Question

In: Math

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.

Solutions

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..


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