Cap 7 Montgomery

Chapter 7 Exercise Solutions



Note: Several exercises in this chapter differ from those in the 4th edition. An “*” indicates that the description has changed. A second exercise number in parentheses indicates that the exercise number has changed. New exercises are denoted with an “(”.


7-1.
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7-2.
In Exercise 5-1, samples 12 and 15 are out of control, and the new processparameters are used in the process capability analysis.

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7-3.
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The process produces product that uses approximately 18% of the total specification band.

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This is an extremely capable process, with an estimated percent defective much less than 1 ppb. Note that the Cpk is less than Cp, indicating that the process is not centered and is not achieving potentialcapability. However, this PCR does not tell where the mean is located within the specification band.

[pic]
Since Cpm is greater than 4/3, the mean ( lies within approximately the middle fourth of the specification band.

[pic]

7-4.
[pic]; tolerances: 0 ( 0.01
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The process produces product that uses approximately 82% of the total specification band.

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This process is notconsidered capable, failing to meet the minimally acceptable definition of capable Cpk ( 1.33

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Since Cpm is greater than 1, the mean ( lies within approximately the middle third of the specification band.

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7-5.
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(a)
Potential: [pic]

(b)
Actual: [pic]

(c)
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7-6(.
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USL = 200 + 8 = 208; LSL = 200 – 8 = 192

(a)
Potential: [pic]
The processproduces product that uses approximately 64% of the total specification band.

(b)
Actual: [pic]

(c)
The current fraction nonconforming is:
[pic]
If the process mean could be centered at the specification target, the fraction nonconforming would be:
[pic]
7-7(.
[pic]
USL = 40 + 5 = 45; LSL = 40 – 5 = 35

(a)
Potential: [pic]

(b)
Actual: [pic]

(c)
[pic]
The closeness ofestimates for Cp, Cpk, Cpm, and Cpkm indicate that the process mean is very close to the specification target.

(d)
The current fraction nonconforming is:
[pic]
7-7 (d) continued
If the process mean could be centered at the specification target, the fraction nonconforming would be:
[pic]


7-8 (7-6).
[pic]

(a)
Potential: [pic]

(b)
Actual: [pic]

(c) Let [pic]
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7-9 (7-7).Assume n = 5

Process A
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Process B
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7-9 continued
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Prefer to use Process B with estimated process fallout of 0.000001 instead of Process A with estimated fallout 0.001726.


7-10 (7-8).
Process A: [pic]
Process B: [pic]

Process B will result in fewer defective assemblies. For the parts [pic] indicates thatmore parts from Process B are within specification than from Process A.
7-11 (7-9).
MTB > Stat > Basic Statistics > Normality Test

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A normal probability plot of the 1-kg container weights shows the distribution is close to normal.

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7-12(.
LSL = 0.985 kg
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[pic]
7-13(.
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 toestimate ( and (.)

[pic]

A normal probability plot of computer disk heights shows the distribution is close to normal.

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7-14(.
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate ( and (.)

[pic]

A normal probability plot of reimbursement cycle times shows the distribution is close to normal.

[pic]

7-15(.
MTB > Stat >Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate ( and (.)

[pic]

A normal probability plot of response times shows the distribution is close to normal.

(a)
[pic]

(b)
USL = 2 hrs = 120 mins
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7-16 (7-10).
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate ( and (.)

[pic]...