In: Math
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.
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 |
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