Question

In: Operations Management

Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in...

Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Management is concerned about whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes were​ taken, each tube was​ weighed, and the weights in the table below were obtained.

the ounces of caulking per tube data.

Tube number

Sample

1

2

3

4

5

6

7

8

1

7.58

8.34

8.02

7.94

8.44

7.67

7.81

8.11

2

8.33

8.17

8.08

8.51

8.41

8.28

8.09

8.16

3

7.89

7.77

7.81

8.04

8.00

7.89

7.93

8.09

4

8.24

8.18

7.83

8.03

7.90

8.16

7.97

8.07

5

7.87

8.13

7.92

7.99

7.94

7.81

8.14

7.88

6

8.13

8.14

8.11

8.13

8.14

7.99

8.13

8.14

the table of factors for calculating​ three-sigma limits for the x over bar x​-chart and​ R-chart.

Factors for calculating​ three-sigma limits for the

x overbarx​-chart

and​ R-chart.

Size of Sample​ (n)

 Factor for UCL and LCL for

x overbarx​-chart

​(Upper A 2A2​)

 Factor for LCL for​ R-Chart

​(Upper D 3D3​)

 Factor for UCL for​ R-Chart

​(Upper D 4D4​)

2

1.880

0

3.267

3

1.023

0

2.575

4

0.729

0

2.282

5

0.577

0

2.115

6

0.483

0

2.004

7

0.419

0.076

1.924

8

0.373

0.136

1.864

9

0.337

0.184

1.816

10

0.308

0.223

1.777

a. Assume that only six samples are sufficient and develop the control charts for the mean and the range.

Set up the​ R-chart by specifying the center line and​ three-sigma control limits below. ​(Enter your responses rounded to three decimal​ places.)

​R-chart

UCL Subscript Upper RUCLR

equals=

Upper R overbarR

equals=

LCL Subscript Upper RLCLR

equals=

Solutions

Expert Solution

# SAMPLE

1

2

3

4

5

6

7

8

AVERAGE

RANGE

1

7.58

8.34

8.02

7.94

8.44

7.67

7.81

8.11

7.99

0.86

2

8.33

8.17

8.08

8.51

8.41

8.28

8.09

8.16

8.25

0.43

3

7.89

7.77

7.81

8.04

8

7.89

7.93

8.09

7.93

0.32

4

8.24

8.18

7.83

8.03

7.9

8.16

7.97

8.07

8.05

0.41

5

7.87

8.13

7.92

7.99

7.94

7.81

8.14

7.88

7.96

0.33

6

8.13

8.14

8.11

8.13

8.14

7.99

8.13

8.14

8.11

0.15

AVERAGE

8.05

0.42

RANGE = LARGEST OBSERVATION - SMALLEST OBSERVATION

XBAR = 8.05

RBAR = 0.42

A2 VALUE CORRESPONDING TO N = 8 = 0.373

D3 & D4 VALUES CORRESPONDING TO N = 8 = 0.136, 1.864

CONTROL LIMITS FOR XBAR

UCL X = XBAR + (A2 * RBAR) = 8.05 + (0.373 * 0.42) = 8.21

LCL X = XBAR - (A2 * RBAR) = 8.05 - (0.373 * 0.42) = 7.89

NO, BECAUSE ONE OR MORE POINTS ARE ABOVE THE UCL

CONTROL LIMITS FOR RBAR

UCL R = RBAR * D4 = 0.42 * 1.864 = 0.78

LCL R = RBAR * D3 = 0.42 * 0.136 = 0.06

NO, BECAUSE ONE OR MORE POINTS ARE ABOVE THE UCL

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keosha answered 1 year ago
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