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

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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