Question

The management of the Diners Delight restaurant chain is in the process of establishing quality control charts for the time that its service people give to each customer. Management thinks the length of time that each customer is given should remain within certain limits to enhance service quality. For simplicity, the data from three samples are displayed below. In each sample, four service people were randomly selected, and the customer service they provided was observed. The time (in seconds) to service each customer was recorded as below.

Service person	Sample 1 (morning)	Sample 2 (afternoon)	Sample 3 (evening)
#1	201	196	197
#2	198	206	210
#3	190	207	211
#4	195	202	206

• Keep two decimals if not exact, do not round. For example, 3.24923... will be kept as 3.24, but the exact value of 0.625 will be kept as 0.625

1. What is the sample size of this study?
2. What is the X-bar value of sample 1?
3. What is the R-value of sample 2?
4. What is 99.73% (3-sigma) the upper control limit of the X-bar chart  and R-chart  ?
5. Assume the established control chart are as follows: X-bar chart is [120, 170], and R-chart is [20, 80]. A sample of six service personnel was observed, and the following customer service times in seconds were recorded: 180, 125, 140, 175, 156 and 160. What is the sample mean here?  Based on the information of sample mean and sample range, and the assumed established control chart, please determine whether the sample process is in control or not. (answer Y or N here)

Answer 1

Sample	1	2	3	4	Mean (X-bar)	Range (R)
1	201	198	190	195	196	11
2	196	206	207	202	202.75	11
3	197	210	211	206	206	14

(A) Sample size = Number of observations in a sample

Sample size = 4

(B) X-bar value of sample 1 = 196

(C) R-value of sample 2 = 11

(D) Overall Mean ($\bar{\bar{X}}$) = Average of means

Overall mean ($\bar{\bar{X}}$) = (196 + 202.75 + 206)/3

Overall mean ($\bar{\bar{X}}$) = 201.5833

Average Range ($\bar{R}$) = Average of range

Average Range ($\bar{R}$) = (11 + 11 + 14)/3

Average Range ($\bar{R}$) = 12

Sample Size (n) = 4

FOR X-BAR CHART:

*Use mean factor table to find value of A2 at n = 4.

Value of A2 = 0.729

$UCL_{\bar{X}}=\bar{\bar{X}}+A_{2}\bar{R}$

UCL = 201.5833 + (0.729 × 12)

UCL = 210.33

$LCL_{\bar{X}}=\bar{\bar{X}}-A_{2}\bar{R}$

LCL = 201.5833 - (0.729 × 12)

LCL = 192.83

FOR RANGE CHART:

*Use mean factor table to find value of D3 and D4.

Value of D4 = 2.282

Value of D3 = 0

$UCL_{\bar{R}} =D_{4}\bar{R}$

UCL = 2.282 × 12

UCL = 27.384

$LCL_{\bar{R}} =D_{3}\bar{R}$

LCL = 0 × 12

LCL = 0

(E) Given limits of X-bar chart = [120, 170]

Given limits of range chart = [20, 80]

Given Observations = 180, 125, 140, 175, 156, 160

Mean of observations = (180+125+140+175+156+160)/6

Mean of observations = 156

Range = 180 - 125 = 55

Since the mean and the range of observations lie within the given mean and range control limits, therefore the process is in control.

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