Question

The following table contains the measurements of the key length dimension from a fuel injector. These samples of size five were taken at one-hour intervals. Use three-sigma control limits. Use Exhibit 10.13.

	OBSERVATIONS
SAMPLE NUMBER	1	2	3	4	5
1	0.480	0.497	0.489	0.515	0.482
2	0.479	0.496	0.513	0.488	0.526
3	0.504	0.496	0.506	0.479	0.518
4	0.503	0.496	0.476	0.478	0.481
5	0.481	0.502	0.519	0.453	0.484
6	0.481	0.496	0.497	0.495	0.501
7	0.503	0.513	0.490	0.475	0.512
8	0.512	0.541	0.492	0.476	0.491
9	0.495	0.541	0.508	0.499	0.512
10	0.505	0.497	0.513	0.515	0.489
11	0.504	0.476	0.452	0.498	0.491
12	0.461	0.467	0.529	0.499	0.517
13	0.513	0.513	0.495	0.518	0.511
14	0.495	0.518	0.497	0.499	0.505
15	0.502	0.513	0.491	0.515	0.512
16	0.479	0.496	0.413	0.471	0.529
17	0.464	0.475	0.412	0.481	0.488
18	0.501	0.489	0.352	0.482	0.475
19	0.500	0.526	0.478	0.489	0.497
20	0.485	0.490	0.326	0.499	0.513

a. Calculate the mean and range for the above samples. (Do not round intermediate calculations. Round your answers to 3 decimal places.)

b. Determine X double bar and R bar . (Do not round intermediate calculations. Round your answers to 3 decimal places.)

c. Determine the UCL and LCL for a X bar chart. (Do not round intermediate calculations. Round your answers to 3 decimal places.)

d. Determine the UCL and LCL for R-chart. (Leave no cells blank - be certain to enter "0" wherever required. Do not round intermediate calculations. Round your answers to 3 decimal places.)

e. What comments can you make about the process?

• The process is out of statistical control.

• The process is in statistical control.

Answer 1

Answer a and b:

SAMPLE NUMBER	1	2	3	4	5	R= Max- Min	X-bar= average of sample
1	0.480	0.497	0.489	0.515	0.482	0.035	0.493
2	0.479	0.496	0.513	0.488	0.526	0.047	0.500
3	0.504	0.496	0.506	0.479	0.518	0.039	0.501
4	0.503	0.496	0.476	0.478	0.481	0.027	0.487
5	0.481	0.502	0.519	0.453	0.484	0.066	0.488
6	0.481	0.496	0.497	0.495	0.501	0.020	0.494
7	0.503	0.513	0.490	0.475	0.512	0.038	0.499
8	0.512	0.541	0.492	0.476	0.491	0.065	0.502
9	0.495	0.541	0.508	0.499	0.512	0.046	0.511
10	0.505	0.497	0.513	0.515	0.489	0.026	0.504
11	0.504	0.476	0.452	0.498	0.491	0.052	0.484
12	0.461	0.467	0.529	0.499	0.517	0.068	0.495
13	0.513	0.513	0.495	0.518	0.511	0.023	0.510
14	0.495	0.518	0.497	0.499	0.505	0.023	0.503
15	0.502	0.513	0.491	0.515	0.512	0.024	0.507
16	0.479	0.496	0.413	0.471	0.529	0.116	0.478
17	0.464	0.475	0.412	0.481	0.488	0.076	0.464
18	0.501	0.489	0.352	0.482	0.475	0.149	0.460
19	0.500	0.526	0.478	0.489	0.497	0.048	0.498
20	0.485	0.490	0.326	0.499	0.513	0.187	0.463
						0.059	0.492
						R-bar (Average of above values)	X-bar-bar (average of above values)

D4, D3 and A2 are taken from table of factors computing 3 sigma control limits, sample size=5

Answer c: X bar chart

A2 (n=5)	0.577
control limits for X bar,
CL or Xbarbar	0.492
UCL=Xbarbar+ (A2)*Rbar	0.526
LCL=Xbarbar- (A2)*Rbar	0.458

Please ask, if you have any doubts through the comment section. Do rate the answer

Answer d: Range chart

D4 (n=5)=	2.115
D3 (n=5)	0
control limits for Range,
CL or R-bar= Average of all sample Range	0.059
UCL=R-bar*D4	0.124
LCL=R-bar*D3	0.000

Answer e: The process is in statistical control. As all sample are withing control limits in chart

X-bar= average of sample	UCL	LCL	Xbar
0.493	0.526	0.458	0.492
0.500	0.526	0.458	0.492
0.501	0.526	0.458	0.492
0.487	0.526	0.458	0.492
0.488	0.526	0.458	0.492
0.494	0.526	0.458	0.492
0.499	0.526	0.458	0.492
0.502	0.526	0.458	0.492
0.511	0.526	0.458	0.492
0.504	0.526	0.458	0.492
0.484	0.526	0.458	0.492
0.495	0.526	0.458	0.492
0.510	0.526	0.458	0.492
0.503	0.526	0.458	0.492
0.507	0.526	0.458	0.492
0.478	0.526	0.458	0.492
0.464	0.526	0.458	0.492
0.460	0.526	0.458	0.492
0.498	0.526	0.458	0.492
0.463	0.526	0.458	0.492