Question

Aspen Plastics produces plastic bottles to customer order. To monitor the
process, statistical process control charts are used. The central line of the chart for
the sample means is set at 8.50 and the range at 0.31in. Assume that the sample
size is 6 and the specification for the bottle neck diameter is 8.50 ± 0.25.
a. Calculate the control limits for the mean and range charts. (3 points)
b. Suppose that the standard deviation of the process distribution is 0.13 in.
If the firm is seeking three-sigma performance, is the process capable of
producing the bottle? (4 points)
c. If the process is not capable, what percent of the output will fall outside the
specification limits? (3 points

Answer 1

Solution:

Let,

(X=) = Overall Mean = 8.50

R- = Range Bar = 0.31

A2 = Constant value derived from the control chart constant value table = 0.483

D3 = Constant value derived from the control chart constant value table = 0

D4 = Constant value derived from the control chart constant value table = 2.004

U = Upper Specification Limit = 8.50 + 0.25 = 8.75

L = Lower Specification Limit = 8.50 - 0.25 = 8.25

Answer a:

Control Limits for Mean Chart:

CL = Average Mean (X=) = 8.50

UCL = (X=) + (A2 * R-)

= 8.50 + (0.483 * 0.31)

= 8.6497 (Rounded to 4 decimal places)

LCL = (X=) - (A2 * R-)

= 8.50 - (0.483 * 0.31)

= 8.3503 (Rounded to 4 decimal places)

Control Limits for Range Chart:

CL = Average Range (R-) = 0.31

UCL = D4 X R- = 2.004 X 0.31 = 0.6212 (Rounded to 4 decimal places)

LCL = D3 X R- = 0 X 0.31 = 0

Answer b:

σ = Std Dev. = 0.13

Here, we will calculate Cp and Cpk as mentioned below:

Where,

So,

Cpk = Min (0.641,0.641) = 0.641

As Cpk < 1, The process is not capable of meeting the desired specifications.

Answer c:

i) % of output falling outside the Upper Specification Limit:

= (8.75 - 8.50) / 0.13

= 1.9231

So,

% of units above USL = Probability value derived from the standard normal table for Z_USL = 0.0272 or 2.72%

ii) % of output falling outside the Lower Specification Limit:

= (8.50 - 8.25) / 0.13

= 1.9231

% of units below LSL = Probability value derived from the standard normal table for Z_LSL = 0.0272 or 2.72%

Thus, % of total units out of the desired specification limits = 2.72 + 2.72 = 5.44 % (Rounded to the two decimal places)

(Kindly raise an upvote for this answer, if you found it useful)