Question

An automobile repair shop is concerned about customer satisfaction in terms of the entire experience customers receive from the repair shop. In order to quantify the customer experience, three critical service characteristics have been identified:

Complaints of a failure to fix the vehicle.

Delay beyond the promised pickup time.

Complaints of damage to the inside/outside of the vehicle during repair.

A level of zero defects is the long-term goal. In order to address this goal, statistics have been collected over the past few months regarding these critical characteristics (see Table).

(A)Based on the Pareto principal, what would be the strategy for decreasing the total number of defects? (

B) With respect to the number of delays from promised completion time, does the process appear in-control? Assuming that special causes can be eliminated, what is the best estimate of the process capability?

(C) How is the process performing with respect to the number of items not fixed properly? What is the process capability?

Day	(A)	(B)	(C)	(D)
1	15	2	2	0
2	23	3	3	1
3	17	1	2	0
4	27	2	3	1
5	18	1	1	1
6	16	1	1	0
7	25	3	3	1
8	19	2	2	1
9	17	1	2	0
10	16	1	1	0
11	23	0	2	1
12	29	2	3	2
13	11	0	1	1
14	15	1	2	1
15	31	3	4	1
16	17	1	2	0
17	21	1	3	1
18	25	2	3	1
19	19	1	2	0
20	27	2	3	1
21	18	1	2	1
22	24	2	3	1
23	21	1	2	0
24	17	0	1	0
25	31	3	10	1
26	23	1	2	2
27	26	2	3	0
28	18	1	2	1
29	15	1	1	0
30	19	0	2	0

(A)Number of Vehicles in sample

(B) Number of items not fixed properly

(C) Number of delays from promised completion times

(D)Number of damaged items in repair

Answer 1

1. The null hypothesis (H0) for this research topic is,

B. There is no difference in customer satisfaction related to the service experience in Dealership A or Dealership B.

That is, the average rating for both the Dealers is same.

2. The mean of the customer satisfaction score for each dealership corresponds to sample mean of given sample data for the two dealers which can be calculated using EXcel function AVERAGE() as follows:

A. Dealership A customer satistaction,

T1AVERAGE(9.1,7.8, 8.1,8.8, 6.7 8.9.8.1) 8.07

B. Dealership B customer satisfaction

T2AVERAGE (9.9,6.1,6.8, 8.3, 7.5., 9.7,8.2) 7.9075

3. Since nothing is known about the population standard deviation therefore here we will perform two sample paired t test for the hypothesis of difference of mean assuming the samples to be taken from the independent populations.

The null hypothesis for the test will be defined as;

$H_{0}:\mu_{1}=\mu_{2}$

(That is, the average rating for both the Dealers is same.)

against the alternative hypothesis,

$H_{1}:\mu_{1} \neq \mu_{2}$

(That is, the average rating for both the Dealers is not equal)

Thus, enter the given data in the Excel sheet, go to Data Analysis from Data ribbon. Choose t-Test: Paired Two Sample for Means from Data Analysis menu.

Enter the fields as below:

-Test: Paired Two Sample for Means Input Variable 1 Range: Variable 2 Range: Hypothesized Mean Difference: OK Cancel Help SAS1:SAS41 SBS1:SBSA1 Labels Alpha: 0.05 Output options O Qutput Range: O New Worksheet Phy: New Workbook SDS1

-Test: Paired Two Sample for Means Input Variable 1 Range: Variable 2 Range: Hypothesized Mean Difference: OK Cancel Help SAS1:SAS41 SBS1:SBSA1 Labels Alpha: 0.05 Output options O Qutput Range: O New Worksheet Phy: New Workbook SDS1

The Excel will generate the following output:

1 Dealership A Dealership B t-Test: Paired Two Sample for Means 9.9 Dealership A Dealership B 7.9075 1.3139487181.919685897 4 6.8 Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<-t one-tail t Critical one-tail P(TK-t) two-tail t Critical two-tail 8.07 8.8 6.7 7.5 10 9.9 0.007248989 0 39 0.57357426 0.284774503 1.684875122 0.569549005 2.02269092 10 9.1 8.9 9.9 6.5 8.5 7.7 8.3 9.3 8.4 7.2 5.3 6.9 12 13 16 17 18 19 20 21 6.9 5.5 7.9 8.6 9.5 9.3 5.2 8.3 6.6 8.2 9.3 23

A. As can be observed from the above output that the p-value for the two tailed test is 0.5695 which is quite large and hence it is not considered significant.

B. Since p-value is greater than the level of significance therefore we cannot reject the null hypothesis and hence it is concluded that the average rating for both the Dealers is same.