Question

A producer of cereal measures part of its product quality by measuring box weight. It has established that each box should be 475 grams ± 5 grams. When the production was known to be under control, it obtained samples of 5 boxes and measured their weight. It then calculated the sample range R and sample mean x¯ (x bar). It then calculated the average of the ranges as R¯ = 5.5 grams and the average of the sample means as x¯¯ (x double bar)= 474.36 grams.

(a) Calculate the UCLR and the LCLR.

(b) Calculate the UCLx¯ and the LCLx¯.

(c) This week, the company obtained 25 samples of box weights, again using 5 boxes per sample. The measurements were:

criteria	measurement
minimum R	0.6 grams
maximum R	9.8 grams
minimum x bar	472.84 grams
maximum x bar	476.12 grams

Is the system still under control?

(d) Given the required box weight specifications, is the current system capable of 6σ quality? (the standard deviation was measured to be σ = 1.2731)

(e) A production engineer believes that a new system will reduce the box weight standard deviation by 45%. Assuming that production will remain under control with the same R¯ (R bar) and x¯¯ (x double bar), should the company implement the proposed system?

Answer 1

Average range = R¯ = 5.5 grams

average of the sample means = x¯¯ (x double bar) = 474.36 grams.

For sample size of 5 units, A₂ = 0.577 , d₃ = 0 , d₄ = 2.115

(a) Calculate the UCLR and the LCLR.

UCL_R = (d₄)*R¯ = 2.115 x 5.5 = 11.6325

LCL_R = (d₃)*R¯ = 0 x 5.5 = 0

UCL_R = 11.6325

LCL_R = 0

(b) Calculate the UCLx¯ and the LCLx¯.

UCL_x¯ = (x¯ ¯ ) + (d₄)*R¯ = 474.36 + (0.577)*5.5   

UCL_x¯ = 477.5335

LCL_x¯ = (x¯ ¯ ) - (d₄)*R¯ = 474.36 - (0.577)*5.5   

LCL_x¯ = 471.1865

Part c:

criteria	measurement	Control Limits to compare	Comment
minimum R	0.6 grams	LCLR = 0	Since min. R is more than LCLR, the process is in control
maximum R	9.8 grams	UCLR = 11.6325	Since max. R is less than UCLR, the process is in control
minimum x bar	472.84 grams	LCLx¯ = 471.1865	Since min. x¯ is more than LCLx¯, the process is in control
maximum x bar	476.12 grams	UCLx¯ = 477.5335	Since max. x¯ is less than UCLx¯, the process is in control

Since all the criteria measurements are within the limits the process is still in control.

Part d.

For the given process:

Product specifications are: 475 grams ± 5 grams.

Upper Specification limit (USL) = 475+5 = 480 grams

Lower Specification Limit (LSL) = 475 – 5 = 470 grams

Process performance is:

average of the sample means = x¯¯ = process average = 474.36 grams.

Process standard deviation = σ = 1.2731

The process capability is given as follows:

Cp = (USL – LSL) / (6* σ) = (480 – 470)/(6*1.2731)

Cp = 1.31

Since for 6 sigma level capability, the process capability ratio should be 2. Since the Cp is less than 2, the process is not capable of 6σ quality.

Part e.

Reduction in the standard deviation of process = 45%

New standard deviation = old SD – (0.45*old SD) = 1.2731 – (0.45*1.2731)

New standard deviation = 0.7

The new process capability is given as follows:

Cp = (USL – LSL) / (6* σ) = (480 – 470)/(6*0.7)

Cp = 2.38

For 6 sigma quality level, the process capability ratio should be 2. Since the Cp is greater than 2, the process is capable of 6σ quality. Yes, company should implement proposed system.