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Question
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)
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Solution
| Sample Number | Observations | |||||
| 1 | 2 | 3 | 4 | `bar"X"` | R | |
| 1 | 12.5 | 12.3 | 12.6 | 12.7 | 12.53 | 0.4 |
| 2 | 12.8 | 12.4 | 12.4 | 12.8 | 12.6 | 0.4 |
| 3 | 12.1 | 12.6 | 12.5 | 12.4 | 12.4 | 0.5 |
| 4 | 12.2 | 12.6 | 12.5 | 12.3 | 12.4 | 0.4 |
| 5 | 12.4 | 12.5 | 12.5 | 12.5 | 12.48 | 0.1 |
| 6 | 12.3 | 12.4 | 12.6 | 12.6 | 12.48 | 0.3 |
| 7 | 12.6 | 12.7 | 12.5 | 12.8 | 12.65 | 0.3 |
| 8 | 12.4 | 12.3 | 12.6 | 12.5 | 12.45 | 0.3 |
| 9 | 12.6 | 12.5 | 12.3 | 12.6 | 12.5 | 0.3 |
| 10 | 12.1 | 12.7 | 12.5 | 12.8 | 12.53 | 0.7 |
| Total | 125.02 | 3.7 | ||||
`\overset{==}{"X"} = (sumbar"X")/10 = 125.02/10` = 12.5
`bar"R" = (sum"R")/10 = 3.7/10` = 0.37
UCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 12.5 + (0.58)(0.37)
= 12.5 + 0.2146 = 12.7146
= 12.71
CL = `\overset{==}{"X"}` = 12.5
LCL = `\overset{==}{"X"} - "A"_2 bar"R"` = 12.5 - (0.58)(0.37)
= 12.5 – 0.2146 = 12.2854
= 12.29
The control limits for Range chart is
UCL = `"D"_4 bar"R"` = (2.115)(0.37) = 0.78255
= 0.78
CL = `bar"R"` = 0.37
LCL = `"D"_3 bar"R"` = (0)(0.37) = 0
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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 |
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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 |
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| Samples number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| `bar"X"` | 12 | 13 | 11 | 12 | 14 | 13 | 16 | 15 |
| R | 2 | 5 | 4 | 2 | 3 | 2 | 4 | 3 |
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 |
Choose the correct alternative:
Variations due to natural disorder is known as
Choose the correct alternative:
`bar"X"` chart is a
