SPC Project A small electronic device is designed to emit a timing signal of 200 milliseconds...

SPC Project

A small electronic device is designed to emit a timing signal of 200 milliseconds (ms) duration. In the production of this device, 20 samples of five units are taken and tested. and R are calculated for each sample and used to plot control charts. The results are shown in the following table.

	Duration of automatic signal, milliseconds
	Observation
Sample	a	b	c	d	e			R
1	195	201	194	201	205		199.2	11
2	204	190	199	195	202		198.0	14
3	195	197	205	201	195		198.6	10
4	211	198	193	199	204		201.0	18
5	204	193	197	200	194		197.6	11

6	200	202	195	200	197		198.8	7
7	196	198	197	196	196		196.6	2
8	201	197	206	207	197		201.6	10
9	200	202	204	192	201		199.8	12
10	203	201	209	192	198		200.6	17

11	195	198	196	204	201		198.8	9
12	193	203	197	198	201		198.4	10
13	200	206	208	199	200		202.6	9
14	199	199	197	204	202		200.2	7
15	189	199	205	197	199		197.8	16

16	198	196	199	205	197		199.0	9
17	198	201	201	206	206		202.4	8
18	206	200	190	202	196		198.8	16
19	197	198	198	195	201		197.8	6
20	196	199	197	198	204		198.8	8

						Sum	3986.4	210.0
						Avg	199.32	10.5

a. Compute the averages, upper control limits, and lower control limits for and R charts for this data. Use the equations on pages 194-195 of your text, and Table 6.1 (page 195) for A2, D3, and D4.

b. Plot the R and charts in Excel. Is the process in statistical control?

c. Estimate the standard deviation of the process (σ) from the range data. σ = /d2, where d2 = 2.326 for a sample size of 5.

d. Assuming that the distribution of the data is approximately normal, what proportion of the devices would you expect to meet specifications of Lower Specification Limit = 190.5 and Upper Specification Limit = 210.5? Use the estimate of the standard deviation you calculated in part c.

Expert Solution

Solution:

formula for R chart

UCL = D4 * R bar

LCL = D3 * R bar

where are constant therefore D3 = 0 because sample size is 5 which is less than 7

and D4 = 2.114

Now Upper control, lower control limit and R chart for given data is as below

and R bar is average

Sample	a	b	c	d	e	x bar	R	R bar	LCL	UCL
1	195	201	194	201	205	199.2	11	10.5	0	22.197
2	204	190	199	195	202	198	14	10.5	0	22.197
3	195	197	205	201	195	198.6	10	10.5	0	22.197
4	211	198	193	199	204	201	18	10.5	0	22.197
5	204	193	197	200	194	197.6	11	10.5	0	22.197
6	200	202	195	200	197	198.8	7	10.5	0	22.197
7	196	198	197	196	196	196.6	2	10.5	0	22.197
8	201	197	206	207	197	201.6	10	10.5	0	22.197
9	200	202	204	192	201	199.8	12	10.5	0	22.197
10	203	201	209	192	198	200.6	17	10.5	0	22.197
11	195	198	196	204	201	198.8	9	10.5	0	22.197
12	193	203	197	198	201	198.4	10	10.5	0	22.197
13	200	206	208	199	200	202.6	9	10.5	0	22.197
14	199	199	197	204	202	200.2	7	10.5	0	22.197
15	189	199	205	197	199	197.8	16	10.5	0	22.197
16	198	196	199	205	197	199	9	10.5	0	22.197
17	198	201	201	206	206	202.4	8	10.5	0	22.197
18	206	200	190	202	196	198.8	16	10.5	0	22.197
19	197	198	198	195	201	197.8	6	10.5	0	22.197
20	196	199	197	198	204	198.8	8	10.5	0	22.197

now we have to find the estimate of standard deviation using this formula

where d2 =2.326 given

therefore estimate of hat is

4.514187

Thank You..!!

please like it..

milcah answered 1 week ago

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