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	0.488	0.485	0.482	0.508	0.485
2	0.501	0.504	0.518	0.491	0.526
3	0.498	0.481	0.513	0.483	0.515
4	0.497	0.504	0.482	0.485	0.486
5	0.474	0.498	0.524	0.460	0.486
6	0.475	0.485	0.505	0.482	0.497
7	0.497	0.512	0.485	0.470	0.496
8	0.521	0.503	0.493	0.471	0.494
9	0.494	0.501	0.514	0.496	0.511
10	0.495	0.505	0.518	0.508	0.492
11	0.493	0.482	0.461	0.481	0.487
12	0.483	0.450	0.512	0.496	0.521
13	0.523	0.512	0.482	0.523	0.509
14	0.485	0.523	0.505	0.506	0.502
15	0.492	0.518	0.494	0.508	0.511
16	0.473	0.504	0.471	0.482	0.518
17	0.465	0.483	0.511	0.484	0.491
18	0.517	0.482	0.482	0.485	0.471
19	0.513	0.526	0.485	0.492	0.481
20	0.499	0.485	0.472	0.496	0.512

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

	Sample Number Mean Range 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

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

	X       R

c. Determine the UCL and LCL for a %media:formula252.mml%-chart. (Do not round intermediate calculations. Round your answers to 3 decimal places.)

	UCL    LCL

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.)

	UCL    LCL

e. What comments can you make about the process?

• The process is in statistical control.

• The process is out of statistical control.

Answer 1

(a)

	OBSERVATIONS
SAMPLE NUMBER	1	2	3	4	5	Sample mean	Sample range
1	0.488	0.485	0.482	0.508	0.485	0.4896	0.026
2	0.501	0.504	0.518	0.491	0.526	0.508	0.035
3	0.498	0.481	0.513	0.483	0.515	0.498	0.034
4	0.497	0.504	0.482	0.485	0.486	0.4908	0.022
5	0.474	0.498	0.524	0.46	0.486	0.4884	0.064
6	0.475	0.485	0.505	0.482	0.497	0.4888	0.03
7	0.497	0.512	0.485	0.47	0.496	0.492	0.042
8	0.521	0.503	0.493	0.471	0.494	0.4964	0.05
9	0.494	0.501	0.514	0.496	0.511	0.5032	0.02
10	0.495	0.505	0.518	0.508	0.492	0.5036	0.026
11	0.493	0.482	0.461	0.481	0.487	0.4808	0.032
12	0.483	0.45	0.512	0.496	0.521	0.4924	0.071
13	0.523	0.512	0.482	0.523	0.509	0.5098	0.041
14	0.485	0.523	0.505	0.506	0.502	0.5042	0.038
15	0.492	0.518	0.494	0.508	0.511	0.5046	0.026
16	0.473	0.504	0.471	0.482	0.518	0.4896	0.047
17	0.465	0.483	0.511	0.484	0.491	0.4868	0.046
18	0.517	0.482	0.482	0.485	0.471	0.4874	0.046
19	0.513	0.526	0.485	0.492	0.481	0.4994	0.045
20	0.499	0.485	0.472	0.496	0.512	0.4928	0.04

(b)

Hence

and

(c)

Here the subgroup size is n=5 and hence from the table, we obtain

For the chart, the control limits are given by

(d)

Here the subgroup size is n=5 and hence from the table, we obtain

For the chart, the control limits are given by

(e)

From the range chart, it is evident that all the subgroups are within the control limits and hence the process variability is in control.

From the mean chart, it is evident that all the subgroups are within the control limits and hence the process central tendency is in control.

Hence the process is in statistical control.

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