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Question
In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the morning.
| Time | Weight in ml | ||||
| 8:00 AM | 43 | 41 | 42 | 43 | 41 |
| 9:00 AM | 40 | 39 | 40 | 39 | 44 |
| 10:00 AM | 42 | 42 | 43 | 38 | 40 |
| 11:00 AM | 39 | 43 | 40 | 39 | 42 |
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Solution
| Time | Weight in ml | `bar"X"` | R | ||||
| 8:00 AM | 43 | 41 | 42 | 43 | 41 | 42 | 2 |
| 9:00 AM | 40 | 39 | 40 | 39 | 44 | 40.4 | 5 |
| 10:00 AM | 42 | 42 | 43 | 38 | 40 | 41 | 5 |
| 11:00 AM | 39 | 43 | 40 | 39 | 42 | 40.6 | 4 |
| Total | 164 | 16 | |||||
The control limits for `bar"X"` chart is
`\overset{==}{"X"} = (sumbar"X")/"Number of samples" = 164/4` = 41
`bar"R" = 16/4` = 4
UCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 41 + (0.58)(4)
41 + 2.32 = 43.32
CL = `\overset{==}{"X"}` = 41
LCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 41 – (0.58)(4)
= 41 – 2.32
= 38.68
The control limits for range chart is
UCL = `"D"_4 bar"R"` = 2.115(4)
= 8.46
CL = `bar"R"` = 4
LCL = `"D"_2 bar"R"` = 0(4) = 0
Conclusion: Since all the points of sample mean and Range are within the control limits, the process is in control.
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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)
Ten samples each of size five are drawn at regular intervals from a manufacturing process. The sample means `(bar"X")` and their ranges (R) are given below:
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| `bar"X"` | 49 | 45 | 48 | 53 | 39 | 47 | 46 | 39 | 51 | 45 |
| R | 7 | 5 | 7 | 9 | 5 | 8 | 8 | 6 | 7 | 6 |
Calculate the control limits in respect of `bar"X"` chart. (Given A2 = 0.58, D3 = 0 and D4 = 2.115) Comment on the state of control
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)
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:
`bar"X"` chart is a
Choose the correct alternative:
The upper control limit for `bar"X"` chart is given by
From the following data, calculate the control limits for the mean and range chart.
| Sample No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Sample Observations |
50 | 21 | 50 | 48 | 46 | 55 | 45 | 50 | 47 | 56 |
| 55 | 50 | 53 | 53 | 50 | 51 | 48 | 56 | 53 | 53 | |
| 52 | 53 | 48 | 50 | 44 | 56 | 53 | 54 | 549 | 55 | |
| 49 | 50 | 52 | 51 | 48 | 47 | 48 | 53 | 52 | 54 | |
| 54 | 46 | 47 | 53 | 47 | 51 | 51 | 47 | 54 | 52 |
