Question

54 A manufacturing process for the company producing wheel bearings was investigated Samples of 4 subgroups where tested each containing 6 wheel bearings The specification limits for the process are given as 40 and 52 Given the following data for random measurements taken on the wheel bearing dunng their three shifts Subgroup 1 59 5 0 48 52 56 Subgroup 2 52 56 5 43 51 42 50 58 57 50 Subgroup 3 46 49 49 Subgroup 4 55 50 59 4 73 52 541 Calculate the Upper Control Limit (UCL) and the Lower Control Limit (LCL) of the sample mean Othere to search De ma

Answer 1

for sample size n=5, A2=0.577, D3=0, D4=2.114, d2=2.236

(5.4.1)

LCL=$\bar{\bar{x}}$+A2*$\bar{R}$=5.17-0.577*1.17=4.495

UCL=$\bar{\bar{x}}$+A2*$\bar{R}$=5.17+0.577*1.17=5.845

(5.4.2)

LCL=D3*$\bar{R}$=0*1.17=0

UCL=D4*$\bar{R}$=2.114*1.17=2.473

(5.4.3)

Cp = (USL-LSL)/(6*Sigma)

Sigma= R̄ / d_{2=1.17/2.236=0.523}

Cr=1/Cp

LSL and USL are needed for this part of question,

following information has been generated

sample	1	2	3	4	5	6	mean($\bar{x}$)	range ( R )
SubGroup1	5.9	5	4.8	5.2	5.6	5.4	5.32	1.1
SubGroup2	5.2	5.1	5	5.7	5.6	5.43	5.34	0.7
SubGroup3	4.6	4.2	5.8	5	4.1	4.9	4.77	1.7
SubGroup4	5.5	5.1	5	5.9	4.73	5.2	5.24	1.17
							$\bar{\bar{x}}$=5.17	$\bar{R}$=1.17

Table of d2, A2, D3, and DA values as a Function of the subgroup sample size, n d2 A2 D DA 2 1.128 1.880 0 3.267 1.693 1.023 2.574 2.059 0.729 0 2.282 2.326 0.577 0 2.114 2.534 0.483 0 2.004 7 2.704 0.419 0.076 1.924 Table 4