lies in between

-4.232685 and -1.423315.