MAN4504-Final Exam Tiger Tools Case Study PLEASE REVIEW ATTACHMENTS!! Read the Tiger Tools Case study located in your week 8 module area. Analyze the information and answer all the questions in the ca

Chapter 10 - Quality Control

Boxwood Tools Case

Boxwood Tools, a division of Drillmore Industries, was about to launch a new product. Production Manager Michelle York asked her assistant, Jim Peterson, to check the capability of the oven used in the process. Jim obtained 18 random samples of 20 pieces each. The results of those samples are shown in the following table. After he analyzed the data, he concluded that the process was not capable based on a specification width of 1.45 cm.

Michelle was quite disappointed when she heard this. She had hoped that with the introduction of the new product her operation could run close to full capacity and regain some of its lost luster. The company had a freeze on capital expenditures of more than $11,500, and a replacement oven would cost many times that amount. Jim Peterson worked with the oven crew to see if perhaps different settings could produce the desired results, but they were unable to achieve any meaningful improvements.

Sample	Mean	Range	Sample	Mean	Range
	46.01	0.86	10	45.97	0.92
	45.99	0.9	11	46.11	0.85
	46.02	0.87	12	45.96	0.88
	46.00	0.92	13	46.00	0.87
	46.04	0.88	14	45.92	0.90
	45.98	0.91	15	46.06	0.88
	45.91	0.87	16	45.94	0.87
	45.04	0.9	17	46.00	0.86
	46.00	0.86	18	46.03	0.87

Still not ready to concede, Michelle contacted one of her former professors and explained the problem. The professor suggested obtaining another set of samples, this time using a smaller sample size and taking more samples. Michelle then conferred with Jim and they agreed that he would take 27 samples of five observations each. The results are shown in the following table.

Sample	Mean	Range	Sample	Mean	Range
	45.96	0.43	15	46.00	0.40
	45.98	0.40	16	45.95	0.42
	45.96	0.42	17	44.94	0.44
	45.97	0.38	18	45.94	0.41
	46.02	0.40	19	45.87	0.39
	46.03	0.41	20	45.95	0.42
	46.04	0.40	21	45.93	0.40
	46.02	0.43	22	45.96	0.42
	46.08	0.39	23	45.99	0.41
10	45.12	0.41	24	46.00	0.45
11	46.07	0.42	25	46.03	0.43
12	46.02	0.39	26	46.04	0.39
13	46.01	0.42	27	46.03	0.41
14	45.98	0.41

Questions

Consider the following questions, and then write a brief report to Michelle summarizing your findings. Be sure to include your calculations for supporting your findings. To guide your report and expectations see the Case Study rubric in the syllabus or in the course information area. Submit your report in your week 8 drop-box labeled Final Exam Tiger Tools Case Study.

1.	How did Jim conclude that the process was not capable based on his first set of samples? (Hint: Estimate the process standard deviation, σ, using )
2.	Does the second set of samples show anything that the first set didn't? Explain what and why.
3.	Assuming the problem can be found and corrected, what impact do you think this would have on the capability of the process? Compute the potential process capability using the second data set.
4.	If small samples can reveal something that large samples might not, why not just take small samples in every situation?

For the first data set = 0.882. From Table 10–3 (Page 431), for n = 20, A2 = 0.18. Using the hint, the estimated standard deviation is .237:

Rearranging terms, we have

Solving, we obtain

When determining process capability, we do not know the USL and LSL. We know the specification width only; therefore, we will use Cp.

The process capability is 1.45/(6)(.237) = 1.02 Because this is less than 1.33, the process is not capable.

2. First set of data (n = 20 per sample):

From Table 10.3 with n = 20: A2 = 0.18.

Mean Chart:

Upper Control Limit (UCL) =

Lower Control Limit (LCL) =

Second set of data (n = 5 per sample):

From Table 10.3 with n = 5: A2 = 0.58.

Mean Chart:

Upper Control Limit (UCL) =

Lower Control Limit (LCL) = 4

The second set of samples (sample size = n = 5) seems to be cycling. The first set of samples (sample size = n = 20) did not reveal the changes that were occurring because large samples combine the results of several different process means. By taking smaller samples (second set), the pattern was easier to discern.

3. If the problem with cycling could be removed, the true process standard deviation probably would be much smaller than the apparent process standard deviation. For the second set of samples, = 0.411. Performing the same calculation as for the first set, we obtain an estimate of the process standard deviation of:

The potential process capability is 145/6(.177) = 1.37

Because this is more than 1.33, the process is capable.

4. Small samples tend to be less reliable than large samples (the standard deviation of the sampling distribution of means decreases as the sample size increases). Also, a manager must weigh the cost of inspecting each item and the cost of taking a sample. If the cost to obtain a sample is high, but the cost to inspect an item is low, larger samples might be the better choice.

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