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 4 1 10 12 12 14 2 12 11 13 16 3 11 13 14 14 4 11 10 7 8 5 13 12 14 13

Answer 1

14.6

Explanation:

(A). STEP ONE: Calculate the mean

(1). Row one = (10 + 12 + 12 + 14 ) = 48/4 = 12.

(2). Row Two: (12 + 11 + 13 + 16 ) = 52/4 = 13.

(3). Row three : (11 + 13 + 14 + 14)/4 = 13.

(4). Row four: (11 + 10 + 7 + 8)/4 = 36/4 = 9.

(5). Row five: (13 +12 + 14 + 13)/4 = 52/4 = 13.

(B). STEP TWO:

- determine the maximum and minimum value for each row.

- for each row, maximum - minimum.

Maximum values for each row:

Row one = 14, row two= 16, row three = 14, row four = 11 and row five = 14.

Minimum value for each row:

Row one = 10, row two = 11, row three = 11, row four =7 and row five = 12.

DIFFERENCES in each row :

row one = 14 - 10 = 4, row two = 16 - 11 = 5, row three = 14 - 11 = 3, row four = 11 - 7 = 4 and row five = 14 -12 = 2.

(C). STEP THREE: Calculate the mean of all the rows = 60/5 = 12.

(D). STEP FOUR : Calculate the Average Range = 18/5 = 3.6.

(E). STEP FIVE : Calculate the UCL.

A = Average rage × 0.729 = 3.6 × 0.729.

B = overall mean = 12.

UCL = A + B = 14.6.