Question

A cereal manufacturer weighs cereal boxes using an automatic scale. The target weight is to put 10 ounces of cereal in each box. Fifteen samples of three boxes each have been weighed for quality control purposes. The fill weight for each box is shown below:

Sample 1 2    3

1 10.01 9.90 10.03

2    9.87    10.20 10.15

3 10.08    9.89 9.76

4 10.17 10.01 9.83

5 10.21 10.13    10.04

6    0.16    10.02    9.85

7    10.14 9.89 9.80

8 9.86    9.91 9.99

9 10.18    10.04 9.96

10    9.91 9.87    10.06

11    10.08 10.14    10.03

12    9.71 9.87 9.92

13    10.14 10.06 9.84

14    10.16    10.17    10.19

15    10.13 9.94 9.92

a) Determine the center line, the upper control limit and lower control limit the and R-charts based on the above data.

b) Is the process in control? Why or why not? [Note: You are NOT required to plot the control charts for and R but doing so helps answer this question.]

Answer 1

The information given in the question is as under:

	Observation
Sample	1	2	3
1	10.01	9.9	10.03
2	9.87	10.2	10.15
3	10.08	9.89	9.76
4	10.17	10.01	9.83
5	10.21	10.13	10.04
6	0.16	10.02	9.85
7	10.14	9.89	9.8
8	9.86	9.91	9.99
9	10.18	10.04	9.96
10	9.91	9.87	10.06
11	10.08	10.14	10.03
12	9.71	9.87	9.92
13	10.14	10.06	9.84
14	10.16	10.17	10.19
15	10.13	9.94	9.92

Answer a) R Bar (i.e. Centre Line), Upper Control Limit (UCL) and Lower Control Limit (LCL) is to be calculated:

Range (R) = Maximum value of individual observation sample value - Minimum value of individual observation sample value

R Bar = Centre Line = Average of all Range of samples

UCL = D4 x R Bar

LCL = D3 x R Bar

Based on the above formula, following values are derived:

	Observation
Sample	1	2	3	Range (R)
1	10.01	9.9	10.03	0.13
2	9.87	10.2	10.15	0.33
3	10.08	9.89	9.76	0.32
4	10.17	10.01	9.83	0.34
5	10.21	10.13	10.04	0.17
6	0.16	10.02	9.85	9.86
7	10.14	9.89	9.8	0.34
8	9.86	9.91	9.99	0.13
9	10.18	10.04	9.96	0.22
10	9.91	9.87	10.06	0.19
11	10.08	10.14	10.03	0.11
12	9.71	9.87	9.92	0.21
13	10.14	10.06	9.84	0.30
14	10.16	10.17	10.19	0.03
15	10.13	9.94	9.92	0.21
			R Bar	0.86

Therefore,

Centre Line = R Bar = 0.86

For Observation size of 3 per sample:

D4	2.574
D3	0

UCL = D4 x R Bar = 2.574 x 0.86 = 2.21

LCL = D3 x R Bar = 0 x 0.86 = 0

Answer b) Based on the above values, the process is out of control

As the Range value of Sample no. 6 is “9.86” which is out of the UCL of 2.21.