Question

A bank with a branch located in a commercial district of a city has the business objective of improving the process for serving customers during the noon-to-1 PM lunch period. To do so, the waiting time (defined as the number of minutes that elapses from when the customer enters the line until he or she reaches the teller window) needs to be shortened to increase customer satisfaction. A random sample of 15 customers is selected and the waiting times were collected and can be seen below: 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 the noon-to-1 PM lunch period. A random sample of 15 customers is selected and the waiting times were collected and can be seen below: 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) Is there evidence of a difference in the variability of the waiting time between the two branches? (Use α = 0.05.)

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

(c) What assumption about the population distribution of each bank is necessary in (a)? Also, is the assumption valid for these data? (Justify your response through with Statistical charts, graphs, or calculations)

(d) Based on the results of (a), is it appropriate to use the pooled-variance t test to compare the means of the two branches?

Answer 1

Soln

a)

From Given data we calculate

Group 1 (Branch 1)

s₁² = 2.682

n₁ = 15

Group 2 (Branch 2)

s₂² = 4.336

n₂ = 15

Null and Alternate Hypothesis

H₀: σ₁² = σ₂² (Variance of waiting of Branch 1 is same as that of Branch 2)

H_a: σ₁² <> σ₂² (Variance of waiting of Branch 1 is not same as that of Branch 2)

alpha = 0.05

df₁ = 15-1 = 14

df₂ = 15-1 = 14

Test Statistic

F = s₁² / s₂² = 0.619

p-value = F.DIST.RT(0.619,14,14) = 0.8099

Result

Since the p-value is greater than 0.05, the data is not statistically significant and we fail to reject the null hypothesis.

Conclusion

Variance of waiting of Branch 1 is same as that of Branch 2

b)

p-value = F.DIST.RT(0.619,14,14) = 0.8099

A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so we reject the null hypothesis

c)

Assumptions of F Test

1. The populations from which the two samples are drawn are normally distributed.
2. The two populations are independent of each other

Branch 1

Range	Freq
0-2	1
2-4	5
4-6	6
6-8	3

The Tail is longer on left side but we can approximate normal distribution assumptions

Branch 2

Range	Freq
2-4	1
4-6	5
6-8	2
8-10	6
10-12	1

The above histogram shows that the data is not normally distributed

Assumption 1 is not satisfied but Assumption 2 is satisfied

d)

Based on the results of (a), it is appropriate to use the pooled-variance t test to compare the means of the two branches