Question

The Goodman Tire and Rubber Company periodically tests its tires for tread wear under simulated road conditions. To study and control the manufacturing process, 20 samples, each containing three radial tires, were chosen from different shifts over several days of operation; the data collected are shown below. Assuming that these data were collected when the manufacturing process was believed to be operating in control, develop the R and charts.

R Chart: (to 2 decimals)
UCL =
LCL =
Chart: (to 1 decimal)
UCL =
LCL =

Answer 1

Range:

UCL = 4.73

LCL = 18.08

MEAN :

UCL = 27.115

LCL = 31.219

Explanation:

Given the data:

The mean and range of each sample :

Sample __ Thread wear __ xbar __ R

1 ___31 __ 42 ___ 28 ____ 33.67 _14

2___ 26 _ 18 ____35____ 26.33 _17

3___25 __30 ___ 34____29.67 _ 9

4 __ 17 __ 25 ___ 21 _____ 21 ___ 8

5 __ 38 _ 29 ___ 35 _____ 34 __ 9

6 __ 41 __42 ___36 _____39.67_ 6

7 __ 21 __ 17 ___29 _____22.33 _12

8 __ 32 __26___28 ____ 28.67 _ 6

9 __ 41 __ 34 __ 33 ______ 36 __8

10__29___17___30 _____25.33_ 13

11 __26 __ 31 __ 40 _____32.33_ 14

12__23 __ 19 __ 45 _____12.33 __6

13 __17 __ 24 __ 32_____24.33__15

14 __43__ 35___17_____ 31.67 _ 26

15__18 ___25__ 29_____ 24 ___ 11

16__30___42___31 ____34.33__ 12

17__28___36 __ 32____ 32 ____8

18__40 __ 29 __ 31 ____33.33 __ 11

19__18 ___29__ 28____ 25 ____11

20_ 22 __ 34 __ 26 ___ 27.33 __12

Size per sample, sample size, n = 3

Number of samples, k = 20

We calculate the sample mean and range average :

Sample mean, x-- = Σxbar/n = 29.167

Range average, Rbar = ΣR/n = 11.4

The mean control limit :

x-- ± A2Rbar

From the x chart ;

A2 for n = 20 is A2 = 0.180

29.167 ± 0.180(11.40)

LCL = 29.167 - 0.180(11.40) = 27.115

UCL = 29.167 + 0.180(11.40) = 31.219

The Range control limit :

Rbar(1 ± 3(d3/d2))

From the R-chart :

d2 at n = 20 ; d2 = 3.735

d3 at n = 20 ; d3 = 0.729

LCL = 11.40(1 - 3(0.729/3.735)) = 4.725

UCL = 11.40(1 + 3(0.729/3.735)) = 18.075