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A quality analyst wants to construct a control chart for determining whether three machines, all producing the same product, are in control with regard to a particular quality variable. Accordingly, he sampled four units of output from each machine, with the following results.

Machine 1 Measirement 17 15 15 17
Machine 2 Measurement 16 25 18 25
Machine 3 Measurement 23 24 23 22

What are the Mean chart three-sigma upper and lower control limits?

a. 22 and 18
b. 23.16 and 16.84
c. 22.29 and 16.71
d. 23.5 and 16.5
e. 24 and 16

Sagot :

Tundexi

Answer:

b. 23.16 and 16.84

Explanation:

Mean (X-bar) = Sum of observations / No of observations

Range (R) = Highest observation - Lowest observation

Machine 1

Mean (X-bar) = (17 + 15 + 15 + 17) / 4

Mean (X-bar) = 16

Range (R) = (17 - 15)

Range (R) = 2

Machine 2

Mean (X-bar) = (16 + 25 + 18 + 25) / 4

Mean (X-bar) = 21

Range (R) = (25 - 16)

Range (R) = 9

Machine 3

Mean (X-bar) = (23 + 24 + 23 + 22) / 4

Mean (X-bar) = 23  

Range (R) = (24 - 22)

Range (R) = 2

Mean of means (X-double bar) = Sum of X-bar / Number of samples = (16 + 21 + 23) / 3 = 20

Mean of ranges (R-bar) = Sum of R / Number of samples = (2 + 9 + 2) / 3  = 4.33

From table of constants for calculating the 3-sigma upper and lower control limits, For n = 4, A2 = 0.729

UCL = X-double bar + (A2 x R-bar)

UCL = 20 + (0.729 x 4.33)

UCL = 23.1566

UCL = 23.16

LCL = X-double bar - (A2 x R-bar)

LCL = 20 - (0.729 x 4.33)

LCL = 16.8434

LCL = 16.84

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