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

Answer:

9.37

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