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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

It is known from previous experience that the standard deviation of a typical machine is measured as 2.1067. 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 :

Answer:

b. 23.16 and 16.84

Explanation:

The computation is shown below:

As we know that

The Mean of each and every sample (X-bar) is

= Sum of observations  Number of observations

And,

Range (R) = (Highest observation - Lowest observation)  

For Machine 1:

X-bar = (17 + 15 + 15 + 17) ÷ 4

= 16

R = (17 - 15)

= 2

For Machine 2:

X-bar = (16 + 25 + 18 + 25) ÷ 4

= 21

R = (25 - 16)

= 9

For Machine 3:

X-bar = (23 + 24 + 23 + 22) ÷ 4

= 23

R = (24 - 22)

= 2

Now, the Mean of means (X-double bar) is determined as follows

X-double bar = Sum of X-bar ÷ Number of samples

 = (16 + 21 + 23) ÷ 3

= 20

The Mean of ranges (R-bar) is determined as follows  

R-bar = Sum of R ÷ Number of samples

= (2 + 9 + 2) ÷  3

= 4.33

with reference to the table of constants for determining the 3-sigma upper and lower control limits,

 n = 4  that represent the Number of units sampled  

A2 = 0.729,

Now

UCL = X-double bar + (A2 × R-bar)

= 20 + (0.729 × 4.33)

= 23.16

LCL = X-double bar - (A2 × R-bar)

 = 20 - (0.729 × 4.33)

= 16.84

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