Question

The following data was collected from the manufacturing of an auto component. It represents the diameter (in mm) of that component. What is the LCL for a control chart using this data (z=3)?Sample Obs 1 Obs 2 Obs 3 Obs 41 10 12 12 142 12 11 13 163 11 13 14 144 11 10 7 85 13 12 14 13

Answer 1

9.37

Step-by-step explanation:

The computation of LCL for a control chart is shown below:-

Sample Obs 1 Obs 2 Obs 3 Obs 4 Mean observation Range

1 10 12 12 14 12 4

2 12 11 13 16 13 5

3 11 13 14 14 13 3

4 11 10 7 8 9 4

5 13 12 14 13 13 2

For computing the mean observation and range we will use the below formulas

Mean observation = ( Obs 1 + Obs 2 + Obs 3 + Obs 4) ÷ 4

Range = Highest value - Lowest value

`LCL = \bar{\bar{X}} - A2 \bar{R}`

`\bar X` = ( 12 + 13 + 13 + 9 + 13 ) ÷ 5

= 12

`\bar R` = ( 4 + 5 + 3 + 4 + 2 ) ÷ 5

= 3.6

Since we found the value of A2 with the help of constants table for control charts for a 4 subgroup size.

A2 = 0.729

`LCL = \bar{\bar{X}} - A2 \bar{R}`

12 - 0.729 × 3.6

= 9.37