Question

A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for serving customers during the noon-to-1 P.M. lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined as the number of minutes that elapses from when the customer enters the line until he or she reaches the teller window. Data are collected from a random sample of 15 customers and stored in Bank1. These data are:

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

Suppose that another branch, located in a residential area, is also concerned with improving the process of serving customers in the noon-to-1 p.m. lunch period. Data are collected from a random sample of 15 customers and stored in Bank2. These data are:

9.66	5.90	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

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

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

• c. In addition to equal variances, what other assumption is necessary in (a)?

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

SHOW EXCEL FUNCTIONS USED TO ANSWER.

Answer 1

For Bank 1 :  

Sample mean using excel function AVERAGE(), x̅1 = 4.2867

Sample standard deviation using excel function STDEV.S, s1 = 1.6380

Sample size, n1 = 15

For Bank 2 :  

Sample mean using excel function AVERAGE(), x̅2 = 7.1147

Sample standard deviation using excel function STDEV.S, s2 = 2.0822

Sample size, n2 = 15

α = 0.05

a) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

b) Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((15-1)*1.638² + (15-1)*2.0822²) / (15+15-2) = 3.5093

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (4.2867 - 7.1147) / √(3.5093*(1/15 + 1/15)) = -4.1343

df = n1+n2-2 = 28

p-value :

Two tailed p-value = T.DIST.2T(ABS(-4.1343), 28) = 0.0003

Decision:

p-value < α, Reject the null hypothesis

c) Assumption:

• Sample are random and independent of each other
• Samples are approximately normal.

d) 95% Confidence interval :

At α = 0.05 and df = n1+n2-2 = 28, two tailed critical value, t-crit = T.INV.2T(0.05, 28) = 2.048

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2))

= (4.2867 - 7.1147) - 2.048*√(3.5093*(1/15 + 1/15)) = -4.2292

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2))

= (4.2867 - 7.1147) + 2.048*√(3.5093*(1/15 + 1/15)) = -1.4268

-4.2292 < µ1 - µ2 < -1.4268

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