Question
In a manufacturing process, 4 items are inspected every hour for 10 consecutive hours. The measured quality characteristic (in mm) is given below:
| Hour | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 50 | 52 | 51 | 49 |
| 2 | 51 | 50 | 53 | 52 |
| 3 | 49 | 50 | 48 | 51 |
| 4 | 52 | 54 | 53 | 51 |
| 5 | 50 | 49 | 51 | 50 |
| 6 | 53 | 52 | 54 | 55 |
| 7 | 48 | 49 | 50 | 47 |
| 8 | 51 | 52 | 50 | 51 |
| 9 | 54 | 55 | 53 | 54 |
| 10 | 50 | 51 | 49 | 50 |
Construct control chart for variability and mean and comment on the state of statistical control. If the process is out of control, obtain revised control limits.
Answer :
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Skip in 10
For each hour, the sample size is n = 4. Hence an (\bar X)-chart and an R-chart are appropriate. First, compute the sample mean and range for each hour. Hour 1: values 50, 52, 51, 49 (\bar X_1 = (50+52+51+49)/4 = 50.5), (R_1 = 52-49 = 3) Hour 2: 51, 50, 53, 52 (\bar X_2 = 51.5), (R_2 = 53-50 = 3) Hour 3: 49, 50, 48, 51 (\bar X_3 = 49.5), (R_3 = 51-48 = 3) Hour 4: 52, 54, 53, 51 (\bar X_4 = 52.5), (R_4 = 54-51 = 3) Hour 5: 50, 49, 51, 50 (\bar X_5 = 50.0), (R_5 = 51-49 = 2) Hour 6: 53, 52, 54, 55 (\bar X_6 = 53.5), (R_6 = 55-52 = 3) Hour 7: 48, 49, 50, 47 (\bar X_7 = 48.5), (R_7 = 50-47 = 3) Hour 8: 51, 52, 50, 51 (\bar X_8 = 51.0), (R_8 = 52-50 = 2) Hour 9: 54, 55, 53, 54 (\bar X_9 = 54.0), (R_9 = 55-53 = 2) Hour 10: 50, 51, 49, 50 (\bar X_{10} = 50.0), (R_{10} = 51-49 = 2) Now compute the overall averages. [ \bar{\bar X} = \frac{50.5+51.5+49.5+52.5+50.0+53.5+48.5+51.0+54.0+50.0}{10} = \frac{510.0}{10} = 51.0 ] [ \bar R = \frac{3+3+3+3+2+3+3+2+2+2}{10} = \frac{26}{10} = 2.6 ]
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