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Question
A machine is set to deliver packets of a given weight. Ten samples of size five each were recorded. Below are given relevant data:
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| `bar"X"` | 15 | 17 | 15 | 18 | 17 | 14 | 18 | 15 | 1 | 16 |
| R | 7 | 7 | 4 | 9 | 8 | 7 | 12 | 4 | 11 | 5 |
Calculate the control limits for mean chart and the range chart and then comment on the state of control, (conversion factors for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
Solution
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Tota |
| Mean `(bar"X")` | 15 | 17 | 15 | 18 | 17 | 14 | 18 | 15 | 1 | 16 | `sum"X"` = 162 |
| Range (R) | 7 | 7 | 4 | 9 | 8 | 7 | 12 | 4 | 11 | 5 | `sum"R"` = 74 |
The control limits for `bar"X"` chart is
`\overset{==}{"X"} = (sum"X")/"No. of samples" = 162/10` = 16.2
`bar"R" = (sum"R")/"No. of samples" = 74/10` = 7.4
UCL = `\overset{==}{"X"} + "A"_2+ bar"R"`
= 16.2 + (0.58)(7.4)
= 16.2 + 4.292
= 20.492
= 20.49
CL = `\overset{=}{"X"}` = 16.2
LCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 16.2 – (0.58)(7.4)
= 16.2 – 4.292
= 11.908
= 11.91
The control limits for Range chart is
CL = `"D"_4bar"R"`
= (2.115)(7.4)
= 15.651
= 15.65
CL = `bar"R"` = 7.4
LCL = `"D"_3 bar"R"` = (0)(7.4) = 0
The above diagram shows all the three control lines with the data points plotted.
We see that all the points of the sample mean are within the control limits.
We now draw the R chart for the given data.
The above diagram shows all the three control lines with the sample range points plotted.
We observe that all the points are within the control limits.
Conclusion: From the above two plots of the sample mean `bar"X"` and sample range R, we conclude that the process is in control.
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RELATED QUESTIONS
Mention the types of causes for variation in a production process
Name the control charts for variables
Write the control limits for the mean chart
Construct `bar"X"` and R charts for the following data:
| Sample Number | Observations | ||
| 1 | 32 | 36 | 42 |
| 2 | 28 | 32 | 40 |
| 3 | 39 | 52 | 28 |
| 4 | 50 | 42 | 31 |
| 5 | 42 | 45 | 34 |
| 6 | 50 | 29 | 21 |
| 7 | 44 | 52 | 35 |
| 8 | 22 | 35 | 44 |
(Given for n = 3, A2 = 1.023, D3 = 0 and D4 = 2.574)
The following data show the values of sample mean `(bar"X")` and its range (R) for the samples of size five each. Calculate the values for control limits for mean, range chart and determine whether the process is in control.
| Sample Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Mean | 11.2 | 11.8 | 10.8 | 11.6 | 11.0 | 9.6 | 10.4 | 9.6 | 10.6 | 10.0 |
| Range | 7 | 4 | 8 | 5 | 7 | 4 | 8 | 4 | 7 | 9 |
(conversion factors for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
A quality control inspector has taken ten ” samples of size four packets each from a potato chips company. The contents of the sample are given below, Calculate the control limits for mean and range chart.
| Sample Number | Observations | |||
| 1 | 2 | 3 | 4 | |
| 1 | 12.5 | 12.3 | 12.6 | 12.7 |
| 2 | 12.8 | 12.4 | 12.4 | 12.8 |
| 3 | 12.1 | 12.6 | 12.5 | 12.4 |
| 4 | 12.2 | 12.6 | 12.5 | 12.3 |
| 5 | 12.4 | 12.5 | 12.5 | 12.5 |
| 6 | 12.3 | 12.4 | 12.6 | 12.6 |
| 7 | 12.6 | 12.7 | 12.5 | 12.8 |
| 8 | 12.4 | 12.3 | 12.6 | 12.5 |
| 9 | 12.6 | 12.5 | 12.3 | 12.6 |
| 10 | 12.1 | 12.7 | 12.5 | 12.8 |
(Given for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
Choose the correct alternative:
Variations due to natural disorder is known as
Choose the correct alternative:
The assignable causes can occur due to
Choose the correct alternative:
The upper control limit for `bar"X"` chart is given by
The following data gives the average life(in hours) and range of 12 samples of 5lamps each. The data are
| Sample No | 1 | 2 | 3 | 4 | 5 | 6 |
| Sample Mean | 1080 | 1390 | 1460 | 1380 | 1230 | 1370 |
| Sample Range | 410 | 670 | 180 | 320 | 690 | 450 |
| Sample No | 7 | 8 | 9 | 10 | 11 | 12 |
| Sample Mean | 1310 | 1630 | 1580 | 1510 | 1270 | 1200 |
| Sample Range | 380 | 350 | 270 | 660 | 440 | 310 |
Construct control charts for mean and range. Comment on the control limits.
