Answer the following: For the Dataset provided below, estimate the following :      
a)     Cp      
b)     Cpk (and make a statement on the approximate ppm level that this process generates)      
2.     For each of the estimates (Cp and Cpk) in the above question, develop the 95% CI for the PCR (15 points).      
3.     Calculate why (see the Santos, Day 1-Session 3 lecture, Montgomery's Table 7-2), for a PCR of 1.20, there are 318ppm defects for a two-sided specification. (Hint: as a way of doing this, create a process that yields a PCR of 1.20 and calculate the area outside of spec limits)      
4.     If PCR = 2.0, then (using Montgomery's Table 7-2) the two-sided specifications yield 0.0018ppm. Why then is it often stated that a "six-sigma" process has a 3.4ppm defect level? For example, see the slide after the Montgomery PCR table that shows various processes and their sigma-levels and shows a 6sigma level equating to a 3.4ppm defect rate .      
       
Data:      
15.0098 To the left are sample data drawn from a process trying to make parts
15.0100 according to these specifications (units in mm):  
15.0101      
15.0099   Target 15.0000    +/-  0.0005
15.0101      
15.0099      
15.0099      
15.0099      
15.0101      
15.0101      
15.0101      
15.0099      
15.0100      
15.0099      
15.0100      
15.0100      
15.0098      
15.0099      
15.0101      
15.0101      
15.0099      
15.0099      
15.0100      
15.0099      
15.0102      
15.0098      
15.0099      
15.0101      
15.0100      
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15.0100      
15.0100      
15.0101      
15.0100      
15.0100      
15.0101      
15.0100      
15.0100      
15.0101      
15.0101      
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15.0101      
15.0101      
15.0100      
15.0100      
15.0098      
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15.0101      
15.0100      
15.0101      
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15.0100      
15.0100      
15.0100      
15.0100      
15.0099      
15.0102      
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15.0100      
15.0101      
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15.0100      
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15.0100