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, with the following results. Tread wear measurements are in hundredths of an inch. Sample Tread Wear 1 31 42 28 2 26 18 35 3 25 30 34 4 17 25 21 5 38 29 35 6 41 42 36 7 21 17 29 8 32 26 28 9 41 34 33 10 29 17 30 11 26 31 40 12 23 19 25 13 17 24 32 14 43 35 17 15 18 25 29 16 30 42 31 17 28 36 32 18 40 29 31 19 18 29 28 20 22 34 26 Assuming that these data were collected when the manufacturing process was believed to be operating in control, develop the R chart. Compute the upper and lower control limits for the R chart. (Round your answers to two decimal places.) UCL = LCL =

Answer 1

Answer:

LCL = 4.72

UCL = 18.08

Step-by-step explanation:

Sample size, n = 20

Sample _ tread wear, x __ Range

1 ______ 33.67 ___ 14

2______ 26.33 ___ 17

3 _____ 29.67 ____ 9

4 ______ 21 ______ 8

5 ______ 34 ______ 9

6 ______ 39.67 ___ 6

7 ______ 22.33 ___ 12

8 ______ 28.67 ___ 6

9 ______ 36 ______8

10 _____ 25.33 ___ 13

11 _____ 32.33 ____14

12 _____ 22.33 ___ 6

13 _____ 24.33 ___ 15

14 _____ 31.67 ___ 26

15 _____ 24 _____ 11

16 ____ 34.33 ____ 12

17 ____ 32 _______ 8

18 ____ 33.33 ____ 11

19 ____ 25 _______11

20 ___ 27.33 _____12

Sample mean, xbar = Σx / n = 29.167

Rbar = ΣR/ n = 11.4

The R chart control limit is given by :

Rbar(1 - 3(d3/d2))

From the R chart :

d3 = 0.729 ; d2 = 3.735

Hence,

LCL = 11.4(1 - 3(0.729/3.735))

LCL = 11.4(0.4144578) =

LCL = 4.72481892 = 4.72

UCL = 11.4(1 - 3(0.729/3.735))

UCL = 11.4(1.5855421)

UCL = 18.075180 = 18.08